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https://tex.stackexchange.com/questions/667378/pgfplots-shrink-an-interval-in-the-x-axis

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# pgfplots: shrink an interval in the x-axis I have a set of coordinates, e.g. (-4, 1) (-3, 2) (-2, 3), (-1, 4) (500, 4) (501, 3) (502, 2) (501, 1) Suppose I plot these points with an interpolation curve through them as follows. \begin{tikzpicture} \begin{axis} (-4, 1) (-3, 2) (-2, 3), (-1, 4) %a gap in the interpolation curve here (500, 4) (501, 3) (502, 2) (501, 1) }; \end{axis} \end{tikzpicture} The problem with this is that the graph looks like 2 vertical lines because of the scaling/compressing of the x axis. The graph is meant to show a curve with positive slope on the left, one with negative slope on the right, and a gap in-between. So I want to shrink/disappear the gap, the interval [5, 499] in the x-axis, and still label things appropriately. \documentclass[tikz, border=1cm]{standalone} \usepackage{pgfplots} \pgfplotsset{compat=1.18} \begin{document} \begin{tikzpicture} \begin{axis}[name=axis1, axis y line*=left, xtick=data] \addplot[smooth, mark=*] coordinates {(-4, 1) (-3, 2) (-2, 3) (-1, 4)}; \end{axis} \begin{axis}[at={(axis1.south east)}, axis y line*=right, yticklabels=\empty, xtick=data] \addplot[smooth, mark=*] coordinates {(500, 4) (501, 3) (502, 2) (501, 1)}; \end{axis} \end{tikzpicture} \end{document} Edit: With correct interval removed and same x tick distance \documentclass[tikz, border=1cm]{standalone} \usepackage{pgfplots} \pgfplotsset{compat=1.18} \begin{document} \begin{tikzpicture} \newcommand{\xinterval}{10} \begin{axis}[ name=axis1, axis y line*=left, xmax=5, xmin=\pgfkeysvalueof{/pgfplots/xmax}-\xinterval, ] \addplot[smooth, mark=*] coordinates {(-4, 1) (-3, 2) (-2, 3) (-1, 4)}; \end{axis} \begin{axis}[ at={(axis1.south east)}, axis y line*=right, yticklabels=\empty, xmin=499, xmax=\pgfkeysvalueof{/pgfplots/xmin}+\xinterval, ] \addplot[smooth, mark=*] coordinates {(500, 4) (501, 3) (502, 2) (501, 1)}; \end{axis} \end{tikzpicture} \end{document}

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# zbMATH — the first resource for mathematics Truncated Newton method for sparse unconstrained optimization using automatic differentiation. (English) Zbl 0632.90060 When solving large complex optimization problems, the user is faced with three major problems. These are: (i) the cost in human time in obtaining accurate expressions for the derivatives involved; (ii) the need to store second derivative information; and (iii) of lessening importance, the time taken to solve the problem on the computer. For many problems, a significant part of the latter can be attributed to solving Newton-like equations. In the algorithm described, the equations are solved using a conjugate direction method which only needs the Hessian at the current point when it is multiplied by a trial vector. In this paper, we present a method that finds this product using automatic differentiation while only requiring vector storage. The method takes advantage of any sparsity in the Hessian matrix and computes exact derivatives. It avoids the complexity of symbolic differentiation, the inaccuracy of numerical differentiation, the labor of finding analytic derivatives, and the need for matrix store. When far from a minimum, an accurate solution to the Newton equations is not justified, so an approximate solution is obtained by using a version of R. Dembo and T. Steihaug’s truncated Newton algorithm [Math. Program. 26, 190-212 (1982; Zbl 0523.90078)]. Reviewer: L.C.W.Dixon ##### MSC: 90C30 Nonlinear programming 49M37 Numerical methods based on nonlinear programming 65K05 Numerical mathematical programming methods Full Text: ##### References: [1] Dembo, R., andSteihaug, T.,Truncated Newton Methods for Large Scale Optimization, Mathematical Programming, Vol. 26, pp. 190-212, 1982. · Zbl 0523.90078 · doi:10.1007/BF02592055 [2] Rall, L. B.,Automatic Differentiation: Techniques and Applications, Springer-Verlag, Berlin, Germany, 1981. · Zbl 0473.68025 [3] Dixon, L. C. W., andPrice, R. C.,Numerical Experience with the Truncated Newton Method, Numerical Optimisation Centre, Hatfield Polytechnic, Technical Report No. 169, 1986. [4] Dixon, L. C. W., Dolan, P. D., andPrice, R. C.,Finite Element Optimization, Proceedings of the Meeting on Simulation and Optimization of Large Systems, Reading, England, 1986, edited by A. J. Osiadacz, Oxford Science Publications, Oxford, England, 