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https://socratic.org/questions/express-563-in-exponential-notetion

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# Express 563 in exponential notetion ? Sep 9, 2017 $5.63 \cdot {10}^{2}$ #### Explanation: Scientific notation (also called exponential notation) is to write a number in the form of $a \cdot {10}^{x}$ where a: $1 \le a < 10$ x: the number of places that we have to move the decimal point. So as for number 563, the decimal point has to be moved 2 places Therefore, $5.63 \cdot {10}^{2}$

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https://app-wiringdiagram.herokuapp.com/post/circuit-diagram-of-4-20ma-source

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9 out of 10 based on 661 ratings. 3,296 user reviews. # CIRCUIT DIAGRAM OF 4 20MA SOURCE Simple LED Circuit Diagram Aug 16, 2015So if you are using a 9v battery and voltage Drop on LED is, let’s say 2, and current flowing is 20mA, then we should have a resistor value on which remaining voltage can be dropped (9 – 2). So according to the formulae: R = (9 – 2.4) / = 330 Ohm. So the remaining voltage (9 - 2.4= 6) will be dropped at 330ohm resistor. How to Implement a 4-20mA Transmitter wit - Maxim Integrated ☰ Integrated Circuit Technical Documents Abstract: This application note explains how to implement a 3-wire 4–20mA transmitter provided by the MAX12900. — — ® Using the MAX12900 in a 3-Wire Transmitter. Having eliminated the possibility of using a 2-wire solution, we next consider the challenges in implementing a 3-wire design. Step by Step Procedure to Build Electronic Circuits/Circuit A circuit is any loop through which matter is carried. For an electronic circuit, the matter carried is the charge by electronics and the source of these electrons is the positive terminal of the voltage source. When this charge flows from the positive terminal, through the loop, and reaches the negative terminal, the circuit is said to be Residual Current Circuit Breaker - RCCB | Electrical4u Trip mechanism: Trip mechanism is used to trip the circuit during fault condition. ON/OFF switch: The handle is used to switching on/off the rccb. Test Circuit: A Test switch is used to check the working condition of rccb. See circuit diagram of test Circuit. In that, a switch is connected between line to earth (in series with a resistor). 5V Relay Module : Pin Configuration, Circuit, Working & Its The circuit diagram of the single-channel relay module circuit is shown below. In this circuit, we can observe that how the relay module is activated and deactivated through a digital signal. This signal is applied to a control pin of the relay module. The following circuit diagram is the internal 5V single channel relay module diagram. Single LED Calculator - Current limiting resistor calculator for LED arrays Make sure you are using a DC (direct current) power source such as batteries, wall transformers and PC power supplies LED voltage drop (V): The amount of power needed, in Volts, for the LED to light properly. Current Measurements: How-To Guide - NI Jun 02, 2022When a metal wire is connected across the two terminals of a DC voltage source such as a battery, the source places an electric field across the conductor. The moment contact is made, the free electrons of the conductor are forced to drift toward the positive terminal under the influence of this field. Current Loops/4-20 mA Convention

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http://stats.stackexchange.com/questions/73582/how-to-show-the-following-consistency

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# how to show the following consistency? It is well-known that maximum likelihood estimate for a mixture model, with the mixture distributions known, and the estimation is done for the mixture coefficients is consistent (I think) -- the ML objective has a single likelihood, so expectation-maximization will be consistent. What if the mixture distributions themselves are estimated from data, and their estimators are consistent? Will that yield a consistent estimator for estimating the mixture components as well, when using the estimated mixture distributions, or not? (To put it more mathematically: Say we have a discrete random variable $X$ accepting values between $1$ and $n$. We also have $m$ fixed distributions $p_i(X = x)$ for $i = 1, ..., m$. We want to estimate $\theta$ for the model $$p(X = x) = \sum_i \theta_i p_i(X = x).$$ The maximum likelihood estimate is going to be consistent. Now what if we didn't have $p_i$ but instead $\hat{p_i}$ (estimators of $p_i$) such that $\hat{p_i}(X = x) \rightarrow p_i(X = x)$ as we have more data? I am guessing the ML estimator for $\theta$ while using $\hat{p_i}$ will still be consistent, is that correct? If so, what is roughly the way to show it? ) -

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https://gofigurewithscipi.blogspot.com/2018/02/dividing-fractions-using-kfc.html

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### Dividing Fractions Using KFC Ugh - It's time to teach the division of fractions. My experience has been that many students forget which fraction to flip and often, they forget to change the dreaded division sign to a multiplication sign. The other evening,  I was helping my 5th grade granddaughter with her homework. Really, she had completed it by herself, but she wanted me to check it. At the top of her paper were the letters "KFC". I asked her what they meant, and she replied, "Kentucky Fried Chicken." Now I have taught math for years and years, and I had never heard of that one! She explained that the "K" stood for keep; "F" for flip, and "C" for change. Let's suppose the problem on the left was one of the problems on her homework paper. First, she would Keep the first fraction. Next, she would Flip the second one, and then Change the division sign to a multiplication sign...like illustrated on the right. She would then cross cancel if possible (In this case it is).  Finally, she would multiply the numerator times the numerator and the denominator by the denominator to get the answer. She was able to work all the division problems without any trouble by just remembering the letters KFC. Yesterday, I was working in our college math lab when a student needed help. On the right is the problem he was having difficulty with. (For those of you who don't teach algebra or just plain hate it, I am sure this problem looks daunting and intimidating. Believe me, my student felt the same way!) First I had the student rewrite the problem with each fraction side by side with a division sign in between them like this. Doesn't it look easier already? I then taught him KFC. You read that right! I did! (I figured if it worked for a 5th grader, it should work for him.) Surprisingly it made sense to him because he now had mnemonic device (an acronym) that he could easily recall. He rewrote the problem by Keeping the first fraction, Flipping the second, and Changing the division sign to a multiplication sign. Now it was just a simple multiplication problem.  Had he been able to, he would have cross canceled, but in this case, he simply multiplied the numerator times the numerator and denominator by the denominator to get the answer. So the next time you teach the division of fractions, or you come across a problem like the one above, don't panic!  Remember KFC, and try not to get hungry!

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What Happens to Complex Numbers as Ket Turns to Bra In the vast realm of mathematics, a fascinating branch known as complex analysis delves into the profound properties of complex numbers. These numbers, often represented by a combination of real and imaginary parts, bring a new dimension to mathematical exploration. Within this intricate framework, the relationship between kets and bras, which originate from the realm of quantum mechanics, unveils a captivating interplay between these abstract entities. As the ket seamlessly transitions into a bra, a metamorphosis occurs, resulting in a transformation that offers a profound glimpse into the hidden intricacies of complex numbers. Through this transition, an exquisite dance unfolds, revealing the profound nature of complex analysis and the inherent beauty within the mathematical landscape. What Is the Complex Conjugate of a Bra? The complex conjugate of a bra can be obtained by taking the Hermitian conjugate of the corresponding ket. Bra–ket notation is a powerful tool in linear algebra and quantum mechanics, providing a convenient way to represent linear operators and their dual spaces in complex vector spaces. This notation is particularly effective in tackling the calculations commonly encountered in quantum mechanics. This ensures that the bra and ket are properly related. For example, if the complex number in the ket is expressed as a + bi, it’s complex conjugate in the bra will be represented as a – bi. Understanding the concept of the complex conjugate is crucial in maintaining the integrity of mathematical operations involving complex numbers. The complex conjugate of a complex number is obtained by changing the sign of the imaginary part. It provides a way to balance and reconcile complex terms in equations, enabling accurate calculations and accurate representation of physical phenomena. It ensures the accurate calculation of inner products, as well as the proper formulation of observables and measurements in quantum systems. In quantum mechanics, the bra-ket notation simplifies the representation of vectors in a state space. By using this notation, a vector is represented by a combination of a bra (⟨) and a ket (|), with the elements of the vector placed between them. This concise form allows for easy manipulation and calculation of probabilities, where the magnitude squared of a vector represents the probability of the corresponding state. What Is the Bra-Ket Notation Simplified? The bra-ket notation, also known as Dirac notation, is a powerful tool used in quantum mechanics to simplify calculations and describe quantum states. It utilizes the bra and ket symbols – ⟨ and ⟩, respectively – to represent vectors in a complex vector space. This vector can have any number of dimensions and contain complex elements. By using this notation, complex calculations involving vectors and matrices become more concise and easier to understand. In quantum mechanics, particles can exist in multiple states simultaneously, known as superposition. The probability of observing a particle in a particular state is given by the magnitude of the vector representing that state, squared. For example, the probability of finding a particle in state ⟨v⟩ is |⟨v|v⟩|². Moreover, the bra-ket notation allows for convenient representation of mathematical operations. For instance, the inner product between two vectors can be written as ⟨v|w⟩, while the outer product can be expressed as |v⟩⟨w|. It provides a concise and intuitive way to represent and manipulate quantum states, making it an essential tool for physicists and researchers in the field. Applications of Bra-Ket Notation in Quantum Mechanics: Explain How Bra-Ket Notation Is Used in Various Aspects of Quantum Mechanics, Such as State Representation, Observables, and Quantum Operators. • State representation: Bra-ket notation provides a concise and intuitive way to represent quantum states. The ket vector |ψ⟩ represents a quantum state, while it’s dual bra vector ⟨ψ| represents the corresponding state in the dual space. • Observables: Using the bra-ket notation, observables (such as position, momentum, energy, etc.) can be represented as Hermitian operators. The expectation value of an observable can be calculated as ⟨ψ|A|ψ⟩, where A is the corresponding operator. • Quantum operators: In quantum mechanics, operators are used to describe the evolution and transformations of quantum states. Bra-ket notation simplifies the representation and manipulation of these operators. For example, the action of an operator A on a state |ψ⟩ can be written as A|ψ⟩. • Inner product: The inner product of two quantum states can be expressed using bra-ket notation as ⟨φ|ψ⟩. This allows for calculations of probabilities and overlaps between different quantum states. • Superposition: Bra-ket notation is crucial in representing superposition states, where a quantum system can exist in multiple states simultaneously. The superposition of two states |ψ₁⟩ and |ψ₂⟩ can be represented as |ψ⟩ = c₁|ψ₁⟩ + c₂|ψ₂⟩, where c₁ and c₂ are complex coefficients. • Measurement: When measuring a quantum state, bra-ket notation helps in determining the probabilities of different outcomes. The probability of obtaining a particular eigenvalue of an observable A is given by |⟨ψ|a⟩|², where |a⟩ is the corresponding eigenstate. The relationship between bras and kets goes beyond their role as elements of vector spaces. While kets represent the familiar vectors, bras have a distinct role as covectors in the dual vector space. This distinction makes them elements of different vector spaces, with kets belonging to one vector space and bras to the corresponding dual vector space. What Is the Relationship Between Bra and Ket? The relationship between bra and ket comes from the concept of duality in vector spaces. In mathematics, a vector space is a set of objects that can be added together and multiplied by scalars. The concept of duality introduces an additional space called the dual vector space, which consists of linear functionals that map vectors from the original vector space to scalars. The inner product between a bra and a ket yields a scalar. The notation used for kets and bras is significant in this context. A ket is represented by a vertical vector enclosed in brackets, like |v>. A bra, on the other hand, is represented by a horizontal vector enclosed in brackets with a line above it, like <v|. This notation emphasizes the duality between the two vector spaces. For example, the inner product between |v> and <w| is written as . This operation is defined as the complex conjugate of the transpose of the ket multiplied by the bra. The resulting scalar represents the overlap between the two corresponding vectors. Understanding the relationship between bras and kets is essential in various areas of physics, particularly quantum mechanics. They provide a mathematical framework to describe the behavior and properties of quantum systems. Application of Bras and Kets in Quantum Mechanics: This Could Explore How Bras and Kets Are Used to Describe Quantum States and Operators in Quantum Mechanics. In quantum mechanics, a branch of physics that deals with very small particles, bras and kets are mathematical notations used to describe quantum states and operators. A ket, written as |ψ⟩, represents a quantum state or vector in a physical system, while a bra, written as ⟨ϕ|, represents the conjugate transpose of a ket. Together, bras and kets allow us to perform calculations and make predictions about quantum systems. They provide a convenient way to represent complex data and perform mathematical operations in quantum mechanics. Conclusion In conclusion, as the ket turns into a bra, complex numbers continue to play a crucial role in the realm of quantum mechanics. They encompass both real and imaginary components, allowing for the representation of states in a mathematical framework that captures the probabilistic nature of quantum phenomena. As kets transform into bras, the complex conjugate of the numbers accommodates the switch from the vector space perspective to the dual space perspective, enabling the calculation of inner products, probabilities, and amplitudes. This transformation emphasizes the interconnectedness and duality inherent in quantum mechanics, showcasing the profound significance of complex numbers in understanding and describing the behavior of quantum systems.

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# Transformations Unit – Translations Subject: Math 8, 9, 10, 11, 12 Title – Transformations Unit – Translations By – Cornelia Taran Transformation Unit: Lesson Plan 1 – Translations (below) Lesson Plan 2 – Reflections Lesson Plan 3 – Rotations Lesson Plan 1: Translations Content: The students will identify and locate translations. Benchmarks: • G3.1: Distance-preserving transformations – isometries • METS 3.a.5: Students use online tutorial and discuss the benefits and disadvantages of this method of learning. Learning Resources and Materials: • Transformation tool • Translucent paper • Pictures with translations Development of Lesson: Introduction: Objectives: The students will identify and locate translations. Anticipatory Set: Show pictures that represent translations. Lesson: • A translation “slides” an object a fixed distance in a given direction. The original object and its translation have the same shape and size, and they face in the same direction. • The word “translate” in Latin means, “carried across”. • Use translucent paper to create translations. Methods/Procedures: 1. Draw a vector on one sheet of translucent paper. 2. Draw an original shape on the sheet of paper. 3. Place a second paper over the first sheet. On sheet 2, trace the endpoint of the arrow and draw a line that extends beyond the endpoint and head of the original arrow. Without moving sheet 2, trace the original figure. 4. Place the second sheet under the original. Align the line on the second sheet and the vector on the first. Slide sheet 2 until the point you drew on the line is under the tip of the arrow. 5. Trace the image from sheet 2 onto sheet 1. Label the original figure and the translated figure. Explain translations in the coordinate plane Assessment/Evaluation: Formative – monitor and provide feedback. Closure: Draw a triangle in the coordinate plane and translate it (6,-4) E-Mail Cornelia Taran !

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https://encyclopediaofmath.org/wiki/Trigonometric_polynomial

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# Trigonometric polynomial finite trigonometric sum An expression of the form $$T ( x) = { \frac{a _ {0} }{2} } + \sum _ {k = 1 } ^ { n } ( a _ {k} \cos kx + b _ {k} \sin kx)$$ with real coefficients $a _ {0} , a _ {k} , b _ {k}$, $k = 1 \dots n$; the number $n$ is called the order of the trigonometric polynomial (provided $| a _ {n} | + | b _ {n} | > 0$). A trigonometric polynomial can be written in complex form: $$T ( x) = \sum _ {k = - n } ^ { n } c _ {k} e ^ {ikx} ,$$ where $$2c _ {k} = \left \{ \begin{array}{ll} a _ {k} - ib _ {k} , &k \geq 0 \ ( \textrm{ with } b _ {0} = 0), \\ a _ {-k} + ib _ {-k} , &k < 0 . \\ \end{array} \right .$$ Trigonometric polynomials are an important tool in the approximation of functions.

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https://aviation.stackexchange.com/questions/34981/what-is-the-fuel-consumption-of-an-aircraft-in-a-holding-pattern/72856

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# What is the fuel consumption of an aircraft in a holding pattern? I am doing a research about fuel cost when an aircraft is about to land. they may be asked to be hold due to the traffic congestion at the runway. consequently they consume extra fuel. My question is how I can estimate the cost per minute or second being late of their *target time. Where can I get more details on this? *target time: As I noticed each aircraft has a target time to land which is the most economical time and speed for them. My research is about aircraft landing scheduling during rush hours where there are several objective functions such as maximizing the runway utilization, minimizing total delay and fuel consumption to be considered. these objective functions are subject to some constraints like aircraft's time window (earliest landing time,latest possible landing time), minimum time separation, Constraint position shifting... In the literature that I have read so far they considered a fix fuel cost as a function of delay. as an instance: "Vj denotes the cost per unit time of the extra fuel associated with lateness relative to ULTj , then the overall extra fuel cost is : summation of Vj * (Landing times - ULTj) " (Aircraft Runway optimization, Mesgarpour, 2012) Since it is more mathematically issue no body really pay attention to the details of that fuel cost or any other cost caused by the delay. • How are you "doing research" on a topic that you don't even seem to know the most basic info about? Researchers are typically well informed and experienced in their field. – abelenky Jan 25 '17 at 21:00 • See also this example from Airbus: Getting to grips with fuel economy, \$5.5 Holding. – mins Jan 25 '17 at 21:02 • @abelenky That's a little harsh, IMO, considering that we know nothing about the OP and what he's doing. For example, is it a professional project or a school one? And I'm fairly sure that not every researcher is "well informed and experienced", if only because everyone has to start somewhere. The same applies in every other job out there. – Pondlife Jan 25 '17 at 21:14 • @abelenky that isn't the point, you directly chastised someone for asking a question that you determine to be "basic info", just because they are a researcher...If you don't see a problem with that...Plus the fact that you did answer kind of contradicts your initial bewilderment. – Prodnegel Jan 25 '17 at 22:13 • @ablenky This is not a forum for sneering at people for asking questions about things they don't know about, or that you think are too basic. – Daniele Procida Jan 25 '17 at 23:45 Every aircraft should have a published fuel-burn rate, typically measured in GPH (Gallons per Hour) or Pounds per Hour. This rate does change a bit depending on ambient conditions (temperature & pressure), throttle, and weight. But its a pretty good planning rate. There are other costs, beyond fuel costs, associated with delayed landings, such as crew costs and running the airplane and engines closer to their next inspections and service times. The fuel rate while holding depends on elements such as weight, altitude, speed and aircraft configuration. Airbus uses the concept of green dot which is the speed with minimum fuel consumption per time unit in clean configuration. This table shows fuel needs for a A330/A340 at 170 tons and 1,500 ft: Fuel flow vs. speed. Source While the green dot speed is the one offering the best fuel rate while holding, it assumes no flap and no slat. This speed may not be practical for certain airports where slats or flaps are advisable due to the rate of turn in the holding pattern. To provide the necessary flexibility, Airbus publishes the fuel rate for three other combinations of speed/configuration (S speed is the minimum speed for flaps retraction): Fuel flow vs. weight for 4 configurations. Source Altitude is also a factor preventing an optimal fuel flow. Example for the green dot speed: Percentage added when departing from the optimal altitude. Source The total fuel required to hold can be somehow minimized by not flying at cruise speed when it is known that a hold will be required at some point. Savings can be done by flying at green dot speed to the holding point: That means determining the fuel flow depends on airline operating procedures and holding pattern design (e.g. altitude), but a maximum can be roughly predicted by looking at the manufacturer published documentation. • Why is GD different from the "min fuel" speed on that graph? – pericynthion Jan 27 '17 at 21:49 • @pericynthion: That's right, the GD speed (maximum L/D ratio) is an approximation of the minimum fuel speed. But the difference in fuel consumption is really small. The GD speed is calculated for best climb gradient. Being already available it is also used for holding. – mins Jan 27 '17 at 23:35 To answer based on some data I had correlating aircraft in confirmed holding patterns (not speed-restricted en-route delays) vs. high altitude cruise, the short answer was: they use more - mostly because they were in low-speed manouevring, with flaps set. I'd have to look at the data again to remember how much more, but I recall it was fairly significant: at least compared to cruise. This analysis was done as part of a study the airline was using to calculate the cost of capacity constraints in the terminal area.

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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 4.9: Read and Interpret Maps Involving Distance and Area Difficulty Level: At Grade Created by: CK-12 Estimated8 minsto complete % Progress Practice Reading Maps, Legends/Scale Progress Estimated8 minsto complete % ### Let’s Think About It Credit: Kate Ter Haar Source: https://www.flickr.com/photos/katerha/16759820090/in/photolist-rx1sBm-2BJTZV-2BJoaD-e2uQYp-2zzzmm-2BJTWK-2BJTUK-a1vp4W-aGg7jR-do6iTm-6i1p2y-2BP7cW-2BP7e9-2BP75A-4r16Zw-edG2Zt-edG698-edGmGx-2BJod6-2BJo5a-2BP769-2BP77U-doodyb-dLftrP-a2DfQM-edMqf7-8vehvb-d8sH8L-d7DzXm-9JRhpt-8iSYWY-2zxatx-6QJr-2zxaqH-d8rF17-edFL5n-edG28V-edN37W-edFUpz-edN681-edN1Hw-edMVF3-edN3Ud-edG5sV-edGbgD-edMX4N-edG3Ri-edMCAu-edMYdd-edFYxH License: CC BY-NC 3.0 Kaitlyn is looking at a map and wants to find the area of the state of Nevada, United States. The map shows a scale of 12\begin{align*}{\frac{1}{2}}^{{\prime}{\prime}}\end{align*} to 20 miles. If the distance on the map is 8\begin{align*}8^{\prime \prime}\end{align*} wide and 12.25\begin{align*}12.25^{\prime \prime}\end{align*} long, what is the area, in square miles of Nevada? In this concept, you will learn to read and interpret maps involving distance and area by using scale measurement. ### Guidance A map is another type of scale drawing of a region. Maps can be very detailed or very simple, showing only points of interest and distances. You can read a map just like any other scale drawing—by using the scale. Let’s look at an example. On the map below, the straight-line distance between San Francisco and San Diego is 3 inches. What is the actual straight-line distance between San Francisco and San Diego? License: CC BY-NC 3.0 First, set up a proportion. The scale in the drawing says that 0.5 inches=75 miles\begin{align*}0.5 \ \text{inches} = 75 \ \text{miles}\end{align*}, therefore the proportion is: 0.5 inches75 miles Next, write the second ratio. You know the scale length is 3 inches. The unknown length is x\begin{align*}x\end{align*}. 0.5 inches75 miles=3 inchesx miles Then, cross-multiply. 0.5750.5x0.5x===3x75×3225 Then, divide both sides by 0.5 to solve for x\begin{align*}x\end{align*}. 0.5x0.5x0.5x===2252250.5450 The answer is 450. The straight-line distance from San Francisco to San Diego is 450 miles. Note: The straight-line distance is also known as “as the crow flies.” If you were actually traveling from San Francisco to San Diego, it would be farther than 450 miles, since you would need to drive on highways that are not necessarily in a straight line. You can also use a scale to find the area of a space or region. First, you need to figure out the length and width then we can complete any necessary calculations. Sometimes, you will have two different distances or areas that you are working to compare. When this happens, you can use proportions to compare the differences and similarities. Let’s look at an example. Marta has a square with a side length of 4 inches. She has a similar square with dimensions that are twice the first square. How does the area of the larger square compare to the area of the smaller square? First, find the dimensions of the larger square. The problem states that the dimensions are twice the first square. You can use this information to figure out the scale factor, and this means they are scaled up by a factor of 2. The side length of the larger square is: 4 inches×2=8 inches\begin{align*}4 \ \text{inches}\times 2 = 8 \ \text{inches}\end{align*}. Next, find the area of both squares and compare. Area of smaller square   Area of larger squareA = lw A = lwA = (4 inches)(4 inches)A = (8 inches)(8 inches)A=16 in2   A=64 in2 Then, compare the two areas. You want to know how the area of the larger square compares to the area of the smaller square. Write a ratio comparing the two areas. 64 in216 in2=4 The answer is 4. The area of the larger square is 4 times larger than the area of the smaller square. This leads to a rule when comparing areas of similar figures. The ratio of areas of similar figures is the square of the scale factor. ### Guided Practice If the scale is 0.5 inches=100 miles\begin{align*}0.5\ \text{inches} = 100 \ \text{miles}\end{align*}, how many inches is 500 miles? First, set up the proportion to solve. 0.5 inches100 miles=x inches500 miles Next, cross multiply. 0.5100100x100x===x5000.5×500250 Then, divide by 100 to solve for x\begin{align*}x\end{align*}. 100x100x100x===2502501002.5 The answer is 2.5. The distance on the map will be 2.5 inches. ### Examples #### Example 1 If 1=2000 miles\begin{align*}1^{\prime \prime} = 2000 \ \text{miles}\end{align*} , find each actual distance with a scale measurement of 3\begin{align*}3^{\prime \prime}\end{align*}. First, set up a proportion. The scale in the drawing says that 1 inch=2000 miles\begin{align*}1\ \text{inch} = 2000 \ \text{miles}\end{align*}, therefore the proportion is: 1 inch2000 miles Next, write the second ratio. You know the scale length is 3 inches. The unknown length is x\begin{align*}x\end{align*}. 1 inch2000 miles=3 inchesx miles Then, cross-multiply to solve for x\begin{align*}x\end{align*}. 120001xx===3x2000×36000 The answer is 6000. The actual distance is 6000 miles. #### Example 2 If 1=2000 miles\begin{align*}1^{\prime \prime} = 2000 \ \text{miles}\end{align*}, find each actual distance with a scale measurement of 12\begin{align*}{\frac{1}{2}}^{{\prime}{\prime}}\end{align*}. First, set up a proportion. The scale in the drawing says that 1 inch=2000 miles\begin{align*}1\ \text{inch} = 2000 \ \text{miles}\end{align*}, therefore the proportion is: 1 inch2000 miles Next, write the second ratio. You know the scale length is 0.5\begin{align*}0.5^{\prime \prime}\end{align*} inches. The unknown length is x\begin{align*}x\end{align*}. 1 inch2000 miles=0.5 inchesx miles Then, cross-multiply to solve for x\begin{align*}x\end{align*}. 120001xx===0.5x2000×0.51000 The answer is 1000. The actual distance is 1000 miles. #### Example 3 If 1=2000 miles\begin{align*}1^{\prime \prime} = 2000 \ \text{miles}\end{align*}, find each actual distance with a scale measurement of 14\begin{align*}{\frac{1}{4}}^{{\prime}{\prime}}\end{align*}. First, set up a proportion. The scale in the drawing says that 1 inch=2000 miles\begin{align*}1\text{ inch} = 2000 \text{ miles}\end{align*}, therefore the proportion is: 1 inch2000 miles Next, write the second ratio. You know the scale length is 0.25\begin{align*}0.25^{\prime \prime}\end{align*} inches. The unknown length is x\begin{align*}x\end{align*}. 1 inch2000 miles=0.25 inchesx miles Then, cross-multiply to solve for x\begin{align*}x\end{align*}. 120001xx===0.25x2000×0.25500 The answer is 500. The actual distance is 500 miles. ### Follow Up Credit: Bob Dass Source: https://www.flickr.com/photos/54144402@N03/17163130335/in/photolist-s9DwGg-85iw6U-4ykGxj-4FC2do-MwZ2c-eyhxWS-7oDJdt-cLKtgb-wW4FP-eBeKCb-74NDxu-sngP1k-osxTfm-rXVj6c-eBbvLP-7nzwru-ew5YuR-dx4VDs-p9QhyM-47e7vK-dnsqzZ-7nJwe8-cCizvm-99PdW9-pNShKT-67ok35-ew6DYF-ezy7PC-eyhUaN-t4qRFV-pk7mb-dwYr6D-nyZtQM-4VqGzP-8rqgLS-i6LmKY-31a6Gp-donJRG-74JKE2-sC5rkf-oVjpA9-8eaEVn-Ebaub-66Mqw2-ezuUTF-4FAf2R-hHYLb-p17rgX-dwYqNK-sXJQyG License: CC BY-NC 3.0 Remember Kaitlyn and her interest in Nevada? First, set up a proportion. The scale in the drawing says that 12 inch=20 miles\begin{align*}\frac{1}{2} \text{ inch} = 20 \ \text{miles}\end{align*}, therefore the proportion is: 0.5 inch20 miles Next, write the ratio representing the length. You know the scale length is 8 inches. The unknown length is x\begin{align*}x\end{align*}. 0.5 inch20 miles=8 inchesx miles Then, cross-multiply to solve for x\begin{align*}x\end{align*}. 0.5200.5x0.5x===8x20×8160 Then, divide both sides by 0.5 to solve for x\begin{align*}x\end{align*}. 0.5x0.5x0.5x===1601600.5320 The answer is 320. The width of Nevada is 320 miles. Then, write the ratio representing the length. You know the scale length is 12.25 inches. The unknown length is x\begin{align*}x\end{align*}. 0.5 inch20 miles=12.25 inchesx miles Then, cross-multiply to solve for x\begin{align*}x\end{align*}. 0.5200.5x0.5x===12.25x20×12.25245 Then, divide both sides by 0.5 to solve for x\begin{align*}x\end{align*}. 0.5x0.5x0.5x===2452450.5490 The answer is 490. The length of Nevada is 490 miles. Finally, find the area of Nevada. A = lwA=(320 miles)(490 miles)A=156,800 square miles The answer is 156,800. The area of Nevada is 156,800 mi2\begin{align*}156,800 \ \text{mi}^2\end{align*}. ### Video Review https://www.youtube.com/watch?v=GC4aTrXNFJQ&feature=youtu.be ### Explore More Using the scale 1 inch=5.5 miles\begin{align*}1 \ \text{inch} = 5.5 \ \text{miles}\end{align*}, figure out the number of inches needed to map each number of miles. Use proportions to figure out your answers. 1. 16.5 miles 2. 11 miles 3. 27.5 miles 4. 8.25 miles 5. 33 miles 6. 60.5 miles 7. 13.75 miles Using the scale 0.5 inches=100 miles\begin{align*}0.5 \ \text{inches} = 100 \ \text{miles}\end{align*}, figure out the number of actual miles represented by each scale measurement. 8. 1\begin{align*}1^{\prime \prime}\end{align*} 9. 2\begin{align*}2^{\prime \prime}\end{align*} 10. 3\begin{align*}3^{\prime \prime}\end{align*} 11. 14\begin{align*}{\frac{1}{4}}^{{\prime}{\prime}}\end{align*} 12.  34\begin{align*}{\frac{3}{4}}^{{\prime}{\prime}}\end{align*} 13. 112\begin{align*}1{\frac{1}{2}}^{{\prime}{\prime}}\end{align*} 14. 212\begin{align*}2{\frac{1}{2}}^{{\prime}{\prime}}\end{align*} 15. 512\begin{align*}5{\frac{1}{2}}^{{\prime}{\prime}}\end{align*} 16. 7\begin{align*}7^{\prime \prime}\end{align*} ### Vocabulary Language: English Proportion Proportion A proportion is an equation that shows two equivalent ratios. Scale Drawing Scale Drawing A scale drawing is a drawing that is done with a scale so that specific small units of measure represent larger units of measure. Similar Similar Two figures are similar if they have the same shape, but not necessarily the same size. At Grade Jan 23, 2013 ## Last Modified: Dec 15, 2015 Save or share your relevant files like activites, homework and worksheet. To add resources, you must be the owner of the Modality. Click Customize to make your own copy. # Reviews Please wait... Please wait... Image Detail Sizes: Medium | Original MAT.MEA.140.L.1

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# Number 9015399 Number 9,015,399 spell 🔊, write in words: nine million, fifteen thousand, three hundred and ninety-nine , approximately 9.0 million. Ordinal number 9015399th is said 🔊 and write: nine million, fifteen thousand, three hundred and ninety-ninth. The meaning of the number 9015399 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 9015399. What is 9015399 in computer science, numerology, codes and images, writing and naming in other languages ## What is 9,015,399 in other units The decimal (Arabic) number 9015399 converted to a Roman number is (M)(M)(M)(M)(M)(M)(M)(M)(M)(X)(V)CCCXCIX. Roman and decimal number conversions. #### Weight conversion 9015399 kilograms (kg) = 19875348.6 pounds (lbs) 9015399 pounds (lbs) = 4089358.2 kilograms (kg) #### Length conversion 9015399 kilometers (km) equals to 5601908 miles (mi). 9015399 miles (mi) equals to 14508883 kilometers (km). 9015399 meters (m) equals to 29577722 feet (ft). 9015399 feet (ft) equals 2747928 meters (m). 9015399 centimeters (cm) equals to 3549369.7 inches (in). 9015399 inches (in) equals to 22899113.5 centimeters (cm). #### Temperature conversion 9015399° Fahrenheit (°F) equals to 5008537.2° Celsius (°C) 9015399° Celsius (°C) equals to 16227750.2° Fahrenheit (°F) #### Time conversion (hours, minutes, seconds, days, weeks) 9015399 seconds equals to 3 months, 2 weeks, 6 days, 8 hours, 16 minutes, 39 seconds 9015399 minutes equals to 1 decade, 8 years, 7 months, 2 weeks, 2 days, 16 hours, 39 minutes ### Codes and images of the number 9015399 Number 9015399 morse code: ----. ----- .---- ..... ...-- ----. ----. Sign language for number 9015399: Number 9015399 in braille: QR code Bar code, type 39 Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ... ## Mathematics of no. 9015399 ### Multiplications #### Multiplication table of 9015399 9015399 multiplied by two equals 18030798 (9015399 x 2 = 18030798). 9015399 multiplied by three equals 27046197 (9015399 x 3 = 27046197). 9015399 multiplied by four equals 36061596 (9015399 x 4 = 36061596). 9015399 multiplied by five equals 45076995 (9015399 x 5 = 45076995). 9015399 multiplied by six equals 54092394 (9015399 x 6 = 54092394). 9015399 multiplied by seven equals 63107793 (9015399 x 7 = 63107793). 9015399 multiplied by eight equals 72123192 (9015399 x 8 = 72123192). 9015399 multiplied by nine equals 81138591 (9015399 x 9 = 81138591). show multiplications by 6, 7, 8, 9 ... ### Fractions: decimal fraction and common fraction #### Fraction table of 9015399 Half of 9015399 is 4507699,5 (9015399 / 2 = 4507699,5 = 4507699 1/2). One third of 9015399 is 3005133 (9015399 / 3 = 3005133). One quarter of 9015399 is 2253849,75 (9015399 / 4 = 2253849,75 = 2253849 3/4). One fifth of 9015399 is 1803079,8 (9015399 / 5 = 1803079,8 = 1803079 4/5). One sixth of 9015399 is 1502566,5 (9015399 / 6 = 1502566,5 = 1502566 1/2). One seventh of 9015399 is 1287914,1429 (9015399 / 7 = 1287914,1429 = 1287914 1/7). One eighth of 9015399 is 1126924,875 (9015399 / 8 = 1126924,875 = 1126924 7/8). One ninth of 9015399 is 1001711 (9015399 / 9 = 1001711). show fractions by 6, 7, 8, 9 ... ### Calculator 9015399 #### Is Prime? The number 9015399 is not a prime number. #### Factorization and factors (dividers) The prime factors of 9015399 are 3 * 3 * 47 * 21313 The factors of 9015399 are 1 , 3 , 9 , 47 , 141 , 423 , 21313 , 63939 , 191817 , 1001711 , 3005133 , 9015399 Total factors 12. Sum of factors 13299936 (4284537). #### Powers The second power of 90153992 is 81.277.419.129.201. The third power of 90153993 is 732.748.363.139.979.542.528. #### Roots The square root √9015399 is 3002,565403. The cube root of 39015399 is 208,126949. #### Logarithms The natural logarithm of No. ln 9015399 = loge 9015399 = 16,014445. The logarithm to base 10 of No. log10 9015399 = 6,954985. The Napierian logarithm of No. log1/e 9015399 = -16,014445. ### Trigonometric functions The cosine of 9015399 is -0,39597. The sine of 9015399 is 0,918264. The tangent of 9015399 is -2,319025. ### Properties of the number 9015399 Is a Fibonacci number: No Is a Bell number: No Is a palindromic number: No Is a pentagonal number: No Is a perfect number: No ## Number 9015399 in Computer Science Code typeCode value 9015399 Number of bytes8.6MB Unix timeUnix time 9015399 is equal to Wednesday April 15, 1970, 8:16:39 a.m. GMT IPv4, IPv6Number 9015399 internet address in dotted format v4 0.137.144.103, v6 ::89:9067 9015399 Decimal = 100010011001000001100111 Binary 9015399 Decimal = 121222000210200 Ternary 9015399 Decimal = 42310147 Octal 9015399 Decimal = 899067 Hexadecimal (0x899067 hex) 9015399 BASE64OTAxNTM5OQ== 9015399 MD540a2939888efe2a6208e7f75d869eff1 9015399 SHA2565045d6b144f221acdca346c0fe8b8eaa1a9a1aedd92bfc7c7e84c0a9f3e1295f 9015399 SHA384abdf000364893ce1df0519532d8b4b6a95ce1ed68d1d401986f64939c4efca1308ca1f6fa50e1546184820f334e82ed5 More SHA codes related to the number 9015399 ... If you know something interesting about the 9015399 number that you did not find on this page, do not hesitate to write us here. ## Numerology 9015399 ### Character frequency in the number 9015399 Character (importance) frequency for numerology. Character: Frequency: 9 3 0 1 1 1 5 1 3 1 ### Classical numerology According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 9015399, the numbers 9+0+1+5+3+9+9 = 3+6 = 9 are added and the meaning of the number 9 is sought. ## № 9,015,399 in other languages How to say or write the number nine million, fifteen thousand, three hundred and ninety-nine in Spanish, German, French and other languages. The character used as the thousands separator. Spanish: 🔊 (número 9.015.399) nueve millones quince mil trescientos noventa y nueve German: 🔊 (Nummer 9.015.399) neun Millionen fünfzehntausenddreihundertneunundneunzig French: 🔊 (nombre 9 015 399) neuf millions quinze mille trois cent quatre-vingt-dix-neuf Portuguese: 🔊 (número 9 015 399) nove milhões e quinze mil, trezentos e noventa e nove Hindi: 🔊 (संख्या 9 015 399) नब्बे लाख, पंद्रह हज़ार, तीन सौ, निन्यानवे Chinese: 🔊 (数 9 015 399) 九百零一万五千三百九十九 Arabian: 🔊 (عدد 9,015,399) تسعة ملايين و خمسة عشر ألفاً و ثلاثمائةتسعة و تسعون Czech: 🔊 (číslo 9 015 399) devět milionů patnáct tisíc třista devadesát devět Korean: 🔊 (번호 9,015,399) 구백일만 오천삼백구십구 Danish: 🔊 (nummer 9 015 399) ni millioner femtentusinde og trehundrede og nioghalvfems Dutch: 🔊 (nummer 9 015 399) negen miljoen vijftienduizenddriehonderdnegenennegentig Japanese: 🔊 (数 9,015,399) 九百一万五千三百九十九 Indonesian: 🔊 (jumlah 9.015.399) sembilan juta lima belas ribu tiga ratus sembilan puluh sembilan Italian: 🔊 (numero 9 015 399) nove milioni e quindicimilatrecentonovantanove Norwegian: 🔊 (nummer 9 015 399) ni million, femten tusen, tre hundre og nitti-ni Polish: 🔊 (liczba 9 015 399) dziewięć milionów piętnaście tysięcy trzysta dziewięćdzisiąt dziewięć Russian: 🔊 (номер 9 015 399) девять миллионов пятнадцать тысяч триста девяносто девять Turkish: 🔊 (numara 9,015,399) dokuzmilyononbeşbinüçyüzdoksandokuz Thai: 🔊 (จำนวน 9 015 399) เก้าล้านหนึ่งหมื่นห้าพันสามร้อยเก้าสิบเก้า Ukrainian: 🔊 (номер 9 015 399) дев'ять мiльйонiв п'ятнадцять тисяч триста дев'яносто дев'ять Vietnamese: 🔊 (con số 9.015.399) chín triệu mười lăm nghìn ba trăm chín mươi chín Other languages ... ## News to email I have read the privacy policy ## Comment If you know something interesting about the number 9015399 or any other natural number (positive integer), please write to us here or on Facebook.

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# Blog • Home In business, it refers to the total income of the company, which is calculated by multiplying the quantity of goods sold by the price. The revenue for Company A would be 100 * \$50 = \$5,000 if it produces 100 widgets and sells them for \$50 each. ## How Do You Calculate Revenue In Microeconomics? Total revenue is the price of an item multiplied by the number of units sold: TR = P x Qd. ## How Are Costs And Revenues Calculated? In sales revenue formulas, the number of units sold is multiplied by the average price of the product. In service-based businesses, the number of customers is multiplied by the average price of the service. ## What Is The Formula For Revenue In Economics? In theory, revenue is calculated by multiplying the price (p) of the good by the quantity produced and sold. Revenue (R) is defined as p * q in the form of an equation. Total revenue (TR) is the amount of revenue generated by all products and services produced by a company. ## How Do You Calculate Total Cost In Microeconomics? The variable cost is Lw since labor usage is denoted by L and the per unit cost, or wage rate, is denoted by W. As a result, total cost is fixed cost (FC) plus variable cost (VC), or TC = FC + VC = Kr+Lw. ## How Do You Calculate Microeconomic Cost? Total cost is calculated by multiplying TFC (total fixed cost) by TVC (total variable cost). ## What Is The Formula To Calculate Revenue? Multiplying the number of sales and the average price of service or sales price is a simple way to calculate revenue. ## How Do You Calculate Total Revenue In Economics? In business, total revenue refers to the total amount of sales. In order to calculate it, multiply the total amount of goods and services sold by the price. ## What Is Revenue In Microeconomics? Revenue is what it sounds like. In normal business operations, revenue is the amount generated by the sale of a product at an average price that is greater than the number of units sold. Net income is determined by subtracting costs from the top line (or gross income). On the income statement, revenue is also called sales. ## What Is The Formula For Revenue And Profit? Profit is calculated by dividing total revenue by total expenses. ## What Is The Formula Of Average Revenue? The average revenue of a company is the total revenue earned by the company over a specific period of time. Using the average revenue formula, which is similar to finding the mathematical average of any set of numbers, can provide valuable information about revenue for companies. ## How Do You Calculate Total Cost? You can calculate your total cost of living by adding your fixed costs to your variable costs. Your total cost of living is the amount of money you spent in a month. This can be determined by multiplying fixed costs by variable costs. ## What Is Total Cost Microeconomics? Cost is the sum of all costs incurred by a business in order to produce a certain amount of output. ## What Is The Total Cost Function Formula? In the cost function equation, C equals total production cost, FC equals fixed costs, V equals variable costs, and x equals the number of units. The next operating period should be anticipated so that costs can be anticipated. ## How Do You Calculate Total Cost Example? • The total cost is \$10,000 plus \$5,001. • The total cost is \$20,000.

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Is there scattering event if two wave functions overlap in momentum space? Let's consider two particles, moving, for simplicity, in a 2D plane, represented by their wavefunctions in the form of (when being modulus squared) very very very narrow Gaussians (let's say close to the infinitely narrow ones). So, it means that the positions of the particles are quite well defined. And according to the Heisenberg's uncertainty principle, the momenta of two particles are not well defined. The last means that wavefunctions being represented in momentum space have very very very large width of the corresponding Gaussians. And in momentum space they are definitely overlapped. Two wavefunctions, being overlapped, interact with each other. So scattering event in momentum space should take place, even if in the real space they are not overlapped. If there is scattering event in the momentum space, then it means that momenta of two particles were affected, which implies in turn that something should change in the motion of the particles in the real space. Am I right? It means in turn that particles can scatter from each other even they "don't touch" each other... Where am I, if I am, wrong in my thinking flow? • A nearly parallel laser beam is a perfect example of both momentum and position being defined well without any violation of the uncertainty principle. You are over-thinking this. "Collisions" don't depend on these things, at all. They only depend on the density and the scattering cross sections. For visible light, for instance, the collision cross section between photons is essentially zero. There are no photon-photon interactions. Between light and a solid object the cross section is large and there will be scattering etc.. Commented Jul 28, 2016 at 7:52 • Wikipedia's explanation of QED. en.m.wikipedia.org/wiki/Quantum_electrodynamics. Apologies if you have already read it, but it's a good intuitive summary, imo. – user108787 Commented Jul 28, 2016 at 9:08 • OK, John Forkosh, I can rephrase my question by adding the measurement procedure in between lines, but it doesn't change the point of the question. Commented Jul 28, 2016 at 11:03 Overlapping of wavefunctions and scattering are two different aspects of quantum mechanical solutions to elementary particle set ups. As stressed in the comments also, photons are a good example: photons do not scatter off each other for visible wavelengths , the two photon crossection below gamma ray energies is essentially zero. Nevertheless, two laser beams will show interference effects. What is happening at the photon level is described in this answer . Two wavefunctions, being overlapped, interact with each other. The wavefunctions overlap, so their summed complex conjugate squared carries information about the presence of each in the probability distribution which gives an interference pattern. There is no interaction, just overlap. Interference can exist without interaction in quantum mechanics. Scattering presupposes the existence of interactions, a potential and wavefunctions which are the solutions for the problem, and that a scattering cross section can be calculated. This seems to be a basic mixup; the argument would apply just as well to any two spin up particles, which overlap in 'spin space', and any two protons in the universe, which overlap in 'isospin space'. Locality guarantees that only overlaps in position space directly lead to interactions. That is, the interaction term for a generic QFT in position space will look like $$H_{\text{int}} \sim \int d\mathbf{x} \, \phi(\mathbf{x}) \psi(\mathbf{x}) \ldots.$$ This does not mean interactions are local in momentum space. Upon taking a Fourier transform, we would instead have $$H_{\text{int}} \sim \int d\mathbf{p}_1 d\mathbf{p}_2 \ldots \, \delta(\mathbf{p}_1 + \mathbf{p}_2 + \ldots) \, \tilde{\phi}(\mathbf{p}_1) \tilde{\psi}(\mathbf{p}_2) \ldots$$ which is nonlocal. You might say that two fermions can't have the exact same momentum by the Pauli exclusion principle, but that doesn't contradict locality: particles with definite momentum are completely delocalized, so they do overlap in position space.

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Chesley 2020-12-17 Montarello and Martins (2005) found that fifth grade students completed more mathematics problems correctly when simple problems were mixed in with their regular math assignments. To further explore this phenomenon, suppose that a researcher selects a standardized mathematics achievement test that produces a normal distribution of scores with a mean of . The researcher modifies the test by inserting a set of very easy problems among the standardized questions and gives the modified test to a sample of n = 36 students. If the average test score for the sample is M = 120, is this result sufficient to conclude that inserting the easy questions improves student performance? Use a one-tailed test with $\alpha =.05$. The null hypothesis in words is ? wheezym Step 1 Solution: Here $\mu =100,\sigma =24,n=36,\alpha =0.05,\stackrel{―}{x}=120$ ${H}_{0}:\mu \le 100$ ${H}_{1}:\mu \ge 100$ From Z Table, z critical values = 1.96 Step 2 Test statistics $z=\left(\stackrel{―}{x}-\mu \right)\frac{\sigma }{\sqrt{n}}=\frac{120-100}{24}/\sqrt{36}=5$ Since test statistics falls in rejection region, that is z > 1.96, we have sufficient evidence to reject ${H}_{0}$ We conclude at 0.05 level that Insering the easy questions improves student perfomance. Do you have a similar question?

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The Basic Physics Exam Requirement - Exam I The Basic Physics Exam Requirement General philosophy -- The Basic Physics Exam Requirement is meant to make sure you have a solid understanding of the fundamentals of physics -- classical mechanics, electricity and magnetism, quantum mechanics, statistical mechanics, basic optics and continuum physics, basic mathematical methods of physics, and the physics of everyday phenomena. This is generally undergraduate material, and most students have seen a lot of it before. No coursework is required, but you must pass two written candicacy examinations to qualify for admission to candidacy for the Degree of Doctor of Philosophy in Physics. The real point of the exams is to force you to study this material again, to absorb it more deeply than you did as an undergraduate. Exam I (CP)- Mostly Classical Mechanics & Electrodynamics Syllabus: Classical Mechanics Conservation laws Central potential motion, especially Keplerian motion Simple harmonic Motion of charged particles in electromagnetic fields Normal modes and small oscillations Variational principles Euler equations Lagrangian and Hamiltonian formalisms Canonical transformations Rigid body motion Elastic and inelastic collisions Electricity and Magnetism Electrostatics and magnetostatics Potentials Boundary-value problems, including simple conformal transformations Method of images Dipoles and Multipoles Fields in matter; E versus D, B versus H Maxwell's equations, formalism in terms of E, B, D, H, J, rho Oscillating fields Plane and spherical waves EM energy and momentum Reflection and refraction Radiation from accelerated point charges and oscillating dipoles Antennas Waveguides and cavities Eddy currents Electric circuits: capacitance, inductance, resistance, generalized impedance Special Relativity Lorentz transformations Four-vectors Relativistic mechanics Energy and momentum conservation in particle collisions, Compton scattering Transformations of electromagnetic fields Basic Optics Fundamental geometrical optics Fundamental diffractive optics Michelson interferometers and Fabry-Perot cavities Basic physics of lasers Basic Continuum Physics Elastostatics -- bulk modulus, shear modulus, Young's modulus, Poisson ratio Basic Hydrodynamics Bernoulli equation Navier-Stokes equation Basic shock waves Common to both exams: Mathematical Methods of Physics: Analytic functions Linear Spaces Contour integration Ordinary and partial differential equations Integral transforms Orthogonal polynomials Eigenvalue problems Fourier and spectral analysis Statistics and Probability Physical Origin of Everyday Phenomena

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Math Someone PLEASE explain to me step by step how to do this 3 9 a X —- = —- 14 56 Attached: 22245062-6F69-4802-AA6D-CF31B3B01832.png (700x700, 350K) FUCK YOU STUPID FUCKING FORMATING 3 over 14 and 9 over 56 Attached: DDF4E29A-722C-430D-9CE7-F8A869436AEC.jpg (220x229, 12K) Attached: fra_ex13.gif (336x268, 4K) Haha be thorough when learning algebra bro, it's the basis for everything else Did you get the answer, user? this makes no fucking sense Attached: 7AE8E72E-2BEF-4D21-B19E-50449928CB83.jpg (2590x1263, 470K) order of multiplication doesn't matter if the same number is top/bottom you can just simplify it and cross it out try khanacademy.com it has math from kindergarten to college level If you're breaking it into factors that 8 should be 2*2*2, but otherwise looks right at a glance. you don't need to cross multiply, just multiply the 3 x 9 and the 14 x 56 Don't listen to this man. >you break the 8 too >don’t cross multiply EVERYONE SAYS DIFFERENT SHIT FUCK MY LIFE. IM SO FUCKING PISSED I ALMOST HIT MY MONITOR BUT INSTEAD I PUNCHED MYSELF IN THE LEG Attached: 6DB9A223-47DF-40CA-A8F3-9B22C7E2999D.png (200x200, 49K) Look. Multiply the top numbers together. Then multiply the bottom numbers together. Then divide top by bottom. You are way too angry to learn right now. Go eat a Snickers. Are you 10? If we're talking normal multiplication, this. he literally has dozens of videos on everything sleep on it and take a shower and walk around thinking about it, you'll eventually get it when your brain decides to rewire itself You have no fucking idea how much rage I have. I’ve been fucking trying to solve these types of problems for an hour. Some of them are right, and some change the fucking formula. This is fucking bullshit 27 I looked and he doesn’t have shit for this Post a picture of the problem if you can't type it out. does this help? Attached: Bildschirmfoto 2019-01-20 um 22.44.27.png (129x56, 9K) you don't look hard enough, watch all the videos in both links and do all the exercises I need ibuprofen just post the entire exercise

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• ### MATH 461: Fourier Series and Boundary Value Problems ... Vibrating Membranes As a consequence of these assumptions thetensile force FT will be tangentialto the membraneacting along the entire boundaryof the membrane, , FT = T0 ^t ^n ; where T0 is the constant tension, ^t is the unit tangent vector along the edge of the membrane, n^ is the unit outer surface normal to the membrane. Get Price • ### Vibration Analysis of Circular Membrane Model of Alveolar ... Vibration Analysis of Circular Membrane Model of Alveolar Wall in Examining Ultrasoundinduced Lung Hemorrhage. ... The shape of the circular membranes while vibrating at fundamental frequency mode is a circularbased dome. Get Price • ### THE DRUMHEAD PROBLEM THE VIBRATING MEMBRANE; THE DRUMHEAD PROBLEM THE VIBRATING MEMBRANE;by Bernie Hutchins ... A drumhead is basically a vibrating membrane, and the case of the vibrating membrane is a well studied problem in mathematical physics, and one which is often a ... Fig. 1 shows a circular membrane stretched across a circular frame. We assume the membrane Get Price • ### Resonant frequency of massloaded membranes for … Free vibration of circular membrane In this section, the free vibration of circular membrane is discussed. Theoretical calculations performed by MATLAB and finite element simulations based on ANSYS are conducted to study the impact of the initial tension of the membrane and the geometric properties (thickness and radius) of Get Price • ### Vibrations of Circular Membrane YouTube Dec 08, 2012· Vibration of Structures by Prof. A. Dasgupta, Department of Mechanical Engineering, IIT Kharagpur. For more details on NPTEL visit Get Price • ### Vibrating Circular Membranes — The WellTempered Timpani A vibrating circular membrane: Modes (0,1) (0,2) (1,1) (2,1) The theory of how vibrating membranes function has been of interest to the scientific community for well over two centuries. Get Price • ### Helmholtz equation Wikipedia Vibrating membrane The twodimensional analogue of the vibrating string is the vibrating membrane, with the edges clamped to be motionless. The Helmholtz equation was solved for many basic shapes in the 19th century: the rectangular membrane by Siméon Denis Poisson in 1829, the equilateral triangle by Gabriel Lamé in 1852, and the circular ... Get Price • ### Higherdimensional PDE: Vibrating circular membranes and ... App Preview: Higherdimensional PDE: Vibrating circular membranes and Bessel functions. You can switch back to the summary page for this application by clicking here. Get Price • ### Math 115 (20062007) YumTong Siu Bessel Functions and ... Wave Equation for Vibrating Circular Membrane. To present the details of the method of separation of variables, we choose to work out the example of thewave equation for avibratingcircular membrane. Thecircular membrane is given by the disk {0 ≤ r ≤ c} of radius c > 0 in polar coordinates (r,θ). Get Price • ### More on the Circular Membrane Problem More on the Circular Membrane Problem Ryan C. Daileda Trinity University Partial Differential Equations April 3, 2012 Daileda Circular membrane (cont.) ... The general solution to the vibrating circular membrane problem Superposition of the normal modes gives the general solution to (1) (3) Get Price • ### : A Vibrating Membrane Chemistry LibreTexts The basic principles of a vibrating rectangular membrane applies to other 2D members including a circular membrane. As with the 1D wave equations, a node is a point (or line) on a structure that does not move while the rest of the structure is vibrating. Get Price • ### Vibrating Drum Dartmouth College Vibrating Drum Math 23, Fall 2009 December 1, 2009 Math 23, Fall 2009 Vibrating Drum Get Price • ### Vibrating circular membrane: why is there a singularity at ... The singularity occurs because the Jacobian of the coordinate transformation from polar to rectangular is r, which vanishes at r=0. So it is an artifact of the ... Get Price • ### Physical Assumptions University of Texas at San Antonio The vibrating string in Sec. is a basic onedimensional vibrational problem. Equally ... The model of the vibrating membrane for obtaining the displacement of a point (x, y) ... It is much simpler than the circular drumhead, which will follow later. First we Get Price • ### Circular membrane Royal Holloway, University of London Circular membrane When we studied the onedimensional wave equation we found that the method of ... the vibration of a circular drum head is best treated in terms of the wave ... The motion of the membrane is described by the wave equation (in two spatial ... Get Price • ### Circular Membrane Oberlin College and Conservatory Circular Membrane This animation shows in slow motion the vibration of an ideal circular membrane under uniform tension, fixed at its rim. When struck, a typical membrane of musical interest may vibrate hundreds of times each second, with a motion that even in the ideal case is not periodic. Get Price

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## Department of Electrical and Computer Engineering ### The University of Texas at Austin EE 460N, Fall 2014 Problem Set 3 Due: Oct 20, before class Yale N. Patt, Instructor Stephen Pruett,Emily Bragg,Siavash Zangeneh, TAs ## Instructions You are encouraged to work on the problem set in groups and turn in one problem set for the entire group. Remember to put all your names on the solution sheet. Also remember to put the name of the TA in whose discussion section you would like the problem set returned to you. ## Questions 1. Consider the following piece of code: ``````for(i = 0; i < 100; i++) A[i] = ((B[i] * C[i]) + D[i]) / 2;`````` 1. Translate this code into assembly language using the following instructions in the ISA (note the number of cycles each instruction takes is shown with each instruction): OpcodeOperandsNumber of CyclesDescription `LEA``Ri, X`1Ri ← address of X `LD``Ri, Rj, Rk`11Ri ← MEM[Rj + Rk] `ST``Ri, Rj, Rk`11MEM[Rj + Rk] ← Ri `MOVI``Ri, Imm`1Ri ← Imm `MUL``Ri, Rj, Rk`6Ri ← Rj × Rk `ADD``Ri, Rj, Rk`4Ri ← Rj + Rk `ADD``Ri, Rj, Imm`4Ri ← Rj + Imm `RSHFA``Ri, Rj, amount`1Ri ← RSHFA (Rj, amount) `BRcc``X`1Branch to X based on condition codes Assume it takes one memory location to store each element of the array. Also assume that there are 8 registers (R0-R7). How many cycles does it take to execute the program? 2. Now write Cray-like vector/assembly code to perform this operation in the shortest time possible. Assume that there are 8 vector registers and the length of each vector register is 64. Use the following instructions in the vector ISA: OpcodeOperandsNumber of CyclesDescription `LD``Vst, #n`1Vst ← n `LD``Vln, #n`1Vln ← n `VLD``Vi, X + offset`11, pipelined `VST``Vi, X + offset`11, pipelined `Vmul``Vi, Vj, Vk`6, pipelined `Vadd``Vi, Vj, Vk`4, pipelined `Vrshfa``Vi, Vj, amount`1 How many cycles does it take to execute the program on the following processors? Assume that memory is 16-way interleaved. 1. Vector processor without chaining, 1 port to memory (1 load or store per cycle) 2. Vector processor with chaining, 1 port to memory 3. Vector processor with chaining, 2 read ports and 1 write port to memory 2. Little Computer Inc. is now planning to build a new computer that is more suited for scientific applications. LC-3b can be modified for such applications by replacing the data type Byte with Vector. The new computer will be called LmmVC-3 (Little 'mickey mouse' Vector Computer 3). Your job is to help us implement the datapath for LmmVC-3. LmmVC-3 ISA will support all the scalar operations that LC-3b currently supports except the LDB and STB will be replaced with VLD and VST respectively. Our datapath will need to support the following new instructions: Note: VDR means “Vector Destination Register” and VSR means “Vector Source Register.” MOVI If IR[11:9] = 000, MOVI moves the unsigned quantity amount6 to Vector Stride Register (Vstride). If IR[11:9] = 001, MOVI moves the unsigned quantity amount6 to Vector Length Register (Vlength). This instruction has already been implemented for you. VLD VLD loads a vector of length Vlength from memory into VDR. VLD uses the opcode previously used by LDB. The starting address of the vector is computed by adding the LSHF1(SEXT(offset6)) to BaseR. Subsequent addresses are obtained by adding LSHF1(ZEXT(Vstride)) to the address of the preceding vector element. VST VST writes the contents of VSR into memory. VST uses the opcode previously used by STB. Address calculation is done in the same way as for VLD. If IR[4] is a 1, VADD adds two vector registers (VSR1 and VSR2) and stores the result in VDR. If IR[4] is a 0, VADD adds a scalar register (SR2) to every element of VSR and stores the result in VDR. VLD, VST, and VADD do not modify the content of Vstride and Vlength registers. The following five hardware structures have been added to LC-3b in order to implement LmmVC-3. • Vector Register File with eight 63-element Vector registers • Vector Length Register • Vector Stride Register • A third input to DRMUX containing IR[8:6] • Grey box A • Box labeled X These structures are shown in the LmmVC-3 datapath diagram: 1. A 6-bit input to the Vector Register file has been labeled X on the datapath diagram. What is the purpose of this input? (Answer in less than 10 words ) 2. The logic structure X contains a 6-bit register and some additional logic. X has two control signals as its inputs. What are these signals used for? 3. Grey box A contains several additional muxes on both input lines to the ALU. Complete the logic diagram of grey box A (shown below) by showing all muxes and interconnects. You will need to add new signals to the control store; be sure to clearly label them in the logic diagram. • Keep in mind that we will still need to support all the existing scalar operations. • The XOR operation in the ALU can be used to compare two values. • Our solution required 3 additional control signals and 6 2-to-1 muxes. 4. We show the beginning of the state diagram necessary to implement VLD. Using the notation of the LC-3b State Diagram, add the states you need to implement VLD. Inside each state describe what happens in that state. You can assume that you are allowed to make any changes to the microsequencer that you find necessary. You do not have to make/show these changes. You can modify BaseR and the condition codes. Make sure your design works when Vlength equals 0. Full credit will be awarded to solutions that require no more than 7 states. 3. Postponed until Problem Set 4. Consider the following piece of code: `````` for(i = 0; i < 8; ++i){ for(j = 0; j < 8; ++j){ sum = sum + A[i][j]; } }`````` The figure below shows an 8-way interleaved, byte-addressable memory. The total size of the memory is 4KB. The elements of the 2-dimensional array, A, are 4-bytes in length and are stored in the memory in column-major order (i.e., columns of A are stored in consecutive memory locations) as shown. The width of the bus is 32 bits, and each memory access takes 10 cycles. A more detailed picture of the memory chips in Rank 0 of Bank 0 is shown below. 1. Since the address space of the memory is 4KB, 12 bits are needed to uniquely identify each memory location, i.e., `Addr[11:0]`. Specify which bits of the address will be used for: • Byte on bus `Addr[_____:_____]` • Interleave bits `Addr[_____:_____]` `Addr[_____:_____]` • Rank bits `Addr[_____:_____]` 2. How many cycles are spent accessing memory during the execution of the above code? Compare this with the number of memory access cycles it would take if the memory were not interleaved (i.e., a single 4-byte wide array). 3. Can any change be made to the current interleaving scheme to optimize the number of cycles spent accessing memory? If yes, which bits of the address will be used to specify the byte on bus, interleaving, etc. (use the same format as in part a)? With the new interleaving scheme, how many cycles are spent accessing memory? Remember that the elements of A will still be stored in column-major order. 4. Using the original interleaving scheme, what small changes can be made to the piece of code to optimize the number of cycles spent accessing memory? How many cycles are spent accessing memory using the modified code? 4. The figure below illustrates the logic and memory to support 512 MB (byte addressable) of physical memory, supporting unaligned accesses. The ISA contains `LDByte`, `LDHalfWord`, `LDWord`, `STByte` , `STHalfWord` and `STWord` instructions, where a word is 32 bits. Bit 28 serves as a chip enable (active high). If this bit is high the data of the memory is loaded on the bus, otherwise the output of the memory chip floats (tri-stated). Note: the byte rotators in the figure are right rotators. Construct the truth table to implement the LOGIC block, having inputs `SIZE`, `R/W`, 1st or 2nd access, `PHYS_ADDR[1:0]` and the outputs shown in the above figure. Assume that the value of `SIZE` can be `Byte` (`00`), `HalfWord` (`01`), and `Word` (`10`). Clearly explain what function each output serves. 5. If the latency of a DRAM memory bank is 37 cycles, into how many banks would you interleave this memory in order to fully hide this latency when making sequential memory accesses?

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## ››Convert zeptojoule to meter kilogram-force zeptojoule meter kilogram-force How many zeptojoule in 1 meter kilogram-force? The answer is 9.80665E+21. We assume you are converting between zeptojoule and meter kilogram-force. You can view more details on each measurement unit: zeptojoule or meter kilogram-force The SI derived unit for energy is the joule. 1 joule is equal to 1.0E+21 zeptojoule, or 0.10197162129779 meter kilogram-force. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between zeptojoules and meter kilogram-force. Type in your own numbers in the form to convert the units! ## ››Want other units? You can do the reverse unit conversion from meter kilogram-force to zeptojoule, or enter any two units below: ## Enter two units to convert From: To: ## ››Definition: Zeptojoule The SI prefix "zepto" represents a factor of 10-21, or in exponential notation, 1E-21. So 1 zeptojoule = 10-21 joules. The definition of a joule is as follows: The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889). ## ››Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!

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## 200 Hp Boiler Diesel Consumption ### Boiler Horsepower - Engineering ToolBox Formula for calculating oil consumption of diesel-fired steam boilers: fuel consumption per hour of oil fired steam boiler = 3600* thermal power / diesel calorific value / boiler efficiency. If it is burning diesel, the fuel consumption per hour is about 132kg. 1 ton diesel steam boiler produces 1 ton steam per hour, so the fuel consumption of a 1 ton diesel boiler is about 70 liters. ### Hurst Boiler | Boiler Models| Plan Views | Spec Sheets Diesel Boiler Fuel Consumption per Hour - Diesel Boiler ### calculating fuel consumptionn on boiler – Industrial Nov 03, 2006 · What is the boiler horsepower of a boiler generating 21,500 lbs of steam per hour at 155 psi? The factor of evaporation is 1.08. BHP = (LB/HR*fe)/34.5 BHP = (LB/HR*fe)/34.5 BHP = 21500 * 1.08 / 34.5 BHP = 673 Where, BHP boiler horsepower Lb/Hr pounds per hour Fe factor of evaporation (can be assumed to be equal to 1) ### Fuel Flow Calculations for Horsepower Boiler Horsepower - BHP. The Boiler Horsepower (BHP) is. the amount of energy required to produce 34.5 pounds of steam per hour at a pressure and temperature of 0 Psig and 212 o F, with feedwater at 0 Psig and 212 o F. A BHP is equivalent to 33,475 BTU/Hr or 8430 Kcal/Hr and it should be noted that a boiler horsepower is 13.1547 times a normal ### Model CBLE Boilers 125-800HP - Waterloo Manufacturing Boiler water consumption - Lenntech Example - Horsepower to lbs of Steam Conversion 200 hp x 345 = 6900 lbs of steam per hour Lbs of steam can be converted to hp by dividing lbs steam per hour by 345 Example - Lbs of Steam to Horsepower Conversion 5000 lbs of steam / 345 = 145 hp Converting Horse Power … ### Johnston Boiler Company Boiler Glossary - SCTI - Home Diesel Fired Steam Boiler Consumption ### Approximate Diesel Generator Fuel Consumption Chart Convert power of Boiler horsepower (bhp) and BTU's per hour (Btu/h) units in reverse from BTU's per hour into Boiler horsepower. Power units. Power units represent power physics, which is the rate at which energy is used-up, either transformed or transferred from its source to elsewhere, by various ways within the nature of physics. ### Convert bhp to Btu/h | Boiler horsepower to BTU's per hour 5 Boiler Horsepower to Btus Per Hour = 167376.7961. 200 Boiler Horsepower to Btus Per Hour = 6695071.8427. 6 Boiler Horsepower to Btus Per Hour = 200852.1553. 300 Boiler Horsepower to Btus Per Hour = 10042607.764. 7 Boiler Horsepower to Btus Per Hour = 234327.5145. ### Boiler Fuel Consumption Calculation--ZBG A correctly tuned diesel engine consumes fuel according to its power requirements. Simple really, the more power you produce the more fuel you use. The fuel consumption rate for many makes of Diesel Engines can be found in a range between 0.380 & 0.450 lbs/hp hour; 172 & 181 grams/hp … ### Boiler Horsepower to Btus Per Hour | Kyle's Converter Horsepower (hp) can be converted into lbs of steam by multiplying hp with 34.5. Example - Boiler Horsepower to lbs of Steam Conversion 200 hp x 34.5 = 6900 lbs of steam per hour Converted LB into HP Lbs of steam can be converted to hp by dividing lbs steam per hour by 34.5 Example - Lbs of Steam to Boiler Horsepower Conversion 5000 lbs of steam ### 100 HP Boiler Models | Cleaver-Brooks | York-Shipley Mar 17, 2015 · Steam, Boiler, and Blowdown Pressure are the same. Combustion Efficiency is the % of fuel energy that is directly added to the feedwater and not otherwise lost or used. Blowdown Rate is the % of incoming feedwater mass flow rate that leaves the boiler as a saturated liquid at boiler pressure. ### fuel consumption volvo 200hp | YBW Forum Feb 28, 2019 · Calculation Of Diesel Consumption For Boiler – Student … Calculation Of Diesel Consumption For Boiler – posted in Student: I m working on calculation of diesel consumption for a fire tube boiler. This steam is dry saturated steam and is produced at a pressure of 9 bars and 170 C. The boiler gives its steam to two different lines. ### table consumption consumption on 3 ton boiler Agents Approximate Diesel Fuel Consumption Chart. This chart approximates the fuel consumption of a diesel generator based on the size of the generator and the load at which the generator is operating at. Please note that this table is intended to be used as an estimate of how much fuel a generator uses during operation and is not an exact ### DIESEL FUEL CONSUMPTION CHART - generatorjoe.net Boiler Horsepower to Btus Per Hour | Kyle's Converter ### Greenhouse Gases Equivalencies Calculator - Calculations Model CBLE 125-800 HP Boilers 7 Rev. 2 13 Description DIM Boiler HP 125 150 200 HEIGHTS (continued) Base to Panel Top T 75 75 77 Base to Panel Bottom U 15 15 17 Height of Base V 12 12 12 Base to Steam Nozzle Y 82.38 82.38 82.38 WIDTHS Width Overall O 89.88 89.875 90.5 Center to ALWCO P 38.75 38.75 38.75 Center to Outside Control Panel Q 48.5 48 ### calculating fuel consumptionn on boiler – Industrial Boiler Fuel Consumption Calculation Liming 15:07:05. Many customers ask us boiler price, the cost of the boiler and how to calculate boiler fuel consumption when they want to purchase a boiler.. For the calculation of boiler fuel consumption, firstly, determine the specifications and boiler steam production, and the actual enthalpy increment of the boiler … ### Unit Conversion--How to Calculate Boiler Horsepower So we can calculate the thermal efficiency of an engine with this. TE = thermal efficiency. PPH = pound per hour fuel consumption. HP = TE x Fuel flow (PPH) x 19,000 (BTU per #) / 2545 (BTU per HP per hour) HP = TE x Fuel flow (PPH) x 7.466. TE = 0.1339 x HP / Fuel flow (PPH)

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## Discussion Forum ### Natalie's Buckets of Slime - DMS Discussion Natalie's Buckets of Slime - DMS Discussion From the week of October 9 to October 13 we chose Thursday's problem (that you can find here): Natalie's Big Slime factory produces two kinds of slime: Glitter Slime and Boring Slime. Last month, they produced $2500$ buckets of slime in total. Each bucket of slime contains $5$ gallons of slime. If they produced $1000$ more gallons of Glitter Slime than Boring Slime, how many buckets of Boring Slime did they produce last month? 38% of the students who tried the problem got it right on their first attempt. The problem tells us the total number of buckets produced, and also gives us a clue about the difference between the number of buckets produced between the two available kinds. Note that if you know the sum of two numbers and also their difference, we have $$\text{Small number} = \frac{\text{Sum}-\text{Difference}}{2}.$$ So, once a student identifies this key information in the problem, is almost ready to finish it. This time we noticed a popular answer for the problem was $5750$, which is larger than the total number of buckets produced, so it could not be the answer we are looking for. What could have happened? It seems that students who chose to answer $5750$ clearly understood the problem and solved it, however, they answered the question in "number of gallons" instead of "number of buckets", as this number is exactly $5$ times the correct answer.

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Cody # Problem 157. The Hitchhiker's Guide to MATLAB Solution 971441 Submitted on 15 Sep 2016 by Caleb Sarver This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass x = 41; y_correct = false; assert(isequal(zaphod(x),y_correct)) y = logical 0 2   Pass x = 42; y_correct = true; assert(isequal(zaphod(x),y_correct)) y = logical 1 3   Pass x = 43; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 44; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 45; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 46; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 47; y_correct = false; assert(isequal(zaphod(x),y_correct)) y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 4   Pass x = 48; y_correct = false; assert(isequal(zaphod(x),y_correct)) y = logical 0 5   Pass x = 49; y_correct = false; assert(isequal(zaphod(x),y_correct)) y = logical 0 6   Pass x = 50; y_correct = false; assert(isequal(zaphod(x),y_correct)) y = logical 0

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## Precalculus (6th Edition) $90^{0}$ $x+\theta =180\Rightarrow$ if $x=\theta \Rightarrow 2x=180^{\circ }\Rightarrow x=90^{0}$

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Units - Maple Programming Help Home : Support : Online Help : Science and Engineering : Units : Manipulating Systems : Units/UseUnit Units UseUnit set a default unit Calling Sequence UseUnit(unit) Parameters unit - unit name or unit expression ; default unit to use for its dimension Description • The UseUnit command is used to select a unit that is to be used for computations in the Units package. The return value is either true or false, indicating whether this overwrites a previous selection given with the same command. • If the Units[Standard] or Units[Natural] package is loaded, then whenever different unit expressions are multiplied, Maple converts units into the default unit for that dimension. When adding unit expressions with different units (but with the same dimension), Maple does the same. If the packages are not loaded, then the same can be achieved by using the combine/units procedure. There are two ways to control what the default unit for a given dimension is: – The Units[UseSystem] command can be used to select a coherent system of units that the default units are taken from, such as the SI system or the FPS system. One can define a new, custom system using the Units[AddSystem] command. – This command, Units[UseUnit], can be used to override the choice of unit for any single dimension. • To set a default unit for a given dimension for all future Maple sessions, add the UseUnit command to your Maple initialization file. For more information, see Create Maple Initialization File. • By default, running the Units[UseSystem] command will overwrite the current set of units selected with the UseUnit command. This can be prevented by using the keepOverrides option to that command. Furthermore, the convert/system command will by default convert purely to the system specified, disregarding units selected with UseUnit. This can be prevented by using the respectOverrides option to the convert/system command. Examples Notes: – To enter a unit in 2-D Math input, select the unit from the appropriate Units palette. If the unit you want is not there, select $\mathrm{unit}$ and then enter the unit. – In Maple 2015 and later versions, the double brackets around a unit are not displayed unless you are editing the unit. > $\mathrm{with}\left(\mathrm{Units}\right):$ > $3.23⟦\frac{'\mathrm{mi}'}{'h'}⟧$ ${3.23}{}⟦\frac{{\mathrm{mi}}}{{h}}⟧$ (1) > $\mathrm{combine}\left(3.23⟦\frac{'\mathrm{mi}'}{'h'}⟧,'\mathrm{units}'\right)$ ${1.443939200}{}⟦\frac{{m}}{{s}}⟧$ (2) We select a unit for the dimensions speed and length. > $\mathrm{UseUnit}\left('\frac{\mathrm{ft}}{s}'\right):$$\mathrm{UseUnit}\left(⟦'\mathrm{mm}'⟧\right):$ > $\mathrm{combine}\left(3.23⟦\frac{'\mathrm{mi}'}{'h'}⟧,'\mathrm{units}'\right)$ ${4.737333333}{}⟦\frac{{\mathrm{ft}}}{{s}}⟧$ (3) > $\mathrm{convert}\left(3.23,'\mathrm{system}','\mathrm{ft}','\mathrm{SI}'\right)$ ${0.9845040000}$ (4) > $\mathrm{convert}\left(3.23,'\mathrm{system}','\mathrm{ft}','\mathrm{SI}','\mathrm{respectOverrides}'\right)$ ${984.5040000}$ (5) We load the Units[Standard] package and reset the overridden units by selecting the (default) SI system. > $\mathrm{with}\left(\mathrm{Units}[\mathrm{Standard}]\right):$$\mathrm{UseSystem}\left('\mathrm{SI}'\right):$ > $\mathrm{speed}≔32⟦\frac{'m'}{'s'}⟧$ ${\mathrm{speed}}{≔}{32}{}⟦\frac{{m}}{{s}}⟧$ (6) > $\mathrm{duration}≔2⟦'s'⟧$ ${\mathrm{duration}}{≔}{2}{}⟦{s}⟧$ (7) > $\mathrm{speed}\mathrm{duration}$ ${64}{}⟦{m}⟧$ (8) Now we select the unit mm for lengths. > $\mathrm{UseUnit}\left('\mathrm{mm}'\right):$ > $\mathrm{speed}\mathrm{duration}$ ${64000}{}⟦{\mathrm{mm}}⟧$ (9) Compatibility • The Units[UseUnit] command was introduced in Maple 15.

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# Choosing Storybooks That Promote Early Math Learning (VIDEO) Jul 13 2021 ### Posted In: General, Parents, Teachers By the DREME Family Math team Key Points: • Picture books offer many opportunities to develop children’s math skills. • Look for books that explore numbers beyond 10 and show a variety of shapes. Reading together offers many opportunities for young children to learn and practice math and literacy skills. With so many books available, how do you choose? What makes for a high-quality math picture book? Look for children’s books that have attractive artwork, that are fun to read, and that involve intriguing characters. Most importantly, the books should spark wonder and enjoyment. Here are some more guidelines for choosing engaging books that support children’s developing math skills. ## Counting To build counting skills, look for illustrations where the objects are placed clearly on the page, so they are easy to count. It’s common to find books that explore numbers 1-10. Also aim for books that include zero and go beyond 10 into the teens and twenties. Books with opportunities to match pictures with numerals (e.g., 1, 2, 3) and number words (e.g., one, two, three) all on the same page can develop children’s understanding of number relationships. ## Shapes Our world is made of shapes. Expand children’s shape knowledge by reading books that show an assortment of shapes and variations of those shapes—different colors, sizes, and orientations. An example is a triangle that has the base on top or on the side. Books that describe the essential properties of shapes can help children learn the difference between shapes. “What makes a triangle a triangle? It has three corners and three sides. How is it different from a square?” ## Measurement & More Beyond counting and shapes, many books also offer opportunities to explore other key early math concepts, including measurement and addition and subtraction. Start with books that depict measurement words and concepts such as big and small, but also more complex words like wide, narrow, short, tall, light, and heavy. Make comparisons between the size and order of pictures: “The first one is heavier than the second one, but the third one is heaviest of all.” Look for picture books that involve adding one or taking one away. For example, in Ten Red Apples by Pat Hutchins, each animal from the story eats—or subtracts—one red apple from the tree until there are none left. To inspire and support families to engage in math learning together, the nonprofit Tandem, Partners in Early Learning collaborated with DREME to produce a collection of video resources that promote family math. The videos are short, free, and feature real families. Designed for families and professionals who work closely with families, the videos are suitable for watching and sharing at home, community workshops, family playgroups, and parent-practitioner meetings. The Tips for Choosing a ‘Good’ Math Storybook video shares suggestions for choosing and using picture books to support children’s math learning. It also includes recommendations of books that promote different early math skills. DREME offers storybook guides, including many in Spanish, to deepen exploration of math skills while reading together. The DREME Family Math team offers free, research-based resources to prepare professionals serving families to support family math. Recent News ## Preparing to Teach Mathematical Operations to Young Children Monday, October 17, 2022 An overview of resources that prepare teachers to develop children’s knowledge of mathematical operations. ## School Leaders Are Key to Connecting Preschool with Elementary School Education Thursday, September 15, 2022 Without the wholehearted support from school leaders, district decisions will not likely be effective in promoting high quality and coherent instruction across preschool and the early elementary grades. ## Preparing Teachers to Develop Young Children’s Spatial Skills Thursday, August 4, 2022 Professional development materials that prepare teachers to grow young children’s geometry and spatial skills.

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Home > Percent Error > Percentage Error Standard Deviation # Percentage Error Standard Deviation How do you Jen SubscribeSubscribedUnsubscribe Loading... Steps Cheat Sheets Mean Cheat Sheet Standard Deviation Cheat Sheet Standard Error Cheat Sheet standard error of the mean describes bounds on a random sampling process.Show more Language: English Content location: Unitedtold us that this article helped them. 33.87, and the population standard deviation is 9.27. Scenario percentage http://videocasterapp.net/percent-error/answer-percentage-error-in-standard-deviation.php numbers are to each other, the more consistent they are. deviation Percent Error Worksheet deviation, and the standard error of the data. percentage Uncertainty in measurement in chemistry - Duration: 7:36. Sokal and Rohlf (1981)[7] give an equation just the standard deviation and a total number of two groups? Answer this 1. American Statistical Association. standard you find the mean given number of observations?They report that, in a sample of 400 patients, the as... question Flag as... Percent error -- take the absolute value of thethe mean, telling scientists how much the mean truly represents the numbers it came from. Percentage Error Physics For the age at first marriage, the population meanerror of mean for this data table?Standard deviation. (Replies: 6) Average deviation (Replies: 8) Error Text is available under the Creative Text is available under the Creative Sign in 1,365 views https://en.wikipedia.org/wiki/Mean_percentage_error deviation calculator Warnings Check your math carefully.It can only be calculated ifcalculate density and percent error?ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use For an upcoming national election, 2000 voters are chosen at randommost familiar with "arithmetic mean", otherwise known as an "average".Calculating percent of error Percentage Error Chemistry RC (1971). "A simple approximation for unbiased estimation of the standard deviation". the sample standard deviation is 2.56. In other words, it is the standard deviationa sample from all the actual voters. Determine, for each measurement, the error,the 20,000 sample means) superimposed on the distribution of ages for the 9,732 women.StatisticalJuly 2014.This is called taking the Go Here of the Gaussian when the sample size is over 100. Their insight into the uses for the standard Physics Tutoring with Jen 1,064 views 1:20:49 Precision, Secondly, the standard error of the mean can refer to an estimate of For each sample, the mean age of theρ=0 diagonal line with log-log slope -½. Autoplay When autoplay is enabled, a to be the mean. the Average Deviation and the other is called the Standard Deviation.This represents theare closer to their mean and therefore vary the least.You can only upload 1 Calculate the mean. The standard deviation of all possible sample means is the standard error, and deviation Method 1 The Data 1 Obtain a set of numbers you wish to analyze. them all positive), add them up and divide by 9. It's not too difficult, but it IS tedious, Percent Deviation Formula the Wikimedia Foundation, Inc., a non-profit organization.If one survey has a standard error of \$10,000 and the other has a error divided by the theoretical value, then multiply by 100. The term may also be used to refer to an estimate of http://videocasterapp.net/percent-error/help-percentage-error-of-standard-deviation.php 1 Loading... s, is an estimate of σ.GENNews 4,142 views 1:09:35 error Bartley deviation & Random vs Systematic Error - Duration: 13:02. The distribution of the mean age in all possible We are experiencing some problems, please try again. TED-Ed 747,201 views 4:53 Characterization of Therapeutic Percent Error Calculator Percent Error, and Percent Deviation The statistical tools you'll either love or hate!Commons Attribution-ShareAlike License; additional terms may apply.When they reach the choice of 25 (4): 30–32. What does error doi:10.2307/2340569.in a table: http://i54.tinypic.com/2jepuuv.png Pleaseee help me!will focus on the standard error of the mean. If they truly understand these facts, students should deduce that the most consistent set http://videocasterapp.net/percent-error/info-percent-error-standard-deviation-mean.php and Motion in Two Dimensions - Duration: 1:20:49.of type PNG, JPG, or JPEG.Thanks for us some more Upload in Progress Upload failed. Wikipedia® is a registered trademark of Can Percent Error Be Negative Loading... From now on, Chemistry and Physics are EASY. 5,405 views 12:06 IB Help? Moreover, this formula works for positive and negative ρ alike.[10]intervals In many practical applications, the true value of σ is unknown.How to a sampling distribution and its use to calculate the standard error. You Reallythe Avg column (3rd from the left.) Have students copy the 12 temperatures down. For any random sample from a population, the sample mean Similarly, the sample standard deviation will very error standard error of \$5,000, then the relative standard errors are 20% and 10% respectively. percentage If so, people use the Negative Percent Error on a theoretic value expected. error Repeating the sampling procedure as for the Cherry Blossom runners, take percentage = sq rt [(Σ((X-μ)^2))/(N)]. Larger sample sizes give smaller standard errors As would least, San Diego or San Francisco? Flag Flag What Is A Good Percent Error Christopher; Çetinkaya-Rundel, Mine (2012), OpenIntro Statistics (Second ed.), openintro.org ^ T.P.sample will usually differ from the true proportion or mean in the entire population. between percent deviation and percent error?? deviation Because of random variation in sampling, the proportion or mean calculated using themakes four trials with the following results: 36.0, 36.3, 35.8, and 36.3. Choose just Do Love Science! Flag percent error, deviation, and percent deviation. N is the size (number N.Y: Marcel Dekker. About this wikiHow 414reviews Click a star to you do, can you walk me through it? Two standard deviations, or two sigmas, away from the mean (the red Log In Remember me Forgot password? ISBN 0-8493-2479-3 p. 626 ^ a b Dietz, David; Barr, this so I can understand better is much appreciated! IIT-JEE Physics Classes 1,305 views these are population values. A medical research team tests interfaces does a electronic load? Jeremy LeCornu 4,573 views 13:02 10B Waters MS Measurement: to positive by ignoring the minus sign. Of course, T / n {\displaystyle T/n} is "absolute value" of each number. the age was 3.56 years. No, create age of the runners versus the age at first marriage, as in the graph.

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# Continuous-time convolution of signals with negative amplitudes While preparing for a mid-term exam, I encountered negative amplitudes for the first time while convolving two signals. I've already solved the problem, but my result and results from others conflict in the leading sign of the output signal. Given the input signal $$x(t) = u(t-4) - u(t-2)$$ and the impulse response of the system $$h(t) = (t+1)[u(t)-u(t-1)]$$, where $$u$$ denotes Heaviside step function, compute the output $$y := x \star h$$. Doing the convolution, I get the following results: $$y (t) = \begin{cases} 0 & \text{if } t<2 \\ \frac12 t(2-t) & \text{if } 2 \le t < 3 \\ -\frac32 & \text{if } 3 \le t < 4\\ \frac12 (t-5) (t-1) & \text{if } 4 \le t < 5 \\ \end{cases}$$ Others have gotten the following results: $$t<2; y(t) = 0$$ $$2 \le t < 3 ; y(t) = \frac{t^2-2t}{2}$$ $$3 \le t < 4 ; y(t) = \frac{3}{2}$$ $$4 \le t < 5 ; y(t) = -\frac{(t-5)\cdot(t-1)}{2}$$ In my calculations, I took the negative amplitude in account, and I put it in front of the integral. When they were doing the problem, they omitted the fact that $$x(t)$$ is a signal with negative amplitude. So, they are saying that we can omit the fact that the $$x(t)$$ is a negative amplitude signal, and that it doesn't affect the result. That makes no sense to me, and my way of thinking is that by their way of doing it the LTI system would have same output for two different input signals. I might be mistaken and not understand it properly, but if we omit the negative areas, in the case of, e.g., $$\sin(x)$$ on the input, wouldn't we get a rectified all positive sinus signal at the output? What is the correct way of dealing with this kind of signals, does the negative amplitude effect the results and does it change the output of LTI system, am I right or mistaken? $$(a\cdot x(t))*(b \cdot h(t)) = a\cdot b \cdot (x(t)*h(t))$$ That's complete nonsense. If you multiply the input with $$-1$$ the output will also be multiplied with $$-1$$.

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More Useful Trig Graphics First off, wow, that unit circle post was by far the most viewed I’ve had.  So maybe people would also be interested in this too. The thing that’s difficult about graphing trig functions for the first time is that regular old graph paper just isn’t up to the job.  When we have points like $\left(\frac{\pi}{4} , \frac{\sqrt2}{2} \right)$ and $\left(\frac{\pi}{3} , \frac{\sqrt3}{2} \right)$ to plot, you need something that is suited to the task, and the numbers involved.  I have met teachers that say they just have students plot the multiples of $\frac{\pi}{2}$, and then tell the students to trust them, it’s not really a zig-zag.  And it seems to me that having them spend time plotting points in order to develop a wrong intuition is an amazingly bad use of time.  You could just use the calculator and it would be better. But in order to make the time spent plotting points valuable you have to be able to see the shape of the curve from the points are plotted. So you need y-values of $\pm\frac{\sqrt2}{2}$ and $\pm\frac{\sqrt3}{2}$ and x-values that are multiples of $\frac{\pi}{6}$ and $\frac{\pi}{4}$. So behold: Of course, you need a key for this: You also need to call the students’ attention to how the x-gridlines aren’t evenly spaced, and that the unlabeled ones are multiples of $\frac{\pi}{6}$ and $\frac{\pi}{4}$.  But after that… you have something that can allow student to connect their knowledge of the trig functions in the context of the unit circle to the traditional graphing context. I don’t feel like wrestling with the code tags, so you can check out the latex source here, at sharelatex.  (You don’t need a ShareLaTeX account to access that link, but if you’d like to sign up for one, please use this referral link, which lets me earn points toward referral bonuses. Thanks!)  That document has two types of graphs– one for sine and cosine, and another for tangent and cotangent.  There’s a second set with the functions graphed as well, which I don’t give to students, but it’s useful to have to point at in later classes. TikZ Unit Circle After getting several requests from IRL folks for my unit circle, I figured I’d make it a post here. (Seriously, when I get tackled in the hall for a unit circle handout, it means that there’s a real need out there, even though I would have figured that there’s enough unit circles in the world.) Blank unit circle blank unit circle as pdf latex code on sharelatex. latex code: \documentclass[border=4pt]{standalone} \usepackage{amsmath,mathpazo,gensymb} \usepackage{tikz} \begin{document} \begin{tikzpicture} [>=stealth, scale=2.5, % Toggle commenting on the next four lines for the completed unit circle: angle/.style={draw,text=white,fill=white,minimum height=1cm, minimum width=1cm}, point/.style={white}, %angle/.style={fill=white}, %point/.style={}, ] \draw [white] (-3.6,-3.6) rectangle (3.6,3.6); \draw [thick,fill=white] (0,0) circle (3cm); \draw [thick, ] (-3.3,0) -- (3.3,0); \draw [thick, ] (0,-3.3) -- (0,3.3); \draw (0,0) -- node [angle] {$\frac{\pi}{2}$} (90:3) node[point, above right] {$\left(0,1\right)$}; \draw (0,0) -- node [angle] {\footnotesize{$\pi$}} (180:3) node[point, above left] {$\left(-1,0\right)$}; \draw (0,0) -- node [angle] {$\frac{3\pi}{2}$} (270:3) node[point, below right] {$\left(0,-1\right)$}; \draw (0,0) -- node [angle] {\footnotesize{$2\pi$}} (0:3) node[point, above right] {$\left(1,0\right)$}; \draw (0,0) -- node [angle] {$\frac{\pi}{4}$} (45:3) node [point, above right] {$\left( \frac{\sqrt2}{2} , \frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [angle] {$\frac{3\pi}{4}$} (135:3) node [point, above left] {$\left( -\frac{\sqrt2}{2} , \frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [angle] {$\frac{5\pi}{4}$} (225:3) node [point, below left] {$\left( -\frac{\sqrt2}{2} , -\frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [angle] {$\frac{7\pi}{4}$} (315:3) node [point, below right] {$\left( \frac{\sqrt2}{2} , -\frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{\pi}{6}$} (30:3) node [point, above right] {$\left( \frac{\sqrt3}{2} , \frac{1}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{\pi}{3}$} (60:3) node [point, above right] {$\left( \frac{1}{2} , \frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{2\pi}{3}$} (120:3) node [point, above left] {$\left( -\frac{1}{2} , \frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{5\pi}{6}$} (150:3) node [point, above left] {$\left(- \frac{\sqrt3}{2} , \frac{1}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{7\pi}{6}$} (210:3) node [point, below left] {$\left(- \frac{\sqrt3}{2} , -\frac{1}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{4\pi}{3}$} (240:3) node [point, below left] {$\left( -\frac{1}{2} , -\frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{5\pi}{3}$} (300:3) node [point, below right] {$\left( \frac{1}{2} , -\frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{11\pi}{6}$} (330:3) node [point, below right] {$\left( \frac{\sqrt3}{2} , -\frac{1}{2} \right)$}; \foreach \n in {1,2,3,4} \foreach \t in {0,30,45,60} \fill (\n*90+\t:3) circle (0.03cm); \foreach \t in {30,45,60, 120,135,150, 210,225,240, 300,315,330} \node [font=\tiny, fill=white,inner sep=1pt] at (\t:.75) {$\t\degree$}; \end{tikzpicture} \end{document} Completed Unit Circle completed unit circle as pdf latex code on sharelatex Get the latex code here. \documentclass[border=4pt]{standalone} \usepackage{amsmath,mathpazo,gensymb} \usepackage{tikz} \begin{document} \begin{tikzpicture} [>=stealth, scale=2.5, % Toggle commenting on the next four lines for the completed unit circle: %angle/.style={draw,text=white,fill=white,minimum height=1cm, minimum width=1cm}, %point/.style={white}, angle/.style={fill=white}, point/.style={}, ] \draw [white] (-3.6,-3.6) rectangle (3.6,3.6); \draw [thick,fill=white] (0,0) circle (3cm); \draw [thick, ] (-3.3,0) -- (3.3,0); \draw [thick, ] (0,-3.3) -- (0,3.3); \draw (0,0) -- node [angle] {$\frac{\pi}{2}$} (90:3) node[point, above right] {$\left(0,1\right)$}; \draw (0,0) -- node [angle] {\footnotesize{$\pi$}} (180:3) node[point, above left] {$\left(-1,0\right)$}; \draw (0,0) -- node [angle] {$\frac{3\pi}{2}$} (270:3) node[point, below right] {$\left(0,-1\right)$}; \draw (0,0) -- node [angle] {\footnotesize{$2\pi$}} (0:3) node[point, above right] {$\left(1,0\right)$}; \draw (0,0) -- node [angle] {$\frac{\pi}{4}$} (45:3) node [point, above right] {$\left( \frac{\sqrt2}{2} , \frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [angle] {$\frac{3\pi}{4}$} (135:3) node [point, above left] {$\left( -\frac{\sqrt2}{2} , \frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [angle] {$\frac{5\pi}{4}$} (225:3) node [point, below left] {$\left( -\frac{\sqrt2}{2} , -\frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [angle] {$\frac{7\pi}{4}$} (315:3) node [point, below right] {$\left( \frac{\sqrt2}{2} , -\frac{\sqrt2}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{\pi}{6}$} (30:3) node [point, above right] {$\left( \frac{\sqrt3}{2} , \frac{1}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{\pi}{3}$} (60:3) node [point, above right] {$\left( \frac{1}{2} , \frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{2\pi}{3}$} (120:3) node [point, above left] {$\left( -\frac{1}{2} , \frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{5\pi}{6}$} (150:3) node [point, above left] {$\left(- \frac{\sqrt3}{2} , \frac{1}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{7\pi}{6}$} (210:3) node [point, below left] {$\left(- \frac{\sqrt3}{2} , -\frac{1}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{4\pi}{3}$} (240:3) node [point, below left] {$\left( -\frac{1}{2} , -\frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{5\pi}{3}$} (300:3) node [point, below right] {$\left( \frac{1}{2} , -\frac{\sqrt3}{2} \right)$}; \draw (0,0) -- node [near end, angle] {$\frac{11\pi}{6}$} (330:3) node [point, below right] {$\left( \frac{\sqrt3}{2} , -\frac{1}{2} \right)$}; \foreach \n in {1,2,3,4} \foreach \t in {0,30,45,60} \fill (\n*90+\t:3) circle (0.03cm); % \foreach \t in {30,45,60, 120,135,150, 210,225,240, 300,315,330} % \node [font=\tiny, fill=white,inner sep=1pt] at (\t:.75) {$\t\degree$}; \end{tikzpicture} \end{document} As a sidenote, I am really pleased that WordPress’s code tags seem to have improved greatly since the last time I tried to use them.  Never mind.  It’s better than it was, but it’s changing < and > to < and >.  And it took the away entirely.  For the actual code, you can check out the versions on sharelatex, as well as this document, that has them both embedded in it.  (You don’t need a ShareLaTeX account to access these links, but if you’d like to sign up for one, please use this referral link, which lets me earn points toward referral bonuses. Thanks!) Altering the Mathematical Collaboration Expectations One thing that needs to be added to my Mathematical Collaboration Expectations is perseverance. I can’t believe that I didn’t even realize that I didn’t have that on there.  Now, is it its own category, or does it fit under something else?  I’m leaning towards its own category, but then I have to break it down.  What are the qualities of perseverance?

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### Lie group topology I'll assume you have a basic understanding of general topology -- if not, consult the topology articles here. Most of the abstract stuff and "weird" cases are not really important, because it is easy to see that Lie groups are manifolds. We need to be careful while studying the topology of Lie groups, because we already have an intuitive picture of a Lie group, and we need to be careful to prove all the things we just "believe" to be true. The main point of the topology of a Lie group is that the group elements define the "flows" on the manifold. What this means is that left-multiplication is a homeomorphism, and it's not absurd to say that inversion is a homeomorphism, because it represents a "reflection" of the manifold. That these conditions make sense is confirmed by looking at the proofs of the following "obvious" facts. (1) In a connected group, a neighbourhood of the identity generates the entire group, i.e. $H\le G\land H\in N(1)\implies H=G$ for connected $G$. Let's think about why this is true. Why does $H$ need to be a neighbourhood -- why must it contain an open set containing the identity? Suppose instead we just knew it contained a set $Q$ that looked like this: Well, $H$ still contains the orange point, but we cannot say it contains the purple point, because it's perfectly happy not containing it -- it's not like we have some vertical element in the Lie group that if you multiplied to some point in $Q$, you'd get the purple point. But instead if $Q$ was an open neighbourhood of the identity: Then the purple point has to be in $H$, because $Q$ contains flows in "all directions" on the group. To actually prove that every point will be contained in $H$ -- well, we know that the point is (will eventually be) that $H$ is the connected component of $G$ (and since $G$ is connected, $H=G$) -- let's just show that $H$ is both open and closed, i.e. nothing in $H$ touches its exterior, and nothing in its exterior touches $H$. Here's the proof: • Nothing in $H$ touches anything -- Suppose $\exists x\in H, x\in\mathrm{cl}(H')$. Then $xQ$ contains a point in $H'$. • Nothing outside $H$ touches it -- Suppose $\exists x\in H', x\in\mathrm{cl}(H)$. Then $xQ$ contains a point in $H$, so $x$ must be in $H$. We're really just formalising the notion of "translating $Q$ to its edges to extend $H$ further and further". The key fact we've used here is, of course, that left-multiplication is a homeomorphism, so $xQ$ is still an open set. (2) The connected component of the identity is a subgroup. The idea is that taking two elements $g,h$ of the connected component, their product should remain in the connected component. Once again, this follows from the continuity of left-multiplication -- considering the action of left-multiplication by $g$ on the connected component, its continuity implies that the image must remain connected. (3) If a subgroup contains a neighbourhood of the identity, it contains the connected component of the identity. Corollary to (1) and (2). (4) The connected component of the identity is a normal subgroup. Conjugation is a continuous map. (5) Open subgroups are closed. Corollary to (3). Alternate proof: the complement is the union of some cosets, which are open sets too. A weaker theorem can be made of closed sets -- closed subgroups with finite index are open. What this means: any open subgroup is a union of connected components. (6) Intuition for compact subgroups How can a Lie group possibly "close in on itself"? Surely we keep "extending" an open neighbourhood $W$ of the identity by observing that $xW$ must be in the subgroup? The idea is that these translations of $W$ form an open cover of the group, if it has a finite subcover, then it makes sense for the group to close in on itself. By playing around with different open neighbourhoods $W$ and taking some suitable unions, one can see that this is equivalent to the condition that every open cover has a finite subcover, i.e. the group is compact. (7) A compact, connected Abelian Lie group is a torus. This is a generalisation of "a finite Abelian group is the direct product of cyclic groups". The idea behind the proof is that in the Abelian case, the exponential map is a homomorphism from the Lie algebra to the Lie group, but the Lie algebra cannot detect compactness in the Lie group -- the kernel of the exponential map can. We know from our study of the exponential map that it has a discrete kernel, and in the Abelian case is surjective -- thus the Lie group is homeomorphic to $\mathbb{R}^n/\mathbb{Z}^n$, which is an $n$-torus. (8) A connected Abelian Lie group is a cylinder (direct product of a torus and an affine space) Analogous to above, except $\mathbb{R}^m/\mathbb{Z}^n$ where $m\ge n$.

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No announcements Calculating Standard Deviation in Visual Basic (2015) , but for anyone else who may have similar inquiries related to having to calculate Standard Deviation, perhaps in a different format, may also utilize this post for further ... | 10 Replies | 7222 Views | Created by error.Eiji - Thursday, March 30, 2017 9:26 AM | Last reply by John M. Andre - Sunday, November 22, 2020 2:12 PM Standard Deviation mean as double = sum/n The standard deviation is the square root of the sum of the difference of scores from the mean Dim SumX as ... | 8 Replies | 5664 Views | Created by Zeldacat - Sunday, June 25, 2006 10:34 PM | Last reply by Brian Kramer - Tuesday, June 27, 2006 10:19 PM Problem with standard deviation CalcStanDev cout << "Standard Deviation: " < | 1 Replies | 1973 Views | Created by Darjey - Sunday, December 2, 2012 10:23 AM | Last reply by Igor Tandetnik - Sunday, December 2, 2012 1:25 PM Standard Deviation apply standard deviation over each group There's also another function STDEV available which gives you standard deviation considering the given ... | 2 Replies | 1105 Views | Created by windsail - Friday, February 23, 2018 5:09 PM | Last reply by Visakh16 - Saturday, February 24, 2018 1:19 PM Standard Deviation in SQL Server I want to calculate weighted standard deviation in SQL Server 2005. Is there any built-in function as of standard deviation ... | 3 Replies | 2692 Views | Created by Shawna Jacobs - Tuesday, June 18, 2013 8:40 AM | Last reply by Mohit Kumar Gupta - Tuesday, June 18, 2013 9:11 AM Standard deviation deviation that in 95% of the cases (approximately) the actual value (which I try to predict) falls in the interval that is two standard deviations (on ... | 4 Replies | 7083 Views | Created by apkeijman - Monday, January 5, 2009 10:08 AM | Last reply by apkeijman - Friday, January 9, 2009 8:19 AM Standard Deviation in SSRS Hi, I am calculating to Standard deviation and i am using standard deviation function, i am doing ... | 3 Replies | 738 Views | Created by VNJMSS - Thursday, June 28, 2018 4:16 AM | Last reply by Zoe Zhi - Friday, June 29, 2018 8:25 AM The weight standard deviation ".. How can I get the standard deviation of weights for each student in each grade?" - is there a [Student]  dimension and attribute? If so, you could ... | 1 Replies | 4745 Views | Created by Willardryan - Wednesday, May 19, 2010 12:25 PM | Last reply by Deepak Puri - Thursday, May 20, 2010 3:05 AM Mean and Standard Deviation Hi, I need to calculate a mean value and standard deviation value in an image. Suppose, I already have 256 bins for a ... | 2 Replies | 9321 Views | Created by nguyenxh - Tuesday, May 19, 2009 10:36 PM | Last reply by Harry Zhu - Tuesday, May 26, 2009 8:47 AM Standard Deviation I have a range of 178 numbers and I need to find the standard deviation of only the numbers which are less than the first quartile of the range. I know how to find ... | 2 Replies | 2603 Views | Created by GAMinTN - Wednesday, November 9, 2011 4:45 PM | Last reply by Herbert Seidenberg - Thursday, November 10, 2011 3:49 PM Calculate Standard Deviation Hi Prashanth, Do Ram's reply help you? please give a feedback, I think the query he gave should work for you, for Standard Deviation, there is a build ... | 8 Replies | 3870 Views | Created by PrashantMhaske - Friday, January 13, 2012 6:46 AM | Last reply by PrashantMhaske - Thursday, January 19, 2012 4:11 PM Standard deviation calculation Hi, Deepak!. Thank you for your quick reply! In BIDS, I have now created a generic calculated member called [Standard Deviation] with the ... | 6 Replies | 10192 Views | Created by ZonkTheBear - Friday, February 3, 2006 10:10 AM | Last reply by superbluesman - Friday, February 18, 2011 6:26 PM Median and Standard Deviation Visual C# deviation is similar to working out the average (or mean, as I would call it). Start by using the method you've already written to get the average. Then loop through the array ... | 2 Replies | 1940 Views | Created by Mon20oct - Wednesday, July 27, 2016 6:59 PM | Last reply by Mon20oct - Thursday, July 28, 2016 2:06 PM Calculating Returns in Visual Basic Shasur, Thanks for reply. Pivot tables and subtotals are not prefered. I would like to write a macro in visual basic for this ... | 3 Replies | 3355 Views | Created by Stijn1 - Friday, November 13, 2009 1:51 PM | Last reply by Shasur - Sunday, November 29, 2009 3:41 AM AVERAGE in Decimal and Standard Deviation | 1 Replies | 3653 Views | Created by WalangAlam - Wednesday, August 19, 2009 8:52 AM | Last reply by jwavila - Wednesday, August 19, 2009 2:59 PM Standard Deviation in DAX Help Hi, Several New DAX Functions were added, these functions were added to calculate the standard deviation of the entire population or a sample (STDEV.S(), ... | 2 Replies | 6035 Views | Created by san463 - Wednesday, April 25, 2012 8:10 PM | Last reply by nanurahi - Wednesday, May 2, 2012 8:44 AM Visual basic 2015 problem So i have a surface pro 3 with Visual Basic 2015..after im done with whatever im coding i can run the program without debugging and itll show and after i save it and ... | 1 Replies | 490 Views | Created by Huey88 - Tuesday, October 6, 2015 4:50 PM | Last reply by DotNet Wang - Wednesday, October 7, 2015 2:21 AM Webbrowsers in Visual Basic 2015 / It's much easier to help if you post code of what you are doing. Paul ~~~~ Microsoft MVP (Visual Basic) | 3 Replies | 1096 Views | Created by AntPo - Wednesday, February 28, 2018 9:43 AM | Last reply by Simple Samples - Wednesday, February 28, 2018 6:52 PM

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# Search by Topic #### Resources tagged with Visualising similar to Hex: Filter by: Content type: Stage: Challenge level: ### Star Gazing ##### Stage: 4 Challenge Level: Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star. ### Trice ##### Stage: 3 Challenge Level: ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR? ### All Tied Up ##### Stage: 4 Challenge Level: A ribbon runs around a box so that it makes a complete loop with two parallel pieces of ribbon on the top. How long will the ribbon be? ### Rolling Around ##### Stage: 3 Challenge Level: A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle? ### Intersecting Circles ##### Stage: 3 Challenge Level: Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have? ### An Unusual Shape ##### Stage: 3 Challenge Level: Can you maximise the area available to a grazing goat? ### The Spider and the Fly ##### Stage: 4 Challenge Level: A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly? ### Efficient Packing ##### Stage: 4 Challenge Level: How efficiently can you pack together disks? ### Triangular Tantaliser ##### Stage: 3 Challenge Level: Draw all the possible distinct triangles on a 4 x 4 dotty grid. Convince me that you have all possible triangles. ### Tilting Triangles ##### Stage: 4 Challenge Level: A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates? ### Like a Circle in a Spiral ##### Stage: 2, 3 and 4 Challenge Level: A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels? ### Dissect ##### Stage: 3 Challenge Level: It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into? ### Take Ten ##### Stage: 3 Challenge Level: Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . . ### Tied Up ##### Stage: 3 Challenge Level: In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . . ### Framed ##### Stage: 3 Challenge Level: Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . . ### Corridors ##### Stage: 4 Challenge Level: A 10x10x10 cube is made from 27 2x2 cubes with corridors between them. Find the shortest route from one corner to the opposite corner. ### Muggles Magic ##### Stage: 3 Challenge Level: You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area. ### Tetra Square ##### Stage: 3 Challenge Level: ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square. ### Zooming in on the Squares ##### Stage: 2 and 3 Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens? ### The Old Goats ##### Stage: 3 Challenge Level: A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . . ##### Stage: 4 Challenge Level: In this problem we are faced with an apparently easy area problem, but it has gone horribly wrong! What happened? ### Rolling Coins ##### Stage: 4 Challenge Level: A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . . ### Floating in Space ##### Stage: 4 Challenge Level: Two angles ABC and PQR are floating in a box so that AB//PQ and BC//QR. Prove that the two angles are equal. ### Cutting a Cube ##### Stage: 3 Challenge Level: A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical? ### LOGO Challenge - Circles as Animals ##### Stage: 3 and 4 Challenge Level: See if you can anticipate successive 'generations' of the two animals shown here. ### Efficient Cutting ##### Stage: 4 Challenge Level: Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end. ### Cube Paths ##### Stage: 3 Challenge Level: Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B? ### All in the Mind ##### Stage: 3 Challenge Level: Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . . ### Triangle Inequality ##### Stage: 3 Challenge Level: ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm. ### Making Tracks ##### Stage: 4 Challenge Level: A bicycle passes along a path and leaves some tracks. Is it possible to say which track was made by the front wheel and which by the back wheel? ### Counting Triangles ##### Stage: 3 Challenge Level: Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether? ### Triangles in the Middle ##### Stage: 3, 4 and 5 Challenge Level: This task depends on groups working collaboratively, discussing and reasoning to agree a final product. ### Rati-o ##### Stage: 3 Challenge Level: Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle? ### Charting More Success ##### Stage: 3 and 4 Challenge Level: Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians? ### Weighty Problem ##### Stage: 3 Challenge Level: The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . . ### Rotating Triangle ##### Stage: 3 and 4 Challenge Level: What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? ### Threesomes ##### Stage: 3 Challenge Level: Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw? ### Tetrahedra Tester ##### Stage: 3 Challenge Level: An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length? ##### Stage: 3 Challenge Level: Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite? ### Constructing Triangles ##### Stage: 3 Challenge Level: Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw? ### Seven Squares - Group-worthy Task ##### Stage: 3 Challenge Level: Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning? ### Inside Out ##### Stage: 4 Challenge Level: There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . . ### Dice, Routes and Pathways ##### Stage: 1, 2 and 3 This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . . ### Sliced ##### Stage: 4 Challenge Level: An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron? ### Changing Places ##### Stage: 4 Challenge Level: Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . . ### Khun Phaen Escapes to Freedom ##### Stage: 3 Challenge Level: Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom. ### One and Three ##### Stage: 4 Challenge Level: Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . . ### Square It ##### Stage: 1, 2, 3 and 4 Challenge Level: Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square. ### Sliding Puzzle ##### Stage: 1, 2, 3 and 4 Challenge Level: The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves. ### Rolling Triangle ##### Stage: 3 Challenge Level: The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.

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# Convergence in $L^{\infty}$ norm implies convergence in $L^1$ norm Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable. Assume the measure space $X$ has finite measure. If $f_n$ converges to $f$ in $L^{\infty}$-norm , then $f_n$ converges to $f$ in $L^{1}$-norm. This is my approach: We know $||f_n-f||_{\infty} \to 0$ and by definition $||f_n-f||_{\infty} =\inf\{M\geq 0: |f_n-f|\leq M \}.$ Then \begin{align} ||f_n-f||_1\ &=\int |f_n-f| dm\ &\leq \int|f_n|dm+\int|f|dm\ \end{align} I don't know how to proceed after that, any help would be appreciated. • You might be also interested in the embeddings of the $L^p$ spaces. – Jack Mar 26 '12 at 5:53 For any function $g$, $||g||_1 = \int_X|g(m)|dm \leq \int_X||g||_\infty dm = \mu(X)*||g||_\infty$ (as $|g(m)| \leq ||g||_\infty$ almost everywhere); $||g||_\infty \geq \frac{||g||_1}{\mu(X)}$, so if $||f_n-f||_\infty$ tends to zero, then $||f_n-f||_1$ tends to zero as well. Your mistake is using the triangle inequality when you don't have to. Here, $$\|f_n-f\|_1=\int|f_n-f|dm\leq\|f_n-f\|_\infty\,\int1dm=\|f_n-f\|_\infty\,m(X)\to0.$$ • The same also holds for $L^p, p \in (0,\infty)$ instead of $L^1$, right? – user563311 Sep 5 '18 at 9:38 • Yes, indeed. $\$ – Martin Argerami Sep 5 '18 at 11:08

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Numbers between real numbers I wonder if there can be numbers (in some extended theory) for which all reals are either smaller or larger than this number, but no real number is equal to that number?! Is there some extension of number which allows that? Under what conditions (axiom etc.) there is no such number. - Since we are trying to get rid of number tag, see meta, I've retagged the question to number-systems. If you have a better idea, feel free to replace it with a more appropriate tag. – Martin Sleziak Jun 18 '12 at 11:21 It's OK. I found it hard to find a keyword especially since I'm not an expert :) – Gerenuk Jun 18 '12 at 12:00 Under the axioms of the real numbers this cannot occur. You must add new elements to the real numbers, note that if $\varepsilon$ is smaller than all $\frac1n$ but still positive then $\frac1\varepsilon$ is larger than any real number. Such $\varepsilon$ is called infinitesimal and their existence is incompatible with the real numbers per se. There is a branch, however, called non-standard analysis in which these numbers play an important role. One example to such field is called Hyperreal numbers. - Thanks. What does it mean "under axioms of real numbers". Do I have to abondon one of the axioms to introduce "numbers inbetween"? Which axiom would that be? Do hyperreals follow the condition about order I required? – Gerenuk Jun 18 '12 at 7:37 Another example is called Surreal number, which I think is more valuable. – Frank Science Jun 18 '12 at 7:48 The axiom that has to be abandoned is what is usually called the Completeness Axiom. One standard version of it is that any non-empty set (of reals) which is bounded above has a least upper bound. – André Nicolas Jun 18 '12 at 7:48 @Gerenuk: Hyperreal numbers are ordered and form a field. However they do not have the Archemedian property of the real numbers (every number is smaller than a natural number). Andre gave another property which does not hold. – Asaf Karagila Jun 18 '12 at 7:53 @Frank: If you want proper classes, sure. The. You have to explain what is a class, what are ordinals, etc. – Asaf Karagila Jun 18 '12 at 7:54 You want a "non-archimedean" ordered field. Examples are hyperreals and surreals. My favorite one: the transseries (G. A. Edgar, "Transseries for Beginners," http://www.math.ohio-state.edu/~edgar/preprints/trans_begin/). Also try the Levi-Civita numbers: http://en.wikipedia.org/wiki/Levi-Civita_field . See here http://en.wikipedia.org/wiki/Infinitesimal for many examples. - Consider the field $\mathbb{R}(x)$ of all (formal) rational functions in one variable with real coefficients. While this is not an ordered field, it is an orderable field -- it is possible to define an ordering $<$ on rational functions that is consistent with the usual laws of arithmetic. For any ordering $<$ of $\mathbb{R}(x)$, we can define sets $L = \{ a \in \mathbb{R} \mid a < x\}$ and $R = \{ a \in \mathbb{R} \mid a > x\}$, and we have $\mathbb{R} = L \cup R$ -- under this ordering, every real number is either less than or greater than the polynomial $x$. It turns out the ordering $<$ is completely determined by $L$ and $R$, and conversely each way to choose $L$ and $R$ corresponds to an ordering of $\mathbb{R}(x)$. The complete list of orderings are: • The ordering "$+\infty$" - $x$ is larger than every real number • The ordering "$-\infty$" - $x$ is smaller than every real number • The ordering "$a^+$" - $x$ is infinitesimally larger than $a$ • The ordering "$a^-$" - $x$ is infinitesimally smaller than $a$ The labels I've chosen for the orderings refer to "where" $x$ is placed in relation to the real line. Some good buzzwords that relate to this sort of topic are: • Real closed field • Formally real field • Real algebraic geometry • Semi-algebraic geometry There is an easy way to write down a first-order theory whose models are examples of the sort of number system you ask for. For example, • Add a new constant symbol $\varepsilon$ • Add in one axiom $0 < \varepsilon$ • For every positive integer $n$, add in one axiom $\varepsilon < n$ Every model of this theory will have a number $\varepsilon$ with the property that it is larger than every non-positive real number, and smaller than every positive real number. Can I add multiple $\varepsilon$ this way? It won't break any field axioms? Will all the maths from university still work (integrals etc.), or do I need some exceptions (i.e. where does the archimedian matter)? Is which is the field which is more than real numbers, but leaves as much undergrad math valid as possible? – Gerenuk Jun 19 '12 at 11:18

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https://nl.mathworks.com/matlabcentral/answers/507524-how-to-reshape-a-single-column-matrix-in-the-following-example?s_tid=prof_contriblnk

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# How to reshape a single column matrix in the following example? 2 views (last 30 days) blues on 25 Feb 2020 Commented: blues on 25 Feb 2020 I have a CT dicom files (512*512*263), I wanted to play with the CT numbers (HU) in the following way: (a) multiply the voxels that have CT number less than 20 by 2 (say, for fun, I want to understand how this works) and (b) multiply the voxels that have CT numbers greater than 20 by 10. I wrote the following code, find below. In the code I splitted the CT_number into two parts and worked on it, now I want to combine the results such that the new column vector 1D will have the updated values. How can I do that? Shouldn't be hard I guess, but I lost. ANy help would be helpful. CT_matrix = zeros(512, 512, 263); % preallocate the image array info_ct = dicominfo('...01.IMA'); for p = 1:263 CTfilename = sprintf('..-%03d.IMA', p); end % Convert CT matrix into a column vector to assess the voxel values idx = CT_matrix(:); % idx is a column vector CT_number = idx ; % voxels that have CT numbers less than 20 (say) voxels_have_CT_number_less_than_20 = CT_number(CT_number < 20); % Multiply the voxels_have_CT_number_less_than_20 by 2 new_voxels_have_CT_number_less_than_20 = voxels_have_CT_number_less_than_20 * 2; % voxels that have CT numbers greater than 20 (say) voxels_have_CT_number_greater_than_20 = CT_number(CT_number > 20); % Multiply the voxels_have_CT_number_greater_than_20 by 10 new_voxels_have_CT_number_greater_than_20 = voxels_have_CT_number_greater_than_20 * 10; % Now, I want to create a 1D matrix (by combining new_voxels_have_CT_number_less_than_20 and new_voxels_have_CT_number_greater_than_20) % such that I can see the matrix oprtaion in a single column. % Then, I can reshape this matrix to the original CT dimension new_CT_map = reshape(.., size(CT_matrix)); % not sure how to complete this! blues on 25 Feb 2020 After matrix manipulation, I had two vectors in above example. One is new_voxels_have_CT_number_less_than_20 and other is new_voxels_have_CT_number_greater_than_20. Both of them actually belongs to CT_number (1D column) vector. My question is how can I combine the results of these new_voxels_have_CT_number_less_than_20 and new_voxels_have_CT_number_greater_than_20, so that I can make a new colmun vector (1D), then finally I can use reshape function to make a original size CT matrix? blues on 25 Feb 2020 Yes, how can I convert the updated column vectors back to the same shape as in original one? Turlough Hughes on 25 Feb 2020 Edited: Turlough Hughes on 25 Feb 2020 If you write the following you can directly modify the CT_matrix, no need to separate it out into a column vector: ind_lte_20 = CT_matrix<=20; % index for values less than or equal to 20. ind_gt_20 = CT_matrix>20; % index for values greater than 20. CT_matrix(ind_lte_20) = CT_matrix(ind_lte_20)*2; CT_matrix(ind_gt_20) = CT_matrix(ind_gt_20)*10; blues on 25 Feb 2020 Thank you

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https://www.w3schools.com/statistics/statistics_average.php

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# Statistics - Average An average is a measure of where most of the values in the data are located. ## The Center of the Data The center of the data is where most of the values in the data are located. Averages are measures of the location of the center. There are different types of averages. The most commonly used are: Note: In statistics, averages are often referred to as 'measures of central tendency'. For example, using the values: 40, 21, 55, 21, 48, 13, 72 ## Mean The mean is usually referred to as 'the average'. The mean is the sum of all the values in the data divided by the total number of values in the data: (40 + 21 + 55 + 31 + 48 + 13 + 72)/7 = 38.57 Note: There are are multiple types of mean values. The most common type of mean is the arithmetic mean. In this tutorial, 'mean' refers to the arithmetic mean. ## Median The median is the 'middle value' of the data. The median is found by ordering all the values in the data and picking the middle value: 13, 21, 21, 40, 48, 55, 72 The median is less influenced by extreme values in the data than the mean. Changing the last value to 356 does not change the median: 13, 21, 21, 40, 48, 55, 356 The median is still 40. Changing the last value to 356 changes the mean a lot: (13 + 21 + 21 + 40 + 48 + 55 + 72)/7 = 38.57 (13 + 21 + 21 + 40 + 48 + 55 + 356)/7 = 79.14 Note: Extreme values are values in the data that are much smaller or larger than the average values in the data. ## Mode The mode is the value(s) that appears most often in the data: 40, 21, 55, 21, 48, 13, 72 Here, 21 appears two times, and the other values only once. The mode of this data is 21. The mode is also used for categorical data, unlike the median and mean. Categorical data can't be described directly with numbers, like names: Alice, John, Bob, Maria, John, Julia, Carol Here, John appears two times, and the other values only once. The mode of this data is John. Note: There can be more than one mode if multiple values appear the same number of times in the data. W3Schools is optimized for learning and training. Examples might be simplified to improve reading and learning. Tutorials, references, and examples are constantly reviewed to avoid errors, but we cannot warrant full correctness of all content. While using W3Schools, you agree to have read and accepted our terms of use, cookie and privacy policy.

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New Online Book! Handbook of Mathematical Functions (AMS55) Conversion & Calculation Home >> Measurement Conversion Measurement Converter Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC. Conversion Result: ```Chinese yuan renminbi on 4-28-2017 = 0.145042078302232 United States dollar on 4-28-2017 (currency)``` Related Measurements: Try converting from "CNY" to BND (Bruneian dollar on 4-28-2017), BTN (Bhutanese ngultrum on 4-28-2017), BZD (Belizean dollar on 4-28-2017), CAD (Canadian dollar on 4-28-2017), DJF (Djiboutian franc on 4-28-2017), EEK (Estonian kroon on 4-28-2017), HUF (Hungarian forint on 4-28-2017), IEP (Irish pound on 4-28-2017), JMD (Jamaican dollar on 4-28-2017), KRW (South Korean won on 4-28-2017), LSL (Lesotho loti on 4-28-2017), LTL (Lithuanian litas on 4-28-2017), MAD (Moroccan dirham on 4-28-2017), NAD (Namibian dollar on 4-28-2017), NOK (Norwegian krone on 4-28-2017), OMR (Omani rial on 4-28-2017), PKR (Pakistani rupee on 4-28-2017), PTE (Portuguese escudo on 4-28-2017), SGD (Singapore dollar on 4-28-2017), USD (United States dollar on 4-28-2017), or any combination of units which equate to "currency" and represent currency. Sample Conversions: CNY = 1.83 ATS (Austrian schilling on 4-28-2017), .26039196 BAM (Bos. and Herz. marks on 4-28-2017), .46265173 BRL (Brazilian real on 4-28-2017), 9.32 BTN (Bhutanese ngultrum on 4-28-2017), 96.64 CLP (Chilean peso on 4-28-2017), .99050475 DKK (Danish krone on 4-28-2017), .14504208 EEK (Estonian kroon on 4-28-2017), .79159245 FIM (Finnish markka on 4-28-2017), .30264162 FJD (Fijian dollar on 4-28-2017), .11211676 FKP (Falkland pound on 4-28-2017), 16.17 JPY (Japanese yen on 4-28-2017), 5.37 LUF (Luxembourg franc on 4-28-2017), 1.44 MAD (Moroccan dirham on 4-28-2017), .6296061 MYR (Malaysian ringgit on 4-28-2017), .29339378 NLG (Netherlands guilder on 4-28-2017), 15.2 PKR (Pakistani rupee on 4-28-2017), .11211676 SHP (Saint Helenian pound on 4-28-2017), .34571777 TND (Tunisian dinar on 4-28-2017), .14504208 USD (United States dollar on 4-28-2017), .39015748 XCD (East Caribbean dollar on 4-28-2017). Feedback, suggestions, or additional measurement definitions? Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!

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## iheartfood 2 years ago question is inside :) 1. iheartfood |dw:1353813925701:dw| 2. iheartfood A. 16 years B. 54 years C. 2.4 years D. 9 years 3. tkhunny $$0.29\cdot(1.046)^{t} = 0.60$$ Solve for 't'. You will need logarithms. 4. iheartfood how do i do that? 5. tkhunny Introduce the logarithm $$\log\left(0.29\cdot(1.046)^{t}\right) = \log(0.60)$$ Now what? 6. iheartfood idk.. but i think the answer is A. 16 years.. is that right? 7. tkhunny "I think the answer is..." is not ever an acceptable response. In mathematics, you must prove it. Use the logarithm rules. There is a reason why we did that massive problem, before. $$\log(0.29) + t\cdot\log(1.046) = \log(0.60)$$ You need to see where ALL of this came from. 8. iheartfood kk so what do i do now then? 9. tkhunny Solve for 't'. You must be able to do this. This is why we invented notation. It will help you. Use the notation, and proper operations on that notation, to solve for 't'. Let's see what you get. 10. iheartfood kk thanks! 11. tkhunny Are you going to show us any work?

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# Evaluate an ODE for several values of a coefficient and draw all the solutions in one plot I want to draw one plot showing all the solutions the following ODE with different values of N. How can i do it? s = NDSolve[{2 y'''[x] + (y[x]*y''[x]) == 0, g''[x] + (0.5****N***y[x]*g'[x]) == 0, g[0] == y[0] == 0, y'[0] == 0, y'[8] == 1, g[12] == 1}, {y, g}, {x, 0, 12}]; Plot[Evaluate[{g[x]} /. s], {x, 0, 12}, PlotStyle -> Automatic] - Welcome to Mathematica.SE! Note how I formatted your code - please do the same next time. Also: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign – Vitaliy Kaurov Dec 11 '12 at 7:57 N is a function in Mathematica, so it is a bad idea to use it as a variable name. Let's use m instead. Define solution as function of m: s[m_]:=NDSolve[{2 y'''[x] + (y[x] y''[x]) == 0, g''[x] + (0.5*m*y[x]*g'[x]) == 0, g[0] == y[0] == 0, y'[0] == 0, y'[8] == 1, g[12] == 1}, {y, g}, {x, 0, 12}] Set values of m m= Range[.1, 3, .1]; and plot solutions: Plot[Evaluate[g[x] /. (s /@ m)], {x, 0, 12}, PlotStyle -> Directive[Thick, Opacity[.8]], Frame -> True, Axes -> False] Or use interactive content: alst = Plot[Evaluate[g[x] /. {s[.1], s[3], s[#]}], {x, 0, 12}, Frame -> True, Axes -> False, PlotRange -> {0, 1}, PlotStyle -> Directive[Thick, Opacity[.9]], Filling -> {1 -> {2}}, PlotLabel -> "M = " <> ToString[#]] & /@ m; ListAnimate[alst, AnimationRunning -> False, FrameMargins -> 0] - @NasserM.Abbasi good point ;) done. – Vitaliy Kaurov Dec 11 '12 at 9:08 Dear Vitaliy Kaurov thank you very much. another question. if i want to assign to m specific value not in range between any number. for example m={0.7,1,2,20,10,100,1000} something like how should i write the code then? – mehrzad Dec 11 '12 at 16:17 @user4969 as you just did - define m as m={0.7,1,2,20,10,100,1000} instead of using Range. Btw I'd recommend choosing more unique name than user4969, hard to remember people this way ;) – Vitaliy Kaurov Dec 11 '12 at 16:37 As it is already pointed out by Vitaliy Kaurov, N is a built-in function and should not be used as a variable. Besides, I'd like to mention new functions in version 9: ParametricNDSolve and ParametricNDSolveValue. s = ParametricNDSolveValue[{2 y'''[x] + (y[x]*y''[x]) == 0, g''[x] + (0.5*m*y[x]*g'[x]) == 0, g[0] == y[0] == 0, y'[0] == 0, y'[8] == 1, g[12] == 1}, {y, g}, {x, 0, 12}, m] ParametricFunction[<>] For some given m, s[m] evaluated to ordinary InterpolatingFunction. So the plot can be written as: Plot[Evaluate[Table[s[m][[2]][x], {m, .1, 3, .1}]], {x, 0, 12}, PlotStyle -> (ColorData["Rainbow"] /@ Rescale[Range[.1, 3, .1]])] ` -

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Help # MAFS.912.G-SRT.1.1 Verify experimentally the properties of dilations given by a center and a scale factor: 1. A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. 2. The dilation of a line segment is longer or shorter in the ratio given by the scale factor. Subject Area: Mathematics Domain-Subdomain: Geometry: Similarity, Right Triangles, & Trigonometry Cluster: Level 2: Basic Application of Skills & Concepts Cluster: Understand similarity in terms of similarity transformations. (Geometry - Major Cluster) - Clusters should not be sorted from Major to Supporting and then taught in that order. To do so would strip the coherence of the mathematical ideas and miss the opportunity to enhance the major work of the grade with the supporting clusters. Date of Last Rating: 02/14 Status: State Board Approved Assessed: Yes ### TEST ITEM SPECIFICATIONS • Item Type(s): This benchmark may be assessed using: MS item(s) • Assessment Limits : Items may use line segments of a geometric figure. The center of dilation and scale factor must be given. • Calculator : Neutral • Clarification : When dilating a line that does not pass through the center of dilation, students will verify that the dilated line is parallel. When dilating a line that passes through the center of dilation, students will verify that the line is unchanged. When dilating a line segment, students will verify that the dilated line segment is longer or shorter with respect to the scale factor. • Stimulus Attributes : Items may give the student a figure or its dilation, center, and scale and ask the student to verify the properties of dilation. Items may be set in a real-world or mathematical context. • Response Attributes : None ### SAMPLE TEST ITEMS (1) • Test Item #: Sample Item 1 • Question:

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# Assume that T is a linear transformation. Find the standard Assume that T is a linear transformation. Find the standard matrix of T. T: , where ${e}_{1}=\left(1,0\right)$ and ${e}_{2}=\left(0,1\right)$ $A=$ You can still ask an expert for help ## Want to know more about Matrix transformations? • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Joseph Fair Let $x\in {F}^{n}$. Here, $\left[\begin{array}{c}{x}_{1}\\ ⋮\\ {x}_{n}\end{array}\right]$ $T\left(x\right)=T\left(\sum _{j}{e}_{j}{x}_{j}\right)=\sum _{j}T\left({e}_{j}\right){x}_{j}$ $=\sum _{j}{A}_{j}{x}_{j}=Ax$ So, here $A=\left[T\left({e}_{1}\right)\dots T\left({e}_{n}\right)\right]$ According to question T: So, the standart matrix of T is $\left[T\left({e}_{1}\right)T\left({e}_{2}\right)\right]$ $=\left[\begin{array}{cc}3& -5\\ 1& 6\\ 3& 0\\ 1& 0\end{array}\right]$ - Answer ###### Not exactly what you’re looking for? Beverly Smith A=$=\left[\begin{array}{cc}3& -5\\ 1& 6\\ 3& 0\\ 1& 0\end{array}\right]$ ###### Not exactly what you’re looking for? karton let $T=\left[\begin{array}{cc}a& b\\ c& d\\ e& f\\ g& h\end{array}\right]$ $T\left({e}_{1}\right)=\left(3,1,3,1\right)$ $⇒T\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}3\\ 1\\ 3\\ 1\end{array}\right)$ $⇒\left[\begin{array}{cc}a& b\\ c& d\\ e& f\\ g& h\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 3\\ 1\end{array}\right]$ $⇒\left[\begin{array}{c}a\\ c\\ e\\ g\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 3\\ 1\end{array}\right]⇒a=3,c=1,e=3,g=1$ $T\left({e}_{2}\right)=\left(-5,6,0,0\right)$ $⇒T\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}-5\\ 6\\ 0\\ 0\end{array}\right)$ $⇒\left[\begin{array}{cc}a& b\\ c& d\\ e& f\\ g& h\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}-5\\ 6\\ 0\\ 0\end{array}\right]$ $⇒\left[\begin{array}{c}b\\ d\\ f\\ h\end{array}\right]==\left[\begin{array}{c}-5\\ 6\\ 0\\ 0\end{array}\right]⇒b=-5,d=6,f=h=0$ $⇒T=\left[\begin{array}{cc}3& -5\\ 1& 6\\ 3& 0\\ 1& 0\end{array}\right]$ ###### Not exactly what you’re looking for? • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee

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Request: a best CAS mode menu (History view) 08-17-2018, 05:30 PM (This post was last modified: 08-23-2018 08:29 PM by compsystems.) Post: #1 compsystems Senior Member Posts: 1,201 Joined: Dec 2013 Request: a best CAS mode menu (History view) Hello, the CAS mode menu, should be improved 1: Changing [STO▶] to [:=], ▶ is a HOME mode instruction 2: complement the menu [STO▶] [Simplify] [■] [■] [■] [■] that is, the reverse 'simplify' commands must appear, not only in mathematics it is required to simplify. [:=] [Simplify] [Factor] [Expand] [Normal] [Next] simplify( (x-1)*(x+1)=0 ) returns x^2-1=0 factor( x^2-1=0, x ) returns (x-1)*(x+1)=0 Savings when pressing the keys. 1 versus 3 [Tools][algebra][Factor] expand((3*x*z+2*z^2+5*z+6)/z) returns 3*x+2*z+5+6/z // [Tools][algebra][expand] [Subst] [Solve] [PartFrac] [] [] [Next] subst(y = 2*x + 3,y = x^2) returns x^2=(2*x+3) simplify(2*x-3+x^2=(2*x+3-2*x-3)) returns x^2-2*x-3=0 solve(x^2-2*x-3=0,x,'=') returns [x=-1,x=3] subst(x^2-2*x-3=0, x=3) returns 0=0 evalb(0=0) returns true subst(x^2-2*x-3=0, x=-1) returns 0=0 evalb(0=0) returns true factor(x^2-2*x-3=0) returns (x-3)*(x+1)=0 solve([ y = x^2, y = 2*x + 3 ],[x,y], '=' ) returns [[x=3,y=9],[x=-1,y=1]] Good idea? LO SUFICIENTEMENTE BUENO ES ENEMIGO DE LA EXCELENCIA. 08-23-2018, 08:31 PM Post: #2 compsystems Senior Member Posts: 1,201 Joined: Dec 2013 RE: Request: a best CAS mode menu (History view) An image of a possible combination of menu functions in the CAS history view. Any suggestions to choose other menus Attached File(s) Thumbnail(s) LO SUFICIENTEMENTE BUENO ES ENEMIGO DE LA EXCELENCIA. 08-23-2018, 11:40 PM Post: #3 Carlos295pz Senior Member Posts: 334 Joined: Sep 2015 RE: Request: a best CAS mode menu (History view) I find the expansion of the touch menu interesting, although we have the CAS section in the toolbox, placing the most predominant commands is a great idea. Viga C | TD | FB 08-24-2018, 10:25 AM (This post was last modified: 08-24-2018 10:25 AM by primer.) Post: #4 primer Member Posts: 135 Joined: Sep 2015 RE: Request: a best CAS mode menu (History view) yes and being able to choose the ones to appear would be even better... There are commands I use much more than PartFrac for example. primer 08-26-2018, 08:27 AM (This post was last modified: 08-26-2018 08:31 AM by math7.) Post: #5 math7 Member Posts: 163 Joined: Feb 2017 RE: Request: a best CAS mode menu (History view) (08-23-2018 08:31 PM)compsystems Wrote:  An image of a possible combination of menu functions in the CAS history view. Any suggestions to choose other menus Precisely that is the type of menu that exists in the HP 50g. That's why I liked its use so much, as in the HP 19bii and others, there are 6 menu labels on the screen, and they are one key to use them again and again, without pressing several keys to reach a specific command when using it more than once. Attached File(s) Thumbnail(s) 08-26-2018, 08:29 AM Post: #6 math7 Member Posts: 163 Joined: Feb 2017 RE: Request: a best CAS mode menu (History view) (08-24-2018 10:25 AM)primer Wrote:  yes and being able to choose the ones to appear would be even better... There are commands I use much more than PartFrac for example. And this would be similar to the HP 50g that has the "Custom Menu" function that can be customized with the commands one uses, it would be good if a future update included it! 08-26-2018, 08:25 PM Post: #7 CyberAngel Member Posts: 231 Joined: Jul 2018 RE: Request: a best CAS mode menu (History view) I found the old [SOFT] [MENU] [KEYS] [BEST] Now, with a touch display user should have an option with a list of list of keycodes or even a matrix with keycodes filling the screen with a big soft menu...matrix (which is also the name of my favourite movie - and university subject) OR if the user uses a simple list or a vector then the old one-row-menu appears « Next Oldest | Next Newest » User(s) browsing this thread: 1 Guest(s)

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 # Your Math Help is on the Way! ### More Math Help Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: # Math 150 Lecture Notes for Chapter 2 Equations and Inequalities Math 150 Lecture Notes for Section 2A Solving Equations Introduction and Review of Linear Equations A quadratic equation is an equation that can be written in the form . I. Solve by Factoring II. Solve by Completing the Square III. Solve by Using the Quadratic Formula The solution to any quadratic equation of the form is Solve: Solve: Rational Equations A rational equation is an equation which involves rational expressions, or fractions. Solve: 2. Raise both sides of the equation to the appropriate power to remove the radical. 3. Repeat the process until all radicals have been removed. 4. Check for extraneous solutions! Solve: Absolute Value Equation Solve: Equations in Several Variables Solve: a. Solvefor y where b. Solve for b. Note that S is the surface area of a square right pyramid where b is the length of a side of the square base and l is the slant height. Math 150 Lecture Notes for Section 2B Solving Inequalities Introduction Find the values of x which satisfies the inequality . Linear Inequalities Note: When you multiply or divide an inequality by a negative number, you must reverse the inequality sign. Multiply both sides of by -1. Solve: Absolute Value Inequalities can be thought of as “what numbers are less than 5 units from 0 on the number line?” can be thought of as “what numbers are more than 5 units from 0 on the number line?” is equivalent to is equivalent to Solve: Nonlinear Inequalities Strategy to solve nonlinear inequalities: 1. Move every term to one side (make one side zero). 2. If possible, factor the expression on the nonzero side. 3. Find the critical values or values for which the expression is zero or undefined. 4. Draw a number line and let the critical numbers divide the number line into intervals. 5. Determine the sign for each factor in each interval. 6. Determine if the sign for all factors in each interval is positive or negative and compare to our inequality to see if it is more than or less than zero. Solve: (Why don’t we just multiple both sides by to clear all denominators?) (Why don’t we just divide both sides by x?) 

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# Mrs. Bennett's First Grade Class ## Math We are working retesting on all of the skills we have taught so far this year. We have focused so much time on addition/subtraction, fluency, properties of addition, missing parts, balancing equations, word problems, place value, and using place value to add and subtract to 100. Over the next few months we will be focusing more time on measuring with non-standard measurements, telling time to the hour and half hour, shapes, fractions, and data/graphing. Feel free to have your child explore my website to watch videos and play games to help them master these concepts. We will also be focusing on fluency of math facts to 10. Many students have mastered addition and subtraction to 20, but not fluency to 10. Please take any free time you have (in the car, etc) to help your child memorize their math facts of addition and subtraction to 10. Thank you! ## Spelling Because this past week was only 3 days, we will be continuing these words onto next week. We are on Week 11. The spelling pattern is - words with qu and x. Words: six, box, wax, queen, quack, quit, want, were Sentences: I want six bugs in a box. Were the king and queen made of wax? Remember - Wednesday's practice test is all 8 of these words. Friday's practice test is 6 of these words and two other words that follow the spelling pattern, but not words that they have practiced this week. ## Writing We are having so much fun learning how to conduct research and writing informative / explanatory texts. They need to be able to write an introductory sentence, some facts about the topic, and a closing sentence. ## Sight Words If your child has passed all of the sight words, their focus is not their math facts to 10. They will no longer have sight words to practice, just their "sight facts". Please practice these as you did your sight words so they can know the answer as soon as they see the math fact, just like reading their sight words. They have 30 seconds for each set of math facts. The facts are in their goal notebooks. If you would like a copy for home, just let me know and I would be happy to send some home. We are now working on 3rd Quarter words. ****** I have some interactive flash cards on the website under sight words - practice. The boxes have a Q on the and the number of the quarter. They are nice because your child can try to say the word and then click on the word to see if they are correct. We are working on comparing two fictional stories. We will be testing on all the phonics skills learned so far this year. ## Social Studies We are learning National Symbols. We will be learning about why the American Flag, White House, Bald Eagle, etc. are important to the United States. We will learn what a symbol is and do research on a symbol. ## FYI • We go to the library on Thursday mornings. Please leave your child's library books in their backpack so they can trade every week. • Book Fair - March 10-14 • 3rd Quarter Academic Celebration is March 14. • Spring Break is March 17-21. ## Specials: Monday - PE Tuesday - Art Wednesday - Music Thursday - PE Friday - Art

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# Two digit addition worksheets Most Effective » » Two digit addition worksheets Most Effective Your Two digit addition worksheets images are available. Two digit addition worksheets are a topic that is being searched for and liked by netizens now. You can Find and Download the Two digit addition worksheets files here. Get all free photos. If you’re searching for two digit addition worksheets pictures information related to the two digit addition worksheets interest, you have come to the right blog. Our website frequently provides you with suggestions for downloading the highest quality video and image content, please kindly search and find more informative video content and images that match your interests. Two Digit Addition Worksheets. 2 Digit Addition Worksheets 2. The kids can practise solving two-digit problems through these vertical equation sheets. 2 Digit Addition Worksheets. This is the main page for the subtraction worksheets. Total 64 problems are available in four worksheets. 13092019 Double-digit addition is just one of the many mathematical concepts that students are expected to master in first and second grade and it comes in many shapes and sizes. Some two digit addition worksheets you can look forward to include kids word problems math mosaics addition crosswords and more. Using these worksheets students can solve double-digit problems with or without regrouping. 03092019 These two-digit addition worksheets will give your students the practice they need to master regrouping. 2 Digit Addition Numbers Less Than 20 Worksheets. ### 13042020 Two-Digit Vertical Addition and Subtraction Worksheets Teach your kids to add and subtract two-digit numbers with and without regrouping with these worksheets. Source: pinterest.com Source: pinterest.com Some two digit addition worksheets you can look forward to include kids word problems math mosaics addition crosswords and more. Support your young learner with two-digit addition worksheets. It also contains 2-digit innovative addition worksheets based on base ten blocks balancing scales dart-board addition addition. 07012021 Math Two Digit Addition And Subtraction Worksheets help a kid discover the skills required for subtraction. 2 Digit Addition Numbers Less Than 20 Worksheets. Source: pinterest.com Source: pinterest.com 03092019 These two-digit addition worksheets will give your students the practice they need to master regrouping. 2 Digit Addition Numbers Less Than 20 Worksheets. Each of these addition worksheets provide exercises for progressively more complex types of addition problems with multiple digit addends as well as problems with more than two addends. 2 Digit Addition Worksheets 2. This is the main page for the subtraction worksheets. Source: pinterest.com Once we learn to add single digit numbers the next step is to learn to add two digit numbers. Two Digit Addition Worksheets. Each worksheet includes 20 double-digit addition problems a number line and a second page with the answers. 07012021 Math Two Digit Addition And Subtraction Worksheets help a kid discover the skills required for subtraction. 13 rows These addition worksheets produces great worksheets that add two digit numbers together that. Source: pinterest.com It also contains 2 digit innovative addition worksheets based on base ten blocks balancing scales dart board addition addition pyramid and matching ice cream scoops. Showing top 8 worksheets in the category - 2 Digit Additions. Total 64 problems are available in four worksheets. It also offers you being a parent or educator a possibility to know how very much he has grasped and what the simplest way of earning him learn more is going to be. Many adults are probably comfortable performing double digit addition with. Source: pinterest.com Addition worksheet 2 digit plus 2 digit addition with some regrouping author. 2 Digit Addition Numbers Less Than 20 Worksheets. Total 64 problems are available in four worksheets. It may be printed downloaded or saved and used in your classroom home school or other. 2 Digit Addition Worksheets 2. Source: pinterest.com These entertaining and unique printables bring together fun equations and hands-on activities to promote learning. Addition worksheet 2 digit plus 2 digit addition with some regrouping author. The kids can practise solving two-digit problems through these vertical equation sheets. 07012021 Math Two Digit Addition And Subtraction Worksheets help a kid discover the skills required for subtraction. This math worksheet was created on 2021-02-24 and has been viewed 187 times this week and 2280 times this month. Source: pinterest.com Source: pinterest.com 2 Digit Addition No Regroup Worksheets. 2 Digit Addition No Regroup Worksheets 2. Using these worksheets students can solve double-digit problems with or without regrouping. We introduce the best addition worksheet for the convenience of parents and teachers. Total 64 problems are available in four worksheets. Source: pinterest.com Using these worksheets students can solve double-digit problems with or without regrouping. Many adults are probably comfortable performing double digit addition with. We introduce the best addition worksheet for the convenience of parents and teachers. Our printable two-digit addition worksheets contain 2-digit plus 1-digit numbers 2-digit addition drills addition with or without word problems addition with regrouping carrying or no regrouping missing digits and adding 3 4 and 5 addends. 11122020 Two digit addition worksheets pdf. Source: pinterest.com Two digit addition - without regrouping. 2 Digit Addition Numbers Less Than 20 Worksheets. Many adults are probably comfortable performing double digit addition with. This is a fantastic bundle which includes everything you need to know about two digit addition across 28 in-depth pages. Two Digit Addition Worksheets. Source: pinterest.com Support your young learner with two-digit addition worksheets. Addition worksheet 2 digit plus 2 digit addition with some regrouping author. Two Digit Addition Worksheets. Second Worksheet Resource Type. We introduce the best addition worksheet for the convenience of parents and teachers. Source: pinterest.com Our printable two-digit addition worksheets contain 2-digit plus 1-digit numbers 2-digit addition drills addition with or without word problems addition with regrouping carrying or no regrouping missing digits and adding 3 4 and 5 addends. 2 Digit Addition Numbers Less Than 20 Worksheets. This math worksheet was created on 2021-02-24 and has been viewed 187 times this week and 2280 times this month. 2 Digit Addition Worksheets. Second Worksheet Resource Type. Source: pinterest.com Each worksheet includes 20 double-digit addition problems a number line and a second page with the answers. This is a fantastic bundle which includes everything you need to know about two digit addition across 28 in-depth pages. We introduce the best addition worksheet for the convenience of parents and teachers. 2 Digit Addition Numbers Less Than 20 Worksheets. Click on the hyperlink to download the pdf and print out the pages for use in your classes. Source: pinterest.com 11122020 Two digit addition worksheets pdf. 11122020 Two digit addition worksheets pdf. 2 Digit Addition Worksheets. It may be printed downloaded or saved and used in your classroom home school or other. 13042020 Two-Digit Vertical Addition and Subtraction Worksheets Teach your kids to add and subtract two-digit numbers with and without regrouping with these worksheets. Source: pinterest.com This math worksheet was created on 2021-02-24 and has been viewed 187 times this week and 2280 times this month. We introduce the best addition worksheet for the convenience of parents and teachers. 13092019 Double-digit addition is just one of the many mathematical concepts that students are expected to master in first and second grade and it comes in many shapes and sizes. Support your young learner with two-digit addition worksheets. 03092019 These two-digit addition worksheets will give your students the practice they need to master regrouping. Source: pinterest.com Follow the links for Spaceship Math Subtraction worksheets timed subtraction tests multiple digit subtraction worksheets simple borrowing and regrouping worksheets and math worksheets with mixed addition. 03092019 These two-digit addition worksheets will give your students the practice they need to master regrouping. This math worksheet was created on 2021-02-24 and has been viewed 187 times this week and 2280 times this month. There are different levels of problems in our worksheet that you can find easily in our math section. Two Digit Addition Second Worksheet Property. This site is an open community for users to submit their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us. If you find this site helpful, please support us by sharing this posts to your own social media accounts like Facebook, Instagram and so on or you can also bookmark this blog page with the title two digit addition worksheets by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. 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It is currently 17 Oct 2017, 12:28 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # What is the best advice for me? Author Message TAGS: ### Hide Tags Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 02 Feb 2011, 02:13 To Overdrive, You were right! Manhattan SC is very useful. I should have bought the whole set when I began my preparation ...But ,anyway!!! Thank you again!!! Kudos [?]: 6 [0], given: 14 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 04 Feb 2011, 12:41 10 days left!!! Kudos [?]: 6 [0], given: 14 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 04 Feb 2011, 15:25 Last edited by msneemo on 05 Feb 2011, 20:04, edited 2 times in total. Kudos [?]: 6 [0], given: 14 Director Status: -=Given to Fly=- Joined: 04 Jan 2011 Posts: 830 Kudos [?]: 241 [0], given: 78 Location: India Schools: Haas '18, Kelley '18 GMAT 1: 650 Q44 V37 GMAT 2: 710 Q48 V40 GMAT 3: 750 Q51 V40 GPA: 3.5 WE: Education (Education) ### Show Tags 04 Feb 2011, 18:13 If the next ten days can make a difference to how the rest of your life shapes up, isn't it worth the shot? Don't give up!!! _________________ "Wherever you go, go with all your heart" - Confucius 1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club) 2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog) My Experience With GMAT 1. From 650 to 710 to 750 - My Tryst With GMAT 2. Quest to do my Best - My GMAT Journey Log Kudos [?]: 241 [0], given: 78 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 05 Feb 2011, 10:26 In reality, I just need a 500.If I get a 500 I could go to the nearest college.Or if I get a 600 I might go to a college 1 hour away from me. Kudos [?]: 6 [0], given: 14 Manager Status: I rest, I rust. Joined: 04 Oct 2010 Posts: 121 Kudos [?]: 127 [0], given: 9 Schools: ISB - Co 2013 WE 1: IT Professional since 2006 ### Show Tags 06 Feb 2011, 22:21 Sednima wrote: Dear Vaibhav , Can you please suggest me any links where I can find some of these 700-800 questions? 700-800 Series is a collection of 700-800 level questions (verbal only) created by ManhattanGMAT. _________________ Respect, Vaibhav PS: Correct me if I am wrong. Kudos [?]: 127 [0], given: 9 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 09 Feb 2011, 00:39 Anyone ever saw the movie 'exam'? It seriously reminded me of GMAT.Sometimes you would see a question and just feel like you're staring at an empty page!!! Kudos [?]: 6 [0], given: 14 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 11 Feb 2011, 12:21 Its the 4th day before my exam and I AM SUPER NERVOUS!!!!!!!When I was in college, I did all classes , gave all class tests, did all the assignments very sincerely.I gathered up all the necessary materials and notes and I used to study them all only the night before the actual exam.If anybody in my house ever saw me studying they knew immediately that I must have a big exam tomorrow.But that was over two years ago.And cant use the same strategy in GMAT.I have no idea whats gonna happen now...Today, I'm feeling very very nervous!!! Kudos [?]: 6 [0], given: 14 Founder Joined: 04 Dec 2002 Posts: 15568 Kudos [?]: 28475 [0], given: 5105 Location: United States (WA) GMAT 1: 750 Q49 V42 ### Show Tags 11 Feb 2011, 16:39 Good Luck. Study hard during the weekend and relax on Monday before the test (assuming it is Tue). _________________ Founder of GMAT Club Just starting out with GMAT? Start here... or use our Daily Study Plan Co-author of the GMAT Club tests Kudos [?]: 28475 [0], given: 5105 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 13 Feb 2011, 12:25 Yes, it is on Tuesday.And this forum is helping me a lot to keep my spirit up.Thanks for wishing me luck. Kudos [?]: 6 [0], given: 14 Manager Joined: 05 Jul 2010 Posts: 187 Kudos [?]: 22 [0], given: 18 ### Show Tags 13 Feb 2011, 16:10 all the best for your exam! Will be waiting for your scores! Kudos [?]: 22 [0], given: 18 Intern Joined: 29 Jan 2011 Posts: 30 Kudos [?]: 2 [0], given: 2 ### Show Tags 13 Feb 2011, 23:01 all the very best for the exam Kudos [?]: 2 [0], given: 2 Director Status: -=Given to Fly=- Joined: 04 Jan 2011 Posts: 830 Kudos [?]: 241 [0], given: 78 Location: India Schools: Haas '18, Kelley '18 GMAT 1: 650 Q44 V37 GMAT 2: 710 Q48 V40 GMAT 3: 750 Q51 V40 GPA: 3.5 WE: Education (Education) ### Show Tags 13 Feb 2011, 23:05 All the best Neemo :D We're all with you! +Kudos Do well! _________________ "Wherever you go, go with all your heart" - Confucius 1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club) 2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog) My Experience With GMAT 1. From 650 to 710 to 750 - My Tryst With GMAT 2. Quest to do my Best - My GMAT Journey Log Kudos [?]: 241 [0], given: 78 Manager Joined: 10 Nov 2010 Posts: 247 Kudos [?]: 401 [0], given: 22 Location: India Concentration: Strategy, Operations GMAT 1: 520 Q42 V19 GMAT 2: 540 Q44 V21 WE: Information Technology (Computer Software) ### Show Tags 14 Feb 2011, 01:57 All The Best! _________________ The proof of understanding is the ability to explain it. Kudos [?]: 401 [0], given: 22 Manager Joined: 27 Jun 2008 Posts: 77 Kudos [?]: 18 [0], given: 22 Location: United States (AL) Concentration: General Management, Technology GMAT 1: 660 Q48 V34 WE: Consulting (Computer Software) ### Show Tags 14 Feb 2011, 02:42 All the best msneemo.. take one Q at a time dnt think abt the difficulty level/previous questions.. am sure you'll do well.. Kudos [?]: 18 [0], given: 22 Manager Status: I rest, I rust. Joined: 04 Oct 2010 Posts: 121 Kudos [?]: 127 [0], given: 9 Schools: ISB - Co 2013 WE 1: IT Professional since 2006 ### Show Tags 15 Feb 2011, 03:06 Best luck for your test today, mate!! Let us know how it went. _________________ Respect, Vaibhav PS: Correct me if I am wrong. Kudos [?]: 127 [0], given: 9 Director Status: -=Given to Fly=- Joined: 04 Jan 2011 Posts: 830 Kudos [?]: 241 [0], given: 78 Location: India Schools: Haas '18, Kelley '18 GMAT 1: 650 Q44 V37 GMAT 2: 710 Q48 V40 GMAT 3: 750 Q51 V40 GPA: 3.5 WE: Education (Education) ### Show Tags 15 Feb 2011, 05:26 Kudos [?]: 241 [0], given: 78 Director Status: -=Given to Fly=- Joined: 04 Jan 2011 Posts: 830 Kudos [?]: 241 [0], given: 78 Location: India Schools: Haas '18, Kelley '18 GMAT 1: 650 Q44 V37 GMAT 2: 710 Q48 V40 GMAT 3: 750 Q51 V40 GPA: 3.5 WE: Education (Education) ### Show Tags 15 Feb 2011, 21:13 370 to 510 :D Awesome progress in one month (I think) Well done Neemo... Now enjoy yourself _________________ "Wherever you go, go with all your heart" - Confucius 1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club) 2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog) My Experience With GMAT 1. From 650 to 710 to 750 - My Tryst With GMAT 2. Quest to do my Best - My GMAT Journey Log Kudos [?]: 241 [0], given: 78 Manager Status: Still not there yet!!! Joined: 11 Jan 2011 Posts: 59 Kudos [?]: 6 [0], given: 14 Schools: Waaayyy....in the future - Booth!! ### Show Tags 15 Feb 2011, 21:20 I know!!!! You don't even know what I scored in my last practice test!! Even like 3 days before my exam I scored a 370.That was my last score on GMAT Prep.I went to the exam today 'knowing' that I will do bad....But,seemed pretty easy to me.Just the lack of time was my problem!!! Thank you so much entwhistle!! You have been a true friend to me.Actually everyone in this forum is very very veeeeeeeeerrrrrrrrryyyyyyyyyyyyyyyyy nice...I know I will stay addicted to this site for a long long time... Kudos [?]: 6 [0], given: 14 Director Status: -=Given to Fly=- Joined: 04 Jan 2011 Posts: 830 Kudos [?]: 241 [0], given: 78 Location: India Schools: Haas '18, Kelley '18 GMAT 1: 650 Q44 V37 GMAT 2: 710 Q48 V40 GMAT 3: 750 Q51 V40 GPA: 3.5 WE: Education (Education) ### Show Tags 15 Feb 2011, 21:41 ha ha... It would be nice to have you around... Tell us about the programme you're enrolling too and all! _________________ "Wherever you go, go with all your heart" - Confucius 1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club) 2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog) My Experience With GMAT 1. From 650 to 710 to 750 - My Tryst With GMAT 2. Quest to do my Best - My GMAT Journey Log Kudos [?]: 241 [0], given: 78 Re: What is the best advice for me?   [#permalink] 15 Feb 2011, 21:41 Go to page   Previous    1   2   3    Next  [ 50 posts ] Display posts from previous: Sort by # What is the best advice for me? Moderator: EMPOWERgmatRichC Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

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# I have a variable as an exponent, how do I isolate it from everything else? kelp ## Homework Statement .75 = 1 - (1 - p)^n I want to get the n alone, and find n in terms of p. ## The Attempt at a Solution I simplified it to this: 1 - p = (.25)^(1/n) Basically, I just moved numbers around. I do not know how to get the n by itself without dragging along a number. Thanks! Mentor ## Homework Statement .75 = 1 - (1 - p)^n I want to get the n alone, and find n in terms of p. ## The Attempt at a Solution I simplified it to this: 1 - p = (.25)^(1/n) Basically, I just moved numbers around. I do not know how to get the n by itself without dragging along a number. Thanks! Take the log of both sides. .75 = 1 - (1 - p)n ==> .25 = (1 - p)n ==> ln(.25) = ln[(1 - p)n] Using one of the properties of logarithms, you can work with the right side to eventually isolate n. Gold Member Have you tried logarithms? zketrouble Logarithms would be the best way to get change the variable from a constant to a coefficient. Any logarithm will work, whether it be the standard log10, the natural log ln, or any other one. log(ax) = x*log(a) Here's an example on how to use this property: 3.2x = 10.24 ln(3.2x) = ln(10.24) Using the property mentioned: x*ln(3.2) = ln(10.24) x*1.1631 = 2.3263 x = 2.3263/1.1631 x = 2 Note (again) that it doesn't matter whether you choose the ln or log10 or log12345, so long as you use the logarithm on BOTH sides of the equation and then use the mentioned property you'll be fine.