1988. [5] Pope, J., andShepherd, J.,On the Integrated Analysis of Catch-at-Age and Groundfish Survey or CPUE Data, Ministry of Fisheries and Food, Lowestoft, England, 1984. [6] Dembo, R., Eisenstat, S. C., andSteihaug, T.,Inexact Newton Methods, SIAM Journal on Numerical Analysis, Vol. 19, pp. 400-408, 1982. · Zbl 0478.65030 · doi:10.1137/0719025 [7] Steihaug, T.,The Conjugate Gradient Method and Trust Regions in Large Scale Optimization, SIAM Journal on Numerical Analysis, Vol. 20, pp. 626-637, 1983. · Zbl 0518.65042 · doi:10.1137/0720042 [8] Anonymous,Optima Manual, Numerical Optimisation Centre, Hatfield Polytechnic, Hatfield, Hertfordshire, England, 1984. This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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CC-MAIN-2021-25

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https://wirings-diagram.com/electric-motor-wiring-diagram/wiring-diagrams-for-electric-motors-today-wiring-diagram-electric-motor-wiring-diagram/

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# Wiring Diagrams For Electric Motors – Today Wiring Diagram – Electric Motor Wiring Diagram Wiring Diagrams For Electric Motors – Today Wiring Diagram – Electric Motor Wiring Diagram Electric Motor Wiring Diagram – electric motor wiring diagram, electric motor wiring diagram 220 to 110, electric motor wiring diagram 3 phase, Every electrical structure consists of various different components. Each component ought to be set and connected with different parts in specific manner. If not, the structure won’t function as it should be. So as to make sure that the electrical circuit is constructed correctly, Electric Motor Wiring Diagram is demanded. How does this diagram aid with circuit construction? The diagram offers visual representation of the electric arrangement. On the other hand, this diagram is a simplified variant of the arrangement. It makes the procedure for building circuit simpler. This diagram provides advice of circuit’s components as well as their placements. ## Components of Electric Motor Wiring Diagram and Some Tips There are two things which are going to be present in almost any Electric Motor Wiring Diagram. The first element is emblem that indicate electric component in the circuit. A circuit is generally composed by several components. The other thing which you will get a circuit diagram would be lines. Lines in the diagram show how every component connects to a another. The rankings of circuit’s parts are comparative, not accurate. The order is also not logical, unlike wiring schematics. Diagram only reveals where to place component at a place relative to other elements inside the circuit. Though it is exemplary, diagram is a great basis for anyone to construct their own circuit. One thing you have to learn before studying a circuit diagram would be your symbols. Every symbol that is shown on the diagram reveals specific circuit element. The most common elements are capacitor, resistor, and battery. There are also other elements such as ground, switch, motor, and inductor. It all rides on circuit that is being assembled. As stated previous, the traces at a Electric Motor Wiring Diagram signifies wires. Sometimes, the wires will cross. However, it does not mean link between the wires. Injunction of two wires is generally indicated by black dot on the junction of two lines. There will be principal lines that are represented by L1, L2, L3, etc. Colours can also be utilised to differentiate wires. Ordinarily, there are two chief types of circuit connections. The first one is called string connection. It’s the easier type of link as circuit’s elements are placed inside a singular line. Because of that the electrical current in each component is comparable while voltage of this circuit is total of voltage in each component. Parallel connection is much more complicated than the string one. Unlike in series connection, the voltage of every component is comparable. It is because the element is directly connected to power resource. This circuit includes branches which are passed by distinct electric current amounts. The current joins together when the branches meet. There are lots of items that an engineer needs to focus on when drawing wirings diagram. To begin with, the symbols used in the diagram ought to be accurate. It must represent the exact element needed to construct an intended circuit. When the logo is wrong or uncertain, the circuit won’t function since it is supposed to. It’s also highly advised that engineer draws