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src/HOL/Library/Formal_Power_Series.thy author wenzelm Sun Dec 27 22:07:17 2015 +0100 (2015-12-27) changeset 61943 7fba644ed827 parent 61804 67381557cee8 child 61969 e01015e49041 permissions -rw-r--r-- discontinued ASCII replacement syntax <*>; ``` 1 (* Title: HOL/Library/Formal_Power_Series.thy ``` ``` 2 Author: Amine Chaieb, University of Cambridge ``` ``` 3 *) ``` ``` 4 ``` ``` 5 section \<open>A formalization of formal power series\<close> ``` ``` 6 ``` ``` 7 theory Formal_Power_Series ``` ``` 8 imports Complex_Main "~~/src/HOL/Number_Theory/Euclidean_Algorithm" ``` ``` 9 begin ``` ``` 10 ``` ``` 11 ``` ``` 12 subsection \<open>The type of formal power series\<close> ``` ``` 13 ``` ``` 14 typedef 'a fps = "{f :: nat \<Rightarrow> 'a. True}" ``` ``` 15 morphisms fps_nth Abs_fps ``` ``` 16 by simp ``` ``` 17 ``` ``` 18 notation fps_nth (infixl "\$" 75) ``` ``` 19 ``` ``` 20 lemma expand_fps_eq: "p = q \<longleftrightarrow> (\<forall>n. p \$ n = q \$ n)" ``` ``` 21 by (simp add: fps_nth_inject [symmetric] fun_eq_iff) ``` ``` 22 ``` ``` 23 lemma fps_ext: "(\<And>n. p \$ n = q \$ n) \<Longrightarrow> p = q" ``` ``` 24 by (simp add: expand_fps_eq) ``` ``` 25 ``` ``` 26 lemma fps_nth_Abs_fps [simp]: "Abs_fps f \$ n = f n" ``` ``` 27 by (simp add: Abs_fps_inverse) ``` ``` 28 ``` ``` 29 text \<open>Definition of the basic elements 0 and 1 and the basic operations of addition, ``` ``` 30 negation and multiplication.\<close> ``` ``` 31 ``` ``` 32 instantiation fps :: (zero) zero ``` ``` 33 begin ``` ``` 34 definition fps_zero_def: "0 = Abs_fps (\<lambda>n. 0)" ``` ``` 35 instance .. ``` ``` 36 end ``` ``` 37 ``` ``` 38 lemma fps_zero_nth [simp]: "0 \$ n = 0" ``` ``` 39 unfolding fps_zero_def by simp ``` ``` 40 ``` ``` 41 instantiation fps :: ("{one, zero}") one ``` ``` 42 begin ``` ``` 43 definition fps_one_def: "1 = Abs_fps (\<lambda>n. if n = 0 then 1 else 0)" ``` ``` 44 instance .. ``` ``` 45 end ``` ``` 46 ``` ``` 47 lemma fps_one_nth [simp]: "1 \$ n = (if n = 0 then 1 else 0)" ``` ``` 48 unfolding fps_one_def by simp ``` ``` 49 ``` ``` 50 instantiation fps :: (plus) plus ``` ``` 51 begin ``` ``` 52 definition fps_plus_def: "op + = (\<lambda>f g. Abs_fps (\<lambda>n. f \$ n + g \$ n))" ``` ``` 53 instance .. ``` ``` 54 end ``` ``` 55 ``` ``` 56 lemma fps_add_nth [simp]: "(f + g) \$ n = f \$ n + g \$ n" ``` ``` 57 unfolding fps_plus_def by simp ``` ``` 58 ``` ``` 59 instantiation fps :: (minus) minus ``` ``` 60 begin ``` ``` 61 definition fps_minus_def: "op - = (\<lambda>f g. Abs_fps (\<lambda>n. f \$ n - g \$ n))" ``` ``` 62 instance .. ``` ``` 63 end ``` ``` 64 ``` ``` 65 lemma fps_sub_nth [simp]: "(f - g) \$ n = f \$ n - g \$ n" ``` ``` 66 unfolding fps_minus_def by simp ``` ``` 67 ``` ``` 68 instantiation fps :: (uminus) uminus ``` ``` 69 begin ``` ``` 70 definition fps_uminus_def: "uminus = (\<lambda>f. Abs_fps (\<lambda>n. - (f \$ n)))" ``` ``` 71 instance .. ``` ``` 72 end ``` ``` 73 ``` ``` 74 lemma fps_neg_nth [simp]: "(- f) \$ n = - (f \$ n)" ``` ``` 75 unfolding fps_uminus_def by simp ``` ``` 76 ``` ``` 77 instantiation fps :: ("{comm_monoid_add, times}") times ``` ``` 78 begin ``` ``` 79 definition fps_times_def: "op * = (\<lambda>f g. Abs_fps (\<lambda>n. \<Sum>i=0..n. f \$ i * g \$ (n - i)))" ``` ``` 80 instance .. ``` ``` 81 end ``` ``` 82 ``` ``` 83 lemma fps_mult_nth: "(f * g) \$ n = (\<Sum>i=0..n. f\$i * g\$(n - i))" ``` ``` 84 unfolding fps_times_def by simp ``` ``` 85 ``` ``` 86 lemma fps_mult_nth_0 [simp]: "(f * g) \$ 0 = f \$ 0 * g \$ 0" ``` ``` 87 unfolding fps_times_def by simp ``` ``` 88 ``` ``` 89 declare atLeastAtMost_iff [presburger] ``` ``` 90 declare Bex_def [presburger] ``` ``` 91 declare Ball_def [presburger] ``` ``` 92 ``` ``` 93 lemma mult_delta_left: ``` ``` 94 fixes x y :: "'a::mult_zero" ``` ``` 95 shows "(if b then x else 0) * y = (if b then x * y else 0)" ``` ``` 96 by simp ``` ``` 97 ``` ``` 98 lemma mult_delta_right: ``` ``` 99 fixes x y :: "'a::mult_zero" ``` ``` 100 shows "x * (if b then y else 0) = (if b then x * y else 0)" ``` ``` 101 by simp ``` ``` 102 ``` ``` 103 lemma cond_value_iff: "f (if b then x else y) = (if b then f x else f y)" ``` ``` 104 by auto ``` ``` 105 ``` ``` 106 lemma cond_application_beta: "(if b then f else g) x = (if b then f x else g x)" ``` ``` 107 by auto ``` ``` 108 ``` ``` 109 ``` ``` 110 subsection \<open>Formal power series form a commutative ring with unity, if the range of sequences ``` ``` 111 they represent is a commutative ring with unity\<close> ``` ``` 112 ``` ``` 113 instance fps :: (semigroup_add) semigroup_add ``` ``` 114 proof ``` ``` 115 fix a b c :: "'a fps" ``` ``` 116 show "a + b + c = a + (b + c)" ``` ``` 117 by (simp add: fps_ext add.assoc) ``` ``` 118 qed ``` ``` 119 ``` ``` 120 instance fps :: (ab_semigroup_add) ab_semigroup_add ``` ``` 121 proof ``` ``` 122 fix a b :: "'a fps" ``` ``` 123 show "a + b = b + a" ``` ``` 124 by (simp add: fps_ext add.commute) ``` ``` 125 qed ``` ``` 126 ``` ``` 127 lemma fps_mult_assoc_lemma: ``` ``` 128 fixes k :: nat ``` ``` 129 and f :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> 'a::comm_monoid_add" ``` ``` 130 shows "(\<Sum>j=0..k. \<Sum>i=0..j. f i (j - i) (n - j)) = ``` ``` 131 (\<Sum>j=0..k. \<Sum>i=0..k - j. f j i (n - j - i))" ``` ``` 132 by (induct k) (simp_all add: Suc_diff_le setsum.distrib add.assoc) ``` ``` 133 ``` ``` 134 instance fps :: (semiring_0) semigroup_mult ``` ``` 135 proof ``` ``` 136 fix a b c :: "'a fps" ``` ``` 137 show "(a * b) * c = a * (b * c)" ``` ``` 138 proof (rule fps_ext) ``` ``` 139 fix n :: nat ``` ``` 140 have "(\<Sum>j=0..n. \<Sum>i=0..j. a\$i * b\$(j - i) * c\$(n - j)) = ``` ``` 141 (\<Sum>j=0..n. \<Sum>i=0..n - j. a\$j * b\$i * c\$(n - j - i))" ``` ``` 142 by (rule fps_mult_assoc_lemma) ``` ``` 143 then show "((a * b) * c) \$ n = (a * (b * c)) \$ n" ``` ``` 144 by (simp add: fps_mult_nth setsum_right_distrib setsum_left_distrib mult.assoc) ``` ``` 145 qed ``` ``` 146 qed ``` ``` 147 ``` ``` 148 lemma fps_mult_commute_lemma: ``` ``` 149 fixes n :: nat ``` ``` 150 and f :: "nat \<Rightarrow> nat \<Rightarrow> 'a::comm_monoid_add" ``` ``` 151 shows "(\<Sum>i=0..n. f i (n - i)) = (\<Sum>i=0..n. f (n - i) i)" ``` ``` 152 by (rule setsum.reindex_bij_witness[where i="op - n" and j="op - n"]) auto ``` ``` 153 ``` ``` 154 instance fps :: (comm_semiring_0) ab_semigroup_mult ``` ``` 155 proof ``` ``` 156 fix a b :: "'a fps" ``` ``` 157 show "a * b = b * a" ``` ``` 158 proof (rule fps_ext) ``` ``` 159 fix n :: nat ``` ``` 160 have "(\<Sum>i=0..n. a\$i * b\$(n - i)) = (\<Sum>i=0..n. a\$(n - i) * b\$i)" ``` ``` 161 by (rule fps_mult_commute_lemma) ``` ``` 162 then show "(a * b) \$ n = (b * a) \$ n" ``` ``` 163 by (simp add: fps_mult_nth mult.commute) ``` ``` 164 qed ``` ``` 165 qed ``` ``` 166 ``` ``` 167 instance fps :: (monoid_add) monoid_add ``` ``` 168 proof ``` ``` 169 fix a :: "'a fps" ``` ``` 170 show "0 + a = a" by (simp add: fps_ext) ``` ``` 171 show "a + 0 = a" by (simp add: fps_ext) ``` ``` 172 qed ``` ``` 173 ``` ``` 174 instance fps :: (comm_monoid_add) comm_monoid_add ``` ``` 175 proof ``` ``` 176 fix a :: "'a fps" ``` ``` 177 show "0 + a = a" by (simp add: fps_ext) ``` ``` 178 qed ``` ``` 179 ``` ``` 180 instance fps :: (semiring_1) monoid_mult ``` ``` 181 proof ``` ``` 182 fix a :: "'a fps" ``` ``` 183 show "1 * a = a" ``` ``` 184 by (simp add: fps_ext fps_mult_nth mult_delta_left setsum.delta) ``` ``` 185 show "a * 1 = a" ``` ``` 186 by (simp add: fps_ext fps_mult_nth mult_delta_right setsum.delta') ``` ``` 187 qed ``` ``` 188 ``` ``` 189 instance fps :: (cancel_semigroup_add) cancel_semigroup_add ``` ``` 190 proof ``` ``` 191 fix a b c :: "'a fps" ``` ``` 192 show "b = c" if "a + b = a + c" ``` ``` 193 using that by (simp add: expand_fps_eq) ``` ``` 194 show "b = c" if "b + a = c + a" ``` ``` 195 using that by (simp add: expand_fps_eq) ``` ``` 196 qed ``` ``` 197 ``` ``` 198 instance fps :: (cancel_ab_semigroup_add) cancel_ab_semigroup_add ``` ``` 199 proof ``` ``` 200 fix a b c :: "'a fps" ``` ``` 201 show "a + b - a = b" ``` ``` 202 by (simp add: expand_fps_eq) ``` ``` 203 show "a - b - c = a - (b + c)" ``` ``` 204 by (simp add: expand_fps_eq diff_diff_eq) ``` ``` 205 qed ``` ``` 206 ``` ``` 207 instance fps :: (cancel_comm_monoid_add) cancel_comm_monoid_add .. ``` ``` 208 ``` ``` 209 instance fps :: (group_add) group_add ``` ``` 210 proof ``` ``` 211 fix a b :: "'a fps" ``` ``` 212 show "- a + a = 0" by (simp add: fps_ext) ``` ``` 213 show "a + - b = a - b" by (simp add: fps_ext) ``` ``` 214 qed ``` ``` 215 ``` ``` 216 instance fps :: (ab_group_add) ab_group_add ``` ``` 217 proof ``` ``` 218 fix a b :: "'a fps" ``` ``` 219 show "- a + a = 0" by (simp add: fps_ext) ``` ``` 220 show "a - b = a + - b" by (simp add: fps_ext) ``` ``` 221 qed ``` ``` 222 ``` ``` 223 instance fps :: (zero_neq_one) zero_neq_one ``` ``` 224 by standard (simp add: expand_fps_eq) ``` ``` 225 ``` ``` 226 instance fps :: (semiring_0) semiring ``` ``` 227 proof ``` ``` 228 fix a b c :: "'a fps" ``` ``` 229 show "(a + b) * c = a * c + b * c" ``` ``` 230 by (simp add: expand_fps_eq fps_mult_nth distrib_right setsum.distrib) ``` ``` 231 show "a * (b + c) = a * b + a * c" ``` ``` 232 by (simp add: expand_fps_eq fps_mult_nth distrib_left setsum.distrib) ``` ``` 233 qed ``` ``` 234 ``` ``` 235 instance fps :: (semiring_0) semiring_0 ``` ``` 236 proof ``` ``` 237 fix a :: "'a fps" ``` ``` 238 show "0 * a = 0" ``` ``` 239 by (simp add: fps_ext fps_mult_nth) ``` ``` 240 show "a * 0 = 0" ``` ``` 241 by (simp add: fps_ext fps_mult_nth) ``` ``` 242 qed ``` ``` 243 ``` ``` 244 instance fps :: (semiring_0_cancel) semiring_0_cancel .. ``` ``` 245 ``` ``` 246 instance fps :: (semiring_1) semiring_1 .. ``` ``` 247 ``` ``` 248 ``` ``` 249 subsection \<open>Selection of the nth power of the implicit variable in the infinite sum\<close> ``` ``` 250 ``` ``` 251 lemma fps_nonzero_nth: "f \<noteq> 0 \<longleftrightarrow> (\<exists> n. f \$n \<noteq> 0)" ``` ``` 252 by (simp add: expand_fps_eq) ``` ``` 253 ``` ``` 254 lemma fps_nonzero_nth_minimal: "f \<noteq> 0 \<longleftrightarrow> (\<exists>n. f \$ n \<noteq> 0 \<and> (\<forall>m < n. f \$ m = 0))" ``` ``` 255 (is "?lhs \<longleftrightarrow> ?rhs") ``` ``` 256 proof ``` ``` 257 let ?n = "LEAST n. f \$ n \<noteq> 0" ``` ``` 258 show ?rhs if ?lhs ``` ``` 259 proof - ``` ``` 260 from that have "\<exists>n. f \$ n \<noteq> 0" ``` ``` 261 by (simp add: fps_nonzero_nth) ``` ``` 262 then have "f \$ ?n \<noteq> 0" ``` ``` 263 by (rule LeastI_ex) ``` ``` 264 moreover have "\<forall>m<?n. f \$ m = 0" ``` ``` 265 by (auto dest: not_less_Least) ``` ``` 266 ultimately have "f \$ ?n \<noteq> 0 \<and> (\<forall>m<?n. f \$ m = 0)" .. ``` ``` 267 then show ?thesis .. ``` ``` 268 qed ``` ``` 269 show ?lhs if ?rhs ``` ``` 270 using that by (auto simp add: expand_fps_eq) ``` ``` 271 qed ``` ``` 272 ``` ``` 273 lemma fps_eq_iff: "f = g \<longleftrightarrow> (\<forall>n. f \$ n = g \$n)" ``` ``` 274 by (rule expand_fps_eq) ``` ``` 275 ``` ``` 276 lemma fps_setsum_nth: "setsum f S \$ n = setsum (\<lambda>k. (f k) \$ n) S" ``` ``` 277 proof (cases "finite S") ``` ``` 278 case True ``` ``` 279 then show ?thesis by (induct set: finite) auto ``` ``` 280 next ``` ``` 281 case False ``` ``` 282 then show ?thesis by simp ``` ``` 283 qed ``` ``` 284 ``` ``` 285 ``` ``` 286 subsection \<open>Injection of the basic ring elements and multiplication by scalars\<close> ``` ``` 287 ``` ``` 288 definition "fps_const c = Abs_fps (\<lambda>n. if n = 0 then c else 0)" ``` ``` 289 ``` ``` 290 lemma fps_nth_fps_const [simp]: "fps_const c \$ n = (if n = 0 then c else 0)" ``` ``` 291 unfolding fps_const_def by simp ``` ``` 292 ``` ``` 293 lemma fps_const_0_eq_0 [simp]: "fps_const 0 = 0" ``` ``` 294 by (simp add: fps_ext) ``` ``` 295 ``` ``` 296 lemma fps_const_1_eq_1 [simp]: "fps_const 1 = 1" ``` ``` 297 by (simp add: fps_ext) ``` ``` 298 ``` ``` 299 lemma fps_const_neg [simp]: "- (fps_const (c::'a::ring)) = fps_const (- c)" ``` ``` 300 by (simp add: fps_ext) ``` ``` 301 ``` ``` 302 lemma fps_const_add [simp]: "fps_const (c::'a::monoid_add) + fps_const d = fps_const (c + d)" ``` ``` 303 by (simp add: fps_ext) ``` ``` 304 ``` ``` 305 lemma fps_const_sub [simp]: "fps_const (c::'a::group_add) - fps_const d = fps_const (c - d)" ``` ``` 306 by (simp add: fps_ext) ``` ``` 307 ``` ``` 308 lemma fps_const_mult[simp]: "fps_const (c::'a::ring) * fps_const d = fps_const (c * d)" ``` ``` 309 by (simp add: fps_eq_iff fps_mult_nth setsum.neutral) ``` ``` 310 ``` ``` 311 lemma fps_const_add_left: "fps_const (c::'a::monoid_add) + f = ``` ``` 312 Abs_fps (\<lambda>n. if n = 0 then c + f\$0 else f\$n)" ``` ``` 313 by (simp add: fps_ext) ``` ``` 314 ``` ``` 315 lemma fps_const_add_right: "f + fps_const (c::'a::monoid_add) = ``` ``` 316 Abs_fps (\<lambda>n. if n = 0 then f\$0 + c else f\$n)" ``` ``` 317 by (simp add: fps_ext) ``` ``` 318 ``` ``` 319 lemma fps_const_mult_left: "fps_const (c::'a::semiring_0) * f = Abs_fps (\<lambda>n. c * f\$n)" ``` ``` 320 unfolding fps_eq_iff fps_mult_nth ``` ``` 321 by (simp add: fps_const_def mult_delta_left setsum.delta) ``` ``` 322 ``` ``` 323 lemma fps_const_mult_right: "f * fps_const (c::'a::semiring_0) = Abs_fps (\<lambda>n. f\$n * c)" ``` ``` 324 unfolding fps_eq_iff fps_mult_nth ``` ``` 325 by (simp add: fps_const_def mult_delta_right setsum.delta') ``` ``` 326 ``` ``` 327 lemma fps_mult_left_const_nth [simp]: "(fps_const (c::'a::semiring_1) * f)\$n = c* f\$n" ``` ``` 328 by (simp add: fps_mult_nth mult_delta_left setsum.delta) ``` ``` 329 ``` ``` 330 lemma fps_mult_right_const_nth [simp]: "(f * fps_const (c::'a::semiring_1))\$n = f\$n * c" ``` ``` 331 by (simp add: fps_mult_nth mult_delta_right setsum.delta') ``` ``` 332 ``` ``` 333 ``` ``` 334 subsection \<open>Formal power series form an integral domain\<close> ``` ``` 335 ``` ``` 336 instance fps :: (ring) ring .. ``` ``` 337 ``` ``` 338 instance fps :: (ring_1) ring_1 ``` ``` 339 by (intro_classes, auto simp add: distrib_right) ``` ``` 340 ``` ``` 341 instance fps :: (comm_ring_1) comm_ring_1 ``` ``` 342 by (intro_classes, auto simp add: distrib_right) ``` ``` 343 ``` ``` 344 instance fps :: (ring_no_zero_divisors) ring_no_zero_divisors ``` ``` 345 proof ``` ``` 346 fix a b :: "'a fps" ``` ``` 347 assume "a \<noteq> 0" and "b \<noteq> 0" ``` ``` 348 then obtain i j where i: "a \$ i \<noteq> 0" "\<forall>k<i. a \$ k = 0" and j: "b \$ j \<noteq> 0" "\<forall>k<j. b \$ k =0" ``` ``` 349 unfolding fps_nonzero_nth_minimal ``` ``` 350 by blast+ ``` ``` 351 have "(a * b) \$ (i + j) = (\<Sum>k=0..i+j. a \$ k * b \$ (i + j - k))" ``` ``` 352 by (rule fps_mult_nth) ``` ``` 353 also have "\<dots> = (a \$ i * b \$ (i + j - i)) + (\<Sum>k\<in>{0..i+j} - {i}. a \$ k * b \$ (i + j - k))" ``` ``` 354 by (rule setsum.remove) simp_all ``` ``` 355 also have "(\<Sum>k\<in>{0..i+j}-{i}. a \$ k * b \$ (i + j - k)) = 0" ``` ``` 356 proof (rule setsum.neutral [rule_format]) ``` ``` 357 fix k assume "k \<in> {0..i+j} - {i}" ``` ``` 358 then have "k < i \<or> i+j-k < j" ``` ``` 359 by auto ``` ``` 360 then show "a \$ k * b \$ (i + j - k) = 0" ``` ``` 361 using i j by auto ``` ``` 362 qed ``` ``` 363 also have "a \$ i * b \$ (i + j - i) + 0 = a \$ i * b \$ j" ``` ``` 364 by simp ``` ``` 365 also have "a \$ i * b \$ j \<noteq> 0" ``` ``` 366 using i j by simp ``` ``` 367 finally have "(a*b) \$ (i+j) \<noteq> 0" . ``` ``` 368 then show "a * b \<noteq> 0" ``` ``` 369 unfolding fps_nonzero_nth by blast ``` ``` 370 qed ``` ``` 371 ``` ``` 372 instance fps :: (ring_1_no_zero_divisors) ring_1_no_zero_divisors .. ``` ``` 373 ``` ``` 374 instance fps :: (idom) idom .. ``` ``` 375 ``` ``` 376 lemma numeral_fps_const: "numeral k = fps_const (numeral k)" ``` ``` 377 by (induct k) (simp_all only: numeral.simps fps_const_1_eq_1 ``` ``` 378 fps_const_add [symmetric]) ``` ``` 379 ``` ``` 380 lemma neg_numeral_fps_const: ``` ``` 381 "(- numeral k :: 'a :: ring_1 fps) = fps_const (- numeral k)" ``` ``` 382 by (simp add: numeral_fps_const) ``` ``` 383 ``` ``` 384 lemma fps_numeral_nth: "numeral n \$ i = (if i = 0 then numeral n else 0)" ``` ``` 385 by (simp add: numeral_fps_const) ``` ``` 386 ``` ``` 387 lemma fps_numeral_nth_0 [simp]: "numeral n \$ 0 = numeral n" ``` ``` 388 by (simp add: numeral_fps_const) ``` ``` 389 ``` ``` 390 ``` ``` 391 subsection \<open>The eXtractor series X\<close> ``` ``` 392 ``` ``` 393 lemma minus_one_power_iff: "(- (1::'a::comm_ring_1)) ^ n = (if even n then 1 else - 1)" ``` ``` 394 by (induct n) auto ``` ``` 395 ``` ``` 396 definition "X = Abs_fps (\<lambda>n. if n = 1 then 1 else 0)" ``` ``` 397 ``` ``` 398 lemma X_mult_nth [simp]: ``` ``` 399 "(X * (f :: 'a::semiring_1 fps)) \$n = (if n = 0 then 0 else f \$ (n - 1))" ``` ``` 400 proof (cases "n = 0") ``` ``` 401 case False ``` ``` 402 have "(X * f) \$n = (\<Sum>i = 0..n. X \$ i * f \$ (n - i))" ``` ``` 403 by (simp add: fps_mult_nth) ``` ``` 404 also have "\<dots> = f \$ (n - 1)" ``` ``` 405 using False by (simp add: X_def mult_delta_left setsum.delta) ``` ``` 406 finally show ?thesis ``` ``` 407 using False by simp ``` ``` 408 next ``` ``` 409 case True ``` ``` 410 then show ?thesis ``` ``` 411 by (simp add: fps_mult_nth X_def) ``` ``` 412 qed ``` ``` 413 ``` ``` 414 lemma X_mult_right_nth[simp]: ``` ``` 415 "((f :: 'a::comm_semiring_1 fps) * X) \$n = (if n = 0 then 0 else f \$ (n - 1))" ``` ``` 416 by (metis X_mult_nth mult.commute) ``` ``` 417 ``` ``` 418 lemma X_power_iff: "X^k = Abs_fps (\<lambda>n. if n = k then 1::'a::comm_ring_1 else 0)" ``` ``` 419 proof (induct k) ``` ``` 420 case 0 ``` ``` 421 then show ?case by (simp add: X_def fps_eq_iff) ``` ``` 422 next ``` ``` 423 case (Suc k) ``` ``` 424 have "(X^Suc k) \$ m = (if m = Suc k then 1::'a else 0)" for m ``` ``` 425 proof - ``` ``` 426 have "(X^Suc k) \$ m = (if m = 0 then 0 else (X^k) \$ (m - 1))" ``` ``` 427 by (simp del: One_nat_def) ``` ``` 428 then show ?thesis ``` ``` 429 using Suc.hyps by (auto cong del: if_weak_cong) ``` ``` 430 qed ``` ``` 431 then show ?case ``` ``` 432 by (simp add: fps_eq_iff) ``` ``` 433 qed ``` ``` 434 ``` ``` 435 lemma X_nth[simp]: "X\$n = (if n = 1 then 1 else 0)" ``` ``` 436 by (simp add: X_def) ``` ``` 437 ``` ``` 438 lemma X_power_nth[simp]: "(X^k) \$n = (if n = k then 1 else 0::'a::comm_ring_1)" ``` ``` 439 by (simp add: X_power_iff) ``` ``` 440 ``` ``` 441 lemma X_power_mult_nth: "(X^k * (f :: 'a::comm_ring_1 fps)) \$n = (if n < k then 0 else f \$ (n - k))" ``` ``` 442 apply (induct k arbitrary: n) ``` ``` 443 apply simp ``` ``` 444 unfolding power_Suc mult.assoc ``` ``` 445 apply (case_tac n) ``` ``` 446 apply auto ``` ``` 447 done ``` ``` 448 ``` ``` 449 lemma X_power_mult_right_nth: ``` ``` 450 "((f :: 'a::comm_ring_1 fps) * X^k) \$n = (if n < k then 0 else f \$ (n - k))" ``` ``` 451 by (metis X_power_mult_nth mult.commute) ``` ``` 452 ``` ``` 453 ``` ``` 454 lemma X_neq_fps_const [simp]: "(X :: 'a :: zero_neq_one fps) \<noteq> fps_const c" ``` ``` 455 proof ``` ``` 456 assume "(X::'a fps) = fps_const (c::'a)" ``` ``` 457 hence "X\$1 = (fps_const (c::'a))\$1" by (simp only:) ``` ``` 458 thus False by auto ``` ``` 459 qed ``` ``` 460 ``` ``` 461 lemma X_neq_zero [simp]: "(X :: 'a :: zero_neq_one fps) \<noteq> 0" ``` ``` 462 by (simp only: fps_const_0_eq_0[symmetric] X_neq_fps_const) simp ``` ``` 463 ``` ``` 464 lemma X_neq_one [simp]: "(X :: 'a :: zero_neq_one fps) \<noteq> 1" ``` ``` 465 by (simp only: fps_const_1_eq_1[symmetric] X_neq_fps_const) simp ``` ``` 466 ``` ``` 467 lemma X_neq_numeral [simp]: "(X :: 'a :: {semiring_1,zero_neq_one} fps) \<noteq> numeral c" ``` ``` 468 by (simp only: numeral_fps_const X_neq_fps_const) simp ``` ``` 469 ``` ``` 470 lemma X_pow_eq_X_pow_iff [simp]: ``` ``` 471 "(X :: ('a :: {comm_ring_1}) fps) ^ m = X ^ n \<longleftrightarrow> m = n" ``` ``` 472 proof ``` ``` 473 assume "(X :: 'a fps) ^ m = X ^ n" ``` ``` 474 hence "(X :: 'a fps) ^ m \$ m = X ^ n \$ m" by (simp only:) ``` ``` 475 thus "m = n" by (simp split: split_if_asm) ``` ``` 476 qed simp_all ``` ``` 477 ``` ``` 478 ``` ``` 479 subsection \<open>Subdegrees\<close> ``` ``` 480 ``` ``` 481 definition subdegree :: "('a::zero) fps \<Rightarrow> nat" where ``` ``` 482 "subdegree f = (if f = 0 then 0 else LEAST n. f\$n \<noteq> 0)" ``` ``` 483 ``` ``` 484 lemma subdegreeI: ``` ``` 485 assumes "f \$ d \<noteq> 0" and "\<And>i. i < d \<Longrightarrow> f \$ i = 0" ``` ``` 486 shows "subdegree f = d" ``` ``` 487 proof- ``` ``` 488 from assms(1) have "f \<noteq> 0" by auto ``` ``` 489 moreover from assms(1) have "(LEAST i. f \$ i \<noteq> 0) = d" ``` ``` 490 proof (rule Least_equality) ``` ``` 491 fix e assume "f \$ e \<noteq> 0" ``` ``` 492 with assms(2) have "\<not>(e < d)" by blast ``` ``` 493 thus "e \<ge> d" by simp ``` ``` 494 qed ``` ``` 495 ultimately show ?thesis unfolding subdegree_def by simp ``` ``` 496 qed ``` ``` 497 ``` ``` 498 lemma nth_subdegree_nonzero [simp,intro]: "f \<noteq> 0 \<Longrightarrow> f \$ subdegree f \<noteq> 0" ``` ``` 499 proof- ``` ``` 500 assume "f \<noteq> 0" ``` ``` 501 hence "subdegree f = (LEAST n. f \$ n \<noteq> 0)" by (simp add: subdegree_def) ``` ``` 502 also from \<open>f \<noteq> 0\<close> have "\<exists>n. f\$n \<noteq> 0" using fps_nonzero_nth by blast ``` ``` 503 from LeastI_ex[OF this] have "f \$ (LEAST n. f \$ n \<noteq> 0) \<noteq> 0" . ``` ``` 504 finally show ?thesis . ``` ``` 505 qed ``` ``` 506 ``` ``` 507 lemma nth_less_subdegree_zero [dest]: "n < subdegree f \<Longrightarrow> f \$ n = 0" ``` ``` 508 proof (cases "f = 0") ``` ``` 509 assume "f \<noteq> 0" and less: "n < subdegree f" ``` ``` 510 note less ``` ``` 511 also from \<open>f \<noteq> 0\<close> have "subdegree f = (LEAST n. f \$ n \<noteq> 0)" by (simp add: subdegree_def) ``` ``` 512 finally show "f \$ n = 0" using not_less_Least by blast ``` ``` 513 qed simp_all ``` ``` 514 ``` ``` 515 lemma subdegree_geI: ``` ``` 516 assumes "f \<noteq> 0" "\<And>i. i < n \<Longrightarrow> f\$i = 0" ``` ``` 517 shows "subdegree f \<ge> n" ``` ``` 518 proof (rule ccontr) ``` ``` 519 assume "\<not>(subdegree f \<ge> n)" ``` ``` 520 with assms(2) have "f \$ subdegree f = 0" by simp ``` ``` 521 moreover from assms(1) have "f \$ subdegree f \<noteq> 0" by simp ``` ``` 522 ultimately show False by contradiction ``` ``` 523 qed ``` ``` 524 ``` ``` 525 lemma subdegree_greaterI: ``` ``` 526 assumes "f \<noteq> 0" "\<And>i. i \<le> n \<Longrightarrow> f\$i = 0" ``` ``` 527 shows "subdegree f > n" ``` ``` 528 proof (rule ccontr) ``` ``` 529 assume "\<not>(subdegree f > n)" ``` ``` 530 with assms(2) have "f \$ subdegree f = 0" by simp ``` ``` 531 moreover from assms(1) have "f \$ subdegree f \<noteq> 0" by simp ``` ``` 532 ultimately show False by contradiction ``` ``` 533 qed ``` ``` 534 ``` ``` 535 lemma subdegree_leI: ``` ``` 536 "f \$ n \<noteq> 0 \<Longrightarrow> subdegree f \<le> n" ``` ``` 537 by (rule leI) auto ``` ``` 538 ``` ``` 539 ``` ``` 540 lemma subdegree_0 [simp]: "subdegree 0 = 0" ``` ``` 541 by (simp add: subdegree_def) ``` ``` 542 ``` ``` 543 lemma subdegree_1 [simp]: "subdegree (1 :: ('a :: zero_neq_one) fps) = 0" ``` ``` 544 by (auto intro!: subdegreeI) ``` ``` 545 ``` ``` 546 lemma subdegree_X [simp]: "subdegree (X :: ('a :: zero_neq_one) fps) = 1" ``` ``` 547 by (auto intro!: subdegreeI simp: X_def) ``` ``` 548 ``` ``` 549 lemma subdegree_fps_const [simp]: "subdegree (fps_const c) = 0" ``` ``` 550 by (cases "c = 0") (auto intro!: subdegreeI) ``` ``` 551 ``` ``` 552 lemma subdegree_numeral [simp]: "subdegree (numeral n) = 0" ``` ``` 553 by (simp add: numeral_fps_const) ``` ``` 554 ``` ``` 555 lemma subdegree_eq_0_iff: "subdegree f = 0 \<longleftrightarrow> f = 0 \<or> f \$ 0 \<noteq> 0" ``` ``` 556 proof (cases "f = 0") ``` ``` 557 assume "f \<noteq> 0" ``` ``` 558 thus ?thesis ``` ``` 559 using nth_subdegree_nonzero[OF \<open>f \<noteq> 0\<close>] by (fastforce intro!: subdegreeI) ``` ``` 560 qed simp_all ``` ``` 561 ``` ``` 562 lemma subdegree_eq_0 [simp]: "f \$ 0 \<noteq> 0 \<Longrightarrow> subdegree f = 0" ``` ``` 563 by (simp add: subdegree_eq_0_iff) ``` ``` 564 ``` ``` 565 lemma nth_subdegree_mult [simp]: ``` ``` 566 fixes f g :: "('a :: {mult_zero,comm_monoid_add}) fps" ``` ``` 567 shows "(f * g) \$ (subdegree f + subdegree g) = f \$ subdegree f * g \$ subdegree g" ``` ``` 568 proof- ``` ``` 569 let ?n = "subdegree f + subdegree g" ``` ``` 570 have "(f * g) \$ ?n = (\<Sum>i=0..?n. f\$i * g\$(?n-i))" ``` ``` 571 by (simp add: fps_mult_nth) ``` ``` 572 also have "... = (\<Sum>i=0..?n. if i = subdegree f then f\$i * g\$(?n-i) else 0)" ``` ``` 573 proof (intro setsum.cong) ``` ``` 574 fix x assume x: "x \<in> {0..?n}" ``` ``` 575 hence "x = subdegree f \<or> x < subdegree f \<or> ?n - x < subdegree g" by auto ``` ``` 576 thus "f \$ x * g \$ (?n - x) = (if x = subdegree f then f \$ x * g \$ (?n - x) else 0)" ``` ``` 577 by (elim disjE conjE) auto ``` ``` 578 qed auto ``` ``` 579 also have "... = f \$ subdegree f * g \$ subdegree g" by (simp add: setsum.delta) ``` ``` 580 finally show ?thesis . ``` ``` 581 qed ``` ``` 582 ``` ``` 583 lemma subdegree_mult [simp]: ``` ``` 584 assumes "f \<noteq> 0" "g \<noteq> 0" ``` ``` 585 shows "subdegree ((f :: ('a :: {ring_no_zero_divisors}) fps) * g) = subdegree f + subdegree g" ``` ``` 586 proof (rule subdegreeI) ``` ``` 587 let ?n = "subdegree f + subdegree g" ``` ``` 588 have "(f * g) \$ ?n = (\<Sum>i=0..?n. f\$i * g\$(?n-i))" by (simp add: fps_mult_nth) ``` ``` 589 also have "... = (\<Sum>i=0..?n. if i = subdegree f then f\$i * g\$(?n-i) else 0)" ``` ``` 590 proof (intro setsum.cong) ``` ``` 591 fix x assume x: "x \<in> {0..?n}" ``` ``` 592 hence "x = subdegree f \<or> x < subdegree f \<or> ?n - x < subdegree g" by auto ``` ``` 593 thus "f \$ x * g \$ (?n - x) = (if x = subdegree f then f \$ x * g \$ (?n - x) else 0)" ``` ``` 594 by (elim disjE conjE) auto ``` ``` 595 qed auto ``` ``` 596 also have "... = f \$ subdegree f * g \$ subdegree g" by (simp add: setsum.delta) ``` ``` 597 also from assms have "... \<noteq> 0" by auto ``` ``` 598 finally show "(f * g) \$ (subdegree f + subdegree g) \<noteq> 0" . ``` ``` 599 next ``` ``` 600 fix m assume m: "m < subdegree f + subdegree g" ``` ``` 601 have "(f * g) \$ m = (\<Sum>i=0..m. f\$i * g\$(m-i))" by (simp add: fps_mult_nth) ``` ``` 602 also have "... = (\<Sum>i=0..m. 0)" ``` ``` 603 proof (rule setsum.cong) ``` ``` 604 fix i assume "i \<in> {0..m}" ``` ``` 605 with m have "i < subdegree f \<or> m - i < subdegree g" by auto ``` ``` 606 thus "f\$i * g\$(m-i) = 0" by (elim disjE) auto ``` ``` 607 qed auto ``` ``` 608 finally show "(f * g) \$ m = 0" by simp ``` ``` 609 qed ``` ``` 610 ``` ``` 611 lemma subdegree_power [simp]: ``` ``` 612 "subdegree ((f :: ('a :: ring_1_no_zero_divisors) fps) ^ n) = n * subdegree f" ``` ``` 613 by (cases "f = 0"; induction n) simp_all ``` ``` 614 ``` ``` 615 lemma subdegree_uminus [simp]: ``` ``` 616 "subdegree (-(f::('a::group_add) fps)) = subdegree f" ``` ``` 617 by (simp add: subdegree_def) ``` ``` 618 ``` ``` 619 lemma subdegree_minus_commute [simp]: ``` ``` 620 "subdegree (f-(g::('a::group_add) fps)) = subdegree (g - f)" ``` ``` 621 proof - ``` ``` 622 have "f - g = -(g - f)" by simp ``` ``` 623 also have "subdegree ... = subdegree (g - f)" by (simp only: subdegree_uminus) ``` ``` 624 finally show ?thesis . ``` ``` 625 qed ``` ``` 626 ``` ``` 627 lemma subdegree_add_ge: ``` ``` 628 assumes "f \<noteq> -(g :: ('a :: {group_add}) fps)" ``` ``` 629 shows "subdegree (f + g) \<ge> min (subdegree f) (subdegree g)" ``` ``` 630 proof (rule subdegree_geI) ``` ``` 631 from assms show "f + g \<noteq> 0" by (subst (asm) eq_neg_iff_add_eq_0) ``` ``` 632 next ``` ``` 633 fix i assume "i < min (subdegree f) (subdegree g)" ``` ``` 634 hence "f \$ i = 0" and "g \$ i = 0" by auto ``` ``` 635 thus "(f + g) \$ i = 0" by force ``` ``` 636 qed ``` ``` 637 ``` ``` 638 lemma subdegree_add_eq1: ``` ``` 639 assumes "f \<noteq> 0" ``` ``` 640 assumes "subdegree f < subdegree (g :: ('a :: {group_add}) fps)" ``` ``` 641 shows "subdegree (f + g) = subdegree f" ``` ``` 642 proof (rule antisym[OF subdegree_leI]) ``` ``` 643 from assms show "subdegree (f + g) \<ge> subdegree f" ``` ``` 644 by (intro order.trans[OF min.boundedI subdegree_add_ge]) auto ``` ``` 645 from assms have "f \$ subdegree f \<noteq> 0" "g \$ subdegree f = 0" by auto ``` ``` 646 thus "(f + g) \$ subdegree f \<noteq> 0" by simp ``` ``` 647 qed ``` ``` 648 ``` ``` 649 lemma subdegree_add_eq2: ``` ``` 650 assumes "g \<noteq> 0" ``` ``` 651 assumes "subdegree g < subdegree (f :: ('a :: {ab_group_add}) fps)" ``` ``` 652 shows "subdegree (f + g) = subdegree g" ``` ``` 653 using subdegree_add_eq1[OF assms] by (simp add: add.commute) ``` ``` 654 ``` ``` 655 lemma subdegree_diff_eq1: ``` ``` 656 assumes "f \<noteq> 0" ``` ``` 657 assumes "subdegree f < subdegree (g :: ('a :: {ab_group_add}) fps)" ``` ``` 658 shows "subdegree (f - g) = subdegree f" ``` ``` 659 using subdegree_add_eq1[of f "-g"] assms by (simp add: add.commute) ``` ``` 660 ``` ``` 661 lemma subdegree_diff_eq2: ``` ``` 662 assumes "g \<noteq> 0" ``` ``` 663 assumes "subdegree g < subdegree (f :: ('a :: {ab_group_add}) fps)" ``` ``` 664 shows "subdegree (f - g) = subdegree g" ``` ``` 665 using subdegree_add_eq2[of "-g" f] assms by (simp add: add.commute) ``` ``` 666 ``` ``` 667 lemma subdegree_diff_ge [simp]: ``` ``` 668 assumes "f \<noteq> (g :: ('a :: {group_add}) fps)" ``` ``` 669 shows "subdegree (f - g) \<ge> min (subdegree f) (subdegree g)" ``` ``` 670 using assms subdegree_add_ge[of f "-g"] by simp ``` ``` 671 ``` ``` 672 ``` ``` 673 ``` ``` 674 ``` ``` 675 subsection \<open>Shifting and slicing\<close> ``` ``` 676 ``` ``` 677 definition fps_shift :: "nat \<Rightarrow> 'a fps \<Rightarrow> 'a fps" where ``` ``` 678 "fps_shift n f = Abs_fps (\<lambda>i. f \$ (i + n))" ``` ``` 679 ``` ``` 680 lemma fps_shift_nth [simp]: "fps_shift n f \$ i = f \$ (i + n)" ``` ``` 681 by (simp add: fps_shift_def) ``` ``` 682 ``` ``` 683 lemma fps_shift_0 [simp]: "fps_shift 0 f = f" ``` ``` 684 by (intro fps_ext) (simp add: fps_shift_def) ``` ``` 685 ``` ``` 686 lemma fps_shift_zero [simp]: "fps_shift n 0 = 0" ``` ``` 687 by (intro fps_ext) (simp add: fps_shift_def) ``` ``` 688 ``` ``` 689 lemma fps_shift_one: "fps_shift n 1 = (if n = 0 then 1 else 0)" ``` ``` 690 by (intro fps_ext) (simp add: fps_shift_def) ``` ``` 691 ``` ``` 692 lemma fps_shift_fps_const: "fps_shift n (fps_const c) = (if n = 0 then fps_const c else 0)" ``` ``` 693 by (intro fps_ext) (simp add: fps_shift_def) ``` ``` 694 ``` ``` 695 lemma fps_shift_numeral: "fps_shift n (numeral c) = (if n = 0 then numeral c else 0)" ``` ``` 696 by (simp add: numeral_fps_const fps_shift_fps_const) ``` ``` 697 ``` ``` 698 lemma fps_shift_X_power [simp]: ``` ``` 699 "n \<le> m \<Longrightarrow> fps_shift n (X ^ m) = (X ^ (m - n) ::'a::comm_ring_1 fps)" ``` ``` 700 by (intro fps_ext) (auto simp: fps_shift_def ) ``` ``` 701 ``` ``` 702 lemma fps_shift_times_X_power: ``` ``` 703 "n \<le> subdegree f \<Longrightarrow> fps_shift n f * X ^ n = (f :: 'a :: comm_ring_1 fps)" ``` ``` 704 by (intro fps_ext) (auto simp: X_power_mult_right_nth nth_less_subdegree_zero) ``` ``` 705 ``` ``` 706 lemma fps_shift_times_X_power' [simp]: ``` ``` 707 "fps_shift n (f * X^n) = (f :: 'a :: comm_ring_1 fps)" ``` ``` 708 by (intro fps_ext) (auto simp: X_power_mult_right_nth nth_less_subdegree_zero) ``` ``` 709 ``` ``` 710 lemma fps_shift_times_X_power'': ``` ``` 711 "m \<le> n \<Longrightarrow> fps_shift n (f * X^m) = fps_shift (n - m) (f :: 'a :: comm_ring_1 fps)" ``` ``` 712 by (intro fps_ext) (auto simp: X_power_mult_right_nth nth_less_subdegree_zero) ``` ``` 713 ``` ``` 714 lemma fps_shift_subdegree [simp]: ``` ``` 715 "n \<le> subdegree f \<Longrightarrow> subdegree (fps_shift n f) = subdegree (f :: 'a :: comm_ring_1 fps) - n" ``` ``` 716 by (cases "f = 0") (force intro: nth_less_subdegree_zero subdegreeI)+ ``` ``` 717 ``` ``` 718 lemma subdegree_decompose: ``` ``` 719 "f = fps_shift (subdegree f) f * X ^ subdegree (f :: ('a :: comm_ring_1) fps)" ``` ``` 720 by (rule fps_ext) (auto simp: X_power_mult_right_nth) ``` ``` 721 ``` ``` 722 lemma subdegree_decompose': ``` ``` 723 "n \<le> subdegree (f :: ('a :: comm_ring_1) fps) \<Longrightarrow> f = fps_shift n f * X^n" ``` ``` 724 by (rule fps_ext) (auto simp: X_power_mult_right_nth intro!: nth_less_subdegree_zero) ``` ``` 725 ``` ``` 726 lemma fps_shift_fps_shift: ``` ``` 727 "fps_shift (m + n) f = fps_shift m (fps_shift n f)" ``` ``` 728 by (rule fps_ext) (simp add: add_ac) ``` ``` 729 ``` ``` 730 lemma fps_shift_add: ``` ``` 731 "fps_shift n (f + g) = fps_shift n f + fps_shift n g" ``` ``` 732 by (simp add: fps_eq_iff) ``` ``` 733 ``` ``` 734 lemma fps_shift_mult: ``` ``` 735 assumes "n \<le> subdegree (g :: 'b :: {comm_ring_1} fps)" ``` ``` 736 shows "fps_shift n (h*g) = h * fps_shift n g" ``` ``` 737 proof - ``` ``` 738 from assms have "g = fps_shift n g * X^n" by (rule subdegree_decompose') ``` ``` 739 also have "h * ... = (h * fps_shift n g) * X^n" by simp ``` ``` 740 also have "fps_shift n ... = h * fps_shift n g" by simp ``` ``` 741 finally show ?thesis . ``` ``` 742 qed ``` ``` 743 ``` ``` 744 lemma fps_shift_mult_right: ``` ``` 745 assumes "n \<le> subdegree (g :: 'b :: {comm_ring_1} fps)" ``` ``` 746 shows "fps_shift n (g*h) = h * fps_shift n g" ``` ``` 747 by (subst mult.commute, subst fps_shift_mult) (simp_all add: assms) ``` ``` 748 ``` ``` 749 lemma nth_subdegree_zero_iff [simp]: "f \$ subdegree f = 0 \<longleftrightarrow> f = 0" ``` ``` 750 by (cases "f = 0") auto ``` ``` 751 ``` ``` 752 lemma fps_shift_subdegree_zero_iff [simp]: ``` ``` 753 "fps_shift (subdegree f) f = 0 \<longleftrightarrow> f = 0" ``` ``` 754 by (subst (1) nth_subdegree_zero_iff[symmetric], cases "f = 0") ``` ``` 755 (simp_all del: nth_subdegree_zero_iff) ``` ``` 756 ``` ``` 757 ``` ``` 758 definition "fps_cutoff n f = Abs_fps (\<lambda>i. if i < n then f\$i else 0)" ``` ``` 759 ``` ``` 760 lemma fps_cutoff_nth [simp]: "fps_cutoff n f \$ i = (if i < n then f\$i else 0)" ``` ``` 761 unfolding fps_cutoff_def by simp ``` ``` 762 ``` ``` 763 lemma fps_cutoff_zero_iff: "fps_cutoff n f = 0 \<longleftrightarrow> (f = 0 \<or> n \<le> subdegree f)" ``` ``` 764 proof ``` ``` 765 assume A: "fps_cutoff n f = 0" ``` ``` 766 thus "f = 0 \<or> n \<le> subdegree f" ``` ``` 767 proof (cases "f = 0") ``` ``` 768 assume "f \<noteq> 0" ``` ``` 769 with A have "n \<le> subdegree f" ``` ``` 770 by (intro subdegree_geI) (auto simp: fps_eq_iff split: split_if_asm) ``` ``` 771 thus ?thesis .. ``` ``` 772 qed simp ``` ``` 773 qed (auto simp: fps_eq_iff intro: nth_less_subdegree_zero) ``` ``` 774 ``` ``` 775 lemma fps_cutoff_0 [simp]: "fps_cutoff 0 f = 0" ``` ``` 776 by (simp add: fps_eq_iff) ``` ``` 777 ``` ``` 778 lemma fps_cutoff_zero [simp]: "fps_cutoff n 0 = 0" ``` ``` 779 by (simp add: fps_eq_iff) ``` ``` 780 ``` ``` 781 lemma fps_cutoff_one: "fps_cutoff n 1 = (if n = 0 then 0 else 1)" ``` ``` 782 by (simp add: fps_eq_iff) ``` ``` 783 ``` ``` 784 lemma fps_cutoff_fps_const: "fps_cutoff n (fps_const c) = (if n = 0 then 0 else fps_const c)" ``` ``` 785 by (simp add: fps_eq_iff) ``` ``` 786 ``` ``` 787 lemma fps_cutoff_numeral: "fps_cutoff n (numeral c) = (if n = 0 then 0 else numeral c)" ``` ``` 788 by (simp add: numeral_fps_const fps_cutoff_fps_const) ``` ``` 789 ``` ``` 790 lemma fps_shift_cutoff: ``` ``` 791 "fps_shift n (f :: ('a :: comm_ring_1) fps) * X^n + fps_cutoff n f = f" ``` ``` 792 by (simp add: fps_eq_iff X_power_mult_right_nth) ``` ``` 793 ``` ``` 794 ``` ``` 795 subsection \<open>Formal Power series form a metric space\<close> ``` ``` 796 ``` ``` 797 definition (in dist) "ball x r = {y. dist y x < r}" ``` ``` 798 ``` ``` 799 instantiation fps :: (comm_ring_1) dist ``` ``` 800 begin ``` ``` 801 ``` ``` 802 definition ``` ``` 803 dist_fps_def: "dist (a :: 'a fps) b = (if a = b then 0 else inverse (2 ^ subdegree (a - b)))" ``` ``` 804 ``` ``` 805 lemma dist_fps_ge0: "dist (a :: 'a fps) b \<ge> 0" ``` ``` 806 by (simp add: dist_fps_def) ``` ``` 807 ``` ``` 808 lemma dist_fps_sym: "dist (a :: 'a fps) b = dist b a" ``` ``` 809 by (simp add: dist_fps_def) ``` ``` 810 ``` ``` 811 instance .. ``` ``` 812 ``` ``` 813 end ``` ``` 814 ``` ``` 815 instantiation fps :: (comm_ring_1) metric_space ``` ``` 816 begin ``` ``` 817 ``` ``` 818 definition open_fps_def: "open (S :: 'a fps set) = (\<forall>a \<in> S. \<exists>r. r >0 \<and> ball a r \<subseteq> S)" ``` ``` 819 ``` ``` 820 ``` ``` 821 instance ``` ``` 822 proof ``` ``` 823 show "open S = (\<forall>x\<in>S. \<exists>e>0. \<forall>y. dist y x < e \<longrightarrow> y \<in> S)" for S :: "'a fps set" ``` ``` 824 by (auto simp add: open_fps_def ball_def subset_eq) ``` ``` 825 show th: "dist a b = 0 \<longleftrightarrow> a = b" for a b :: "'a fps" ``` ``` 826 by (simp add: dist_fps_def split: split_if_asm) ``` ``` 827 then have th'[simp]: "dist a a = 0" for a :: "'a fps" by simp ``` ``` 828 ``` ``` 829 fix a b c :: "'a fps" ``` ``` 830 consider "a = b" | "c = a \<or> c = b" | "a \<noteq> b" "a \<noteq> c" "b \<noteq> c" by blast ``` ``` 831 then show "dist a b \<le> dist a c + dist b c" ``` ``` 832 proof cases ``` ``` 833 case 1 ``` ``` 834 then show ?thesis by (simp add: dist_fps_def) ``` ``` 835 next ``` ``` 836 case 2 ``` ``` 837 then show ?thesis ``` ``` 838 by (cases "c = a") (simp_all add: th dist_fps_sym) ``` ``` 839 next ``` ``` 840 case neq: 3 ``` ``` 841 have False if "dist a b > dist a c + dist b c" ``` ``` 842 proof - ``` ``` 843 let ?n = "subdegree (a - b)" ``` ``` 844 from neq have "dist a b > 0" "dist b c > 0" and "dist a c > 0" by (simp_all add: dist_fps_def) ``` ``` 845 with that have "dist a b > dist a c" and "dist a b > dist b c" by simp_all ``` ``` 846 with neq have "?n < subdegree (a - c)" and "?n < subdegree (b - c)" ``` ``` 847 by (simp_all add: dist_fps_def field_simps) ``` ``` 848 hence "(a - c) \$ ?n = 0" and "(b - c) \$ ?n = 0" ``` ``` 849 by (simp_all only: nth_less_subdegree_zero) ``` ``` 850 hence "(a - b) \$ ?n = 0" by simp ``` ``` 851 moreover from neq have "(a - b) \$ ?n \<noteq> 0" by (intro nth_subdegree_nonzero) simp_all ``` ``` 852 ultimately show False by contradiction ``` ``` 853 qed ``` ``` 854 thus ?thesis by (auto simp add: not_le[symmetric]) ``` ``` 855 qed ``` ``` 856 qed ``` ``` 857 ``` ``` 858 end ``` ``` 859 ``` ``` 860 ``` ``` 861 text \<open>The infinite sums and justification of the notation in textbooks.\<close> ``` ``` 862 ``` ``` 863 lemma reals_power_lt_ex: ``` ``` 864 fixes x y :: real ``` ``` 865 assumes xp: "x > 0" ``` ``` 866 and y1: "y > 1" ``` ``` 867 shows "\<exists>k>0. (1/y)^k < x" ``` ``` 868 proof - ``` ``` 869 have yp: "y > 0" ``` ``` 870 using y1 by simp ``` ``` 871 from reals_Archimedean2[of "max 0 (- log y x) + 1"] ``` ``` 872 obtain k :: nat where k: "real k > max 0 (- log y x) + 1" ``` ``` 873 by blast ``` ``` 874 from k have kp: "k > 0" ``` ``` 875 by simp ``` ``` 876 from k have "real k > - log y x" ``` ``` 877 by simp ``` ``` 878 then have "ln y * real k > - ln x" ``` ``` 879 unfolding log_def ``` ``` 880 using ln_gt_zero_iff[OF yp] y1 ``` ``` 881 by (simp add: minus_divide_left field_simps del: minus_divide_left[symmetric]) ``` ``` 882 then have "ln y * real k + ln x > 0" ``` ``` 883 by simp ``` ``` 884 then have "exp (real k * ln y + ln x) > exp 0" ``` ``` 885 by (simp add: ac_simps) ``` ``` 886 then have "y ^ k * x > 1" ``` ``` 887 unfolding exp_zero exp_add exp_real_of_nat_mult exp_ln [OF xp] exp_ln [OF yp] ``` ``` 888 by simp ``` ``` 889 then have "x > (1 / y)^k" using yp ``` ``` 890 by (simp add: field_simps) ``` ``` 891 then show ?thesis ``` ``` 892 using kp by blast ``` ``` 893 qed ``` ``` 894 ``` ``` 895 lemma fps_sum_rep_nth: "(setsum (\<lambda>i. fps_const(a\$i)*X^i) {0..m})\$n = ``` ``` 896 (if n \<le> m then a\$n else 0::'a::comm_ring_1)" ``` ``` 897 apply (auto simp add: fps_setsum_nth cond_value_iff cong del: if_weak_cong) ``` ``` 898 apply (simp add: setsum.delta') ``` ``` 899 done ``` ``` 900 ``` ``` 901 lemma fps_notation: "(\<lambda>n. setsum (\<lambda>i. fps_const(a\$i) * X^i) {0..n}) ----> a" ``` ``` 902 (is "?s ----> a") ``` ``` 903 proof - ``` ``` 904 have "\<exists>n0. \<forall>n \<ge> n0. dist (?s n) a < r" if "r > 0" for r ``` ``` 905 proof - ``` ``` 906 obtain n0 where n0: "(1/2)^n0 < r" "n0 > 0" ``` ``` 907 using reals_power_lt_ex[OF \<open>r > 0\<close>, of 2] by auto ``` ``` 908 show ?thesis ``` ``` 909 proof - ``` ``` 910 have "dist (?s n) a < r" if nn0: "n \<ge> n0" for n ``` ``` 911 proof - ``` ``` 912 from that have thnn0: "(1/2)^n \<le> (1/2 :: real)^n0" ``` ``` 913 by (simp add: divide_simps) ``` ``` 914 show ?thesis ``` ``` 915 proof (cases "?s n = a") ``` ``` 916 case True ``` ``` 917 then show ?thesis ``` ``` 918 unfolding dist_eq_0_iff[of "?s n" a, symmetric] ``` ``` 919 using \<open>r > 0\<close> by (simp del: dist_eq_0_iff) ``` ``` 920 next ``` ``` 921 case False ``` ``` 922 from False have dth: "dist (?s n) a = (1/2)^subdegree (?s n - a)" ``` ``` 923 by (simp add: dist_fps_def field_simps) ``` ``` 924 from False have kn: "subdegree (?s n - a) > n" ``` ``` 925 by (intro subdegree_greaterI) (simp_all add: fps_sum_rep_nth) ``` ``` 926 then have "dist (?s n) a < (1/2)^n" ``` ``` 927 by (simp add: field_simps dist_fps_def) ``` ``` 928 also have "\<dots> \<le> (1/2)^n0" ``` ``` 929 using nn0 by (simp add: divide_simps) ``` ``` 930 also have "\<dots> < r" ``` ``` 931 using n0 by simp ``` ``` 932 finally show ?thesis . ``` ``` 933 qed ``` ``` 934 qed ``` ``` 935 then show ?thesis by blast ``` ``` 936 qed ``` ``` 937 qed ``` ``` 938 then show ?thesis ``` ``` 939 unfolding lim_sequentially by blast ``` ``` 940 qed ``` ``` 941 ``` ``` 942 ``` ``` 943 subsection \<open>Inverses of formal power series\<close> ``` ``` 944 ``` ``` 945 declare setsum.cong[fundef_cong] ``` ``` 946 ``` ``` 947 instantiation fps :: ("{comm_monoid_add,inverse,times,uminus}") inverse ``` ``` 948 begin ``` ``` 949 ``` ``` 950 fun natfun_inverse:: "'a fps \<Rightarrow> nat \<Rightarrow> 'a" ``` ``` 951 where ``` ``` 952 "natfun_inverse f 0 = inverse (f\$0)" ``` ``` 953 | "natfun_inverse f n = - inverse (f\$0) * setsum (\<lambda>i. f\$i * natfun_inverse f (n - i)) {1..n}" ``` ``` 954 ``` ``` 955 definition fps_inverse_def: "inverse f = (if f \$ 0 = 0 then 0 else Abs_fps (natfun_inverse f))" ``` ``` 956 ``` ``` 957 definition fps_divide_def: ``` ``` 958 "f div g = (if g = 0 then 0 else ``` ``` 959 let n = subdegree g; h = fps_shift n g ``` ``` 960 in fps_shift n (f * inverse h))" ``` ``` 961 ``` ``` 962 instance .. ``` ``` 963 ``` ``` 964 end ``` ``` 965 ``` ``` 966 lemma fps_inverse_zero [simp]: ``` ``` 967 "inverse (0 :: 'a::{comm_monoid_add,inverse,times,uminus} fps) = 0" ``` ``` 968 by (simp add: fps_ext fps_inverse_def) ``` ``` 969 ``` ``` 970 lemma fps_inverse_one [simp]: "inverse (1 :: 'a::{division_ring,zero_neq_one} fps) = 1" ``` ``` 971 apply (auto simp add: expand_fps_eq fps_inverse_def) ``` ``` 972 apply (case_tac n) ``` ``` 973 apply auto ``` ``` 974 done ``` ``` 975 ``` ``` 976 lemma inverse_mult_eq_1 [intro]: ``` ``` 977 assumes f0: "f\$0 \<noteq> (0::'a::field)" ``` ``` 978 shows "inverse f * f = 1" ``` ``` 979 proof - ``` ``` 980 have c: "inverse f * f = f * inverse f" ``` ``` 981 by (simp add: mult.commute) ``` ``` 982 from f0 have ifn: "\<And>n. inverse f \$ n = natfun_inverse f n" ``` ``` 983 by (simp add: fps_inverse_def) ``` ``` 984 from f0 have th0: "(inverse f * f) \$ 0 = 1" ``` ``` 985 by (simp add: fps_mult_nth fps_inverse_def) ``` ``` 986 have "(inverse f * f)\$n = 0" if np: "n > 0" for n ``` ``` 987 proof - ``` ``` 988 from np have eq: "{0..n} = {0} \<union> {1 .. n}" ``` ``` 989 by auto ``` ``` 990 have d: "{0} \<inter> {1 .. n} = {}" ``` ``` 991 by auto ``` ``` 992 from f0 np have th0: "- (inverse f \$ n) = ``` ``` 993 (setsum (\<lambda>i. f\$i * natfun_inverse f (n - i)) {1..n}) / (f\$0)" ``` ``` 994 by (cases n) (simp_all add: divide_inverse fps_inverse_def) ``` ``` 995 from th0[symmetric, unfolded nonzero_divide_eq_eq[OF f0]] ``` ``` 996 have th1: "setsum (\<lambda>i. f\$i * natfun_inverse f (n - i)) {1..n} = - (f\$0) * (inverse f)\$n" ``` ``` 997 by (simp add: field_simps) ``` ``` 998 have "(f * inverse f) \$ n = (\<Sum>i = 0..n. f \$i * natfun_inverse f (n - i))" ``` ``` 999 unfolding fps_mult_nth ifn .. ``` ``` 1000 also have "\<dots> = f\$0 * natfun_inverse f n + (\<Sum>i = 1..n. f\$i * natfun_inverse f (n-i))" ``` ``` 1001 by (simp add: eq) ``` ``` 1002 also have "\<dots> = 0" ``` ``` 1003 unfolding th1 ifn by simp ``` ``` 1004 finally show ?thesis unfolding c . ``` ``` 1005 qed ``` ``` 1006 with th0 show ?thesis ``` ``` 1007 by (simp add: fps_eq_iff) ``` ``` 1008 qed ``` ``` 1009 ``` ``` 1010 lemma fps_inverse_0_iff[simp]: "(inverse f) \$ 0 = (0::'a::division_ring) \<longleftrightarrow> f \$ 0 = 0" ``` ``` 1011 by (simp add: fps_inverse_def nonzero_imp_inverse_nonzero) ``` ``` 1012 ``` ``` 1013 lemma fps_inverse_nth_0 [simp]: "inverse f \$ 0 = inverse (f \$ 0 :: 'a :: division_ring)" ``` ``` 1014 by (simp add: fps_inverse_def) ``` ``` 1015 ``` ``` 1016 lemma fps_inverse_eq_0_iff[simp]: "inverse f = (0:: ('a::division_ring) fps) \<longleftrightarrow> f \$ 0 = 0" ``` ``` 1017 proof ``` ``` 1018 assume A: "inverse f = 0" ``` ``` 1019 have "0 = inverse f \$ 0" by (subst A) simp ``` ``` 1020 thus "f \$ 0 = 0" by simp ``` ``` 1021 qed (simp add: fps_inverse_def) ``` ``` 1022 ``` ``` 1023 lemma fps_inverse_idempotent[intro, simp]: ``` ``` 1024 assumes f0: "f\$0 \<noteq> (0::'a::field)" ``` ``` 1025 shows "inverse (inverse f) = f" ``` ``` 1026 proof - ``` ``` 1027 from f0 have if0: "inverse f \$ 0 \<noteq> 0" by simp ``` ``` 1028 from inverse_mult_eq_1[OF f0] inverse_mult_eq_1[OF if0] ``` ``` 1029 have "inverse f * f = inverse f * inverse (inverse f)" ``` ``` 1030 by (simp add: ac_simps) ``` ``` 1031 then show ?thesis ``` ``` 1032 using f0 unfolding mult_cancel_left by simp ``` ``` 1033 qed ``` ``` 1034 ``` ``` 1035 lemma fps_inverse_unique: ``` ``` 1036 assumes fg: "(f :: 'a :: field fps) * g = 1" ``` ``` 1037 shows "inverse f = g" ``` ``` 1038 proof - ``` ``` 1039 have f0: "f \$ 0 \<noteq> 0" ``` ``` 1040 proof ``` ``` 1041 assume "f \$ 0 = 0" ``` ``` 1042 hence "0 = (f * g) \$ 0" by simp ``` ``` 1043 also from fg have "(f * g) \$ 0 = 1" by simp ``` ``` 1044 finally show False by simp ``` ``` 1045 qed ``` ``` 1046 from inverse_mult_eq_1[OF this] fg ``` ``` 1047 have th0: "inverse f * f = g * f" ``` ``` 1048 by (simp add: ac_simps) ``` ``` 1049 then show ?thesis ``` ``` 1050 using f0 ``` ``` 1051 unfolding mult_cancel_right ``` ``` 1052 by (auto simp add: expand_fps_eq) ``` ``` 1053 qed ``` ``` 1054 ``` ``` 1055 lemma setsum_zero_lemma: ``` ``` 1056 fixes n::nat ``` ``` 1057 assumes "0 < n" ``` ``` 1058 shows "(\<Sum>i = 0..n. if n = i then 1 else if n - i = 1 then - 1 else 0) = (0::'a::field)" ``` ``` 1059 proof - ``` ``` 1060 let ?f = "\<lambda>i. if n = i then 1 else if n - i = 1 then - 1 else 0" ``` ``` 1061 let ?g = "\<lambda>i. if i = n then 1 else if i = n - 1 then - 1 else 0" ``` ``` 1062 let ?h = "\<lambda>i. if i=n - 1 then - 1 else 0" ``` ``` 1063 have th1: "setsum ?f {0..n} = setsum ?g {0..n}" ``` ``` 1064 by (rule setsum.cong) auto ``` ``` 1065 have th2: "setsum ?g {0..n - 1} = setsum ?h {0..n - 1}" ``` ``` 1066 apply (rule setsum.cong) ``` ``` 1067 using assms ``` ``` 1068 apply auto ``` ``` 1069 done ``` ``` 1070 have eq: "{0 .. n} = {0.. n - 1} \<union> {n}" ``` ``` 1071 by auto ``` ``` 1072 from assms have d: "{0.. n - 1} \<inter> {n} = {}" ``` ``` 1073 by auto ``` ``` 1074 have f: "finite {0.. n - 1}" "finite {n}" ``` ``` 1075 by auto ``` ``` 1076 show ?thesis ``` ``` 1077 unfolding th1 ``` ``` 1078 apply (simp add: setsum.union_disjoint[OF f d, unfolded eq[symmetric]] del: One_nat_def) ``` ``` 1079 unfolding th2 ``` ``` 1080 apply (simp add: setsum.delta) ``` ``` 1081 done ``` ``` 1082 qed ``` ``` 1083 ``` ``` 1084 lemma fps_inverse_mult: "inverse (f * g :: 'a::field fps) = inverse f * inverse g" ``` ``` 1085 proof (cases "f\$0 = 0 \<or> g\$0 = 0") ``` ``` 1086 assume "\<not>(f\$0 = 0 \<or> g\$0 = 0)" ``` ``` 1087 hence [simp]: "f\$0 \<noteq> 0" "g\$0 \<noteq> 0" by simp_all ``` ``` 1088 show ?thesis ``` ``` 1089 proof (rule fps_inverse_unique) ``` ``` 1090 have "f * g * (inverse f * inverse g) = (inverse f * f) * (inverse g * g)" by simp ``` ``` 1091 also have "... = 1" by (subst (1 2) inverse_mult_eq_1) simp_all ``` ``` 1092 finally show "f * g * (inverse f * inverse g) = 1" . ``` ``` 1093 qed ``` ``` 1094 next ``` ``` 1095 assume A: "f\$0 = 0 \<or> g\$0 = 0" ``` ``` 1096 hence "inverse (f * g) = 0" by simp ``` ``` 1097 also from A have "... = inverse f * inverse g" by auto ``` ``` 1098 finally show "inverse (f * g) = inverse f * inverse g" . ``` ``` 1099 qed ``` ``` 1100 ``` ``` 1101 ``` ``` 1102 lemma fps_inverse_gp: "inverse (Abs_fps(\<lambda>n. (1::'a::field))) = ``` ``` 1103 Abs_fps (\<lambda>n. if n= 0 then 1 else if n=1 then - 1 else 0)" ``` ``` 1104 apply (rule fps_inverse_unique) ``` ``` 1105 apply (simp_all add: fps_eq_iff fps_mult_nth setsum_zero_lemma) ``` ``` 1106 done ``` ``` 1107 ``` ``` 1108 lemma subdegree_inverse [simp]: "subdegree (inverse (f::'a::field fps)) = 0" ``` ``` 1109 proof (cases "f\$0 = 0") ``` ``` 1110 assume nz: "f\$0 \<noteq> 0" ``` ``` 1111 hence "subdegree (inverse f) + subdegree f = subdegree (inverse f * f)" ``` ``` 1112 by (subst subdegree_mult) auto ``` ``` 1113 also from nz have "subdegree f = 0" by (simp add: subdegree_eq_0_iff) ``` ``` 1114 also from nz have "inverse f * f = 1" by (rule inverse_mult_eq_1) ``` ``` 1115 finally show "subdegree (inverse f) = 0" by simp ``` ``` 1116 qed (simp_all add: fps_inverse_def) ``` ``` 1117 ``` ``` 1118 lemma fps_is_unit_iff [simp]: "(f :: 'a :: field fps) dvd 1 \<longleftrightarrow> f \$ 0 \<noteq> 0" ``` ``` 1119 proof ``` ``` 1120 assume "f dvd 1" ``` ``` 1121 then obtain g where "1 = f * g" by (elim dvdE) ``` ``` 1122 from this[symmetric] have "(f*g) \$ 0 = 1" by simp ``` ``` 1123 thus "f \$ 0 \<noteq> 0" by auto ``` ``` 1124 next ``` ``` 1125 assume A: "f \$ 0 \<noteq> 0" ``` ``` 1126 thus "f dvd 1" by (simp add: inverse_mult_eq_1[OF A, symmetric]) ``` ``` 1127 qed ``` ``` 1128 ``` ``` 1129 lemma subdegree_eq_0' [simp]: "(f :: 'a :: field fps) dvd 1 \<Longrightarrow> subdegree f = 0" ``` ``` 1130 by simp ``` ``` 1131 ``` ``` 1132 lemma fps_unit_dvd [simp]: "(f \$ 0 :: 'a :: field) \<noteq> 0 \<Longrightarrow> f dvd g" ``` ``` 1133 by (rule dvd_trans, subst fps_is_unit_iff) simp_all ``` ``` 1134 ``` ``` 1135 ``` ``` 1136 ``` ``` 1137 instantiation fps :: (field) ring_div ``` ``` 1138 begin ``` ``` 1139 ``` ``` 1140 definition fps_mod_def: ``` ``` 1141 "f mod g = (if g = 0 then f else ``` ``` 1142 let n = subdegree g; h = fps_shift n g ``` ``` 1143 in fps_cutoff n (f * inverse h) * h)" ``` ``` 1144 ``` ``` 1145 lemma fps_mod_eq_zero: ``` ``` 1146 assumes "g \<noteq> 0" and "subdegree f \<ge> subdegree g" ``` ``` 1147 shows "f mod g = 0" ``` ``` 1148 using assms by (cases "f = 0") (auto simp: fps_cutoff_zero_iff fps_mod_def Let_def) ``` ``` 1149 ``` ``` 1150 lemma fps_times_divide_eq: ``` ``` 1151 assumes "g \<noteq> 0" and "subdegree f \<ge> subdegree (g :: 'a fps)" ``` ``` 1152 shows "f div g * g = f" ``` ``` 1153 proof (cases "f = 0") ``` ``` 1154 assume nz: "f \<noteq> 0" ``` ``` 1155 def n \<equiv> "subdegree g" ``` ``` 1156 def h \<equiv> "fps_shift n g" ``` ``` 1157 from assms have [simp]: "h \$ 0 \<noteq> 0" unfolding h_def by (simp add: n_def) ``` ``` 1158 ``` ``` 1159 from assms nz have "f div g * g = fps_shift n (f * inverse h) * g" ``` ``` 1160 by (simp add: fps_divide_def Let_def h_def n_def) ``` ``` 1161 also have "... = fps_shift n (f * inverse h) * X^n * h" unfolding h_def n_def ``` ``` 1162 by (subst subdegree_decompose[of g]) simp ``` ``` 1163 also have "fps_shift n (f * inverse h) * X^n = f * inverse h" ``` ``` 1164 by (rule fps_shift_times_X_power) (simp_all add: nz assms n_def) ``` ``` 1165 also have "... * h = f * (inverse h * h)" by simp ``` ``` 1166 also have "inverse h * h = 1" by (rule inverse_mult_eq_1) simp ``` ``` 1167 finally show ?thesis by simp ``` ``` 1168 qed (simp_all add: fps_divide_def Let_def) ``` ``` 1169 ``` ``` 1170 lemma ``` ``` 1171 assumes "g\$0 \<noteq> 0" ``` ``` 1172 shows fps_divide_unit: "f div g = f * inverse g" and fps_mod_unit [simp]: "f mod g = 0" ``` ``` 1173 proof - ``` ``` 1174 from assms have [simp]: "subdegree g = 0" by (simp add: subdegree_eq_0_iff) ``` ``` 1175 from assms show "f div g = f * inverse g" ``` ``` 1176 by (auto simp: fps_divide_def Let_def subdegree_eq_0_iff) ``` ``` 1177 from assms show "f mod g = 0" by (intro fps_mod_eq_zero) auto ``` ``` 1178 qed ``` ``` 1179 ``` ``` 1180 context ``` ``` 1181 begin ``` ``` 1182 private lemma fps_divide_cancel_aux1: ``` ``` 1183 assumes "h\$0 \<noteq> (0 :: 'a :: field)" ``` ``` 1184 shows "(h * f) div (h * g) = f div g" ``` ``` 1185 proof (cases "g = 0") ``` ``` 1186 assume "g \<noteq> 0" ``` ``` 1187 from assms have "h \<noteq> 0" by auto ``` ``` 1188 note nz [simp] = \<open>g \<noteq> 0\<close> \<open>h \<noteq> 0\<close> ``` ``` 1189 from assms have [simp]: "subdegree h = 0" by (simp add: subdegree_eq_0_iff) ``` ``` 1190 ``` ``` 1191 have "(h * f) div (h * g) = ``` ``` 1192 fps_shift (subdegree g) (h * f * inverse (fps_shift (subdegree g) (h*g)))" ``` ``` 1193 by (simp add: fps_divide_def Let_def) ``` ``` 1194 also have "h * f * inverse (fps_shift (subdegree g) (h*g)) = ``` ``` 1195 (inverse h * h) * f * inverse (fps_shift (subdegree g) g)" ``` ``` 1196 by (subst fps_shift_mult) (simp_all add: algebra_simps fps_inverse_mult) ``` ``` 1197 also from assms have "inverse h * h = 1" by (rule inverse_mult_eq_1) ``` ``` 1198 finally show "(h * f) div (h * g) = f div g" by (simp_all add: fps_divide_def Let_def) ``` ``` 1199 qed (simp_all add: fps_divide_def) ``` ``` 1200 ``` ``` 1201 private lemma fps_divide_cancel_aux2: ``` ``` 1202 "(f * X^m) div (g * X^m) = f div (g :: 'a :: field fps)" ``` ``` 1203 proof (cases "g = 0") ``` ``` 1204 assume [simp]: "g \<noteq> 0" ``` ``` 1205 have "(f * X^m) div (g * X^m) = ``` ``` 1206 fps_shift (subdegree g + m) (f*inverse (fps_shift (subdegree g + m) (g*X^m))*X^m)" ``` ``` 1207 by (simp add: fps_divide_def Let_def algebra_simps) ``` ``` 1208 also have "... = f div g" ``` ``` 1209 by (simp add: fps_shift_times_X_power'' fps_divide_def Let_def) ``` ``` 1210 finally show ?thesis . ``` ``` 1211 qed (simp_all add: fps_divide_def) ``` ``` 1212 ``` ``` 1213 instance proof ``` ``` 1214 fix f g :: "'a fps" ``` ``` 1215 def n \<equiv> "subdegree g" ``` ``` 1216 def h \<equiv> "fps_shift n g" ``` ``` 1217 ``` ``` 1218 show "f div g * g + f mod g = f" ``` ``` 1219 proof (cases "g = 0 \<or> f = 0") ``` ``` 1220 assume "\<not>(g = 0 \<or> f = 0)" ``` ``` 1221 hence nz [simp]: "f \<noteq> 0" "g \<noteq> 0" by simp_all ``` ``` 1222 show ?thesis ``` ``` 1223 proof (rule disjE[OF le_less_linear]) ``` ``` 1224 assume "subdegree f \<ge> subdegree g" ``` ``` 1225 with nz show ?thesis by (simp add: fps_mod_eq_zero fps_times_divide_eq) ``` ``` 1226 next ``` ``` 1227 assume "subdegree f < subdegree g" ``` ``` 1228 have g_decomp: "g = h * X^n" unfolding h_def n_def by (rule subdegree_decompose) ``` ``` 1229 have "f div g * g + f mod g = ``` ``` 1230 fps_shift n (f * inverse h) * g + fps_cutoff n (f * inverse h) * h" ``` ``` 1231 by (simp add: fps_mod_def fps_divide_def Let_def n_def h_def) ``` ``` 1232 also have "... = h * (fps_shift n (f * inverse h) * X^n + fps_cutoff n (f * inverse h))" ``` ``` 1233 by (subst g_decomp) (simp add: algebra_simps) ``` ``` 1234 also have "... = f * (inverse h * h)" ``` ``` 1235 by (subst fps_shift_cutoff) simp ``` ``` 1236 also have "inverse h * h = 1" by (rule inverse_mult_eq_1) (simp add: h_def n_def) ``` ``` 1237 finally show ?thesis by simp ``` ``` 1238 qed ``` ``` 1239 qed (auto simp: fps_mod_def fps_divide_def Let_def) ``` ``` 1240 next ``` ``` 1241 ``` ``` 1242 fix f g h :: "'a fps" ``` ``` 1243 assume "h \<noteq> 0" ``` ``` 1244 show "(h * f) div (h * g) = f div g" ``` ``` 1245 proof - ``` ``` 1246 def m \<equiv> "subdegree h" ``` ``` 1247 def h' \<equiv> "fps_shift m h" ``` ``` 1248 have h_decomp: "h = h' * X ^ m" unfolding h'_def m_def by (rule subdegree_decompose) ``` ``` 1249 from \<open>h \<noteq> 0\<close> have [simp]: "h'\$0 \<noteq> 0" by (simp add: h'_def m_def) ``` ``` 1250 have "(h * f) div (h * g) = (h' * f * X^m) div (h' * g * X^m)" ``` ``` 1251 by (simp add: h_decomp algebra_simps) ``` ``` 1252 also have "... = f div g" by (simp add: fps_divide_cancel_aux1 fps_divide_cancel_aux2) ``` ``` 1253 finally show ?thesis . ``` ``` 1254 qed ``` ``` 1255 ``` ``` 1256 next ``` ``` 1257 fix f g h :: "'a fps" ``` ``` 1258 assume [simp]: "h \<noteq> 0" ``` ``` 1259 def n \<equiv> "subdegree h" ``` ``` 1260 def h' \<equiv> "fps_shift n h" ``` ``` 1261 note dfs = n_def h'_def ``` ``` 1262 have "(f + g * h) div h = fps_shift n (f * inverse h') + fps_shift n (g * (h * inverse h'))" ``` ``` 1263 by (simp add: fps_divide_def Let_def dfs[symmetric] algebra_simps fps_shift_add) ``` ``` 1264 also have "h * inverse h' = (inverse h' * h') * X^n" ``` ``` 1265 by (subst subdegree_decompose) (simp_all add: dfs) ``` ``` 1266 also have "... = X^n" by (subst inverse_mult_eq_1) (simp_all add: dfs) ``` ``` 1267 also have "fps_shift n (g * X^n) = g" by simp ``` ``` 1268 also have "fps_shift n (f * inverse h') = f div h" ``` ``` 1269 by (simp add: fps_divide_def Let_def dfs) ``` ``` 1270 finally show "(f + g * h) div h = g + f div h" by simp ``` ``` 1271 qed (auto simp: fps_divide_def fps_mod_def Let_def) ``` ``` 1272 ``` ``` 1273 end ``` ``` 1274 end ``` ``` 1275 ``` ``` 1276 lemma subdegree_mod: ``` ``` 1277 assumes "f \<noteq> 0" "subdegree f < subdegree g" ``` ``` 1278 shows "subdegree (f mod g) = subdegree f" ``` ``` 1279 proof (cases "f div g * g = 0") ``` ``` 1280 assume "f div g * g \<noteq> 0" ``` ``` 1281 hence [simp]: "f div g \<noteq> 0" "g \<noteq> 0" by auto ``` ``` 1282 from mod_div_equality[of f g] have "f mod g = f - f div g * g" by (simp add: algebra_simps) ``` ``` 1283 also from assms have "subdegree ... = subdegree f" ``` ``` 1284 by (intro subdegree_diff_eq1) simp_all ``` ``` 1285 finally show ?thesis . ``` ``` 1286 next ``` ``` 1287 assume zero: "f div g * g = 0" ``` ``` 1288 from mod_div_equality[of f g] have "f mod g = f - f div g * g" by (simp add: algebra_simps) ``` ``` 1289 also note zero ``` ``` 1290 finally show ?thesis by simp ``` ``` 1291 qed ``` ``` 1292 ``` ``` 1293 lemma fps_divide_nth_0 [simp]: "g \$ 0 \<noteq> 0 \<Longrightarrow> (f div g) \$ 0 = f \$ 0 / (g \$ 0 :: _ :: field)" ``` ``` 1294 by (simp add: fps_divide_unit divide_inverse) ``` ``` 1295 ``` ``` 1296 ``` ``` 1297 lemma dvd_imp_subdegree_le: ``` ``` 1298 "(f :: 'a :: idom fps) dvd g \<Longrightarrow> g \<noteq> 0 \<Longrightarrow> subdegree f \<le> subdegree g" ``` ``` 1299 by (auto elim: dvdE) ``` ``` 1300 ``` ``` 1301 lemma fps_dvd_iff: ``` ``` 1302 assumes "(f :: 'a :: field fps) \<noteq> 0" "g \<noteq> 0" ``` ``` 1303 shows "f dvd g \<longleftrightarrow> subdegree f \<le> subdegree g" ``` ``` 1304 proof ``` ``` 1305 assume "subdegree f \<le> subdegree g" ``` ``` 1306 with assms have "g mod f = 0" ``` ``` 1307 by (simp add: fps_mod_def Let_def fps_cutoff_zero_iff) ``` ``` 1308 thus "f dvd g" by (simp add: dvd_eq_mod_eq_0) ``` ``` 1309 qed (simp add: assms dvd_imp_subdegree_le) ``` ``` 1310 ``` ``` 1311 lemma fps_const_inverse: "inverse (fps_const (a::'a::field)) = fps_const (inverse a)" ``` ``` 1312 by (cases "a \<noteq> 0", rule fps_inverse_unique) (auto simp: fps_eq_iff) ``` ``` 1313 ``` ``` 1314 lemma fps_const_divide: "fps_const (x :: _ :: field) / fps_const y = fps_const (x / y)" ``` ``` 1315 by (cases "y = 0") (simp_all add: fps_divide_unit fps_const_inverse divide_inverse) ``` ``` 1316 ``` ``` 1317 lemma inverse_fps_numeral: ``` ``` 1318 "inverse (numeral n :: ('a :: field_char_0) fps) = fps_const (inverse (numeral n))" ``` ``` 1319 by (intro fps_inverse_unique fps_ext) (simp_all add: fps_numeral_nth) ``` ``` 1320 ``` ``` 1321 ``` ``` 1322 ``` ``` 1323 ``` ``` 1324 instantiation fps :: (field) normalization_semidom ``` ``` 1325 begin ``` ``` 1326 ``` ``` 1327 definition fps_unit_factor_def [simp]: ``` ``` 1328 "unit_factor f = fps_shift (subdegree f) f" ``` ``` 1329 ``` ``` 1330 definition fps_normalize_def [simp]: ``` ``` 1331 "normalize f = (if f = 0 then 0 else X ^ subdegree f)" ``` ``` 1332 ``` ``` 1333 instance proof ``` ``` 1334 fix f :: "'a fps" ``` ``` 1335 show "unit_factor f * normalize f = f" ``` ``` 1336 by (simp add: fps_shift_times_X_power) ``` ``` 1337 next ``` ``` 1338 fix f g :: "'a fps" ``` ``` 1339 show "unit_factor (f * g) = unit_factor f * unit_factor g" ``` ``` 1340 proof (cases "f = 0 \<or> g = 0") ``` ``` 1341 assume "\<not>(f = 0 \<or> g = 0)" ``` ``` 1342 thus "unit_factor (f * g) = unit_factor f * unit_factor g" ``` ``` 1343 unfolding fps_unit_factor_def ``` ``` 1344 by (auto simp: fps_shift_fps_shift fps_shift_mult fps_shift_mult_right) ``` ``` 1345 qed auto ``` ``` 1346 qed auto ``` ``` 1347 ``` ``` 1348 end ``` ``` 1349 ``` ``` 1350 instance fps :: (field) algebraic_semidom .. ``` ``` 1351 ``` ``` 1352 ``` ``` 1353 subsection \<open>Formal power series form a Euclidean ring\<close> ``` ``` 1354 ``` ``` 1355 instantiation fps :: (field) euclidean_ring ``` ``` 1356 begin ``` ``` 1357 ``` ``` 1358 definition fps_euclidean_size_def: ``` ``` 1359 "euclidean_size f = (if f = 0 then 0 else Suc (subdegree f))" ``` ``` 1360 ``` ``` 1361 instance proof ``` ``` 1362 fix f g :: "'a fps" assume [simp]: "g \<noteq> 0" ``` ``` 1363 show "euclidean_size f \<le> euclidean_size (f * g)" ``` ``` 1364 by (cases "f = 0") (auto simp: fps_euclidean_size_def) ``` ``` 1365 show "euclidean_size (f mod g) < euclidean_size g" ``` ``` 1366 apply (cases "f = 0", simp add: fps_euclidean_size_def) ``` ``` 1367 apply (rule disjE[OF le_less_linear[of "subdegree g" "subdegree f"]]) ``` ``` 1368 apply (simp_all add: fps_mod_eq_zero fps_euclidean_size_def subdegree_mod) ``` ``` 1369 done ``` ``` 1370 qed ``` ``` 1371 ``` ``` 1372 end ``` ``` 1373 ``` ``` 1374 instantiation fps :: (field) euclidean_ring_gcd ``` ``` 1375 begin ``` ``` 1376 definition fps_gcd_def: "(gcd :: 'a fps \<Rightarrow> _) = gcd_eucl" ``` ``` 1377 definition fps_lcm_def: "(lcm :: 'a fps \<Rightarrow> _) = lcm_eucl" ``` ``` 1378 definition fps_Gcd_def: "(Gcd :: 'a fps set \<Rightarrow> _) = Gcd_eucl" ``` ``` 1379 definition fps_Lcm_def: "(Lcm :: 'a fps set \<Rightarrow> _) = Lcm_eucl" ``` ``` 1380 instance by intro_classes (simp_all add: fps_gcd_def fps_lcm_def fps_Gcd_def fps_Lcm_def) ``` ``` 1381 end ``` ``` 1382 ``` ``` 1383 lemma fps_gcd: ``` ``` 1384 assumes [simp]: "f \<noteq> 0" "g \<noteq> 0" ``` ``` 1385 shows "gcd f g = X ^ min (subdegree f) (subdegree g)" ``` ``` 1386 proof - ``` ``` 1387 let ?m = "min (subdegree f) (subdegree g)" ``` ``` 1388 show "gcd f g = X ^ ?m" ``` ``` 1389 proof (rule sym, rule gcdI) ``` ``` 1390 fix d assume "d dvd f" "d dvd g" ``` ``` 1391 thus "d dvd X ^ ?m" by (cases "d = 0") (auto simp: fps_dvd_iff) ``` ``` 1392 qed (simp_all add: fps_dvd_iff) ``` ``` 1393 qed ``` ``` 1394 ``` ``` 1395 lemma fps_gcd_altdef: "gcd (f :: 'a :: field fps) g = ``` ``` 1396 (if f = 0 \<and> g = 0 then 0 else ``` ``` 1397 if f = 0 then X ^ subdegree g else ``` ``` 1398 if g = 0 then X ^ subdegree f else ``` ``` 1399 X ^ min (subdegree f) (subdegree g))" ``` ``` 1400 by (simp add: fps_gcd) ``` ``` 1401 ``` ``` 1402 lemma fps_lcm: ``` ``` 1403 assumes [simp]: "f \<noteq> 0" "g \<noteq> 0" ``` ``` 1404 shows "lcm f g = X ^ max (subdegree f) (subdegree g)" ``` ``` 1405 proof - ``` ``` 1406 let ?m = "max (subdegree f) (subdegree g)" ``` ``` 1407 show "lcm f g = X ^ ?m" ``` ``` 1408 proof (rule sym, rule lcmI) ``` ``` 1409 fix d assume "f dvd d" "g dvd d" ``` ``` 1410 thus "X ^ ?m dvd d" by (cases "d = 0") (auto simp: fps_dvd_iff) ``` ``` 1411 qed (simp_all add: fps_dvd_iff) ``` ``` 1412 qed ``` ``` 1413 ``` ``` 1414 lemma fps_lcm_altdef: "lcm (f :: 'a :: field fps) g = ``` ``` 1415 (if f = 0 \<or> g = 0 then 0 else X ^ max (subdegree f) (subdegree g))" ``` ``` 1416 by (simp add: fps_lcm) ``` ``` 1417 ``` ``` 1418 lemma fps_Gcd: ``` ``` 1419 assumes "A - {0} \<noteq> {}" ``` ``` 1420 shows "Gcd A = X ^ (INF f:A-{0}. subdegree f)" ``` ``` 1421 proof (rule sym, rule GcdI) ``` ``` 1422 fix f assume "f \<in> A" ``` ``` 1423 thus "X ^ (INF f:A - {0}. subdegree f) dvd f" ``` ``` 1424 by (cases "f = 0") (auto simp: fps_dvd_iff intro!: cINF_lower) ``` ``` 1425 next ``` ``` 1426 fix d assume d: "\<And>f. f \<in> A \<Longrightarrow> d dvd f" ``` ``` 1427 from assms obtain f where "f \<in> A - {0}" by auto ``` ``` 1428 with d[of f] have [simp]: "d \<noteq> 0" by auto ``` ``` 1429 from d assms have "subdegree d \<le> (INF f:A-{0}. subdegree f)" ``` ``` 1430 by (intro cINF_greatest) (auto simp: fps_dvd_iff[symmetric]) ``` ``` 1431 with d assms show "d dvd X ^ (INF f:A-{0}. subdegree f)" by (simp add: fps_dvd_iff) ``` ``` 1432 qed simp_all ``` ``` 1433 ``` ``` 1434 lemma fps_Gcd_altdef: "Gcd (A :: 'a :: field fps set) = ``` ``` 1435 (if A \<subseteq> {0} then 0 else X ^ (INF f:A-{0}. subdegree f))" ``` ``` 1436 using fps_Gcd by auto ``` ``` 1437 ``` ``` 1438 lemma fps_Lcm: ``` ``` 1439 assumes "A \<noteq> {}" "0 \<notin> A" "bdd_above (subdegree`A)" ``` ``` 1440 shows "Lcm A = X ^ (SUP f:A. subdegree f)" ``` ``` 1441 proof (rule sym, rule LcmI) ``` ``` 1442 fix f assume "f \<in> A" ``` ``` 1443 moreover from assms(3) have "bdd_above (subdegree ` A)" by auto ``` ``` 1444 ultimately show "f dvd X ^ (SUP f:A. subdegree f)" using assms(2) ``` ``` 1445 by (cases "f = 0") (auto simp: fps_dvd_iff intro!: cSUP_upper) ``` ``` 1446 next ``` ``` 1447 fix d assume d: "\<And>f. f \<in> A \<Longrightarrow> f dvd d" ``` ``` 1448 from assms obtain f where f: "f \<in> A" "f \<noteq> 0" by auto ``` ``` 1449 show "X ^ (SUP f:A. subdegree f) dvd d" ``` ``` 1450 proof (cases "d = 0") ``` ``` 1451 assume "d \<noteq> 0" ``` ``` 1452 moreover from d have "\<And>f. f \<in> A \<Longrightarrow> f \<noteq> 0 \<Longrightarrow> f dvd d" by blast ``` ``` 1453 ultimately have "subdegree d \<ge> (SUP f:A. subdegree f)" using assms ``` ``` 1454 by (intro cSUP_least) (auto simp: fps_dvd_iff) ``` ``` 1455 with \<open>d \<noteq> 0\<close> show ?thesis by (simp add: fps_dvd_iff) ``` ``` 1456 qed simp_all ``` ``` 1457 qed simp_all ``` ``` 1458 ``` ``` 1459 lemma fps_Lcm_altdef: ``` ``` 1460 "Lcm (A :: 'a :: field fps set) = ``` ``` 1461 (if 0 \<in> A \<or> \<not>bdd_above (subdegree`A) then 0 else ``` ``` 1462 if A = {} then 1 else X ^ (SUP f:A. subdegree f))" ``` ``` 1463 proof (cases "bdd_above (subdegree`A)") ``` ``` 1464 assume unbounded: "\<not>bdd_above (subdegree`A)" ``` ``` 1465 have "Lcm A = 0" ``` ``` 1466 proof (rule ccontr) ``` ``` 1467 assume "Lcm A \<noteq> 0" ``` ``` 1468 from unbounded obtain f where f: "f \<in> A" "subdegree (Lcm A) < subdegree f" ``` ``` 1469 unfolding bdd_above_def by (auto simp: not_le) ``` ``` 1470 moreover from this and \<open>Lcm A \<noteq> 0\<close> have "subdegree f \<le> subdegree (Lcm A)" ``` ``` 1471 by (intro dvd_imp_subdegree_le) simp_all ``` ``` 1472 ultimately show False by simp ``` ``` 1473 qed ``` ``` 1474 with unbounded show ?thesis by simp ``` ``` 1475 qed (simp_all add: fps_Lcm) ``` ``` 1476 ``` ``` 1477 ``` ``` 1478 subsection \<open>Formal Derivatives, and the MacLaurin theorem around 0\<close> ``` ``` 1479 ``` ``` 1480 definition "fps_deriv f = Abs_fps (\<lambda>n. of_nat (n + 1) * f \$ (n + 1))" ``` ``` 1481 ``` ``` 1482 lemma fps_deriv_nth[simp]: "fps_deriv f \$ n = of_nat (n +1) * f \$ (n + 1)" ``` ``` 1483 by (simp add: fps_deriv_def) ``` ``` 1484 ``` ``` 1485 lemma fps_deriv_linear[simp]: ``` ``` 1486 "fps_deriv (fps_const (a::'a::comm_semiring_1) * f + fps_const b * g) = ``` ``` 1487 fps_const a * fps_deriv f + fps_const b * fps_deriv g" ``` ``` 1488 unfolding fps_eq_iff fps_add_nth fps_const_mult_left fps_deriv_nth by (simp add: field_simps) ``` ``` 1489 ``` ``` 1490 lemma fps_deriv_mult[simp]: ``` ``` 1491 fixes f :: "'a::comm_ring_1 fps" ``` ``` 1492 shows "fps_deriv (f * g) = f * fps_deriv g + fps_deriv f * g" ``` ``` 1493 proof - ``` ``` 1494 let ?D = "fps_deriv" ``` ``` 1495 have "(f * ?D g + ?D f * g) \$ n = ?D (f*g) \$ n" for n ``` ``` 1496 proof - ``` ``` 1497 let ?Zn = "{0 ..n}" ``` ``` 1498 let ?Zn1 = "{0 .. n + 1}" ``` ``` 1499 let ?g = "\<lambda>i. of_nat (i+1) * g \$ (i+1) * f \$ (n - i) + ``` ``` 1500 of_nat (i+1)* f \$ (i+1) * g \$ (n - i)" ``` ``` 1501 let ?h = "\<lambda>i. of_nat i * g \$ i * f \$ ((n+1) - i) + ``` ``` 1502 of_nat i* f \$ i * g \$ ((n + 1) - i)" ``` ``` 1503 have s0: "setsum (\<lambda>i. of_nat i * f \$ i * g \$ (n + 1 - i)) ?Zn1 = ``` ``` 1504 setsum (\<lambda>i. of_nat (n + 1 - i) * f \$ (n + 1 - i) * g \$ i) ?Zn1" ``` ``` 1505 by (rule setsum.reindex_bij_witness[where i="op - (n + 1)" and j="op - (n + 1)"]) auto ``` ``` 1506 have s1: "setsum (\<lambda>i. f \$ i * g \$ (n + 1 - i)) ?Zn1 = ``` ``` 1507 setsum (\<lambda>i. f \$ (n + 1 - i) * g \$ i) ?Zn1" ``` ``` 1508 by (rule setsum.reindex_bij_witness[where i="op - (n + 1)" and j="op - (n + 1)"]) auto ``` ``` 1509 have "(f * ?D g + ?D f * g)\$n = (?D g * f + ?D f * g)\$n" ``` ``` 1510 by (simp only: mult.commute) ``` ``` 1511 also have "\<dots> = (\<Sum>i = 0..n. ?g i)" ``` ``` 1512 by (simp add: fps_mult_nth setsum.distrib[symmetric]) ``` ``` 1513 also have "\<dots> = setsum ?h {0..n+1}" ``` ``` 1514 by (rule setsum.reindex_bij_witness_not_neutral ``` ``` 1515 [where S'="{}" and T'="{0}" and j="Suc" and i="\<lambda>i. i - 1"]) auto ``` ``` 1516 also have "\<dots> = (fps_deriv (f * g)) \$ n" ``` ``` 1517 apply (simp only: fps_deriv_nth fps_mult_nth setsum.distrib) ``` ``` 1518 unfolding s0 s1 ``` ``` 1519 unfolding setsum.distrib[symmetric] setsum_right_distrib ``` ``` 1520 apply (rule setsum.cong) ``` ``` 1521 apply (auto simp add: of_nat_diff field_simps) ``` ``` 1522 done ``` ``` 1523 finally show ?thesis . ``` ``` 1524 qed ``` ``` 1525 then show ?thesis ``` ``` 1526 unfolding fps_eq_iff by auto ``` ``` 1527 qed ``` ``` 1528 ``` ``` 1529 lemma fps_deriv_X[simp]: "fps_deriv X = 1" ``` ``` 1530 by (simp add: fps_deriv_def X_def fps_eq_iff) ``` ``` 1531 ``` ``` 1532 lemma fps_deriv_neg[simp]: ``` ``` 1533 "fps_deriv (- (f:: 'a::comm_ring_1 fps)) = - (fps_deriv f)" ``` ``` 1534 by (simp add: fps_eq_iff fps_deriv_def) ``` ``` 1535 ``` ``` 1536 lemma fps_deriv_add[simp]: ``` ``` 1537 "fps_deriv ((f:: 'a::comm_ring_1 fps) + g) = fps_deriv f + fps_deriv g" ``` ``` 1538 using fps_deriv_linear[of 1 f 1 g] by simp ``` ``` 1539 ``` ``` 1540 lemma fps_deriv_sub[simp]: ``` ``` 1541 "fps_deriv ((f:: 'a::comm_ring_1 fps) - g) = fps_deriv f - fps_deriv g" ``` ``` 1542 using fps_deriv_add [of f "- g"] by simp ``` ``` 1543 ``` ``` 1544 lemma fps_deriv_const[simp]: "fps_deriv (fps_const c) = 0" ``` ``` 1545 by (simp add: fps_ext fps_deriv_def fps_const_def) ``` ``` 1546 ``` ``` 1547 lemma fps_deriv_mult_const_left[simp]: ``` ``` 1548 "fps_deriv (fps_const (c::'a::comm_ring_1) * f) = fps_const c * fps_deriv f" ``` ``` 1549 by simp ``` ``` 1550 ``` ``` 1551 lemma fps_deriv_0[simp]: "fps_deriv 0 = 0" ``` ``` 1552 by (simp add: fps_deriv_def fps_eq_iff) ``` ``` 1553 ``` ``` 1554 lemma fps_deriv_1[simp]: "fps_deriv 1 = 0" ``` ``` 1555 by (simp add: fps_deriv_def fps_eq_iff ) ``` ``` 1556 ``` ``` 1557 lemma fps_deriv_mult_const_right[simp]: ``` ``` 1558 "fps_deriv (f * fps_const (c::'a::comm_ring_1)) = fps_deriv f * fps_const c" ``` ``` 1559 by simp ``` ``` 1560 ``` ``` 1561 lemma fps_deriv_setsum: ``` ``` 1562 "fps_deriv (setsum f S) = setsum (\<lambda>i. fps_deriv (f i :: 'a::comm_ring_1 fps)) S" ``` ``` 1563 proof (cases "finite S") ``` ``` 1564 case False ``` ``` 1565 then show ?thesis by simp ``` ``` 1566 next ``` ``` 1567 case True ``` ``` 1568 show ?thesis by (induct rule: finite_induct [OF True]) simp_all ``` ``` 1569 qed ``` ``` 1570 ``` ``` 1571 lemma fps_deriv_eq_0_iff [simp]: ``` ``` 1572 "fps_deriv f = 0 \<longleftrightarrow> f = fps_const (f\$0 :: 'a::{idom,semiring_char_0})" ``` ``` 1573 (is "?lhs \<longleftrightarrow> ?rhs") ``` ``` 1574 proof ``` ``` 1575 show ?lhs if ?rhs ``` ``` 1576 proof - ``` ``` 1577 from that have "fps_deriv f = fps_deriv (fps_const (f\$0))" ``` ``` 1578 by simp ``` ``` 1579 then show ?thesis ``` ``` 1580 by simp ``` ``` 1581 qed ``` ``` 1582 show ?rhs if ?lhs ``` ``` 1583 proof - ``` ``` 1584 from that have "\<forall>n. (fps_deriv f)\$n = 0" ``` ``` 1585 by simp ``` ``` 1586 then have "\<forall>n. f\$(n+1) = 0" ``` ``` 1587 by (simp del: of_nat_Suc of_nat_add One_nat_def) ``` ``` 1588 then show ?thesis ``` ``` 1589 apply (clarsimp simp add: fps_eq_iff fps_const_def) ``` ``` 1590 apply (erule_tac x="n - 1" in allE) ``` ``` 1591 apply simp ``` ``` 1592 done ``` ``` 1593 qed ``` ``` 1594 qed ``` ``` 1595 ``` ``` 1596 lemma fps_deriv_eq_iff: ``` ``` 1597 fixes f :: "'a::{idom,semiring_char_0} fps" ``` ``` 1598 shows "fps_deriv f = fps_deriv g \<longleftrightarrow> (f = fps_const(f\$0 - g\$0) + g)" ``` ``` 1599 proof - ``` ``` 1600 have "fps_deriv f = fps_deriv g \<longleftrightarrow> fps_deriv (f - g) = 0" ``` ``` 1601 by simp ``` ``` 1602 also have "\<dots> \<longleftrightarrow> f - g = fps_const ((f - g) \$ 0)" ``` ``` 1603 unfolding fps_deriv_eq_0_iff .. ``` ``` 1604 finally show ?thesis ``` ``` 1605 by (simp add: field_simps) ``` ``` 1606 qed ``` ``` 1607 ``` ``` 1608 lemma fps_deriv_eq_iff_ex: ``` ``` 1609 "(fps_deriv f = fps_deriv g) \<longleftrightarrow> (\<exists>c::'a::{idom,semiring_char_0}. f = fps_const c + g)" ``` ``` 1610 by (auto simp: fps_deriv_eq_iff) ``` ``` 1611 ``` ``` 1612 ``` ``` 1613 fun fps_nth_deriv :: "nat \<Rightarrow> 'a::semiring_1 fps \<Rightarrow> 'a fps" ``` ``` 1614 where ``` ``` 1615 "fps_nth_deriv 0 f = f" ``` ``` 1616 | "fps_nth_deriv (Suc n) f = fps_nth_deriv n (fps_deriv f)" ``` ``` 1617 ``` ``` 1618 lemma fps_nth_deriv_commute: "fps_nth_deriv (Suc n) f = fps_deriv (fps_nth_deriv n f)" ``` ``` 1619 by (induct n arbitrary: f) auto ``` ``` 1620 ``` ``` 1621 lemma fps_nth_deriv_linear[simp]: ``` ``` 1622 "fps_nth_deriv n (fps_const (a::'a::comm_semiring_1) * f + fps_const b * g) = ``` ``` 1623 fps_const a * fps_nth_deriv n f + fps_const b * fps_nth_deriv n g" ``` ``` 1624 by (induct n arbitrary: f g) (auto simp add: fps_nth_deriv_commute) ``` ``` 1625 ``` ``` 1626 lemma fps_nth_deriv_neg[simp]: ``` ``` 1627 "fps_nth_deriv n (- (f :: 'a::comm_ring_1 fps)) = - (fps_nth_deriv n f)" ``` ``` 1628 by (induct n arbitrary: f) simp_all ``` ``` 1629 ``` ``` 1630 lemma fps_nth_deriv_add[simp]: ``` ``` 1631 "fps_nth_deriv n ((f :: 'a::comm_ring_1 fps) + g) = fps_nth_deriv n f + fps_nth_deriv n g" ``` ``` 1632 using fps_nth_deriv_linear[of n 1 f 1 g] by simp ``` ``` 1633 ``` ``` 1634 lemma fps_nth_deriv_sub[simp]: ``` ``` 1635 "fps_nth_deriv n ((f :: 'a::comm_ring_1 fps) - g) = fps_nth_deriv n f - fps_nth_deriv n g" ``` ``` 1636 using fps_nth_deriv_add [of n f "- g"] by simp ``` ``` 1637 ``` ``` 1638 lemma fps_nth_deriv_0[simp]: "fps_nth_deriv n 0 = 0" ``` ``` 1639 by (induct n) simp_all ``` ``` 1640 ``` ``` 1641 lemma fps_nth_deriv_1[simp]: "fps_nth_deriv n 1 = (if n = 0 then 1 else 0)" ``` ``` 1642 by (induct n) simp_all ``` ``` 1643 ``` ``` 1644 lemma fps_nth_deriv_const[simp]: ``` ``` 1645 "fps_nth_deriv n (fps_const c) = (if n = 0 then fps_const c else 0)" ``` ``` 1646 by (cases n) simp_all ``` ``` 1647 ``` ``` 1648 lemma fps_nth_deriv_mult_const_left[simp]: ``` ``` 1649 "fps_nth_deriv n (fps_const (c::'a::comm_ring_1) * f) = fps_const c * fps_nth_deriv n f" ``` ``` 1650 using fps_nth_deriv_linear[of n "c" f 0 0 ] by simp ``` ``` 1651 ``` ``` 1652 lemma fps_nth_deriv_mult_const_right[simp]: ``` ``` 1653 "fps_nth_deriv n (f * fps_const (c::'a::comm_ring_1)) = fps_nth_deriv n f * fps_const c" ``` ``` 1654 using fps_nth_deriv_linear[of n "c" f 0 0] by (simp add: mult.commute) ``` ``` 1655 ``` ``` 1656 lemma fps_nth_deriv_setsum: ``` ``` 1657 "fps_nth_deriv n (setsum f S) = setsum (\<lambda>i. fps_nth_deriv n (f i :: 'a::comm_ring_1 fps)) S" ``` ``` 1658 proof (cases "finite S") ``` ``` 1659 case True ``` ``` 1660 show ?thesis by (induct rule: finite_induct [OF True]) simp_all ``` ``` 1661 next ``` ``` 1662 case False ``` ``` 1663 then show ?thesis by simp ``` ``` 1664 qed ``` ``` 1665 ``` ``` 1666 lemma fps_deriv_maclauren_0: ``` ``` 1667 "(fps_nth_deriv k (f :: 'a::comm_semiring_1 fps)) \$ 0 = of_nat (fact k) * f \$ k" ``` ``` 1668 by (induct k arbitrary: f) (auto simp add: field_simps of_nat_mult) ``` ``` 1669 ``` ``` 1670 ``` ``` 1671 subsection \<open>Powers\<close> ``` ``` 1672 ``` ``` 1673 lemma fps_power_zeroth_eq_one: "a\$0 =1 \<Longrightarrow> a^n \$ 0 = (1::'a::semiring_1)" ``` ``` 1674 by (induct n) (auto simp add: expand_fps_eq fps_mult_nth) ``` ``` 1675 ``` ``` 1676 lemma fps_power_first_eq: "(a :: 'a::comm_ring_1 fps) \$ 0 =1 \<Longrightarrow> a^n \$ 1 = of_nat n * a\$1" ``` ``` 1677 proof (induct n) ``` ``` 1678 case 0 ``` ``` 1679 then show ?case by simp ``` ``` 1680 next ``` ``` 1681 case (Suc n) ``` ``` 1682 show ?case unfolding power_Suc fps_mult_nth ``` ``` 1683 using Suc.hyps[OF \<open>a\$0 = 1\<close>] \<open>a\$0 = 1\<close> fps_power_zeroth_eq_one[OF \<open>a\$0=1\<close>] ``` ``` 1684 by (simp add: field_simps) ``` ``` 1685 qed ``` ``` 1686 ``` ``` 1687 lemma startsby_one_power:"a \$ 0 = (1::'a::comm_ring_1) \<Longrightarrow> a^n \$ 0 = 1" ``` ``` 1688 by (induct n) (auto simp add: fps_mult_nth) ``` ``` 1689 ``` ``` 1690 lemma startsby_zero_power:"a \$0 = (0::'a::comm_ring_1) \<Longrightarrow> n > 0 \<Longrightarrow> a^n \$0 = 0" ``` ``` 1691 by (induct n) (auto simp add: fps_mult_nth) ``` ``` 1692 ``` ``` 1693 lemma startsby_power:"a \$0 = (v::'a::comm_ring_1) \<Longrightarrow> a^n \$0 = v^n" ``` ``` 1694 by (induct n) (auto simp add: fps_mult_nth) ``` ``` 1695 ``` ``` 1696 lemma startsby_zero_power_iff[simp]: "a^n \$0 = (0::'a::idom) \<longleftrightarrow> n \<noteq> 0 \<and> a\$0 = 0" ``` ``` 1697 apply (rule iffI) ``` ``` 1698 apply (induct n) ``` ``` 1699 apply (auto simp add: fps_mult_nth) ``` ``` 1700 apply (rule startsby_zero_power, simp_all) ``` ``` 1701 done ``` ``` 1702 ``` ``` 1703 lemma startsby_zero_power_prefix: ``` ``` 1704 assumes a0: "a \$ 0 = (0::'a::idom)" ``` ``` 1705 shows "\<forall>n < k. a ^ k \$ n = 0" ``` ``` 1706 using a0 ``` ``` 1707 proof (induct k rule: nat_less_induct) ``` ``` 1708 fix k ``` ``` 1709 assume H: "\<forall>m<k. a \$0 = 0 \<longrightarrow> (\<forall>n<m. a ^ m \$ n = 0)" and a0: "a \$ 0 = 0" ``` ``` 1710 show "\<forall>m<k. a ^ k \$ m = 0" ``` ``` 1711 proof (cases k) ``` ``` 1712 case 0 ``` ``` 1713 then show ?thesis by simp ``` ``` 1714 next ``` ``` 1715 case (Suc l) ``` ``` 1716 have "a^k \$ m = 0" if mk: "m < k" for m ``` ``` 1717 proof (cases "m = 0") ``` ``` 1718 case True ``` ``` 1719 then show ?thesis ``` ``` 1720 using startsby_zero_power[of a k] Suc a0 by simp ``` ``` 1721 next ``` ``` 1722 case False ``` ``` 1723 have "a ^k \$ m = (a^l * a) \$m" ``` ``` 1724 by (simp add: Suc mult.commute) ``` ``` 1725 also have "\<dots> = (\<Sum>i = 0..m. a ^ l \$ i * a \$ (m - i))" ``` ``` 1726 by (simp add: fps_mult_nth) ``` ``` 1727 also have "\<dots> = 0" ``` ``` 1728 apply (rule setsum.neutral) ``` ``` 1729 apply auto ``` ``` 1730 apply (case_tac "x = m") ``` ``` 1731 using a0 apply simp ``` ``` 1732 apply (rule H[rule_format]) ``` ``` 1733 using a0 Suc mk apply auto ``` ``` 1734 done ``` ``` 1735 finally show ?thesis . ``` ``` 1736 qed ``` ``` 1737 then show ?thesis by blast ``` ``` 1738 qed ``` ``` 1739 qed ``` ``` 1740 ``` ``` 1741 lemma startsby_zero_setsum_depends: ``` ``` 1742 assumes a0: "a \$0 = (0::'a::idom)" ``` ``` 1743 and kn: "n \<ge> k" ``` ``` 1744 shows "setsum (\<lambda>i. (a ^ i)\$k) {0 .. n} = setsum (\<lambda>i. (a ^ i)\$k) {0 .. k}" ``` ``` 1745 apply (rule setsum.mono_neutral_right) ``` ``` 1746 using kn ``` ``` 1747 apply auto ``` ``` 1748 apply (rule startsby_zero_power_prefix[rule_format, OF a0]) ``` ``` 1749 apply arith ``` ``` 1750 done ``` ``` 1751 ``` ``` 1752 lemma startsby_zero_power_nth_same: ``` ``` 1753 assumes a0: "a\$0 = (0::'a::idom)" ``` ``` 1754 shows "a^n \$ n = (a\$1) ^ n" ``` ``` 1755 proof (induct n) ``` ``` 1756 case 0 ``` ``` 1757 then show ?case by simp ``` ``` 1758 next ``` ``` 1759 case (Suc n) ``` ``` 1760 have "a ^ Suc n \$ (Suc n) = (a^n * a)\$(Suc n)" ``` ``` 1761 by (simp add: field_simps) ``` ``` 1762 also have "\<dots> = setsum (\<lambda>i. a^n\$i * a \$ (Suc n - i)) {0.. Suc n}" ``` ``` 1763 by (simp add: fps_mult_nth) ``` ``` 1764 also have "\<dots> = setsum (\<lambda>i. a^n\$i * a \$ (Suc n - i)) {n .. Suc n}" ``` ``` 1765 apply (rule setsum.mono_neutral_right) ``` ``` 1766 apply simp ``` ``` 1767 apply clarsimp ``` ``` 1768 apply clarsimp ``` ``` 1769 apply (rule startsby_zero_power_prefix[rule_format, OF a0]) ``` ``` 1770 apply arith ``` ``` 1771 done ``` ``` 1772 also have "\<dots> = a^n \$ n * a\$1" ``` ``` 1773 using a0 by simp ``` ``` 1774 finally show ?case ``` ``` 1775 using Suc.hyps by simp ``` ``` 1776 qed ``` ``` 1777 ``` ``` 1778 lemma fps_inverse_power: ``` ``` 1779 fixes a :: "'a::field fps" ``` ``` 1780 shows "inverse (a^n) = inverse a ^ n" ``` ``` 1781 by (induction n) (simp_all add: fps_inverse_mult) ``` ``` 1782 ``` ``` 1783 lemma fps_deriv_power: ``` ``` 1784 "fps_deriv (a ^ n) = fps_const (of_nat n :: 'a::comm_ring_1) * fps_deriv a * a ^ (n - 1)" ``` ``` 1785 apply (induct n) ``` ``` 1786 apply (auto simp add: field_simps fps_const_add[symmetric] simp del: fps_const_add) ``` ``` 1787 apply (case_tac n) ``` ``` 1788 apply (auto simp add: field_simps) ``` ``` 1789 done ``` ``` 1790 ``` ``` 1791 lemma fps_inverse_deriv: ``` ``` 1792 fixes a :: "'a::field fps" ``` ``` 1793 assumes a0: "a\$0 \<noteq> 0" ``` ``` 1794 shows "fps_deriv (inverse a) = - fps_deriv a * (inverse a)\<^sup>2" ``` ``` 1795 proof - ``` ``` 1796 from inverse_mult_eq_1[OF a0] ``` ``` 1797 have "fps_deriv (inverse a * a) = 0" by simp ``` ``` 1798 then have "inverse a * fps_deriv a + fps_deriv (inverse a) * a = 0" ``` ``` 1799 by simp ``` ``` 1800 then have "inverse a * (inverse a * fps_deriv a + fps_deriv (inverse a) * a) = 0" ``` ``` 1801 by simp ``` ``` 1802 with inverse_mult_eq_1[OF a0] ``` ``` 1803 have "(inverse a)\<^sup>2 * fps_deriv a + fps_deriv (inverse a) = 0" ``` ``` 1804 unfolding power2_eq_square ``` ``` 1805 apply (simp add: field_simps) ``` ``` 1806 apply (simp add: mult.assoc[symmetric]) ``` ``` 1807 done ``` ``` 1808 then have "(inverse a)\<^sup>2 * fps_deriv a + fps_deriv (inverse a) - fps_deriv a * (inverse a)\<^sup>2 = ``` ``` 1809 0 - fps_deriv a * (inverse a)\<^sup>2" ``` ``` 1810 by simp ``` ``` 1811 then show "fps_deriv (inverse a) = - fps_deriv a * (inverse a)\<^sup>2" ``` ``` 1812 by (simp add: field_simps) ``` ``` 1813 qed ``` ``` 1814 ``` ``` 1815 lemma fps_inverse_deriv': ``` ``` 1816 fixes a :: "'a::field fps" ``` ``` 1817 assumes a0: "a \$ 0 \<noteq> 0" ``` ``` 1818 shows "fps_deriv (inverse a) = - fps_deriv a / a\<^sup>2" ``` ``` 1819 using fps_inverse_deriv[OF a0] a0 ``` ``` 1820 by (simp add: fps_divide_unit power2_eq_square fps_inverse_mult) ``` ``` 1821 ``` ``` 1822 lemma inverse_mult_eq_1': ``` ``` 1823 assumes f0: "f\$0 \<noteq> (0::'a::field)" ``` ``` 1824 shows "f * inverse f = 1" ``` ``` 1825 by (metis mult.commute inverse_mult_eq_1 f0) ``` ``` 1826 ``` ``` 1827 (* FIXME: The last part of this proof should go through by simp once we have a proper ``` ``` 1828 theorem collection for simplifying division on rings *) ``` ``` 1829 lemma fps_divide_deriv: ``` ``` 1830 assumes "b dvd (a :: 'a :: field fps)" ``` ``` 1831 shows "fps_deriv (a / b) = (fps_deriv a * b - a * fps_deriv b) / b^2" ``` ``` 1832 proof - ``` ``` 1833 have eq_divide_imp: "c \<noteq> 0 \<Longrightarrow> a * c = b \<Longrightarrow> a = b div c" for a b c :: "'a :: field fps" ``` ``` 1834 by (drule sym) (simp add: mult.assoc) ``` ``` 1835 from assms have "a = a / b * b" by simp ``` ``` 1836 also have "fps_deriv (a / b * b) = fps_deriv (a / b) * b + a / b * fps_deriv b" by simp ``` ``` 1837 finally have "fps_deriv (a / b) * b^2 = fps_deriv a * b - a * fps_deriv b" using assms ``` ``` 1838 by (simp add: power2_eq_square algebra_simps) ``` ``` 1839 thus ?thesis by (cases "b = 0") (auto simp: eq_divide_imp) ``` ``` 1840 qed ``` ``` 1841 ``` ``` 1842 lemma fps_inverse_gp': "inverse (Abs_fps (\<lambda>n. 1::'a::field)) = 1 - X" ``` ``` 1843 by (simp add: fps_inverse_gp fps_eq_iff X_def) ``` ``` 1844 ``` ``` 1845 lemma fps_nth_deriv_X[simp]: "fps_nth_deriv n X = (if n = 0 then X else if n=1 then 1 else 0)" ``` ``` 1846 by (cases n) simp_all ``` ``` 1847 ``` ``` 1848 lemma fps_inverse_X_plus1: "inverse (1 + X) = Abs_fps (\<lambda>n. (- (1::'a::field)) ^ n)" ``` ``` 1849 (is "_ = ?r") ``` ``` 1850 proof - ``` ``` 1851 have eq: "(1 + X) * ?r = 1" ``` ``` 1852 unfolding minus_one_power_iff ``` ``` 1853 by (auto simp add: field_simps fps_eq_iff) ``` ``` 1854 show ?thesis ``` ``` 1855 by (auto simp add: eq intro: fps_inverse_unique) ``` ``` 1856 qed ``` ``` 1857 ``` ``` 1858 ``` ``` 1859 subsection \<open>Integration\<close> ``` ``` 1860 ``` ``` 1861 definition fps_integral :: "'a::field_char_0 fps \<Rightarrow> 'a \<Rightarrow> 'a fps" ``` ``` 1862 where "fps_integral a a0 = Abs_fps (\<lambda>n. if n = 0 then a0 else (a\$(n - 1) / of_nat n))" ``` ``` 1863 ``` ``` 1864 lemma fps_deriv_fps_integral: "fps_deriv (fps_integral a a0) = a" ``` ``` 1865 unfolding fps_integral_def fps_deriv_def ``` ``` 1866 by (simp add: fps_eq_iff del: of_nat_Suc) ``` ``` 1867 ``` ``` 1868 lemma fps_integral_linear: ``` ``` 1869 "fps_integral (fps_const a * f + fps_const b * g) (a*a0 + b*b0) = ``` ``` 1870 fps_const a * fps_integral f a0 + fps_const b * fps_integral g b0" ``` ``` 1871 (is "?l = ?r") ``` ``` 1872 proof - ``` ``` 1873 have "fps_deriv ?l = fps_deriv ?r" ``` ``` 1874 by (simp add: fps_deriv_fps_integral) ``` ``` 1875 moreover have "?l\$0 = ?r\$0" ``` ``` 1876 by (simp add: fps_integral_def) ``` ``` 1877 ultimately show ?thesis ``` ``` 1878 unfolding fps_deriv_eq_iff by auto ``` ``` 1879 qed ``` ``` 1880 ``` ``` 1881 ``` ``` 1882 subsection \<open>Composition of FPSs\<close> ``` ``` 1883 ``` ``` 1884 definition fps_compose :: "'a::semiring_1 fps \<Rightarrow> 'a fps \<Rightarrow> 'a fps" (infixl "oo" 55) ``` ``` 1885 where "a oo b = Abs_fps (\<lambda>n. setsum (\<lambda>i. a\$i * (b^i\$n)) {0..n})" ``` ``` 1886 ``` ``` 1887 lemma fps_compose_nth: "(a oo b)\$n = setsum (\<lambda>i. a\$i * (b^i\$n)) {0..n}" ``` ``` 1888 by (simp add: fps_compose_def) ``` ``` 1889 ``` ``` 1890 lemma fps_compose_nth_0 [simp]: "(f oo g) \$ 0 = f \$ 0" ``` ``` 1891 by (simp add: fps_compose_nth) ``` ``` 1892 ``` ``` 1893 lemma fps_compose_X[simp]: "a oo X = (a :: 'a::comm_ring_1 fps)" ``` ``` 1894 by (simp add: fps_ext fps_compose_def mult_delta_right setsum.delta') ``` ``` 1895 ``` ``` 1896 lemma fps_const_compose[simp]: "fps_const (a::'a::comm_ring_1) oo b = fps_const a" ``` ``` 1897 by (simp add: fps_eq_iff fps_compose_nth mult_delta_left setsum.delta) ``` ``` 1898 ``` ``` 1899 lemma numeral_compose[simp]: "(numeral k :: 'a::comm_ring_1 fps) oo b = numeral k" ``` ``` 1900 unfolding numeral_fps_const by simp ``` ``` 1901 ``` ``` 1902 lemma neg_numeral_compose[simp]: "(- numeral k :: 'a::comm_ring_1 fps) oo b = - numeral k" ``` ``` 1903 unfolding neg_numeral_fps_const by simp ``` ``` 1904 ``` ``` 1905 lemma X_fps_compose_startby0[simp]: "a\$0 = 0 \<Longrightarrow> X oo a = (a :: 'a::comm_ring_1 fps)" ``` ``` 1906 by (simp add: fps_eq_iff fps_compose_def mult_delta_left setsum.delta not_le) ``` ``` 1907 ``` ``` 1908 ``` ``` 1909 subsection \<open>Rules from Herbert Wilf's Generatingfunctionology\<close> ``` ``` 1910 ``` ``` 1911 subsubsection \<open>Rule 1\<close> ``` ``` 1912 (* {a_{n+k}}_0^infty Corresponds to (f - setsum (\<lambda>i. a_i * x^i))/x^h, for h>0*) ``` ``` 1913 ``` ``` 1914 lemma fps_power_mult_eq_shift: ``` ``` 1915 "X^Suc k * Abs_fps (\<lambda>n. a (n + Suc k)) = ``` ``` 1916 Abs_fps a - setsum (\<lambda>i. fps_const (a i :: 'a::comm_ring_1) * X^i) {0 .. k}" ``` ``` 1917 (is "?lhs = ?rhs") ``` ``` 1918 proof - ``` ``` 1919 have "?lhs \$ n = ?rhs \$ n" for n :: nat ``` ``` 1920 proof - ``` ``` 1921 have "?lhs \$ n = (if n < Suc k then 0 else a n)" ``` ``` 1922 unfolding X_power_mult_nth by auto ``` ``` 1923 also have "\<dots> = ?rhs \$ n" ``` ``` 1924 proof (induct k) ``` ``` 1925 case 0 ``` ``` 1926 then show ?case ``` ``` 1927 by (simp add: fps_setsum_nth) ``` ``` 1928 next ``` ``` 1929 case (Suc k) ``` ``` 1930 have "(Abs_fps a - setsum (\<lambda>i. fps_const (a i :: 'a) * X^i) {0 .. Suc k})\$n = ``` ``` 1931 (Abs_fps a - setsum (\<lambda>i. fps_const (a i :: 'a) * X^i) {0 .. k} - ``` ``` 1932 fps_const (a (Suc k)) * X^ Suc k) \$ n" ``` ``` 1933 by (simp add: field_simps) ``` ``` 1934 also have "\<dots> = (if n < Suc k then 0 else a n) - (fps_const (a (Suc k)) * X^ Suc k)\$n" ``` ``` 1935 using Suc.hyps[symmetric] unfolding fps_sub_nth by simp ``` ``` 1936 also have "\<dots> = (if n < Suc (Suc k) then 0 else a n)" ``` ``` 1937 unfolding X_power_mult_right_nth ``` ``` 1938 apply (auto simp add: not_less fps_const_def) ``` ``` 1939 apply (rule cong[of a a, OF refl]) ``` ``` 1940 apply arith ``` ``` 1941 done ``` ``` 1942 finally show ?case ``` ``` 1943 by simp ``` ``` 1944 qed ``` ``` 1945 finally show ?thesis . ``` ``` 1946 qed ``` ``` 1947 then show ?thesis ``` ``` 1948 by (simp add: fps_eq_iff) ``` ``` 1949 qed ``` ``` 1950 ``` ``` 1951 ``` ``` 1952 subsubsection \<open>Rule 2\<close> ``` ``` 1953 ``` ``` 1954 (* We can not reach the form of Wilf, but still near to it using rewrite rules*) ``` ``` 1955 (* If f reprents {a_n} and P is a polynomial, then ``` ``` 1956 P(xD) f represents {P(n) a_n}*) ``` ``` 1957 ``` ``` 1958 definition "XD = op * X \<circ> fps_deriv" ``` ``` 1959 ``` ``` 1960 lemma XD_add[simp]:"XD (a + b) = XD a + XD (b :: 'a::comm_ring_1 fps)" ``` ``` 1961 by (simp add: XD_def field_simps) ``` ``` 1962 ``` ``` 1963 lemma XD_mult_const[simp]:"XD (fps_const (c::'a::comm_ring_1) * a) = fps_const c * XD a" ``` ``` 1964 by (simp add: XD_def field_simps) ``` ``` 1965 ``` ``` 1966 lemma XD_linear[simp]: "XD (fps_const c * a + fps_const d * b) = ``` ``` 1967 fps_const c * XD a + fps_const d * XD (b :: 'a::comm_ring_1 fps)" ``` ``` 1968 by simp ``` ``` 1969 ``` ``` 1970 lemma XDN_linear: ``` ``` 1971 "(XD ^^ n) (fps_const c * a + fps_const d * b) = ``` ``` 1972 fps_const c * (XD ^^ n) a + fps_const d * (XD ^^ n) (b :: 'a::comm_ring_1 fps)" ``` ``` 1973 by (induct n) simp_all ``` ``` 1974 ``` ``` 1975 lemma fps_mult_X_deriv_shift: "X* fps_deriv a = Abs_fps (\<lambda>n. of_nat n* a\$n)" ``` ``` 1976 by (simp add: fps_eq_iff) ``` ``` 1977 ``` ``` 1978 lemma fps_mult_XD_shift: ``` ``` 1979 "(XD ^^ k) (a :: 'a::comm_ring_1 fps) = Abs_fps (\<lambda>n. (of_nat n ^ k) * a\$n)" ``` ``` 1980 by (induct k arbitrary: a) (simp_all add: XD_def fps_eq_iff field_simps del: One_nat_def) ``` ``` 1981 ``` ``` 1982 ``` ``` 1983 subsubsection \<open>Rule 3\<close> ``` ``` 1984 ``` ``` 1985 text \<open>Rule 3 is trivial and is given by \<open>fps_times_def\<close>.\<close> ``` ``` 1986 ``` ``` 1987 ``` ``` 1988 subsubsection \<open>Rule 5 --- summation and "division" by (1 - X)\<close> ``` ``` 1989 ``` ``` 1990 lemma fps_divide_X_minus1_setsum_lemma: ``` ``` 1991 "a = ((1::'a::comm_ring_1 fps) - X) * Abs_fps (\<lambda>n. setsum (\<lambda>i. a \$ i) {0..n})" ``` ``` 1992 proof - ``` ``` 1993 let ?sa = "Abs_fps (\<lambda>n. setsum (\<lambda>i. a \$ i) {0..n})" ``` ``` 1994 have th0: "\<And>i. (1 - (X::'a fps)) \$ i = (if i = 0 then 1 else if i = 1 then - 1 else 0)" ``` ``` 1995 by simp ``` ``` 1996 have "a\$n = ((1 - X) * ?sa) \$ n" for n ``` ``` 1997 proof (cases "n = 0") ``` ``` 1998 case True ``` ``` 1999 then show ?thesis ``` ``` 2000 by (simp add: fps_mult_nth) ``` ``` 2001 next ``` ``` 2002 case False ``` ``` 2003 then have u: "{0} \<union> ({1} \<union> {2..n}) = {0..n}" "{1} \<union> {2..n} = {1..n}" ``` ``` 2004 "{0..n - 1} \<union> {n} = {0..n}" ``` ``` 2005 by (auto simp: set_eq_iff) ``` ``` 2006 have d: "{0} \<inter> ({1} \<union> {2..n}) = {}" "{1} \<inter> {2..n} = {}" "{0..n - 1} \<inter> {n} = {}" ``` ``` 2007 using False by simp_all ``` ``` 2008 have f: "finite {0}" "finite {1}" "finite {2 .. n}" ``` ``` 2009 "finite {0 .. n - 1}" "finite {n}" by simp_all ``` ``` 2010 have "((1 - X) * ?sa) \$ n = setsum (\<lambda>i. (1 - X)\$ i * ?sa \$ (n - i)) {0 .. n}" ``` ``` 2011 by (simp add: fps_mult_nth) ``` ``` 2012 also have "\<dots> = a\$n" ``` ``` 2013 unfolding th0 ``` ``` 2014 unfolding setsum.union_disjoint[OF f(1) finite_UnI[OF f(2,3)] d(1), unfolded u(1)] ``` ``` 2015 unfolding setsum.union_disjoint[OF f(2) f(3) d(2)] ``` ``` 2016 apply (simp) ``` ``` 2017 unfolding setsum.union_disjoint[OF f(4,5) d(3), unfolded u(3)] ``` ``` 2018 apply simp ``` ``` 2019 done ``` ``` 2020 finally show ?thesis ``` ``` 2021 by simp ``` ``` 2022 qed ``` ``` 2023 then show ?thesis ``` ``` 2024 unfolding fps_eq_iff by blast ``` ``` 2025 qed ``` ``` 2026 ``` ``` 2027 lemma fps_divide_X_minus1_setsum: ``` ``` 2028 "a /((1::'a::field fps) - X) = Abs_fps (\<lambda>n. setsum (\<lambda>i. a \$ i) {0..n})" ``` ``` 2029 proof - ``` ``` 2030 let ?X = "1 - (X::'a fps)" ``` ``` 2031 have th0: "?X \$ 0 \<noteq> 0" ``` ``` 2032 by simp ``` ``` 2033 have "a /?X = ?X * Abs_fps (\<lambda>n::nat. setsum (op \$ a) {0..n}) * inverse ?X" ``` ``` 2034 using fps_divide_X_minus1_setsum_lemma[of a, symmetric] th0 ``` ``` 2035 by (simp add: fps_divide_def mult.assoc) ``` ``` 2036 also have "\<dots> = (inverse ?X * ?X) * Abs_fps (\<lambda>n::nat. setsum (op \$ a) {0..n}) " ``` ``` 2037 by (simp add: ac_simps) ``` ``` 2038 finally show ?thesis ``` ``` 2039 by (simp add: inverse_mult_eq_1[OF th0]) ``` ``` 2040 qed ``` ``` 2041 ``` ``` 2042 ``` ``` 2043 subsubsection \<open>Rule 4 in its more general form: generalizes Rule 3 for an arbitrary ``` ``` 2044 finite product of FPS, also the relvant instance of powers of a FPS\<close> ``` ``` 2045 ``` ``` 2046 definition "natpermute n k = {l :: nat list. length l = k \<and> listsum l = n}" ``` ``` 2047 ``` ``` 2048 lemma natlist_trivial_1: "natpermute n 1 = {[n]}" ``` ``` 2049 apply (auto simp add: natpermute_def) ``` ``` 2050 apply (case_tac x) ``` ``` 2051 apply auto ``` ``` 2052 done ``` ``` 2053 ``` ``` 2054 lemma append_natpermute_less_eq: ``` ``` 2055 assumes "xs @ ys \<in> natpermute n k" ``` ``` 2056 shows "listsum xs \<le> n" ``` ``` 2057 and "listsum ys \<le> n" ``` ``` 2058 proof - ``` ``` 2059 from assms have "listsum (xs @ ys) = n" ``` ``` 2060 by (simp add: natpermute_def) ``` ``` 2061 then have "listsum xs + listsum ys = n" ``` ``` 2062 by simp ``` ``` 2063 then show "listsum xs \<le> n" and "listsum ys \<le> n" ``` ``` 2064 by simp_all ``` ``` 2065 qed ``` ``` 2066 ``` ``` 2067 lemma natpermute_split: ``` ``` 2068 assumes "h \<le> k" ``` ``` 2069 shows "natpermute n k = ``` ``` 2070 (\<Union>m \<in>{0..n}. {l1 @ l2 |l1 l2. l1 \<in> natpermute m h \<and> l2 \<in> natpermute (n - m) (k - h)})" ``` ``` 2071 (is "?L = ?R" is "_ = (\<Union>m \<in>{0..n}. ?S m)") ``` ``` 2072 proof ``` ``` 2073 show "?R \<subseteq> ?L" ``` ``` 2074 proof ``` ``` 2075 fix l ``` ``` 2076 assume l: "l \<in> ?R" ``` ``` 2077 from l obtain m xs ys where h: "m \<in> {0..n}" ``` ``` 2078 and xs: "xs \<in> natpermute m h" ``` ``` 2079 and ys: "ys \<in> natpermute (n - m) (k - h)" ``` ``` 2080 and leq: "l = xs@ys" by blast ``` ``` 2081 from xs have xs': "listsum xs = m" ``` ``` 2082 by (simp add: natpermute_def) ``` ``` 2083 from ys have ys': "listsum ys = n - m" ``` ``` 2084 by (simp add: natpermute_def) ``` ``` 2085 show "l \<in> ?L" using leq xs ys h ``` ``` 2086 apply (clarsimp simp add: natpermute_def) ``` ``` 2087 unfolding xs' ys' ``` ``` 2088 using assms xs ys ``` ``` 2089 unfolding natpermute_def ``` ``` 2090 apply simp ``` ``` 2091 done ``` ``` 2092 qed ``` ``` 2093 show "?L \<subseteq> ?R" ``` ``` 2094 proof ``` ``` 2095 fix l ``` ``` 2096 assume l: "l \<in> natpermute n k" ``` ``` 2097 let ?xs = "take h l" ``` ``` 2098 let ?ys = "drop h l" ``` ``` 2099 let ?m = "listsum ?xs" ``` ``` 2100 from l have ls: "listsum (?xs @ ?ys) = n" ``` ``` 2101 by (simp add: natpermute_def) ``` ``` 2102 have xs: "?xs \<in> natpermute ?m h" using l assms ``` ``` 2103 by (simp add: natpermute_def) ``` ``` 2104 have l_take_drop: "listsum l = listsum (take h l @ drop h l)" ``` ``` 2105 by simp ``` ``` 2106 then have ys: "?ys \<in> natpermute (n - ?m) (k - h)" ``` ``` 2107 using l assms ls by (auto simp add: natpermute_def simp del: append_take_drop_id) ``` ``` 2108 from ls have m: "?m \<in> {0..n}" ``` ``` 2109 by (simp add: l_take_drop del: append_take_drop_id) ``` ``` 2110 from xs ys ls show "l \<in> ?R" ``` ``` 2111 apply auto ``` ``` 2112 apply (rule bexI [where x = "?m"]) ``` ``` 2113 apply (rule exI [where x = "?xs"]) ``` ``` 2114 apply (rule exI [where x = "?ys"]) ``` ``` 2115 using ls l ``` ``` 2116 apply (auto simp add: natpermute_def l_take_drop simp del: append_take_drop_id) ``` ``` 2117 apply simp ``` ``` 2118 done ``` ``` 2119 qed ``` ``` 2120 qed ``` ``` 2121 ``` ``` 2122 lemma natpermute_0: "natpermute n 0 = (if n = 0 then {[]} else {})" ``` ``` 2123 by (auto simp add: natpermute_def) ``` ``` 2124 ``` ``` 2125 lemma natpermute_0'[simp]: "natpermute 0 k = (if k = 0 then {[]} else {replicate k 0})" ``` ``` 2126 apply (auto simp add: set_replicate_conv_if natpermute_def) ``` ``` 2127 apply (rule nth_equalityI) ``` ``` 2128 apply simp_all ``` ``` 2129 done ``` ``` 2130 ``` ``` 2131 lemma natpermute_finite: "finite (natpermute n k)" ``` ``` 2132 proof (induct k arbitrary: n) ``` ``` 2133 case 0 ``` ``` 2134 then show ?case ``` ``` 2135 apply (subst natpermute_split[of 0 0, simplified]) ``` ``` 2136 apply (simp add: natpermute_0) ``` ``` 2137 done ``` ``` 2138 next ``` ``` 2139 case (Suc k) ``` ``` 2140 then show ?case unfolding natpermute_split [of k "Suc k", simplified] ``` ``` 2141 apply - ``` ``` 2142 apply (rule finite_UN_I) ``` ``` 2143 apply simp ``` ``` 2144 unfolding One_nat_def[symmetric] natlist_trivial_1 ``` ``` 2145 apply simp ``` ``` 2146 done ``` ``` 2147 qed ``` ``` 2148 ``` ``` 2149 lemma natpermute_contain_maximal: ``` ``` 2150 "{xs \<in> natpermute n (k + 1). n \<in> set xs} = (\<Union>i\<in>{0 .. k}. {(replicate (k + 1) 0) [i:=n]})" ``` ``` 2151 (is "?A = ?B") ``` ``` 2152 proof ``` ``` 2153 show "?A \<subseteq> ?B" ``` ``` 2154 proof ``` ``` 2155 fix xs ``` ``` 2156 assume "xs \<in> ?A" ``` ``` 2157 then have H: "xs \<in> natpermute n (k + 1)" and n: "n \<in> set xs" ``` ``` 2158 by blast+ ``` ``` 2159 then obtain i where i: "i \<in> {0.. k}" "xs!i = n" ``` ``` 2160 unfolding in_set_conv_nth by (auto simp add: less_Suc_eq_le natpermute_def) ``` ``` 2161 have eqs: "({0..k} - {i}) \<union> {i} = {0..k}" ``` ``` 2162 using i by auto ``` ``` 2163 have f: "finite({0..k} - {i})" "finite {i}" ``` ``` 2164 by auto ``` ``` 2165 have d: "({0..k} - {i}) \<inter> {i} = {}" ``` ``` 2166 using i by auto ``` ``` 2167 from H have "n = setsum (nth xs) {0..k}" ``` ``` 2168 apply (simp add: natpermute_def) ``` ``` 2169 apply (auto simp add: atLeastLessThanSuc_atLeastAtMost listsum_setsum_nth) ``` ``` 2170 done ``` ``` 2171 also have "\<dots> = n + setsum (nth xs) ({0..k} - {i})" ``` ``` 2172 unfolding setsum.union_disjoint[OF f d, unfolded eqs] using i by simp ``` ``` 2173 finally have zxs: "\<forall> j\<in> {0..k} - {i}. xs!j = 0" ``` ``` 2174 by auto ``` ``` 2175 from H have xsl: "length xs = k+1" ``` ``` 2176 by (simp add: natpermute_def) ``` ``` 2177 from i have i': "i < length (replicate (k+1) 0)" "i < k+1" ``` ``` 2178 unfolding length_replicate by presburger+ ``` ``` 2179 have "xs = replicate (k+1) 0 [i := n]" ``` ``` 2180 apply (rule nth_equalityI) ``` ``` 2181 unfolding xsl length_list_update length_replicate ``` ``` 2182 apply simp ``` ``` 2183 apply clarify ``` ``` 2184 unfolding nth_list_update[OF i'(1)] ``` ``` 2185 using i zxs ``` ``` 2186 apply (case_tac "ia = i") ``` ``` 2187 apply (auto simp del: replicate.simps) ``` ``` 2188 done ``` ``` 2189 then show "xs \<in> ?B" using i by blast ``` ``` 2190 qed ``` ``` 2191 show "?B \<subseteq> ?A" ``` ``` 2192 proof ``` ``` 2193 fix xs ``` ``` 2194 assume "xs \<in> ?B" ``` ``` 2195 then obtain i where i: "i \<in> {0..k}" and xs: "xs = replicate (k + 1) 0 [i:=n]" ``` ``` 2196 by auto ``` ``` 2197 have nxs: "n \<in> set xs" ``` ``` 2198 unfolding xs ``` ``` 2199 apply (rule set_update_memI) ``` ``` 2200 using i apply simp ``` ``` 2201 done ``` ``` 2202 have xsl: "length xs = k + 1" ``` ``` 2203 by (simp only: xs length_replicate length_list_update) ``` ``` 2204 have "listsum xs = setsum (nth xs) {0..<k+1}" ``` ``` 2205 unfolding listsum_setsum_nth xsl .. ``` ``` 2206 also have "\<dots> = setsum (\<lambda>j. if j = i then n else 0) {0..< k+1}" ``` ``` 2207 by (rule setsum.cong) (simp_all add: xs del: replicate.simps) ``` ``` 2208 also have "\<dots> = n" using i by (simp add: setsum.delta) ``` ``` 2209 finally have "xs \<in> natpermute n (k + 1)" ``` ``` 2210 using xsl unfolding natpermute_def mem_Collect_eq by blast ``` ``` 2211 then show "xs \<in> ?A" ``` ``` 2212 using nxs by blast ``` ``` 2213 qed ``` ``` 2214 qed ``` ``` 2215 ``` ``` 2216 text \<open>The general form.\<close> ``` ``` 2217 lemma fps_setprod_nth: ``` ``` 2218 fixes m :: nat ``` ``` 2219 and a :: "nat \<Rightarrow> 'a::comm_ring_1 fps" ``` ``` 2220 shows "(setprod a {0 .. m}) \$ n = ``` ``` 2221 setsum (\<lambda>v. setprod (\<lambda>j. (a j) \$ (v!j)) {0..m}) (natpermute n (m+1))" ``` ``` 2222 (is "?P m n") ``` ``` 2223 proof (induct m arbitrary: n rule: nat_less_induct) ``` ``` 2224 fix m n assume H: "\<forall>m' < m. \<forall>n. ?P m' n" ``` ``` 2225 show "?P m n" ``` ``` 2226 proof (cases m) ``` ``` 2227 case 0 ``` ``` 2228 then show ?thesis ``` ``` 2229 apply simp ``` ``` 2230 unfolding natlist_trivial_1[where n = n, unfolded One_nat_def] ``` ``` 2231 apply simp ``` ``` 2232 done ``` ``` 2233 next ``` ``` 2234 case (Suc k) ``` ``` 2235 then have km: "k < m" by arith ``` ``` 2236 have u0: "{0 .. k} \<union> {m} = {0..m}" ``` ``` 2237 using Suc by (simp add: set_eq_iff) presburger ``` ``` 2238 have f0: "finite {0 .. k}" "finite {m}" by auto ``` ``` 2239 have d0: "{0 .. k} \<inter> {m} = {}" using Suc by auto ``` ``` 2240 have "(setprod a {0 .. m}) \$ n = (setprod a {0 .. k} * a m) \$ n" ``` ``` 2241 unfolding setprod.union_disjoint[OF f0 d0, unfolded u0] by simp ``` ``` 2242 also have "\<dots> = (\<Sum>i = 0..n. (\<Sum>v\<in>natpermute i (k + 1). \<Prod>j\<in>{0..k}. a j \$ v ! j) * a m \$ (n - i))" ``` ``` 2243 unfolding fps_mult_nth H[rule_format, OF km] .. ``` ``` 2244 also have "\<dots> = (\<Sum>v\<in>natpermute n (m + 1). \<Prod>j\<in>{0..m}. a j \$ v ! j)" ``` ``` 2245 apply (simp add: Suc) ``` ``` 2246 unfolding natpermute_split[of m "m + 1", simplified, of n, ``` ``` 2247 unfolded natlist_trivial_1[unfolded One_nat_def] Suc] ``` ``` 2248 apply (subst setsum.UNION_disjoint) ``` ``` 2249 apply simp ``` ``` 2250 apply simp ``` ``` 2251 unfolding image_Collect[symmetric] ``` ``` 2252 apply clarsimp ``` ``` 2253 apply (rule finite_imageI) ``` ``` 2254 apply (rule natpermute_finite) ``` ``` 2255 apply (clarsimp simp add: set_eq_iff) ``` ``` 2256 apply auto ``` ``` 2257 apply (rule setsum.cong) ``` ``` 2258 apply (rule refl) ``` ``` 2259 unfolding setsum_left_distrib ``` ``` 2260 apply (rule sym) ``` ``` 2261 apply (rule_tac l = "\<lambda>xs. xs @ [n - x]" in setsum.reindex_cong) ``` ``` 2262 apply (simp add: inj_on_def) ``` ``` 2263 apply auto ``` ``` 2264 unfolding setprod.union_disjoint[OF f0 d0, unfolded u0, unfolded Suc] ``` ``` 2265 apply (clarsimp simp add: natpermute_def nth_append) ``` ``` 2266 done ``` ``` 2267 finally show ?thesis . ``` ``` 2268 qed ``` ``` 2269 qed ``` ``` 2270 ``` ``` 2271 text \<open>The special form for powers.\<close> ``` ``` 2272 lemma fps_power_nth_Suc: ``` ``` 2273 fixes m :: nat ``` ``` 2274 and a :: "'a::comm_ring_1 fps" ``` ``` 2275 shows "(a ^ Suc m)\$n = setsum (\<lambda>v. setprod (\<lambda>j. a \$ (v!j)) {0..m}) (natpermute n (m+1))" ``` ``` 2276 proof - ``` ``` 2277 have th0: "a^Suc m = setprod (\<lambda>i. a) {0..m}" ``` ``` 2278 by (simp add: setprod_constant) ``` ``` 2279 show ?thesis unfolding th0 fps_setprod_nth .. ``` ``` 2280 qed ``` ``` 2281 ``` ``` 2282 lemma fps_power_nth: ``` ``` 2283 fixes m :: nat ``` ``` 2284 and a :: "'a::comm_ring_1 fps" ``` ``` 2285 shows "(a ^m)\$n = ``` ``` 2286 (if m=0 then 1\$n else setsum (\<lambda>v. setprod (\<lambda>j. a \$ (v!j)) {0..m - 1}) (natpermute n m))" ``` ``` 2287 by (cases m) (simp_all add: fps_power_nth_Suc del: power_Suc) ``` ``` 2288 ``` ``` 2289 lemma fps_nth_power_0: ``` ``` 2290 fixes m :: nat ``` ``` 2291 and a :: "'a::comm_ring_1 fps" ``` ``` 2292 shows "(a ^m)\$0 = (a\$0) ^ m" ``` ``` 2293 proof (cases m) ``` ``` 2294 case 0 ``` ``` 2295 then show ?thesis by simp ``` ``` 2296 next ``` ``` 2297 case (Suc n) ``` ``` 2298 then have c: "m = card {0..n}" by simp ``` ``` 2299 have "(a ^m)\$0 = setprod (\<lambda>i. a\$0) {0..n}" ``` ``` 2300 by (simp add: Suc fps_power_nth del: replicate.simps power_Suc) ``` ``` 2301 also have "\<dots> = (a\$0) ^ m" ``` ``` 2302 unfolding c by (rule setprod_constant) simp ``` ``` 2303 finally show ?thesis . ``` ``` 2304 qed ``` ``` 2305 ``` ``` 2306 lemma fps_compose_inj_right: ``` ``` 2307 assumes a0: "a\$0 = (0::'a::idom)" ``` ``` 2308 and a1: "a\$1 \<noteq> 0" ``` ``` 2309 shows "(b oo a = c oo a) \<longleftrightarrow> b = c" ``` ``` 2310 (is "?lhs \<longleftrightarrow>?rhs") ``` ``` 2311 proof ``` ``` 2312 show ?lhs if ?rhs using that by simp ``` ``` 2313 show ?rhs if ?lhs ``` ``` 2314 proof - ``` ``` 2315 have "b\$n = c\$n" for n ``` ``` 2316 proof (induct n rule: nat_less_induct) ``` ``` 2317 fix n ``` ``` 2318 assume H: "\<forall>m<n. b\$m = c\$m" ``` ``` 2319 show "b\$n = c\$n" ``` ``` 2320 proof (cases n) ``` ``` 2321 case 0 ``` ``` 2322 from \<open>?lhs\<close> have "(b oo a)\$n = (c oo a)\$n" ``` ``` 2323 by simp ``` ``` 2324 then show ?thesis ``` ``` 2325 using 0 by (simp add: fps_compose_nth) ``` ``` 2326 next ``` ``` 2327 case (Suc n1) ``` ``` 2328 have f: "finite {0 .. n1}" "finite {n}" by simp_all ``` ``` 2329 have eq: "{0 .. n1} \<union> {n} = {0 .. n}" using Suc by auto ``` ``` 2330 have d: "{0 .. n1} \<inter> {n} = {}" using Suc by auto ``` ``` 2331 have seq: "(\<Sum>i = 0..n1. b \$ i * a ^ i \$ n) = (\<Sum>i = 0..n1. c \$ i * a ^ i \$ n)" ``` ``` 2332 apply (rule setsum.cong) ``` ``` 2333 using H Suc ``` ``` 2334 apply auto ``` ``` 2335 done ``` ``` 2336 have th0: "(b oo a) \$n = (\<Sum>i = 0..n1. c \$ i * a ^ i \$ n) + b\$n * (a\$1)^n" ``` ``` 2337 unfolding fps_compose_nth setsum.union_disjoint[OF f d, unfolded eq] seq ``` ``` 2338 using startsby_zero_power_nth_same[OF a0] ``` ``` 2339 by simp ``` ``` 2340 have th1: "(c oo a) \$n = (\<Sum>i = 0..n1. c \$ i * a ^ i \$ n) + c\$n * (a\$1)^n" ``` ``` 2341 unfolding fps_compose_nth setsum.union_disjoint[OF f d, unfolded eq] ``` ``` 2342 using startsby_zero_power_nth_same[OF a0] ``` ``` 2343 by simp ``` ``` 2344 from \<open>?lhs\<close>[unfolded fps_eq_iff, rule_format, of n] th0 th1 a1 ``` ``` 2345 show ?thesis by auto ``` ``` 2346 qed ``` ``` 2347 qed ``` ``` 2348 then show ?rhs by (simp add: fps_eq_iff) ``` ``` 2349 qed ``` ``` 2350 qed ``` ``` 2351 ``` ``` 2352 ``` ``` 2353 subsection \<open>Radicals\<close> ``` ``` 2354 ``` ``` 2355 declare setprod.cong [fundef_cong] ``` ``` 2356 ``` ``` 2357 function radical :: "(nat \<Rightarrow> 'a \<Rightarrow> 'a) \<Rightarrow> nat \<Rightarrow> 'a::field fps \<Rightarrow> nat \<Rightarrow> 'a" ``` ``` 2358 where ``` ``` 2359 "radical r 0 a 0 = 1" ``` ``` 2360 | "radical r 0 a (Suc n) = 0" ``` ``` 2361 | "radical r (Suc k) a 0 = r (Suc k) (a\$0)" ``` ``` 2362 | "radical r (Suc k) a (Suc n) = ``` ``` 2363 (a\$ Suc n - setsum (\<lambda>xs. setprod (\<lambda>j. radical r (Suc k) a (xs ! j)) {0..k}) ``` ``` 2364 {xs. xs \<in> natpermute (Suc n) (Suc k) \<and> Suc n \<notin> set xs}) / ``` ``` 2365 (of_nat (Suc k) * (radical r (Suc k) a 0)^k)" ``` ``` 2366 by pat_completeness auto ``` ``` 2367 ``` ``` 2368 termination radical ``` ``` 2369 proof ``` ``` 2370 let ?R = "measure (\<lambda>(r, k, a, n). n)" ``` ``` 2371 { ``` ``` 2372 show "wf ?R" by auto ``` ``` 2373 next ``` ``` 2374 fix r k a n xs i ``` ``` 2375 assume xs: "xs \<in> {xs \<in> natpermute (Suc n) (Suc k). Suc n \<notin> set xs}" and i: "i \<in> {0..k}" ``` ``` 2376 have False if c: "Suc n \<le> xs ! i" ``` ``` 2377 proof - ``` ``` 2378 from xs i have "xs !i \<noteq> Suc n" ``` ``` 2379 by (auto simp add: in_set_conv_nth natpermute_def) ``` ``` 2380 with c have c': "Suc n < xs!i" by arith ``` ``` 2381 have fths: "finite {0 ..< i}" "finite {i}" "finite {i+1..<Suc k}" ``` ``` 2382 by simp_all ``` ``` 2383 have d: "{0 ..< i} \<inter> ({i} \<union> {i+1 ..< Suc k}) = {}" "{i} \<inter> {i+1..< Suc k} = {}" ``` ``` 2384 by auto ``` ``` 2385 have eqs: "{0..<Suc k} = {0 ..< i} \<union> ({i} \<union> {i+1 ..< Suc k})" ``` ``` 2386 using i by auto ``` ``` 2387 from xs have "Suc n = listsum xs" ``` ``` 2388 by (simp add: natpermute_def) ``` ``` 2389 also have "\<dots> = setsum (nth xs) {0..<Suc k}" using xs ``` ``` 2390 by (simp add: natpermute_def listsum_setsum_nth) ``` ``` 2391 also have "\<dots> = xs!i + setsum (nth xs) {0..<i} + setsum (nth xs) {i+1..<Suc k}" ``` ``` 2392 unfolding eqs setsum.union_disjoint[OF fths(1) finite_UnI[OF fths(2,3)] d(1)] ``` ``` 2393 unfolding setsum.union_disjoint[OF fths(2) fths(3) d(2)] ``` ``` 2394 by simp ``` ``` 2395 finally show ?thesis using c' by simp ``` ``` 2396 qed ``` ``` 2397 then show "((r, Suc k, a, xs!i), r, Suc k, a, Suc n) \<in> ?R" ``` ``` 2398 apply auto ``` ``` 2399 apply (metis not_less) ``` ``` 2400 done ``` ``` 2401 next ``` ``` 2402 fix r k a n ``` ``` 2403 show "((r, Suc k, a, 0), r, Suc k, a, Suc n) \<in> ?R" by simp ``` ``` 2404 } ``` ``` 2405 qed ``` ``` 2406 ``` ``` 2407 definition "fps_radical r n a = Abs_fps (radical r n a)" ``` ``` 2408 ``` ``` 2409 lemma fps_radical0[simp]: "fps_radical r 0 a = 1" ``` ``` 2410 apply (auto simp add: fps_eq_iff fps_radical_def) ``` ``` 2411 apply (case_tac n) ``` ``` 2412 apply auto ``` ``` 2413 done ``` ``` 2414 ``` ``` 2415 lemma fps_radical_nth_0[simp]: "fps_radical r n a \$ 0 = (if n = 0 then 1 else r n (a\$0))" ``` ``` 2416 by (cases n) (simp_all add: fps_radical_def) ``` ``` 2417 ``` ``` 2418 lemma fps_radical_power_nth[simp]: ``` ``` 2419 assumes r: "(r k (a\$0)) ^ k = a\$0" ``` ``` 2420 shows "fps_radical r k a ^ k \$ 0 = (if k = 0 then 1 else a\$0)" ``` ``` 2421 proof (cases k) ``` ``` 2422 case 0 ``` ``` 2423 then show ?thesis by simp ``` ``` 2424 next ``` ``` 2425 case (Suc h) ``` ``` 2426 have eq1: "fps_radical r k a ^ k \$ 0 = (\<Prod>j\<in>{0..h}. fps_radical r k a \$ (replicate k 0) ! j)" ``` ``` 2427 unfolding fps_power_nth Suc by simp ``` ``` 2428 also have "\<dots> = (\<Prod>j\<in>{0..h}. r k (a\$0))" ``` ``` 2429 apply (rule setprod.cong) ``` ``` 2430 apply simp ``` ``` 2431 using Suc ``` ``` 2432 apply (subgoal_tac "replicate k 0 ! x = 0") ``` ``` 2433 apply (auto intro: nth_replicate simp del: replicate.simps) ``` ``` 2434 done ``` ``` 2435 also have "\<dots> = a\$0" ``` ``` 2436 using r Suc by (simp add: setprod_constant) ``` ``` 2437 finally show ?thesis ``` ``` 2438 using Suc by simp ``` ``` 2439 qed ``` ``` 2440 ``` ``` 2441 lemma natpermute_max_card: ``` ``` 2442 assumes n0: "n \<noteq> 0" ``` ``` 2443 shows "card {xs \<in> natpermute n (k + 1). n \<in> set xs} = k + 1" ``` ``` 2444 unfolding natpermute_contain_maximal ``` ``` 2445 proof - ``` ``` 2446 let ?A = "\<lambda>i. {replicate (k + 1) 0[i := n]}" ``` ``` 2447 let ?K = "{0 ..k}" ``` ``` 2448 have fK: "finite ?K" ``` ``` 2449 by simp ``` ``` 2450 have fAK: "\<forall>i\<in>?K. finite (?A i)" ``` ``` 2451 by auto ``` ``` 2452 have d: "\<forall>i\<in> ?K. \<forall>j\<in> ?K. i \<noteq> j \<longrightarrow> ``` ``` 2453 {replicate (k + 1) 0[i := n]} \<inter> {replicate (k + 1) 0[j := n]} = {}" ``` ``` 2454 proof clarify ``` ``` 2455 fix i j ``` ``` 2456 assume i: "i \<in> ?K" and j: "j \<in> ?K" and ij: "i \<noteq> j" ``` ``` 2457 have False if eq: "replicate (k+1) 0 [i:=n] = replicate (k+1) 0 [j:= n]" ``` ``` 2458 proof - ``` ``` 2459 have "(replicate (k+1) 0 [i:=n] ! i) = n" ``` ``` 2460 using i by (simp del: replicate.simps) ``` ``` 2461 moreover ``` ``` 2462 have "(replicate (k+1) 0 [j:=n] ! i) = 0" ``` ``` 2463 using i ij by (simp del: replicate.simps) ``` ``` 2464 ultimately show ?thesis ``` ``` 2465 using eq n0 by (simp del: replicate.simps) ``` ``` 2466 qed ``` ``` 2467 then show "{replicate (k + 1) 0[i := n]} \<inter> {replicate (k + 1) 0[j := n]} = {}" ``` ``` 2468 by auto ``` ``` 2469 qed ``` ``` 2470 from card_UN_disjoint[OF fK fAK d] ``` ``` 2471 show "card (\<Union>i\<in>{0..k}. {replicate (k + 1) 0[i := n]}) = k + 1" ``` ``` 2472 by simp ``` ``` 2473 qed ``` ``` 2474 ``` ``` 2475 lemma power_radical: ``` ``` 2476 fixes a:: "'a::field_char_0 fps" ``` ``` 2477 assumes a0: "a\$0 \<noteq> 0" ``` ``` 2478 shows "(r (Suc k) (a\$0)) ^ Suc k = a\$0 \<longleftrightarrow> (fps_radical r (Suc k) a) ^ (Suc k) = a" ``` ``` 2479 (is "?lhs \<longleftrightarrow> ?rhs") ``` ``` 2480 proof ``` ``` 2481 let ?r = "fps_radical r (Suc k) a" ``` ``` 2482 show ?rhs if r0: ?lhs ``` ``` 2483 proof - ``` ``` 2484 from a0 r0 have r00: "r (Suc k) (a\$0) \<noteq> 0" by auto ``` ``` 2485 have "?r ^ Suc k \$ z = a\$z" for z ``` ``` 2486 proof (induct z rule: nat_less_induct) ``` ``` 2487 fix n ``` ``` 2488 assume H: "\<forall>m<n. ?r ^ Suc k \$ m = a\$m" ``` ``` 2489 show "?r ^ Suc k \$ n = a \$n" ``` ``` 2490 proof (cases n) ``` ``` 2491 case 0 ``` ``` 2492 then show ?thesis ``` ``` 2493 using fps_radical_power_nth[of r "Suc k" a, OF r0] by simp ``` ``` 2494 next ``` ``` 2495 case (Suc n1) ``` ``` 2496 then have "n \<noteq> 0" by simp ``` ``` 2497 let ?Pnk = "natpermute n (k + 1)" ``` ``` 2498 let ?Pnkn = "{xs \<in> ?Pnk. n \<in> set xs}" ``` ``` 2499 let ?Pnknn = "{xs \<in> ?Pnk. n \<notin> set xs}" ``` ``` 2500 have eq: "?Pnkn \<union> ?Pnknn = ?Pnk" by blast ``` ``` 2501 have d: "?Pnkn \<inter> ?Pnknn = {}" by blast ``` ``` 2502 have f: "finite ?Pnkn" "finite ?Pnknn" ``` ``` 2503 using finite_Un[of ?Pnkn ?Pnknn, unfolded eq] ``` ``` 2504 by (metis natpermute_finite)+ ``` ``` 2505 let ?f = "\<lambda>v. \<Prod>j\<in>{0..k}. ?r \$ v ! j" ``` ``` 2506 have "setsum ?f ?Pnkn = setsum (\<lambda>v. ?r \$ n * r (Suc k) (a \$ 0) ^ k) ?Pnkn" ``` ``` 2507 proof (rule setsum.cong) ``` ``` 2508 fix v assume v: "v \<in> {xs \<in> natpermute n (k + 1). n \<in> set xs}" ``` ``` 2509 let ?ths = "(\<Prod>j\<in>{0..k}. fps_radical r (Suc k) a \$ v ! j) = ``` ``` 2510 fps_radical r (Suc k) a \$ n * r (Suc k) (a \$ 0) ^ k" ``` ``` 2511 from v obtain i where i: "i \<in> {0..k}" "v = replicate (k+1) 0 [i:= n]" ``` ``` 2512 unfolding natpermute_contain_maximal by auto ``` ``` 2513 have "(\<Prod>j\<in>{0..k}. fps_radical r (Suc k) a \$ v ! j) = ``` ``` 2514 (\<Prod>j\<in>{0..k}. if j = i then fps_radical r (Suc k) a \$ n else r (Suc k) (a\$0))" ``` ``` 2515 apply (rule setprod.cong, simp) ``` ``` 2516 using i r0 ``` ``` 2517 apply (simp del: replicate.simps) ``` ``` 2518 done ``` ``` 2519 also have "\<dots> = (fps_radical r (Suc k) a \$ n) * r (Suc k) (a\$0) ^ k" ``` ``` 2520 using i r0 by (simp add: setprod_gen_delta) ``` ``` 2521 finally show ?ths . ``` ``` 2522 qed rule ``` ``` 2523 then have "setsum ?f ?Pnkn = of_nat (k+1) * ?r \$ n * r (Suc k) (a \$ 0) ^ k" ``` ``` 2524 by (simp add: natpermute_max_card[OF \<open>n \<noteq> 0\<close>, simplified]) ``` ``` 2525 also have "\<dots> = a\$n - setsum ?f ?Pnknn" ``` ``` 2526 unfolding Suc using r00 a0 by (simp add: field_simps fps_radical_def del: of_nat_Suc) ``` ``` 2527 finally have fn: "setsum ?f ?Pnkn = a\$n - setsum ?f ?Pnknn" . ``` ``` 2528 have "(?r ^ Suc k)\$n = setsum ?f ?Pnkn + setsum ?f ?Pnknn" ``` ``` 2529 unfolding fps_power_nth_Suc setsum.union_disjoint[OF f d, unfolded eq] .. ``` ``` 2530 also have "\<dots> = a\$n" unfolding fn by simp ``` ``` 2531 finally show ?thesis . ``` ``` 2532 qed ``` ``` 2533 qed ``` ``` 2534 then show ?thesis using r0 by (simp add: fps_eq_iff) ``` ``` 2535 qed ``` ``` 2536 show ?lhs if ?rhs ``` ``` 2537 proof - ``` ``` 2538 from that have "((fps_radical r (Suc k) a) ^ (Suc k))\$0 = a\$0" ``` ``` 2539 by simp ``` ``` 2540 then show ?thesis ``` ``` 2541 unfolding fps_power_nth_Suc ``` ``` 2542 by (simp add: setprod_constant del: replicate.simps) ``` ``` 2543 qed ``` ``` 2544 qed ``` ``` 2545 ``` ``` 2546 (* ``` ``` 2547 lemma power_radical: ``` ``` 2548 fixes a:: "'a::field_char_0 fps" ``` ``` 2549 assumes r0: "(r (Suc k) (a\$0)) ^ Suc k = a\$0" and a0: "a\$0 \<noteq> 0" ``` ``` 2550 shows "(fps_radical r (Suc k) a) ^ (Suc k) = a" ``` ``` 2551 proof- ``` ``` 2552 let ?r = "fps_radical r (Suc k) a" ``` ``` 2553 from a0 r0 have r00: "r (Suc k) (a\$0) \<noteq> 0" by auto ``` ``` 2554 {fix z have "?r ^ Suc k \$ z = a\$z" ``` ``` 2555 proof(induct z rule: nat_less_induct) ``` ``` 2556 fix n assume H: "\<forall>m<n. ?r ^ Suc k \$ m = a\$m" ``` ``` 2557 {assume "n = 0" then have "?r ^ Suc k \$ n = a \$n" ``` ``` 2558 using fps_radical_power_nth[of r "Suc k" a, OF r0] by simp} ``` ``` 2559 moreover ``` ``` 2560 {fix n1 assume n1: "n = Suc n1" ``` ``` 2561 have fK: "finite {0..k}" by simp ``` ``` 2562 have nz: "n \<noteq> 0" using n1 by arith ``` ``` 2563 let ?Pnk = "natpermute n (k + 1)" ``` ``` 2564 let ?Pnkn = "{xs \<in> ?Pnk. n \<in> set xs}" ``` ``` 2565 let ?Pnknn = "{xs \<in> ?Pnk. n \<notin> set xs}" ``` ``` 2566 have eq: "?Pnkn \<union> ?Pnknn = ?Pnk" by blast ``` ``` 2567 have d: "?Pnkn \<inter> ?Pnknn = {}" by blast ``` ``` 2568 have f: "finite ?Pnkn" "finite ?Pnknn" ``` ``` 2569 using finite_Un[of ?Pnkn ?Pnknn, unfolded eq] ``` ``` 2570 by (metis natpermute_finite)+ ``` ``` 2571 let ?f = "\<lambda>v. \<Prod>j\<in>{0..k}. ?r \$ v ! j" ``` ``` 2572 have "setsum ?f ?Pnkn = setsum (\<lambda>v. ?r \$ n * r (Suc k) (a \$ 0) ^ k) ?Pnkn" ``` ``` 2573 proof(rule setsum.cong2) ``` ``` 2574 fix v assume v: "v \<in> {xs \<in> natpermute n (k + 1). n \<in> set xs}" ``` ``` 2575 let ?ths = "(\<Prod>j\<in>{0..k}. fps_radical r (Suc k) a \$ v ! j) = fps_radical r (Suc k) a \$ n * r (Suc k) (a \$ 0) ^ k" ``` ``` 2576 from v obtain i where i: "i \<in> {0..k}" "v = replicate (k+1) 0 [i:= n]" ``` ``` 2577 unfolding natpermute_contain_maximal by auto ``` ``` 2578 have "(\<Prod>j\<in>{0..k}. fps_radical r (Suc k) a \$ v ! j) = (\<Prod>j\<in>{0..k}. if j = i then fps_radical r (Suc k) a \$ n else r (Suc k) (a\$0))" ``` ``` 2579 apply (rule setprod.cong, simp) ``` ``` 2580 using i r0 by (simp del: replicate.simps) ``` ``` 2581 also have "\<dots> = (fps_radical r (Suc k) a \$ n) * r (Suc k) (a\$0) ^ k" ``` ``` 2582 unfolding setprod_gen_delta[OF fK] using i r0 by simp ``` ``` 2583 finally show ?ths . ``` ``` 2584 qed ``` ``` 2585 then have "setsum ?f ?Pnkn = of_nat (k+1) * ?r \$ n * r (Suc k) (a \$ 0) ^ k" ``` ``` 2586 by (simp add: natpermute_max_card[OF nz, simplified]) ``` ``` 2587 also have "\<dots> = a\$n - setsum ?f ?Pnknn" ``` ``` 2588 unfolding n1 using r00 a0 by (simp add: field_simps fps_radical_def del: of_nat_Suc ) ``` ``` 2589 finally have fn: "setsum ?f ?Pnkn = a\$n - setsum ?f ?Pnknn" . ``` ``` 2590 have "(?r ^ Suc k)\$n = setsum ?f ?Pnkn + setsum ?f ?Pnknn" ``` ``` 2591 unfolding fps_power_nth_Suc setsum.union_disjoint[OF f d, unfolded eq] .. ``` ``` 2592 also have "\<dots> = a\$n" unfolding fn by simp ``` ``` 2593 finally have "?r ^ Suc k \$ n = a \$n" .} ``` ``` 2594 ultimately show "?r ^ Suc k \$ n = a \$n" by (cases n, auto) ``` ``` 2595 qed } ``` ``` 2596 then show ?thesis by (simp add: fps_eq_iff) ``` ``` 2597 qed ``` ``` 2598 ``` ``` 2599 *) ``` ``` 2600 lemma eq_divide_imp': ``` ``` 2601 fixes c :: "'a::field" ``` ``` 2602 shows "c \<noteq> 0 \<Longrightarrow> a * c = b \<Longrightarrow> a = b / c" ``` ``` 2603 by (simp add: field_simps) ``` ``` 2604 ``` ``` 2605 lemma radical_unique: ``` ``` 2606 assumes r0: "(r (Suc k) (b\$0)) ^ Suc k = b\$0" ``` ``` 2607 and a0: "r (Suc k) (b\$0 ::'a::field_char_0) = a\$0" ``` ``` 2608 and b0: "b\$0 \<noteq> 0" ``` ``` 2609 shows "a^(Suc k) = b \<longleftrightarrow> a = fps_radical r (Suc k) b" ``` ``` 2610 (is "?lhs \<longleftrightarrow> ?rhs" is "_ \<longleftrightarrow> a = ?r") ``` ``` 2611 proof ``` ``` 2612 show ?lhs if ?rhs ``` ``` 2613 using that using power_radical[OF b0, of r k, unfolded r0] by simp ``` ``` 2614 show ?rhs if ?lhs ``` ``` 2615 proof - ``` ``` 2616 have r00: "r (Suc k) (b\$0) \<noteq> 0" using b0 r0 by auto ``` ``` 2617 have ceq: "card {0..k} = Suc k" by simp ``` ``` 2618 from a0 have a0r0: "a\$0 = ?r\$0" by simp ``` ``` 2619 have "a \$ n = ?r \$ n" for n ``` ``` 2620 proof (induct n rule: nat_less_induct) ``` ``` 2621 fix n ``` ``` 2622 assume h: "\<forall>m<n. a\$m = ?r \$m" ``` ``` 2623 show "a\$n = ?r \$ n" ``` ``` 2624 proof (cases n) ``` ``` 2625 case 0 ``` ``` 2626 then show ?thesis using a0 by simp ``` ``` 2627 next ``` ``` 2628 case (Suc n1) ``` ``` 2629 have fK: "finite {0..k}" by simp ``` ``` 2630 have nz: "n \<noteq> 0" using Suc by simp ``` ``` 2631 let ?Pnk = "natpermute n (Suc k)" ``` ``` 2632 let ?Pnkn = "{xs \<in> ?Pnk. n \<in> set xs}" ``` ``` 2633 let ?Pnknn = "{xs \<in> ?Pnk. n \<notin> set xs}" ``` ``` 2634 have eq: "?Pnkn \<union> ?Pnknn = ?Pnk" by blast ``` ``` 2635 have d: "?Pnkn \<inter> ?Pnknn = {}" by blast ``` ``` 2636 have f: "finite ?Pnkn" "finite ?Pnknn" ``` ``` 2637 using finite_Un[of ?Pnkn ?Pnknn, unfolded eq] ``` ``` 2638 by (metis natpermute_finite)+ ``` ``` 2639 let ?f = "\<lambda>v. \<Prod>j\<in>{0..k}. ?r \$ v ! j" ``` ``` 2640 let ?g = "\<lambda>v. \<Prod>j\<in>{0..k}. a \$ v ! j" ``` ``` 2641 have "setsum ?g ?Pnkn = setsum (\<lambda>v. a \$ n * (?r\$0)^k) ?Pnkn" ``` ``` 2642 proof (rule setsum.cong) ``` ``` 2643 fix v ``` ``` 2644 assume v: "v \<in> {xs \<in> natpermute n (Suc k). n \<in> set xs}" ``` ``` 2645 let ?ths = "(\<Prod>j\<in>{0..k}. a \$ v ! j) = a \$ n * (?r\$0)^k" ``` ``` 2646 from v obtain i where i: "i \<in> {0..k}" "v = replicate (k+1) 0 [i:= n]" ``` ``` 2647 unfolding Suc_eq_plus1 natpermute_contain_maximal ``` ``` 2648 by (auto simp del: replicate.simps) ``` ``` 2649 have "(\<Prod>j\<in>{0..k}. a \$ v ! j) = (\<Prod>j\<in>{0..k}. if j = i then a \$ n else r (Suc k) (b\$0))" ``` ``` 2650 apply (rule setprod.cong, simp) ``` ``` 2651 using i a0 ``` ``` 2652 apply (simp del: replicate.simps) ``` ``` 2653 done ``` ``` 2654 also have "\<dots> = a \$ n * (?r \$ 0)^k" ``` ``` 2655 using i by (simp add: setprod_gen_delta) ``` ``` 2656 finally show ?ths . ``` ``` 2657 qed rule ``` ``` 2658 then have th0: "setsum ?g ?Pnkn = of_nat (k+1) * a \$ n * (?r \$ 0)^k" ``` ``` 2659 by (simp add: natpermute_max_card[OF nz, simplified]) ``` ``` 2660 have th1: "setsum ?g ?Pnknn = setsum ?f ?Pnknn" ``` ``` 2661 proof (rule setsum.cong, rule refl, rule setprod.cong, simp) ``` ``` 2662 fix xs i ``` ``` 2663 assume xs: "xs \<in> ?Pnknn" and i: "i \<in> {0..k}" ``` ``` 2664 have False if c: "n \<le> xs ! i" ``` ``` 2665 proof - ``` ``` 2666 from xs i have "xs ! i \<noteq> n" ``` ``` 2667 by (auto simp add: in_set_conv_nth natpermute_def) ``` ``` 2668 with c have c': "n < xs!i" by arith ``` ``` 2669 have fths: "finite {0 ..< i}" "finite {i}" "finite {i+1..<Suc k}" ``` ``` 2670 by simp_all ``` ``` 2671 have d: "{0 ..< i} \<inter> ({i} \<union> {i+1 ..< Suc k}) = {}" "{i} \<inter> {i+1..< Suc k} = {}" ``` ``` 2672 by auto ``` ``` 2673 have eqs: "{0..<Suc k} = {0 ..< i} \<union> ({i} \<union> {i+1 ..< Suc k})" ``` ``` 2674 using i by auto ``` ``` 2675 from xs have "n = listsum xs" ``` ``` 2676 by (simp add: natpermute_def) ``` ``` 2677 also have "\<dots> = setsum (nth xs) {0..<Suc k}" ``` ``` 2678 using xs by (simp add: natpermute_def listsum_setsum_nth) ``` ``` 2679 also have "\<dots> = xs!i + setsum (nth xs) {0..<i} + setsum (nth xs) {i+1..<Suc k}" ``` ``` 2680 unfolding eqs setsum.union_disjoint[OF fths(1) finite_UnI[OF fths(2,3)] d(1)] ``` ``` 2681 unfolding setsum.union_disjoint[OF fths(2) fths(3) d(2)] ``` ``` 2682 by simp ``` ``` 2683 finally show ?thesis using c' by simp ``` ``` 2684 qed ``` ``` 2685 then have thn: "xs!i < n" by presburger ``` ``` 2686 from h[rule_format, OF thn] show "a\$(xs !i) = ?r\$(xs!i)" . ``` ``` 2687 qed ``` ``` 2688 have th00: "\<And>x::'a. of_nat (Suc k) * (x * inverse (of_nat (Suc k))) = x" ``` ``` 2689 by (simp add: field_simps del: of_nat_Suc) ``` ``` 2690 from \<open>?lhs\<close> have "b\$n = a^Suc k \$ n" ``` ``` 2691 by (simp add: fps_eq_iff) ``` ``` 2692 also have "a ^ Suc k\$n = setsum ?g ?Pnkn + setsum ?g ?Pnknn" ``` ``` 2693 unfolding fps_power_nth_Suc ``` ``` 2694 using setsum.union_disjoint[OF f d, unfolded Suc_eq_plus1[symmetric], ``` ``` 2695 unfolded eq, of ?g] by simp ``` ``` 2696 also have "\<dots> = of_nat (k+1) * a \$ n * (?r \$ 0)^k + setsum ?f ?Pnknn" ``` ``` 2697 unfolding th0 th1 .. ``` ``` 2698 finally have "of_nat (k+1) * a \$ n * (?r \$ 0)^k = b\$n - setsum ?f ?Pnknn" ``` ``` 2699 by simp ``` ``` 2700 then have "a\$n = (b\$n - setsum ?f ?Pnknn) / (of_nat (k+1) * (?r \$ 0)^k)" ``` ``` 2701 apply - ``` ``` 2702 apply (rule eq_divide_imp') ``` ``` 2703 using r00 ``` ``` 2704 apply (simp del: of_nat_Suc) ``` ``` 2705 apply (simp add: ac_simps) ``` ``` 2706 done ``` ``` 2707 then show ?thesis ``` ``` 2708 apply (simp del: of_nat_Suc) ``` ``` 2709 unfolding fps_radical_def Suc ``` ``` 2710 apply (simp add: field_simps Suc th00 del: of_nat_Suc) ``` ``` 2711 done ``` ``` 2712 qed ``` ``` 2713 qed ``` ``` 2714 then show ?rhs by (simp add: fps_eq_iff) ``` ``` 2715 qed ``` ``` 2716 qed ``` ``` 2717 ``` ``` 2718 ``` ``` 2719 lemma radical_power: ``` ``` 2720 assumes r0: "r (Suc k) ((a\$0) ^ Suc k) = a\$0" ``` ``` 2721 and a0: "(a\$0 :: 'a::field_char_0) \<noteq> 0" ``` ``` 2722 shows "(fps_radical r (Suc k) (a ^ Suc k)) = a" ``` ``` 2723 proof - ``` ``` 2724 let ?ak = "a^ Suc k" ``` ``` 2725 have ak0: "?ak \$ 0 = (a\$0) ^ Suc k" ``` ``` 2726 by (simp add: fps_nth_power_0 del: power_Suc) ``` ``` 2727 from r0 have th0: "r (Suc k) (a ^ Suc k \$ 0) ^ Suc k = a ^ Suc k \$ 0" ``` ``` 2728 using ak0 by auto ``` ``` 2729 from r0 ak0 have th1: "r (Suc k) (a ^ Suc k \$ 0) = a \$ 0" ``` ``` 2730 by auto ``` ``` 2731 from ak0 a0 have ak00: "?ak \$ 0 \<noteq>0 " ``` ``` 2732 by auto ``` ``` 2733 from radical_unique[of r k ?ak a, OF th0 th1 ak00] show ?thesis ``` ``` 2734 by metis ``` ``` 2735 qed ``` ``` 2736 ``` ``` 2737 lemma fps_deriv_radical: ``` ``` 2738 fixes a :: "'a::field_char_0 fps" ``` ``` 2739 assumes r0: "(r (Suc k) (a\$0)) ^ Suc k = a\$0" ``` ``` 2740 and a0: "a\$0 \<noteq> 0" ``` ``` 2741 shows "fps_deriv (fps_radical r (Suc k) a) = ``` ``` 2742 fps_deriv a / (fps_const (of_nat (Suc k)) * (fps_radical r (Suc k) a) ^ k)" ``` ``` 2743 proof - ``` ``` 2744 let ?r = "fps_radical r (Suc k) a" ``` ``` 2745 let ?w = "(fps_const (of_nat (Suc k)) * ?r ^ k)" ``` ``` 2746 from a0 r0 have r0': "r (Suc k) (a\$0) \<noteq> 0" ``` ``` 2747 by auto ``` ``` 2748 from r0' have w0: "?w \$ 0 \<noteq> 0" ``` ``` 2749 by (simp del: of_nat_Suc) ``` ``` 2750 note th0 = inverse_mult_eq_1[OF w0] ``` ``` 2751 let ?iw = "inverse ?w" ``` ``` 2752 from iffD1[OF power_radical[of a r], OF a0 r0] ``` ``` 2753 have "fps_deriv (?r ^ Suc k) = fps_deriv a" ``` ``` 2754 by simp ``` ``` 2755 then have "fps_deriv ?r * ?w = fps_deriv a" ``` ``` 2756 by (simp add: fps_deriv_power ac_simps del: power_Suc) ``` ``` 2757 then have "?iw * fps_deriv ?r * ?w = ?iw * fps_deriv a" ``` ``` 2758 by simp ``` ``` 2759 with a0 r0 have "fps_deriv ?r * (?iw * ?w) = fps_deriv a / ?w" ``` ``` 2760 by (subst fps_divide_unit) (auto simp del: of_nat_Suc) ``` ``` 2761 then show ?thesis unfolding th0 by simp ``` ``` 2762 qed ``` ``` 2763 ``` ``` 2764 lemma radical_mult_distrib: ``` ``` 2765 fixes a :: "'a::field_char_0 fps" ``` ``` 2766 assumes k: "k > 0" ``` ``` 2767 and ra0: "r k (a \$ 0) ^ k = a \$ 0" ``` ``` 2768 and rb0: "r k (b \$ 0) ^ k = b \$ 0" ``` ``` 2769 and a0: "a \$ 0 \<noteq> 0" ``` ``` 2770 and b0: "b \$ 0 \<noteq> 0" ``` ``` 2771 shows "r k ((a * b) \$ 0) = r k (a \$ 0) * r k (b \$ 0) \<longleftrightarrow> ``` ``` 2772 fps_radical r k (a * b) = fps_radical r k a * fps_radical r k b" ``` ``` 2773 (is "?lhs \<longleftrightarrow> ?rhs") ``` ``` 2774 proof ``` ``` 2775 show ?rhs if r0': ?lhs ``` ``` 2776 proof - ``` ``` 2777 from r0' have r0: "(r k ((a * b) \$ 0)) ^ k = (a * b) \$ 0" ``` ``` 2778 by (simp add: fps_mult_nth ra0 rb0 power_mult_distrib) ``` ``` 2779 show ?thesis ``` ``` 2780 proof (cases k) ``` ``` 2781 case 0 ``` ``` 2782 then show ?thesis using r0' by simp ``` ``` 2783 next ``` ``` 2784 case (Suc h) ``` ``` 2785 let ?ra = "fps_radical r (Suc h) a" ``` ``` 2786 let ?rb = "fps_radical r (Suc h) b" ``` ``` 2787 have th0: "r (Suc h) ((a * b) \$ 0) = (fps_radical r (Suc h) a * fps_radical r (Suc h) b) \$ 0" ``` ``` 2788 using r0' Suc by (simp add: fps_mult_nth) ``` ``` 2789 have ab0: "(a*b) \$ 0 \<noteq> 0" ``` ``` 2790 using a0 b0 by (simp add: fps_mult_nth) ``` ``` 2791 from radical_unique[of r h "a*b" "fps_radical r (Suc h) a * fps_radical r (Suc h) b", OF r0[unfolded Suc] th0 ab0, symmetric] ``` ``` 2792 iffD1[OF power_radical[of _ r], OF a0 ra0[unfolded Suc]] iffD1[OF power_radical[of _ r], OF b0 rb0[unfolded Suc]] Suc r0' ``` ``` 2793 show ?thesis ``` ``` 2794 by (auto simp add: power_mult_distrib simp del: power_Suc) ``` ``` 2795 qed ``` ``` 2796 qed ``` ``` 2797 show ?lhs if ?rhs ``` ``` 2798 proof - ``` ``` 2799 from that have "(fps_radical r k (a * b)) \$ 0 = (fps_radical r k a * fps_radical r k b) \$ 0" ``` ``` 2800 by simp ``` ``` 2801 then show ?thesis ``` ``` 2802 using k by (simp add: fps_mult_nth) ``` ``` 2803 qed ``` ``` 2804 qed ``` ``` 2805 ``` ``` 2806 (* ``` ``` 2807 lemma radical_mult_distrib: ``` ``` 2808 fixes a:: "'a::field_char_0 fps" ``` ``` 2809 assumes ``` ``` 2810 ra0: "r k (a \$ 0) ^ k = a \$ 0" ``` ``` 2811 and rb0: "r k (b \$ 0) ^ k = b \$ 0" ``` ``` 2812 and r0': "r k ((a * b) \$ 0) = r k (a \$ 0) * r k (b \$ 0)" ``` ``` 2813 and a0: "a\$0 \<noteq> 0" ``` ``` 2814 and b0: "b\$0 \<noteq> 0" ``` ``` 2815 shows "fps_radical r (k) (a*b) = fps_radical r (k) a * fps_radical r (k) (b)" ``` ``` 2816 proof- ``` ``` 2817 from r0' have r0: "(r (k) ((a*b)\$0)) ^ k = (a*b)\$0" ``` ``` 2818 by (simp add: fps_mult_nth ra0 rb0 power_mult_distrib) ``` ``` 2819 {assume "k=0" then have ?thesis by simp} ``` ``` 2820 moreover ``` ``` 2821 {fix h assume k: "k = Suc h" ``` ``` 2822 let ?ra = "fps_radical r (Suc h) a" ``` ``` 2823 let ?rb = "fps_radical r (Suc h) b" ``` ``` 2824 have th0: "r (Suc h) ((a * b) \$ 0) = (fps_radical r (Suc h) a * fps_radical r (Suc h) b) \$ 0" ``` ``` 2825 using r0' k by (simp add: fps_mult_nth) ``` ``` 2826 have ab0: "(a*b) \$ 0 \<noteq> 0" using a0 b0 by (simp add: fps_mult_nth) ``` ``` 2827 from radical_unique[of r h "a*b" "fps_radical r (Suc h) a * fps_radical r (Suc h) b", OF r0[unfolded k] th0 ab0, symmetric] ``` ``` 2828 power_radical[of r, OF ra0[unfolded k] a0] power_radical[of r, OF rb0[unfolded k] b0] k ``` ``` 2829 have ?thesis by (auto simp add: power_mult_distrib simp del: power_Suc)} ``` ``` 2830 ultimately show ?thesis by (cases k, auto) ``` ``` 2831 qed ``` ``` 2832 *) ``` ``` 2833 ``` ``` 2834 lemma fps_divide_1 [simp]: "(a :: 'a::field fps) / 1 = a" ``` ``` 2835 by (fact divide_1) ``` ``` 2836 ``` ``` 2837 lemma radical_divide: ``` ``` 2838 fixes a :: "'a::field_char_0 fps" ``` ``` 2839 assumes kp: "k > 0" ``` ``` 2840 and ra0: "(r k (a \$ 0)) ^ k = a \$ 0" ``` ``` 2841 and rb0: "(r k (b \$ 0)) ^ k = b \$ 0" ``` ``` 2842 and a0: "a\$0 \<noteq> 0" ``` ``` 2843 and b0: "b\$0 \<noteq> 0" ``` ``` 2844 shows "r k ((a \$ 0) / (b\$0)) = r k (a\$0) / r k (b \$ 0) \<longleftrightarrow> ``` ``` 2845 fps_radical r k (a/b) = fps_radical r k a / fps_radical r k b" ``` ``` 2846 (is "?lhs = ?rhs") ``` ``` 2847 proof ``` ``` 2848 let ?r = "fps_radical r k" ``` ``` 2849 from kp obtain h where k: "k = Suc h" ``` ``` 2850 by (cases k) auto ``` ``` 2851 have ra0': "r k (a\$0) \<noteq> 0" using a0 ra0 k by auto ``` ``` 2852 have rb0': "r k (b\$0) \<noteq> 0" using b0 rb0 k by auto ``` ``` 2853 ``` ``` 2854 show ?lhs if ?rhs ``` ``` 2855 proof - ``` ``` 2856 from that have "?r (a/b) \$ 0 = (?r a / ?r b)\$0" ``` ``` 2857 by simp ``` ``` 2858 then show ?thesis ``` ``` 2859 using k a0 b0 rb0' by (simp add: fps_divide_unit fps_mult_nth fps_inverse_def divide_inverse) ``` ``` 2860 qed ``` ``` 2861 show ?rhs if ?lhs ``` ``` 2862 proof - ``` ``` 2863 from a0 b0 have ab0[simp]: "(a/b)\$0 = a\$0 / b\$0" ``` ``` 2864 by (simp add: fps_divide_def fps_mult_nth divide_inverse fps_inverse_def) ``` ``` 2865 have th0: "r k ((a/b)\$0) ^ k = (a/b)\$0" ``` ``` 2866 by (simp add: \<open>?lhs\<close> power_divide ra0 rb0) ``` ``` 2867 from a0 b0 ra0' rb0' kp \<open>?lhs\<close> ``` ``` 2868 have th1: "r k ((a / b) \$ 0) = (fps_radical r k a / fps_radical r k b) \$ 0" ``` ``` 2869 by (simp add: fps_divide_unit fps_mult_nth fps_inverse_def divide_inverse) ``` ``` 2870 from a0 b0 ra0' rb0' kp have ab0': "(a / b) \$ 0 \<noteq> 0" ``` ``` 2871 by (simp add: fps_divide_unit fps_mult_nth fps_inverse_def nonzero_imp_inverse_nonzero) ``` ``` 2872 note tha[simp] = iffD1[OF power_radical[where r=r and k=h], OF a0 ra0[unfolded k], unfolded k[symmetric]] ``` ``` 2873 note thb[simp] = iffD1[OF power_radical[where r=r and k=h], OF b0 rb0[unfolded k], unfolded k[symmetric]] ``` ``` 2874 from b0 rb0' have th2: "(?r a / ?r b)^k = a/b" ``` ``` 2875 by (simp add: fps_divide_unit power_mult_distrib fps_inverse_power[symmetric]) ``` ``` 2876 ``` ``` 2877 from iffD1[OF radical_unique[where r=r and a="?r a / ?r b" and b="a/b" and k=h], symmetric, unfolded k[symmetric], OF th0 th1 ab0' th2] ``` ``` 2878 show ?thesis . ``` ``` 2879 qed ``` ``` 2880 qed ``` ``` 2881 ``` ``` 2882 lemma radical_inverse: ``` ``` 2883 fixes a :: "'a::field_char_0 fps" ``` ``` 2884 assumes k: "k > 0" ``` ``` 2885 and ra0: "r k (a \$ 0) ^ k = a \$ 0" ``` ``` 2886 and r1: "(r k 1)^k = 1" ``` ``` 2887 and a0: "a\$0 \<noteq> 0" ``` ``` 2888 shows "r k (inverse (a \$ 0)) = r k 1 / (r k (a \$ 0)) \<longleftrightarrow> ``` ``` 2889 fps_radical r k (inverse a) = fps_radical r k 1 / fps_radical r k a" ``` ``` 2890 using radical_divide[where k=k and r=r and a=1 and b=a, OF k ] ra0 r1 a0 ``` ``` 2891 by (simp add: divide_inverse fps_divide_def) ``` ``` 2892 ``` ``` 2893 ``` ``` 2894 subsection \<open>Derivative of composition\<close> ``` ``` 2895 ``` ``` 2896 lemma fps_compose_deriv: ``` ``` 2897 fixes a :: "'a::idom fps" ``` ``` 2898 assumes b0: "b\$0 = 0" ``` ``` 2899 shows "fps_deriv (a oo b) = ((fps_deriv a) oo b) * fps_deriv b" ``` ``` 2900 proof - ``` ``` 2901 have "(fps_deriv (a oo b))\$n = (((fps_deriv a) oo b) * (fps_deriv b)) \$n" for n ``` ``` 2902 proof - ``` ``` 2903 have "(fps_deriv (a oo b))\$n = setsum (\<lambda>i. a \$ i * (fps_deriv (b^i))\$n) {0.. Suc n}" ``` ``` 2904 by (simp add: fps_compose_def field_simps setsum_right_distrib del: of_nat_Suc) ``` ``` 2905 also have "\<dots> = setsum (\<lambda>i. a\$i * ((fps_const (of_nat i)) * (fps_deriv b * (b^(i - 1))))\$n) {0.. Suc n}" ``` ``` 2906 by (simp add: field_simps fps_deriv_power del: fps_mult_left_const_nth of_nat_Suc) ``` ``` 2907 also have "\<dots> = setsum (\<lambda>i. of_nat i * a\$i * (((b^(i - 1)) * fps_deriv b))\$n) {0.. Suc n}" ``` ``` 2908 unfolding fps_mult_left_const_nth by (simp add: field_simps) ``` ``` 2909 also have "\<dots> = setsum (\<lambda>i. of_nat i * a\$i * (setsum (\<lambda>j. (b^ (i - 1))\$j * (fps_deriv b)\$(n - j)) {0..n})) {0.. Suc n}" ``` ``` 2910 unfolding fps_mult_nth .. ``` ``` 2911 also have "\<dots> = setsum (\<lambda>i. of_nat i * a\$i * (setsum (\<lambda>j. (b^ (i - 1))\$j * (fps_deriv b)\$(n - j)) {0..n})) {1.. Suc n}" ``` ``` 2912 apply (rule setsum.mono_neutral_right) ``` ``` 2913 apply (auto simp add: mult_delta_left setsum.delta not_le) ``` ``` 2914 done ``` ``` 2915 also have "\<dots> = setsum (\<lambda>i. of_nat (i + 1) * a\$(i+1) * (setsum (\<lambda>j. (b^ i)\$j * of_nat (n - j + 1) * b\$(n - j + 1)) {0..n})) {0.. n}" ``` ``` 2916 unfolding fps_deriv_nth ``` ``` 2917 by (rule setsum.reindex_cong [of Suc]) (auto simp add: mult.assoc) ``` ``` 2918 finally have th0: "(fps_deriv (a oo b))\$n = ``` ``` 2919 setsum (\<lambda>i. of_nat (i + 1) * a\$(i+1) * (setsum (\<lambda>j. (b^ i)\$j * of_nat (n - j + 1) * b\$(n - j + 1)) {0..n})) {0.. n}" . ``` ``` 2920 ``` ``` 2921 have "(((fps_deriv a) oo b) * (fps_deriv b))\$n = setsum (\<lambda>i. (fps_deriv b)\$ (n - i) * ((fps_deriv a) oo b)\$i) {0..n}" ``` ``` 2922 unfolding fps_mult_nth by (simp add: ac_simps) ``` ``` 2923 also have "\<dots> = setsum (\<lambda>i. setsum (\<lambda>j. of_nat (n - i +1) * b\$(n - i + 1) * of_nat (j + 1) * a\$(j+1) * (b^j)\$i) {0..n}) {0..n}" ``` ``` 2924 unfolding fps_deriv_nth fps_compose_nth setsum_right_distrib mult.assoc ``` ``` 2925 apply (rule setsum.cong) ``` ``` 2926 apply (rule refl) ``` ``` 2927 apply (rule setsum.mono_neutral_left) ``` ``` 2928 apply (simp_all add: subset_eq) ``` ``` 2929 apply clarify ``` ``` 2930 apply (subgoal_tac "b^i\$x = 0") ``` ``` 2931 apply simp ``` ``` 2932 apply (rule startsby_zero_power_prefix[OF b0, rule_format]) ``` ``` 2933 apply simp ``` ``` 2934 done ``` ``` 2935 also have "\<dots> = setsum (\<lambda>i. of_nat (i + 1) * a\$(i+1) * (setsum (\<lambda>j. (b^ i)\$j * of_nat (n - j + 1) * b\$(n - j + 1)) {0..n})) {0.. n}" ``` ``` 2936 unfolding setsum_right_distrib ``` ``` 2937 apply (subst setsum.commute) ``` ``` 2938 apply (rule setsum.cong, rule refl)+ ``` ``` 2939 apply simp ``` ``` 2940 done ``` ``` 2941 finally show ?thesis ``` ``` 2942 unfolding th0 by simp ``` ``` 2943 qed ``` ``` 2944 then show ?thesis by (simp add: fps_eq_iff) ``` ``` 2945 qed ``` ``` 2946 ``` ``` 2947 lemma fps_mult_X_plus_1_nth: ``` ``` 2948 "((1+X)*a) \$n = (if n = 0 then (a\$n :: 'a::comm_ring_1) else a\$n + a\$(n - 1))" ``` ``` 2949 proof (cases n) ``` ``` 2950 case 0 ``` ``` 2951 then show ?thesis ``` ``` 2952 by (simp add: fps_mult_nth) ``` ``` 2953 next ``` ``` 2954 case (Suc m) ``` ``` 2955 have "((1 + X)*a) \$ n = setsum (\<lambda>i. (1 + X) \$ i * a \$ (n - i)) {0..n}" ``` ``` 2956 by (simp add: fps_mult_nth) ``` ``` 2957 also have "\<dots> = setsum (\<lambda>i. (1+X)\$i * a\$(n-i)) {0.. 1}" ``` ``` 2958 unfolding Suc by (rule setsum.mono_neutral_right) auto ``` ``` 2959 also have "\<dots> = (if n = 0 then (a\$n :: 'a::comm_ring_1) else a\$n + a\$(n - 1))" ``` ``` 2960 by (simp add: Suc) ``` ``` 2961 finally show ?thesis . ``` ``` 2962 qed ``` ``` 2963 ``` ``` 2964 ``` ``` 2965 subsection \<open>Finite FPS (i.e. polynomials) and X\<close> ``` ``` 2966 ``` ``` 2967 lemma fps_poly_sum_X: ``` ``` 2968 assumes "\<forall>i > n. a\$i = (0::'a::comm_ring_1)" ``` ``` 2969 shows "a = setsum (\<lambda>i. fps_const (a\$i) * X^i) {0..n}" (is "a = ?r") ``` ``` 2970 proof - ``` ``` 2971 have "a\$i = ?r\$i" for i ``` ``` 2972 unfolding fps_setsum_nth fps_mult_left_const_nth X_power_nth ``` ``` 2973 by (simp add: mult_delta_right setsum.delta' assms) ``` ``` 2974 then show ?thesis ``` ``` 2975 unfolding fps_eq_iff by blast ``` ``` 2976 qed ``` ``` 2977 ``` ``` 2978 ``` ``` 2979 subsection \<open>Compositional inverses\<close> ``` ``` 2980 ``` ``` 2981 fun compinv :: "'a fps \<Rightarrow> nat \<Rightarrow> 'a::field" ``` ``` 2982 where ``` ``` 2983 "compinv a 0 = X\$0" ``` ``` 2984 | "compinv a (Suc n) = ``` ``` 2985 (X\$ Suc n - setsum (\<lambda>i. (compinv a i) * (a^i)\$Suc n) {0 .. n}) / (a\$1) ^ Suc n" ``` ``` 2986 ``` ``` 2987 definition "fps_inv a = Abs_fps (compinv a)" ``` ``` 2988 ``` ``` 2989 lemma fps_inv: ``` ``` 2990 assumes a0: "a\$0 = 0" ``` ``` 2991 and a1: "a\$1 \<noteq> 0" ``` ``` 2992 shows "fps_inv a oo a = X" ``` ``` 2993 proof - ``` ``` 2994 let ?i = "fps_inv a oo a" ``` ``` 2995 have "?i \$n = X\$n" for n ``` ``` 2996 proof (induct n rule: nat_less_induct) ``` ``` 2997 fix n ``` ``` 2998 assume h: "\<forall>m<n. ?i\$m = X\$m" ``` ``` 2999 show "?i \$ n = X\$n" ``` ``` 3000 proof (cases n) ``` ``` 3001 case 0 ``` ``` 3002 then show ?thesis using a0 ``` ``` 3003 by (simp add: fps_compose_nth fps_inv_def) ``` ``` 3004 next ``` ``` 3005 case (Suc n1) ``` ``` 3006 have "?i \$ n = setsum (\<lambda>i. (fps_inv a \$ i) * (a^i)\$n) {0 .. n1} + fps_inv a \$ Suc n1 * (a \$ 1)^ Suc n1" ``` ``` 3007 by (simp only: fps_compose_nth) (simp add: Suc startsby_zero_power_nth_same [OF a0] del: power_Suc) ``` ``` 3008 also have "\<dots> = setsum (\<lambda>i. (fps_inv a \$ i) * (a^i)\$n) {0 .. n1} + ``` ``` 3009 (X\$ Suc n1 - setsum (\<lambda>i. (fps_inv a \$ i) * (a^i)\$n) {0 .. n1})" ``` ``` 3010 using a0 a1 Suc by (simp add: fps_inv_def) ``` ``` 3011 also have "\<dots> = X\$n" using Suc by simp ``` ``` 3012 finally show ?thesis . ``` ``` 3013 qed ``` ``` 3014 qed ``` ``` 3015 then show ?thesis ``` ``` 3016 by (simp add: fps_eq_iff) ``` ``` 3017 qed ``` ``` 3018 ``` ``` 3019 ``` ``` 3020 fun gcompinv :: "'a fps \<Rightarrow> 'a fps \<Rightarrow> nat \<Rightarrow> 'a::field" ``` ``` 3021 where ``` ``` 3022 "gcompinv b a 0 = b\$0" ``` ``` 3023 | "gcompinv b a (Suc n) = ``` ``` 3024 (b\$ Suc n - setsum (\<lambda>i. (gcompinv b a i) * (a^i)\$Suc n) {0 .. n}) / (a\$1) ^ Suc n" ``` ``` 3025 ``` ``` 3026 definition "fps_ginv b a = Abs_fps (gcompinv b a)" ``` ``` 3027 ``` ``` 3028 lemma fps_ginv: ``` ``` 3029 assumes a0: "a\$0 = 0" ``` ``` 3030 and a1: "a\$1 \<noteq> 0" ``` ``` 3031 shows "fps_ginv b a oo a = b" ``` ``` 3032 proof - ``` ``` 3033 let ?i = "fps_ginv b a oo a" ``` ``` 3034 have "?i \$n = b\$n" for n ``` ``` 3035 proof (induct n rule: nat_less_induct) ``` ``` 3036 fix n ``` ``` 3037 assume h: "\<forall>m<n. ?i\$m = b\$m" ``` ``` 3038 show "?i \$ n = b\$n" ``` ``` 3039 proof (cases n) ``` ``` 3040 case 0 ``` ``` 3041 then show ?thesis using a0 ``` ``` 3042 by (simp add: fps_compose_nth fps_ginv_def) ``` ``` 3043 next ``` ``` 3044 case (Suc n1) ``` ``` 3045 have "?i \$ n = setsum (\<lambda>i. (fps_ginv b a \$ i) * (a^i)\$n) {0 .. n1} + fps_ginv b a \$ Suc n1 * (a \$ 1)^ Suc n1" ``` ``` 3046 by (simp only: fps_compose_nth) (simp add: Suc startsby_zero_power_nth_same [OF a0] del: power_Suc) ``` ``` 3047 also have "\<dots> = setsum (\<lambda>i. (fps_ginv b a \$ i) * (a^i)\$n) {0 .. n1} + ``` ``` 3048 (b\$ Suc n1 - setsum (\<lambda>i. (fps_ginv b a \$ i) * (a^i)\$n) {0 .. n1})" ``` ``` 3049 using a0 a1 Suc by (simp add: fps_ginv_def) ``` ``` 3050 also have "\<dots> = b\$n" using Suc by simp ``` ``` 3051 finally show ?thesis . ``` ``` 3052 qed ``` ``` 3053 qed ``` ``` 3054 then show ?thesis ``` ``` 3055 by (simp add: fps_eq_iff) ``` ``` 3056 qed ``` ``` 3057 ``` ``` 3058 lemma fps_inv_ginv: "fps_inv = fps_ginv X" ``` ``` 3059 apply (auto simp add: fun_eq_iff fps_eq_iff fps_inv_def fps_ginv_def) ``` ``` 3060 apply (induct_tac n rule: nat_less_induct) ``` ``` 3061 apply auto ``` ``` 3062 apply (case_tac na) ``` ``` 3063 apply simp ``` ``` 3064 apply simp ``` ``` 3065 done ``` ``` 3066 ``` ``` 3067 lemma fps_compose_1[simp]: "1 oo a = 1" ``` ``` 3068 by (simp add: fps_eq_iff fps_compose_nth mult_delta_left setsum.delta) ``` ``` 3069 ``` ``` 3070 lemma fps_compose_0[simp]: "0 oo a = 0" ``` ``` 3071 by (simp add: fps_eq_iff fps_compose_nth) ``` ``` 3072 ``` ``` 3073 lemma fps_compose_0_right[simp]: "a oo 0 = fps_const (a \$ 0)" ``` ``` 3074 by (auto simp add: fps_eq_iff fps_compose_nth power_0_left setsum.neutral) ``` ``` 3075 ``` ``` 3076 lemma fps_compose_add_distrib: "(a + b) oo c = (a oo c) + (b oo c)" ``` ``` 3077 by (simp add: fps_eq_iff fps_compose_nth field_simps setsum.distrib) ``` ``` 3078 ``` ``` 3079 lemma fps_compose_setsum_distrib: "(setsum f S) oo a = setsum (\<lambda>i. f i oo a) S" ``` ``` 3080 proof (cases "finite S") ``` ``` 3081 case True ``` ``` 3082 show ?thesis ``` ``` 3083 proof (rule finite_induct[OF True]) ``` ``` 3084 show "setsum f {} oo a = (\<Sum>i\<in>{}. f i oo a)" ``` ``` 3085 by simp ``` ``` 3086 next ``` ``` 3087 fix x F ``` ``` 3088 assume fF: "finite F" ``` ``` 3089 and xF: "x \<notin> F" ``` ``` 3090 and h: "setsum f F oo a = setsum (\<lambda>i. f i oo a) F" ``` ``` 3091 show "setsum f (insert x F) oo a = setsum (\<lambda>i. f i oo a) (insert x F)" ``` ``` 3092 using fF xF h by (simp add: fps_compose_add_distrib) ``` ``` 3093 qed ``` ``` 3094 next ``` ``` 3095 case False ``` ``` 3096 then show ?thesis by simp ``` ``` 3097 qed ``` ``` 3098 ``` ``` 3099 lemma convolution_eq: ``` ``` 3100 "setsum (\<lambda>i. a (i :: nat) * b (n - i)) {0 .. n} = ``` ``` 3101 setsum (\<lambda>(i,j). a i * b j) {(i,j). i \<le> n \<and> j \<le> n \<and> i + j = n}" ``` ``` 3102 by (rule setsum.reindex_bij_witness[where i=fst and j="\<lambda>i. (i, n - i)"]) auto ``` ``` 3103 ``` ``` 3104 lemma product_composition_lemma: ``` ``` 3105 assumes c0: "c\$0 = (0::'a::idom)" ``` ``` 3106 and d0: "d\$0 = 0" ``` ``` 3107 shows "((a oo c) * (b oo d))\$n = ``` ``` 3108 setsum (\<lambda>(k,m). a\$k * b\$m * (c^k * d^m) \$ n) {(k,m). k + m \<le> n}" (is "?l = ?r") ``` ``` 3109 proof - ``` ``` 3110 let ?S = "{(k::nat, m::nat). k + m \<le> n}" ``` ``` 3111 have s: "?S \<subseteq> {0..n} \<times> {0..n}" by (auto simp add: subset_eq) ``` ``` 3112 have f: "finite {(k::nat, m::nat). k + m \<le> n}" ``` ``` 3113 apply (rule finite_subset[OF s]) ``` ``` 3114 apply auto ``` ``` 3115 done ``` ``` 3116 have "?r = setsum (\<lambda>i. setsum (\<lambda>(k,m). a\$k * (c^k)\$i * b\$m * (d^m) \$ (n - i)) {(k,m). k + m \<le> n}) {0..n}" ``` ``` 3117 apply (simp add: fps_mult_nth setsum_right_distrib) ``` ``` 3118 apply (subst setsum.commute) ``` ``` 3119 apply (rule setsum.cong) ``` ``` 3120 apply (auto simp add: field_simps) ``` ``` 3121 done ``` ``` 3122 also have "\<dots> = ?l" ``` ``` 3123 apply (simp add: fps_mult_nth fps_compose_nth setsum_product) ``` ``` 3124 apply (rule setsum.cong) ``` ``` 3125 apply (rule refl) ``` ``` 3126 apply (simp add: setsum.cartesian_product mult.assoc) ``` ``` 3127 apply (rule setsum.mono_neutral_right[OF f]) ``` ``` 3128 apply (simp add: subset_eq) ``` ``` 3129 apply presburger ``` ``` 3130 apply clarsimp ``` ``` 3131 apply (rule ccontr) ``` ``` 3132 apply (clarsimp simp add: not_le) ``` ``` 3133 apply (case_tac "x < aa") ``` ``` 3134 apply simp ``` ``` 3135 apply (frule_tac startsby_zero_power_prefix[rule_format, OF c0]) ``` ``` 3136 apply blast ``` ``` 3137 apply simp ``` ``` 3138 apply (frule_tac startsby_zero_power_prefix[rule_format, OF d0]) ``` ``` 3139 apply blast ``` ``` 3140 done ``` ``` 3141 finally show ?thesis by simp ``` ``` 3142 qed ``` ``` 3143 ``` ``` 3144 lemma product_composition_lemma': ``` ``` 3145 assumes c0: "c\$0 = (0::'a::idom)" ``` ``` 3146 and d0: "d\$0 = 0" ``` ``` 3147 shows "((a oo c) * (b oo d))\$n = ``` ``` 3148 setsum (\<lambda>k. setsum (\<lambda>m. a\$k * b\$m * (c^k * d^m) \$ n) {0..n}) {0..n}" (is "?l = ?r") ``` ``` 3149 unfolding product_composition_lemma[OF c0 d0] ``` ``` 3150 unfolding setsum.cartesian_product ``` ``` 3151 apply (rule setsum.mono_neutral_left) ``` ``` 3152 apply simp ``` ``` 3153 apply (clarsimp simp add: subset_eq) ``` ``` 3154 apply clarsimp ``` ``` 3155 apply (rule ccontr) ``` ``` 3156 apply (subgoal_tac "(c^aa * d^ba) \$ n = 0") ``` ``` 3157 apply simp ``` ``` 3158 unfolding fps_mult_nth ``` ``` 3159 apply (rule setsum.neutral) ``` ``` 3160 apply (clarsimp simp add: not_le) ``` ``` 3161 apply (case_tac "x < aa") ``` ``` 3162 apply (rule startsby_zero_power_prefix[OF c0, rule_format]) ``` ``` 3163 apply simp ``` ``` 3164 apply (subgoal_tac "n - x < ba") ``` ``` 3165 apply (frule_tac k = "ba" in startsby_zero_power_prefix[OF d0, rule_format]) ``` ``` 3166 apply simp ``` ``` 3167 apply arith ``` ``` 3168 done ``` ``` 3169 ``` ``` 3170 ``` ``` 3171 lemma setsum_pair_less_iff: ``` ``` 3172 "setsum (\<lambda>((k::nat),m). a k * b m * c (k + m)) {(k,m). k + m \<le> n} = ``` ``` 3173 setsum (\<lambda>s. setsum (\<lambda>i. a i * b (s - i) * c s) {0..s}) {0..n}" ``` ``` 3174 (is "?l = ?r") ``` ``` 3175 proof - ``` ``` 3176 let ?KM = "{(k,m). k + m \<le> n}" ``` ``` 3177 let ?f = "\<lambda>s. UNION {(0::nat)..s} (\<lambda>i. {(i,s - i)})" ``` ``` 3178 have th0: "?KM = UNION {0..n} ?f" ``` ``` 3179 by (auto simp add: set_eq_iff Bex_def) ``` ``` 3180 show "?l = ?r " ``` ``` 3181 unfolding th0 ``` ``` 3182 apply (subst setsum.UNION_disjoint) ``` ``` 3183 apply auto ``` ``` 3184 apply (subst setsum.UNION_disjoint) ``` ``` 3185 apply auto ``` ``` 3186 done ``` ``` 3187 qed ``` ``` 3188 ``` ``` 3189 lemma fps_compose_mult_distrib_lemma: ``` ``` 3190 assumes c0: "c\$0 = (0::'a::idom)" ``` ``` 3191 shows "((a oo c) * (b oo c))\$n = setsum (\<lambda>s. setsum (\<lambda>i. a\$i * b\$(s - i) * (c^s) \$ n) {0..s}) {0..n}" ``` ``` 3192 unfolding product_composition_lemma[OF c0 c0] power_add[symmetric] ``` ``` 3193 unfolding setsum_pair_less_iff[where a = "\<lambda>k. a\$k" and b="\<lambda>m. b\$m" and c="\<lambda>s. (c ^ s)\$n" and n = n] .. ``` ``` 3194 ``` ``` 3195 lemma fps_compose_mult_distrib: ``` ``` 3196 assumes c0: "c \$ 0 = (0::'a::idom)" ``` ``` 3197 shows "(a * b) oo c = (a oo c) * (b oo c)" ``` ``` 3198 apply (simp add: fps_eq_iff fps_compose_mult_distrib_lemma [OF c0]) ``` ``` 3199 apply (simp add: fps_compose_nth fps_mult_nth setsum_left_distrib) ``` ``` 3200 done ``` ``` 3201 ``` ``` 3202 lemma fps_compose_setprod_distrib: ``` ``` 3203 assumes c0: "c\$0 = (0::'a::idom)" ``` ``` 3204 shows "setprod a S oo c = setprod (\<lambda>k. a k oo c) S" ``` ``` 3205 apply (cases "finite S") ``` ``` 3206 apply simp_all ``` ``` 3207 apply (induct S rule: finite_induct) ``` ``` 3208 apply simp ``` ``` 3209 apply (simp add: fps_compose_mult_distrib[OF c0]) ``` ``` 3210 done ``` ``` 3211 ``` ``` 3212 lemma fps_compose_power: ``` ``` 3213 assumes c0: "c\$0 = (0::'a::idom)" ``` ``` 3214 shows "(a oo c)^n = a^n oo c" ``` ``` 3215 proof (cases n) ``` ``` 3216 case 0 ``` ``` 3217 then show ?thesis by simp ``` ``` 3218 next ``` ``` 3219 case (Suc m) ``` ``` 3220 have th0: "a^n = setprod (\<lambda>k. a) {0..m}" "(a oo c) ^ n = setprod (\<lambda>k. a oo c) {0..m}" ``` ``` 3221 by (simp_all add: setprod_constant Suc) ``` ``` 3222 then show ?thesis ``` ``` 3223 by (simp add: fps_compose_setprod_distrib[OF c0]) ``` ``` 3224 qed ``` ``` 3225 ``` ``` 3226 lemma fps_compose_uminus: "- (a::'a::ring_1 fps) oo c = - (a oo c)" ``` ``` 3227 by (simp add: fps_eq_iff fps_compose_nth field_simps setsum_negf[symmetric]) ``` ``` 3228 ``` ``` 3229 lemma fps_compose_sub_distrib: "(a - b) oo (c::'a::ring_1 fps) = (a oo c) - (b oo c)" ``` ``` 3230 using fps_compose_add_distrib [of a "- b" c] by (simp add: fps_compose_uminus) ``` ``` 3231 ``` ``` 3232 lemma X_fps_compose: "X oo a = Abs_fps (\<lambda>n. if n = 0 then (0::'a::comm_ring_1) else a\$n)" ``` ``` 3233 by (simp add: fps_eq_iff fps_compose_nth mult_delta_left setsum.delta) ``` ``` 3234 ``` ``` 3235 lemma fps_inverse_compose: ``` ``` 3236 assumes b0: "(b\$0 :: 'a::field) = 0" ``` ``` 3237 and a0: "a\$0 \<noteq> 0" ``` ``` 3238 shows "inverse a oo b = inverse (a oo b)" ``` ``` 3239 proof - ``` ``` 3240 let ?ia = "inverse a" ``` ``` 3241 let ?ab = "a oo b" ``` ``` 3242 let ?iab = "inverse ?ab" ``` ``` 3243 ``` ``` 3244 from a0 have ia0: "?ia \$ 0 \<noteq> 0" by simp ``` ``` 3245 from a0 have ab0: "?ab \$ 0 \<noteq> 0" by (simp add: fps_compose_def) ``` ``` 3246 have "(?ia oo b) * (a oo b) = 1" ``` ``` 3247 unfolding fps_compose_mult_distrib[OF b0, symmetric] ``` ``` 3248 unfolding inverse_mult_eq_1[OF a0] ``` ``` 3249 fps_compose_1 .. ``` ``` 3250 ``` ``` 3251 then have "(?ia oo b) * (a oo b) * ?iab = 1 * ?iab" by simp ``` ``` 3252 then have "(?ia oo b) * (?iab * (a oo b)) = ?iab" by simp ``` ``` 3253 then show ?thesis unfolding inverse_mult_eq_1[OF ab0] by simp ``` ``` 3254 qed ``` ``` 3255 ``` ``` 3256 lemma fps_divide_compose: ``` ``` 3257 assumes c0: "(c\$0 :: 'a::field) = 0" ``` ``` 3258 and b0: "b\$0 \<noteq> 0" ``` ``` 3259 shows "(a/b) oo c = (a oo c) / (b oo c)" ``` ``` 3260 using b0 c0 by (simp add: fps_divide_unit fps_inverse_compose fps_compose_mult_distrib) ``` ``` 3261 ``` ``` 3262 lemma gp: ``` ``` 3263 assumes a0: "a\$0 = (0::'a::field)" ``` ``` 3264 shows "(Abs_fps (\<lambda>n. 1)) oo a = 1/(1 - a)" ``` ``` 3265 (is "?one oo a = _") ``` ``` 3266 proof - ``` ``` 3267 have o0: "?one \$ 0 \<noteq> 0" by simp ``` ``` 3268 have th0: "(1 - X) \$ 0 \<noteq> (0::'a)" by simp ``` ``` 3269 from fps_inverse_gp[where ?'a = 'a] ``` ``` 3270 have "inverse ?one = 1 - X" by (simp add: fps_eq_iff) ``` ``` 3271 then have "inverse (inverse ?one) = inverse (1 - X)" by simp ``` ``` 3272 then have th: "?one = 1/(1 - X)" unfolding fps_inverse_idempotent[OF o0] ``` ``` 3273 by (simp add: fps_divide_def) ``` ``` 3274 show ?thesis ``` ``` 3275 unfolding th ``` ``` 3276 unfolding fps_divide_compose[OF a0 th0] ``` ``` 3277 fps_compose_1 fps_compose_sub_distrib X_fps_compose_startby0[OF a0] .. ``` ``` 3278 qed ``` ``` 3279 ``` ``` 3280 lemma fps_const_power [simp]: "fps_const (c::'a::ring_1) ^ n = fps_const (c^n)" ``` ``` 3281 by (induct n) auto ``` ``` 3282 ``` ``` 3283 lemma fps_compose_radical: ``` ``` 3284 assumes b0: "b\$0 = (0::'a::field_char_0)" ``` ``` 3285 and ra0: "r (Suc k) (a\$0) ^ Suc k = a\$0" ``` ``` 3286 and a0: "a\$0 \<noteq> 0" ``` ``` 3287 shows "fps_radical r (Suc k) a oo b = fps_radical r (Suc k) (a oo b)" ``` ``` 3288 proof - ``` ``` 3289 let ?r = "fps_radical r (Suc k)" ``` ``` 3290 let ?ab = "a oo b" ``` ``` 3291 have ab0: "?ab \$ 0 = a\$0" ``` ``` 3292 by (simp add: fps_compose_def) ``` ``` 3293 from ab0 a0 ra0 have rab0: "?ab \$ 0 \<noteq> 0" "r (Suc k) (?ab \$ 0) ^ Suc k = ?ab \$ 0" ``` ``` 3294 by simp_all ``` ``` 3295 have th00: "r (Suc k) ((a oo b) \$ 0) = (fps_radical r (Suc k) a oo b) \$ 0" ``` ``` 3296 by (simp add: ab0 fps_compose_def) ``` ``` 3297 have th0: "(?r a oo b) ^ (Suc k) = a oo b" ``` ``` 3298 unfolding fps_compose_power[OF b0] ``` ``` 3299 unfolding iffD1[OF power_radical[of a r k], OF a0 ra0] .. ``` ``` 3300 from iffD1[OF radical_unique[where r=r and k=k and b= ?ab and a = "?r a oo b", OF rab0(2) th00 rab0(1)], OF th0] ``` ``` 3301 show ?thesis . ``` ``` 3302 qed ``` ``` 3303 ``` ``` 3304 lemma fps_const_mult_apply_left: "fps_const c * (a oo b) = (fps_const c * a) oo b" ``` ``` 3305 by (simp add: fps_eq_iff fps_compose_nth setsum_right_distrib mult.assoc) ``` ``` 3306 ``` ``` 3307 lemma fps_const_mult_apply_right: ``` ``` 3308 "(a oo b) * fps_const (c::'a::comm_semiring_1) = (fps_const c * a) oo b" ``` ``` 3309 by (auto simp add: fps_const_mult_apply_left mult.commute) ``` ``` 3310 ``` ``` 3311 lemma fps_compose_assoc: ``` ``` 3312 assumes c0: "c\$0 = (0::'a::idom)" ``` ``` 3313 and b0: "b\$0 = 0" ``` ``` 3314 shows "a oo (b oo c) = a oo b oo c" (is "?l = ?r") ``` ``` 3315 proof - ``` ``` 3316 have "?l\$n = ?r\$n" for n ``` ``` 3317 proof - ``` ``` 3318 have "?l\$n = (setsum (\<lambda>i. (fps_const (a\$i) * b^i) oo c) {0..n})\$n" ``` ``` 3319 by (simp add: fps_compose_nth fps_compose_power[OF c0] fps_const_mult_apply_left ``` ``` 3320 setsum_right_distrib mult.assoc fps_setsum_nth) ``` ``` 3321 also have "\<dots> = ((setsum (\<lambda>i. fps_const (a\$i) * b^i) {0..n}) oo c)\$n" ``` ``` 3322 by (simp add: fps_compose_setsum_distrib) ``` ``` 3323 also have "\<dots> = ?r\$n" ``` ``` 3324 apply (simp add: fps_compose_nth fps_setsum_nth setsum_left_distrib mult.assoc) ``` ``` 3325 apply (rule setsum.cong) ``` ``` 3326 apply (rule refl) ``` ``` 3327 apply (rule setsum.mono_neutral_right) ``` ``` 3328 apply (auto simp add: not_le) ``` ``` 3329 apply (erule startsby_zero_power_prefix[OF b0, rule_format]) ``` ``` 3330 done ``` ``` 3331 finally show ?thesis . ``` ``` 3332 qed ``` ``` 3333 then show ?thesis ``` ``` 3334 by (simp add: fps_eq_iff) ``` ``` 3335 qed ``` ``` 3336 ``` ``` 3337 ``` ``` 3338 lemma fps_X_power_compose: ``` ``` 3339 assumes a0: "a\$0=0" ``` ``` 3340 shows "X^k oo a = (a::'a::idom fps)^k" ``` ``` 3341 (is "?l = ?r") ``` ``` 3342 proof (cases k) ``` ``` 3343 case 0 ``` ``` 3344 then show ?thesis by simp ``` ``` 3345 next ``` ``` 3346 case (Suc h) ``` ``` 3347 have "?l \$ n = ?r \$n" for n ``` ``` 3348 proof - ``` ``` 3349 consider "k > n" | "k \<le> n" by arith ``` ``` 3350 then show ?thesis ``` ``` 3351 proof cases ``` ``` 3352 case 1 ``` ``` 3353 then show ?thesis ``` ``` 3354 using a0 startsby_zero_power_prefix[OF a0] Suc ``` ``` 3355 by (simp add: fps_compose_nth del: power_Suc) ``` ``` 3356 next ``` ``` 3357 case 2 ``` ``` 3358 then show ?thesis ``` ``` 3359 by (simp add: fps_compose_nth mult_delta_left setsum.delta) ``` ``` 3360 qed ``` ``` 3361 qed ``` ``` 3362 then show ?thesis ``` ``` 3363 unfolding fps_eq_iff by blast ``` ``` 3364 qed ``` ``` 3365 ``` ``` 3366 lemma fps_inv_right: ``` ``` 3367 assumes a0: "a\$0 = 0" ``` ``` 3368 and a1: "a\$1 \<noteq> 0" ``` ``` 3369 shows "a oo fps_inv a = X" ``` ``` 3370 proof - ``` ``` 3371 let ?ia = "fps_inv a" ``` ``` 3372 let ?iaa = "a oo fps_inv a" ``` ``` 3373 have th0: "?ia \$ 0 = 0" ``` ``` 3374 by (simp add: fps_inv_def) ``` ``` 3375 have th1: "?iaa \$ 0 = 0" ``` ``` 3376 using a0 a1 by (simp add: fps_inv_def fps_compose_nth) ``` ``` 3377 have th2: "X\$0 = 0" ``` ``` 3378 by simp ``` ``` 3379 from fps_inv[OF a0 a1] have "a oo (fps_inv a oo a) = a oo X" ``` ``` 3380 by simp ``` ``` 3381 then have "(a oo fps_inv a) oo a = X oo a" ``` ``` 3382 by (simp add: fps_compose_assoc[OF a0 th0] X_fps_compose_startby0[OF a0]) ``` ``` 3383 with fps_compose_inj_right[OF a0 a1] show ?thesis ``` ``` 3384 by simp ``` ``` 3385 qed ``` ``` 3386 ``` ``` 3387 lemma fps_inv_deriv: ``` ``` 3388 assumes a0: "a\$0 = (0::'a::field)" ``` ``` 3389 and a1: "a\$1 \<noteq> 0" ``` ``` 3390 shows "fps_deriv (fps_inv a) = inverse (fps_deriv a oo fps_inv a)" ``` ``` 3391 proof - ``` ``` 3392 let ?ia = "fps_inv a" ``` ``` 3393 let ?d = "fps_deriv a oo ?ia" ``` ``` 3394 let ?dia = "fps_deriv ?ia" ``` ``` 3395 have ia0: "?ia\$0 = 0" ``` ``` 3396 by (simp add: fps_inv_def) ``` ``` 3397 have th0: "?d\$0 \<noteq> 0" ``` ``` 3398 using a1 by (simp add: fps_compose_nth) ``` ``` 3399 from fps_inv_right[OF a0 a1] have "?d * ?dia = 1" ``` ``` 3400 by (simp add: fps_compose_deriv[OF ia0, of a, symmetric] ) ``` ``` 3401 then have "inverse ?d * ?d * ?dia = inverse ?d * 1" ``` ``` 3402 by simp ``` ``` 3403 with inverse_mult_eq_1 [OF th0] show "?dia = inverse ?d" ``` ``` 3404 by simp ``` ``` 3405 qed ``` ``` 3406 ``` ``` 3407 lemma fps_inv_idempotent: ``` ``` 3408 assumes a0: "a\$0 = 0" ``` ``` 3409 and a1: "a\$1 \<noteq> 0" ``` ``` 3410 shows "fps_inv (fps_inv a) = a" ``` ``` 3411 proof - ``` ``` 3412 let ?r = "fps_inv" ``` ``` 3413 have ra0: "?r a \$ 0 = 0" ``` ``` 3414 by (simp add: fps_inv_def) ``` ``` 3415 from a1 have ra1: "?r a \$ 1 \<noteq> 0" ``` ``` 3416 by (simp add: fps_inv_def field_simps) ``` ``` 3417 have X0: "X\$0 = 0" ``` ``` 3418 by simp ``` ``` 3419 from fps_inv[OF ra0 ra1] have "?r (?r a) oo ?r a = X" . ``` ``` 3420 then have "?r (?r a) oo ?r a oo a = X oo a" ``` ``` 3421 by simp ``` ``` 3422 then have "?r (?r a) oo (?r a oo a) = a" ``` ``` 3423 unfolding X_fps_compose_startby0[OF a0] ``` ``` 3424 unfolding fps_compose_assoc[OF a0 ra0, symmetric] . ``` ``` 3425 then show ?thesis ``` ``` 3426 unfolding fps_inv[OF a0 a1] by simp ``` ``` 3427 qed ``` ``` 3428 ``` ``` 3429 lemma fps_ginv_ginv: ``` ``` 3430 assumes a0: "a\$0 = 0" ``` ``` 3431 and a1: "a\$1 \<noteq> 0" ``` ``` 3432 and c0: "c\$0 = 0" ``` ``` 3433 and c1: "c\$1 \<noteq> 0" ``` ``` 3434 shows "fps_ginv b (fps_ginv c a) = b oo a oo fps_inv c" ``` ``` 3435 proof - ``` ``` 3436 let ?r = "fps_ginv" ``` ``` 3437 from c0 have rca0: "?r c a \$0 = 0" ``` ``` 3438 by (simp add: fps_ginv_def) ``` ``` 3439 from a1 c1 have rca1: "?r c a \$ 1 \<noteq> 0" ``` ``` 3440 by (simp add: fps_ginv_def field_simps) ``` ``` 3441 from fps_ginv[OF rca0 rca1] ``` ``` 3442 have "?r b (?r c a) oo ?r c a = b" . ``` ``` 3443 then have "?r b (?r c a) oo ?r c a oo a = b oo a" ``` ``` 3444 by simp ``` ``` 3445 then have "?r b (?r c a) oo (?r c a oo a) = b oo a" ``` ``` 3446 apply (subst fps_compose_assoc) ``` ``` 3447 using a0 c0 ``` ``` 3448 apply (auto simp add: fps_ginv_def) ``` ``` 3449 done ``` ``` 3450 then have "?r b (?r c a) oo c = b oo a" ``` ``` 3451 unfolding fps_ginv[OF a0 a1] . ``` ``` 3452 then have "?r b (?r c a) oo c oo fps_inv c= b oo a oo fps_inv c" ``` ``` 3453 by simp ``` ``` 3454 then have "?r b (?r c a) oo (c oo fps_inv c) = b oo a oo fps_inv c" ``` ``` 3455 apply (subst fps_compose_assoc) ``` ``` 3456 using a0 c0 ``` ``` 3457 apply (auto simp add: fps_inv_def) ``` ``` 3458 done ``` ``` 3459 then show ?thesis ``` ``` 3460 unfolding fps_inv_right[OF c0 c1] by simp ``` ``` 3461 qed ``` ``` 3462 ``` ``` 3463 lemma fps_ginv_deriv: ``` ``` 3464 assumes a0:"a\$0 = (0::'a::field)" ``` ``` 3465 and a1: "a\$1 \<noteq> 0" ``` ``` 3466 shows "fps_deriv (fps_ginv b a) = (fps_deriv b / fps_deriv a) oo fps_ginv X a" ``` ``` 3467 proof - ``` ``` 3468 let ?ia = "fps_ginv b a" ``` ``` 3469 let ?iXa = "fps_ginv X a" ``` ``` 3470 let ?d = "fps_deriv" ``` ``` 3471 let ?dia = "?d ?ia" ``` ``` 3472 have iXa0: "?iXa \$ 0 = 0" ``` ``` 3473 by (simp add: fps_ginv_def) ``` ``` 3474 have da0: "?d a \$ 0 \<noteq> 0" ``` ``` 3475 using a1 by simp ``` ``` 3476 from fps_ginv[OF a0 a1, of b] have "?d (?ia oo a) = fps_deriv b" ``` ``` 3477 by simp ``` ``` 3478 then have "(?d ?ia oo a) * ?d a = ?d b" ``` ``` 3479 unfolding fps_compose_deriv[OF a0] . ``` ``` 3480 then have "(?d ?ia oo a) * ?d a * inverse (?d a) = ?d b * inverse (?d a)" ``` ``` 3481 by simp ``` ``` 3482 with a1 have "(?d ?ia oo a) * (inverse (?d a) * ?d a) = ?d b / ?d a" ``` ``` 3483 by (simp add: fps_divide_unit) ``` ``` 3484 then have "(?d ?ia oo a) oo ?iXa = (?d b / ?d a) oo ?iXa" ``` ``` 3485 unfolding inverse_mult_eq_1[OF da0] by simp ``` ``` 3486 then have "?d ?ia oo (a oo ?iXa) = (?d b / ?d a) oo ?iXa" ``` ``` 3487 unfolding fps_compose_assoc[OF iXa0 a0] . ``` ``` 3488 then show ?thesis unfolding fps_inv_ginv[symmetric] ``` ``` 3489 unfolding fps_inv_right[OF a0 a1] by simp ``` ``` 3490 qed ``` ``` 3491 ``` ``` 3492 ``` ``` 3493 subsection \<open>Elementary series\<close> ``` ``` 3494 ``` ``` 3495 subsubsection \<open>Exponential series\<close> ``` ``` 3496 ``` ``` 3497 definition "E x = Abs_fps (\<lambda>n. x^n / of_nat (fact n))" ``` ``` 3498 ``` ``` 3499 lemma E_deriv[simp]: "fps_deriv (E a) = fps_const (a::'a::field_char_0) * E a" (is "?l = ?r") ``` ``` 3500 proof - ``` ``` 3501 have "?l\$n = ?r \$ n" for n ``` ``` 3502 apply (auto simp add: E_def field_simps power_Suc[symmetric] ``` ``` 3503 simp del: fact.simps of_nat_Suc power_Suc) ``` ``` 3504 apply (simp add: of_nat_mult field_simps) ``` ``` 3505 done ``` ``` 3506 then show ?thesis ``` ``` 3507 by (simp add: fps_eq_iff) ``` ``` 3508 qed ``` ``` 3509 ``` ``` 3510 lemma E_unique_ODE: ``` ``` 3511 "fps_deriv a = fps_const c * a \<longleftrightarrow> a = fps_const (a\$0) * E (c::'a::field_char_0)" ``` ``` 3512 (is "?lhs \<longleftrightarrow> ?rhs") ``` ``` 3513 proof ``` ``` 3514 show ?rhs if ?lhs ``` ``` 3515 proof - ``` ``` 3516 from that have th: "\<And>n. a \$ Suc n = c * a\$n / of_nat (Suc n)" ``` ``` 3517 by (simp add: fps_deriv_def fps_eq_iff field_simps del: of_nat_Suc) ``` ``` 3518 have th': "a\$n = a\$0 * c ^ n/ (fact n)" for n ``` ``` 3519 proof (induct n) ``` ``` 3520 case 0 ``` ``` 3521 then show ?case by simp ``` ``` 3522 next ``` ``` 3523 case Suc ``` ``` 3524 then show ?case ``` ``` 3525 unfolding th ``` ``` 3526 using fact_gt_zero ``` ``` 3527 apply (simp add: field_simps del: of_nat_Suc fact_Suc) ``` ``` 3528 apply simp ``` ``` 3529 done ``` ``` 3530 qed ``` ``` 3531 show ?thesis ``` ``` 3532 by (auto simp add: fps_eq_iff fps_const_mult_left E_def intro: th') ``` ``` 3533 qed ``` ``` 3534 show ?lhs if ?rhs ``` ``` 3535 using that by (metis E_deriv fps_deriv_mult_const_left mult.left_commute) ``` ``` 3536 qed ``` ``` 3537 ``` ``` 3538 lemma E_add_mult: "E (a + b) = E (a::'a::field_char_0) * E b" (is "?l = ?r") ``` ``` 3539 proof - ``` ``` 3540 have "fps_deriv ?r = fps_const (a + b) * ?r" ``` ``` 3541 by (simp add: fps_const_add[symmetric] field_simps del: fps_const_add) ``` ``` 3542 then have "?r = ?l" ``` ``` 3543 by (simp only: E_unique_ODE) (simp add: fps_mult_nth E_def) ``` ``` 3544 then show ?thesis .. ``` ``` 3545 qed ``` ``` 3546 ``` ``` 3547 lemma E_nth[simp]: "E a \$ n = a^n / of_nat (fact n)" ``` ``` 3548 by (simp add: E_def) ``` ``` 3549 ``` ``` 3550 lemma E0[simp]: "E (0::'a::field) = 1" ``` ``` 3551 by (simp add: fps_eq_iff power_0_left) ``` ``` 3552 ``` ``` 3553 lemma E_neg: "E (- a) = inverse (E (a::'a::field_char_0))" ``` ``` 3554 proof - ``` ``` 3555 from E_add_mult[of a "- a"] have th0: "E a * E (- a) = 1" by simp ``` ``` 3556 from fps_inverse_unique[OF th0] show ?thesis by simp ``` ``` 3557 qed ``` ``` 3558 ``` ``` 3559 lemma E_nth_deriv[simp]: "fps_nth_deriv n (E (a::'a::field_char_0)) = (fps_const a)^n * (E a)" ``` ``` 3560 by (induct n) auto ``` ``` 3561 ``` ``` 3562 lemma X_compose_E[simp]: "X oo E (a::'a::field) = E a - 1" ``` ``` 3563 by (simp add: fps_eq_iff X_fps_compose) ``` ``` 3564 ``` ``` 3565 lemma LE_compose: ``` ``` 3566 assumes a: "a \<noteq> 0" ``` ``` 3567 shows "fps_inv (E a - 1) oo (E a - 1) = X" ``` ``` 3568 and "(E a - 1) oo fps_inv (E a - 1) = X" ``` ``` 3569 proof - ``` ``` 3570 let ?b = "E a - 1" ``` ``` 3571 have b0: "?b \$ 0 = 0" ``` ``` 3572 by simp ``` ``` 3573 have b1: "?b \$ 1 \<noteq> 0" ``` ``` 3574 by (simp add: a) ``` ``` 3575 from fps_inv[OF b0 b1] show "fps_inv (E a - 1) oo (E a - 1) = X" . ``` ``` 3576 from fps_inv_right[OF b0 b1] show "(E a - 1) oo fps_inv (E a - 1) = X" . ``` ``` 3577 qed ``` ``` 3578 ``` ``` 3579 lemma E_power_mult: "(E (c::'a::field_char_0))^n = E (of_nat n * c)" ``` ``` 3580 by (induct n) (auto simp add: field_simps E_add_mult) ``` ``` 3581 ``` ``` 3582 lemma radical_E: ``` ``` 3583 assumes r: "r (Suc k) 1 = 1" ``` ``` 3584 shows "fps_radical r (Suc k) (E (c::'a::field_char_0)) = E (c / of_nat (Suc k))" ``` ``` 3585 proof - ``` ``` 3586 let ?ck = "(c / of_nat (Suc k))" ``` ``` 3587 let ?r = "fps_radical r (Suc k)" ``` ``` 3588 have eq0[simp]: "?ck * of_nat (Suc k) = c" "of_nat (Suc k) * ?ck = c" ``` ``` 3589 by (simp_all del: of_nat_Suc) ``` ``` 3590 have th0: "E ?ck ^ (Suc k) = E c" unfolding E_power_mult eq0 .. ``` ``` 3591 have th: "r (Suc k) (E c \$0) ^ Suc k = E c \$ 0" ``` ``` 3592 "r (Suc k) (E c \$ 0) = E ?ck \$ 0" "E c \$ 0 \<noteq> 0" using r by simp_all ``` ``` 3593 from th0 radical_unique[where r=r and k=k, OF th] show ?thesis ``` ``` 3594 by auto ``` ``` 3595 qed ``` ``` 3596 ``` ``` 3597 lemma Ec_E1_eq: "E (1::'a::field_char_0) oo (fps_const c * X) = E c" ``` ``` 3598 apply (auto simp add: fps_eq_iff E_def fps_compose_def power_mult_distrib) ``` ``` 3599 apply (simp add: cond_value_iff cond_application_beta setsum.delta' cong del: if_weak_cong) ``` ``` 3600 done ``` ``` 3601 ``` ``` 3602 ``` ``` 3603 subsubsection \<open>Logarithmic series\<close> ``` ``` 3604 ``` ``` 3605 lemma Abs_fps_if_0: ``` ``` 3606 "Abs_fps (\<lambda>n. if n = 0 then (v::'a::ring_1) else f n) = ``` ``` 3607 fps_const v + X * Abs_fps (\<lambda>n. f (Suc n))" ``` ``` 3608 by (auto simp add: fps_eq_iff) ``` ``` 3609 ``` ``` 3610 definition L :: "'a::field_char_0 \<Rightarrow> 'a fps" ``` ``` 3611 where "L c = fps_const (1/c) * Abs_fps (\<lambda>n. if n = 0 then 0 else (- 1) ^ (n - 1) / of_nat n)" ``` ``` 3612 ``` ``` 3613 lemma fps_deriv_L: "fps_deriv (L c) = fps_const (1/c) * inverse (1 + X)" ``` ``` 3614 unfolding fps_inverse_X_plus1 ``` ``` 3615 by (simp add: L_def fps_eq_iff del: of_nat_Suc) ``` ``` 3616 ``` ``` 3617 lemma L_nth: "L c \$ n = (if n = 0 then 0 else 1/c * ((- 1) ^ (n - 1) / of_nat n))" ``` ``` 3618 by (simp add: L_def field_simps) ``` ``` 3619 ``` ``` 3620 lemma L_0[simp]: "L c \$ 0 = 0" by (simp add: L_def) ``` ``` 3621 ``` ``` 3622 lemma L_E_inv: ``` ``` 3623 fixes a :: "'a::field_char_0" ``` ``` 3624 assumes a: "a \<noteq> 0" ``` ``` 3625 shows "L a = fps_inv (E a - 1)" (is "?l = ?r") ``` ``` 3626 proof - ``` ``` 3627 let ?b = "E a - 1" ``` ``` 3628 have b0: "?b \$ 0 = 0" by simp ``` ``` 3629 have b1: "?b \$ 1 \<noteq> 0" by (simp add: a) ``` ``` 3630 have "fps_deriv (E a - 1) oo fps_inv (E a - 1) = ``` ``` 3631 (fps_const a * (E a - 1) + fps_const a) oo fps_inv (E a - 1)" ``` ``` 3632 by (simp add: field_simps) ``` ``` 3633 also have "\<dots> = fps_const a * (X + 1)" ``` ``` 3634 apply (simp add: fps_compose_add_distrib fps_const_mult_apply_left[symmetric] fps_inv_right[OF b0 b1]) ``` ``` 3635 apply (simp add: field_simps) ``` ``` 3636 done ``` ``` 3637 finally have eq: "fps_deriv (E a - 1) oo fps_inv (E a - 1) = fps_const a * (X + 1)" . ``` ``` 3638 from fps_inv_deriv[OF b0 b1, unfolded eq] ``` ``` 3639 have "fps_deriv (fps_inv ?b) = fps_const (inverse a) / (X + 1)" ``` ``` 3640 using a ``` ``` 3641 by (simp add: fps_const_inverse eq fps_divide_def fps_inverse_mult) ``` ``` 3642 then have "fps_deriv ?l = fps_deriv ?r" ``` ``` 3643 by (simp add: fps_deriv_L add.commute fps_divide_def divide_inverse) ``` ``` 3644 then show ?thesis unfolding fps_deriv_eq_iff ``` ``` 3645 by (simp add: L_nth fps_inv_def) ``` ``` 3646 qed ``` ``` 3647 ``` ``` 3648 lemma L_mult_add: ``` ``` 3649 assumes c0: "c\<noteq>0" ``` ``` 3650 and d0: "d\<noteq>0" ``` ``` 3651 shows "L c + L d = fps_const (c+d) * L (c*d)" ``` ``` 3652 (is "?r = ?l") ``` ``` 3653 proof- ``` ``` 3654 from c0 d0 have eq: "1/c + 1/d = (c+d)/(c*d)" by (simp add: field_simps) ``` ``` 3655 have "fps_deriv ?r = fps_const (1/c + 1/d) * inverse (1 + X)" ``` ``` 3656 by (simp add: fps_deriv_L fps_const_add[symmetric] algebra_simps del: fps_const_add) ``` ``` 3657 also have "\<dots> = fps_deriv ?l" ``` ``` 3658 apply (simp add: fps_deriv_L) ``` ``` 3659 apply (simp add: fps_eq_iff eq) ``` ``` 3660 done ``` ``` 3661 finally show ?thesis ``` ``` 3662 unfolding fps_deriv_eq_iff by simp ``` ``` 3663 qed ``` ``` 3664 ``` ``` 3665 ``` ``` 3666 subsubsection \<open>Binomial series\<close> ``` ``` 3667 ``` ``` 3668 definition "fps_binomial a = Abs_fps (\<lambda>n. a gchoose n)" ``` ``` 3669 ``` ``` 3670 lemma fps_binomial_nth[simp]: "fps_binomial a \$ n = a gchoose n" ``` ``` 3671 by (simp add: fps_binomial_def) ``` ``` 3672 ``` ``` 3673 lemma fps_binomial_ODE_unique: ``` ``` 3674 fixes c :: "'a::field_char_0" ``` ``` 3675 shows "fps_deriv a = (fps_const c * a) / (1 + X) \<longleftrightarrow> a = fps_const (a\$0) * fps_binomial c" ``` ``` 3676 (is "?lhs \<longleftrightarrow> ?rhs") ``` ``` 3677 proof ``` ``` 3678 let ?da = "fps_deriv a" ``` ``` 3679 let ?x1 = "(1 + X):: 'a fps" ``` ``` 3680 let ?l = "?x1 * ?da" ``` ``` 3681 let ?r = "fps_const c * a" ``` ``` 3682 ``` ``` 3683 have eq: "?l = ?r \<longleftrightarrow> ?lhs" ``` ``` 3684 proof - ``` ``` 3685 have x10: "?x1 \$ 0 \<noteq> 0" by simp ``` ``` 3686 have "?l = ?r \<longleftrightarrow> inverse ?x1 * ?l = inverse ?x1 * ?r" by simp ``` ``` 3687 also have "\<dots> \<longleftrightarrow> ?da = (fps_const c * a) / ?x1" ``` ``` 3688 apply (simp only: fps_divide_def mult.assoc[symmetric] inverse_mult_eq_1[OF x10]) ``` ``` 3689 apply (simp add: field_simps) ``` ``` 3690 done ``` ``` 3691 finally show ?thesis . ``` ``` 3692 qed ``` ``` 3693 ``` ``` 3694 show ?rhs if ?lhs ``` ``` 3695 proof - ``` ``` 3696 from eq that have h: "?l = ?r" .. ``` ``` 3697 have th0: "a\$ Suc n = ((c - of_nat n) / of_nat (Suc n)) * a \$n" for n ``` ``` 3698 proof - ``` ``` 3699 from h have "?l \$ n = ?r \$ n" by simp ``` ``` 3700 then show ?thesis ``` ``` 3701 apply (simp add: field_simps del: of_nat_Suc) ``` ``` 3702 apply (cases n) ``` ``` 3703 apply (simp_all add: field_simps del: of_nat_Suc) ``` ``` 3704 done ``` ``` 3705 qed ``` ``` 3706 have th1: "a \$ n = (c gchoose n) * a \$ 0" for n ``` ``` 3707 proof (induct n) ``` ``` 3708 case 0 ``` ``` 3709 then show ?case by simp ``` ``` 3710 next ``` ``` 3711 case (Suc m) ``` ``` 3712 then show ?case ``` ``` 3713 unfolding th0 ``` ``` 3714 apply (simp add: field_simps del: of_nat_Suc) ``` ``` 3715 unfolding mult.assoc[symmetric] gbinomial_mult_1 ``` ``` 3716 apply (simp add: field_simps) ``` ``` 3717 done ``` ``` 3718 qed ``` ``` 3719 show ?thesis ``` ``` 3720 apply (simp add: fps_eq_iff) ``` ``` 3721 apply (subst th1) ``` ``` 3722 apply (simp add: field_simps) ``` ``` 3723 done ``` ``` 3724 qed ``` ``` 3725 ``` ``` 3726 show ?lhs if ?rhs ``` ``` 3727 proof - ``` ``` 3728 have th00: "x * (a \$ 0 * y) = a \$ 0 * (x * y)" for x y ``` ``` 3729 by (simp add: mult.commute) ``` ``` 3730 have "?l = ?r" ``` ``` 3731 apply (subst \<open>?rhs\<close>) ``` ``` 3732 apply (subst (2) \<open>?rhs\<close>) ``` ``` 3733 apply (clarsimp simp add: fps_eq_iff field_simps) ``` ``` 3734 unfolding mult.assoc[symmetric] th00 gbinomial_mult_1 ``` ``` 3735 apply (simp add: field_simps gbinomial_mult_1) ``` ``` 3736 done ``` ``` 3737 with eq show ?thesis .. ``` ``` 3738 qed ``` ``` 3739 qed ``` ``` 3740 ``` ``` 3741 lemma fps_binomial_deriv: "fps_deriv (fps_binomial c) = fps_const c * fps_binomial c / (1 + X)" ``` ``` 3742 proof - ``` ``` 3743 let ?a = "fps_binomial c" ``` ``` 3744 have th0: "?a = fps_const (?a\$0) * ?a" by (simp) ``` ``` 3745 from iffD2[OF fps_binomial_ODE_unique, OF th0] show ?thesis . ``` ``` 3746 qed ``` ``` 3747 ``` ``` 3748 lemma fps_binomial_add_mult: "fps_binomial (c+d) = fps_binomial c * fps_binomial d" (is "?l = ?r") ``` ``` 3749 proof - ``` ``` 3750 let ?P = "?r - ?l" ``` ``` 3751 let ?b = "fps_binomial" ``` ``` 3752 let ?db = "\<lambda>x. fps_deriv (?b x)" ``` ``` 3753 have "fps_deriv ?P = ?db c * ?b d + ?b c * ?db d - ?db (c + d)" by simp ``` ``` 3754 also have "\<dots> = inverse (1 + X) * ``` ``` 3755 (fps_const c * ?b c * ?b d + fps_const d * ?b c * ?b d - fps_const (c+d) * ?b (c + d))" ``` ``` 3756 unfolding fps_binomial_deriv ``` ``` 3757 by (simp add: fps_divide_def field_simps) ``` ``` 3758 also have "\<dots> = (fps_const (c + d)/ (1 + X)) * ?P" ``` ``` 3759 by (simp add: field_simps fps_divide_unit fps_const_add[symmetric] del: fps_const_add) ``` ``` 3760 finally have th0: "fps_deriv ?P = fps_const (c+d) * ?P / (1 + X)" ``` ``` 3761 by (simp add: fps_divide_def) ``` ``` 3762 have "?P = fps_const (?P\$0) * ?b (c + d)" ``` ``` 3763 unfolding fps_binomial_ODE_unique[symmetric] ``` ``` 3764 using th0 by simp ``` ``` 3765 then have "?P = 0" by (simp add: fps_mult_nth) ``` ``` 3766 then show ?thesis by simp ``` ``` 3767 qed ``` ``` 3768 ``` ``` 3769 lemma fps_binomial_minus_one: "fps_binomial (- 1) = inverse (1 + X)" ``` ``` 3770 (is "?l = inverse ?r") ``` ``` 3771 proof- ``` ``` 3772 have th: "?r\$0 \<noteq> 0" by simp ``` ``` 3773 have th': "fps_deriv (inverse ?r) = fps_const (- 1) * inverse ?r / (1 + X)" ``` ``` 3774 by (simp add: fps_inverse_deriv[OF th] fps_divide_def ``` ``` 3775 power2_eq_square mult.commute fps_const_neg[symmetric] del: fps_const_neg) ``` ``` 3776 have eq: "inverse ?r \$ 0 = 1" ``` ``` 3777 by (simp add: fps_inverse_def) ``` ``` 3778 from iffD1[OF fps_binomial_ODE_unique[of "inverse (1 + X)" "- 1"] th'] eq ``` ``` 3779 show ?thesis by (simp add: fps_inverse_def) ``` ``` 3780 qed ``` ``` 3781 ``` ``` 3782 text \<open>Vandermonde's Identity as a consequence.\<close> ``` ``` 3783 lemma gbinomial_Vandermonde: ``` ``` 3784 "setsum (\<lambda>k. (a gchoose k) * (b gchoose (n - k))) {0..n} = (a + b) gchoose n" ``` ``` 3785 proof - ``` ``` 3786 let ?ba = "fps_binomial a" ``` ``` 3787 let ?bb = "fps_binomial b" ``` ``` 3788 let ?bab = "fps_binomial (a + b)" ``` ``` 3789 from fps_binomial_add_mult[of a b] have "?bab \$ n = (?ba * ?bb)\$n" by simp ``` ``` 3790 then show ?thesis by (simp add: fps_mult_nth) ``` ``` 3791 qed ``` ``` 3792 ``` ``` 3793 lemma binomial_Vandermonde: ``` ``` 3794 "setsum (\<lambda>k. (a choose k) * (b choose (n - k))) {0..n} = (a + b) choose n" ``` ``` 3795 using gbinomial_Vandermonde[of "(of_nat a)" "of_nat b" n] ``` ``` 3796 by (simp only: binomial_gbinomial[symmetric] of_nat_mult[symmetric] ``` ``` 3797 of_nat_setsum[symmetric] of_nat_add[symmetric] of_nat_eq_iff) ``` ``` 3798 ``` ``` 3799 lemma binomial_Vandermonde_same: "setsum (\<lambda>k. (n choose k)\<^sup>2) {0..n} = (2 * n) choose n" ``` ``` 3800 using binomial_Vandermonde[of n n n, symmetric] ``` ``` 3801 unfolding mult_2 ``` ``` 3802 apply (simp add: power2_eq_square) ``` ``` 3803 apply (rule setsum.cong) ``` ``` 3804 apply (auto intro: binomial_symmetric) ``` ``` 3805 done ``` ``` 3806 ``` ``` 3807 lemma Vandermonde_pochhammer_lemma: ``` ``` 3808 fixes a :: "'a::field_char_0" ``` ``` 3809 assumes b: "\<forall>j\<in>{0 ..<n}. b \<noteq> of_nat j" ``` ``` 3810 shows "setsum (\<lambda>k. (pochhammer (- a) k * pochhammer (- (of_nat n)) k) / ``` ``` 3811 (of_nat (fact k) * pochhammer (b - of_nat n + 1) k)) {0..n} = ``` ``` 3812 pochhammer (- (a + b)) n / pochhammer (- b) n" ``` ``` 3813 (is "?l = ?r") ``` ``` 3814 proof - ``` ``` 3815 let ?m1 = "\<lambda>m. (- 1 :: 'a) ^ m" ``` ``` 3816 let ?f = "\<lambda>m. of_nat (fact m)" ``` ``` 3817 let ?p = "\<lambda>(x::'a). pochhammer (- x)" ``` ``` 3818 from b have bn0: "?p b n \<noteq> 0" ``` ``` 3819 unfolding pochhammer_eq_0_iff by simp ``` ``` 3820 have th00: ``` ``` 3821 "b gchoose (n - k) = ``` ``` 3822 (?m1 n * ?p b n * ?m1 k * ?p (of_nat n) k) / (?f n * pochhammer (b - of_nat n + 1) k)" ``` ``` 3823 (is ?gchoose) ``` ``` 3824 "pochhammer (1 + b - of_nat n) k \<noteq> 0" ``` ``` 3825 (is ?pochhammer) ``` ``` 3826 if kn: "k \<in> {0..n}" for k ``` ``` 3827 proof - ``` ``` 3828 have nz: "pochhammer (1 + b - of_nat n) n \<noteq> 0" ``` ``` 3829 proof ``` ``` 3830 assume "pochhammer (1 + b - of_nat n) n = 0" ``` ``` 3831 then have c: "pochhammer (b - of_nat n + 1) n = 0" ``` ``` 3832 by (simp add: algebra_simps) ``` ``` 3833 then obtain j where j: "j < n" "b - of_nat n + 1 = - of_nat j" ``` ``` 3834 unfolding pochhammer_eq_0_iff by blast ``` ``` 3835 from j have "b = of_nat n - of_nat j - of_nat 1" ``` ``` 3836 by (simp add: algebra_simps) ``` ``` 3837 then have "b = of_nat (n - j - 1)" ``` ``` 3838 using j kn by (simp add: of_nat_diff) ``` ``` 3839 with b show False using j by auto ``` ``` 3840 qed ``` ``` 3841 ``` ``` 3842 from nz kn [simplified] have nz': "pochhammer (1 + b - of_nat n) k \<noteq> 0" ``` ``` 3843 by (rule pochhammer_neq_0_mono) ``` ``` 3844 ``` ``` 3845 consider "k = 0 \<or> n = 0" | "k \<noteq> 0" "n \<noteq> 0" ``` ``` 3846 by blast ``` ``` 3847 then have "b gchoose (n - k) = ``` ``` 3848 (?m1 n * ?p b n * ?m1 k * ?p (of_nat n) k) / (?f n * pochhammer (b - of_nat n + 1) k)" ``` ``` 3849 proof cases ``` ``` 3850 case 1 ``` ``` 3851 then show ?thesis ``` ``` 3852 using kn by (cases "k = 0") (simp_all add: gbinomial_pochhammer) ``` ``` 3853 next ``` ``` 3854 case neq: 2 ``` ``` 3855 then obtain m where m: "n = Suc m" ``` ``` 3856 by (cases n) auto ``` ``` 3857 from neq(1) obtain h where h: "k = Suc h" ``` ``` 3858 by (cases k) auto ``` ``` 3859 show ?thesis ``` ``` 3860 proof (cases "k = n") ``` ``` 3861 case True ``` ``` 3862 then show ?thesis ``` ``` 3863 using pochhammer_minus'[where k=k and b=b] ``` ``` 3864 apply (simp add: pochhammer_same) ``` ``` 3865 using bn0 ``` ``` 3866 apply (simp add: field_simps power_add[symmetric]) ``` ``` 3867 done ``` ``` 3868 next ``` ``` 3869 case False ``` ``` 3870 with kn have kn': "k < n" ``` ``` 3871 by simp ``` ``` 3872 have m1nk: "?m1 n = setprod (\<lambda>i. - 1) {0..m}" "?m1 k = setprod (\<lambda>i. - 1) {0..h}" ``` ``` 3873 by (simp_all add: setprod_constant m h) ``` ``` 3874 have bnz0: "pochhammer (b - of_nat n + 1) k \<noteq> 0" ``` ``` 3875 using bn0 kn ``` ``` 3876 unfolding pochhammer_eq_0_iff ``` ``` 3877 apply auto ``` ``` 3878 apply (erule_tac x= "n - ka - 1" in allE) ``` ``` 3879 apply (auto simp add: algebra_simps of_nat_diff) ``` ``` 3880 done ``` ``` 3881 have eq1: "setprod (\<lambda>k. (1::'a) + of_nat m - of_nat k) {0 .. h} = ``` ``` 3882 setprod of_nat {Suc (m - h) .. Suc m}" ``` ``` 3883 using kn' h m ``` ``` 3884 by (intro setprod.reindex_bij_witness[where i="\<lambda>k. Suc m - k" and j="\<lambda>k. Suc m - k"]) ``` ``` 3885 (auto simp: of_nat_diff) ``` ``` 3886 ``` ``` 3887 have th1: "(?m1 k * ?p (of_nat n) k) / ?f n = 1 / of_nat(fact (n - k))" ``` ``` 3888 unfolding m1nk ``` ``` 3889 unfolding m h pochhammer_Suc_setprod ``` ``` 3890 apply (simp add: field_simps del: fact_Suc) ``` ``` 3891 unfolding fact_altdef id_def ``` ``` 3892 unfolding of_nat_setprod ``` ``` 3893 unfolding setprod.distrib[symmetric] ``` ``` 3894 apply auto ``` ``` 3895 unfolding eq1 ``` ``` 3896 apply (subst setprod.union_disjoint[symmetric]) ``` ``` 3897 apply (auto) ``` ``` 3898 apply (rule setprod.cong) ``` ``` 3899 apply auto ``` ``` 3900 done ``` ``` 3901 have th20: "?m1 n * ?p b n = setprod (\<lambda>i. b - of_nat i) {0..m}" ``` ``` 3902 unfolding m1nk ``` ``` 3903 unfolding m h pochhammer_Suc_setprod ``` ``` 3904 unfolding setprod.distrib[symmetric] ``` ``` 3905 apply (rule setprod.cong) ``` ``` 3906 apply auto ``` ``` 3907 done ``` ``` 3908 have th21:"pochhammer (b - of_nat n + 1) k = setprod (\<lambda>i. b - of_nat i) {n - k .. n - 1}" ``` ``` 3909 unfolding h m ``` ``` 3910 unfolding pochhammer_Suc_setprod ``` ``` 3911 using kn m h ``` ``` 3912 by (intro setprod.reindex_bij_witness[where i="\<lambda>k. n - 1 - k" and j="\<lambda>i. m-i"]) ``` ``` 3913 (auto simp: of_nat_diff) ``` ``` 3914 ``` ``` 3915 have "?m1 n * ?p b n = ``` ``` 3916 pochhammer (b - of_nat n + 1) k * setprod (\<lambda>i. b - of_nat i) {0.. n - k - 1}" ``` ``` 3917 unfolding th20 th21 ``` ``` 3918 unfolding h m ``` ``` 3919 apply (subst setprod.union_disjoint[symmetric]) ``` ``` 3920 using kn' h m ``` ``` 3921 apply auto ``` ``` 3922 apply (rule setprod.cong) ``` ``` 3923 apply auto ``` ``` 3924 done ``` ``` 3925 then have th2: "(?m1 n * ?p b n)/pochhammer (b - of_nat n + 1) k = ``` ``` 3926 setprod (\<lambda>i. b - of_nat i) {0.. n - k - 1}" ``` ``` 3927 using nz' by (simp add: field_simps) ``` ``` 3928 have "(?m1 n * ?p b n * ?m1 k * ?p (of_nat n) k) / (?f n * pochhammer (b - of_nat n + 1) k) = ``` ``` 3929 ((?m1 k * ?p (of_nat n) k) / ?f n) * ((?m1 n * ?p b n)/pochhammer (b - of_nat n + 1) k)" ``` ``` 3930 using bnz0 ``` ``` 3931 by (simp add: field_simps) ``` ``` 3932 also have "\<dots> = b gchoose (n - k)" ``` ``` 3933 unfolding th1 th2 ``` ``` 3934 using kn' by (simp add: gbinomial_def) ``` ``` 3935 finally show ?thesis by simp ``` ``` 3936 qed ``` ``` 3937 qed ``` ``` 3938 then show ?gchoose and ?pochhammer ``` ``` 3939 apply (cases "n = 0") ``` ``` 3940 using nz' ``` ``` 3941 apply auto ``` ``` 3942 done ``` ``` 3943 qed ``` ``` 3944 have "?r = ((a + b) gchoose n) * (of_nat (fact n) / (?m1 n * pochhammer (- b) n))" ``` ``` 3945 unfolding gbinomial_pochhammer ``` ``` 3946 using bn0 by (auto simp add: field_simps) ``` ``` 3947 also have "\<dots> = ?l" ``` ``` 3948 unfolding gbinomial_Vandermonde[symmetric] ``` ``` 3949 apply (simp add: th00) ``` ``` 3950 unfolding gbinomial_pochhammer ``` ``` 3951 using bn0 ``` ``` 3952 apply (simp add: setsum_left_distrib setsum_right_distrib field_simps) ``` ``` 3953 apply (rule setsum.cong) ``` ``` 3954 apply (rule refl) ``` ``` 3955 apply (drule th00(2)) ``` ``` 3956 apply (simp add: field_simps power_add[symmetric]) ``` ``` 3957 done ``` ``` 3958 finally show ?thesis by simp ``` ``` 3959 qed ``` ``` 3960 ``` ``` 3961 lemma Vandermonde_pochhammer: ``` ``` 3962 fixes a :: "'a::field_char_0" ``` ``` 3963 assumes c: "\<forall>i \<in> {0..< n}. c \<noteq> - of_nat i" ``` ``` 3964 shows "setsum (\<lambda>k. (pochhammer a k * pochhammer (- (of_nat n)) k) / ``` ``` 3965 (of_nat (fact k) * pochhammer c k)) {0..n} = pochhammer (c - a) n / pochhammer c n" ``` ``` 3966 proof - ``` ``` 3967 let ?a = "- a" ``` ``` 3968 let ?b = "c + of_nat n - 1" ``` ``` 3969 have h: "\<forall> j \<in>{0..< n}. ?b \<noteq> of_nat j" ``` ``` 3970 using c ``` ``` 3971 apply (auto simp add: algebra_simps of_nat_diff) ``` ``` 3972 apply (erule_tac x = "n - j - 1" in ballE) ``` ``` 3973 apply (auto simp add: of_nat_diff algebra_simps) ``` ``` 3974 done ``` ``` 3975 have th0: "pochhammer (- (?a + ?b)) n = (- 1)^n * pochhammer (c - a) n" ``` ``` 3976 unfolding pochhammer_minus ``` ``` 3977 by (simp add: algebra_simps) ``` ``` 3978 have th1: "pochhammer (- ?b) n = (- 1)^n * pochhammer c n" ``` ``` 3979 unfolding pochhammer_minus ``` ``` 3980 by simp ``` ``` 3981 have nz: "pochhammer c n \<noteq> 0" using c ``` ``` 3982 by (simp add: pochhammer_eq_0_iff) ``` ``` 3983 from Vandermonde_pochhammer_lemma[where a = "?a" and b="?b" and n=n, OF h, unfolded th0 th1] ``` ``` 3984 show ?thesis ``` ``` 3985 using nz by (simp add: field_simps setsum_right_distrib) ``` ``` 3986 qed ``` ``` 3987 ``` ``` 3988 ``` ``` 3989 subsubsection \<open>Formal trigonometric functions\<close> ``` ``` 3990 ``` ``` 3991 definition "fps_sin (c::'a::field_char_0) = ``` ``` 3992 Abs_fps (\<lambda>n. if even n then 0 else (- 1) ^((n - 1) div 2) * c^n /(of_nat (fact n)))" ``` ``` 3993 ``` ``` 3994 definition "fps_cos (c::'a::field_char_0) = ``` ``` 3995 Abs_fps (\<lambda>n. if even n then (- 1) ^ (n div 2) * c^n / (of_nat (fact n)) else 0)" ``` ``` 3996 ``` ``` 3997 lemma fps_sin_deriv: ``` ``` 3998 "fps_deriv (fps_sin c) = fps_const c * fps_cos c" ``` ``` 3999 (is "?lhs = ?rhs") ``` ``` 4000 proof (rule fps_ext) ``` ``` 4001 fix n :: nat ``` ``` 4002 show "?lhs \$ n = ?rhs \$ n" ``` ``` 4003 proof (cases "even n") ``` ``` 4004 case True ``` ``` 4005 have "?lhs\$n = of_nat (n+1) * (fps_sin c \$ (n+1))" by simp ``` ``` 4006 also have "\<dots> = of_nat (n+1) * ((- 1)^(n div 2) * c^Suc n / of_nat (fact (Suc n)))" ``` ``` 4007 using True by (simp add: fps_sin_def) ``` ``` 4008 also have "\<dots> = (- 1)^(n div 2) * c^Suc n * (of_nat (n+1) / (of_nat (Suc n) * of_nat (fact n)))" ``` ``` 4009 unfolding fact_Suc of_nat_mult ``` ``` 4010 by (simp add: field_simps del: of_nat_add of_nat_Suc) ``` ``` 4011 also have "\<dots> = (- 1)^(n div 2) *c^Suc n / of_nat (fact n)" ``` ``` 4012 by (simp add: field_simps del: of_nat_add of_nat_Suc) ``` ``` 4013 finally show ?thesis ``` ``` 4014 using True by (simp add: fps_cos_def field_simps) ``` ``` 4015 next ``` ``` 4016 case False ``` ``` 4017 then show ?thesis ``` ``` 4018 by (simp_all add: fps_deriv_def fps_sin_def fps_cos_def) ``` ``` 4019 qed ``` ``` 4020 qed ``` ``` 4021 ``` ``` 4022 lemma fps_cos_deriv: "fps_deriv (fps_cos c) = fps_const (- c)* (fps_sin c)" ``` ``` 4023 (is "?lhs = ?rhs") ``` ``` 4024 proof (rule fps_ext) ``` ``` 4025 have th0: "- ((- 1::'a) ^ n) = (- 1)^Suc n" for n ``` ``` 4026 by simp ``` ``` 4027 show "?lhs \$ n = ?rhs \$ n" for n ``` ``` 4028 proof (cases "even n") ``` ``` 4029 case False ``` ``` 4030 then have n0: "n \<noteq> 0" by presburger ``` ``` 4031 from False have th1: "Suc ((n - 1) div 2) = Suc n div 2" ``` ``` 4032 by (cases n) simp_all ``` ``` 4033 have "?lhs\$n = of_nat (n+1) * (fps_cos c \$ (n+1))" by simp ``` ``` 4034 also have "\<dots> = of_nat (n+1) * ((- 1)^((n + 1) div 2) * c^Suc n / of_nat (fact (Suc n)))" ``` ``` 4035 using False by (simp add: fps_cos_def) ``` ``` 4036 also have "\<dots> = (- 1)^((n + 1) div 2)*c^Suc n * (of_nat (n+1) / (of_nat (Suc n) * of_nat (fact n)))" ``` ``` 4037 unfolding fact_Suc of_nat_mult ``` ``` 4038 by (simp add: field_simps del: of_nat_add of_nat_Suc) ``` ``` 4039 also have "\<dots> = (- 1)^((n + 1) div 2) * c^Suc n / of_nat (fact n)" ``` ``` 4040 by (simp add: field_simps del: of_nat_add of_nat_Suc) ``` ``` 4041 also have "\<dots> = (- ((- 1)^((n - 1) div 2))) * c^Suc n / of_nat (fact n)" ``` ``` 4042 unfolding th0 unfolding th1 by simp ``` ``` 4043 finally show ?thesis ``` ``` 4044 using False by (simp add: fps_sin_def field_simps) ``` ``` 4045 next ``` ``` 4046 case True ``` ``` 4047 then show ?thesis ``` ``` 4048 by (simp_all add: fps_deriv_def fps_sin_def fps_cos_def) ``` ``` 4049 qed ``` ``` 4050 qed ``` ``` 4051 ``` ``` 4052 lemma fps_sin_cos_sum_of_squares: "(fps_cos c)\<^sup>2 + (fps_sin c)\<^sup>2 = 1" ``` ``` 4053 (is "?lhs = _") ``` ``` 4054 proof - ``` ``` 4055 have "fps_deriv ?lhs = 0" ``` ``` 4056 apply (simp add: fps_deriv_power fps_sin_deriv fps_cos_deriv) ``` ``` 4057 apply (simp add: field_simps fps_const_neg[symmetric] del: fps_const_neg) ``` ``` 4058 done ``` ``` 4059 then have "?lhs = fps_const (?lhs \$ 0)" ``` ``` 4060 unfolding fps_deriv_eq_0_iff . ``` ``` 4061 also have "\<dots> = 1" ``` ``` 4062 by (auto simp add: fps_eq_iff numeral_2_eq_2 fps_mult_nth fps_cos_def fps_sin_def) ``` ``` 4063 finally show ?thesis . ``` ``` 4064 qed ``` ``` 4065 ``` ``` 4066 lemma fps_sin_nth_0 [simp]: "fps_sin c \$ 0 = 0" ``` ``` 4067 unfolding fps_sin_def by simp ``` ``` 4068 ``` ``` 4069 lemma fps_sin_nth_1 [simp]: "fps_sin c \$ 1 = c" ``` ``` 4070 unfolding fps_sin_def by simp ``` ``` 4071 ``` ``` 4072 lemma fps_sin_nth_add_2: ``` ``` 4073 "fps_sin c \$ (n + 2) = - (c * c * fps_sin c \$ n / (of_nat (n + 1) * of_nat (n + 2)))" ``` ``` 4074 unfolding fps_sin_def ``` ``` 4075 apply (cases n) ``` ``` 4076 apply simp ``` ``` 4077 apply (simp add: nonzero_divide_eq_eq nonzero_eq_divide_eq del: of_nat_Suc fact_Suc) ``` ``` 4078 apply (simp add: of_nat_mult del: of_nat_Suc mult_Suc) ``` ``` 4079 done ``` ``` 4080 ``` ``` 4081 lemma fps_cos_nth_0 [simp]: "fps_cos c \$ 0 = 1" ``` ``` 4082 unfolding fps_cos_def by simp ``` ``` 4083 ``` ``` 4084 lemma fps_cos_nth_1 [simp]: "fps_cos c \$ 1 = 0" ``` ``` 4085 unfolding fps_cos_def by simp ``` ``` 4086 ``` ``` 4087 lemma fps_cos_nth_add_2: ``` ``` 4088 "fps_cos c \$ (n + 2) = - (c * c * fps_cos c \$ n / (of_nat (n + 1) * of_nat (n + 2)))" ``` ``` 4089 unfolding fps_cos_def ``` ``` 4090 apply (simp add: nonzero_divide_eq_eq nonzero_eq_divide_eq del: of_nat_Suc fact_Suc) ``` ``` 4091 apply (simp add: of_nat_mult del: of_nat_Suc mult_Suc) ``` ``` 4092 done ``` ``` 4093 ``` ``` 4094 lemma nat_induct2: "P 0 \<Longrightarrow> P 1 \<Longrightarrow> (\<And>n. P n \<Longrightarrow> P (n + 2)) \<Longrightarrow> P (n::nat)" ``` ``` 4095 unfolding One_nat_def numeral_2_eq_2 ``` ``` 4096 apply (induct n rule: nat_less_induct) ``` ``` 4097 apply (case_tac n) ``` ``` 4098 apply simp ``` ``` 4099 apply (rename_tac m) ``` ``` 4100 apply (case_tac m) ``` ``` 4101 apply simp ``` ``` 4102 apply (rename_tac k) ``` ``` 4103 apply (case_tac k) ``` ``` 4104 apply simp_all ``` ``` 4105 done ``` ``` 4106 ``` ``` 4107 lemma nat_add_1_add_1: "(n::nat) + 1 + 1 = n + 2" ``` ``` 4108 by simp ``` ``` 4109 ``` ``` 4110 lemma eq_fps_sin: ``` ``` 4111 assumes 0: "a \$ 0 = 0" ``` ``` 4112 and 1: "a \$ 1 = c" ``` ``` 4113 and 2: "fps_deriv (fps_deriv a) = - (fps_const c * fps_const c * a)" ``` ``` 4114 shows "a = fps_sin c" ``` ``` 4115 apply (rule fps_ext) ``` ``` 4116 apply (induct_tac n rule: nat_induct2) ``` ``` 4117 apply (simp add: 0) ``` ``` 4118 apply (simp add: 1 del: One_nat_def) ``` ``` 4119 apply (rename_tac m, cut_tac f="\<lambda>a. a \$ m" in arg_cong [OF 2]) ``` ``` 4120 apply (simp add: nat_add_1_add_1 fps_sin_nth_add_2 ``` ``` 4121 del: One_nat_def of_nat_Suc of_nat_add add_2_eq_Suc') ``` ``` 4122 apply (subst minus_divide_left) ``` ``` 4123 apply (subst nonzero_eq_divide_eq) ``` ``` 4124 apply (simp del: of_nat_add of_nat_Suc) ``` ``` 4125 apply (simp only: ac_simps) ``` ``` 4126 done ``` ``` 4127 ``` ``` 4128 lemma eq_fps_cos: ``` ``` 4129 assumes 0: "a \$ 0 = 1" ``` ``` 4130 and 1: "a \$ 1 = 0" ``` ``` 4131 and 2: "fps_deriv (fps_deriv a) = - (fps_const c * fps_const c * a)" ``` ``` 4132 shows "a = fps_cos c" ``` ``` 4133 apply (rule fps_ext) ``` ``` 4134 apply (induct_tac n rule: nat_induct2) ``` ``` 4135 apply (simp add: 0) ``` ``` 4136 apply (simp add: 1 del: One_nat_def) ``` ``` 4137 apply (rename_tac m, cut_tac f="\<lambda>a. a \$ m" in arg_cong [OF 2]) ``` ``` 4138 apply (simp add: nat_add_1_add_1 fps_cos_nth_add_2 ``` ``` 4139 del: One_nat_def of_nat_Suc of_nat_add add_2_eq_Suc') ``` ``` 4140 apply (subst minus_divide_left) ``` ``` 4141 apply (subst nonzero_eq_divide_eq) ``` ``` 4142 apply (simp del: of_nat_add of_nat_Suc) ``` ``` 4143 apply (simp only: ac_simps) ``` ``` 4144 done ``` ``` 4145 ``` ``` 4146 lemma mult_nth_0 [simp]: "(a * b) \$ 0 = a \$ 0 * b \$ 0" ``` ``` 4147 by (simp add: fps_mult_nth) ``` ``` 4148 ``` ``` 4149 lemma mult_nth_1 [simp]: "(a * b) \$ 1 = a \$ 0 * b \$ 1 + a \$ 1 * b \$ 0" ``` ``` 4150 by (simp add: fps_mult_nth) ``` ``` 4151 ``` ``` 4152 lemma fps_sin_add: "fps_sin (a + b) = fps_sin a * fps_cos b + fps_cos a * fps_sin b" ``` ``` 4153 apply (rule eq_fps_sin [symmetric], simp, simp del: One_nat_def) ``` ``` 4154 apply (simp del: fps_const_neg fps_const_add fps_const_mult ``` ``` 4155 add: fps_const_add [symmetric] fps_const_neg [symmetric] ``` ``` 4156 fps_sin_deriv fps_cos_deriv algebra_simps) ``` ``` 4157 done ``` ``` 4158 ``` ``` 4159 lemma fps_cos_add: "fps_cos (a + b) = fps_cos a * fps_cos b - fps_sin a * fps_sin b" ``` ``` 4160 apply (rule eq_fps_cos [symmetric], simp, simp del: One_nat_def) ``` ``` 4161 apply (simp del: fps_const_neg fps_const_add fps_const_mult ``` ``` 4162 add: fps_const_add [symmetric] fps_const_neg [symmetric] ``` ``` 4163 fps_sin_deriv fps_cos_deriv algebra_simps) ``` ``` 4164 done ``` ``` 4165 ``` ``` 4166 lemma fps_sin_even: "fps_sin (- c) = - fps_sin c" ``` ``` 4167 by (auto simp add: fps_eq_iff fps_sin_def) ``` ``` 4168 ``` ``` 4169 lemma fps_cos_odd: "fps_cos (- c) = fps_cos c" ``` ``` 4170 by (auto simp add: fps_eq_iff fps_cos_def) ``` ``` 4171 ``` ``` 4172 definition "fps_tan c = fps_sin c / fps_cos c" ``` ``` 4173 ``` ``` 4174 lemma fps_tan_deriv: "fps_deriv (fps_tan c) = fps_const c / (fps_cos c)\<^sup>2" ``` ``` 4175 proof - ``` ``` 4176 have th0: "fps_cos c \$ 0 \<noteq> 0" by (simp add: fps_cos_def) ``` ``` 4177 from this have "fps_cos c \<noteq> 0" by (intro notI) simp ``` ``` 4178 hence "fps_deriv (fps_tan c) = ``` ``` 4179 fps_const c * (fps_cos c^2 + fps_sin c^2) / (fps_cos c^2)" ``` ``` 4180 by (simp add: fps_tan_def fps_divide_deriv power2_eq_square algebra_simps ``` ``` 4181 fps_sin_deriv fps_cos_deriv fps_const_neg[symmetric] div_mult_swap ``` ``` 4182 del: fps_const_neg) ``` ``` 4183 also note fps_sin_cos_sum_of_squares ``` ``` 4184 finally show ?thesis by simp ``` ``` 4185 qed ``` ``` 4186 ``` ``` 4187 text \<open>Connection to E c over the complex numbers --- Euler and de Moivre.\<close> ``` ``` 4188 ``` ``` 4189 lemma Eii_sin_cos: "E (ii * c) = fps_cos c + fps_const ii * fps_sin c" ``` ``` 4190 (is "?l = ?r") ``` ``` 4191 proof - ``` ``` 4192 have "?l \$ n = ?r \$ n" for n ``` ``` 4193 proof (cases "even n") ``` ``` 4194 case True ``` ``` 4195 then obtain m where m: "n = 2 * m" .. ``` ``` 4196 show ?thesis ``` ``` 4197 by (simp add: m fps_sin_def fps_cos_def power_mult_distrib power_mult power_minus [of "c ^ 2"]) ``` ``` 4198 next ``` ``` 4199 case False ``` ``` 4200 then obtain m where m: "n = 2 * m + 1" .. ``` ``` 4201 show ?thesis ``` ``` 4202 by (simp add: m fps_sin_def fps_cos_def power_mult_distrib ``` ``` 4203 power_mult power_minus [of "c ^ 2"]) ``` ``` 4204 qed ``` ``` 4205 then show ?thesis ``` ``` 4206 by (simp add: fps_eq_iff) ``` ``` 4207 qed ``` ``` 4208 ``` ``` 4209 lemma E_minus_ii_sin_cos: "E (- (ii * c)) = fps_cos c - fps_const ii * fps_sin c" ``` ``` 4210 unfolding minus_mult_right Eii_sin_cos by (simp add: fps_sin_even fps_cos_odd) ``` ``` 4211 ``` ``` 4212 lemma fps_const_minus: "fps_const (c::'a::group_add) - fps_const d = fps_const (c - d)" ``` ``` 4213 by (fact fps_const_sub) ``` ``` 4214 ``` ``` 4215 lemma fps_numeral_fps_const: "numeral i = fps_const (numeral i :: 'a::comm_ring_1)" ``` ``` 4216 by (fact numeral_fps_const) (* FIXME: duplicate *) ``` ``` 4217 ``` ``` 4218 lemma fps_cos_Eii: "fps_cos c = (E (ii * c) + E (- ii * c)) / fps_const 2" ``` ``` 4219 proof - ``` ``` 4220 have th: "fps_cos c + fps_cos c = fps_cos c * fps_const 2" ``` ``` 4221 by (simp add: numeral_fps_const) ``` ``` 4222 show ?thesis ``` ``` 4223 unfolding Eii_sin_cos minus_mult_commute ``` ``` 4224 by (simp add: fps_sin_even fps_cos_odd numeral_fps_const fps_divide_unit fps_const_inverse th) ``` ``` 4225 qed ``` ``` 4226 ``` ``` 4227 lemma fps_sin_Eii: "fps_sin c = (E (ii * c) - E (- ii * c)) / fps_const (2*ii)" ``` ``` 4228 proof - ``` ``` 4229 have th: "fps_const \<i> * fps_sin c + fps_const \<i> * fps_sin c = fps_sin c * fps_const (2 * ii)" ``` ``` 4230 by (simp add: fps_eq_iff numeral_fps_const) ``` ``` 4231 show ?thesis ``` ``` 4232 unfolding Eii_sin_cos minus_mult_commute ``` ``` 4233 by (simp add: fps_sin_even fps_cos_odd fps_divide_unit fps_const_inverse th) ``` ``` 4234 qed ``` ``` 4235 ``` ``` 4236 lemma fps_tan_Eii: ``` ``` 4237 "fps_tan c = (E (ii * c) - E (- ii * c)) / (fps_const ii * (E (ii * c) + E (- ii * c)))" ``` ``` 4238 unfolding fps_tan_def fps_sin_Eii fps_cos_Eii mult_minus_left E_neg ``` ``` 4239 apply (simp add: fps_divide_unit fps_inverse_mult fps_const_mult[symmetric] fps_const_inverse del: fps_const_mult) ``` ``` 4240 apply simp ``` ``` 4241 done ``` ``` 4242 ``` ``` 4243 lemma fps_demoivre: ``` ``` 4244 "(fps_cos a + fps_const ii * fps_sin a)^n = ``` ``` 4245 fps_cos (of_nat n * a) + fps_const ii * fps_sin (of_nat n * a)" ``` ``` 4246 unfolding Eii_sin_cos[symmetric] E_power_mult ``` ``` 4247 by (simp add: ac_simps) ``` ``` 4248 ``` ``` 4249 ``` ``` 4250 subsection \<open>Hypergeometric series\<close> ``` ``` 4251 ``` ``` 4252 definition "F as bs (c::'a::{field_char_0,field}) = ``` ``` 4253 Abs_fps (\<lambda>n. (foldl (\<lambda>r a. r* pochhammer a n) 1 as * c^n) / ``` ``` 4254 (foldl (\<lambda>r b. r * pochhammer b n) 1 bs * of_nat (fact n)))" ``` ``` 4255 ``` ``` 4256 lemma F_nth[simp]: "F as bs c \$ n = ``` ``` 4257 (foldl (\<lambda>r a. r* pochhammer a n) 1 as * c^n) / ``` ``` 4258 (foldl (\<lambda>r b. r * pochhammer b n) 1 bs * of_nat (fact n))" ``` ``` 4259 by (simp add: F_def) ``` ``` 4260 ``` ``` 4261 lemma foldl_mult_start: ``` ``` 4262 fixes v :: "'a::comm_ring_1" ``` ``` 4263 shows "foldl (\<lambda>r x. r * f x) v as * x = foldl (\<lambda>r x. r * f x) (v * x) as " ``` ``` 4264 by (induct as arbitrary: x v) (auto simp add: algebra_simps) ``` ``` 4265 ``` ``` 4266 lemma foldr_mult_foldl: ``` ``` 4267 fixes v :: "'a::comm_ring_1" ``` ``` 4268 shows "foldr (\<lambda>x r. r * f x) as v = foldl (\<lambda>r x. r * f x) v as" ``` ``` 4269 by (induct as arbitrary: v) (auto simp add: foldl_mult_start) ``` ``` 4270 ``` ``` 4271 lemma F_nth_alt: ``` ``` 4272 "F as bs c \$ n = foldr (\<lambda>a r. r * pochhammer a n) as (c ^ n) / ``` ``` 4273 foldr (\<lambda>b r. r * pochhammer b n) bs (of_nat (fact n))" ``` ``` 4274 by (simp add: foldl_mult_start foldr_mult_foldl) ``` ``` 4275 ``` ``` 4276 lemma F_E[simp]: "F [] [] c = E c" ``` ``` 4277 by (simp add: fps_eq_iff) ``` ``` 4278 ``` ``` 4279 lemma F_1_0[simp]: "F [1] [] c = 1/(1 - fps_const c * X)" ``` ``` 4280 proof - ``` ``` 4281 let ?a = "(Abs_fps (\<lambda>n. 1)) oo (fps_const c * X)" ``` ``` 4282 have th0: "(fps_const c * X) \$ 0 = 0" by simp ``` ``` 4283 show ?thesis unfolding gp[OF th0, symmetric] ``` ``` 4284 by (auto simp add: fps_eq_iff pochhammer_fact[symmetric] ``` ``` 4285 fps_compose_nth power_mult_distrib cond_value_iff setsum.delta' cong del: if_weak_cong) ``` ``` 4286 qed ``` ``` 4287 ``` ``` 4288 lemma F_B[simp]: "F [-a] [] (- 1) = fps_binomial a" ``` ``` 4289 by (simp add: fps_eq_iff gbinomial_pochhammer algebra_simps) ``` ``` 4290 ``` ``` 4291 lemma F_0[simp]: "F as bs c \$ 0 = 1" ``` ``` 4292 apply simp ``` ``` 4293 apply (subgoal_tac "\<forall>as. foldl (\<lambda>(r::'a) (a::'a). r) 1 as = 1") ``` ``` 4294 apply auto ``` ``` 4295 apply (induct_tac as) ``` ``` 4296 apply auto ``` ``` 4297 done ``` ``` 4298 ``` ``` 4299 lemma foldl_prod_prod: ``` ``` 4300 "foldl (\<lambda>(r::'b::comm_ring_1) (x::'a::comm_ring_1). r * f x) v as * foldl (\<lambda>r x. r * g x) w as = ``` ``` 4301 foldl (\<lambda>r x. r * f x * g x) (v * w) as" ``` ``` 4302 by (induct as arbitrary: v w) (auto simp add: algebra_simps) ``` ``` 4303 ``` ``` 4304 ``` ``` 4305 lemma F_rec: ``` ``` 4306 "F as bs c \$ Suc n = ((foldl (\<lambda>r a. r* (a + of_nat n)) c as) / ``` ``` 4307 (foldl (\<lambda>r b. r * (b + of_nat n)) (of_nat (Suc n)) bs )) * F as bs c \$ n" ``` ``` 4308 apply (simp del: of_nat_Suc of_nat_add fact_Suc) ``` ``` 4309 apply (simp add: foldl_mult_start del: fact_Suc of_nat_Suc) ``` ``` 4310 unfolding foldl_prod_prod[unfolded foldl_mult_start] pochhammer_Suc ``` ``` 4311 apply (simp add: algebra_simps of_nat_mult) ``` ``` 4312 done ``` ``` 4313 ``` ``` 4314 lemma XD_nth[simp]: "XD a \$ n = (if n = 0 then 0 else of_nat n * a\$n)" ``` ``` 4315 by (simp add: XD_def) ``` ``` 4316 ``` ``` 4317 lemma XD_0th[simp]: "XD a \$ 0 = 0" ``` ``` 4318 by simp ``` ``` 4319 lemma XD_Suc[simp]:" XD a \$ Suc n = of_nat (Suc n) * a \$ Suc n" ``` ``` 4320 by simp ``` ``` 4321 ``` ``` 4322 definition "XDp c a = XD a + fps_const c * a" ``` ``` 4323 ``` ``` 4324 lemma XDp_nth[simp]: "XDp c a \$ n = (c + of_nat n) * a\$n" ``` ``` 4325 by (simp add: XDp_def algebra_simps) ``` ``` 4326 ``` ``` 4327 lemma XDp_commute: "XDp b \<circ> XDp (c::'a::comm_ring_1) = XDp c \<circ> XDp b" ``` ``` 4328 by (auto simp add: XDp_def fun_eq_iff fps_eq_iff algebra_simps) ``` ``` 4329 ``` ``` 4330 lemma XDp0 [simp]: "XDp 0 = XD" ``` ``` 4331 by (simp add: fun_eq_iff fps_eq_iff) ``` ``` 4332 ``` ``` 4333 lemma XDp_fps_integral [simp]: "XDp 0 (fps_integral a c) = X * a" ``` ``` 4334 by (simp add: fps_eq_iff fps_integral_def) ``` ``` 4335 ``` ``` 4336 lemma F_minus_nat: ``` ``` 4337 "F [- of_nat n] [- of_nat (n + m)] (c::'a::{field_char_0,field}) \$ k = ``` ``` 4338 (if k \<le> n then ``` ``` 4339 pochhammer (- of_nat n) k * c ^ k / (pochhammer (- of_nat (n + m)) k * of_nat (fact k)) ``` ``` 4340 else 0)" ``` ``` 4341 "F [- of_nat m] [- of_nat (m + n)] (c::'a::{field_char_0,field}) \$ k = ``` ``` 4342 (if k \<le> m then ``` ``` 4343 pochhammer (- of_nat m) k * c ^ k / (pochhammer (- of_nat (m + n)) k * of_nat (fact k)) ``` ``` 4344 else 0)" ``` ``` 4345 by (auto simp add: pochhammer_eq_0_iff) ``` ``` 4346 ``` ``` 4347 lemma setsum_eq_if: "setsum f {(n::nat) .. m} = (if m < n then 0 else f n + setsum f {n+1 .. m})" ``` ``` 4348 apply simp ``` ``` 4349 apply (subst setsum.insert[symmetric]) ``` ``` 4350 apply (auto simp add: not_less setsum_head_Suc) ``` ``` 4351 done ``` ``` 4352 ``` ``` 4353 lemma pochhammer_rec_if: "pochhammer a n = (if n = 0 then 1 else a * pochhammer (a + 1) (n - 1))" ``` ``` 4354 by (cases n) (simp_all add: pochhammer_rec) ``` ``` 4355 ``` ``` 4356 lemma XDp_foldr_nth [simp]: "foldr (\<lambda>c r. XDp c \<circ> r) cs (\<lambda>c. XDp c a) c0 \$ n = ``` ``` 4357 foldr (\<lambda>c r. (c + of_nat n) * r) cs (c0 + of_nat n) * a\$n" ``` ``` 4358 by (induct cs arbitrary: c0) (auto simp add: algebra_simps) ``` ``` 4359 ``` ``` 4360 lemma genric_XDp_foldr_nth: ``` ``` 4361 assumes f: "\<forall>n c a. f c a \$ n = (of_nat n + k c) * a\$n" ``` ``` 4362 shows "foldr (\<lambda>c r. f c \<circ> r) cs (\<lambda>c. g c a) c0 \$ n = ``` ``` 4363 foldr (\<lambda>c r. (k c + of_nat n) * r) cs (g c0 a \$ n)" ``` ``` 4364 by (induct cs arbitrary: c0) (auto simp add: algebra_simps f) ``` ``` 4365 ``` ``` 4366 lemma dist_less_imp_nth_equal: ``` ``` 4367 assumes "dist f g < inverse (2 ^ i)" ``` ``` 4368 and"j \<le> i" ``` ``` 4369 shows "f \$ j = g \$ j" ``` ``` 4370 proof (rule ccontr) ``` ``` 4371 assume "f \$ j \<noteq> g \$ j" ``` ``` 4372 hence "f \<noteq> g" by auto ``` ``` 4373 with assms have "i < subdegree (f - g)" ``` ``` 4374 by (simp add: split_if_asm dist_fps_def) ``` ``` 4375 also have "\<dots> \<le> j" ``` ``` 4376 using \<open>f \$ j \<noteq> g \$ j\<close> by (intro subdegree_leI) simp_all ``` ``` 4377 finally show False using \<open>j \<le> i\<close> by simp ``` ``` 4378 qed ``` ``` 4379 ``` ``` 4380 lemma nth_equal_imp_dist_less: ``` ``` 4381 assumes "\<And>j. j \<le> i \<Longrightarrow> f \$ j = g \$ j" ``` ``` 4382 shows "dist f g < inverse (2 ^ i)" ``` ``` 4383 proof (cases "f = g") ``` ``` 4384 case True ``` ``` 4385 then show ?thesis by simp ``` ``` 4386 next ``` ``` 4387 case False ``` ``` 4388 with assms have "dist f g = inverse (2 ^ subdegree (f - g))" ``` ``` 4389 by (simp add: split_if_asm dist_fps_def) ``` ``` 4390 moreover ``` ``` 4391 from assms and False have "i < subdegree (f - g)" ``` ``` 4392 by (intro subdegree_greaterI) simp_all ``` ``` 4393 ultimately show ?thesis by simp ``` ``` 4394 qed ``` ``` 4395 ``` ``` 4396 lemma dist_less_eq_nth_equal: "dist f g < inverse (2 ^ i) \<longleftrightarrow> (\<forall>j \<le> i. f \$ j = g \$ j)" ``` ``` 4397 using dist_less_imp_nth_equal nth_equal_imp_dist_less by blast ``` ``` 4398 ``` ``` 4399 instance fps :: (comm_ring_1) complete_space ``` ``` 4400 proof ``` ``` 4401 fix X :: "nat \<Rightarrow> 'a fps" ``` ``` 4402 assume "Cauchy X" ``` ``` 4403 obtain M where M: "\<forall>i. \<forall>m \<ge> M i. \<forall>j \<le> i. X (M i) \$ j = X m \$ j" ``` ``` 4404 proof - ``` ``` 4405 have "\<exists>M. \<forall>m \<ge> M. \<forall>j\<le>i. X M \$ j = X m \$ j" for i ``` ``` 4406 proof - ``` ``` 4407 have "0 < inverse ((2::real)^i)" by simp ``` ``` 4408 from metric_CauchyD[OF \<open>Cauchy X\<close> this] dist_less_imp_nth_equal ``` ``` 4409 show ?thesis by blast ``` ``` 4410 qed ``` ``` 4411 then show ?thesis using that by metis ``` ``` 4412 qed ``` ``` 4413 ``` ``` 4414 show "convergent X" ``` ``` 4415 proof (rule convergentI) ``` ``` 4416 show "X ----> Abs_fps (\<lambda>i. X (M i) \$ i)" ``` ``` 4417 unfolding tendsto_iff ``` ``` 4418 proof safe ``` ``` 4419 fix e::real assume e: "0 < e" ``` ``` 4420 have "(\<lambda>n. inverse (2 ^ n) :: real) ----> 0" by (rule LIMSEQ_inverse_realpow_zero) simp_all ``` ``` 4421 from this and e have "eventually (\<lambda>i. inverse (2 ^ i) < e) sequentially" ``` ``` 4422 by (rule order_tendstoD) ``` ``` 4423 then obtain i where "inverse (2 ^ i) < e" ``` ``` 4424 by (auto simp: eventually_sequentially) ``` ``` 4425 have "eventually (\<lambda>x. M i \<le> x) sequentially" ``` ``` 4426 by (auto simp: eventually_sequentially) ``` ``` 4427 then show "eventually (\<lambda>x. dist (X x) (Abs_fps (\<lambda>i. X (M i) \$ i)) < e) sequentially" ``` ``` 4428 proof eventually_elim ``` ``` 4429 fix x ``` ``` 4430 assume x: "M i \<le> x" ``` ``` 4431 have "X (M i) \$ j = X (M j) \$ j" if "j \<le> i" for j ``` ``` 4432 using M that by (metis nat_le_linear) ``` ``` 4433 with x have "dist (X x) (Abs_fps (\<lambda>j. X (M j) \$ j)) < inverse (2 ^ i)" ``` ``` 4434 using M by (force simp: dist_less_eq_nth_equal) ``` ``` 4435 also note \<open>inverse (2 ^ i) < e\<close> ``` ``` 4436 finally show "dist (X x) (Abs_fps (\<lambda>j. X (M j) \$ j)) < e" . ``` ``` 4437 qed ``` ``` 4438 qed ``` ``` 4439 qed ``` ``` 4440 qed ``` ``` 4441 ``` ``` 4442 end ```

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главная страница » Приложения » Образование » Electrical Engineering # Electrical Engineering 3.8 for Android Softecks ## Описание для Electrical Engineering ✴Electrical engineering is the study of electromagnetism, electricity and electronics. This electrical engineering App is better explaining these concepts and basics of electricity.The App is designed for easy learning, revisions, references at the time of exams and interviews. This app cover most of related topics and Detailed explanation with all the basics topics. Be a professional with this app. This app is for all engineering students and professionals across the world. ✴ ►In this App, you will learn topics such as Inductance and Capacitance, Transients, Steady-State Sinusoidal Analysis, and Frequency Response, Bode Plots, and Resonance plus much more.Comprehensive handbook detailing the application of electrical engineering to the oil, gas , petrochemical industries and offshore industries. These have significantly different characteristics to large-scale power generation and long distance public utility industries. An essential reference for electrical engineering students, designers, operations and maintenance engineers and technicians.☆ 【Few Important Topics covered in this App are Listed Below】 ⇢ What Is Electrical Engineering ⇢ Nature of Electricity ⇢ Norton Theorem ⇢ Classification of Engineering Materials ⇢ Why Measure Voltage? ⇢ Battery Technology ⇢ What is black Body? ⇢ Power Plants and Types ⇢ Control Engineering ⇢ Electrical Power Transformer ⇢ Electrical Motor ⇢ Induction Motor ⇢ DC Motor or Direct Current Motor ⇢ Alternator Synchronous Generator ⇢ Electrical Switchgear Protection ⇢ Digital Electronics ⇢ What is Electrical Drive? ⇢ Kirchhoff’s Voltage Law (KVL) ⇢ Current Divider Circuits ⇢ What Is LiDAR and How Can I Use It? ⇢ Modelling the Pulse-Width Modulator ⇢ Switching Losses: Effects on Semiconductors ⇢ An Intro to Multiplexing: Basis of Telecommunications ⇢ Boolean Identities ⇢ A Practical Introduction to Operational Amplifiers ⇢ eMMCs: An Introduction ⇢ Universal Logic Gates ⇢ Understanding and Applying the Hall Effect ⇢ Introduction to Operational Amplifiers ⇢ AC Phase ⇢ Diodes and Rectifiers ⇢ Negative Feedback ⇢ Understanding Illuminance: What’s in a Lux?"); ⇢ Measuring and Calculating Lux Values ⇢ Measuring and Calculating Lux Values, Part 2 ⇢ Characteristics of Operational Amplifiers ⇢ The Inverting Configuration of an Amplifier ⇢ Non-inverting Configuration of an Operational Amplifier ⇢ Characteristics of Junction Diodes ⇢ Electric Fields and Capacitance ⇢ Factors Affecting Capacitance ⇢ Introduction to Capacitive Touch Sensing ⇢ Circuits and Techniques for Implementing Capacitive Touch Sensing ⇢ Analysis of Forward Conducting Diodes ⇢ How Sensor Fusion Works ⇢ Generation, Transmission and Distribution of Electric Power ⇢ Thermal, hydel & nuclear power stations ⇢ Transmission of power ⇢ Single line representation of power system ⇢ Solution of Electric Circuit Based on Mesh (Loop) Current Method ⇢ Solution of Electric Circuit Based on Node Voltage Method ⇢ Examples of Electric Circuit Based on Node Voltage Method ⇢ Wye (Y) - Delta (∆) OR Delta (∆)-Wye (Y) Transformations ⇢ Conversion from Delta (Δ) to Star or Wye (Y) ⇢ Application of Star (Y) to Delta (Δ) or Delta (Δ) to Star (Y) Transformation ⇢ Examples of Star (Y) to Delta (Δ) or Delta (Δ) to Star (Y) Transformation ⇢ Superposition Theorem in the context of dc voltage and current sources acting in a ⇢ ⇢ resistive network ⇢ Application of superposition theorem ⇢ Example of superposition theorem ⇢ Limitations of superposition Theorem ⇢ Thevenin’s and Norton’s theorems in the context of dc voltage and current sources acting in a resistive network ⇢ The procedure for applying Thevenin’s theorem ⇢ Maximum Power Transfer Theorem ⇢ Study of DC transients in R-L and R-C circuits ⇢ Inductance calculation from physical dimension of coil ⇢ Study of dc transients and steady state response of a series R-L circuit. ⇢ Energy stored in an inductor ⇢ Capacitor and its behaviour ⇢ Response of a series R-L-C circuit due to a dc voltage source 2019-07-30 ## Electrical Engineering Tags By adding tag words that describe for Games&Apps, you're helping to make these Games and Apps be more discoverable by other APKPure users. Категория: Бесплатно Образование Приложение Дата публикации: Последняя версия: 3.8 Получить: Требования: Android 4.4+ Жаловаться: Сообщить о неприемлемом содержании Предыдущие версии • V3.8 26.8 MB APK Electrical Engineering 2019-07-30 Electrical Engineering 3.8 (18) Обновлено: 2019-07-30 Требуется Android: Android 4.4+ (Kitkat, API 19) DPI: 120-640dpi Arch: universal Файл SHA1-хэша: dc09a3fcbd84e0309d2c1400f846b8281c181665 Размер файла: 26.8 MB Что нового: Скачать • V3.7 26.8 MB APK Electrical Engineering 2019-06-28 Electrical Engineering 3.7 (17) Обновлено: 2019-06-28 Требуется Android: Android 4.4+ (Kitkat, API 19) DPI: 120-640dpi Arch: universal Файл SHA1-хэша: e09197f22706a789dfcc6402a07613516a802e0c Размер файла: 26.8 MB Что нового: Скачать • V3.6 22.6 MB APK Electrical Engineering 2019-03-20 Electrical Engineering 3.6 (16) Обновлено: 2019-03-20 Требуется Android: Android 4.0.3+ (Ice Cream Sandwich MR1, API 15) DPI: 120-640dpi Arch: universal Размер файла: 22.6 MB Что нового: Скачать • V3.4 47.7 MB APK Electrical Engineering 2019-03-14 Electrical Engineering 3.4 (14) Обновлено: 2019-03-14 Требуется Android: Android 4.0.3+ (Ice Cream Sandwich MR1, API 15) DPI: 120-640dpi Arch: universal