favorable supply and damaging source symbols for clearer interpretation. Ordinarily positive supply symbol (+) is located above the line. Meanwhile the negative supply emblem is set under it. The current flows in the left to right. In addition to that, diagram drawer is advised to limit the number of line crossing. The line and element placement should be made to minimize it. However, if it is inevitable, use universal emblem to indicate if there’s a intersection or when the lines aren’t actually connected. Since you can begin drawing and interpreting Electric Motor Wiring Diagram can be a complicated job on itself. The advice and ideas which were elaborated above ought to be a wonderful kick start, though. Electric Motor Wiring Diagram Wiring Diagrams For Electric Motors – Today Wiring Diagram – Electric Motor Wiring Diagram Uploaded by Hadir on Friday, February 15th, 2019 in category Wiring Diagram. Here we have another image Baldor Motor Capacitor Wiring Diagram L1410T Electric Motors 15 8 – Electric Motor Wiring Diagram featured under Wiring Diagrams For Electric Motors – Today Wiring Diagram – Electric Motor Wiring Diagram. We hope you enjoyed it and if you want to download the pictures in high quality, simply right click the image and choose "Save As". Thanks for reading Wiring Diagrams For Electric Motors – Today Wiring Diagram – Electric Motor Wiring Diagram.

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https://www.mathworks.com/matlabcentral/fileexchange/53482-omeda-x-filtersize-plotmode

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## omeda(x,filterSize,​plotMode) Version 1.0.0.0 (7.32 KB) by Non-iterative optimal solution for Minimum Entropy Deconvolution (MED) and convolution fix. Updated 25 Oct 2015 OPTIMAL NON-ITERATIVE MINIMUM ENTROPY DECONVOLUTION SOLUTION code by Geoff McDonald (glmcdona@gmail.com), 2015 omeda(x,filterSize,overlapMode,plotMode) A noniterative optimal MED (OMED) computation algorithm. This method solves a similar deconvolution problem to MED, and is able to directly solve for the optimal filter solution as proposed by Carlos Cabrelli. This can be used is rotating machine fault detection from vibration signals to detect gear and bearing faults, and OMED is applied to extract the impulse-like features in the vibration. This implementation uses the convolution adjustment proposed by myself in the second paper reference, which is important to prevent this method from reaching the trivial solution of deconvolving the convolution discontinuity. Note to readers, you may want to refer to some of my other MED submissions: MED: The iterative non-optimal solution is often better for vibration fault detections. It is often better than this optimal solution, since the MED problem aims to deconvolve only a single-impulse as the result. As a result, OMED is able to better-extract the solution of deconvolving only the single-impulse, whereas MED more commonly reaches the solution of deconvolving the desired impulse train. MOMEDA: This is the optimal solution to the periodic impulses and is recommended for rotating machine faults instead of MED or OMED. Since it is non-iterative, it is able to quickly generate spectrum's to diagnose machine health. Algorithm Reference: Original derivation: C A. Cabrelli, Minimum entropy deconvolution and simplicity: A non-iterative algorithm, Geophysics, Vol. 50. No. 3. March 1984. G.L. McDonald, <others>, Multipoint Optimal Minimum Entropy Deconvolution and Convolution Fix: Application to Vibration Fault Detection, unpublished Inputs: x: Signal to perform Minimum Entropy Deconvolution on. filterSize: This is the length of the finite inpulse filter filter to design. Using a value of around 30 is appropriate depending on the data. Investigate the performance difference using different values. plotMode: If this value is > 0, plots will be generated of the iterative performance and of the resulting signal. Outputs: y_final: The input signal(s) x, filtered by the resulting MED filter. This is obtained simply as: y_final = filter(f_final,1,x); f_final: The final 1d MED filter in finite impulse response format. d_norm: Final D-Norm of the filtered signal. Example: % Simple vibration fault model close all n = 0:1999; h = [-0.05 0.1 -0.4 -0.8 1 -0.8 -0.4 0.1 -0.05]; faultn = 0.05*(mod(n,50)==0); fault = filter(h,1,faultn); noise = wgn(1,length(n),-40); x = sin(2*pi*n/30) + 0.2*sin(2*pi*n/60) + 0.1*sin(2*pi*n/15) + fault; xn = x + noise; % 20-sample FIR filters will be designed L = 20; % Recover the fault signal [y, f, d_norm] = omeda(xn,L,1); ### Cite As Geoff McDonald (2024). omeda(x,filterSize,plotMode) (https://www.mathworks.com/matlabcentral/fileexchange/53482-omeda-x-filtersize-plotmode), MATLAB Central File Exchange. Retrieved . ##### MATLAB Release Compatibility Created with R2014a Compatible with any release ##### Platform Compatibility Windows macOS Linux