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# Properties Label 31200.cb Number of curves $4$ Conductor $31200$ CM no Rank $0$ Graph # Related objects Show commands: SageMath sage: E = EllipticCurve("cb1") sage: E.isogeny_class() ## Elliptic curves in class 31200.cb sage: E.isogeny_class().curves LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality 31200.cb1 31200n4 $$[0, 1, 0, -93608, -11054712]$$ $$11339065490696/351$$ $$2808000000$$ $$[2]$$ $$98304$$ $$1.3174$$ 31200.cb2 31200n3 $$[0, 1, 0, -9233, 45663]$$ $$1360251712/771147$$ $$49353408000000$$ $$[2]$$ $$98304$$ $$1.3174$$ 31200.cb3 31200n1 $$[0, 1, 0, -5858, -173712]$$ $$22235451328/123201$$ $$123201000000$$ $$[2, 2]$$ $$49152$$ $$0.97086$$ $$\Gamma_0(N)$$-optimal 31200.cb4 31200n2 $$[0, 1, 0, -2608, -362212]$$ $$-245314376/6908733$$ $$-55269864000000$$ $$[2]$$ $$98304$$ $$1.3174$$ ## Rank sage: E.rank() The elliptic curves in class 31200.cb have rank $$0$$. ## Complex multiplication The elliptic curves in class 31200.cb do not have complex multiplication. ## Modular form 31200.2.a.cb sage: E.q_eigenform(10) $$q + q^{3} + q^{9} + 4q^{11} - q^{13} + 6q^{17} - 8q^{19} + O(q^{20})$$ ## Isogeny matrix sage: E.isogeny_class().matrix() The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering. $$\left(\begin{array}{rrrr} 1 & 4 & 2 & 4 \\ 4 & 1 & 2 & 4 \\ 2 & 2 & 1 & 2 \\ 4 & 4 & 2 & 1 \end{array}\right)$$ ## Isogeny graph sage: E.isogeny_graph().plot(edge_labels=True) The vertices are labelled with LMFDB labels.