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http://gmatclub.com/forum/slope-and-intercept-of-lines-104676.html

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Find all School-related info fast with the new School-Specific MBA Forum It is currently 24 Aug 2016, 20:24 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 Slope and Intercept of lines Author Message TAGS: Hide Tags Intern Joined: 26 Nov 2009 Posts: 46 Location: India Followers: 3 Kudos [?]: 4 [0], given: 8 Slope and Intercept of lines [#permalink] Show Tags 11 Nov 2010, 11:15 I found this link to be very helpful in understanding the concept of slope and intercepts of line..... Check how a line with negative slope has both intercepts in either I or III quadrant etc.... http://www.analyzemath.com/Slope_Intercept_Line/Slope_Intercept_Line.html _________________ When going gets tough, tough gets going......... Manager Joined: 01 Nov 2010 Posts: 181 Location: Zürich, Switzerland Followers: 2 Kudos [?]: 36 [0], given: 20 Re: Slope and Intercept of lines [#permalink] Show Tags 11 Nov 2010, 13:10 Quote: I found this link to be very helpful in understanding the concept of slope and intercepts of line..... Check how a line with negative slope has both intercepts in either I or III quadrant etc.... http://www.analyzemath.com/Slope_Interc ... _Line.html Re: Slope and Intercept of lines   [#permalink] 11 Nov 2010, 13:10 Similar topics Replies Last post Similar Topics: Coordinate Geometry: Slope of a line 3 10 Mar 2016, 19:27 Slope Intercept form and Exponentail Growth Doubt!!! 2 20 May 2011, 09:08 1 Slope of two point 4 14 Mar 2011, 19:05 Property of Lines 1 25 Sep 2010, 01:46 2 Meaning of Slope and y-Intercept in word problems 2 17 Feb 2010, 00:18 Display posts from previous: Sort by

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https://convertoctopus.com/312-7-feet-to-kilometers

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## Conversion formula The conversion factor from feet to kilometers is 0.0003048, which means that 1 foot is equal to 0.0003048 kilometers: 1 ft = 0.0003048 km To convert 312.7 feet into kilometers we have to multiply 312.7 by the conversion factor in order to get the length amount from feet to kilometers. We can also form a simple proportion to calculate the result: 1 ft → 0.0003048 km 312.7 ft → L(km) Solve the above proportion to obtain the length L in kilometers: L(km) = 312.7 ft × 0.0003048 km L(km) = 0.09531096 km The final result is: 312.7 ft → 0.09531096 km We conclude that 312.7 feet is equivalent to 0.09531096 kilometers: 312.7 feet = 0.09531096 kilometers ## Alternative conversion We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilometer is equal to 10.491972801449 × 312.7 feet. Another way is saying that 312.7 feet is equal to 1 ÷ 10.491972801449 kilometers. ## Approximate result For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred twelve point seven feet is approximately zero point zero nine five kilometers: 312.7 ft ≅ 0.095 km An alternative is also that one kilometer is approximately ten point four nine two times three hundred twelve point seven feet. ## Conversion table ### feet to kilometers chart For quick reference purposes, below is the conversion table you can use to convert from feet to kilometers feet (ft) kilometers (km) 313.7 feet 0.096 kilometers 314.7 feet 0.096 kilometers 315.7 feet 0.096 kilometers 316.7 feet 0.097 kilometers 317.7 feet 0.097 kilometers 318.7 feet 0.097 kilometers 319.7 feet 0.097 kilometers 320.7 feet 0.098 kilometers 321.7 feet 0.098 kilometers 322.7 feet 0.098 kilometers

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https://interviewing.io/recordings/Java-Google-1/