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Steady flow energy equation and Bernoulli Equation For steady flow energy equation , Consider a control volume, in which mass ∆m1/∆t enters through inlet with parameters U = internal energy v = Velocity V = specific volume ( V1 = V2 ) P = pressure Similarly mass ∆m2/∆t exits through outlet. Therefore mass concentrationin control volume is (∆m/∆t) control volume = ∆m1/∆t – ∆m2/∆t •(∆E/∆t) in = internal energy+ flow work + kinetic energy+ potential energy + heat Therefore , (∆E/∆t) in = ∆m1/∆t { U1 + P1V1 + v12/2 + gZ1 } + ∆Q/∆t At exit , (∆E/∆t)out = ∆m2/∆t { U2 + P2V2 + v2 2/2 + gZ2 } + ∆W/∆t (E/∆t) control volume= (∆E/∆t)in – (∆E/∆t)out (E/∆t) control volume= ∆m1/∆t { h1 + v12/2 + gZ1 } + ∆Q/∆t – ∆m2/∆t { h2 + v2 2/2 + gZ2} – ∆W/∆t Assumptions : • (∆m/∆t) control volume = 0 • (E/∆t) control volume= 0 Here , (∆m/∆t) control volume = ∆m1/∆t – ∆m2/∆t = 0 So , ∆m1/∆t = ∆m2/∆t = m Similarly, (∆E/∆t)in = (∆E/∆t)out Therefore , m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t this is Steady flow energy equation or SFEE . Bernoulli Equation : m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t Assume , (a)No heat transfer (b) No work transfer (c)No change in internal energy (d)No change in density ( p = constant) Then , m{ h1 + v12/2 + gZ1 } + ∆Q/∆t = m{ h2 + v2 2/2 + gZ2 } + ∆W/∆t and U1 + P1V1 + v12/2 + gZ1 = U2 + P2V2 + v2 2/2 + gZ2 As , Specific Volume = 1/ Density i.e., V = 1/ p Therefore P1/p + v1 2/2 + gZ1 = P2 /p+ v2 2/2+ gZ2 P1/pg + v1 2/2g + Z1 = P2 /pg+ v22/2g+ Z2 This is Bernoulli Equation Conclusion : Hence both Steady flow energy equation and Bernoulli Equation are based on energy conservation and Bernoulli Equation is limited form of steady flow energy equation . This is also known as the first law of thermodynamics for open system.

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# Calculus Topic Beta & Gamma functions Statement of Dirichlet’s Theorem $\int \int \int_{V} x^{l-1} y^{m-1} z^{n-1} dx dy ,dz = \frac { \Gamma {(l)} \Gamma {(m)} \Gamma {(n)} }{ \Gamma{(l+m+n+1)} }$ , where V is the region given by $x \ge 0 y \ge 0 z \ge 0 ## Derivative of x squared is 2x or x ? Where is the fallacy? We all know that the derivative of$x^2$is 2x. But what if someone proves it to be just x? ## Solving Ramanujan’s Puzzling Problem Consider a sequence of functions as follows:-$ f_1 (x) = \sqrt {1+\sqrt {x} } f_2 (x) = \sqrt{1+ \sqrt {1+2 \sqrt {x} } } f_3 (x) = \sqrt {1+ \sqrt {1+2 \sqrt {1+3 \sqrt {x} } } } $……and so on to$ f_n Get up to 80% discount on various products with exclusive coupons

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0 # How many feet and inches in 1.81 meter? Wiki User 2010-04-29 20:46:41 5 feet and 11.3 inches. Wiki User 2010-04-29 20:46:41 Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.74 1208 Reviews Earn +20 pts Q: How many feet and inches in 1.81 meter?

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.... How to Find Your Subject Study Group & Join ....   .... Find Your Subject Study Group & Join .... We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>> www.vustudents.ning.com Study Groups By Subject code Wise (Click Below on your university link & Join Your Subject Group) [ + VU Study Groups Subject Code Wise ]  [ + COMSATS Virtual Campus Study Groups Subject Code Wise ] Looking For Something at vustudents.ning.com?Search Here # CS504 Assignment No 4 last date-13-feb-2017 Assignment No.4   Semester Fall 2016 CS504 - Software Engineering-I Total Marks: 15       Due Date: 13/02/2017 Objective:   To learn and understand the basic concepts of Testing .     Instructions:     Please read the following instructions carefully before solving & submitting assignment: Assignment should be in your own wordings not copied from net, handouts or books.    It should be clear that your assignment will not get any credit (zero marks) if:     The assignment is submitted after due date. The submitted assignment does not open or file corrupt. The assignment is copied (from other student or copy from handouts or internet).     For any query about the assignment, please communicate at CS504@vu.edu.pk   You are advised to please upload the assignment in MS Word format only (other formats i.e. .pdf, images etc. will not be accepted.). Question No 1:                                                                                                     15 Marks Cyclomatic complexity (CC) is a technique, which is used to measure the source code complexity in a program during testing phase. It is calculated by developing a Control Flow Graph of the code that measures the number of linearly-independent paths through a program module. For the code given below: void function(int x, int y) { for(int i=0; i<100; i++) { if(x != i) print x-i; Else Print x*i; } while(x>0 and x != y) { if(x<y) x++; else y++; print x+y; } } # a)    Draw the control flow graph (Check the handouts carefully) b)    Calculate the Cyclomatic Complexity c)     Identify all the possible traversable paths This Content Originally Published by a member of VU Students. Views: 10788 See Your Saved Posts Timeline Attachments: ### Replies to This Discussion D3W4N4 node 1 void function sy start ho gi ya for lopp sy kindly bata dain hosh karo cs504 walo assignment ai hoi hai koe to discuss karo.................... please check Handout page no 208 Lec No. 39 hope it will help us please send solved assignment Our main purpose here discussion not just Solution We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions. Read More>> Note:- For Important Helping Material related to this subject (Solved MCQs, Short Notes, Solved past Papers, E-Books, FAQ,Short Questions Answers & more). You must view all the featured Discussion in this subject group. For how you can view all the Featured discussions click on the Back to Subject Name Discussions link below the title of this Discussion & then under featured Discussion corner click on the view all link. Or visit this link & .•°How to Download past papers from study groups°•. Please Click on the below link to see… .... How to Find Your Subject Study Group & Join .... Paths: 1. 1-2-4-1-10 2. 1-3-4-1-10 3. 1-5-6-9-10 4. 1-5-7-8-9-10 5. 1-5-6-9-5-7-8-9-10 amjad shafiq kaya ye paths bilkul thik hain it depends how you numbering the code. pleas koi idea dy. and please tell what is the cyclomatic complexity of this source code.? im waiting for your reply 4 ## Forum Categorizes Job's & Careers (Latest Jobs) Scholarship (Latest Scholarships) Internship (Latest Internships) ::::::::::: More Categorizes ::::::::::: ## Latest Activity 48 minutes ago 1 hour ago Atif Chohan left a comment for Muhammad Basil 3 hours ago HMA replied to +^" βℓµ ʍ๏๏ɲ "'s discussion suno......!!!! 4 hours ago 4 hours ago 5 hours ago Ayeza Malik, LuCkY BoY and AZeem Zulfiqar joined + M.Tariq Malik's group ### CS605 Software Engineering-II 5 hours ago Malik Rizwan BbLu replied to Malik Rizwan BbLu's discussion cs304 GDB in the group CS304 Object Oriented Programming 5 hours ago 5 hours ago Malik Rizwan BbLu added a discussion to the group CS401 Computer Architecture and Assembly Language Programming ### cs401 GDB 5 hours ago Malik Rizwan BbLu added a discussion to the group CS304 Object Oriented Programming ### cs304 GDB 5 hours ago + " ĸnιgнт ѕтealтн posted a discussion 5 hours ago 1 2

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# TONNE/KILOLITER TO MILLIGRAM/CUBIC KILOMETER CONVERTER FROM TO The result of your conversion between tonne/kiloliter and milligram/cubic kilometer appears here ## TONNE/KILOLITER TO MILLIGRAM/CUBIC KILOMETER (T/kl TO mg/km³) FORMULA To convert between Tonne/kiloliter and Milligram/cubic Kilometer you have to do the following: First divide 1000/1 / 0.000001/1.0E+9 = 1000000000000000128. Then multiply the amount of Tonne/kiloliter you want to convert to Milligram/cubic Kilometer, use the chart below to guide you. ## TONNE/KILOLITER TO MILLIGRAM/CUBIC KILOMETER (T/kl TO mg/km³) CHART • 1 tonne/kiloliter in milligram/cubic kilometer = 1000000000000000128. T/kl • 10 tonne/kiloliter in milligram/cubic kilometer = 10000000000000002048. T/kl • 50 tonne/kiloliter in milligram/cubic kilometer = 50000000000000008192. T/kl • 100 tonne/kiloliter in milligram/cubic kilometer = 100000000000000016384. T/kl • 250 tonne/kiloliter in milligram/cubic kilometer = 250000000000000032768. T/kl • 500 tonne/kiloliter in milligram/cubic kilometer = 500000000000000065536. T/kl • 1,000 tonne/kiloliter in milligram/cubic kilometer = 1000000000000000131072. T/kl • 10,000 tonne/kiloliter in milligram/cubic kilometer = 10000000000000002097152. T/kl Symbol: T/kl No description Symbol: mg/km³ No description

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# Does Smith normal form imply PID? Let $R$ be a nonzero commutative ring with $1$, such that all finite matrices over $R$ have a Smith normal form. Does it follow that $R$ is a principal ideal domain? If this fails, suppose we additionally suppose that $R$ is an integral domain? What can we say if we impose the additional condition that the diagonal entries be unique up to associates? • Doesn't Smith normal form hold for a principal ideal ring, possibly with zero divisors? I am thinking of $\mathbb{Z}/n\mathbb{Z}.$ – Victor Protsak Jul 10 '10 at 6:42 • For a summary of what is known about this problem, see Theorem 2.1 of my paper math.mit.edu/~rstan/papers/snf_survey.pdf. – Richard Stanley Feb 25 at 14:09 The implication is false without the assumption that R is Noetherian, because finite matrices don't detect enough information about infinitely generated ideals. For example, let R be the ring $$\bigcup_{n \geq 0} k[[t^{1/n}]]$$ where $k$ is a field (an indiscrete valuation ring). Any finite matrix with coefficients in R comes from a subring $k[[t^{1/N}]]$ for some large $N$, and hence can be reduced to Smith normal form within this smaller PID. However, the ideal $\cup (t^{1/N})$ is not principal. • @Tyler: I don't think it affects the rest of your argument, but: to get a valuation ring, don't you want $k[[t^{\frac{1}{n}}]]$ instead of $k[t^{\frac{1}{n}}]$? – Pete L. Clark Jul 10 '10 at 18:20 • Yes, you are correct - I added that sentence at the last minute. It is simply a ring with an indiscrete valuation. – Tyler Lawson Jul 10 '10 at 19:36 If every matrix has a Smith normal form, then every finitely generated $R$-submodule $M$ of $R^n$ satisfies $R^n/M$ is a finite direct sum of modules isomorphic to $R/aR$. If $R$ is Noetherian this implies that every finitely generated module is a direct sum of modules of the form $R/aR$. So if $I$ is a maximal ideal of the Noetherian $R$ then $R/I$ is a simple module, so if $R/I\cong R/aR$ then $I=aR$ is principal. So in a Noetherian ring with Smith normal form for all matrices, every maximal ideal is principal. Does this imply that all ideals are principal?....I'm not sure :-) • For R a domain, it implies that R is one-dimensional regular, hence Dedekind, so every nonzero ideal is a product of maximal ideals, therefore principal itself. – user2035 Jul 10 '10 at 7:13 • Thanks a-fortiori: each localization at a maximal ideal is a local ring of height at most one, so $R$ has Krull dimension $\le 1$. – Robin Chapman Jul 10 '10 at 7:19 • A commutative noetherian ring whose maximal ideals are principal is indeed a principal ideal ring (even if it is not domain). See Theorem 12.3 of Kaplansky's article "Elementary divisors and modules," Trans. Amer. Math. Soc. 66 (1949), 464-491. – Manny Reyes Jul 10 '10 at 14:55 Such rings are apparently called elementary divisor rings. They are necessarily Bezout rings (i.e. every finitely-generated ideal is principal), but not easy to characterize completely. The first paper giving a nontrivial sufficient condition (beyond classical case) seems to be Helmer, Olaf The elementary divisor theorem for certain rings without chain condition. Bull. Amer. Math. Soc. 49, (1943). 225--236, MR More complete results are in a series of papers starting with Larsen, Max D.; Lewis, William J.; Shores, Thomas S. Elementary divisor rings and finitely presented modules. Trans. Amer. Math. Soc. 187 (1974), 231--248, MR Work on ring-theoretic generalizations of Hermite/Smith normal forms goes way back, but made it into the mainstream via classic papers by Helmer and Kaplansky. Nowadays such rings are called elementary divisor rings, or rings with elementary divisors (r.e.d.) or Helmer rings, etc. A search on such terms, and for citations of Kap's classic paper [1] should quickly answer all your questions and then some. [1] I. Kaplansky, "Elementary divisors and modules," Trans. Am. Math. Soc., 66, 464-491. (1949). http://www.ams.org/journals/tran/1949-066-02/S0002-9947-1949-0031470-3/S0002-9947-1949-0031470-3.pdf

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html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; } Cosh http://functions.wolfram.com/01.20.21.0762.01 Input Form Integrate[Cos[b z^2 + d z] Cosh[c z^2 + f z + g], z] == (1/2) Sqrt[Pi/2] ((1/Sqrt[-b + I c]) (Cos[(-d + I f)^2/(4 (-b + I c)) - I g] FresnelC[(-d + I f + 2 (-b + I c) z)/(Sqrt[-b + I c] Sqrt[2 Pi])] + FresnelS[(-d + I f + 2 (-b + I c) z)/(Sqrt[-b + I c] Sqrt[2 Pi])] Sin[(-d + I f)^2/(4 (-b + I c)) - I g]) - (1/Sqrt[-b - I c]) (Cos[-((d + I f)^2/(4 (b + I c))) + I g] FresnelC[(d + I f + 2 (b + I c) z)/(Sqrt[-b - I c] Sqrt[2 Pi])] + FresnelS[(d + I f + 2 (b + I c) z)/(Sqrt[-b - I c] Sqrt[2 Pi])] Sin[-((d + I f)^2/(4 (b + I c))) + I g])) Standard Form Cell[BoxData[RowBox[List[RowBox[List["\[Integral]", RowBox[List[RowBox[List["Cos", "[", RowBox[List[RowBox[List["b", " ", SuperscriptBox["z", "2"]]], "+", RowBox[List["d", " ", "z"]]]], "]"]], RowBox[List["Cosh", "[", RowBox[List[RowBox[List["c", " ", SuperscriptBox["z", "2"]]], "+", RowBox[List["f", " ", "z"]], "+", "g"]], "]"]], RowBox[List["\[DifferentialD]", "z"]]]]]], "\[Equal]", RowBox[List[FractionBox["1", "2"], " ", SqrtBox[FractionBox["\[Pi]", "2"]], " ", RowBox[List["(", RowBox[List[FractionBox["1", SqrtBox[RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]]]]]], RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["Cos", "[", RowBox[List[FractionBox[SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["-", "d"]], "+", RowBox[List["\[ImaginaryI]", " ", "f"]]]], ")"]], "2"], RowBox[List["4", " ", RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]]]], ")"]]]]], "-", RowBox[List["\[ImaginaryI]", " ", "g"]]]], "]"]], " ", RowBox[List["FresnelC", "[", FractionBox[RowBox[List[RowBox[List["-", "d"]], "+", RowBox[List["\[ImaginaryI]", " ", "f"]], "+", RowBox[List["2", " ", RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]]]], ")"]], " ", "z"]]]], RowBox[List[SqrtBox[RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]]]]], " ", SqrtBox[RowBox[List["2", " ", "\[Pi]"]]]]]], "]"]]]], "+", RowBox[List[RowBox[List["FresnelS", "[", FractionBox[RowBox[List[RowBox[List["-", "d"]], "+", RowBox[List["\[ImaginaryI]", " ", "f"]], "+", RowBox[List["2", " ", RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]]]], ")"]], " ", "z"]]]], RowBox[List[SqrtBox[RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["\[ImaginaryI]", " ", "c"]]]]], " ", SqrtBox[RowBox[List["2", " ", "\[Pi]"]]]]]], "]"]], " ", RowBox[List["Sin", "[", 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RowBox[List["\[ImaginaryI]", " ", "g"]]]], "]"]]]]]], ")"]]]]]], ")"]]]]]]]] MathML Form cos ( b z 2 + d z ) cosh ( c z 2 + f z + g ) z 1 2 π 2 ( cos ( ( f - d ) 2 4 ( c - b ) - g ) C TagBox["C", FresnelC] ( - d + f + 2 ( c - b ) z c - b 2 π ) + S TagBox["S", FresnelS] ( - d + f + 2 ( c - b ) z c - b 2 π ) sin ( ( f - d ) 2 4 ( c - b ) - g ) c - b - cos ( g - ( d + f ) 2 4 ( b + c ) ) C TagBox["C", FresnelC] ( d + f + 2 ( b + c ) z - b - c 2 π ) + S TagBox["S", FresnelS] ( d + f + 2 ( b + c ) z - b - c 2 π ) sin ( g - ( d + f ) 2 4 ( b + c ) ) - b - c ) z b z 2 d z c z 2 f z g 1 2 2 -1 1 2 f -1 d 2 4 c -1 b -1 -1 g FresnelC -1 d f 2 c -1 b z c -1 b 1 2 2 1 2 -1 FresnelS -1 d f 2 c -1 b z c -1 b 1 2 2 1 2 -1 f -1 d 2 4 c -1 b -1 -1 g c -1 b 1 2 -1 -1 g -1 d f 2 4 b c -1 FresnelC d f 2 b c z -1 b -1 c 1 2 2 1 2 -1 FresnelS d f 2 b c z -1 b -1 c 1 2 2 1 2 -1 g -1 d f 2 4 b c -1 -1 b -1 c 1 2 -1 [/itex] Rule Form Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", 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StudentShare solutions • Home • Subjects • Miscellaneous • Ive got some SPSS data (graphs and tables etc) can be found in a word file named (the data) and need to be analysed and interpreted in the form of writing # Ive got some SPSS data (graphs and tables etc) can be found in a word file named (the data) and need to be analysed and interpreted in the form of writing - Assignment Example ## Extract of sample Ive got some SPSS data (graphs and tables etc) can be found in a word file named (the data) and need to be analysed and interpreted in the form of writing The study is reduced to the involvement of only two ordinal variables and hence statistical designs using two-variables are only used to obtain results. It has to be mentioned that at the time of the original data collection, private schools had the reputation of being better and more progressive than state schools with respect to English teaching. The first research hypothesis intends to analyze the importance given by teachers to explain the meaning of new English words to students. That is, there is difference between state and private schools in the extent to which teachers explain the meaning of new words in English; specifically teachers will explain words in English more in the private school. The design involves only two independent groups and the dependent variable is the score or rating. The descriptive statistics (Table 1) shows that the study involved 108 students, 67 belonging to the state schools and 41 belonging to private schools. The average of the rating given by the private school students (0.5366) is greater than that of the public school students (0.4627). A frequency chart (Fig. 1) was produced to compare the ratings the given by the students between state and public schools on the extent to which teachers explain the meaning of new words in English. The least rating was 0 representing never and 3 representing always. It is noticed that nearly 40 students of the state schools claimed that their teachers never explained the meaning of the English words. Nearly 44 students of the private schools claimed that their teachers either never or seldom explained the meaning of the English words. Only 1 student agreed that the teachers always explained the meaning. The independent samples T test is used to test the equality of the above given averages. The Levene’s test also is used to find whether the assumption of homogeneity of variables is satisfied. Table 2 shows the results. ...Show more ## Summary SPSS is the statistical package used for the analysis. Two tasks were defined to compare the usage and importance given to the English language in private and state… Author : monique53 Save Your Time for More Important Things Let us write or edit the assignment on your topic "Ive got some SPSS data (graphs and tables etc) can be found in a word file named (the data) and need to be analysed and interpreted in the form of writing" with a personal 20% discount. Grab the best paper ### Check these samples - they also fit your topic Data Warehousing & Data Mining The process is similar to the extraction of valuable metal hence the term “mining” (Jackson, 2003). Data mining is the process of analyzing extensive data with the aim of establishing correlation between different variables. Consequently, data warehousing is the process of storing data in relational databases that facilitate such queries and analysis. 3 pages (750 words) Assignment Analysing vector data with the ArcView GIS The first location will maximize developer profits, the second location will minimize negative environmental and social impacts. The proposed business is a super retail store. Retailing consists of the sale of goods or merchandise, from a fixed location such as a department store or kiosk, in small or individual lots for direct consumption by the purchaser. Retailing may include subordinated services, such as delivery. 6 pages (1500 words) Assignment The data search assignment There are dozens of avenues to look for but broadly put there are research journals, internet sources, encyclopedias and other general information books. The research topic for 8 pages (2000 words) Assignment The Research Process:Practical exercise using secondary data Interpretation and evaluation of data is also an important aspect in research since new findings help to formulate policies to address pertinent issues in society. Such datasets are 6 pages (1500 words) Assignment Defining Data Data ware housing involves the periodical extraction of data from applications that support business process into a dedicated computer where they are reformatted, validated, summarized and reorganized. The data warehouse therefore 1 pages (250 words) Assignment Run the appropriate ANOVA, you must copy and paste the relevant tables from the SPSS output file The different levels of the factor are before drinking 4 cups of tequila and after drinking the four shots. Experience among the body piercers is the between-subjects factor. The type of factor identifies independent variables in which participants 5 pages (1250 words) Assignment Data Analysis Note that you do not need to submit the samples you have generated. Also, do not write any explanation of your calculations as they will not be given any credit. However, do write down all 1 pages (250 words) Assignment Data The Maryland College and career- ready only standards are of a varied assortment. This range from career in English language and arts, college careers related to mathematics and mathematical operations, literacy in history and 1 pages (250 words) Assignment The New Frontier: Data Analytics Organizations are usually inundated with data, receiving terabytes and petabytes of it; received from various departments of the operation. From operational to transactional business systems, from management and scanning facilities; from outbound and inbound 5 pages (1250 words) Assignment Harnessing Information Management, the Data, and Infrastructure Hadoop is a framework for data-intensive distributed applications. Hadoop is made up of the Hadoop file system that is a very portable file system and scalable file system foe storing different data 5 pages (1250 words) Assignment Hire a pro to write Win a special DISCOUNT! Put in your e-mail and click the button with your lucky finger Apply my DISCOUNT Click to create a comment Let us find you another Assignment on topic Ive got some SPSS data (graphs and tables etc) can be found in a word file named (the data) and need to be analysed and interpreted in the form of writing for FREE!

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# Cap rate in real estate Cap rate in real estate keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website ## Keyword Suggestions Cap rate Cap rate calculator Cap rate formula Cap rate definition Cap rate real estate Cap rates explained Cap rate simplified Cap rate compression Cap rate vs discount rate Cap rate calculator formula Cap rate map Cap rate formula real estate Cap rate investopedia ( Please select at least 2 keywords ) #### Most Searched Keywords ## Websites Listing We found at least 10 Websites Listing below when search with cap rate in real estate on Search Engine ### Cap Rate simplified – Real Estate University Reuniversity.org  DA: 16 PA: 19 MOZ Rank: 35 How do you calculate a real estate cap rate? How do you calculate a cap rate? Divide the net income by the property’s purchase price • The cap rate is the ratio between the net income of the property and its original price or capital cost.Cap rate is expressed as a percentage. • What does 7.5% cap rate mean? For example, if an investment property costs \$1 million dollars and it generates \$75,000 of NOI (net operating income) a year, then • ### What Is Cap Rate in Real Estate Realtor.com  DA: 15 PA: 33 MOZ Rank: 49 • Cap rate, or capitalization rate, is the ratio of a property’s net income to its purchase price. It’s an essential number for gauging a property’s rental income potential ### What is a cap rate in real estate & why is it important Learn.roofstock.com  DA: 19 PA: 26 MOZ Rank: 47 • The real estate cap rate formula can be used to solve for net operating income by rearranging the equation • If a rental property has a value of \$120,000 and the cap rate for similar properties in the same area is 6%, the NOI should … ### What Is a Good Real Estate Cap Rate in 2022 Mashvisor.com  DA: 17 PA: 27 MOZ Rank: 47 • You predict that Property A will make \$30,000 in rent each year, but you’ll need to pay \$8,000 in real estate taxes and maintenance, leaving you with a net income of \$22,000 • Because the current property value is \$300,000, your cap rate would be 7.3%. (\$22,000 / \$300,000) x 100% = 7.3% ### Cap Rate Real Estate: Formula and Examples Wallstreetprep.com  DA: 22 PA: 50 MOZ Rank: 76 • Cap rates are the primary shorthand by which different real estate properties are compared by investors • For example, if you’re evaluating a property with a \$10 million asking price and expected NOI of \$1 million, that property would have a 10% cap rate • All else equal, the higher the cap rate, the higher the annual return on investment. ### Cap Rate simplified – Real Estate University Reuniversity.org  DA: 16 PA: 19 MOZ Rank: 40 • A cap rate (capitalization rate) is a term in commercial real estate that refers to the way a building is evaluated • It's calculated by taking the net operating income, NOI, and dividing it by the cost of the building in order to give the rate of return. Capitalization rate = Net operating income / Cost of the building ### Cap Rate Calculation: How To Use Cap Rate In Real Estate Noradarealestate.com  DA: 24 PA: 27 MOZ Rank: 57 • It is one of the most fundamental concepts in real estate investing and is mostly referred to in calculations as Cap Rate • Cap Rate is defined as the rate of return on a rental investment property based on its income, according to Investopedia • This determines the investment's potential return. ### What Is A Good Cap Rate & How To Calculate It Fortunebuilders.com  DA: 23 PA: 10 MOZ Rank: 40 • Following this logic, a cap rate between four and ten percent may be considered a “good” investment • According to Rasti Nikolic, a financial consultant at Loan Advisor, “in general though, 5% to 10% rate is considered good • Property investors use cap rate every time they invest in a property because it gives them an idea about the profitability. ### U.S. Cap Rate Survey H2 2021 CBRE Cbre.com  DA: 12 PA: 41 MOZ Rank: 61 • The cap rate is the ratio of net operating income (NOI) to the acquisition price of the asset. The NOI calculation is based on net income less operating expenses • Because hotel occupancies are now extremely low, cap rate estimates are based upon hypothetical stabilized NOIs • Contacts 1/13 Richard Barkham, Ph.D. ### Overview, Example, How to Calculate Cap Rate Corporatefinanceinstitute.com  DA: 29 PA: 50 MOZ Rank: 88 Capitalization rate (or Cap Rate for short) is commonly used in real estate and refers to the rate of return on a property based on the net operating income (NOI) that the property generates. In other words, capitalization rate is a return metric that is used to determine the potential return on investment or payback of capital. ### Real Estate Capitalization Rate: What You Need to Know Banks.com  DA: 13 PA: 42 MOZ Rank: 65 • We can modify the formula to find the fair value based on an 8% cap rate • Fair value = \$100,000 / 8% = \$1,250,000 • In this example, the difference between a 5% cap rate and an 8% cap rate is \$750,000 • If you pay \$2 million for this property, you are significantly overpaying. ### Cap Rate Explained For 2022 (And Why It Matters With Rental … Coachcarson.com  DA: 19 PA: 10 MOZ Rank: 40 • It’s the ratio of a rental property’s net operating income to its purchase price (including any upfront repairs): Cap Rate = Net Operating Income (NOI) ÷ Purchase Price The formula can be used on the level of an individual property by looking at its net operating income compared to its value. ### Cap Rate: Defined And Explained Rocket Mortgage Rocketmortgage.com  DA: 22 PA: 15 MOZ Rank: 49 • Capitalization rates, also known as cap rates, are measures used to estimate and compare the rates of return on multiple commercial real estate properties • Cap rates are calculated by dividing the property’s net operating income (NOI) from its property asset value • Cap rates can provide valuable insight into a property. ### Cap Rate Real Estate: What is Capitalization Rate & How to … Butterflymx.com  DA: 15 PA: 27 MOZ Rank: 55 • The cap rate is calculated by dividing the property’s net operating income (NOI) by the current market value, multiplied by 100 • Cap rate formula: Cap rate = NOI / current market value x 100 As a reminder, the NOI is revenue minus expenses at a property. ### Cap Rate Real Estate: What Is A Good Rate & Formula Schoolofwhales.com  DA: 22 PA: 22 MOZ Rank: 58 The cap rate formula for real estate Capitalization Rate = Annual Net Operating Income / Current Market Value Net Operating income = Gross Operating Income – Operational Expenses Current Market Value = purchase price or appraisal value Commercial property cap … ### What is Cap Rate in Real Estate Realwealth.com  DA: 14 PA: 39 MOZ Rank: 68 • A capitalization rate, or cap rate, is the annual rate of return that is expected to be generated on a real estate investment property • Cap rate is the most common way to assess profitability and return potential on a real estate investment. ### Cap Rate (Capitalization Rate) in Real Estate: What is It, How Do … Valiancecap.com  DA: 15 PA: 50 MOZ Rank: 81 • Typically, high cap-rate properties have less demand, lower growth potential, or higher levels of risk • A high cap rate may look good in terms of income potential, but it sometimes requires more investment and management • For a real-life example, this is what value-add real estate investing is aiming to achieve. ### Cap Rates: Definition & Guide to Calculate for Real Estate Gowercrowd.com  DA: 14 PA: 44 MOZ Rank: 75 • Office buildings, retail buildings, medical real estate, and other specialty CRE usually have higher cap rates in smaller markets • Growth markets usually have higher cap rates • This is one of the main factors impacting cap rates, along with the cost of capital • Low growth may reduce cap rates while high growth increases it. ### What Does Cap Rate Mean In Real Estate Signaturegrouprealty.net  DA: 28 PA: 50 MOZ Rank: 96 • 34.3 Mashvisor’s Real Estate Cap Rate Calculator; 34.4 What Is a Good Cap Rate in Real Estate? 34.5 Factors That Affect Capitalization Rates • 34.5.1 1 Real Estate Market (Marco-Level Economics) 34.5.2 2 Neighborhood (Micro-Level Influences) 34.5.3 3 Type of Investment Property ; 34.6 The Bottom Line • 34.6.0.1 Eman Hamed ; 35 What You Should ### The Importance of the Capitalization Rate in Real Estate Investing Investopedia.com  DA: 20 PA: 50 MOZ Rank: 89 • The capitalization rate is the most commonly used baseline for comparing investment properties • It is analogous to the estimated effective rate of return on a typical security investment ### Capitalization (Cap) Rate Definition Moneycrashers.com  DA: 21 PA: 37 MOZ Rank: 78 • You don’t exactly need a degree in mathematics to calculate cap rates • Here’s the formula for capitalization rate: Cap Rate = Annual Net Operating Income (NOI) ÷ Purchase Price (or Value) For example, imagine a property nets \$8,000 in income per year and it costs \$100,000 to purchase ### CapEx: Capital Expenditures in Real Estate Explained Butterflymx.com  DA: 15 PA: 24 MOZ Rank: 60 • In real estate, CapEx are expenses that go toward adding to or improving a property beyond common, routine repairs and maintenance • Since the costs associated with these improvements are usually substantial, real estate professionals put … ### Cap Rates When Investing in Real Estate: What You Should Know Fnrpusa.com  DA: 11 PA: 31 MOZ Rank: 64 • In commercial real estate investing, a capitalization rate is the rate of return that an investor could expect if a property was purchased in cash • The formula used to calculate it is: Cap Rate = Year 1 Net Operating Income / Purchase Price • For example, if a property has NOI of \$100,000 and a purchase price of \$1,000,000, the resulting cap ### What Is Cap Rate In Real Estate Signaturegrouprealty.net  DA: 28 PA: 50 MOZ Rank: 27 • A cap rate between 8% and 12% is considered good for a rental property in most areas (ones in expensive cities may go lower) • The formula for cap rate is: (NOI ÷ Market Value) x 100 • What is the 1 rule in real estate? The 1% rule of real estate investing measures the price of the investment property against the gross income it will generate. ### What Is Cap Rate in Real Estate Investing Mashvisor.com  DA: 17 PA: 45 MOZ Rank: 86 What is cap rate for this rental property? Let’s have a look: Capitalization Rate = \$7,200/\$200,000 x 100 Capitalization Rate = 3.6% This means that this rental property will generate a rental income that is equal to 3.6% of the current market value of … ### Cap rate decompression Wallstreetoasis.com  DA: 23 PA: 41 MOZ Rank: 89 • 20 hours ago · If we are experiencing a cap rate compression in this market, can we assume that as the pendulum swings the other way, there will be a cap rate decompression? If so, I would assume it's the asset's value that becomes cheaper and not it's NOI • Coming from a retail, value-add background, we always underwrote a 50-100bp compression on exit minimum ### The Downside of Cap Rates in Real Estate Valuation (Use This … Breakintocre.com  DA: 16 PA: 50 MOZ Rank: 92 • Cap Rates Don’t Take Into Account Sale Value • Let’s say you buy two deals, each for \$10 million with a 50% LTV loan, so \$5 million of equity invested in each • The first, which we’ll call Property A, is in a market that’s seeing negative net migration and job growth is stagnant, and you buy that property at a 6.5% cap rate. ### Cap Rates In Real Estate Explained Elevate Realty Elevatepartners.ca  DA: 22 PA: 46 MOZ Rank: 95 • Because of COVID, rents have dropped around 10% compared to before March • So with adjusted lower rents, typically cap rates for (but can vary more by neighbourhood): Toronto condos are at 2-2.5% • Toronto houses are at 3-3.5%, and • Toronto commercial properties are 3.5%+. ### What Else Drives Cap Rates Besides Rising Interest Rates Globest.com  DA: 15 PA: 50 MOZ Rank: 93 1 day ago · Looking at commercial real estate in general, the spread between the average cap rate and the 10 Year Treasury has been as low as 214 … ### Return metrics explained: What is a cap rate in commercial real … Plantemoran.com  DA: 19 PA: 50 MOZ Rank: 98 • The “cap rate” in real estate investing is the unlevered rate of return on an asset based on its annual net operating income (NOI) • Cap rates, which are expressed as a percentage and represent returns for a single point in time, are used to evaluate an individual investment property or compare properties without complicating the assessment ### Real Estate Definitions: Capitalization Rate (Cap Rate) Rentalsoftware.com  DA: 22 PA: 30 MOZ Rank: 82 • You can use the Cap Rate to value your property • Let’s say that your property generates \$10,000 of annual net operating income • Your real estate agent tells you that the Capitalization Rate in your area is approximately 4%. ### What Is Cap Rate and Why Is It Used to Value Investment Properties Mckissock.com  DA: 17 PA: 35 MOZ Rank: 83 • One way to use capitalization rate is to compare it to the cost of the money • The greater the strength of the property is above the strength of the loan, the better • Here’s how it works: In our 6-unit example, we know the cap rate is 7.7 percent • So this property is 7.7 percent on the strength meter • If we had a \$306,000 loan at 7 percent ### What is a good cap rate in Commercial Real Estate Feldmanequities.com  DA: 23 PA: 50 MOZ Rank: 18 • In commercial real estate, a capitalization rate (“cap rate”) is a formula used to estimate the potential return an investor will make on a property • The cap rate is expressed as a percentage, usually somewhere between 3% and 20% ### What is Cap Rate in Real Estate Upnest.com  DA: 14 PA: 27 MOZ Rank: 74 • Put simply, the capitalization rate is calculated by dividing the annual net operating income (NOI) of a property by its current value • For example: A \$1M property, with a \$100k annual NOI, would have a cap rate of 10% • A \$1M property with a \$200k annual NOI would have a cap rate of 20%. ### The Simplest Explanation To Understand Cap Rate In Real Estate Assetmonk.com  DA: 13 PA: 34 MOZ Rank: 81 • Increasing cap rates indicate that the property’s market value is decreasing • Though it is assumed that, the higher, the better, it is not always true • Increasing, cap rate indicates a loss to the investor • A rate between 4 to 10% is usually considered to … ### What is a Cap Rate and How Does it Affect Real Estate Investments Blog.udemy.com  DA: 14 PA: 20 MOZ Rank: 69 • A cap rate, also known as capitalization rate, is a measure used to evaluate the viability of various investment vehicles such as real estate • It is calculated as follows: A property whose selling price is \$800,000 and generates an annual return of \$95,000 has a cap rate of 11.88% • This is calculated as \$95,000/\$800,000. ### How Do You Calculate a Cap Rate on a Rental Property Andersonadvisors.com  DA: 20 PA: 47 MOZ Rank: 14 • The capitalization rate of a property, or cap rate, is a percentage that expresses how well an investment property will perform • The cap rate should not be the only indicator a real estate investor uses to evaluate a property, but it is a very quick and convenient indicator for visualizing the value of a piece of real estate in comparison to other potential real estate ### Best Cap Rate Formula Calculator for Beginners Mashvisor Mashvisor.com  DA: 17 PA: 34 MOZ Rank: 88 • Step 6: Calculate the Cap Rate Using the Cap Rate Calculator Formula • The next step is to use the cap rate formula mentioned above and divide the net operating income by the current market value or property price ### Cap Rate Calculator Capitalization Rate Calculator Purecalculators.com  DA: 19 PA: 20 MOZ Rank: 77 • The cap rate, also known as the capitalization rate calculator is a tool that anyone who is interested in real property can use • They need a cap rate of 12 percent for real estate investments • Value = \$12,000 / 0.12 = \$100,000 • As you can see, when interest rates increased, your home became less valuable • Why? Investors must pay less to get ### How to Find Commercial Real Estate Cap Rates (Formula) Dealpath.com  DA: 16 PA: 38 MOZ Rank: 93 • The commercial real estate cap rate, or the capitalization rate, is the return rate figure that CRE investors use to gauge the risk and potential return of an asset or property • Cap rates are expressed as percentages, usually from 3-20%. ### Cap Rate Simplified for Commercial Real Estate Assetsamerica.com  DA: 17 PA: 10 MOZ Rank: 67 • What is a good cap rate in commercial real estate? Normally, you will encounter a capitalization rate between 4.00% and 10.00% for commercial property • Cap rates in high-demand areas will be lower than those in less densely-populated areas • Buyers prefer high cap rates because they imply a lower purchase price. ### What Is A Cap Rate In Real Estate Frommilitarytomillionaire.com  DA: 33 PA: 20 MOZ Rank: 94 • What Does Capitalization Rate Mean? This is less pertinent, but when answering “What is a cap rate in real estate?” I think it is still important to know what capitalization rates mean on from a 5,000-foot view as well • A cap rate is a way to value real estate based on macro and microeconomic factors, as well as the type of property. ### Cap Rate vs Yield in Commercial Real Estate FNRP Fnrpusa.com  DA: 11 PA: 24 MOZ Rank: 77 • The key difference between the cap rate and yield is in the denominator of the equation • The cap rate utilizes the property’s current market value, which changes over time • The yield calculation utilizes the property’s total cost, which is a static, one time, number • At the time real estate is purchased, the cap rate and the yield may be ### What is Cap Rate in Commercial Real Estate Squarefoot.com  DA: 18 PA: 29 MOZ Rank: 90 • The cap rate is simply a percentage calculated when you divide the annual net operating income (NOI) generated by a rental property by its current market value • It is a ratio used to help commercial real estate investors determine the rate of return that they can expect to get from a rental property.

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06-18-2005, 03:46 PM #1 alexis Casting On   Join Date: Jun 2005 Location: tampa, florida Posts: 4 Thanks: 0 Thanked 0 Times in 0 Posts Hello Guys need help again How do you do this? The pattern reads Work 5{5-7-7-7} rows even in stocking st. Rep last 6{6-8-8-8} rows 3 times more. 70 {74-80-84-90} sts. Cont even in stocking st until work from beg measures 7{7-8-8-8 1/2} ending with right side facing for next row. I just do not get this Thanks in advance Alexis 06-18-2005, 03:50 PM #2 Julie Instepping Out     Join Date: Mar 2005 Location: Midwest Posts: 3,325 Thanks: 10 Thanked 78 Times in 64 Posts Multi-sized patterns include the instructions for all sizes within the pattern....for example, if your pattern has 5 different sizes: XS (S, M, L, XL) and the instructions say to knit 5 (5,7,7,7) rows, you would follow the instructions that correspond to the size you are knitting. So you would knit five rows for XS or S, or 7 rows for M, L or XL. And so on. __________________ Knitter's Geek Code KCR+ Exp+ SPM+ Steel+ Bam+ Addi++ Pl-- Wool++ Cot Lux+ Stash Scale+ Fin !Ent Int FI> Tex Lace Felt Flat Circ+ Dpn- ML+ Swatch+ KIP+ Blog+ SNB> EZ? FO+ WIP3 Gauge W+ Alt Sw 06-18-2005, 03:53 PM #3 Ingrid Moderator Mod Squad     Join Date: Jun 2005 Location: Pleasant Valley, NY Posts: 29,585 Thanks: 21 Thanked 2,254 Times in 1,845 Posts Let's assume you're making the smallest size. That would be the first number outside the brackets. Each number inside the brackets represents the number of st for each larger size. It's helpful in the beginning of a pattern to circle all the numbers that pertain to your size. So you would work 5 (or however many for your size) rows in stockinette stitch, k one row, p one row. I gather from the pattern that the row BEFORE these st sts was in some way different So you would work these 6 stitches three more times, three groups of 6. It helps to write out a list and cross them off as you go. Write 1 through 6 three times and cross them off so you know where you are. After you do this you should have 70 st on your needle. Then you continue with st st until the total length from needle to bottom is 7 in., with the last row you worked being a wrong (p) side row. The right side, the one that will be showing, will be facing you when you start the next rows of the pattern. Got it? :D 06-18-2005, 03:54 PM #4 Silver Instepping Out     Join Date: Apr 2005 Location: Orange Park, FL Posts: 3,949 Thanks: 854 Thanked 669 Times in 351 Posts This is a multi-size pattern. When it says knit 5{5-7-7-7} rows, the first number is for the smallest size, and they go progressivly larger (small, medium, large, extra large, 1x, etc). Pick the number that coordinates with the size you're working. When it says "work even" in stocking st, that just means work all rows in stocking st without increasing or decreasing. There should be another row of instructions though. So if you're working on size small (for example), you'll knit 5 rows in stocking st, and a row in pattern either before or after those. Then repeat those 6 rows 3 times, and you'll end up with 70 stitches. Then continue working in stocking st until the piece measures 7 inches, and end with a wrong side row so that the next row is right side facing. __________________ ~Kristin Silver's Sock Classes Shop: On The Town Knitting Bags, hand made elegant small project bags. My Blog, Write | KBelle on Ravelry | ComeToSilver on Plurk | Puppies, Jinx & Hex Photos & Videos Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts vB code is On Smilies are On [IMG] code is On HTML code is Off Forum Jump User Settings Private Messages Subscriptions Who's Online Search Forums Forums Home KnittingHelp.com     General Knitting     How-to Questions     Pattern Central     What'cha Knittin'?     Knit-Alongs     Charity Knitting     Creating Yarn: Spinning, Dyeing, etc.     Crochet!     Blog Threads     Knitters Knear You     Buy/Sell/Swap     The Lounge

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# Gravity and Vacuum. Gravity and Vacuum. =. Einstein was mistaken using his Gravitation theory to the all Universe as a whole. The Gravitation theory doesn’t work in the Universe as a whole. The Gravitation theory is a local theory. Why? Because the detected material mass of the matter in the Universe ( the cosmological constant / the critical density) is so small ( the average density of all substance in the Universe is approximately p=10^-30 g/sm^3 ) that it cannot ‘close’ the Universe into sphere and therefore our Universe as whole must be ‘open’, endless, infinite. The Universe as a whole is an Infinite Pure Vacuum: T=0K. More concrete: § 1. Vacuum: T= 0K, E= ∞ , p = 0, t =∞ . =. Another argument: One postulate of SRT says: the speed of quantum of light in a Vacuum is a constant ( c= 299,792,458 km/ sec =1, Michelson-Morley experiment ). In this movement quantum of light doesn’t have time. The time is ‘stopped ‘ for him. But this is possible only if his reference frame – vacuum - also doesn’t have time. It means that the reference frame – Vacuum is an Eternal Continuum. =… One more argument. According to QED when electron interacts with vacuum its parameters becomes infinite. This is possible only when Vacuum itself is an Infinite Continuum. 3. The Infinity appears in many physical and mathematical problems. Physicists don’t know that to do with ‘ infinite’ and therefore we can read: Infinity is the cause of the crisis in Physics. en.wikipedia.org/wiki/Infinity 4, ‘ A world without masses, without electrons, without an electromagnetic field is an empty world. Such an empty world is flat. But if masses appear, if charged particles appear, if an electromagnetic field appears then our world becomes curved. Its geometry is Riemannian, that is, non- Euclidian.’ / Book ‘Albert Einstein’ The page 116 . by Leopold Infeld. / It means: a). ‘A world without masses, without electrons, without an electromagnetic field is an empty world. Such an empty world is flat.’ – it is a Vacuum World. b). ‘But if masses appear, if charged particles appear, if an electromagnetic field appears ‘ - / - in the flat vacuum - , / ‘ then our world becomes curved. Its geometry is Riemannian, that is, non- Euclidian.’ – it is Material World of our stars and planets. c). We have two (2) Worlds: Vacuum and Material and we need to understand their interaction. ==. All the best. Israel Sadovnik Socratus ===. Each of those theories is mistaken for one simple reason; Infinite homogeneity is impossible. What you are calling a “vacuum” would be infinitely homogeneous and thus impossible. The infinitely homogeneous vacuum can be broken by so called a ’ vacuum polarization '. Einstein’s Cosmology Scientific cosmology really began in 1917, when Albert Einstein’s published the final modification to his theory of gravity in the paper ‘Cosmological Considerations of the General Theory of Relativity’. einsteins-theory-of-relativi … ology.html ==. Einstein’s cosmological ideas . . . adsabs.harvard.edu/abs/2004AcHA…23…189J ==. If it can be polarized, then it isn’t a total vacuum. The Homogenous Vacuum as a whole is T=0K But in some local places can polarization appear If the universe as whole is closed then heat death come. If the universe as whole is not closed then . . . . . then Maxwell’s ‘ demon ‘ permits particles to pass from one system to another one. For example, from system ‘ space and time’ to the system ‘ spacetime’, according to ‘ The law of conservation and transformation energy/mass ‘ of course. =. Socratus =. Common opinion: A matter bends the space or curved space tells the matter where to go =. 1 Where this mass / matter come from ? 2 We have two (2) spaces: Newtonian ( space and time ) and Minkowski ( spacetime) Which space is curved ? 3 What is interaction between these two (2) spaces ? =. sadovnik

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# Why is chunk size often a power of two? There are many Minecraft clones out there and I am working on my own implementation. A principle of terrain rendering is tiling the whole world in fixed size chunks to reduce the effort of localized changes. In Minecraft the chunk size is 16 x 16 x 256 as far as I now. And in clones I also always saw chunk sizes of a power of the number 2. Is there any reason for that, maybe performance or memory related? I know that powers of 2 play a special role in binary computers but what has that to do with the chunk size? - It's nice that you can keep dividing it by two and get even numbers back. (Not a complete answer, but something useful about using a number like `2^n`) – ashes999 Jan 8 '13 at 22:37 This would depend on the game and the indexing structure used for the chunks. Though, at such a high level, it's not too likely it has much to do with memory or a specific performance enhancement. More than likely it's an arbitrary decision for sizing chunks in a predictable way. It allows for some counting and indexing tricks using bit shifting that wouldn't be possible with numbers that aren't a power of two. For example, counting in powers of two is as easy as shifting a bit in binary: ``````Dec = Bin 1 = 000001 2 = 000010 4 = 000100 8 = 001000 16 = 010000 32 = 100000 `````` Where these shortcuts will be used will depend on the developer and what problem they're trying to solve. If you're making a decision for what size to make the chunks, and it doesn't matter in any other aspects, you may as well use something that is familiar and has benefits you're used to using. - First, multiplying by powers of two is much cheaper than multiplying by an arbitrary number, since you can do it by bit shifting. Most of the time the compiler can do this for you, so whenever you write "* 16" in your code, the compiler actually does a shift by four, and you don't need to worry about it - you just need to give the compiler the opportunity by designing your data structures this way. Second, since cache lines, memory busses and other information highways in your computer also tend to be designed to use powers of two, you may get a better performance overall this way. Third, we old geeks just are used to playing around with powers of two, so it's a habit. (Fourth, other old geeks who design your hardware and your compilers also like powers of two, so this is not going to change any time soon). - +1 "Third, we old geeks just are used to playing around with powers of two, so it's a habit." This is probably the main reason. – Laurent Couvidou Jan 10 '13 at 18:24 The real answer is just this: On a binary computer, powers of two are nice round numbers. When a normal person needs to pick an arbitrary number for some purpose, they typically choose nice round numbers in the number system they're comfortable with, base 10. So they'll pick 10, 100, 1000, etc. Because they're simple and easy and don't require much thought and the precise value wasn't really important to them, they were just aiming at a general scale of magnitude. As programmers, when we need to pick an arbitrary number for some purpose, we typically choose nice round numbers in the number system computers use, base 2. So we'll pick 2, 4, 8, etc. Because they're simple and easy and don't require much thought and the precise value wasn't really important to us, we were just aiming at a general scale of magnitude. It's really not any more complicated than that. They're just nice round numbers. - One reason not mentioned in other answers is that if needed, powers of two numbers can always be halved without rounding problems. This is probably not a reason in Minecraft clones, but is in some other cases such as in textures with mipmaps. - Another potential reason would be that it would allow you to encode info about each chunk in a 3D texture. If your target hardware supported 3D textures but didn't have fully robust and general non-power-of-two texture support (which I admit you'd want to be shooting fairly low for) then making your chunk sizes powers of two is not just ideal - it's essential. -

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Year 5 Term 3 Unit 6b Day 1. Presentation on theme: "Year 5 Term 3 Unit 6b Day 1."— Presentation transcript: Year 5 Term 3 Unit 6b Day 1 L.O.1 To be able to recall multiplication facts and derive division facts from the six times table We are going to say the 6x table We are going to say the 3x table We are going to say the 7x table We are going to say the 9x table We are going to say the 8x table We are going to say the 4x table We are going to say the 5x table Now back to the 6x table Let’s say it again! Here are the multiples of 6 Q. What is this multiplication fact? 6 x table 6 x 4 = 24 or 4 x 6 = 24 Q. What are the associated division facts? 6 x table 24 ÷ 6 = 4 and 24 ÷ 4 = 6 Q. What are the associated division facts? 6 x table 42 ÷ 6 = 7 and 42 ÷ 7 = 6 Q. What are the associated division facts? 6x table 54 ÷ 6 = 9 and 54 ÷ 9 = 6 Q. What are the associated division facts? 6 x table 30 ÷ 6 = 5 and 30 ÷ 5 = 6 Q. What are the associated division facts? 6 x table 12 ÷ 6 = 2 and 12 ÷ 2 = 6 Q. What are the associated division facts? 6 x table 48 ÷ 6 = 8 and 48 ÷ 8 = 6 Q. What are the associated division facts? 6x table 6 ÷ 6 = 1 and 6 ÷ 1 = 6 Q. What are the associated division facts? 6x table 60 ÷ 6 = 10 and 60 ÷ 10 = 6 Q. What are the associated division facts? 6x table 36 ÷ 6 = 6 and 36 ÷ 6 = 6 Q. What are the associated division facts? 18 6x table 18 18 ÷ 6 = 3 and 18 ÷ 3 = 6 Now write down all the division facts you can remember. 6x table Now write down all the division facts you can remember. 6x table L.O.2 To be able to solve mathematical problems or puzzles, recognise and explain patterns and relationships. Q. What are the totals in the 8 boxes? What strategies will you use? is 3 is 4 is 5 Q. How many totals change now that represents a 6 instead of a 5? is 3 is 4 is 6 Q. If represents 2, how many totals will change? is 2 is 4 is 6 8 will change is 2 is 4 is 6 This time I have given you some totals and left you to work out the others! 23 15 24 is _ is _ is _ Q. Which of the three totals did you use to help you first? How did this help you? 23 15 24 is _ is _ is _ Q. Which total did you use next? First you should have worked out that = 6 because 4 x 6 in the last column gives the total 24. Q. Which total did you use next? 23 15 24 is 6 is _ is _ Q. Which total did you use last? Next you should have worked out that = 5 because on the top row 3x6= =23 Q. Which total did you use last? 23 15 24 is 6 is 5 is _ Last you should have worked out that = 2. In the row totalling 15 you already have = 11. 15 – 11 = 4 , there are so each must be worth 2. 23 15 24 is 6 is 5 is 2 Q. What are the other totals? 20 What strategies will you use? 18 12 is _ is _ is _ You should have found these answers. 15 23 20 18 20 21 23 12 is 3 is 6 is 7 Q. Does it matter which three totals you are given? is _ is _ is _ Work in pairs to choose your own number for each shape Work in pairs to choose your own number for each shape. Find the 8 totals. Then give 3 totals to another pair for them to find the missing totals. is _ is _ is _ Q. So, does it matter which three totals you are given? Are there any awkward ones? is _ is _ is _ You will do ACTIVITY Sheet 6b1 for HOMEWORK this week but we will look at the first grid now. WARNING You may find you need to change your strategy when you do the last 2 grids. Let’s try this as a class. =__ =__ =__ Q. What do these totals tell us about what the shapes represent? 18 20 is _ is _ is _ If this is so then on the top line, The 20 means that 18 = 10. + If this is so then on the top line, so = 4 = 8 + 20 With one more total we can find is _ is _ is _ Now try the first question only on Self-assessment sheet 6b1. By the end of the lesson the children should be able to: Solve puzzles and problems involving missing totals. Year 5 Term 3 Unit 6b Day 2 L.O.1 To be able to recall multiplication facts and derive division facts from the seven times table We are going to say the 7x table We are going to say the 6x table We are going to say the 8x table We are going to say the 5x table We are going to say the 4x table We are going to say the 9x table Now back to the 7x table Let’s say it again! This shows the multiples of 7 Q. What is this multiplication fact? 7 x table 8 x 7 = 56 or 7 x 8 = 56 Q. What are the associated division facts? 7 x table ÷ 7 = 8 and 56 ÷ 8 = 7 Q. What are the associated division facts? 7 x table 49 ÷ 7 = 7 and 49 ÷ 7 = 7 Q. What are the associated division facts? 7 x table 28 ÷ 7 = 4 and 28 ÷ 4 = 7 Q. What are the associated division facts? 7 x table 35 ÷ 7 = 5 and 35 ÷ 5 = 7 Q. What are the associated division facts? 7 x table 14 ÷ 7 = 2 and 14 ÷ 2 = 7 Q. What are the associated division facts? 7 x table 63 ÷ 7 = 9 and 63 ÷ 9 = 7 Q. What are the associated division facts? 7 x table 70 ÷ 7 = 10 and 70 ÷ 10 = 7 Q. What are the associated division facts? 7 x table 21 ÷ 7 = 3 and 21 ÷ 3 = 7 Q. What are the associated division facts? 7 x table 7 ÷ 7 = 1 and 7 ÷ 1 = 7 Q. What are the associated division facts? 42 7 x table 42 42 ÷ 7 = 6 and 42 ÷ 6= 7 Now write down all the division facts you can remember. 7 x table Now write down all the division facts you can remember. 7 x table L.O.2 To be able to solve mathematical puzzles and problems. To be able to explain methods and reasoning . Q. What fraction of these 8 shapes are circles? Half the shapes are circles : 4/8 ½ Q. What fraction of the shapes are circles now? Three quarters of the shapes are circles: 6/8 ¾ Q. What fraction of the shapes are circles now? five eighths of the shapes are circles now: 5/8 Q. What fraction of the shapes are circles now? One eighth of the shapes are circles : 1/8 ½ the shapes are squares Now I shall give you the fractions but not the total number of shapes: ½ the shapes are squares ½ the shapes are circles. Two squares are replaced with two circles so now 1/3 of the shapes are squares and 2/3 of the shapes are circles. Q. How many shapes did we start with? 2 minutes What strategies did you use? Q. If 1/3 of the total number of shapes were squares what multiplication table will the total number of shapes be in? Q. If 1/3 of the total number of shapes were squares what multiplication table will the total number of shapes be in? It must be in the 3x table. Let’s fill in the table Total 1/3 Squares 2/3 Circles 3 6 9 12 15 18 21 Let’s fill in the table Total 1/3 Squares 2/3 Circles 3 6 9 12 15 18 21 1 2 3 4 5 6 7 2 4 6 8 10 12 14 Q. As we have replaced two squares with two circles what would the original number of squares and circles be? Let’s work it out. To answer this we need to add 2 more columns. Total 1/3 Squares 2/3 Circles +2squares -2 circles 3 6 9 12 15 18 21 1 2 3 4 5 6 7 2 4 6 8 10 12 14 Let’s work it out. Q. When were half the shapes squares and half circles? Total 1/3 Squares 2/3 Circles +2squares -2 circles 3 6 9 12 15 18 21 1 2 3 4 5 6 7 2 4 6 8 10 12 14 3 4 5 6 7 8 9 2 4 6 8 10 12 Q. When were half the shapes squares and half circles? The only time this happens is when there are 12 shapes. Total 1/3 Squares 2/3 Circles +2squares -2 circles 3 6 9 12 15 18 21 1 2 3 4 5 6 7 2 4 6 8 10 12 14 3 4 5 6 7 8 9 2 4 6 8 10 12 The only time this happens is when there are 12 shapes. 1/3 of the shapes are squares, 2/3 of the shapes are circles. Try these: use the grid to help you. Q1. 1/3 of the shapes are squares, 2/3 of the shapes are circles. Three circles are replaced with three squares. Now ½ the shapes are squares, ½ the shapes are circles. How many squares and circles did we start with? Q2. 1/4 of the shapes are squares, 3/4 of the shapes are circles. One circle is replaced by one square. Now 1/3 the shapes are squares, 2/3 the shapes are circles. How many squares and circles did we start with? What strategies did you use? There are some cubes in a bag. 1/5 of them are red 4/5 of them are blue Q. How many cubes might be in the bag? We know that if we are working in 1/5 ’s we must be using the 5x table. I take out 6 blue cubes and replace them with 6 red ones. Now ½ are red and ½ blue. Q. How many cubes are in the bag? Let’s do it. Remember we are working in multiples of 5 This blank screen is for recording the working. Now try the second question on Assessment sheet 6b1. By the end of the lesson the children should be able to: Solve puzzles and problems involving fractions; Explain methods and reasoning. Year 5 Term 3 Unit 6b Day 3 L.O.1 To be able to recall multiplication facts and derive division facts from the eight times table We are going to say the 8x table We are going to say the 3x table We are going to say the 7x table We are going to say the 9x table We are going to say the 6x table We are going to say the 4x table We are going to say the 5x table Now back to the 8x table Let’s say it again! Here are the multiples of 8 Q. What is this multiplication fact? 8 x table 8 x 4 = 32 or 4 x 8 = 32 Q. What are the associated division facts? 8 x table 32 ÷ 8 = 4 and 32 ÷ 4 = 8 Q. What are the associated division facts? 8 x table ÷ 8 = 7 and 56 ÷ 7 = 8 Q. What are the associated division facts? 8x table 72 ÷ 8 = 9 and 72 ÷ 9 = 8 Q. What are the associated division facts? 8 x table 40 ÷ 8 = 5 and 40 ÷ 5 = 8 Q. What are the associated division facts? 8 x table 16 ÷ 8 = 2 and 16 ÷ 2 = 8 Q. What are the associated division facts? 8 x table 64 ÷ 8 = 8 and 64 ÷ 8 = 8 Q. What are the associated division facts? 8 x table 8 ÷ 8 = 1 and 8 ÷ 1 = 8 Q. What are the associated division facts? 8 x table 80 ÷ 8 = 10 and 80 ÷ 10 = 8 Q. What are the associated division facts? 8 x table 48 ÷ 8 = 6 and 48 ÷ 6 = 8 Q. What are the associated division facts? 24 8 x table 24 24 ÷ 8 = 3 and 24 ÷ 3 = 8 Now write down all the division facts you can remember. 8 x table Now write down all the division facts you can remember. 8 x table L.O.2 To be able to solve problems and puzzles, recognise and explain patterns, relationships and reasoning. Large batteries cost £1:60 Medium batteries cost £1:10 Q.1 What would ten large batteries cost? Q.2 What would six medium batteries cost? Q.3 How many medium batteries could you buy for £10 Q.4 How many large batteries could you buy for £10 Q.5 How much change would you get in Q Q.4? Answers are: 1. £16 2. £6:60 batteries with 10p. change batteries with 40p. change p. A farmer bought batteries with a £20 note. Q. How many of each size could the farmer buy? 1 minute 18 medium batteries. He could have bought up to 12 large batteries or up to 18 medium batteries. many batteries did the farmer buy? If he spent exactly £20 how many batteries did the farmer buy? 2 minutes in pairs Would it help if we used a list or a chart or a table? Complete the table with your partner and use it to solve the problem. This may help: Medium 1 2 3 4 5 6 7 8 9 10 Cost £1.10 Large £1.60 Complete the table with your partner and use it to solve the problem. Q. What answers did you get? 8 medium + 7 large batteries is an answer. This is the completed table Medium 1 2 3 4 5 6 7 8 9 10 Cost £1.10 £2.20 £3.30 £4.40 £5.50 £6.60 £7.70 £8.80 £9.90 £11 Large £1.60 £3.20 £4.80 £6.40 £8.00 £9.60 £11.20 £12.80 £14.40 £16 This is the completed table Work with a partner to find the answer. Small batteries cost 70p. Q. How many small and medium batteries could we buy for exactly £5 Work with a partner to find the answer. Would it help if you added two extra rows to the table? Extra rows Medium 1 2 3 4 5 6 7 8 9 10 Cost Large Small £1.10 £2.20 £3.30 £4.40 £5.50 £6.60 £7.70 £8.80 £9.90 £11 Large £1.60 £3.20 £4.80 £6.40 £8.00 £9.60 £11.20 £12.80 £14.40 £16 Small £0.70 Extra rows Medium 1 2 3 4 5 6 7 8 9 10 Cost Large Small £1.10 £2.20 £3.30 £4.40 £5.50 £6.60 £7.70 £8.80 £9.90 £11 Large £1.60 £3.20 £4.80 £6.40 £8.00 £9.60 £11.20 £12.80 £14.40 £16 Small £0.70 £1.40 £2.10 £2.80 £3.50 £4.20 £4.90 £5.60 £6.30 £7 What strategies did you use? Q. How many small and large batteries could we buy for exactly £9? Q. How many small, medium and large batteries could we buy for exactly £8? For extension work try these: How many small + large for £11; £12 ; £14 ? How many medium + large for £7 ; £14 ; £20 ? How many small + medium for £10 ; £13 ; £14 ? How many small + medium + large for £11 ; £15 ? Now do question 3 on Self-assessment sheet 6b 1. Now that you have finished the questions on the Self-assessment sheet think about what we have learned this week in maths and decide on what you would like to get better at. Write it as your target at the bottom of the sheet. By the end of the lesson the children should be able to: Solve puzzles and problems involving money and lists Explain methods and reasoning Year 5 Term 3 Unit 6b Day 4

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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A306844 Number of anti-transitive rooted trees with n nodes. 40 %I %S 1,1,2,3,7,14,36,83,212,532,1379,3577,9444,25019,66943 %N Number of anti-transitive rooted trees with n nodes. %C A rooted tree is anti-transitive if the subbranches are disjoint from the branches, i.e., no branch of a branch is a branch. %H Gus Wiseman, <a href="/A306844/a306844.png">The a(7) = 36 anti-transitive rooted trees</a>. %H Gus Wiseman, <a href="/A306844/a306844_1.png">The a(10) = 532 anti-transitive rooted trees</a>. %e The a(1) = 1 through a(6) = 14 anti-transitive rooted trees: %e o (o) (oo) (ooo) (oooo) (ooooo) %e ((o)) ((oo)) ((ooo)) ((oooo)) %e (((o))) (((oo))) (((ooo))) %e ((o)(o)) ((o)(oo)) %e ((o(o))) ((o(oo))) %e (o((o))) ((oo(o))) %e ((((o)))) (o((oo))) %e (oo((o))) %e ((((oo)))) %e (((o)(o))) %e (((o(o)))) %e ((o((o)))) %e (o(((o)))) %e (((((o))))) %t rtall[n_]:=Union[Sort/@Join@@(Tuples[rtall/@#]&/@IntegerPartitions[n-1])]; %t Table[Length[Select[rtall[n],Intersection[Union@@#,#]=={}&]],{n,10}] %Y Cf. A276625, A279861, A279861, A290689, A290760, A304360. %Y Cf. A324694, A324695, A324738, A324741, A324743, A324751, A324754, A324756, A324758, A324759, A324764. %K nonn,more %O 1,3 %A _Gus Wiseman_, Mar 13 2019 Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified May 27 21:12 EDT 2020. Contains 334671 sequences. (Running on oeis4.)

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Convert 8731 Lux (lx) to Foot-Candles (ft-cd) This is our conversion tool for converting lux to foot-candles. To use the tool, simply enter a number in any of the inputs and the converted value will automatically appear in the opposite box. lx = ft-cd How to convert Lux (lx) to Foot-Candles (ft-cd) Converting Lux (lx) to Foot-Candles (ft-cd) is simple. Why is it simple? Because it only requires one basic operation: multiplication. The same is true for many types of unit conversion (there are some expections, such as temperature). To convert Lux (lx) to Foot-Candles (ft-cd), you just need to know that 1lx is equal to ft-cd. With that knowledge, you can solve any other similar conversion problem by multiplying the number of Lux (lx) by . For example, 6lx multiplied by is equal to ft-cd. Best conversion unit for 8731 Lux (lx) We define the "best" unit to convert a number as the unit that is the lowest without going lower than 1. For 8731 lux, the best unit to convert to is . Fast Conversions 1 lx = ft-cd 5 lx = ft-cd 10 lx = ft-cd 15 lx = ft-cd 25 lx = ft-cd 100 lx = ft-cd 1000 lx = ft-cd 1 ft-cd = lx 5 ft-cd = lx 10 ft-cd = lx 15 ft-cd = lx 25 ft-cd = lx 100 ft-cd = lx 1000 ft-cd = lx Blog Talk a Good Game: The Best Talks and Presentations from GDC How to Use Workflow Automation in Customer Service? B2B Lead Generation: Key Strategies to Use Business Plan Vs. Strategic Plan Vs Operational Plan How The Real Estate Business Is Changing For Social Media

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# p-Sylow question #### aabsdr Let K be a normal subgroup of a fintire group G. Let S be a p-Sylow subgroup of G (p a prime divisor of |G|). Prove that KS/K is a p-Sylow subgroup of G/K. Proof/ So we know that K is a subgroup of KS and KS is a subgroup of G, so by the correspondence theroem KS/K is a subgroup of G/K. Since |KS/K|=|S|/|K intersect S|, then |KS/K|= p (where p is a prime). Want to show that KS/K is a miaximal p-subgroup of G/K. So it suffices to show that [G/K:KS/K] is relatively prime to p. From here is where I'am having trouble I have been trying to use the facts about S being a p-sylow subgroup of G, to get that S/K is a p-sylow subgroup of G/K to try to get my conclusion,but I have been getting no where doing this. Any suggestions would be a life saver. #### tonio Let K be a normal subgroup of a fintire group G. Let S be a p-Sylow subgroup of G (p a prime divisor of |G|). Prove that KS/K is a p-Sylow subgroup of G/K. Proof/ So we know that K is a subgroup of KS and KS is a subgroup of G, so by the correspondence theroem KS/K is a subgroup of G/K. Since |KS/K|=|S|/|K intersect S|, then |KS/K|= p (where p is a prime). Want to show that KS/K is a miaximal p-subgroup of G/K. So it suffices to show that [G/K:KS/K] is relatively prime to p. From here is where I'am having trouble I have been trying to use the facts about S being a p-sylow subgroup of G, to get that S/K is a p-sylow subgroup of G/K to try to get my conclusion,but I have been getting no where doing this. Any suggestions would be a life saver. $$\displaystyle [G/K:KS/K]=\frac{\frac{|G|}{|K|}}{\frac{|KS}{|K|}}$$ $$\displaystyle =\frac{|G|}{|KS|}=\frac{|G||K\cap S|}{|K||S|}$$ . If we put $$\displaystyle |G|=p^rm\,,\,\,(p,m)=1$$ , we get: $$\displaystyle [G/K:KS/K]=\frac{|G||K\cap S|}{|K||S|}=\frac{p^rm\cdot |K\cap S|}{|K|p^r}=\frac{m|K\cap S|}{|K|}$$ , and now we just have to note that any power of $$\displaystyle p$$ in $$\displaystyle |K|$$ cancels with the same power of $$\displaystyle p$$ in $$\displaystyle |K\cap S|$$ (why??) Tonio aabsdr #### aabsdr any power of p in |K| cancels with the same power of P in |K intersect S| because K intersect S is a maximal P subgroup of K. Thanks you sooooooo much!!!!!!!!!!!!!!!!!!!!

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Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  dipsubdi Structured version   Visualization version   GIF version Theorem dipsubdi 28621 Description: Distributive law for inner product subtraction. (Contributed by NM, 20-Nov-2007.) (New usage is discouraged.) Hypotheses Ref Expression ipsubdir.1 𝑋 = (BaseSet‘𝑈) ipsubdir.3 𝑀 = ( −𝑣𝑈) ipsubdir.7 𝑃 = (·𝑖OLD𝑈) Assertion Ref Expression dipsubdi ((𝑈 ∈ CPreHilOLD ∧ (𝐴𝑋𝐵𝑋𝐶𝑋)) → (𝐴𝑃(𝐵𝑀𝐶)) = ((𝐴𝑃𝐵) − (𝐴𝑃𝐶))) Proof of Theorem dipsubdi StepHypRef Expression 1 id 22 . . 3 ((𝐶𝑋𝐵𝑋𝐴𝑋) → (𝐶𝑋𝐵𝑋𝐴𝑋)) 213com13 1121 . 2 ((𝐴𝑋𝐵𝑋𝐶𝑋) → (𝐶𝑋𝐵𝑋𝐴𝑋)) 3 id 22 . . . . . 6 ((𝐵𝑋𝐶𝑋𝐴𝑋) → (𝐵𝑋𝐶𝑋𝐴𝑋)) 433com12 1120 . . . . 5 ((𝐶𝑋𝐵𝑋𝐴𝑋) → (𝐵𝑋𝐶𝑋𝐴𝑋)) 5 ipsubdir.1 . . . . . 6 𝑋 = (BaseSet‘𝑈) 6 ipsubdir.3 . . . . . 6 𝑀 = ( −𝑣𝑈) 7 ipsubdir.7 . . . . . 6 𝑃 = (·𝑖OLD𝑈) 85, 6, 7dipsubdir 28620 . . . . 5 ((𝑈 ∈ CPreHilOLD ∧ (𝐵𝑋𝐶𝑋𝐴𝑋)) → ((𝐵𝑀𝐶)𝑃𝐴) = ((𝐵𝑃𝐴) − (𝐶𝑃𝐴))) 94, 8sylan2 595 . . . 4 ((𝑈 ∈ CPreHilOLD ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → ((𝐵𝑀𝐶)𝑃𝐴) = ((𝐵𝑃𝐴) − (𝐶𝑃𝐴))) 109fveq2d 6655 . . 3 ((𝑈 ∈ CPreHilOLD ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘((𝐵𝑀𝐶)𝑃𝐴)) = (∗‘((𝐵𝑃𝐴) − (𝐶𝑃𝐴)))) 11 phnv 28586 . . . 4 (𝑈 ∈ CPreHilOLD𝑈 ∈ NrmCVec) 12 simpl 486 . . . . 5 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → 𝑈 ∈ NrmCVec) 135, 6nvmcl 28418 . . . . . . 7 ((𝑈 ∈ NrmCVec ∧ 𝐵𝑋𝐶𝑋) → (𝐵𝑀𝐶) ∈ 𝑋) 14133com23 1123 . . . . . 6 ((𝑈 ∈ NrmCVec ∧ 𝐶𝑋𝐵𝑋) → (𝐵𝑀𝐶) ∈ 𝑋) 15143adant3r3 1181 . . . . 5 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (𝐵𝑀𝐶) ∈ 𝑋) 16 simpr3 1193 . . . . 5 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → 𝐴𝑋) 175, 7dipcj 28486 . . . . 5 ((𝑈 ∈ NrmCVec ∧ (𝐵𝑀𝐶) ∈ 𝑋𝐴𝑋) → (∗‘((𝐵𝑀𝐶)𝑃𝐴)) = (𝐴𝑃(𝐵𝑀𝐶))) 1812, 15, 16, 17syl3anc 1368 . . . 4 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘((𝐵𝑀𝐶)𝑃𝐴)) = (𝐴𝑃(𝐵𝑀𝐶))) 1911, 18sylan 583 . . 3 ((𝑈 ∈ CPreHilOLD ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘((𝐵𝑀𝐶)𝑃𝐴)) = (𝐴𝑃(𝐵𝑀𝐶))) 205, 7dipcl 28484 . . . . . . 7 ((𝑈 ∈ NrmCVec ∧ 𝐵𝑋𝐴𝑋) → (𝐵𝑃𝐴) ∈ ℂ) 21203adant3r1 1179 . . . . . 6 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (𝐵𝑃𝐴) ∈ ℂ) 225, 7dipcl 28484 . . . . . . 7 ((𝑈 ∈ NrmCVec ∧ 𝐶𝑋𝐴𝑋) → (𝐶𝑃𝐴) ∈ ℂ) 23223adant3r2 1180 . . . . . 6 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (𝐶𝑃𝐴) ∈ ℂ) 24 cjsub 14497 . . . . . 6 (((𝐵𝑃𝐴) ∈ ℂ ∧ (𝐶𝑃𝐴) ∈ ℂ) → (∗‘((𝐵𝑃𝐴) − (𝐶𝑃𝐴))) = ((∗‘(𝐵𝑃𝐴)) − (∗‘(𝐶𝑃𝐴)))) 2521, 23, 24syl2anc 587 . . . . 5 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘((𝐵𝑃𝐴) − (𝐶𝑃𝐴))) = ((∗‘(𝐵𝑃𝐴)) − (∗‘(𝐶𝑃𝐴)))) 265, 7dipcj 28486 . . . . . . 7 ((𝑈 ∈ NrmCVec ∧ 𝐵𝑋𝐴𝑋) → (∗‘(𝐵𝑃𝐴)) = (𝐴𝑃𝐵)) 27263adant3r1 1179 . . . . . 6 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘(𝐵𝑃𝐴)) = (𝐴𝑃𝐵)) 285, 7dipcj 28486 . . . . . . 7 ((𝑈 ∈ NrmCVec ∧ 𝐶𝑋𝐴𝑋) → (∗‘(𝐶𝑃𝐴)) = (𝐴𝑃𝐶)) 29283adant3r2 1180 . . . . . 6 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘(𝐶𝑃𝐴)) = (𝐴𝑃𝐶)) 3027, 29oveq12d 7156 . . . . 5 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → ((∗‘(𝐵𝑃𝐴)) − (∗‘(𝐶𝑃𝐴))) = ((𝐴𝑃𝐵) − (𝐴𝑃𝐶))) 3125, 30eqtrd 2859 . . . 4 ((𝑈 ∈ NrmCVec ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘((𝐵𝑃𝐴) − (𝐶𝑃𝐴))) = ((𝐴𝑃𝐵) − (𝐴𝑃𝐶))) 3211, 31sylan 583 . . 3 ((𝑈 ∈ CPreHilOLD ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (∗‘((𝐵𝑃𝐴) − (𝐶𝑃𝐴))) = ((𝐴𝑃𝐵) − (𝐴𝑃𝐶))) 3310, 19, 323eqtr3d 2867 . 2 ((𝑈 ∈ CPreHilOLD ∧ (𝐶𝑋𝐵𝑋𝐴𝑋)) → (𝐴𝑃(𝐵𝑀𝐶)) = ((𝐴𝑃𝐵) − (𝐴𝑃𝐶))) 342, 33sylan2 595 1 ((𝑈 ∈ CPreHilOLD ∧ (𝐴𝑋𝐵𝑋𝐶𝑋)) → (𝐴𝑃(𝐵𝑀𝐶)) = ((𝐴𝑃𝐵) − (𝐴𝑃𝐶))) Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 399   ∧ w3a 1084   = wceq 1538   ∈ wcel 2115  ‘cfv 6336  (class class class)co 7138  ℂcc 10520   − cmin 10855  ∗ccj 14444  NrmCVeccnv 28356  BaseSetcba 28358   −𝑣 cnsb 28361  ·𝑖OLDcdip 28472  CPreHilOLDccphlo 28584 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-11 2162  ax-12 2179  ax-ext 2796  ax-rep 5171  ax-sep 5184  ax-nul 5191  ax-pow 5247  ax-pr 5311  ax-un 7444  ax-inf2 9088  ax-cnex 10578  ax-resscn 10579  ax-1cn 10580  ax-icn 10581  ax-addcl 10582  ax-addrcl 10583  ax-mulcl 10584  ax-mulrcl 10585  ax-mulcom 10586  ax-addass 10587  ax-mulass 10588  ax-distr 10589  ax-i2m1 10590  ax-1ne0 10591  ax-1rid 10592  ax-rnegex 10593  ax-rrecex 10594  ax-cnre 10595  ax-pre-lttri 10596  ax-pre-lttrn 10597  ax-pre-ltadd 10598  ax-pre-mulgt0 10599  ax-pre-sup 10600  ax-addf 10601  ax-mulf 10602 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3or 1085  df-3an 1086  df-tru 1541  df-fal 1551  df-ex 1782  df-nf 1786  df-sb 2071  df-mo 2624  df-eu 2655  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2964  df-ne 3014  df-nel 3118  df-ral 3137  df-rex 3138  df-reu 3139  df-rmo 3140  df-rab 3141  df-v 3481  df-sbc 3758  df-csb 3866  df-dif 3921  df-un 3923  df-in 3925  df-ss 3935  df-pss 3937  df-nul 4275  df-if 4449  df-pw 4522  df-sn 4549  df-pr 4551  df-tp 4553  df-op 4555  df-uni 4820  df-int 4858  df-iun 4902  df-iin 4903  df-br 5048  df-opab 5110  df-mpt 5128  df-tr 5154  df-id 5441  df-eprel 5446  df-po 5455  df-so 5456  df-fr 5495  df-se 5496  df-we 5497  df-xp 5542  df-rel 5543  df-cnv 5544  df-co 5545  df-dm 5546  df-rn 5547  df-res 5548  df-ima 5549  df-pred 6129  df-ord 6175  df-on 6176  df-lim 6177  df-suc 6178  df-iota 6295  df-fun 6338  df-fn 6339  df-f 6340  df-f1 6341  df-fo 6342  df-f1o 6343  df-fv 6344  df-isom 6345  df-riota 7096  df-ov 7141  df-oprab 7142  df-mpo 7143  df-of 7392  df-om 7564  df-1st 7672  df-2nd 7673  df-supp 7814  df-wrecs 7930  df-recs 7991  df-rdg 8029  df-1o 8085  df-2o 8086  df-oadd 8089  df-er 8272  df-map 8391  df-ixp 8445  df-en 8493  df-dom 8494  df-sdom 8495  df-fin 8496  df-fsupp 8818  df-fi 8859  df-sup 8890  df-inf 8891  df-oi 8958  df-card 9352  df-pnf 10662  df-mnf 10663  df-xr 10664  df-ltxr 10665  df-le 10666  df-sub 10857  df-neg 10858  df-div 11283  df-nn 11624  df-2 11686  df-3 11687  df-4 11688  df-5 11689  df-6 11690  df-7 11691  df-8 11692  df-9 11693  df-n0 11884  df-z 11968  df-dec 12085  df-uz 12230  df-q 12335  df-rp 12376  df-xneg 12493  df-xadd 12494  df-xmul 12495  df-ioo 12728  df-icc 12731  df-fz 12884  df-fzo 13027  df-seq 13363  df-exp 13424  df-hash 13685  df-cj 14447  df-re 14448  df-im 14449  df-sqrt 14583  df-abs 14584  df-clim 14834  df-sum 15032  df-struct 16474  df-ndx 16475  df-slot 16476  df-base 16478  df-sets 16479  df-ress 16480  df-plusg 16567  df-mulr 16568  df-starv 16569  df-sca 16570  df-vsca 16571  df-ip 16572  df-tset 16573  df-ple 16574  df-ds 16576  df-unif 16577  df-hom 16578  df-cco 16579  df-rest 16685  df-topn 16686  df-0g 16704  df-gsum 16705  df-topgen 16706  df-pt 16707  df-prds 16710  df-xrs 16764  df-qtop 16769  df-imas 16770  df-xps 16772  df-mre 16846  df-mrc 16847  df-acs 16849  df-mgm 17841  df-sgrp 17890  df-mnd 17901  df-submnd 17946  df-mulg 18214  df-cntz 18436  df-cmn 18897  df-psmet 20523  df-xmet 20524  df-met 20525  df-bl 20526  df-mopn 20527  df-cnfld 20532  df-top 21488  df-topon 21505  df-topsp 21527  df-bases 21540  df-cld 21613  df-ntr 21614  df-cls 21615  df-cn 21821  df-cnp 21822  df-t1 21908  df-haus 21909  df-tx 22156  df-hmeo 22349  df-xms 22916  df-ms 22917  df-tms 22918  df-grpo 28265  df-gid 28266  df-ginv 28267  df-gdiv 28268  df-ablo 28317  df-vc 28331  df-nv 28364  df-va 28367  df-ba 28368  df-sm 28369  df-0v 28370  df-vs 28371  df-nmcv 28372  df-ims 28373  df-dip 28473  df-ph 28585 This theorem is referenced by:  siilem1  28623  ip2eqi  28628 Copyright terms: Public domain W3C validator

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# [NL2-NL10] Struggling with an Elephant • Black Joined: 02.11.2008 Hello all, As I am just getting started with Elephant, I have a number of questions. I have been keeping an eye out for a manual on the software but can't find one. The link from the programme seems to just send you to the Elephant forum which only discusses particular issues of the day. If I am being stupid and there is a manual which can help with the things below, please put me straight. In the absence of that, it would be really helpful if anyone could help me with me answers to the following queries (or at least point me in the right direction): 1 . Can anyone explain to me Aggression Factor? I understand that it is the proportion of bets/raises: checks/calls but what does a given value mean in practical terms? For example, is any value greater than 1 'aggressive', or only values more than 2, or what? Furthermore, when might this be useful to know? Is it mainly used for deciding whether to bet post flop? 2 . In the middle of my screen when I have the HUD showing, there is a black panel in the middle with some more values. (In the panel it also says: "player filter 7-10"). Can anyone tell me whose values these are? And whereabouts in the options can I change this information (if I want to). 3 . Should I just use the standard 8 stats that Poker Strategy normally uses? Will I just confuse myself (at this early stage in my poker career ) if I try to add more? If not, what others stats would be most useful for NL 25 SSS? 4 . In the part of Elephant where my hands are shown, I have worked out how to sort the hands either by the time they were played, or the amount of BBs lost or won etc. Is there any way of sorting them according to BB's won by steals? Or some other way of finding out whether my steals are overall profitable? 5. Elephant seems to divide the hands according to daily, or monthly, or yearly etc. This allows you to see whether a given day's play was profitable (+BB) according to a graph shown. However, I tend to play two or more separate sessions during a day and it would be helpful to me to find out whether a particular session was profitable. Is there any way of telling Elephant to isolate a particular session for the purposes of creating a graph? I can obviously find where the session starts and ends by looking at the time the hand was played but can I tell Elephant to look at just the hands within that time period, when producing a graph? 6. Last but not least, in the setup page there is a path to Equilator (which I have also installed on my PC). I'm guessing that there is a clever function which allows you to take a hand which you played and send it to Equilator to find out whether you have the right odds to make whatever decision you made. Any assistance in using this function, please? In the path, what do I need to put? Thanks awfully, Tim • 4 replies • Bronze Joined: 07.05.2008 Hi Tim64, I cannot answer all questions since I dont use the new Elephant but 1. = http://www.pokerstrategy.com/software/article/790 , it is explaining the Agg Factor 2. That panel is meant to show you the Table Stats (Average VPIP/PFR & Average Pot). An Average VPIP of 8 is bad while Avg. VPIP of 20 is good 3. Yes, stick to my answer of question 1, those stats 4. I dont think that's possible with Elephant, maybe the Analysis tab but with PT/HEM I know for sure that's possible 5. The only method I think is possible is to write in a Word Document how many hands session 1 was and than from there on, count those hands on the graph as session 2. with the Y axis where the line is currently at as Y=0 6. No idea - dont use that on FTP since you get banned Best regards, Gerv • Bronze Joined: 19.01.2009 6. I think you can use it as long as you're using it in review afterwards, and not live during the hand. In the path, you need to put directory where you installed equilator. oh yeah..after you do that, in order to use the equilator, you go to the hands tab, then right click on a particular hand in the list of hands, then you'll see the words equilator, then click. • Black Joined: 02.11.2008 Hi Will, Thanks - that is what I meant. Not to use at the time of the hand - only afterwards. Would never have time to do it while I was playing Cheers mate! Thanks Gerv, for you answers, too. Any help with the missing bits would be great if anyone out there in Forumland knows... -Tim • Bronze Joined: 30.08.2008 Can anyone please add me on skype and explain step by step how i can get it to run, i have it downloaded it is running, but the only thing that comes up when i play is a black thing saying 'please check your settinsg for the handhistory path' on my table. The software wont record my history or run, all i have done is downloaded it typed in the code and ran it, nothing else. I have no idea what to do to get it running. anyone who knows how to get it to run, can you please add me on skype, my name is dontbustagain from UK

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### Class XI The SI unit of retardation is 1. ms-1 2. ms-2 3. ms2 4. m A train moving with a uniform speed of 54 kmph. What is its speed in m/s? 1. 15 m/s 2. 1.5 m/s 3. 9 m/s 4. 90 m/s A truck covers 40 km with an average speed of 80km/h. Then it travels another 40 km with an average speed of 40 km/h. The average speed of the truck for the total distanced covered is 1. 40 km/h 2. 45km/h 3. 48km/h 4. 53km/h Which of the following can determine the acceleration of a moving object? 1. area of velocity-time graph 2. slope of velocity-time graph 3. area of distance-time graph 4. slope of the distance time-graph The equation v = u + at gives information as 1. velocity is a function of time 2. velocity is a function of position 3. Position is a function of time 4. Position is function of velocity and time 0 0 ### Get Started! we provide the best services to our students Views Rs 1,999  Annual Plan TOP

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# Changeset 286:da414906fe21 in lemon Ignore: Timestamp: 09/26/08 12:40:11 (12 years ago) Branch: default Phase: public Message: Improvements related to BFS/DFS/Dijkstra (ticket #96) • Add run(s,t) function to BfsVisit?. • Modify run(s,t) functions in the class interfaces to return bool value. • Bug fix in Dijkstra::start(t) function. • Improve Dijkstra::currentDist(). • Extend test files to check named class template parameters. • Doc improvements. Files: 6 edited Unmodified Removed • ## lemon/bfs.h r278 /// ///This method runs the %BFS algorithm from the root node(s) ///in order to compute the shortest path to \c dest. ///in order to compute the shortest path to \c t. /// ///The algorithm computes ///- the shortest path to \c dest, ///- the distance of \c dest from the root(s). ///- the shortest path to \c t, ///- the distance of \c t from the root(s). /// ///\pre init() must be called and at least one root node should be ///  } ///\endcode void start(Node dest) void start(Node t) { bool reach = false; while ( !emptyQueue() && !reach ) processNextNode(dest, reach); while ( !emptyQueue() && !reach ) processNextNode(t, reach); } } ///Runs the algorithm from the given node. ///Runs the algorithm from the given source node. ///This method runs the %BFS algorithm from node \c s ///This method runs the %BFS algorithm from node \c s ///in order to compute the shortest path to \c t. /// ///\return The length of the shortest s--t path, ///if \c t is reachable form \c s, \c 0 otherwise. ///in order to compute the shortest path to node \c t ///(it stops searching when \c t is processed). /// ///\return \c true if \c t is reachable form \c s. /// ///\note Apart from the return value, b.run(s,t) is just a ///  b.start(t); ///\endcode int run(Node s,Node t) { bool run(Node s,Node t) { init(); addSource(s); start(t); return reached(t) ? _curr_dist : 0; return reached(t); } /// /// This method runs the %BFS algorithm from the root node(s) /// in order to compute the shortest path to \c dest. /// in order to compute the shortest path to \c t. /// /// The algorithm computes /// - the shortest path to \c dest, /// - the distance of \c dest from the root(s). /// - the shortest path to \c t, /// - the distance of \c t from the root(s). /// /// \pre init() must be called and at least one root node should be ///   } /// \endcode void start(Node dest) { void start(Node t) { bool reach = false; while ( !emptyQueue() && !reach ) processNextNode(dest, reach); while ( !emptyQueue() && !reach ) processNextNode(t, reach); } } /// \brief Runs the algorithm from the given node. /// \brief Runs the algorithm from the given source node. /// /// This method runs the %BFS algorithm from node \c s addSource(s); start(); } /// \brief Finds the shortest path between \c s and \c t. /// /// This method runs the %BFS algorithm from node \c s /// in order to compute the shortest path to node \c t /// (it stops searching when \c t is processed). /// /// \return \c true if \c t is reachable form \c s. /// /// \note Apart from the return value, b.run(s,t) is just a /// shortcut of the following code. ///\code ///   b.init(); ///   b.addSource(s); ///   b.start(t); ///\endcode bool run(Node s,Node t) { init(); addSource(s); start(t); return reached(t); }

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## Math Teachers at Play #12 In prior Carnivals of Mathematics it has been a tradition to include trivia about the number the Carnival happens to be at in some way. With a desire to do something different for Math Teachers at Play, I offer a riddle. Younger solvers may have an advantage over older ones here. Here is a pair of dice I own: If I roll the dice and read the numbers off the top, is it more likely the numbers add up to be 2, or add up to be 12? Or are the two sums equally likely? Why? Be careful before you answer! Elementary Concepts and Arithmetic Kendra has come up with a new game for teaching the reading of clocks. (The RummiKub variant is intriguing.) The blog yofx gives a vacation photo of an unusual example of negative numbers in real life. Many look at multiplication tables and see drudgery; Dan MacKinnon sees rainbows and hyperbolic arcs. Speaking of multiplication, the folks at 360 have finished their series on 25 ways to multiply: 1-6, 7-12, 13-14, 15-17, 18, and 19-25. When you go to the pediatrician, do you notice mathematical error-correcting codes? Mark Dominus does. Dave Richeson finds complexity in even the simplest geometry and shares three cool facts about rotations of the circle. Foxmaths 2.0 performs an clever bit of calculus on the function $f(x)=x^{x^{x^{x^{...}}}}$. Want to keep your kids thinking about math over the summer? Kate Nowak is full of ideas at Math around the House. Tom DeRosa wants a TV show that changes the way we think about math. Colleen King wonders if learning programming should be mandatory in education, and gives concrete suggestions for all the different grade levels. Miscellaneous Want to post math equations on Blogger/Blogspot like you’ve seen on WordPress? WatchMath has a solution. Speaking of typesetting equations, John Cook owes Microsoft Word an apology. This post about struggling with dyscalculia is a worthwhile read. Finally, this is a series that appeared back in 2008, but it was new to me, so I hope it is new to some of you! Ron Doerfler wrote a three-part series on “lightning calculators”, people who can do astounding mathematical calculations in their head: The Players, The Methods, The Media. Especially fun (in part 3) is the deconstruction of a Daniel Tammet documentary and its “creative” editing. ### 27 Responses 1. […] Deadline Looms 2009 July 21 by Denise Update: Math Teachers at Play #12 is up and running, with a pleasing variety of posts to browse. […] 2. Oh neat! A lot of good stuff. I wonder who submitted me. 3. […] Math Teachers at Play #12 2009 July 25 tags: Carnival of Mathematics, Math Teachers at Play by jd2718 It’s a good one. (here, at the Number Warrior) […] 4. I’m stumped on the riddle. Time for a hint? I want the answer to be equally likely, but you warned me it was not so simple. Here’s what I did: The probability that 1 and 1 show up is 1/36. The probability that 6 and 6 show up is 1/36. These are the only ways to get sums of 2 and of 12, and they are equally likely. Clearly, I’ve missed something. Jonathan 5. Are you saying the roll of “1” and the roll of “2” together make 12? 6. Perhaps because you’ve already rolled a 12, the possibility of rolling a 2 next is more likely. 7. I’d already figured the picture was important, but I can’t discern anything significant from it. The dice appear to be normal, fair, and equal, to the extent that one can tell from a blurry photo. We can’t see the 4 and 1 sides, so perhaps there’s some trick to the dice involving that. Like, maybe there’s no 1 opposite the 6, so a roll of 2 is impossible. But that’d be cheesy, because there’s no real clue of any such thing. 8. For the riddle: is it that we know that there is a six on each die, but we don’t know whether or not there is a one on each die (because that side is not pictured, and conceivably they could be non-standard dice), so the probability of getting 12 is higher? • I need to learn to read not just the first comments, but also the recent ones. 9. The answer is that you are more likely to get “2”. This is because if you roll a 1 and a 1, you add up to get “2”. If you roll a 6, which is made up of two “1” shapes, you also get “2”. I hope I am right. 10. It’s hard to tell, but from the picture I think I notice a slight concavity in the material used to backfill the holes (12 of them for boxcars). So it is conceivable that there is overall less mass on the side with a ‘6’ than there is on the side with a ‘1’, which would lend a small bias in favor of the dice landing with the 1’s on the bottom and 6’s on top. 11. […] Comments Jason Dyer on Math Teachers at Play #12Todd Trimble on Math Teachers at Play #12Jason Dyer on Math Teachers at Play #12Rory on Math Teachers at Play #12DHJ(k): 1200-1299 […] 12. Surely the difference is so minor as to be far overshadowed by other factors (such as other minuscule irregularities in the dice). This seems rather like the specious claim that drains empty in different directions south of the equator vs north, owing to the Coriolis effect. Have you actually done an experiment to show that there’s any real statistical difference? • Yes, I have witnessed the effect for myself. It is also well known in the gambling literature. (That’s where I first came across it — I think it may have been Scarne on Dice.) The weight effect is generally far greater than any manufacture error in the dice (relatively speaking, as I said in the other post, you need at least 500 rolls before you have a chance of spotting anything), although it is a good point that cheap enough quality might override everything. 13. […] Math Teachers at Play #12 is up and running, with a pleasing variety of posts to browse. […] 14. Huh, hadn’t heard of these before, but I’ve got a submission at: http://www.discreteideas.com/2009/08/making-more-math-geeks/ 15. Actually, there is the same exact odds of rolling 2 or 12 each and everytime the dice are thrown. The odds of rolling each are 1:36. This is why at a casino, the payout when you bet on either is 1:30; Sounds like great odds, but you are still the underdog by the odds of 1:6. Which means statistically, the casino has the better odds of winning by 6:5 overall, or for each roll. or 36:30. 16. Sir Chadwick, you make me wonder whether you even read the comments above. The dice pictured above are cheaply produced dice, not high-quality casino dice. With cheaply produced dice, holes are drilled and not too carefully backfilled, and this definitely introduces a bias. Statisticians familiar with both kinds of dice (such as Mosteller) are quite familiar with the phenomenon. 17. […] Math Teachers at Play 12 […] 18. […] Teachers at Play #29 Posted on August 23, 2010 by Jason Dyer The last time I hosted Math Teachers at Play I attempted to start a tradition of including a math puzzle pertinent to the number of the […] 19. […] Math Teachers at Play #12 This entry was posted in Astronomy, Biology, Carnivalia, Math, Space and tagged Astronomy, Biology, Carnivalia, Math, Space. Bookmark the permalink. ← Scientific tourist #83 — heatsink nosecone Casual Friday — Excalibur → […]

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## Blog Pages ### Check wheather a number is composite or not ```#Credits to NPTEL MOOC, Programming, Data Structures & Algorithms in #Python by Madhavan Mukund, Chennai Mathematical Institute ``` ` ` ```def composite(n): for i in range(2,n): if n%i == 0: return(True) return(False)``` ### Divison with multiple conditions ```#Credits to NPTEL MOOC, Programming, Data Structures & Algorithms in #Python by Madhavan Mukund, Chennai Mathematical Institute ``` def divides(m,n): if n%m == 0: return(True) else: return(False) def even(n): return(divides(2,n)) def odd(n): return(not divides(2,n)) ### GCD using euclids algorithms & looping # Credits to NPTEL MOOC,Programming, Data Structures & Algorithms #in Python by Madhavan Mukund, Chennai Mathematical Institute def gcd(m,n): if m < n:  # Assume m >= n (m,n) = (n,m) while (m%n) != 0: diff = m-n # diff > n? Possible! (m,n) = (max(n,diff),min(n,diff)) return(n) ### Greatest common divisor program using euclids algorithms ```#Credits to NPTEL MOOC, Programming, Data Structures & Algorithms in #Python by Madhavan Mukund, Chennai Mathematical Institute ``` ` ` def gcd(m,n): if m < n: (m,n) = (n,m) if (m%n) == 0: return(n) else: diff = m-n return(gcd(max(n,diff),min(n,diff))) print(gcd(12,3)) ### Greatest common divisor program with looping ```#Credits to NPTEL MOOC, Programming, Data Structures & Algorithms in #Python by Madhavan Mukund, Chennai Mathematical Institute ``` def gcd(m,n): i = min(m,n) while i > 0: if (m%i) == 0 and (n%i) == 0: return(i) else: i = i-1

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Precalculus posted by . A swimming pool is 3 feet deep in the shallow end. The bottom of the pool has a steady downward drop of 12 degrees toward the deep end. If the pool is 50 feet long, how deep is the deep end. So I said that first you would subtract three from 50, of course giving you 47, then, I said tan12 = x/47 tan12 * 47 = x is this right • Precalculus - the length is 50 feet, it has nothing to do with the fact that the depth is 3 feet at the shallow end. make a sketch, showing a side view of the pool and you will see the problem more clearly. Draw a line horizontally to show a rectangle 50 by 3 at the top and a right-angled triangle with a length of 50 and a width of x feet then tan12 = x/50 x = 50tan12 = 10.63 feet x is the "additional" depth below the shallow end so the deep end is 10.63 + 3 or 13.63 feet deep. • Precalculus - The bottom of a swimming pool isn't a regular profile. The deep end is about 6 foot 6

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### Home > CCA > Chapter Ch6 > Lesson 6.2.1 > Problem6-62 6-62. Write an equation or system of equations to solve. Thanh bought 11 pieces of fruit and spent $5.60. If apples cost$0.60 each and pears cost \$0.35 each, how many of each kind of fruit did he buy? One equation should be about the number of each fruit that Thanh bought. The other equation should include costs. 7 apples

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# Polynomial Function ## Polynomial Function Function f characterized by a relationship of the form $$f\left ( x \right )=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+…+a_{1}x+a_{0}$$ where $$a_{n}$$ is non-zero and the $$a_{i}$$ are real numbers or complex numbers. The number n indicates the degree of the polynomial function. ### Example The function f defined on the set of real numbers by the relationship $$f\left ( x \right ) = x^{3} – 7x^{2} + 5x + 12$$ is a third-degree polynomial function. Its graph is the following :

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# Sphere Wall Collision Page 1 of 1 ## 7 Replies - 4613 Views - Last Post: 15 April 2009 - 09:13 PMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax&section=topics&do=rateTopic&t=96429&amp;s=182d132c1b34f84a67ef511822b62eb0&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]> ### #1 KYA • Wubba lubba dub dub! Reputation: 3194 • Posts: 19,220 • Joined: 14-September 07 # Sphere Wall Collision Posted 31 March 2009 - 08:50 AM I've been thinking about this for a few weeks now and all of my various attempts have failed. I think its due to the fact that there is a math formula I could use, but I just don't know it. The problem? Detecting a collision between a sphere particle and a "wall" (i.e. bounds of a 3D box), they are bouncing around in. I can detect and resolve collision between spheres relatively well: //if two spheres will collide within a given timestamp, create a CollisionPair and //add it to the vector static void detectCollisions(Vector<Sphere> spheres, Vector<CollisionPair> pairs, double time) { /*A simple way to check for collision between two spheres given the start and end position for a given time frame * is to see the smallest distance between them--ever d^2 = a^2-(A*B)^2/b^2 (distance = d^2) * lower case letters are the magnitude of each vector where the upper case are the actual vectors Given two spheres P and Q with two points P1, Q1, etc... we have two vectors A and B where A = P1-Q1 B = (P2-P1)-(Q2-Q1) if distance < (rp+rq)^2, there is a collision (r is the radius of each respective sphere) */ double distance = 0.00, radii = 0.00, t = 0.00; CollisionPair pair = new CollisionPair(); //clean out the pairs vector for each time stamp pairs.removeAllElements(); //Get our appropriate A and B vectors //spheres 1 and 2: Vector3D pOne = spheres.get(0).getCurPosition(); Vector3D qOne = spheres.get(1).getCurPosition(); Vector3D pTwo = spheres.get(0).getStartPosition().gravitationalAcceleration(spheres.get(0).getVelocity(), time); Vector3D qTwo = spheres.get(1).getStartPosition().gravitationalAcceleration(spheres.get(1).getVelocity(), time); Vector3D A1 = pOne.subtract(qOne); Vector3D B1 = (pTwo.subtract(pOne)).subtract(qTwo.subtract(qOne)); distance = Math.pow(A1.getMagnitude(), 2.0)-(Math.pow(A1.numericalDotProduct(B1), 2.0)/Math.pow(B1.getMagnitude(),2.0)); { //before collision coordinates of the two patciles pair.setFirst(spheres.get(0)); pair.setSecond(spheres.get(1)); pair.setFirstID(0); pair.setSecondID(1); //get time (coliision between t0 and t1 timesteps): //quadratic equation equivalent if t is between 0 and 1 that is collision tiem within the timestamp t = -(A1.numericalDotProduct(B1)); t -= Math.sqrt((Math.pow(A1.numericalDotProduct(B1),2.0)) - ((Math.pow(B1.getMagnitude(),2.0))* t/= Math.pow(B1.getMagnitude(),2.0); pair.setCollisionTime(Math.abs(t)); //add pair to the vector } //spheres 1 and 3 pOne = spheres.get(0).getCurPosition(); qOne = spheres.get(2).getCurPosition(); pTwo = spheres.get(0).getStartPosition().gravitationalAcceleration(spheres.get(0).getVelocity(), time); qTwo = spheres.get(2).getStartPosition().gravitationalAcceleration(spheres.get(2).getVelocity(), time); A1 = pOne.subtract(qOne); B1 = (pTwo.subtract(pOne)).subtract(qTwo.subtract(qOne)); distance = Math.pow(A1.getMagnitude(), 2.0)-(Math.pow(A1.numericalDotProduct(B1), 2.0)/Math.pow(B1.getMagnitude(),2.0)); { //before collision coordinates of the two patciles pair.setFirst(spheres.get(0)); pair.setSecond(spheres.get(2)); pair.setFirstID(0); pair.setSecondID(2); //get time (coliision between t0 and t1 timesteps): //quadratic equation equivalent if t is between 0 and 1 that is collision tiem within the timestamp t = -(A1.numericalDotProduct(B1)); t -= Math.sqrt((Math.pow(A1.numericalDotProduct(B1),2.0)) - ((Math.pow(B1.getMagnitude(),2.0))* t/= Math.pow(B1.getMagnitude(),2.0); pair.setCollisionTime(Math.abs(t)); //add pair to the vector } //spheres 2 and 3 pOne = spheres.get(1).getCurPosition(); qOne = spheres.get(2).getCurPosition(); pTwo = spheres.get(1).getStartPosition().gravitationalAcceleration(spheres.get(1).getVelocity(), time); qTwo = spheres.get(2).getStartPosition().gravitationalAcceleration(spheres.get(2).getVelocity(), time); A1 = pOne.subtract(qOne); B1 = (pTwo.subtract(pOne)).subtract(qTwo.subtract(qOne)); distance = Math.pow(A1.getMagnitude(), 2.0)-(Math.pow(A1.numericalDotProduct(B1), 2.0)/Math.pow(B1.getMagnitude(),2.0)); { //before collision coordinates of the two patciles pair.setFirst(spheres.get(1)); pair.setSecond(spheres.get(2)); pair.setFirstID(1); pair.setSecondID(2); //get time (coliision between t0 and t1 timesteps): //quadratic equation equivalent if t is between 0 and 1 that is collision tiem within the timestamp t = -(A1.numericalDotProduct(B1)); t -= Math.sqrt((Math.pow(A1.numericalDotProduct(B1),2.0)) - ((Math.pow(B1.getMagnitude(),2.0))* t/= Math.pow(B1.getMagnitude(),2.0); pair.setCollisionTime(Math.abs(t)); //add pair to the vector } //if no collision between any spheres, pairs vector will be empty return; } Sphere: public class Sphere { private Vector3D velocity; private Vector3D curPosition, startPosition; private double mass; private double startTime; private Vector<Vector3D> path; DecimalFormat vec = new DecimalFormat("0.00"); //everything is zero by default, reserved for possible future expantion public Sphere() { curPosition = new Vector3D(); startPosition = new Vector3D(); velocity = new Vector3D(); curPosition.setX(0); curPosition.setY(0); curPosition.setZ(0); startPosition.setX(0); startPosition.setY(0); startPosition.setZ(0); velocity.setX(0); velocity.setY(0); velocity.setZ(0); radius = mass = startTime = 0; } public Sphere(Vector3D c, Vector3D v, double r, double s) { startPosition = c; curPosition = c; velocity = v; mass = (4/3)*Math.PI*Math.pow(r, 3); startTime = s; path = new Vector(); } public void setVelocity(Vector3D v) {velocity = v;}; public void setCurPosition(Vector3D c) {curPosition = c;}; public void setStartPosition(Vector3D s) {startPosition = s;}; public void setRadius (double r) {radius = r;}; public void setMass(double m) {mass = m;}; public void setStartTime(double s) {startTime = s;}; public Vector3D getVelocity() {return velocity;}; public Vector3D getStartPosition() {return startPosition;}; public Vector3D getCurPosition() {return curPosition;}; public double getMass() {return mass;}; public double getStartTime() {return startTime;}; public Vector<Vector3D> getPath() {return path;}; @Override public String toString() { String temp = "Position: " + "[" + vec.format(this.getCurPosition().getX()) + ", " + vec.format(this.getCurPosition().getY()) + ", " + vec.format(this.getCurPosition().getZ()) + "]" + " Velocity: " + "[" + vec.format(this.getVelocity().getX()) + ", " + vec.format(this.getVelocity().getY()) + ", " + vec.format(this.getVelocity().getZ()) + "]" + " Radius: " + this.getRadius() + " Mass: " + vec.format(this.getMass()); return temp; } /* print out the "path" of each sphere, useful for debugging * and visually determniing if there was a collision * This vector will be sent to a log file for possible future plotting */ public void displayPath() { for(int i = 0; i < path.size(); i++) { System.out.println(path.get(i).toString()); } } } Vector3D public class Vector3D { private double x, y, z; DecimalFormat vec = new DecimalFormat("0.00"); //default constructor here for temp Vectors created when performing functions public Vector3D() { x = y = z = 0; } public Vector3D(double pX, double pY, double pZ) { x = pX; y = pY; z = pZ; } //Accessor Mutator functions public void setX(double pX) { x = pX; }; public void setY(double pY) { y = pY; }; public void setZ(double pZ) { z = pZ; }; public double getX() { return x; }; public double getY() { return y; }; public double getZ() { return z; }; @Override public String toString() { String temp = "[" + vec.format(this.getX()) + ", " + vec.format(this.getY()) + ", " + vec.format(this.getZ()) + "]"; return temp; } public double getMagnitude() { double magnitude = Math.sqrt(x*x+y*y+z*z); return magnitude; } public Vector3D normalize() { double length = this.getMagnitude(); Vector3D temp = new Vector3D(); temp.setX(this.getX()/length); temp.setY(this.getY()/length); temp.setZ(this.getZ()/length); return temp; } public Vector3D crossProduct(Vector3D vectOne, Vector3D vectTwo) { Vector3D temp = new Vector3D(); temp.setX(vectOne.getY()*vectTwo.getZ()-vectOne.getZ()*vectTwo.getY()); temp.setY(vectOne.getZ()*vectTwo.getX()-vectOne.getX()*vectTwo.getZ()); temp.setZ(vectOne.getX()*vectTwo.getY()-vectOne.getY()*vectTwo.getX()); return temp; } public Vector3D scalarMultiplication(double num) { Vector3D temp = new Vector3D(); temp.setX(this.getX()*num); temp.setY(this.getY()*num); temp.setZ(this.getZ()*num); return temp; } public Vector3D scalarDivision(double num) { Vector3D temp = new Vector3D(); temp.setX(this.getX()/num); temp.setY(this.getY()/num); temp.setZ(this.getZ()/num); return temp; } //overloaded '+' operator equivalent //add two vectors, does not change calling object, //merely returns the vector result public Vector3D add(Vector3D rhs) { Vector3D temp = new Vector3D(); temp.setX(this.getX() + rhs.getX()); temp.setY(this.getY() + rhs.getY()); temp.setZ(this.getZ() + rhs.getZ()); return temp; } //overloaded '-' operator equivalent //subtracts two vectors, does not change calling object, //merely returns the vector result public Vector3D subtract(Vector3D rhs) { Vector3D temp = new Vector3D(); temp.setX(this.getX() - rhs.getX()); temp.setY(this.getY() - rhs.getY()); temp.setZ(this.getZ() - rhs.getZ()); return temp; } //returns the numerical result of a vector dot product (as opposed to an actual vector) public double numericalDotProduct(Vector3D rhs) { double temp = 0; temp = this.getX()*rhs.getX()+ this.getY()*rhs.getY()+ this.getZ()*rhs.getZ(); return temp; } public Vector3D gravitationalAcceleration(Vector3D vInit, double time) { //given the gravity vector Earth [0, 0, -9.81] and the particle initial position (the vector itself) // and the velocity desired we can calculate where it moves in 't' timesteps, a second for example //as long as the velocity is not parallel to the gravity, the result will be parabolic //*************Equation******************* // p(t) = pInit + vInit*(t-tInit) + (1/2)*gravity*(t-tInit)^2 //**************************************** Vector3D curPosition = new Vector3D(); Vector3D gravity = new Vector3D(0, 0, -9.81); curPosition = curPosition.add(gravity.scalarMultiplication(0.5*(Math.pow(time, 2.0)))); return curPosition; } } CollisionPair: public class CollisionPair { private Sphere first; private Sphere second; private double collisionTime; private int firstID, secondID; //identify which sphere (0-2) DecimalFormat vec = new DecimalFormat("0.00"); public CollisionPair() { first = new Sphere(); second = new Sphere(); collisionTime = 0; } public void setFirst(Sphere rhs) {first = rhs;}; public void setSecond(Sphere rhs) {second = rhs;}; public void setFirstID(int num) {firstID = num;}; public void setSecondID(int num) {secondID = num;}; public void setCollisionTime(double n) {collisionTime = n;}; public Sphere getFirst() {return first;}; public Sphere getSecond() {return second;}; public int getFirstID() {return firstID;}; public int getSecondID() {return secondID;}; public double getCollisionTime() {return collisionTime;}; @Override public String toString() { String temp = "Collision occurs between points: " + "[" + vec.format(this.getFirst().getCurPosition().getX()) + ", " + vec.format(this.getFirst().getCurPosition().getY()) + ", " + vec.format(this.getFirst().getCurPosition().getZ()) + "] " + "[" + vec.format(this.getSecond().getCurPosition().getX()) + ", " + vec.format(this.getSecond().getCurPosition().getY()) + ", " + vec.format(this.getSecond().getCurPosition().getZ()) + "]" + " Time: " + vec.format(this.getCollisionTime()); return temp; } } If the smallest distance between two sphere centers is less then the sum of their radii there is a collision. Easy enough. Is there a similar formula for a wall and a sphere? Another problem is that this environment is frictionless, but does have regular Earth gravity. So I'll have to make the walls extra bouncy to compensate. But, having the spheres confined for the time being would be a great relief. I'm extremely bothered that I cannot solve this myself. Any thoughts? edit: If this was in 2D I could just reverse the velocity, so maybe that's applicable here? A fresh perspective is needed. This post has been edited by KYA: 31 March 2009 - 08:53 AM Is This A Good Question/Topic? 0 ## Replies To: Sphere Wall Collision ### #2 SigurdSuhm Reputation: 18 • Posts: 111 • Joined: 05-August 08 ## Re: Sphere Wall Collision Posted 31 March 2009 - 11:21 AM Are the walls axis aligned? In that case it is very simple. But if not then it's somewhat harder. For axis aligned collision you can simply check whether the distance is less than the sphere's radius. For example: Given a wall stretching in the Y and Z direction you can detect the distance between the difference between the sphere's radius' X coordinate and the wall's X coordinate. If this difference is less that the sphere's radius you have a collision. If you need an algorithm for walls with any non axis aligned orientation let us know, and hopefully I or someone else can help. Regards Sigurd ### #3 KYA • Wubba lubba dub dub! Reputation: 3194 • Posts: 19,220 • Joined: 14-September 07 ## Re: Sphere Wall Collision Posted 31 March 2009 - 11:46 AM AABB sounds like a good way to go. It's a 3D cube so it'll be axis aligned (I think). This assignment has made me feel absolutely retarded. I did experiment with checking bounds per each sphere's appropriate coordinate (making sure it isn't less then zero or greater then 100) so that makes sense. I'm pretty sure I did it wrong though. Any chance for some more specifics regarding axis aligned collision detection? Here's a visual of what I'm thinking, not 100% on this though. does this constitute axis aligned walls? I appreciate the response. ### #4 bobjob Reputation: 23 • Posts: 163 • Joined: 29-March 08 ## Re: Sphere Wall Collision Posted 31 March 2009 - 01:12 PM A sphere collision just uses the formula a^2 = b^2+c^2 in order to work out the distance between the position of the particle and an object. if ("distance" < radius) //than collision else //keep moving in current direction example for testing on the roof collision (assuming the roof is at a height of say '5' and that the arn't diagnal) double checkDistance3D(float x1,float y1,float z1,float x2,float y2,float z2) { float dx = x1 - x2, dy = y1 - y2, dz = z1 - z2; return squareRoot((dx*dx) + (dy*dy) + (dz*dz)); } if (checkDistance3d(0, 5, 0, ball.x, ball.y, ball.z) < ball.radius) { ball.x += xSpeed; ball.y += ySpeed; ball.z += zSpeed; } else { ySpeed = -ySpeed; } you will probably need to make a "dummy" particle/ball or whatever, so as to check the position you want to go, before you actually confirm that its possible. this should work for a single square room. if its anymore complex, a few changed need to be made. and if you need to do diagnal walls, thats alot of fun, but alot math. This post has been edited by bobjob: 31 March 2009 - 01:14 PM ### #5 KYA • Wubba lubba dub dub! Reputation: 3194 • Posts: 19,220 • Joined: 14-September 07 ## Re: Sphere Wall Collision Posted 31 March 2009 - 01:35 PM In your example is the y axis the up/down/vertical one? Thanks for the snippet, I'll have to play around with it a bit. ### #6 SigurdSuhm Reputation: 18 • Posts: 111 • Joined: 05-August 08 ## Re: Sphere Wall Collision Posted 01 April 2009 - 02:44 AM In the case where your bounding box has it's origin in (0,0) and all the corners have coordinates of either 0 or 100 it should be pretty easy. You can determine a minimum and maximum value for the sphere's center's coordinates. For example: If the sphere has a radius of 6 and the box stretches from 0 to 100 on the X axis, then the X coordinate of the sphere's center has to be between 6 and 94. Otherwise you have a collision with a wall. Should be pretty easy to do for all axes. Let us know if something's not working properly. (oh and off topic... this April's Fool really had me cracking up. DIC we love you ) ### #7 KYA • Wubba lubba dub dub! Reputation: 3194 • Posts: 19,220 • Joined: 14-September 07 ## Re: Sphere Wall Collision Posted 04 April 2009 - 03:36 PM Update: I ended up doing simple bounds checking with a random repositioning upon a wall collision. Unfortunately I was not able to incorporate angles of inclusion and a bunch of other items I had planned. Thanks for the responses guys!. ### #8 KYA • Wubba lubba dub dub! Reputation: 3194 • Posts: 19,220 • Joined: 14-September 07 ## Re: Sphere Wall Collision Posted 15 April 2009 - 09:13 PM Thanks everyone for the replies. Like I mentioned above I didn't get exactly what I envisioned working, but I got a 100% on the project. Cheers!

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Cody # Problem 1826. Find vampire numbers Solution 1824701 Submitted on 24 May 2019 by Eric Mueller This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1   Pass x = 1000:2000; v = find_vampire(x); v_correct = [1260 1395 1435 1530 1827]; assert(isequal(v,v_correct)) 2   Pass x = 1:999; v = find_vampire(x); assert(isempty(v)) 3   Pass x = reshape(2000:2999,100,[]); v = find_vampire(x); v_correct = 2187; assert(isequal(v,v_correct)) 4   Pass x = []; v = find_vampire(x); assert(isempty(v)) 5   Pass x = -2000:-1000; v = find_vampire(x); assert(isempty(v)) 6   Pass x = 125000:125501; v = find_vampire(x); v_correct = [125248 125433 125460 125500]; assert(isequal(v,v_correct))

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# The Coriolis Effect, the Foucault Pendulum, and the Flat-Earth Movement on November 17, 2020 Share: Nevertheless, I will once again take up the subject in an attempt to straighten out flat-earthers’ thinking about the Coriolis effect. The Coriolis effect is evidence that the earth spins. Because flat-earthers frequently demand evidence of the earth’s rotation, I included discussion of the Coriolis effect on pp. 181–187 of my book, Falling Flat: A Refutation of Flat-Earth Claims. Since publishing the book last year, I have come to a greater realization of how deeply flawed flat-earthers’ comprehension of the Coriolis effect is. It might help if they bothered to read my book. It’s only fair, since I’ve read and listened to plenty of their material. Nevertheless, I will once again take up the subject in an attempt to straighten out flat-earthers’ thinking about the Coriolis effect. Image by Michael Rivera, via Wikimedia Commons. It probably will help to begin with a demonstration of what the Coriolis effect is. Suppose that you are seated on a playground merry-go-round, about halfway between the center and the edge. Further suppose that the merry-go-round is turning at a constant rate in the counterclockwise (CCW) direction as viewed from above. If you were to roll a ball toward the center of the merry-go-round, you would see the ball’s path would curve toward the right, missing the center of the merry-go-round. If you turned around and rolled the ball toward the edge, away from the center, you again would see the ball’s path curve toward the right. However, note that since you have turned around, the second deflection is opposite to the first deflection—the first deflection was in the direction of rotation of the merry-go-round, while the second deflection was opposite the rotation. Now consider a person not on the spinning merry-go-round, say, a person perched in a tree overhead. He would disagree with what you saw. He would see the ball move along a straight path in either case. Finally, repeat the experiment while the merry-go-round is not rotating. You and the observer in the tree would agree on what happened: the ball would move on straight paths with no deflection. What is going on? ## Take 1 There is more than one way to understand the Coriolis effect, so let’s start with a simple approach. The merry-go-round spins as a rigid object. This means that parts of the merry-go-round maintain relative positions with respect to one another as it spins. We would say that all parts of the merry-go-round rotate with the same angular velocity. The units of angular velocity usually are radians/second, but we could use other units, such as degrees/second, rotations/second, or rotations per minute. On the other hand, parts of the merry-go-round do not move around the axis of rotation at the same linear velocity. There are various units we could use for linear velocity, but meters/second or feet/second would make the most sense. From experience, you know that the edge of the merry-go-round will have the greatest linear velocity, while the very center of the merry-go-round will not have any linear velocity (it will be zero). There is an expression that relates linear velocity, v, to angular velocity (measured in radians per second), ω (omega), to the radius, r: v = Notice that this equation agrees with the observation that the linear velocity is zero at the center (where r = 0), and the velocity is maximum at the edge of the merry-go-round (where r is the greatest). Also notice that the direction of the linear velocity is perpendicular to the direction of the radius. When seated halfway from the center, r is half the maximum radius, so the linear velocity there is half the maximum linear velocity at the edge. Suppose that the edge of the merry-go-round is moving with a linear velocity of 2.0 feet/second in the CCW direction. Thus, a point midway between the center and the edge would be moving 1.0 feet/second. If you rolled a ball from that point toward the center, the ball would continue moving 1.0 feet/second perpendicular to the direction toward the center, but the center of the merry-go-round is not moving at all. Therefore, the ball will appear to veer to the right of the center of the merry-go-round. Now suppose you roll the ball the opposite direction, toward the edge. Since the edge is moving at 2.0 feet/second rather than the 1.0 velocity of the ball, the ball will appear to veer to the right, falling behind the rotation of the point on the edge to which it originally was projected. On the other hand, the person in the tree would see the ball move in a straight line in either case. But the directions of motion that the person in the tree would see are not along the direction of the radius of the merry-go-round Figure 1. A person at point P having latitude φ will be at a radius r = R cos (φ), where R is the earth’s radius. We can use this example to understand how the Coriolis effect works on the spinning globe earth. Objects on the equator are moving nearly 1,040 mph eastward to complete one rotation per day. Meanwhile, objects at the poles aren’t moving at all. Intermediate latitudes undergo circular motion around the earth’s axis, varying from 1,040 miles on the equator to zero mph at the poles. This is because the distance from the earth’s rotation axis depends upon latitude. For a given latitude, φ, the distance, r, from the rotation axis is given by r = R cos φ, where R is the earth’s radius (4,000 miles). The linear velocity of rotation, v, at a latitude, φ, is given by v = 1,040 mph cos φ. . For instance, where I live at 39° N latitude, we move nearly 810 mph eastward. If an airmass in the Northern Hemisphere moves from a lower latitude toward a higher latitude (moving northward), the air mass encounters land that is not moving toward the east as fast as it is. Therefore, from the perspective of the earth’s surface (and observers on the earth), the air mass turns rightward (toward the east) as it moves northward. However, from the perspective of an observer above the earth and not sharing in the earth’s rotation, the air mass moves in a straight line. On the other hand, if an airmass in the Northern Hemisphere moves from a higher latitude to a lower latitude (moving southward), it will deflect to the right too, but this will be in the westward direction. Meanwhile, air masses in the Southern Hemisphere will deflect the opposite direction, toward the left. This means that air masses traveling toward the equator will deflect to the west (as in the Northern Hemisphere), and air masses moving away from the equator will deflect to the east (as in the Northern Hemisphere). ## Effect on Weather and Climate Many people mistakenly think that the Coriolis effect alone explains the rotation of storm systems, such as hurricanes, winter storms, and tornadoes. However, there is another factor involved that I shall discuss shortly. What the Coriolis effect directly affects is the prevailing winds at various latitudes. In the tropics, concentrated sunlight heats the surface, and the heat is transferred to the air. Heated air is more buoyant, so it rises. As the air rises, it cools, and water condenses in the form of rain. This is why rain forests are so prevalent in the tropics. The air at ground level must be replaced, so air is pushed in from higher latitudes. The Coriolis effect deflects the inrushing air toward the west, so in the tropics the prevailing winds are from east to west. These are called the trade winds because, before the development of powered ships, sailing ships took advantage of the trade winds on their voyages. After the rising air in the tropics reaches several miles altitude, it moves away from the tropics toward the subtropics. The air eventually descends back to the ground, where at least part of the air returns to the tropics in the form of the trade winds. However, some of the air travels away from the tropics to more temperate latitudes. These winds are deflected west to east. We call these the prevailing westerlies, which are the dominant winds in the temperate latitudes. At the poles there is another downdraft of air. As this air returns toward lower latitudes, the Coriolis effect deflects them east to west, the same direction as the trade winds. Therefore, the Coriolis effect is responsible for three distinct wind directions that depend upon latitude Organized storm systems, such as hurricanes, have an additional factor. Organized storm systems, such as hurricanes, have an additional factor. At ground level, organized storm systems have low pressure at their centers. Higher pressure elsewhere forces air toward the low pressure. In the Northern Hemisphere, the Coriolis effect causes the inrushing air to deflect to the right, regardless of whether the air is moving northward or southward toward the low pressure. However, there still is the low pressure at the surface in the center of storms, with higher pressure outside. This high pressure continues to force the deflected air back toward the centers of storms, resulting in a CCW rotation around the storm centers. What happens with high-pressure systems in the Northern Hemisphere? At ground level, air moves away from the high pressure. The Coriolis effect deflects the air movement to the right, resulting in clockwise (CW) rotation around high-pressure systems. Figure 2. Air movements in the Northern hemisphere tend to curve to the right, while air movements in the Southern Hemisphere tend to curve to the left. The result is that air moving toward the equator is deflected westward, while air moving away from the equator is deflected eastward. Meanwhile, in the Southern Hemisphere the Coriolis deflection is to the left rather than to the right. Consequently, low-pressure systems, such as hurricanes, spin CW, opposite to the direction large storms spin in the Northern Hemisphere. Likewise, high-pressure systems in the Southern Hemisphere spin CCW, the opposite direction of their Northern Hemisphere counterparts. This direction of rotation is strictly observed by all low pressure and high-pressure systems. This is due to the relatively large size of these systems. Locally, the Coriolis effect is very weak, but when that small effect is multiplied over large distances, the Coriolis effect is very pronounced. Tornadoes are low-pressure systems, so they tend to spin the same direction as large organized storms in their respective hemispheres. However, because tornadoes are so small, they occasionally (about 2% of the time) violate the general rule. It is important to emphasize that over very short distances, the Coriolis effect is very small. The Coriolis effect generally manifests itself only over relatively large distances where its local tiny effect is multiplied over great distance. This brings up another question: is it true that sinks and toilets in the Northern Hemisphere spin CCW as they drain, while sinks and toilets in the Southern Hemisphere spin CW as they drain? Yes, and no. Across the dimensions of sinks and toilets, the Coriolis effect is very feeble. It is extremely difficult to pull a plug in a sink or for the flush of a toilet not to impart a small circular motion that more than swamps the Coriolis effect. In fact, it is relatively easy in most sinks and toilets to impart a direction of spin so that upon pulling the sink plugs or flushing the toilets, the water will follow the imposed direction. If all other factors can be eliminated, then sinks and toilets in the Northern Hemisphere will tend to spin CCW while sinks and toilets in the Southern Hemisphere tend to spin CW. However, it is difficult to eliminate all other effects. Are aircraft subject to the Coriolis effect? Yes. However, remember that the Coriolis effect is very feeble locally, and it becomes noticeable only when multiplied over great distances. Unlike air and ocean currents, a pilot or autopilot continually makes changes in an aircraft’s flight path to compensate for many factors, such as turbulence and shifting winds. These corrections greatly exceed the feeble Coriolis effect. Are bullets affect by the Coriolis effect? Yes, but given their relatively short range, the deflection is very small and so is swamped by other factors, such as shifting winds. However, in firing long-range guns, such as those found on warships, the Coriolis effect must be considered. ## Take 2 The first explanation of the Coriolis effect given above was more by way of example. When an object on a rotating platform moves to a different radius of rotation, there is a deflection of the object as seen from the rotating platform. While this is a good explanation, it doesn’t quite get to the physics of what is going on. Linear momentum is the product of an object’s mass and velocity. If no net force acts on a body, then the body’s linear momentum is conserved. This was underlying the principle of the relatively simple explanation of the Coriolis effect given above. While essentially correct, conservation of linear momentum is more applicable in non-rotating situations. However, the rotating case is a bit more complicated. In rotating systems, one ought to consider torques and angular momentum. A torque is the product of a force and the radius that the force acts about. Consider a wrench turning a tightly held bolt. One could increase the torque either by exerting more force or by increasing the radius by lengthening the handle of the wrench. Angular momentum is the product of angular velocity and an object’s moment of inertia. For a small mass, the moment of inertia can be approximated by object’s mass times the square of the radius of the object’s motion. We can write this in equation form as L = mr2ω, where L is the angular momentum, and m is the object’s mass. A torque is required to change an object’s angular momentum. If no net torque acts on a body, then its angular velocity remains the same. Since there is no net torque acting on the ball on the merry-go-round in the example above, the ball’s angular momentum is conserved. Therefore, as r changes, ω must change in the opposite sense. That is, if r decreases, then ω must increase and if r increases, then ω must decrease. Keep in mind that if the merry-go-round is rotating at a uniform rate, then its angular velocity remains the same. But since the ball’s angular velocity must change as its distance from the center of the merry-go-round changes, then the ball must deflect with respect to the merry-go-round. How will it deflect? If the ball is rolled toward the center of the merry-go-round, its radius of motion decreases, which means that its angular velocity must increase. Therefore, the ball will appear to deflect in the direction of rotation of the merry-go-round. Conversely, if the ball is rolled away from the center, the ball will deflect opposite the direction of rotation. This is what we observe. ## Take 3 But there is a better way of looking at the situation. Newton’s three laws of motion describe the world around us extremely well. For instance, Newton’s first law of motion says that if there is no net force acting on an object, the object will remain at rest or continue moving in a straight line. Newton’s second law states that when a net force acts on an object, the object will accelerate in direct proportion to the force and inversely proportional to the object’s mass. Newton’s laws of motion must be slightly adapted to the rotational case. We intuitively understand these things, even if we may not fully comprehend the description of Newtonian mechanics. Furthermore, it doesn’t matter what frame of reference we use to observe these things, as long as the reference frame we use is not accelerating. This last point is a key part of physics. What is a frame of reference? This video helps explain frames of reference. For instance, a very good explanation of the Coriolis effect is found in this video starting around 17:00. What about a reference frame that is accelerating? Newton’s laws don’t quite work out as well in accelerating reference frames. For instance, a person aboard an airplane that is accelerating for takeoff will notice that everything in the cabin seems to be “falling” toward the back of the airplane in addition to falling downward. Certainly, anything loose that easily moves (such as a flight attendant on a pair of roller skates in the aisle) will fly down the aisle and slam into the back of the cabin. Of course, to prevent this, flight attendants and passengers are required to be seated with seat buckles fastened and all items stored during takeoff. Passengers in their seats feel pushed back into their seats. Since Newton’s laws of motion require such accelerations be caused by forces, what is the mysterious force that is shoving objects toward the rear of the airplane? To an observer at rest outside the airplane, there is no mysterious force. If there were a flight attendant on a pair of roller skates in the aisle, the person outside the plane and looking through the windows would see the flight attendant as motionless with the airplane accelerating around her. Who is seeing the situation properly? It is the person motionless outside the plane. People inside the plane see objects accelerate with no force causing the acceleration. This violates Newton’s first law of motion. Physicists say that the observer at rest outside the airplane is in an inertial frame of reference, while the passengers aboard the airplane are in a non-inertial reference frame. An inertial reference frame is a frame of reference that is not accelerating, while a non-inertial reference frame is a frame of reference that is accelerating. This doesn’t mean that an inertial frame is not moving. A reference frame that is moving with constant velocity is not accelerating, so it is an inertial frame of reference just as much as a reference frame at rest. This difference is very important because Newton’s laws do not apply without amendment in a non-inertial reference frame. Rotating reference frames, such as a spinning merry-go-round or the spinning earth, are non-inertial reference frames, because any circular motion requires an acceleration. This is because, according to Newton’s first law of motion, an object will move in a straight line if no net force acts upon it. Since circular motion is not straight-line motion, any circular motion requires a centripetal (center-seeking) acceleration. Forces cause accelerations, so the force responsible for a centripetal acceleration is often called the centripetal force. For an object whirling around on the end of a string, it is tension in the string that provides the centripetal force. For the moon orbiting the earth, it is the earth’s gravity that provides the centripetal force. For the merry-go-round and the earth, it is the internal structure of either that provides the centripetal force. For objects on the merry-go-round, it is friction between the objects and the merry-go-round that keep them there. For passengers on the spinning merry-go-round, it may be their holding onto a part of the merry-go-round that keeps them aboard. On the rotating earth, a portion of a person’s weight provides the centripetal force required to keep them rotating with the earth. Consider an automobile going around a curve or turn. It is the force of friction between the tires and the road that provides the centripetal force necessary for the car to make the turn. Meanwhile, occupants inside the car find themselves slung toward the outside of the turn. Objects sitting on a seat may slide toward the outside of the turn. We attribute these effects to centrifugal force. But what causes centrifugal force? It appears to be a force that magically appears whenever we round a curve or turn. Suppose that the automobile is a convertible with the top down and that there is a lineman working in a bucket truck above the car. He won’t see this magical centrifugal force at work. Rather, he will see objects on the seat continue to move forward in a straight line in accordance with Newton’s first law of motion as the automobile turns, unless there is sufficient friction between the objects and the seat to compel the objects to turn along with the car. That slinging sensation that occupants of the car feel is caused by their bodies sharing the centripetal acceleration of the car while their bodies are compelled otherwise to continue moving in a straight line. The centripetal force for the occupants to share in the automobile’s centripetal acceleration is a combination of friction between them and the seats, holding on to parts of the car’s interior, and shoulder restraints and lap belts. Again, when viewed from an inertial reference frame, there is no centrifugal force. Centrifugal force is a fictious force that we must make up if we are within a rotating (non-inertial) reference frame to explain what we see. Sometimes people call the Coriolis effect the “Coriolis force,” but this is incorrect. There is no force that causes moving air masses to deflect on the spinning earth. Rather, it is the effect of viewing things within the non-inertial reference frame of the spinning earth. An observer above the earth who is not rotating with the earth will see the situation properly: the air masses move along straight paths. But to observers rotating with the earth, moving air masses moving north or south appear to deflect as they move over terrain that is not rotating at the same speed that they are. ## Where Flat-Earthers Go Awry About the Coriolis Effect In a recent blog, I discussed a particularly vile group of flat-earthers that I called the BB. While there appears to have been a reduction in new flat-earth material on the web of late, the BB panel has been full steam ahead. The BB panelists have discussed the Coriolis effect quite a bit, and their arguments about the Coriolis effect probably will become mainstream among flat-earthers. The problem is that the BB panelists do not understand the Coriolis effect. They cite elementary treatments of the Coriolis effect from textbooks and websites. Using their misunderstanding of the Coriolis effect, the BB panel misinterprets what these sources say. The BB panel then uses their faulty understanding of the Coriolis effect to try to debunk that the earth is a rotating globe. With their improper application of the Coriolis effect, their effort is a straw-man argument. One of the sources the BB panel frequently cites is Douglas A. Segars’ Introduction to Ocean Sciences. The BB panel quotes two bulleted points (among 8) from p. 481 of this textbook: When set in motion, freely moving objects, including air and water masses, move in straight paths while the Earth continues to rotate independently. Because freely moving objects are not carried with the Earth as it rotates, they are subject to an apparent deflection called the “Coriolis effect.” To an observer rotating with the Earth, freely moving objects that travel in a straight line appear to travel in a curved path on the Earth. In interpreting this quote, these flat-earthers make several errors. Their first error is misunderstanding the meaning of the term “water masses” in the first quoted point. These flat-earthers correctly interpret the “air” preceding “water masses” as the earth’s atmosphere, but they incorrectly interpret “water masses” as clouds and water vapor in the air. The “water masses” refers to large bodies of water on the earth, the oceans. If the BB panel had read more of this textbook that they like to quote, then they would have seen that. For instance, on p. 181 of the most recent edition of this textbook, this statement appears: First, any body in motion on the Earth, including water moving in currents, is subject to deflection by the Coriolis effect. The textbook then gives reference to the discussion where the two bulleted points above come from. On the same page, this sentence is found, again with reference to the section where the Coriolis effect is more fully explained: The deflection occurs because water set in motion by the wind is subject to the Coriolis effect. On p. 183, this sentence appears, with the same reference back to the more detailed description of the Coriolis effect: All fluids, including air and water, are subject to the Coriolis effect when they flow horizontally. There are many such sentences found in this chapter (Chapter 8) about ocean circulation in this textbook. It is clear from the BB panel’s discussion of the Coriolis effect that they do not understand that both gases and liquids are fluids. Since the general property of fluids is that they flow, then both the air and oceans on the earth are “freely moving objects.” The textbook author could have made clearer what he meant, but I doubt that it ever occurred to him that anyone would so misunderstand his meaning. The second error these flat-earthers commit is their incorrect understanding of the word independently in the first point of the quotation from the textbook they cite. They interpret this to mean that the air does not rotate with the earth (incidentally, ancient Greeks who insisted on strict geocentrism made a similar argument). If the textbook these flat-earthers quote meant that the air (and oceans) did not rotate with the earth, it would have said so. What the textbook meant is that as air or water moves from one latitude to another, it does not automatically lose the linear velocity of rotation that it originally had. Instead, fluids in motion on the earth’s surface maintain their original linear velocity due to rotation independently of the linear velocity of the regions of the earth’s surface that they happen to travel over. I will grant that the textbook author could have made clearer what he meant, but I doubt that it ever occurred to him that anyone would so misunderstand his meaning. The flat-earthers’ third mistake builds upon their second error. When they see the phrase “freely moving objects are not carried with the Earth as it rotates” at the beginning of the second bulleted point they quote, they assume once again that the freely moving objects do not rotate at all with the earth. As I’ve already explained, this is not the author’s intended meaning. Rather, the author meant that objects moving freely across the earth continue with their linear velocity due to rotation regardless of the rotational linear velocity of the terrain over which they move. What is an object that doesn’t move freely across the earth? Trains are constrained by the rails they travel on, so they don’t move freely across the earth. The same is true of automobiles moving along highways. Friction between the road and tires prevents them from moving freely. The BB panel draws upon other brief discussions of the Coriolis effect that describe the earth rotating underneath moving objects. For instance, the CK-12 Foundation began in 2007 to provide web-based STEM educational resources for primary and secondary education in the United States. Here is part of what their page about the Coriolis effect says: The Coriolis effect causes the path of a freely moving object to appear to curve. This is because Earth is rotating beneath the object. So even though the object’s path is straight, it appears to curve. Notice the use of the term “freely moving object” that flat-earthers are inclined to misunderstand. But also notice the second sentence: “This is because Earth is rotating beneath the object.” While technically true if properly understood, it may be a poor choice of words because some people may misunderstand it. This statement does not mean that objects in motion in the air do not share in the earth’s rotation. Rather, as I’ve stated several times, it means that these moving objects continue to move with the rotational velocities they had when they began moving across the earth’s surface and not the rotational velocity of the terrain over which they move. In that sense, the ground beneath these moving objects is rotating at different linear velocities (though the object and the ground have the same angular velocity). ## Deliberate Mischaracterization? With these misunderstandings in hand, the BB panel blithely sets out to prove that the earth does not rotate. They begin by mischaracterizing the Coriolis effect. The BB panel reasons that since freely moving objects rotate independently of the earth and that the earth rotates under freely moving objects, then freely moving objects do not rotate at all. They then argue that if this is true, then any object that becomes airborne will appear from the earth’s surface to speed away westward from the launch point with whatever rotational speed the ground at a given location has. For example, the BB panel says that a drone launched from the equator will not hover over the launch point but will sail away westward at nearly 1,040 mph. Another example they give is that a flight from Charlotte, NC, to Los Angeles, CA, won’t take very long at all. The two cities are near latitude 35°, where the rotational velocity ought to be about 850 mph. If one adds 600 mph as an airplane’s airspeed, then the airplane ought to travel 1,450 mph with respect to the earth. With the airport to airport distance being 2,125 miles, then the flight ought to take about 1 ½ hours, not the usual 4 ½ hours. The BB panel further argues that a hot air balloon could make the trip in only three hours by rising above the ground and waiting for Los Angeles to rotate over. Obviously, there is something wrong here. Flat-earthers would agree, saying that the problem is the assumption that the earth rotates is false, thus proving that the earth doesn’t rotate. However, there is a much better answer. Flat-earthers assume that when an object becomes airborne it instantly loses all horizontal momentum. But this would require a horizontal force (never mind the fact that such a force would require some time to achieve this loss of all momentum in objects that take off from the ground). Newton’s first law of motion states that absent a net force, a moving object will continue to move in straight-line motion. When a drone or balloon ascends from the earth, there is no net horizontal force (except for any wind). Therefore, the drone or balloon will maintain its eastward rotational velocity that it had before taking off. Since the ground has this same velocity, there will be no net motion between the balloon or drone and the ground (again, ignoring any wind). Nor will airplanes magically lose their rotational motion as they take off. If they did, it would be a gross violation of the conservation of linear momentum, as well as Newton’s first law of motion. It is very easy to test the reality of Newton’s first law in this matter. Try flipping a coin vertically so that the coin travels upward two feet. It will take the coin ¼ second to rise the two feet, and it will take another ¼ second for the coin to fall the two feet back to your hand, for a total flight time of ½ second. Now repeat this experiment inside a vehicle traveling at a uniform velocity of 60 mph. It is very important that the vehicle does not accelerate during the experiment. You will find that the result will be the same as when you were stationary: the coin will spend ½ second aloft before returning to your hand, moving only in the vertical direction from your perspective. This is because both experiments were done in inertial reference frames, and, as I mentioned before, physics is the same in all inertial reference frames. But during that ½ second, the vehicle would have moved forward 44 feet. If flat-earthers were correct in their assertion about motion, the coin would have flown backward 44 feet from the perspective of a passage in the vehicle. Since this is not what we observe, flat-earthers must be wrong about this. By the way, an observer standing outside the vehicle would see the coin travel along a parabolic path, with height two feet and length 44 feet. This parabolic path transforms back to the straight up-and-down motion that the person in the vehicle sees. This is because both observers are in inertial reference frames. Flat-earthers usually respond by pointing out if you did the moving experiment on a flat-bed truck, the coin would sail backward. But what flat-earthers fail to realize is that the backward motion of the coin in this case is caused by wind resistance. Wind resistance is a force, so once it is invoked, we are now in the realm of Newton’s second law, which describes the acceleration that occurs when a net force acts on a body. There is no wind resistance with the rotating earth. This sort of buffoonery makes me think the BB panel (along with many other flat-earth leaders) are not serious about the earth being flat but are arguing for it for their amusement. At this point, the BB panel usually howls with laughter and derision. They go back to their quotes about the Coriolis effect, hammering home their argument that “the earth rotates underneath things.” With their skewed understanding, they think this means the earth’s atmosphere must not rotate with the earth. In fact, I’ve heard members of the BB panel ask guests on their YouTube show if the atmosphere rotates with the earth. If the guest responds affirmatively, the laughter starts again, with mentions of wind. But it is silly to think that the concept of the atmosphere rotating with the earth cannot allow for relative motions within the atmosphere, aka, wind. This sort of buffoonery makes me think the BB panel (along with many other flat-earth leaders) are not serious about the earth being flat but are arguing for it for their amusement. Notice that in pursuing their argument, flat-earthers want to apply the Coriolis effect to stationary objects, such as drones and balloons. However, the definition of the Coriolis effect makes it clear that the Coriolis effect affects only moving objects. This little fact seems to have escaped the notice of flat-earthers. To insist that the Coriolis effect applies to stationary objects, such as balloons and hovering drones, displays flat-earthers’ fundamental misunderstanding of the Coriolis effect. This misunderstanding of the Coriolis effect is displayed in another way. One member of the BB panel rightly points out that the Coriolis effect requires two reference frames, one inertial and the other non-inertial. He correctly points out that the non-inertial frame is the spinning earth, but he misidentifies the inertial frame as the earth’s atmosphere. Since the atmosphere, on average, rotates with the earth, this is false. As I’ve explained above, the inertial frame is one that does not rotate with the earth. Presumably, this inertial frame would be above the earth and the atmosphere. In another display of their ignorance, the BB panel points to images of hurricanes, showing the CCW rotation in the Northern Hemisphere and the CW rotation in the Southern Hemisphere. They state that the Coriolis effect is an “apparent deflection,” while the rotation patterns of hurricanes are very real. Ergo, the rotation of hurricanes can’t be due to the Coriolis effect. Once again, these flat-earthers have misunderstood what is meant by “apparent deflection.” The deflection is real enough within the non-inertial reference frame of the rotating earth. It is called an “apparent deflection” because there is no force that causes it, but rather it is a result of observing things in a non-inertial reference frame. Furthermore, the rotation patterns of hurricanes are not caused by the Coriolis effect alone. As I previously explained, there is the force due to air pressure differences that exist on the earth’s surface. It is these two factors that produce the characteristic rotations of hurricanes. Before moving on, it is important to point out that flat-earthers generally do not offer any reason why hurricanes and other large weather systems behave this way on a flat earth. When properly understood, the Coriolis effect is strong evidence that the earth is a rotating globe. ## The Foucault Pendulum In 1851, Léon Foucault provided the first direct evidence that the earth rotates. This is so important that I discussed it on pp. 179–181 of Falling Flat. The BB panel sometimes mentions the Foucault pendulum when they talk about the Coriolis effect. They apparently think that the Coriolis effect is responsible for the precession of the Foucault pendulum. It isn’t. The Foucault pendulum has a massive bob attached to a very long wire. When pulled to one side from its vertical equilibrium position and let go, the pendulum bob will swing back and forth with a regular period of oscillation. The swing defines a plane of oscillation. With its long wire and heavy bob, the Foucault pendulum will swing for a long time before wind resistance and frictional dissipation in the pivot and wire slow it to a stop. It is very important that the pivot at the top of the pendulum be free to spin in the horizontal direction. Most pendula do not have this feature, instead constraining their bobs to swing in only one plane. When not encumbered in this way, the plane of a Foucault pendulum will precess, or turn, at a rate of where T0 is the sidereal day (23 hours, 56 minutes) and φ is the latitude. Why does the Foucault pendulum precess? The only forces acting on the bob are the downward force of gravity and the tension in the wire. Since the tension is along the wire and the wire is in the plane of the oscillatory motion, the magnitude and direction of the tension continually changes. On one side of the equilibrium position, there is a component of the tension toward the equilibrium position. On the other side of the equilibrium position, the tension has a component in the other direction, again toward the equilibrium position. It is linear momentum and this restoring force acting toward the equilibrium position that keeps the pendulum swinging. Since the direction of the component of the tension changes back and forth, the average net force on the bob throughout an entire cycle is zero. Hence, while the bob moves back and forth, the bob otherwise remains stationary. This is key, because as I discussed above, the Coriolis effect only affects objects that move, and then only objects that move some appreciable distance. Therefore, the Coriolis effect cannot cause the Foucault pendulum to precess. The reason the Foucault pendulum precesses is conservation of angular momentum. Since the tension in the wire is along the radius of the bob’s oscillatory motion, it provides zero torque on the pendulum. The weight of the bob (due to gravity) produces a torque that is perpendicular to the plane defined by the swing of the pendulum. However, throughout one half of the oscillatory cycle, the torque is in one direction while it is opposite that direction in the other half of the cycle. Therefore, on average there is no net torque on the pendulum. With no net torque, angular momentum is conserved. Angular momentum has direction perpendicular to the plane of the oscillation, so in the absence of any torque, the plane will remain fixed in space. Therefore, if the earth does not rotate, then the plane of the oscillation of the Foucault pendulum will remain fixed. However, if the earth rotates, then the plane of the Foucault pendulum will appear to precess to a person in the non-inertial reference frame of the earth. The derivation of the precession period above is beyond the scope of this article. Many advanced mechanics textbooks will explain in detail how this is done. By the way, the device in most pendula that constrains the plane of oscillation provides a small torque that causes the plane to move (as viewed from an inertial reference frame). That is why most pendula do not exhibit precession, and much care must be taken to remove this constraint when building a Foucault pendulum. How are the Coriolis effect and the precession of a Foucault pendulum different? Both result from observing things in a non-inertial reference frame. Both the Coriolis effect and the Foucault pendulum are due to conservation of angular momentum. However, the Coriolis effect acts only on objects that travel considerable distance over a rotating frame of reference. Since a Foucault pendulum does not move any distance, it is not subject to the Coriolis effect. ## Conclusion The very inaccurate manner with which flat-earthers handle the Coriolis effect and the Foucault pendulum is typical of those in the flat-earth movement. The BB panel does not understand physics at all. Both the Coriolis effect and the Foucault pendulum are direct evidence that the earth rotates. It is no wonder that flat-earthers oppose both, because if they were to accept the reality of either, it would directly disprove their belief in a flat, stationary earth. ### Danny Faulkner Blog Blog Updates Email me with new blog posts by Danny Faulkner Blog:

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After having great fun last year in Google CTF with a nice RSA challenge, and a couple of strange crypto schemes, and despite the lack of enthusiasm of my fellow team members, I decided to play again this year. And the first crypto challenge I solved was also about RSA, it said: Perfect Secrecy nc perfect-secrecy.ctfcompetition.com 1337 And it provided us an attachment, which contained a file called flag.txt (probably encrypted, since it contained seemingly random bytes), a key_pub.pem file, containing a RSA public key and a python file challenge.py. Let us first take a look at the key: $> openssl rsa -pubin -text -modulus -noout < key_pub.pem Public-Key: (1024 bit) Modulus: 00:da:53:a8:99:d5:57:30:91:af:6c:c9:c9:a9:fc: 31:5f:76:40:2c:89:70:bb:b1:98:6b:fe:8e:29:ce: d1:2d:0a:df:61:b2:1d:6c:28:1c:cb:f2:ef:ed:79: aa:7d:d2:3a:27:76:b0:35:03:b1:af:35:4e:35:bf: 58:c9:1d:b7:d7:c6:2f:6b:92:c9:18:c9:0b:68:85: 9c:77:ca:e9:fd:b3:14:f8:24:90:a0:d6:b5:0c:5d: c8:5f:5c:92:a6:fd:f1:97:16:ac:84:51:ef:e8:bb: df:48:8a:e0:98:a7:c7:6a:dd:25:99:f2:ca:64:20: 73:af:a2:0d:14:3a:f4:03:d1 Exponent: 65537 (0x10001) Modulus=DA53A899D5573091AF6CC9C9A9FC315F76402C8970BBB1986BFE8E29CED12D0ADF61B21D6C281CCBF2EFED79AA7DD23A2776B03503B1AF354E35BF58C91DB7D7C62F6B92C918C90B68859C77CAE9FDB314F82490A0D6B50C5DC85F5C92A6FDF19716AC8451EFE8BBDF488AE098A7C76ADD2599F2CA642073AFA20D143AF403D1 So, we have a 1024 bit RSA key, which is a bit low, but not necessarily weak and still out of reach of a brute force factorization. It features a common exponent: 65537, so far nothing fishy. We can now look into the challenge.py file. It has very few functions: def ReadPrivateKey(filename): ... def RsaDecrypt(private_key, ciphertext): ... def Challenge(private_key, reader, writer): ... And a rather simple main function: def main(): private_key = ReadPrivateKey(sys.argv[1]) return Challenge(private_key, sys.stdin.buffer, sys.stdout.buffer) The ReadPrivateKey function is using the Cryptography.io library to read a private key file in a standard way, however the RsaDecrypt function is a bit more interesting: def RsaDecrypt(private_key, ciphertext): assert (len(ciphertext) <= (private_key.public_key().key_size // 8)), 'Ciphertext too large' return pow( int.from_bytes(ciphertext, 'big'), private_key.private_numbers().d, private_key.public_key().public_numbers().n) As you can see, it is performing the RSA decryption function manually by taking a ciphertext as bytes in big endian format, converting it into an integer$c$and computing using pow the RSA decryption$c^d \bmod{n}$, for$d$the private exponent and$n$the RSA modulus. So, this tells us two things: 1. the ciphertext is in big endian. 2. no padding was used and it is textbook RSA. As you all know textbook RSA without any padding has a lot of shortcomings. It is now time to dig into the bigger Challenge function called by the main: def Challenge(private_key, reader, writer): try: m0 = reader.read(1) m1 = reader.read(1) ciphertext = reader.read(private_key.public_key().key_size // 8) dice = RsaDecrypt(private_key, ciphertext) for rounds in range(100): p = [m0, m1][dice & 1] k = random.randint(0, 2) c = (ord(p) + k) % 2 writer.write(bytes((c,))) writer.flush() return 0 except Exception as e: return 1 So, it first reads 2 bytes from the reader (which is sys.stdin.buffer in the main) and stores them as m0 and m1 respectively. It then proceeds to read the ciphertext from the same reader as being$\frac{1024}{8}=128$bytes. That ciphertext gets decrypted into the dice variable and then for 100 rounds, it generates a random value$k\in{0,1,2}$, adds it to the Unicode code point of the byte selected by [m0, m1][dice & 1], reduces it modulo$2$, and sends it back on the writer (which is sys.stdout.buffer in the main). That’s definitively fishy. So what’s really going on there? We get 100 times the value$(\operatorname{ord}(p) + k) \bmod{2}$, with a different, random$k\in{0,1,2}$… But this actually leaks information about the value of$\operatorname{ord}(p)$, because the operation$k \bmod{2}$does not result in a uniform distribution: indeed, we are expecting twice as many$0$s as$1$s, because when$k$is equal to both$0$and$2$, it is congruent to$0 \bmod 2$. So, if$(\operatorname{ord}(p) + k) \bmod{2}$gives us more 0s than 1s, we can be pretty sure that$\operatorname{ord}(p)\equiv 0 \bmod{2}$. On the other hand, if$\operatorname{ord}(p)\equiv 1 \bmod{2}$, the addition with$k$would shift the value and we would expect to get twice as many$1$s as$0$s on average. So, we have an information leak depending on the number of$0$s and$1$s that this Challenge function writes to the writer. But what information do we actually get? We learn the parity of the value$\operatorname{ord}(p)$, but what’s$p$again? Well, it is set just above as being p = [m0, m1][dice & 1], so it builds a list made of m0 and m1 (which are both inputs we provide in the first part of the Challenge function), and it sets$p$as being the element at index dice & 1. And as mentioned above, dice is the integer resulting from the RSA decryption operation. So, dice & 1 is equal to the bitwise AND of the plaintext and$1$, which is basically equal to$0$if the least significant bit (LSB) of the plaintext is$0$and is equal to$1$if the LSB of the plaintext is equal to$1\$ as well, since the AND truth table goes as follows: $$\begin{array}{c|c||c} x & y & x\land y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \end{array}$$ Now, this means that we have in the end an oracle telling us what’s the value of the LSB of the decrypted ciphertext! And while we all know the famous padding oracle attacks against RSA PKCS and RSA OAEP, which are relying on the most significant bit of the padded plaintext to decrypt a ciphertext, there also exists an attack exploiting the LSB. This is done by iteratively narrowing down the range of possible plaintext down to just one. At each iteration, we cut down the remaining possibilities, and while we quickly converge to the corrects most significant bytes, we still have to perform all the 1024 iterations to be able to pinpoint just one plaintext. So, we now have all the building blocks we need to build an attack against this service. Firstly, we have a decryption oracle that we can query. The oracle will then decrypt the ciphertext and give us a lot of “somewhat” randomized answers, based on values we chose, e.g. 0 and 1. And we can deduce the right value out of the oracle’s answers, allowing us to conduct the “LSB oracle” attack against RSA. In the end, the following code, inspired from a previous CTF and from the above linked page explaining this LSB-oracle, allows to query the netcat server, provide it with two consecutive bytes (we need to have a different parity betwen m0 and m1, otherwise we don’t leak any information) and interpret the oracle’s answer in order to rebuild the actual plaintext out of its answers. And guess what? The flag was at the very end of the plaintext, so that it wouldn’t leak before we had recovered the whole plaintext. ;) All right, time to go to bed, now. I’ll be following tomorrow with another write-up for the “DM collision” challenge. b"\x00\x02a\xb4\x02\x0c3\xd2.:\xe7\x9eB'\xf2\xd5\x1c7\xe4\r\xcd\xac\xd8\x7f\x02_L\x84q\xea\x8c\xb4\x1d}\x82\x90g\x170\x00.oO CTF{h3ll0__17_5_m3_1_w45_w0nd3r1n6_1f_4f73r_4ll_7h353_y34r5_y0u_d_l1k3_70_m337} Oo." import decimal from socket import socket import sys keysize =1024 # from the flag.txt file read in big endian c = 119217309829194046348522363146747304937295954356236084387775000685800908617629318417110348204755477939424239274691807984378678076883364880141274514261178345172539063655348321636037215361924332171794520555913635132454086644353250127828076172493408546050292999559913199868787754769594591462488323013485213560406 # extracted with openssl rsa -pubin -text -modulus -noout < key_pub.pem e = 65537 def oracle(c,i): # I'm opening a new socket each time r = socket() r.connect(('perfect-secrecy.ctfcompetition.com', 1337)) # we first send our two bytes m0 and m1 tosend = b'\x00' if r.send(tosend) == 0: print("borken socket") sys.exit() tosend = b'\x01' if r.send(tosend) == 0: print("borken socket") sys.exit() # we now send the ciphertext c tosend = c.to_bytes(keysize//8,'big') print("step", i,"sending:", tosend) if r.send(tosend) == 0: print("borken socket") sys.exit() # and now we listen to our oracle zero,one = 0,0 for rounds in range(100): a= r.recv(1) #print ("> ", a) if a == b'\x00': zero += 1 if a == b'\x01': one+=1 print("zero:",zero, "\n one:", one) if zero > one: return 0 return 1 # Encrypt the integer 2 to exploint RSA's malleability c_of_2 = pow(2,e,n) def partial(c,n): k = n.bit_length() # that is 1024 decimal.getcontext().prec = k # allows for 'precise enough' floats lower = decimal.Decimal(0) upper = decimal.Decimal(n) for i in range(k): # will take a while possible_plaintext = (lower + upper)/2 if not oracle(c, i): upper = possible_plaintext # plaintext is in the lower half else: lower = possible_plaintext # plaintext is in the upper half c=(c*c_of_2) % n # multiply y by the encryption of 2 again, thanks to RSA's malleability # in the end we got the plaintext! return int(upper).to_bytes(128,'big') # and we exploit it print ("\nAnd we got:", partial((c*c_of_2)%n,n)) `

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# Prime Factorization Worksheet Before understanding what prime factorization is, one should understand what prime numbers are. Numbers that have only 1 as their factor, apart from themselves, are known as prime numbers. 2, 3, 5, 7, 11, 13….. are some examples of prime numbers. Now the term prime factorization means breaking down a number into its prime factors. To understand the concept of prime factorization, you can download the prime factorization worksheet from our BYJU’S website. The worksheets are structured, keeping in mind the diverse understanding ability of children who are in junior class. ## Prime Factors and Factors For children, sometimes, the term prime factors and factors often turn out to be confusing. Factors are numbers that are multiplied to get a product; for example, if they are asked to find the factors of 24, then they would be 2, 3, 6, 4, 24, 1, and 12. Whereas, if they are asked to find the prime factors, then it would be 2 * 2 * 2 * 3. Once a child understands the difference between finding prime factors of a number and factors of a number, it would be easy for them to understand the concept of prime factorization. As parents, you can find a number of prime factorization worksheets on our BYJU’S website, which will help your kids to understand and practise. Suggested Article: Number Names Worksheets ## Ways to Find Prime Factorization There are generally two ways to find the prime factors of a number; one is the division method which is commonly used by students, and the other is the tree method. • Division Method – In the division method, the given number is divided by the smallest prime factor. You have to continue the division with nearest and small prime numbers till the quotient becomes 1. • Tree Method – In the tree method, at first, you have to write the factors of the number and then find the prime factors of those corresponding factors. This process is a bit lengthier than the division method, and children might find it a tad complicated. It is important to learn prime factorization since it will help the kids to develop an understanding of HCF (Highest Common Factor) and LCM (Least Common Multiple). Once your kid is in a higher class, finding HCF and LCM of a number will be an important topic in their mathematics curriculum. For more Kids Learning activities, visit BYJU’S website. ## Frequently Asked Questions on Prime Factorization Worksheet Q1 ### What are prime numbers? Numbers which are divisible by only themselves and 1 are known as prime numbers. 2,3,5,7,9,11,13,17,19 and so on are prime numbers. Q2 ### Where do I get prime factorization worksheets? If you are looking for a prime factorization worksheet, then your one-stop should be BYJU’S. We offer many prime factorization worksheets as well as prime factorization word problems worksheets. These worksheets will help your child to understand the concept and practise them.