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# Watch a technical mock interview with a Google engineer Intergalactic Avenger, a Google engineer, interviewed Supersonic Taco in Java Share Summary Problem type Order statistic of an unsorted array Given an unsorted array, find the nth smallest element in the array • Order statistic of unsorted array Feedback Feedback about Supersonic Taco (the interviewee) Advance this person to the next round? Yes How were their technical skills? 4/4 How was their problem solving ability? 3/4 4/4 Great job! I needed to give you a few hints on the algorithm and time complexity, but really it was quite minor. I appreciated how you verbalized your thought process in working through the problem, which gave me confidence you understood the problem and the challenge. Your ability to code up algorithm into a working solution was very solid. Keep up the good work. Feedback about Intergalactic Avenger (the interviewer) Would you want to work with this person? Yes How excited would you be to work with them? 4/4 How good were the questions? 4/4 Intergalactic Avenger: Hello? Supersonic Taco: Hi! Intergalactic Avenger: How's it going? Supersonic Taco: Good, how are you? Intergalactic Avenger: Doing good, doing good. Yeah sorry, couldn't hear you for a second so I just wrote this on the screen there. Have you ever used the platform before? Supersonic Taco: Yes, a couple times. Intergalactic Avenger: Okay so then I guess you're familiar with how it all works, so I'm just going to jump straight into a coding question. Supersonic Taco: Alright. Intergalactic Avenger: So the first question is: are you familiar with the concept of order statistics. Have you heard that term before? Supersonic Taco: No Intergalactic Avenger: It's actually a very simple thing with an overly complicated name. So, given an unordered set of numbers like what I've written here: `1, 6, 3, 9, 8, 5,` the order statistics are like the `nth` element in the list if it were sorted. So, the first smallest number in this list would be `1`, and the second smallest number is `2`, and basically the `nth` order statistic is then the `nth` smallest. That make sense? Supersonic Taco: Yeah Intergalactic Avenger: Alright so we're just going to have some algorithmic questions about finding some of these nth order statistics. Just to get started as sort of a warm-up problem: how about you write a function that, given a list of numbers that are out of order, you just find the smallest. So basically the first order statistic. Supersonic Taco: Okay sounds good. [2:29] Supersonic Taco: So I'll just declare it here: `public static int` is the return and I'll just call it `min`, and this takes an int I'll call it `n`. Intergalactic Avenger: So this should be like an array. Supersonic Taco: Yeah, sorry. So then I'll loop through the array, and the for each of those iterations I'll compare it against a `min` that I've already set up. So I'll set the `min` to be `arr[0]`, and so what should I return if the array is `null`, or if there's no value? Intergalactic Avenger: Don't worry about that case. We'll just assume it'll always have some values in it. Supersonic Taco: Okay, sounds good. So then we'll set the `min` to the first one. So then moving through the array, if the element at the index of `i` is less than `min` then we make that the new min. And at the very end of the for loop you return. So just to test it to make sure, I'll write a litte test here: `int[] input = {2, 3, 0, 6}` should give us `0`. Intergalactic Avenger: Yup, sounds good. Supersonic Taco: And let's try that then. Alright there we go: it returned `0`. Intergalactic Avenger: Perfect, and what's the runtime of this algorithm? Supersonic Taco: `O(n)` Intergalactic Avenger: And is there any faster way for you to do it? Supersonic Taco: For an unsorted array that I know of, no. Intergalactic Avenger: Correct. Okay, so now you get the idea, let's make this a little harder, a little trickier. How about you give me the second smallest number? [5:28] Supersonic Taco: Okay, so for the second smallest number, then I think what we would need to do is maintain two variables and check accross both of them to see...like a larger `min` and a smaller `min` so we can check if there's a number that's smaller than both of them then we would put it into the smaller min, and if it's only smaller than the larger one then we can put it into the larger min. So once again are we assuming that the array's length is two or greater? Intergalactic Avenger: Yep Supersonic Taco: Okay, then I'll just call this `min2` and we'll put this at the second element of the `arr[]`, and if the `arr[]` is greater than the `min` then it's the very smallest one so the `min` becomes `arr[1]`. `min2` should be set to `min` and `min` is set to the new minimum. And then otherwise, if `arr[i]` is less than `min2` whereas it's still greater than `min` then this means `min2` becomes the `arr[i]` and instead `min` becomes `min2`. Intergalactic Avenger: `min2` becomes `arr[i]`, but what happened to `min` then? Supersonic Taco: Oh sorry yeah, this is maintained the same I think, because if it's less than `min2` then `min` is still the smallest so it should stay the same, but `min2` gets updated Intergalactic Avenger: Okay, let's run this. Supersonic Taco: I have to return `min2` this time. Okay this should give me `2`. There, it gave me `2`. Intergalactic Avenger: Perfect. Okay, I'm trying to think if there is a corner case here. What if these were the numbers here: `{3, 2, 5, 6}`. It should actually be `3`, because `3` is the second smallest. Supersonic Taco: Yeah, it returns `3`. Intergalactic Avenger: What if...Oh because you start back at `0` again, then you flip it all around, I got it. Alright, yeah perfect. And what's the runtime of this one? Supersonic Taco: This one is still `O(n)`. Intergalactic Avenger: Okay, so we're going to want to move this and turn it into `n`. So we're going to want to add another parameter here that says I want to find the nth smallest. One way you could do it is you could just expand on this idea and you could just create a list of all of the `min`s up to `n`. Supersonic Taco: Right. Intergalactic Avenger: Now, what would the runtime of that algorithm be? Supersonic Taco: Well if we're storing them all, then we need to check across each of them each time which would become `O(n^2)` because we're running through all of the previous values every time we look through the array. Intergalactic Avenger: Yep, so to get the 1st, it just took `O(n)`. To get `2` it took `O(n)`. But now if you expand this in that same sort of pattern it actually becomes `O(n^2)`. So is there a way you can improve on that `O(n^2)` time? Supersonic Taco: Alright let's see. One way to improve would just be to sort the array and that would make it `O(nlogn)` immediately and then all we have to do is find the `nth` index in the sorted array. Intergalactic Avenger: Okay perfect. Now here's the real challenge. Can we do better than `O(nlogn)` for any arbitrary `m` we're trying to find? So we saw that when `m` is `1`, we can do it in `O(n)` time. When `m` is `2`, we can still do it in `O(n)` time. You think there's a way that you could extrapolate on that sort of patter and say: oh, I can always do it in `O(n)` time, no matter which of these I'm trying to find. Supersonic Taco: Right. So the reason that we could do it in `O(n)` time for this `m = 1` is because we could just go through and compare them all and see which one is smallest. For `m = 2`, we had to compare it with each of them and also the previous minimum. Without having to compare it to all the previous minimums I'm trying to think if there's a way to figure out the `nth smallest`. Supersonic Taco: Maybe if we delete for an object but then that wouldn't help either because if we go through… Intergalactic Avenger: But if you delete? So certainly it's tricky. If you get stuck, I have some hints for you. Supersonic Taco: Yes, I'd appreciate a hint. Intergalactic Avenger: Okay so, you mentioned the `O(nlogn)` case of sorting and then picking the `mth` item in the list. So in that sense, once we sort this input, you have: `[2, 3, 5, 6]`. And then let's say `m = 3`, you go straight for the `5`, and you know that `5` is in the correct place, that everything of the left is below it and everything to the right is above it, and it's in position number `3`. So that means it is necessarily correct. If we expand this out a little bit more `[2, 3, 5, 6, 10, 15, 21]`, what you can see is that number `5` is in the correct position, but also every single other number is in the correct position. So you've also done work to make sure the `3` is the 2nd position and the `2` is in the 1st position and the `10` is in the 5th position and you've put them all in the right places even though at the end of the all you really cared about is that the `5` is in the right place. Because whatever order any of these other ones are in, it doesn't really matter, you don't really care because you're only really interested in this one. Supersonic Taco: Right, okay. So basically we're looking for a way to ensure that a certain value is in a certain spot but all the other values don't have to be sorted. Intergalactic Avenger: Correct. Supersonic Taco: And the one we want to be in the right place is at the `m` index. Intergalactic Avenger: Right. Supersonic Taco: So if we did this for the first minimum then basically we would have to go through each of the