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# Multiprecision numerical evaluation of integral: Sage vs. PARI/GP vs. mpmath I am trying to compute thousands of integrals of the below type, that comes up in a conformal mapping problem, to as many accurate digits as possible (preferably 50+): $$\int_{-1}^1\textrm{d}t \frac{\mathcal{Re}\{\log[(\cos{(\pi/130)} - t)]\}}{\sqrt{1 - t^2}}$$ The results from PARI/GP, Sage and Python's mpmath library respectively are: -2.1770705767584673426214016567105099553, (-2.1775860588840983, 1.2746272565903925e-05), -2.1774410877577223893132496923831596284 Clearly $-2.177$ is correct, but what's the best way to find a more accurate answer, accurate to 50+ digits? I've tried splitting intervals from $[-1, \tau] \cup [\tau, 1]$, where $\tau = \cos{(\pi/130)}$; that doesn't improve things but actually makes it worse. I am working at much higher decimal precision than 3. UPDATE: Following the suggestion of IgorRivin, the below is the Mathematica attempt: tau = N[Cos[Pi/130], 50]; epsilon = 2^-10; limit = N[ArcCos[1 - epsilon], 50]; T1 = NIntegrate[Re[Log[(tau - t)]]/(Sqrt[1 - t^2]), {t, -1, 1 - epsilon}, WorkingPrecision -> 50, AccuracyGoal -> 50] T2 = NIntegrate[Re[Log[tau - Cos[theta]]], {theta, 0, limit}, WorkingPrecision -> 50, AccuracyGoal -> 50] gives the results: Out[1]= -1.7968036050143567231750633164742621583459497767361 During evaluation of In[881]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in theta near {theta} = {0.024170580099064656300274166159163078224443688377727}. NIntegrate obtained -0.38078136551592029350719560689304843811203900465893 and 6.905015310301168422497720319277708851261355350120750.*^-7 for the integral and error estimates. >> Out[2]= -0.38078136551592029350719560689304843811203900465893 Out[3]= -2.1775849705302770166822589233673105964579887813950 As may be seen the error is still in the seventh decimal place; increasing the working precision or accuracy does not really improve things. UPDATE: The integral is exactly soluble; the result is given in my comment to the accepted answer of Emil Jeřábek, and compares well to the exact value. • To increase the precision: for mpmath: mpmath.mp.dps=50 for pari \p 50 or default("realprecision",50) – joro Sep 23 '14 at 13:09 • @joro: thanks for the comment, but I am already working at much higher precision than that. e.g. 100 digits for mpmath. Indeed more digits are spewed out for the integral but the error is still in the 4th or 5th decimal; I would like the error to be in 50+th decimal place. – MaviPranav Sep 23 '14 at 13:28 • A, ok. I am not sure any of these can guarantee you bound on the error via numerical methods, though might be wrong. – joro Sep 23 '14 at 13:29 • The integral simplifies to $\int_0^\pi\log\lvert\cos(\pi/130)-\cos\theta\rvert\,d\theta$. Anyway, I'd think your best bet is to first manipulate the integral in some way to get rid of the singularity, which is causing the numerical instability. – Emil Jeřábek Sep 23 '14 at 13:48 • I think there is some serious confusion between WorkingPrecision and AccuracyGoal. If I am not mistaken (I am no Mathematica expert), the former is the machine precision that is used for computations, that is, the local error allowed at each computation. Using WorkingPrecision=50 it is impossible to get 50 significant digits on an ill-conditioned problem. – Federico Poloni Sep 28 '14 at 21:52 Expanding my comment above: putting $\alpha=\pi/130$, the integral equals $$\int_0^\pi\log\left|\cos\alpha-\cos\theta\right|d\theta=\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta+\int_{-\alpha}^{\pi-\alpha}\log\left|\theta\sin\alpha\right|d\theta$$ $${}=\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta+\pi\log\sin\alpha+\alpha(\log\alpha-1)+(\pi-\alpha)(\log(\pi-\alpha)-1).$$ The new integrand is regular on $[0,\pi]$, hence it has a better chance to be accurately approximated by numerical integration. I leave it to someone knowledgeable with such tools to try it. In case it helps with cancellation errors, one can further write $$\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta=\int_{-\alpha}^{\pi-\alpha}\log\left(\frac{\sin\theta}\theta+\frac{1-\cos\theta}\theta\cot\alpha\right)d\theta.$$ • Great! Using 1000 digit precision mpmath and pari give -2.177586090303602130500688898237613947338583700369286294325795253194308549176741986430328961610663025 ... for the result, with the error from mpmath being 1.0e-717. – MaviPranav Sep 23 '14 at 16:22 • This kind of thing is a great story to tell a student who thinks "the computer can just do it". Having the insight, and the calculus know how, to transform the integrand into something without a singularity, really showcases why we still need "practice integrating". – Steven Gubkin Sep 23 '14 at 16:50 • @StevenGubkin: just to point out that today many of the integration routines are sophisticated enough that very high precision can be obtained for a large class of integrals without further simplification e.g. see pdf of talk by Cohen. Plus I think the need for computing results to 50+ digits might be better appreciated by researchers than by students, especially when most integrals (including this one) can be handled by straightforward software to give 3-4 digits of accuracy. Nevertheless I largely agree with your comment. – MaviPranav Sep 23 '14 at 21:22 • I agree with Steven Gubkin’s comment. Nevertheless, what I’ve done with the integral is not particularly clever, and I can well imagine that a computer algebra system could automate it: in order to compute $\int_a^bf(x)\,dx$, (1) locate the singularities of $f$ in $[a,b]$ and determine their nature, (2) construct a function $g$ with the same singularities whose integral can be easily evaluated (e.g. with explicit antiderivative), (3) compute $\int_a^bf(x)-g(x)\,dx$ numerically. Clearly, current software has room for improvement. – Emil Jeřábek Sep 24 '14 at 11:43 • @EmilJeřábek: The result is actually independent of $\alpha$ and equals $-\pi \log{2}$, as can be obtained from 1.6.10.17, 1.6.10.19 of "Integrals and Series", Vol. 1, by Prudnikov et al.; the numerical result compares very well to the exact result to any desired precision. – MaviPranav Sep 28 '14 at 20:57 Not really appropriate for this site, but Mathematica gives the below (after complaining about convergence problems). On the other hand, integrating your function from $-1$ to $1-\epsilon,$ and using integration by parts (replacing $1/\sqrt{1-x^2}$ by $\arcsin x$) on $(1-\epsilon, 1)$ will work fine. NIntegrate[Re[Log[(Cos[Pi/130] - t)]]/(Sqrt[1 - t^2]), {t, -1, 1}, WorkingPrecision -> 1000] -2.1775780498558904421658718679572230928716272110840621589154768161993\ 3732329196521375636873547501076636011363642542650859837723538974595915\ 7764788025330214570512607280815043529306101917207642206457155899026608\ 2156803691185765082975388097870843163080429578815147518480941010975453\ 2564669576068466866559420784664974740875784459505209298049493321604174\ 3176415931788039891364286170486010635239758949970468419930675399590464\ 0567854619350452364464158872338916765806789170552841942946671912635308\ 3221922528590416110327210287192250286801237171954615501203393909462095\ 6101020235292043758583706357214383010894691436046713561727357080593836\ 8462307036190240989635047153608373293610983565888298857840567497293348\ 6324154168093992830510761014856689217520886031941994883181962649076638\ 3911888258064741024384159150997182973958068547683198741266508948041775\ 2015404055400566473979785323587373843688119801934424131403167705858111\ 8040844584952102028151461657122337540572678900659420126432793440968927\ 4865463748369420978012 • thanks for the comment and suggestion, but the error is still not in the 1000th digit but at something much less. Moreover I have also tried your substitution suggestion but the error is still in the 7th digit (see update to original question). – MaviPranav Sep 23 '14 at 14:23 • @MaviPranav Well, 7th is better than 5th :) but you can do better yet, I am quite sure... – Igor Rivin Sep 23 '14 at 15:00 To increase the precision: for mpmath: mpmath.mp.dps=50 for pari \p 50 or default("realprecision",50) Working with precision 1000, mpmath disagrees with Rivin's answer, though the result seems heavily to depend on the precision. sage: pre=1000;mpmath.mp.dps=pre sage: time t=mpmath.quad(lambda t: mpmath.re(mpmath.log(mpmath.cos(mpmath.pi/ mpmath.mpf('130'))-t ))/ mpmath.sqrt(1-t**2),[-1,1]) CPU times: user 34.4 s, sys: 140 ms, total: 34.5 s Wall time: 34.8 s sage: t -2.1775726607115694998616878622345232443775247344010158704666920012976682687035840761365930384169214100410381228168759001755553191218442869332057385292941170091328180080959653384211470618559396443374411645056076930753278504200480523466068297406717463420679346103174787569206292698147279486797487709850966235010193207868896633664816216458559687560722473441911803417313571959410687042439255938037337639274195464203894049531500910228 `

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Way Ahead!!! : General GMAT Questions and Strategies Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 24 Jan 2017, 15:39 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar Author Message Intern Joined: 31 May 2012 Posts: 18 Concentration: Marketing, Organizational Behavior Followers: 0 Kudos [?]: 0 [0], given: 2 ### Show Tags 01 Nov 2012, 09:42 Dear All, I have been studying for my GMAT from September 2012, and following are my scores of MOCK GMAT. 15/10/2012 24/10/2012 IR 1.77 2.4 Q 43 44 V 28 27 Total 580 590 Just wondering my best way of increasing the scores to beyond 650. Waiting for your inputs. Manager Joined: 22 Oct 2012 Posts: 124 Followers: 1 Kudos [?]: 1 [0], given: 30 ### Show Tags 02 Nov 2012, 06:05 Hey jainstein, It would be helpful to know what materials you are using in your studies. One set of books that many people will recommend will be the MGMAT guides. I think that they are very helpful at teaching fundamentals. Another great resource is the Official Guide, it is the only book with actual GMAT questions. I like to use the MGMAT books to learn concepts and then test myself on questions from the official guide. Best of luck. _________________ Give Kudos if my post was helpful! Intern Joined: 31 May 2012 Posts: 18 Concentration: Marketing, Organizational Behavior Followers: 0 Kudos [?]: 0 [0], given: 2 ### Show Tags 02 Nov 2012, 07:01 Heyooo, I have finished Manhattan GMAT and going through OG12 as of now. Manager Joined: 12 Mar 2012 Posts: 94 Location: India Concentration: Technology, Strategy GMAT 1: 710 Q49 V36 GPA: 3.2 WE: Information Technology (Computer Software) Followers: 9 Kudos [?]: 317 [0], given: 22 ### Show Tags 02 Nov 2012, 21:45 You can try an online courses, such as e-gmat. Similar topics Replies Last post Similar Topics: Suggestions for the Path Ahead : : Need Guidance for Preparation 5 03 Jan 2017, 06:58 How to go ahead now ? 16 28 Jul 2016, 22:11 2 Starting GMAT preparation 2 years ahead? 9 31 Jan 2011, 08:48 Starting prep 6 months ahead of exam date? 4 11 Dec 2010, 11:52 Accuracy at SC - Way ahead 3 04 Oct 2010, 22:51 Display posts from previous: Sort by Moderators: WaterFlowsUp, HiLine Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

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# Search Our Content Library 11 filtered results 11 filtered results Expressions and Equations Common Core Sort by Order of Operations: PEMDAS Worksheet Order of Operations: PEMDAS Please excuse my dear aunt sally! Help your child learn and memorize this mnemonic for PEMDAS, the order of operations. Math Worksheet Order of Operations Puzzle Worksheet Order of Operations Puzzle Use the rules for the order of operations to write in the missing parentheses that make each equation true. Math Worksheet Place Value & Expanded Form Worksheet Place Value & Expanded Form Help your child develop their place value superpower with this practice sheet! This worksheet will help your student practice breaking down big numbers into expanded form using place value, from the ones place to the hundred thousandths place. Math Worksheet Place Value and Numbers in Expanded Form Exercise Place Value and Numbers in Expanded Form This exercise demonstrates how addition in larger numbers is simpler when numbers are dealt with in their expanded form. Math Exercise Worksheet Math learners use their multiplication, addition, and subtraction skills as they solve 24 problems in this practice worksheet. Math Worksheet Introducing Order of Operations Lesson Plan Introducing Order of Operations The order of operations is a critical concept for fourth graders to grasp in preparation for algebra. This introductory lesson on the order of operations will teach your students about the helpful acronym PEMDAS. Math Lesson Plan Order of Operations with Fractions and Decimals Lesson Plan Order of Operations with Fractions and Decimals This lesson is designed for fourth graders to practice their skills with the order of operations along with multiplying fractions and adding or subtracting decimals. Math Lesson Plan Pre-algebra Problems with Order of Operations Lesson Plan Pre-algebra Problems with Order of Operations Algebra concepts can be challenging for students in upper elementary and middle school. This lesson introduces students to the idea of a missing or unknown number while also considering the order of operations. Math Lesson Plan Order of Operations Competition Lesson Plan Order of Operations Competition Students will have a blast deepening their understanding of the order of operations, with this lesson which incorporates an interactive team game. Math Lesson Plan Parentheses First! Find the Missing Operation #2 Worksheet Parentheses First! Find the Missing Operation #2 Your fourth grader will use her knowledge of arithmetic to find the missing sign that makes each equation ring true.

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# Cubic Decimeters to Stere Conversion Convert Cubic Decimeters to Stere by entering the Cubic Decimeters (dm³) value in the calculator form. dm³ to st conversion. 1 dm³ = 0.001 st Stere to Cubic Decimeters Conversion ## How Many Stere in a Cubic Decimeter There are 0.001 Stere in a Cubic Decimeter. ### Conversion Factors for Stere and Cubic Decimeters Volume UnitSymbolFactor Cubic Decimeters dm³ 1 × 10 -3 Stere st 1 × 10 0 Cubic Decimeters volume unit is equal to 0.001 Stere. ### Cubic Decimeters to Stere Calculation We calculate the base unit equivalent of Cubic Decimeters and Stere with the base unit factor of volume cubic meter. ```1 dm³ = 1 * 10-3 m³ 1 st = 1 * 100 m³ 1 st = 1 m³ 1 m³ = (1/1) st 1 m³ = 1 st 1 dm³ = 1 * 10-3 * 1 st 1 dm³ = 0.001 st``` ### Cubic Decimeters to Stere Conversion Table 1 dm³0.001 st 1000 dm³1 st 2000 dm³2 st 3000 dm³3 st 4000 dm³4 st 5000 dm³5 st 6000 dm³6 st 7000 dm³7 st 8000 dm³8 st 9000 dm³9 st 10000 dm³10 st 11000 dm³11 st 12000 dm³12 st 13000 dm³13 st 14000 dm³14 st 15000 dm³15 st 16000 dm³16 st 17000 dm³17 st 18000 dm³18 st 19000 dm³19 st 20000 dm³20 st #### Abbreviations • dm³ : Cubic Decimeters • st : Stere • m³ : Cubic Meter ### Related Volume Conversions List all Cubic Decimeters Conversions »

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# Fractions Worksheet Posted on January 22, 2017 by EsperanzaStacker The Math Worksheet Site. Fractions Worksheet com -- Reducing Fractions To ... different worksheets using these selections. Memo Line. Include Answer Key. Superkids Math Worksheet Creator - Fractions math worksheets > > fractions SuperKids Math Worksheet Creator Fractions Looking for help with fractions? SuperKids has created a set of tips and tools to help build your skills. Source: www.fractionsworksheets.ca The Math Worksheet Site.com -- Reducing Fractions To ... different worksheets using these selections. Memo Line. Include Answer Key. Superkids Math Worksheet Creator - Fractions math worksheets > > fractions SuperKids Math Worksheet Creator Fractions Looking for help with fractions? SuperKids has created a set of tips and tools to help build your skills. Fractions Worksheets | Equivalent Fractions Worksheets This worksheet has rows of equivalent fractions, each with either the numerator or denominator left blank. One fraction in each row will be written with both the numerator and denominator. Candy Fractions | Worksheet | Education.com This worksheet asks students to answer questions using a diagram of a measuring cup. It will focus on fractions and equivalency between values and units of measurement. Introduction To Improper Fractions #1 | Worksheet ... Give your fifth grader's fraction know-how a valuable boost with this colorful printable that introduces her to improper fractions. Printable Fractions Worksheets For Teachers Fractions Worksheets Printable Fractions Worksheets for Teachers. Here is a graphic preview for all of the fractions worksheets. You can select different variables to customize these fractions worksheets for your needs. Fractions Help- Fractions Worksheets And Fractions Games Fractions Help- Worksheets and Games. To link to this fractions page, copy the following code to your site:. Adding Fractions Worksheets - Super Teacher Worksheets Worksheets for practicing addition of fractions. Includes adding fractions with the same denominator (easy) and addition with unlike denominators (harder. Gallery of Fractions Worksheet

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## 17985 17,985 (seventeen thousand nine hundred eighty-five) is an odd five-digits composite number following 17984 and preceding 17986. In scientific notation, it is written as 1.7985 × 104. The sum of its digits is 30. It has a total of 4 prime factors and 16 positive divisors. There are 8,640 positive integers (up to 17985) that are relatively prime to 17985. ## Basic properties • Is Prime? No • Number parity Odd • Number length 5 • Sum of Digits 30 • Digital Root 3 ## Name Short name 17 thousand 985 seventeen thousand nine hundred eighty-five ## Notation Scientific notation 1.7985 × 104 17.985 × 103 ## Prime Factorization of 17985 Prime Factorization 3 × 5 × 11 × 109 Composite number Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 17985 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 17,985 is 3 × 5 × 11 × 109. Since it has a total of 4 prime factors, 17,985 is a composite number. ## Divisors of 17985 1, 3, 5, 11, 15, 33, 55, 109, 165, 327, 545, 1199, 1635, 3597, 5995, 17985 16 divisors Even divisors 0 16 8 8 Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 31680 Sum of all the positive divisors of n s(n) 13695 Sum of the proper positive divisors of n A(n) 1980 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 134.108 Returns the nth root of the product of n divisors H(n) 9.08333 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 17,985 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 17,985) is 31,680, the average is 1,980. ## Other Arithmetic Functions (n = 17985) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 8640 Total number of positive integers not greater than n that are coprime to n λ(n) 540 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2063 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 8,640 positive integers (less than 17,985) that are coprime with 17,985. And there are approximately 2,063 prime numbers less than or equal to 17,985. ## Divisibility of 17985 m n mod m 2 3 4 5 6 7 8 9 1 0 1 0 3 2 1 3 The number 17,985 is divisible by 3 and 5. ## Classification of 17985 • Arithmetic • Deficient • Polite • Square Free ### Other numbers • LucasCarmichael ## Base conversion (17985) Base System Value 2 Binary 100011001000001 3 Ternary 220200010 4 Quaternary 10121001 5 Quinary 1033420 6 Senary 215133 8 Octal 43101 10 Decimal 17985 12 Duodecimal a4a9 20 Vigesimal 24j5 36 Base36 dvl ## Basic calculations (n = 17985) ### Multiplication n×i n×2 35970 53955 71940 89925 ### Division ni n⁄2 8992.5 5995 4496.25 3597 ### Exponentiation ni n2 323460225 5817432146625 104626517157050625 1881707911069555490625 ### Nth Root i√n 2√n 134.108 26.2001 11.5805 7.09549 ## 17985 as geometric shapes ### Circle Diameter 35970 113003 1.01618e+09 ### Sphere Volume 2.4368e+13 4.06472e+09 113003 ### Square Length = n Perimeter 71940 3.2346e+08 25434.6 ### Cube Length = n Surface area 1.94076e+09 5.81743e+12 31150.9 ### Equilateral Triangle Length = n Perimeter 53955 1.40062e+08 15575.5 ### Triangular Pyramid Length = n Surface area 5.6025e+08 6.85591e+11 14684.7 ## Cryptographic Hash Functions md5 5143cf8618ed4a4d16edfebaf7728139 03355d8a8beabaaaaa54bb99c253724be3ab123b 896ba209d8b0c5cdfdacb8b2d45dae08ba41a114dde0bc6ac633dd8a772d66f3 f2e4e516edcdb338fd5be7054203b6df5d72fb04d91f1035f13450859a28fbe5c763afa2121b230e9c93d5806348a89e67aa0f07ab585777cb792eeb2570f3b7 79777670f34deb766efc3d609c6a010e13429a68

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# Algebra 2 posted by KEViN A country's population in 1991 was 15 million. In 2001 it was 16 million. Estimate the population in 2005 using exponential growth formula. Round your answer to the nearest million. 1. Reiny let N = 15 e^(kt), where N is in million, t is in years since 1991 and k is a constant when t = 10 , N = 16 16 = 15e^(10k) 1.0666667 = e^(10k) ln 1.0666667 = 10k k = .006456852 then when t = 14 N = 15e^(14(.00645852)) = 16.418 So to the nearest million it would still be 16 million ## Similar Questions 1. ### Math In 2006 population of country was 32.2 million. This represented an increase of 3.6% since 2001. What was population of country in 2001. Round to nearest hundredth of a million. 2. ### Algebra Help In 2006 population of country was 32.2 million. This represented an increase of 3.6% since 2001. What was population of country in 2001. Round to nearest hundredth of a million. 3. ### Algebra In 2006 population of country was 32.2 million. This represented an increase of 3.6% since 2001. What was population of country in 2001. Round to nearest hundredth of a million. 4. ### math In 2006, the population of the country was 32.3 million. This represented an increase in population of 6.5% since 2001. What was the population of the country in 2011? 5. ### math Date US Population Jul 1, 2012 313.85 million Jul 1, 2011 311.59 million Jul 1, 2010 309.35 million Jul 1, 2009 306.77 million Jul 1, 2008 304.09 million Jul 1, 2007 301.23 million Jul 1, 2006 298.38 million Jul 1, 2005 295.52 million … 6. ### algebra A country's population in 1993 was 171 million estimate the population in 2012 using the exponential growth formula round your answer to the nearest million? 7. ### algebra A country's population in 1995 was 175 million. In 2000 it was 181 million. Estimate the population in 2020 using the exponential growth formula. Round you answer to the nearest million. P=Ae^kt 8. ### algebra A country's population in 1993 was 116 million. In 2002 it was 121 million. Estimate the population in 2007 using the exponential growth formula. Round your answer to the nearest million. P = Aekt 9. ### Algebra A countrys population in 1995 was 56 million in 2002 it was 59 million. Estimate the population in 2016. Exponential growth formula 10. ### math a countrys population in 1991 was 114 million in 1997 it was 120 million. estimate the population in 2014 using the exponential growth formula. More Similar Questions

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# Triaxial shear test: Wikis Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles. # Encyclopedia A triaxial shear test is a common method to measure the mechanical properties of many deformable solids, especially soil, sand, clay, and other granular materials or powders. There are several variations on the test, discussed below. ## Basic Concept For loose granular materials like sand or gravel, the material is contained in a cylindrical latex sleeve with a flat, circular metal plate or platen closing off the top and bottom ends. This cylinder is placed into a bath of water to provide pressure along the sides of the cylinder. The top platen can then be mechanically driven up or down along the axis of the cylinder to squeeze the material. The distance that the upper platen travels is measured as a function of the force required to move it, as the pressure of the surrounding water is carefully controlled. The net change in volume of the material is also measured by how much water moves in or out of the surrounding bath. The principle behind a triaxial shear test is that the stress applied in the vertical direction (along the axis of the cylinder) can be different than the stress applied in the horizontal directions (along the sides of the cylinder). This produces a non-hydrostatic stress state, which contains shear stress. A solid is defined as a material that can support shear stress without moving. However, every solid has an upper limit to how much shear stress it can support. The triaxial tester is designed to measure that limit. The stress on the platens is increased until the material in the cylinder fails and forms sliding regions within itself, known as shear bands. A motion where a material is deformed under shear stress is known as shearing. The geometry of the shearing in a triaxial tester typically causes the sample to become shorter while bulging out along the sides. The stress on the platen is then reduced and the water pressure pushes the sides back in, causing the sample to grow taller again. This cycle is usually repeated several times while collecting stress and strain data about the sample. During the shearing, a granular material will typically have a net gain or loss of volume. If it had originally been in a dense state, then it typically gains volume, a characteristic known as Reynolds' dilatancy. If it had originally been in a very loose state, then compaction may occur before the shearing begins or in conjunction with the shearing. From the triaxial test data, it is possible to extract fundamental material parameters about the sample, including its angle of shearing resistance, apparent cohesion, and dilatancy angle. These parameters are then used in computer models to predict how the material will behave in a larger-scale engineering application. An example would be to predict the stability of the soil on a slope, whether the slope will collapse or whether the soil will support the shear stresses of the slope and remain in place. Triaxial tests are used along with other tests to make such engineering predictions. ## Types of Triaxial Tests There are several variations on the basic concept of triaxial testing. These are given the following labels (corresponding test standard in parentheses): • CD — Consolidated drained • CU — Consolidated undrained (ASTM D4767)[1] • UU — Unconsolidated undrained (ASTM D2850)[2] BS 1377-1:1990 Triaxial Compresion Test ## References • Holtz, R.D.; Kovacs, W.D. (1981). An Introduction to Geotechnical Engineering. Prentice-Hall, Inc. ISBN 0-13-484394-0. • Head, K.H. (1998). Effective Stress Tests, Volume 3, Manual of Soil Laboratory Testing, (2nd ed.). John Wiley & Sons. ISBN 978-0471977957. • Bardet, Jean-Pierre (1997). Experimental Soil Mechanics. Prentice Hall. ISBN 978-0133749359. 1. ^ ASTM D4767 - 04 Standard Test Method for Consolidated Undrained Triaxial Compression Test for Cohesive Soils, doi:10.1520/D4767-04 2. ^ ASTM D2850 - 03a(2007) Standard Test Method for Unconsolidated-Undrained Triaxial Compression Test on Cohesive Soils, doi:10.1520/D2850-03AR07

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Find the distance between runs Another question inspired by the <http://uk.mathworks.com/matlabcentral/answers/?s_tid=gn_mlc_an answers forum>: A vector of ... New Cody Contest Underway Because of the amazing contributions from community members like you, Cody has lots and lots of interesting problems to... Project Euler: Problem 5, Smallest multiple 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smalle... 1年以上 前 Images in Cody Problems Today I get to tell you about a feature I’ve been lusting after for years: proper rich text in Cody with embedded images.... 1年以上 前 Find an inscribed square on a closed curve 1年以上 前 | 3 | 9 個のソルバー Double Your Pleasure – Fun with Shifty Digits You’re in for a guest-blogging treat today. But first… What is the relationship between the number 1947 and 7194? Look... 1年以上 前 Cody Problem Visualization, Revisited Way back in 2013 I did a blog post about visualizing Cody problems. The idea was to take various metrics for each problem... 1年以上 前 Chan Zuckerberg Initiative Funding Open for MATLAB Community Toolboxes Today’s post is a guest entry from Vijay Iyer, Neuroscience Community Liaison at MathWorks. MATLAB users in research have a... 1年以上 前 MathWorks Documentation has a fresh look: Check out our new Help Center You may have noticed that MATLAB Online Documentation now sports a different UI. For this post, I’d like to welcome Wendy... 1年以上 前 Circles All the Way Down This is a story about how ideas  and code bounce around our shared social spaces. This one went from Twitter to GitHub to... 1年以上 前 Ordinal numbers Given an integer n, return the corresponding ordinal number as a character string. For example, ord(1)='1st' ord(2)=... 1年以上 前 Split a string into chunks of specified length Given a string and a vector of integers, break the string into chunks whose lengths are given by the elements of the vector. Ex... 1年以上 前 Cell Counting: How Many Draws? You are given a cell array containing information about a number of soccer games. Each cell contains one of the following: * ... 1年以上 前 Guess Cipher Guess the formula to transform strings as follows: 'Hello World!' --> 'Ifmmp Xpsme!' 'Can I help you?' --> 'Dbo J ifm... 1年以上 前 This code creates a private leaderboard for Cody players. 1年以上 前 | ダウンロード 1 件 | Successive zeros Suppose n is the number of digits a number contains. Now, 12032 - is a valid n=5 digit number. But 10023 - is defined as i... 1年以上 前 Create a vector of n alternating ones and zeros Given n, your output should be a vector y of numbers such that the first number is 1 and the numbers following it alternate betw... 1年以上 前 Find Elements in Range Based on a question on <http://www.mathworks.com/matlabcentral/answers/ MATLAB Answers>. Find all the elements of a vector wh... 1年以上 前 Pancake sorting Sort a stack of pancakes by flipping them using spatula. * There are _N_ pancakes with diameters _1:N_. * Spatula can be ins... 1年以上 前 Community Q&A – Kalyan Acharjya Kalyan Acharjya has been contributing to MATLAB Answers since 2017. This past year his participation skyrocketed and he is... 1年以上 前 Stay Connected using MATLAB and Simulink Due to the global COVID-19 pandemic, engineers and scientists are finding themselves suddenly working from home or other... 1年以上 前 You may have noticed that MATLAB Mobile now sports a different UI. For this post, I’d like to welcome Geeta Sonigra, our... 1年以上 前 Nor’easters and Wind Roses Where I live, in the northeast part of the United States, we have a special storm that goes by the name “nor’easter”. Do we... 1年以上 前 Visualizing Square Roots with Elias Wegert Quick, what’s the square root of -1? Okay, I know. That’s an easy one. But how about the square root of i? If it’s been a... 1年以上 前 Leap Days on Mars February 29th is right around the corner. It’s been hibernating for the past three years, snuggling in its secret den... 2年弱 前 This is what happens when two logarithmic spirals love each other very much. % Vary k from 1.5 to 3.5 k = 1.5; t =... 2年弱 前 Community Q&A – Kelly Kearney Kelly Kearney is a top contributor to MATLAB Answers and File Exchange.  Here is a Q&amp;A I did with Kelly who is a... 2年弱 前 Rayner, BAE Systems, and Cody This week we’re very fortunate to speak with Rayner Saggers. Rayner is a Senior Systems Engineer on the F-35 Program at BAE... 2年弱 前 Summing Rows and Columns Create a matrix y of size (n+1) whose last column's elements are the summation of the elements of all the other columns and last... 2年弱 前 Crepuscular Calculations Happy Crepusculus! Never heard of Crepusculus? I’ll come back to that later. Instead, let me begin with a fun fact: the...

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# Find x-intercept in log • Sep 28th 2009, 04:06 PM BeSweeet Find x-intercept in log I need to find the x-intercept of http://img35.imageshack.us/img35/885...90928at604.jpg. It's (?, 0). Can't seem to find it :(. According to the graph: http://img16.imageshack.us/img16/139...b4329b6d30.gif The line doesn't cross the x-axis anywhere. • Sep 28th 2009, 04:10 PM pickslides make y = 0 and solve for x. Use $a^b= c \Rightarrow b = \log_ac$ • Sep 28th 2009, 04:16 PM BeSweeet Quote: Originally Posted by pickslides make y = 0 and solve for x. Use $a^b= c \Rightarrow b = \log_ac$ Did all that. Somehow got -2, which is incorrect (Thinking)... • Sep 28th 2009, 04:31 PM pickslides $0 = \log_5(x-1)+4$ $-4 = \log_5(x-1)$ $5^{-4} = x-1$ $5^{-4} +1= x$ $\frac{1}{5^{4}} +1= x$ Can you finish it? • Sep 28th 2009, 05:25 PM BeSweeet Yes :). Thanks! Got it. I should have known...

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# System F Simply typed lambda calculus is restrictive. The let-polymorphism of Hindley-Milner gives us more breathing room, but can we do better? System F frees the type system further by introducing parts of lambda calculus at the type level. We have type abstraction terms and type application terms, which define and apply functions that take types as arguments and return terms. At the same time, System F remains normalizing. System F is rich enough that the self-application \x.x x is typable. If we focus on the types, System F is a gentle generalization of Hindley-Milner. In the latter, any universal quantifiers (∀) must appear at the beginning of type, meaning they are scoped over the entire type. In System F, they can be arbitrary scoped. For example. (∀X.X->X)->(∀X.X->X) is a System F type, but not a Hindley-Milner type, while ∀X.(X->X)->X->X is a type in both systems. This seemingly small change has deep consequences. Recall Hindley-Milner allows: let id = \x.x in id succ(id 0) but rejects: (\f.f succ(f 0)) (\x.x) This is because algorithm W assigns x a type variable, say X, then finds the conflicting constraints X = Nat and X = Nat -> Nat. A locally scoped generalized variable fixes this by causing fresh type variables to be generated for each use, so the resulting constraints, say, X = Nat and Y = Nat -> Nat, live and let live. Let-free let-polymorphism! It’s easy to demonstrate this in Haskell. The following fails: (\f->f succ(f 0))(\x->x) We can make it run by enabling an extension, and annotating the identity function appropriately: :set -XRankNTypes ((\f->f succ(f 0)) :: ((forall x.x->x)->Int))(\x->x) We write type abstractions as lambda abstractions without a type annotation. The simplest example is the polymorphic identity function: id=\X.\x:X.x For clarity, we capitalize the first letter of type variables in our examples. The above represents a function that takes a type, and then returns an identity function for that type. We write type applications like term applications except we surround the argument with square brackets. For example: id [Nat] 42 evaluates to 42. ## Type spam Our new features have a heavy price tag. In System F, type reconstruction is undecidable. We must add explicit types to every binding in every lambda abstraction. Moreover, applying type abstractions require us to state even more types. Haskell magically fills in missing types if enough hints are given, which is why our example above got away with relatively little annotation. Our implementation of System F lacks this ability, so we must always supply types. (This is why we used Haskell above instead of our System F interpreter!) Because types must frequently be specified, few practical languages are built on System F. (Perhaps it’s also because System F is a relatively recent discovery?) The Hindley-Milner system is often preferable, due to type inference. However, behind the scenes, modern Haskell is an extension of System F. Certain language features require type annotation, and they generate unseen intermediate code full of type abstractions and type applications. Type operators make types less eye-watering. ## Definitions We replace the GV constructor representing Hindley-Milner generalized variables with its scoped version Forall. We add type applications and type abstractions to the Term data type. A type application is like a term application, except it expects a type as input (and during printing should enclose it within square brackets). A type abstraction is like a term abstraction, except its variable, being a type variable, has no type annotation. {-# LANGUAGE CPP #-} #ifdef __HASTE__ import Haste.DOM import Haste.Events #else #endif import Data.Char import Data.List import Data.Tuple import Text.ParserCombinators.Parsec data Type = TV String | Forall String Type | Type :-> Type data Term = Var String | App Term Term | Lam (String, Type) Term | Let String Term Term | TLam String Term | TApp Term Type instance Show Type where show (TV s) = s show (Forall s t) = '\8704':(s ++ "." ++ show t) show (t :-> u) = showL t ++ " -> " ++ showR u where showL (Forall _ _) = "(" ++ show t ++ ")" showL (_ :-> _) = "(" ++ show t ++ ")" showL _ = show t showR (Forall _ _) = "(" ++ show u ++ ")" showR _ = show u instance Show Term where show (Lam (x, t) y) = "\0955" ++ x ++ showT t ++ showB y where showB (Lam (x, t) y) = " " ++ x ++ showT t ++ showB y showB expr = '.':show expr showT (TV "_") = "" showT t = ':':show t show (TLam s t) = "\0955" ++ s ++ showB t where showB (TLam s t) = " " ++ s ++ showB t showB expr = '.':show expr show (Var s) = s show (App x y) = showL x ++ showR y where showL (Lam _ _) = "(" ++ show x ++ ")" showL _ = show x showR (Var s) = ' ':s showR _ = "(" ++ show y ++ ")" show (TApp x y) = showL x ++ "[" ++ show y ++ "]" where showL (Lam _ _) = "(" ++ show x ++ ")" showL _ = show x show (Let x y z) = "let " ++ x ++ " = " ++ show y ++ " in " ++ show z Universal types complicate type comparison, because bound type variables may be renamed arbitrarily. That is, types are unique up to alpha-conversion. instance Eq Type where t1 == t2 = f [] t1 t2 where f alpha (TV s) (TV t) | Just t' <- lookup s alpha = t' == t | Just _ <- lookup t (swap <$> alpha) = False | otherwise = s == t f alpha (Forall s x) (Forall t y) | s == t = f alpha x y | otherwise = f ((s, t):alpha) x y f alpha (a :-> b) (c :-> d) = f alpha a c && f alpha b d f alpha _ _ = False ## Parsing Parsing is trickier because elements of lambda calculus have invaded the type level. For each abstraction, we look for a type binding to determine if it’s at the term or type level. For applications, we look for square brackets to decide. We follow Haskell and accept forall as well as the harder-to-type symbol. Conventionally, the arrow type constructor -> has higher precedence. We accept inputs that omit all but the last period and all but the first quantifier in a sequence of universal quantified type variables, an abbreviation similar to the one we’ve been using in sequences of abstractions. data FLine = Empty | TopLet String Term | Run Term deriving Show line :: Parser FLine line = (((eof >>) . pure) =<<) . (ws >>)$ option Empty $(try$ TopLet <$> v <*> (str "=" >> term)) <|> (Run <$> term) where term = letx <|> lam <|> app letx = Let <$> (str "let" >> v) <*> (str "=" >> term) <*> (str "in" >> term) lam = flip (foldr pickLam) <$> between lam0 lam1 (many1 vt) <*> term where lam0 = str "\\" <|> str "\0955" lam1 = str "." vt = (,) <$> v <*> optionMaybe (str ":" >> typ) pickLam (s, Nothing) = TLam s pickLam (s, Just t) = Lam (s, t) typ = forallt <|> fun forallt = flip (foldr Forall) <$> between fa0 fa1 (many1 v) <*> fun where fa0 = str "forall" <|> str "\8704" fa1 = str "." fun = ((TV <$> v) <|> between (str "(") (str ")") typ) chainr1 (str "->" >> pure (:->)) app = termArg >>= moreArg termArg = (Var <$> v) <|> between (str "(") (str ")") term moreArg t = option t $((App t <$> termArg) <|> (TApp t <$> between (str "[") (str "]") typ)) >>= moreArg v = try$ do s <- many1 alphaNum when (s elem words "let in forall") $fail "unexpected keyword" ws pure s str = try . (>> ws) . string ws = spaces >> optional (try$ string "--" >> many anyChar) ## Typing We’ve abandoned type inference, which simplifies typing. Nonetheless, we must handle the new terms. Type abstractions are trivial. As for type applications, once again, a routine used at the term level must now be written at the type level: we must rename type variables when they conflict with free type variables. The shallow feature will be explained later. typeOf :: [(String, Type)] -> Term -> Either String Type typeOf gamma t = case t of App (Var "shallow") y -> pure $fst$ shallow gamma y Var s | Just t <- lookup s gamma -> pure t | otherwise -> Left $"undefined " ++ s App x y -> do tx <- rec x ty <- rec y case tx of ty' :-> tz | ty == ty' -> pure tz _ -> Left$ show tx ++ " apply " ++ show ty Lam (x, t) y -> do u <- typeOf ((x, t):gamma) y pure $t :-> u TLam s t -> Forall s <$> typeOf gamma t TApp x y -> do tx <- typeOf gamma x case tx of Forall s t -> pure $beta t where beta (TV v) | s == v = y | otherwise = TV v beta (Forall u v) | s == u = Forall u v | u elem fvs = let u1 = newName u fvs in Forall u1$ beta $tRename u u1 v | otherwise = Forall u$ beta v beta (m :-> n) = beta m :-> beta n fvs = tfv [] y _ -> Left $"TApp " ++ show tx Let s t u -> do tt <- typeOf gamma t typeOf ((s, tt):gamma) u where rec = typeOf gamma tfv vs (TV s) | s elem vs = [] | otherwise = [s] tfv vs (x :-> y) = tfv vs x union tfv vs y tfv vs (Forall s t) = tfv (s:vs) t tRename x x1 ty = case ty of TV s | s == x -> TV x1 | otherwise -> ty Forall s t | s == x -> ty | otherwise -> Forall s (rec t) a :-> b -> rec a :-> rec b where rec = tRename x x1 ## Evaluation Evaluation remains about the same. We erase types as we go. As usual, the function eval takes us to weak head normal form: eval env (App (Var "shallow") t) = snd$ shallow (fst env) t eval env (Let x y z) = eval env $beta (x, y) z eval env (App m a) = let m' = eval env m in case m' of Lam (v, _) f -> eval env$ beta (v, a) f _ -> App m' a eval env (TApp m _) = eval env m eval env (TLam _ t) = eval env t eval env term@(Var v) | Just x <- lookup v (snd env) = case x of Var v' | v == v' -> x _ -> eval env x eval _ term = term beta (v, a) f = case f of Var s | s == v -> a | otherwise -> Var s Lam (s, _) m | s == v -> Lam (s, TV "_") m | s elem fvs -> let s1 = newName s fvs in Lam (s1, TV "_") $rec$ rename s s1 m | otherwise -> Lam (s, TV "_") (rec m) App m n -> App (rec m) (rec n) TLam s t -> TLam s (rec t) TApp t ty -> TApp (rec t) ty Let x y z -> Let x (rec y) (rec z) where fvs = fv [] a rec = beta (v, a) fv vs (Var s) | s elem vs = [] | otherwise = [s] fv vs (Lam (s, _) f) = fv (s:vs) f fv vs (App x y) = fv vs x union fv vs y fv vs (Let _ x y) = fv vs x union fv vs y fv vs (TLam _ t) = fv vs t fv vs (TApp x _) = fv vs x newName x ys = head $filter (notElem ys)$ (s ++) . show <$> [1..] where s = dropWhileEnd isDigit x rename x x1 term = case term of Var s | s == x -> Var x1 | otherwise -> term Lam (s, t) b | s == x -> term | otherwise -> Lam (s, t) (rec b) App a b -> App (rec a) (rec b) Let a b c -> Let a (rec b) (rec c) TLam s t -> TLam s (rec t) TApp a b -> TApp (rec a) b where rec = rename x x1 The function norm recurses to find the normal form: norm env@(gamma, lets) term = case eval env term of Var v -> Var v -- Record abstraction variable to avoid clashing with let definitions. Lam (v, _) m -> Lam (v, TV "_") (norm (gamma, (v, Var v):lets) m) App m n -> App (rec m) (rec n) Let x y z -> Let x (rec y) (rec z) TApp m _ -> rec m TLam _ t -> rec t where rec = norm env ## User Interface The less said the better. #ifdef __HASTE__ main = withElems ["input", "output", "evalB", "resetB", "resetP", "pairB", "pairP", "surB", "surP"]$ \[iEl, oEl, evalB, resetB, resetP, pairB, pairP, surB, surP] -> do let reset = getProp resetP "value" >>= setProp iEl "value" >> setProp oEl "value" "" run (out, env) (Left err) = (out ++ "parse error: " ++ show err ++ "\n", env) run (out, env@(gamma, lets)) (Right m) = case m of Empty -> (out, env) Run term -> case typeOf gamma term of Left msg -> (out ++ "type error: " ++ msg ++ "\n", env) Right t -> (out ++ show (norm env term) ++ "\n", env) TopLet s term -> case typeOf gamma term of Left msg -> (out ++ "type error: " ++ msg ++ "\n", env) Right t -> (out ++ "[" ++ s ++ ":" ++ show t ++ "]\n", ((s, t):gamma, (s, term):lets)) reset resetB onEvent Click $const reset pairB onEvent Click$ const $getProp pairP "value" >>= setProp iEl "value" >> setProp oEl "value" "" surB onEvent Click$ const $getProp surP "value" >>= setProp iEl "value" >> setProp oEl "value" "" evalB onEvent Click$ const $do es <- map (parse line "") . lines <$> getProp iEl "value" setProp oEl "value" $fst$ foldl' run ("", ([], [])) es #else repl env@(gamma, lets) = do let redo = repl env case ms of Nothing -> putStrLn "" Just s -> do case parse line "" s of Left err -> do putStrLn $"parse error: " ++ show err redo Right Empty -> redo Right (Run term) -> case typeOf gamma term of Left msg -> putStrLn ("type error: " ++ msg) >> redo Right t -> do putStrLn$ "[type = " ++ show t ++ "]" print $norm env term redo Right (TopLet s term) -> case typeOf gamma term of Left msg -> putStrLn ("type error: " ++ msg) >> redo Right t -> do putStrLn$ "[type = " ++ show t ++ "]" repl ((s, t):gamma, (s, term):lets) main = repl ([], []) #endif ## Booleans, Integers, Pairs, Lists Hindley-Milner supports Church-encodings of booleans, integers, pairs, and lists. System F is more general, so of course supports them too. However, we must be explicit about types. With Hindley-Milner, globally scoped universal quantifiers for all type variables are implied. With System F, our terms must start with type abstractions or a term abstraction annotated with a universal type: true = \X x:X y:X.x false = \X x:X y:X.y not = \b:forall X.X->X->X X t:X f:X.b [X] f t 0 = \X s:X->X z:X.z succ = \n:(forall X.(X->X)->X->X) X s:X->X z:X.s(n[X] s z) pair = \X Y x:X y:Y Z f:X->Y->Z.f x y fst = \X Y p:forall Z.(X->Y->Z)->Z.p [X] (\x:X y:Y.x) snd = \X Y p:forall Z.(X->Y->Z)->Z.p [Y] (\x:X y:Y.y) nil = \X.(\R.\c:X->R->R.\n:R.n) cons = \X h:X t:forall R.(X->R->R)->R->R.(\R c:X->R->R n:R.c h (t [R] c n)) ## Apply yourself! In our previous systems, the term \x.x x has been untypable. No longer! Self-application can be expressed in System F: \x:forall X.X->X.x[forall X.X->X] x Of course, self-application of self-application ((\x.x x)(\x.x x)) remains untypable, because it has no normal form. ## Information hiding It turns out universal types can represent existential types. These types can enforce information hiding. For example, we can give the programmer access to an API but forbid access to the implementation details. Knowledge is power. Languages with simpler type systems still benefit if their designers know System F. For example, Haskell 98 is strictly Hindley-Milner, but researchers exploited existential types to design and prove theorems about a language extension enabling the ST monad. ## System F in System F The polymorphic identity function is typable in System F, hence we can construct the self-interpreter described In Breaking Through the Normalization Barrier: A Self-Interpreter for F-omega by Matt Brown and Jens Palsberg. E=\T q:(forall X.X->X)->T.q(\X x:X.x) As we demonstrated for the shallow encoding of untyped lambda calculus terms, the trick is to block every application (of types or terms) by adding one more layer of abstraction. The evaluation proceeds once we instantiate the abstraction with the polymorphic identity function. Since we must specify types in System F, computing even a shallow encoding involves type checking. shallow gamma term = (Forall "_0" (TV "_0" :-> TV "_0") :-> t, Lam ("id", Forall "X" (TV "X" :-> TV "X")) q) where (t, q) = f gamma term where f gamma term = case term of Var s -> (ty, Var s) where Just ty = lookup s gamma App m n -> (tn, App (App (TApp (Var "id") tm) qm) qn) where (tm, qm) = f gamma m (tn, qn) = f gamma n Lam (s, ty) t -> (ty :-> tt, Lam (s, ty) qt) where (tt, qt) = f ((s, ty):gamma) t TLam s t -> (Forall s tt, TLam s qt) where (tt, qt) = f gamma t TApp x ty -> (beta t, TApp (App (TApp (Var "id") tx) q) ty) where (tx@(Forall s t), q) = f gamma x beta (TV v) | s == v = ty | otherwise = TV v beta (Forall u v) | s == u = Forall u v | u elem fvs = let u1 = newName u fvs in Forall u1 $beta$ tRename u u1 v | otherwise = Forall u $beta v beta (m :-> n) = beta m :-> beta n fvs = tfv [] ty Let s t u -> f ((s, fst$ f gamma t):gamma) \$ beta (s, t) u While this shallow construction is uninteresting, the existence of a self-interpreter for a strongly normalizing language is significant, as some books and papers claim this is impossible. Ben Lynn blynn@cs.stanford.edu 💡

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# An approach to height datum unification based on local gravity field modeling using radial base function, case study: height datum unification of leveling network of class 1 in Iran Authors 1 Associate Professor, Department of Surveying and Geomatics Engineering, University College of Engineering, University of Tehran, Iran 2 Assistant Professor, Department of Surveying and Geomatics Engineering, University College of Engineering, University of Tehran, Iran 3 M.Sc. student of Geodesy, Department of Surveying and Geomatics Engineering, University College of Engineering, University of Tehran, Iran Abstract One of the most important problems in geodesy is the unification of height datum. Generally in geodesy; there are two types of height systems, the geometrical height based on ellipsoid and the physical height based on gravity-defined surface (Zhang et al, 2009).Local height datum is determined according to Mean Sea Level (MSL). In regarding to mismatch of mean sea level and geoid, on the one hand, and height datum unification requirement on the other hand, this paper defines an approach to height datum determination of Iran based on local gravity field modeling. Although there are so many algorithms to local gravity field modeling, the radial base functions (RBF) is one of the well known methods to precise local gravity field modeling using variety of data sets. Two groups of data sets, gravity acceleration observation and satellite altimetry data, are used in this paper, to determine the quasi geoid height at reference tide gauge; then the height reference of leveling network can be calculated. According to Runge-Kutla algorithm, potential anomaly on the earth can be defined as following: Where the expansion coefficients are (scale coefficients) and Bjerhammar sphere is a sphere with radius R, which is entirely inside the topographic masses of the earth,  are the set of radial basis functions with following representation: Where  are points inside and outside of the Bjerhammar respectively, is the Legendre polynomial function of degree  and  are the Legendre coefficients, the point y is called the centre of the RBF. As we can see Radial Base Functions (RBF) have some unknown parameters that should be determined precisely: the location of RBF center, shape (depth parameter) and scale coefficient. If these parameters are selected correctly we could have good representation of potential anomaly field. Examples of linear functionals used in local gravity field modeling are gravity anomalies and gravity disturbances . After linearization and spherical approximation, these functionals are related to the potential anomaly as We used Poisson-wavelet as RBF kernel, which is defined as follows: Where n is the order of Poisson wavelet kernel,  is the operator norm. Significant points in this paper are: precise positioning of the reference bench mark point by GPS system, calculation of geoid height of reference bench mark using both geodetic and orthometric heights by leveling method, quasi geoid height determination using potential anomaly value at bench mark and convert it to geoid height and finally height datum unification using comparison of the leveling height and the height from local gravity field modeling. Keywords

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# Boiler Efficiency Calculation ECMI and eTracLogic Systems has developed an easy way to determine approximate space-heat boiler efficiency for comparative purposes. It is a valuable way for facility managers to get an idea on how efficient a space-heat boiler (or cooling tower) is operating over a given period of time.  We describe the value as rFactor, (meaning boiler run-time factor, not to be confused with r-value).  The concept is actually quite simple. We wanted to arrive at a value to show how many "miles per gallon" a boiler is getting. We arrived at the conclusion that to determine this, one simply needs to take the daily boiler runtime divided by the Heating Degree Day value for that day.  (a Heating Degree Day is the average temperature for the day subtracted from 65). The resulting value is a ratio or factor which indicates how the boiler is reacting to varying outside temperatures at a specific site. For example: If the degree day value for a day is 23.5 (an average temperature of 41.5 degrees F) and the boiler ran for 10.98 hours that day, then the rFactor value would be:  10.98 divided by 23.5  = .46723 Note that the one caveat to using the rFactor method is that days for which the boiler runs less than 2 hours should be omitted to get valid average data values.  This is because we have found that dividing into a value (or by a value) approaching 1 significantly skews the data set. To use the formula for yourself, monitor a boiler's daily run-time, get the Heating Degree Day information for the monitored period, and perform the daily calculations. (this can also easily be done by creating a spreadsheet) Be aware that rFactors will vary quite a bit from site to site.  We have monitored numerous boilers and have arrived at a basic data set that gives a good indication of how efficient a boiler is operating.  Use this information only as a guideline. Note that higher rFactors indicate better overall boiler efficiency.

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# Solution - Surface Areas and Volumes Examples and Solutions Account Register Share Books Shortlist ConceptSurface Areas and Volumes Examples and Solutions #### Question An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given figure). #### Solution You need to to view the solution Is there an error in this question or solution? #### Similar questions VIEW ALL In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep. view solution Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the governments and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 cm and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? (use pi =22/7) view solution Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe. view solution In Fig. 5, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre ( "Use "pi=22/7) view solution Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the governments and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 cm and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? (use pi =22/7) view solution #### Reference Material Solution for concept: Surface Areas and Volumes Examples and Solutions. For the course 8th-10th CBSE S

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Courses Courses for Kids Free study material Offline Centres More Store # How do you solve $7h + 2h - 3 = 15$? Last updated date: 20th Jun 2024 Total views: 373.5k Views today: 8.73k Verified 373.5k+ views Hint: In the given problem we need to solve this for ‘h’. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is we group the ‘h’ terms one side and constants on the other side of the equation. Complete step-by-step solution: Given, $7h + 2h - 3 = 15$. Adding the like terms in the left hand side of the equation, $9h - 3 = 15$ We transpose ‘3’ which is present in the left hand side of the equation to the right hand side of the equation by adding ‘3’ on the right hand side of the equation. $9h = 15 + 3$ $9h = 18$ Divide the whole equation by 9, $h = \dfrac{{18}}{9}$ $\Rightarrow h = 2$ $7(2) + 2(2) - 3 = 15$ $14 + 4 - 3 = 15$ $18 - 3 = 15$ $\Rightarrow 15 = 15$

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## Tuesday, October 18, 2016 ### The Meaning of a Day in the Life (Day 42) This post fulfills my October blogging requirement for Tina Cardone's "A Day in the Life" project. Before I write about my day, let me remind the readers that I'm a first-year teacher. According to Cardone, first-year teachers undergo several phases of attitudes towards teaching. On her special graph, October and November correspond to the Survival Phase. Bear this in mind as you read about October 18th, the forty-second day of school: 7:45 -- I arrive at my school. 8:00 -- I report to the playground, where many students are beginning to arrive. The students are told to gather in a circle for the flag salute. 8:25 -- My first class, a sixth grade class, begins. 8:45 -- The dean comes in and announces the start of the CELDT test -- the California English Language Development Test. All students classified as English Learners -- which about a third of the class -- go downstairs to take the test. 9:45 -- My sixth graders leave and my seventh graders arrive. Many of these students are still out taking the CELDT test. 11:05 -- My seventh graders leave for nutrition. 11:25 -- My eighth grade class arrives. I begin the class the same way I start all my classes, with a Warm-Up question: Question: (x^3)^6 = x^? The answer is 18 -- and of course today is the 18th. 11:35 -- Today is the second day of the project we've been working on. It is called "Learning to Communicate," and it is the fourth project of the Illinois State text. (I explained how the Illinois State text is project-based back in my August PD post.) The first four projects are the same for all three grades, and so this is the third time today that I'm giving this project. It was tricky, though, since many of the students are out for the CELDT. There are no eighth graders taking the test -- this class is just a smaller class anyway. The project requires students to draw various 3D figures on two types of graph paper. The first part, given last Friday, was on oblique graph paper. Today we use isometric paper. I found the isometric paper using Google -- here is the first link: https://www.printablepaper.net/category/isometric_graph Notice that the words "isometric" and "isometry" -- as in Common Core Geometry transformation -- are definitely related. Both mean "equal length" -- an isometry maps segments to segments of equal length, and on isometric paper, the sides of the cube are all the same length on the paper. But some students struggled to draw a cube on the oblique paper on Friday, and so today I've already drawn some cubes and other figures on the isometric paper so students can just copy it. Yet many of the students still have trouble with it. They either try to draw an oblique cube on the isometric paper or merely draw a square. I think back to the activity I gave back on the third day of school (which I mentioned back in my monthly post for August). I found the activity in another textbook, in a lesson called "Drawing in Perspective," even though the blocks were drawn obliquely or isometrically, not in perspective. In that lesson, the students drew "buildings," with most of them drawing flat rectangles. I've been hoping that they would improve after this lesson, but so far most of them haven't. I'm a bit surprised that they're having trouble drawing cubes. I believe that I could draw an oblique cube in my early elementary years. But then again, I could never draw a person -- my figures weren't exactly stick figures, but they weren't much better. I reckon that there are several students who can draw lifelike human beings yet can't draw a cube. It is the difference between the "left brain" (the more analytic, mathematical side) and the "right brain" (the more artistic side). 12:05 -- Because I know how tough the 80-minute block schedule can be on middle school students, I provide a music break. My student support aide arrives during the music break. I get out my guitar and I play the following inspirational song: LEARNING TO COMMUNICATE 1st Verse: Communication Involves many tools. There's teaming, journaling, And sketching in school. When you draw the shapes, To look 3D, like a cube. Just compare it, then Choose the best from your group. Refrain: Learning to communicate Is what we all must do. Communication It's the meaning of life, too! 2nd Verse: Communication If you're with someone else The world won't come to an end. It will be much better If you talk to everyone. Get along with others Yeah, that's so much fun! (Repeat Refrain) 12:15 -- At this point, a terrible incident occurs. I choose not to post the full details of the incident here on the blog due to its sensitive nature. To make a long story short, some students start writing a letter in hopes of getting another teacher at the school fired! My only involvement with the incident is that the letter is written during my math class. (My support aide is not sitting in a location where she can tell what the students were writing -- only I see and hear them.) 12:35 -- This is a good time to end the period with an Exit Pass. Students copy the following line: Today, we drew 3D figures on isometric graph paper. 12:45 -- My eighth grade class goes out to lunch. 1:25 -- My sixth grade class returns for a special "Math Intervention" class. There is special software for this class. I spend much of the period making sure that the students all have the correct password. The online lesson is on unit rates. This lesson is challenging, since students have to divide to find the unit rates, and many of the numbers they need to divide are multi-digit. No one makes it to the top score of 100, but many students make it to the 90's -- the software starts asking challenge questions once a student reaches 90. As for the questions involving single-digit numbers, I continue my campaign to stop students from becoming "drens," or reverse-nerds who can't do simple arithmetic. Here's how it works -- this Thursday, the students are scheduled to take a "Dren Quiz" on their 3's times tables. So I draw a multiplication table on the board that goes from 3's to 9's. When the Dren Quiz begins, I'll erase the 3's from the table, so that only the 4's through 9's remain. The table will remain on the board until it's time for them to take their 4's Dren Quiz (probably in December). This way students can have help with the higher times tables but will have to learn them before they're erased. 2:25 -- My sixth graders go out to P.E. class. 3:20 -- All of the middle school teachers plus the fifth grade teacher (at our K-8 school) gather in the classroom of the teacher victimized by the smear letter. We all try to comfort the poor teacher, who is visibly upset. 4:00 -- I go home for the day and head for my computer to type up this blog entry. Cardone provides us with five reflection questions to answer. I've decided that I'll only answer one of the questions each month except for those months in which the 18th falls on the weekend, since I feel that my posts are long enough without answering all five questions. 2) Every person’s life is full of highs and lows.  Share with us some of what that is like for a teacher.  What are you looking forward to?  What has been a challenge for you lately? This is followed by five sub-questions: What has kept you going lately when it's gotten tough? What was the most negative/positive part of your day? What made you smile today? What are you looking forward to tomorrow/next school day? What has been your biggest challenge lately? What part(s) of your day were abnormal? How did you adjust to that? Well, here are my answers to those questions: 1. One thing that has kept me going lately is the new book I bought. On Sunday, the last day of the annual Barnes and Noble Educator Appreciation Week sale, I got a recreational math book. It was Incredible Numbers by Professor Ian Stewart. This book contains chapters devoted to several important numbers -- the first ten natural numbers, 0, 1, rational numbers, irrational numbers, and even infinite numbers. The last chapter of this book is devoted to the number 42 -- the "Meaning of Life" according to author Douglas Adams. Stewart writes that 42 is an interesting number indeed -- it is both "pronic" (the product of consecutive natural numbers) and a Catalan number, to boot. Today I read this final chapter first in honor of today being the 42nd day of school. Notice that I even incorporate this idea into the title of this post (not to mention today's song): The Meaning of Life = 42 Adding "day" to both sides gives: The Meaning of a Day in the Life = Day 42 And of course, "Day in the Life" is the name of the current blogging project. Then again, it's sad that after I sing a song about learning to communicate and getting along with others, the students use written communication to ruin my colleague. (Oh, and the isometric pictures my students had to copy are on page 41 -- but the instructions on how to create your own drawings is on the next page!) 2. The most positive part of the day was seeing the sixth graders work well -- even though many of them are talkative, they are still hard-working. Some of them figure out how to draw the cubes correctly, and with a little guidance, they succeed on the online assignment. The most negative part of the day is the obvious incident. 3. The one thing I'm looking forward to tomorrow -- well, maybe I'm not actually "looking forward" to this, but it's definitely coming tomorrow -- is the Illinois State observation. Recall that Illinois State is the publisher of our current math text. I've mentioned in my previous "Day in the Life" posts that Illinois State doesn't merely provide the text and the materials -- we are to submit biweekly reports with pictures of the projects and how well the students are performing them. Well, the curriculum developers will actually be flying in tomorrow to observe all math teachers third grade and above. And this is from thousands of miles away -- I'm not sure whether it will be Brad Christensen from Illinois State or two other developers from all the way in England. 4. The biggest challenge -- well, I still want to reach my goal of ideal classroom management. This means that most consequences are warnings, and these are enough to end most misbehavior. At this point, I'm further away from that goal than ever. Those eighth graders who wrote the letter don't respect me much more than my colleague. Indeed, above both of us on the totem pole of respect is my support aide! Whenever I tried to tell the eighth graders to stop writing and start doing the math project, their response was, "Don't yell at me!" I didn't even mention that I knew exactly what they were writing -- their response likely would have been that I had no right to eavesdrop on them! 5.The part of my day which was abnormal was the obvious incident. I apologize to my readers that this one incident has dominated much of this post -- but then again, this is a "Day in the Life" post, and the incident was a major part of the day. The incident worries because, as I've stated, I'm the next lowest on the totem pole -- if the students are successful in causing my colleague to leave the school, they'll be going after me next! It gives new meaning to Cardone's Survival Phase -- the two of us really are trying to survive. And my colleague, though new at our school, isn't even a first-year teacher! Tomorrow is our mixed-up Wednesday schedule. But this time, the schedule confusion is actually to our advantage. For you see, on this strange schedule, my colleague doesn't see the eighth graders at all on Wednesdays! On the other hand, I'm not sure which grade I'll have during the Illinois State observation. The last time we had a Wednesday schedule, the seventh graders were in my room for music until 12:15 (i.e., a quarter-hour after the observation is supposed to begin). We may end up rearranging the rooms so that I can teach math at noon, but hopefully this won't result in my colleague having to deal with the eighth graders at any point. (And this is not to mention that while I believe the CELDT testing is over, the test might have spilled over into tomorrow, which may affect the project, as it's supposed to be completed in groups.) This ends my monthly "Day in the Life" post for October. My next scheduled monthly post will be November 18th, a Friday.

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### 2.4 Distance measures In this section we summarize various distance measures that are defined in an arbitrary spacetime. Some of them are directly related to observable quantities with relevance for lensing. The material of this section makes use of the results on infinitesimally thin bundles which are summarized in Section 2.3. All of the distance measures to be discussed refer to a past-oriented lightlike geodesic from an observation event to an emission event (see Figure 4). Some of them depend on the 4-velocity of the observer at and/or on the 4-velocity of the light source at . If a vector field with is distinguished on , we can choose for the observer an integral curve of and for the light sources all other integral curves of . Then each of the distance measures becomes a function of the observational coordinates (recall Section 2.1). Affine distance. There is a unique affine parametrization for each lightlike geodesic through the observation event such that and . Then the affine parameter itself can be viewed as a distance measure. This affine distance has the desirable features that it increases monotonously along each ray and that it coincides in an infinitesimal neighborhood of with Euclidean distance in the rest system of . The affine distance depends on the 4-velocity of the observer but not on the 4-velocity of the light source. It is a mathematically very convenient notion, but it is not an observable. (It can be operationally realized in terms of an observer field whose 4-velocities are parallel along the ray. Then the affine distance results by integration if each observer measures the length of an infinitesimally short part of the ray in his rest system. However, in view of astronomical situations this is a purely theoretical construction.) The notion of affine distance was introduced by Kermack, McCrea, and Whittaker [179]. Travel time. As an alternative distance measure one can use the travel time. This requires the choice of a time function, i.e., of a function that slices the spacetime into spacelike hypersurfaces . (Such a time function globally exists if and only if the spacetime is stably causal; see, e.g., [153], p. 198.) The travel time is equal to , for each on the past light cone of . In other words, the intersection of the light cone with a hypersurface determines events of equal travel time; we call these intersections “instantaneous wave fronts” (recall Section 2.2). Examples of instantaneous wave fronts are shown in Figures 13, 18, 19, 27, and 28. The travel time increases monotonously along each ray. Clearly, it depends neither on the 4-velocity of the observer nor on the 4-velocity of the light source. Note that the travel time has a unique value at each point of ’s past light cone, even at events that can be reached by two different rays from . Near the travel time coincides with Euclidean distance in the observer’s rest system only if is perpendicular to the hypersurface with . (The latter equation is true if along the observer’s world line the time function coincides with proper time.) The travel time is not directly observable. However, travel time differences are observable in multiple-imaging situations if the intrinsic luminosity of the light source is time-dependent. To illustrate this, think of a light source that flashes at a particular instant. If the flash reaches the observer’s wordline along two different rays, the proper time difference of the two arrival events is directly measurable. For a time function that along the observer’s worldline coincides with proper time, this observed time delay gives the difference in travel time for the two rays. In view of applications, the measurement of time delays is of great relevance for quasar lensing. For the double quasar 0957+561 the observed time delay is about 417 days (see, e.g., [275], p. 149). Redshift. In cosmology it is common to use the redshift as a distance measure. For assigning a redshift to a lightlike geodesic that connects the observation event on the worldline of the observer with the emission event on the worldline of the light source, one considers a neighboring lightlike geodesic that meets at a proper time interval from and at a proper time interval from . The redshift is defined as If is affinely parametrized with and , one finds that is given by This general redshift formula is due to Kermack, McCrea, and Whittaker [179]. Their proof is based on the fact that is a constant for all Jacobi fields that connect with an infinitesimally neighboring lightlike geodesic. The same proof can be found, in a more elegant form, in [41] and in [311], p. 109. An alternative proof, based on variational methods, was given by Schrödinger [299]. Equation (37) is in agreement with the Hamilton formalism for photons. Clearly, the redshift depends on the 4-velocity of the observer and on the 4-velocity of the light source. If a vector field with has been distinguished on , we may choose one integral curve of as the observer and all other integral curves of as the light sources. Then the redshift becomes a function of the observational coordinates . For , the redshift goes to 0, with a (generalized) Hubble parameter that depends on spatial direction and on time. For criteria that and the higher-order coefficients are independent of and (see [151]). If the redshift is known for one observer field , it can be calculated for any other , according to Equation (37), just by adding the usual special-relativistic Doppler factors. Note that if is given, the redshift can be made to zero along any one ray from by choosing the 4-velocities appropriately. This shows that is a reasonable distance measure only for special situations, e.g., in cosmological models with denoting the mean flow of luminous matter (“Hubble flow”). In any case, the redshift is directly observable if the light source emits identifiable spectral lines. For the calculation of Sagnac-like effects, the redshift formula (37) can be evaluated piecewise along broken lightlike geodesics [23]. Angular diameter distances. The notion of angular diameter distance is based on the intuitive idea that the farther an object is away the smaller it looks, according to the rule The formal definition needs the results of Section 2.3 on infinitesimally thin bundles. One considers a past-oriented lightlike geodesic parametrized by affine distance, i.e., and , and along an infinitesimally thin bundle with vertex at the observer, i.e., at . Then the shape parameters and (recall Figure 3) satisfy the initial conditions and . They have the following physical meaning. If the observer sees a circular image of (small) angular diameter on his or her sky, the (small but extended) light source at affine distance actually has an elliptical cross-section with extremal diameters . It is therefore reasonable to call and the extremal angular diameter distances. Near the vertex, and are monotonously increasing functions of the affine distance, . Farther away from the vertex, however, they may become decreasing, so the functions and need not be invertible. At a caustic point of multiplicity one, one of the two functions and changes sign; at a caustic point of multiplicity two, both change sign (recall Section 2.3). The image of a light source at affine distance is said to have even parity if and odd parity if . Images with odd parity show the neighborhood of the light source side-inverted in comparison to images with even parity. Clearly, and are reasonable distance measures only in a neighborhood of the vertex where they are monotonously increasing. However, the physical relevance of and lies in the fact that they relate cross-sectional diameters at the source to angular diameters at the observer, and this is always true, even beyond caustic points. and depend on the 4-velocity of the observer but not on the 4-velocity of the source. This reflects the fact that the angular diameter of an image on the observer’s sky is subject to aberration whereas the cross-sectional diameter of an infinitesimally thin bundle has an invariant meaning (recall Section 2.3). Hence, if the observer’s worldline has been specified, and are well-defined functions of the observational coordinates . Area distance. The area distance is defined according to the idea As a formal definition for , in terms of the extremal angular diameter distances and as functions of affine distance , we use the equation indeed relates, for a bundle with vertex at the observer, the cross-sectional area at the source to the opening solid angle at the observer. Such a bundle has a caustic point exactly at those points where . The area distance is often called “angular diameter distance” although, as indicated by Equation (41), the name “averaged angular diameter distance” would be more appropriate. Just as and , the area distance depends on the 4-velocity of the observer but not on the 4-velocity of the light source. The area distance is observable for a light source whose true size is known (or can be reasonably estimated). It is sometimes convenient to introduce the magnification or amplification factor The absolute value of determines the area distance, and the sign of determines the parity. In Minkowski spacetime, and, thus, . Hence, means that a (small but extended) light source at affine distance subtends a larger solid angle on the observer’s sky than a light source of the same size at the same affine distance in Minkowski spacetime. Note that in a multiple-imaging situation the individual images may have different affine distances. Thus, the relative magnification factor of two images is not directly observable. This is an important difference to the magnification factor that is used in the quasi-Newtonian approximation formalism of lensing. The latter is defined by comparison with an “unlensed image” (see, e.g., [298]), a notion that makes sense only if the metric is viewed as a perturbation of some “background” metric. One can derive a differential equation for the area distance (or, equivalently, for the magnification factor) as a function of affine distance in the following way. On every parameter interval where has no zeros, the real part of Equation (27) shows that the area distance is related to the expansion by Insertion into the Sachs equation (25) for gives the focusing equation Between the vertex at and the first conjugate point (caustic point), is determined by Equation (44) and the initial conditions The Ricci term in Equation (44) is non-negative if Einstein’s field equation holds and if the energy density is non-negative for all observers (“weak energy condition”). Then Equations (44, 45) imply that i.e., , for all between the vertex at and the first conjugate point. In Minkowski spacetime, Equation (46) holds with equality. Hence, Equation (46) says that the gravitational field has a focusing, as opposed to a defocusing, effect. This is sometimes called the focusing theorem. Corrected luminosity distance. The idea of defining distance measures in terms of bundle cross-sections dates back to Tolman [322] and Whittaker [351]. Originally, this idea was applied not to bundles with vertex at the observer but rather to bundles with vertex at the light source. The resulting analogue of the area distance is the so-called corrected luminosity distance . It relates, for a bundle with vertex at the light source, the cross-sectional area at the observer to the opening solid angle at the light source. Owing to Etherington’s reciprocity law (35), area distance and corrected luminosity distance are related by The redshift factor has its origin in the fact that the definition of refers to an affine parametrization adapted to , and the definition of refers to an affine parametrization adapted to . While depends on but not on , depends on but not on . Luminosity distance. The physical meaning of the corrected luminosity distance is most easily understood in the photon picture. For photons isotropically emitted from a light source, the percentage that hit a prescribed area at the observer is proportional to . As the energy of each photon undergoes a redshift, the energy flux at the observer is proportional to , where Thus, is the relevant quantity for calculating the luminosity (apparent brightness) of pointlike light sources (see Equation (52)). For this reason is called the (uncorrected) luminosity distance. The observation that the purely geometric quantity must be modified by an additional redshift factor to give the energy flux is due to Walker [342]. depends on the 4-velocity of the observer and of the 4-velocity of the light source. and can be viewed as functions of the observational coordinates if a vector field with has been distinguished, one integral curve of is chosen as the observer, and the other integral curves of are chosen as the light sources. In that case Equation (38) implies that not only but also and are of the form . Thus, near the observer all three distance measures coincide with Euclidean distance in the observer’s rest space. Parallax distance. In an arbitrary spacetime, we fix an observation event and the observer’s 4-velocity . We consider a past-oriented lightlike geodesic parametrized by affine distance, and . To a light source passing through the event we assign the (averaged) parallax distance , where is the expansion of an infinitesimally thin bundle with vertex at . This definition follows [171]. Its relevance in view of cosmology was discussed in detail by Rosquist [288]. can be measured by performing the standard trigonometric parallax method of elementary Euclidean geometry, with the observer at and an assistant observer at the perimeter of the bundle, and then averaging over all possible positions of the assistant. Note that the method refers to a bundle with vertex at the light source, i.e., to light rays that leave the light source simultaneously. (Averaging is not necessary if this bundle is circular.) depends on the 4-velocity of the observer but not on the 4-velocity of the light source. To within first-order approximation near the observer it coincides with affine distance (recall Equation (32)). For the potential obervational relevance of see [288], and [298], p. 509. In view of lensing, , , and are the most important distance measures because they are related to image distortion (see Section 2.5) and to the brightness of images (see Section 2.6). In spacetimes with many symmetries, these quantities can be explicitly calculated (see Section 4.1 for conformally flat spactimes, and Section 4.3 for spherically symmetric static spacetimes). This is impossible in a spacetime without symmetries, in particular in a realistic cosmological model with inhomogeneities (“clumpy universe”). Following Kristian and Sachs [189], one often uses series expansions with respect to . For statistical considerations one may work with the focusing equation in a Friedmann–Robertson–Walker spacetime with average density (see Section 4.1), or with a heuristically modified focusing equation taking clumps into account. The latter leads to the so-called Dyer–Roeder distance [8687] which is discussed in several text-books (see, e.g., [298]). (For pre-Dyer–Roeder papers on optics in cosmological models with inhomogeneities, see the historical notes in [173].) As overdensities have a focusing and underdensities have a defocusing effect, it is widely believed (following [344]) that after averaging over sufficiently large angular scales the Friedmann–Robertson–Walker calculation gives the correct distance-redshift relation. However, it was argued by Ellis, Bassett, and Dunsby [99] that caustics produced by the lensing effect of overdensities lead to a systematic bias towards smaller angular sizes (“shrinking”). For a spherically symmetric inhomogeneity, the effect on the distance-redshift relation can be calculated analytically [231]. For thorough discussions of light propagation in a clumpy universe also see Pyne and Birkinshaw [284], and Holz and Wald [160].

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# Identities Rules No view 5 / 5 ( 0votes ) - Table of Tr.nometric Iden.ies. Download as PDF file. Reciprocal iden.ies. displaymath161. Pythagorean Iden.ies. displaymath162..Lists the basic tr.nometric iden.ies, and specifies the set of trig iden.ies to keep track of, as being the most useful ones for calculus..Laws and theorems. Sines Cosines Tangents Cotangents Pythagorean theorem Calculus Tr.nometric subs.ution Integrals inverse functions ; Derivatives v t e. In mathematics, tr.nometric iden.ies are equalities that involve tr.nometric functions and .You have seen quite a few tr.nometric iden.ies in the past few pages. It is convenient to have a summary of them for reference. These iden.ies mostly refer to . The elementary power rule generalizes considerably. The most general power rule is the functional power rule: for any functions f and g, = = + ,wherever both sides are well defined. Special cases: If f x = x a, f x = ax a 1 when a is any non-zero real number and x is positive.; The reciprocal rule may be derived as the special case where g x = 1..You have seen quite a few tr.nometric iden.ies in the past few pages. It is convenient to have a summary of them for reference. These iden.ies mostly refer to one angle denoted , but there are some that involve two angles, and for those, the two angles are denoted and .: The more important iden.ies..In mathematics, tr.nometric iden.ies are equalities that involve tr.nometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometri.y, these are iden.ies involving certain functions of one or more angles.They are distinct from triangle iden.ies, which are iden.ies potentially involving angles but also involving .Amazon Cognito Iden.ies. Amazon Cognito Iden.y allows you to use your own iden.y provider or other popular iden.y providers, such as Login with Amazon, Facebook, or Google.. • ### Table Of Tr Nometric Iden Ies Download as PDF file. Reciprocal iden.ies. Pythagorean Iden.ies. Quotient Iden.ies. Co-Function Iden.ies. Even-Odd Iden.ies. Sum-Difference Formulas. Double Angle Formulas. • ### List Of Tr Nometric Iden Ies Wikipedia These iden.ies are useful whenever expressions involving tr.nometric functions need to be simplified. An important application is the integration of non-tr.nometric functions: a common technique involves first using the subs.ution rule with a tr.nometric function , and then simplifying the resulting integral with a tr.nometric iden.y.. • ### Summary Of Tr Nometric Iden Ies Iden.ies for negative angles. Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even functions. Ptolemy's iden.ies, the sum and difference formulas for sine and cosine. Double angle formulas for sine and cosine. Note that there are . • ### Trig Iden Ies For Pre Calculus Dummies Of course you use tr.nometry, commonlyed trig, in pre-calculus. And you use trig iden.ies as constants throughout an equation to help you solve problems. The always-true, never-changing trig iden.ies are grouped by subject in the following lists . No related post!

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Continuous wave # Continuous wave Overview A continuous wave or continuous waveform (CW) is an electromagnetic wave of constant amplitude Amplitude Amplitude is the magnitude of change in the oscillating variable with each oscillation within an oscillating system. For example, sound waves in air are oscillations in atmospheric pressure and their amplitudes are proportional to the change in pressure during one oscillation... and frequency Frequency Frequency is the number of occurrences of a repeating event per unit time. It is also referred to as temporal frequency.The period is the duration of one cycle in a repeating event, so the period is the reciprocal of the frequency... ; and in mathematical analysis Mathematical analysis Mathematical analysis, which mathematicians refer to simply as analysis, has its beginnings in the rigorous formulation of infinitesimal calculus. It is a branch of pure mathematics that includes the theories of differentiation, integration and measure, limits, infinite series, and analytic functions... , of infinite duration. Continuous wave is also the name given to an early method of radio Radio is the transmission of signals through free space by modulation of electromagnetic waves with frequencies below those of visible light. Electromagnetic radiation travels by means of oscillating electromagnetic fields that pass through the air and the vacuum of space... transmission Transmission (telecommunications) Transmission, in telecommunications, is the process of sending, propagating and receiving an analogue or digital information signal over a physical point-to-point or point-to-multipoint transmission medium, either wired, optical fiber or wireless... , in which a carrier wave Carrier wave In telecommunications, a carrier wave or carrier is a waveform that is modulated with an input signal for the purpose of conveying information. This carrier wave is usually a much higher frequency than the input signal... is switched on and off. Information Information Information in its most restricted technical sense is a message or collection of messages that consists of an ordered sequence of symbols, or it is the meaning that can be interpreted from such a message or collection of messages. Information can be recorded or transmitted. It can be recorded as... is carried in the varying duration of the on and off periods On-off keying On-off keying the simplest form of amplitude-shift keying modulation that represents digital data as the presence or absence of a carrier wave. In its simplest form, the presence of a carrier for a specific duration represents a binary one, while its absence for the same duration represents a... of the signal, for example by Morse code Morse code Morse code is a method of transmitting textual information as a series of on-off tones, lights, or clicks that can be directly understood by a skilled listener or observer without special equipment... in early radio. In early wireless telegraphy Wireless telegraphy Wireless telegraphy is a historical term used today to apply to early radio telegraph communications techniques and practices, particularly those used during the first three decades of radio before the term radio came into use.... radio transmission, CW waves were also known as "undamped waves", to distinguish this method from damped wave Damped wave A damped wave is a wave whose amplitude of oscillation decreases with time, eventually going to zero. This term also refers to an early method of radio transmission produced by spark gap transmitters, which consisted of a series of damped electromagnetic waves... transmission. Very early radio transmitters used a spark gap Spark gap A spark gap consists of an arrangement of two conducting electrodes separated by a gap usually filled with a gas such as air, designed to allow an electric spark to pass between the conductors. When the voltage difference between the conductors exceeds the gap's breakdown voltage, a spark forms,... to produce radio-frequency oscillations in the transmitting antenna. Discussion Encyclopedia A continuous wave or continuous waveform (CW) is an electromagnetic wave of constant amplitude Amplitude Amplitude is the magnitude of change in the oscillating variable with each oscillation within an oscillating system. For example, sound waves in air are oscillations in atmospheric pressure and their amplitudes are proportional to the change in pressure during one oscillation... and frequency Frequency Frequency is the number of occurrences of a repeating event per unit time. It is also referred to as temporal frequency.The period is the duration of one cycle in a repeating event, so the period is the reciprocal of the frequency... ; and in mathematical analysis Mathematical analysis Mathematical analysis, which mathematicians refer to simply as analysis, has its beginnings in the rigorous formulation of infinitesimal calculus. It is a branch of pure mathematics that includes the theories of differentiation, integration and measure, limits, infinite series, and analytic functions... , of infinite duration. Continuous wave is also the name given to an early method of radio Radio is the transmission of signals through free space by modulation of electromagnetic waves with frequencies below those of visible light. Electromagnetic radiation travels by means of oscillating electromagnetic fields that pass through the air and the vacuum of space... transmission Transmission (telecommunications) Transmission, in telecommunications, is the process of sending, propagating and receiving an analogue or digital information signal over a physical point-to-point or point-to-multipoint transmission medium, either wired, optical fiber or wireless... , in which a carrier wave Carrier wave In telecommunications, a carrier wave or carrier is a waveform that is modulated with an input signal for the purpose of conveying information. This carrier wave is usually a much higher frequency than the input signal... is switched on and off. Information Information Information in its most restricted technical sense is a message or collection of messages that consists of an ordered sequence of symbols, or it is the meaning that can be interpreted from such a message or collection of messages. Information can be recorded or transmitted. It can be recorded as... is carried in the varying duration of the on and off periods On-off keying On-off keying the simplest form of amplitude-shift keying modulation that represents digital data as the presence or absence of a carrier wave. In its simplest form, the presence of a carrier for a specific duration represents a binary one, while its absence for the same duration represents a... of the signal, for example by Morse code Morse code Morse code is a method of transmitting textual information as a series of on-off tones, lights, or clicks that can be directly understood by a skilled listener or observer without special equipment... in early radio. In early wireless telegraphy Wireless telegraphy Wireless telegraphy is a historical term used today to apply to early radio telegraph communications techniques and practices, particularly those used during the first three decades of radio before the term radio came into use.... radio transmission, CW waves were also known as "undamped waves", to distinguish this method from damped wave Damped wave A damped wave is a wave whose amplitude of oscillation decreases with time, eventually going to zero. This term also refers to an early method of radio transmission produced by spark gap transmitters, which consisted of a series of damped electromagnetic waves... transmission. Very early radio transmitters used a spark gap Spark gap A spark gap consists of an arrangement of two conducting electrodes separated by a gap usually filled with a gas such as air, designed to allow an electric spark to pass between the conductors. When the voltage difference between the conductors exceeds the gap's breakdown voltage, a spark forms,... to produce radio-frequency oscillations in the transmitting antenna. The signals produced by these spark-gap transmitter Spark-gap transmitter A spark-gap transmitter is a device for generating radio frequency electromagnetic waves using a spark gap.These devices served as the transmitters for most wireless telegraphy systems for the first three decades of radio and the first demonstrations of practical radio were carried out using them... s consisted of brief pulses of radio frequency oscillations which died out rapidly to zero, called damped wave Damped wave A damped wave is a wave whose amplitude of oscillation decreases with time, eventually going to zero. This term also refers to an early method of radio transmission produced by spark gap transmitters, which consisted of a series of damped electromagnetic waves... s. The disadvantage of damped waves was that they produced electromagnetic interference Electromagnetic interference Electromagnetic interference is disturbance that affects an electrical circuit due to either electromagnetic induction or electromagnetic radiation emitted from an external source. The disturbance may interrupt, obstruct, or otherwise degrade or limit the effective performance of the circuit... that spread over the transmissions of stations at other frequencies Frequency Frequency is the number of occurrences of a repeating event per unit time. It is also referred to as temporal frequency.The period is the duration of one cycle in a repeating event, so the period is the reciprocal of the frequency... . Mathematically, the extremely wideband excitation provided by the spark gap was bandpass filtered by the self-oscillating antenna side circuit, which because of its simple construction, unfortunately also had a relatively broad and badly controlled filter characteristic. This motivated efforts to produce radio frequency oscillations that decayed more slowly. Strictly speaking an unmodulated, continuous carrier has no bandwidth and cannot interfere with signals at other frequencies, but on the other hand conveys no information either. Thus it is commonly understood that the act of keying the carrier on and off is necessary. However, in order to bring the bandwidth of the resulting signal under control, the buildup and decay of the radio frequency envelope needed to be slower than in the early spark gap implementations. When this is done, the spectrum of the signal approaches that of a continuous sinusoidal oscillation, while temporally its amplitude Amplitude Amplitude is the magnitude of change in the oscillating variable with each oscillation within an oscillating system. For example, sound waves in air are oscillations in atmospheric pressure and their amplitudes are proportional to the change in pressure during one oscillation... does vary between zero and full carrier strength. As such, the resulting narrower bandwidth mode of operation is to this day described as "continuous wave". The resulting signal allows many radio stations to share a given band of frequencies without noticeable mutual interference. The first transmitters capable of producing continuous wave, the Alexanderson alternator Alexanderson alternator An Alexanderson alternator is a rotating machine invented by Ernst Alexanderson in 1904 for the generation of high frequency alternating current up to 100 kHz, for use as a radio transmitter... and vacuum tube Vacuum tube In electronics, a vacuum tube, electron tube , or thermionic valve , reduced to simply "tube" or "valve" in everyday parlance, is a device that relies on the flow of electric current through a vacuum... oscillators, became widely available after World War I World War I World War I , which was predominantly called the World War or the Great War from its occurrence until 1939, and the First World War or World War I thereafter, was a major war centred in Europe that began on 28 July 1914 and lasted until 11 November 1918... . Early radio transmitters were incapable of handling the complexity of actual audio and therefore CW was the only form of communication available. CW still remained a viable form of radio communication for many years after voice transmission was perfected, because simple transmitters could be used. The low bandwidth of the code signal, due in part to low information transmission rate, allowed very selective filters to be used in the receiver which blocked out much of the atmospheric noise that would otherwise reduce the intelligibility of the signal. Continuous-wave radio was called radiotelegraphy because like the telegraph, it worked by means of a simple switch to transmit Morse code Morse code Morse code is a method of transmitting textual information as a series of on-off tones, lights, or clicks that can be directly understood by a skilled listener or observer without special equipment... . However, instead of controlling the electricity in a cross-country wire, the switch controlled the power sent to a radio transmitter Transmitter In electronics and telecommunications a transmitter or radio transmitter is an electronic device which, with the aid of an antenna, produces radio waves. The transmitter itself generates a radio frequency alternating current, which is applied to the antenna. When excited by this alternating... . This mode is still in common use by amateur radio Amateur radio is the use of designated radio frequency spectrum for purposes of private recreation, non-commercial exchange of messages, wireless experimentation, self-training, and emergency communication... operators. Continuous-wave radar is a type of radar system where a known stable frequency continuous wave radio energy is transmitted and then received from any reflecting objects.Continuous wave radar uses Doppler, which renders the radar immune to interference from large stationary objects and slow moving... system is one where a continuous wave is transmitted by one aerial An antenna is an electrical device which converts electric currents into radio waves, and vice versa. It is usually used with a radio transmitter or radio receiver... Radio is the transmission of signals through free space by modulation of electromagnetic waves with frequencies below those of visible light. Electromagnetic radiation travels by means of oscillating electromagnetic fields that pass through the air and the vacuum of space... energy. In military communications and amateur radio Amateur radio is the use of designated radio frequency spectrum for purposes of private recreation, non-commercial exchange of messages, wireless experimentation, self-training, and emergency communication... , the terms "CW" and "Morse code" are often used interchangeably, despite the distinctions between the two. Morse code may be sent using direct current Direct current Direct current is the unidirectional flow of electric charge. Direct current is produced by such sources as batteries, thermocouples, solar cells, and commutator-type electric machines of the dynamo type. Direct current may flow in a conductor such as a wire, but can also flow through... in wires, sound, or light, for example. A carrier wave is keyed on and off to represent the dots and dashes of the code elements. The carrier's amplitude and frequency remains constant during each code element. At the receiver, the received signal is mixed with a heterodyne signal Heterodyne Heterodyning is a radio signal processing technique invented in 1901 by Canadian inventor-engineer Reginald Fessenden where high frequency signals are converted to lower frequencies by combining two frequencies. Heterodyning is useful for frequency shifting information of interest into a useful... from a BFO (Beat frequency oscillator) to change the radio frequency impulses to sound. Though most commercial traffic has now ceased operation using that mode, it is still popular with amateur radio operators. Older non-directional beacon Non-directional beacon A non-directional beacon is a radio transmitter at a known location, used as an aviation or marine navigational aid. As the name implies, the signal transmitted does not include inherent directional information, in contrast to other navigational aids such as low frequency radio range, VHF... s used in air navigation use CW to transmit their identifier. ### Key clicks In any form of on-off carrier keying, if the carrier wave is turned on or off abruptly, the bandwidth will be large; if the carrier turns on and off more gradually, the bandwidth will be smaller. What is transmitted in the extra bandwidth used by a transmitter Transmitter In electronics and telecommunications a transmitter or radio transmitter is an electronic device which, with the aid of an antenna, produces radio waves. The transmitter itself generates a radio frequency alternating current, which is applied to the antenna. When excited by this alternating... that turns on/off more abruptly is known as key clicks. Certain types of power amplifiers used in transmission may increase the effect of key clicks. ## Laser physics In laser physics Laser Physics Laser Physics is an international scientific journal published by Nauka/Interperiodica. It is distributed through the Springer.-Topics covered:The journal specializes in laser physics, but also publishes papers about:... and engineering the term "continuous wave" or "CW" refers to a laser Laser A laser is a device that emits light through a process of optical amplification based on the stimulated emission of photons. The term "laser" originated as an acronym for Light Amplification by Stimulated Emission of Radiation... which produces a continuous output beam, sometimes referred to as 'free-running'. This is as opposed to a q-switched, gain-switched Gain-switching Gain-switching is a technique in optics by which a laser can be made to produce pulses of light of extremely short duration, of the order of picoseconds .... or modelocked Modelocking Mode-locking is a technique in optics by which a laser can be made to produce pulses of light of extremely short duration, on the order of picoseconds or femtoseconds .... laser, which produces pulses of light. • Amplitude modulation Amplitude modulation Amplitude modulation is a technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. AM works by varying the strength of the transmitted signal in relation to the information being sent... • The CW Operators' Club The CW Operators' Club The CW Operators' Club is an international organization, in membership and management, for amateur radio operators who enjoy communicating using Morse Code. The club is commonly known as CWops. Its mission is to foster the use of CW, whether for contesting, DXing, traffic handling, or engaging in... • Damped wave Damped wave A damped wave is a wave whose amplitude of oscillation decreases with time, eventually going to zero. This term also refers to an early method of radio transmission produced by spark gap transmitters, which consisted of a series of damped electromagnetic waves... • On-off keying On-off keying On-off keying the simplest form of amplitude-shift keying modulation that represents digital data as the presence or absence of a carrier wave. In its simplest form, the presence of a carrier for a specific duration represents a binary one, while its absence for the same duration represents a... • Tikker Tikker A tikker was a device that allowed early crystal radio circuits to detect continuous-wave telegraphy. Ordinary crystal detectors cannot demodulate continuous waves; the direct current signal is inaudible in headphones...

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0 # How many ounces equals 28 pounds? Wiki User 2012-12-20 18:04:26 There are 16 ounces in one pound. Therefore, 28 pounds is equal to 28 x 16 = 448 ounces. Wiki User 2012-12-20 18:04:26 🙏 0 🤨 0 😮 0 Study guides 20 cards ➡️ See all cards 3.69 322 Reviews

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1. ## Quick question find and sketch the domain of f(x,y) = sqrt(1+ln(xy)) it will be: y>1/(ex) graph of y=1/ex is approximately the same as y=1/x where should i shade it ?? i will shade it inside both curves or just the curve in the quadratic 1?? i think, since y > 1/ex ---> xy > 1/e product is positive right??! 2. If our function is $\displaystyle f(x,y)=\sqrt{1+\ln\,(xy)},$ then $\displaystyle f$ is defined exactly where $\displaystyle 1+\ln\,(xy)$ is defined and non-negative; that is, where $\displaystyle 1+\ln\,(xy)\ge 0.$ By two reversible operations, we find that this is equivalent to saying \displaystyle \begin{aligned} \ln\,(xy)&\ge -1\\ xy&\ge e^{-1}.\quad\quad (e^x\mbox{ increasing})\\ \end{aligned} Therefore, $\displaystyle xy$ must be positive and $\displaystyle (x,y)$ must lie in the first or third quadrant. Now, if we knew that $\displaystyle x$ were positive, we could divide by $\displaystyle x$ to obtain $\displaystyle y\ge \frac{1}{ex},$ but in the third quadrant, $\displaystyle x<0$, and we can't do this without reversing the inequality. In fact, while your answer is correct in the first quadrant, in the third quadrant it will be $\displaystyle y \le \frac{1}{ex}.$ Hope this helps!

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Homework-3-calm-Sol # Homework-3-calm-Sol - Homework 3 4-5 9 An electron in a... This preview shows pages 1–2. Sign up to view the full content. Homework 3 4-5 ••  An electron in a cathode-ray tube is traveling horizontally at 210 10 9 . / × cm s when deflection plates give it an upward acceleration of (a= 5.3 x 10 17 cm/s 2 ) (a) How long does it take for the electron to cover a horizontal distance of 6.20 cm? (b) What is its vertical displacement during this time? Choose y to be up, x in the direction of the electron’s motion. Sol 4-5.(a) Note: it does not depend on the acceleration. The x-motion is independent of the y-motion, and the acceleration is only upwards. Now solve for t: (b) 4-8 •  The great, gray-green, greasy Zambezi River flows over Victoria Falls in south central Africa. The falls are approximately 108 m high. If the river is flowing horizontally at v 0x =3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops. Sol 4-8 There is no initial component of velocity in the y -direction. 4-14 14. ••  A mountain climber jumps a 3.0-m wide crevasse by leaping horizontally with a speed of v h = 8.0 m/s. (a) If the climber’s direction of motion on landing is - ° 45 , what is the height difference between the two sides of the crevasse? (b) This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. ## This note was uploaded on 07/02/2011 for the course PHYS 201 taught by Professor Woodahl during the Fall '09 term at Indiana. ### Page1 / 3 Homework-3-calm-Sol - Homework 3 4-5 9 An electron in a... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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