elements and then put all the ones that are larger than a certain minimum to the right of it if so that is would be at the beginning. So basically we would have to go through and figure out how many numbers are lower than that number and how many numbers are larger than that number. Intergalactic Avenger: Right. Supersonic Taco: So what if we do that then. So say we want `m` to be `2` like in this case where the `index` is `2` that means we want two numbers to be less than the one at `m`, and four numbers to be to the right. Somehow there should be a way that we can go through and figure out and make it balanced so that it's like equal on both sides even though those aren't sorted. But we would need to know what the number `5` already is in order to go through and actually put that in the right spot. So there needs to be a way where we can figure that out without already knowing `5`. Intergalactic Avenger: Right. Supersonic Taco: Yeah and if we just chose a random number or when one by one like we did here then it wouldn't work because then we would still go back to being `O(n^2)`. We just still have to go through all the numbers to see which one was right. Intergalactic Avenger: But let's take a look at example. You could have described this process where you take this input array, you take the first number you found, and you made sure it went in the right spot. So you described a process of saying: take this number `3`, and put it (just that one number) where it goes. So everything to the left is less than it, and everything to the right is greater than it. Supersonic Taco: A binary search tree? Intergalactic Avenger: Yeah. That's kind of a part of it. So describe how you might do that. You don't really much information about this array coming into it, because it's unsorted, in any order. So you might as well just start with the first number you find and say: I'm just going to put the first one where it belongs and I'll tackle the rest of the problem after that. For example, you can think of what it would look like afterwards: `[2, 3, 5, 6, 10, 21, 15]`. If you can imagine that this might be what the array would look like after you found the first place. So `3` goes in this spot and everything to the left is less than `3` and everything to the right is greater than `3`, and you don't really know what's going on on either side and you haven't really sorted it altogether, but you do know that `3` is in the right spot. Intergalactic Avenger: So does this then give you some additional information in terms of how you would proceed from here? So keep in mind we're still looking for the third largest number and you just figured out that the number `3` goes in position number 1. Supersonic Taco: So then I know that the number `3` is the second largest number and at this point I would have enough information to find the third largest number using the technique that I just did before where I find the previous minimum then I just have to run through the array one more time to find the next minimum. Intergalactic Avenger: Right, that's true. And which part of that array would you run that algorithm on? The whole thing, or a part of it? Supersonic Taco: The right, because the right is unsorted. Intergalactic Avenger: Well, in a sense the left is also unsorted kind of. Because we don't really have any guarantees about what's going on to the left. In this case you do because there's only one element so there's not really much sorting to do. But what else is an interesting property? So now that you have this `3` in the right spot, in term of what you're looking for are you interested in anything on the left, or are you interested in anything on the right, or are you interested in anything on both sides? Supersonic Taco: Well, right now looking at this array, I would be interested in the right because the `3` is only at index `1` and we're looking for index `3`. But the `2` could be the `5` too, like what if there was a `5` here instead of the `2`. Intergalactic Avenger: It couldn't be. It couldn't be because you've split this array up. Let me start by rearranging this more: `{10, 2, 5, 6, 11, 3, 15}`.So you start with this `10` and you first make sure that everything on your left is less than `10` and everything on your right is greater than `10`. So this might look like this: `[2, 5, 6, 3, 10, 15, 11]`. So you've shifted things around but you've shifted them around in a way so that everything on the left is less than `10` and everything on the right is greater than `10`. So it's not just that you've found the one right place for it, but you've done a little bit extra to make sure that you split the array in two pieces. Supersonic Taco: Yeah, so then this is actually pretty helpful. If the `10` is over here and this is in the 4th index and `m` has to be to the left of it, so we don't really need to worry to the right any more because those are greater than `10` anyways. Intergalactic Avenger: Exactly. So then when you repeat this and you go for another iteration of it, you're only going to be looking at the left side, right? Supersonic Taco: Yeah. Intergalactic Avenger: Okay so, does that give you some ideas about how this algorithm might work? Supersonic Taco: Yeah it does. Because then in the first time around I'd be iterating through all the numbers to find where it belongs and the second time around I'm only iterating through the left side. So this kind of reminds me a little bit of binary search so the runtime might get a little better than `O(n^2)` when we do it this way. Intergalactic Avenger: Okay Supersonic Taco: Alright so, should I start implementing it then? Intergalactic Avenger: Yeah let's see what happens. [25:16] Supersonic Taco: Alright so we're outputting just the number, and this is `NthSmallest()`. I'm just going to write the code for the first time and we can probably iterate through and make that repeat. So the first time around I'll have to go through the entire array. So if I look at the first one, I'll have to put it in the right spot so then I'll take `int index = 0` before this loop. We're going through them and if we say: `if(arr[index] > arr[i])` then we need to move it to the right, or swap it. Yeah so if the `arr[]` at `index` is greater than the `arr[]` at `i` then we need to do some swapping here. So should I just abbreviate it and just write swap here, or do you want me to write the whole swap? Intergalactic Avenger: You can just write swap. Supersonic Taco: Okay. Well actually I'm going to need to test it later so I might as well just do it now. So then we'll write `int temp = arr[i]` and then `arr[i] = arr[index]` and `arr[index] = temp`. So this switches them around if its greater than, but if it's less than, then essentially we need to keep moving so we don't need to do anything there. But if they're equal to each other then we stop. So we start with `index = 0`, oh there's a problem here because if it's starting with first one then it'll just stop immediately because it's the same one. So then we'll start at the first one, then `index = 0` and the first element is `10`, and then it checks to see `arr[1]` which is `2` and `10 > 2` so then it swaps and now it's `2` and then `10` and then it compares it again, `5`: `10 > 5` so it swaps, `6`: `6 > 5` so it swaps and it looks at `11`. This breaks now because it'll stop since it's greater than, but it needs to continue and look for that `3`. Intergalactic Avenger: Wait, if those numbers don't equal each other… Supersonic Taco: Yeah it wouldn't stop but it still needs to...yeah it would stay there and then it would go through and check against `3` and now that it's less than `3` it'll swap with that one instead and `10` will be where the `3` was and then it'll check against `15`, which is greater than so it'll stay there. Intergalactic Avenger: So that would actually work for this specific input but think about what would happen if the input was something like this: `[10, 2, 5, 6, 11, 3, 0]`. First you swap the `10` with the `2` so that's right, and then you swap the `10` with the `5` which is totally right then you swap the `10` with the `6` then you leave that. Then you swap the `3` with the `10`, so that's cool, and then you swap the `0` with the `10`. Supersonic Taco: Yeah that's not right. Intergalactic Avenger: Not quite right because when you skipped over the `11` now you have something on the left that is bigger than it should be. Supersonic Taco: So then I think maybe we have to move the `11` along with the `10` every time, so maybe we can swap the `11` with whatever's next to it so it keeps moving away. Intergalactic Avenger: So you had this point with the `10` and the `0` like this, then what were you going to do? Move up the `0` and the `11`? Supersonic Taco: No it was like the `10` was here: `[2, 5, 6, 10, 11, 3, 0]` and then you would check the `10` and the `11`, and since the `11` is greater you could swap it away. Yeah you could actually just swap it with the end of the array so this becomes `0` and this becomes `11`. And then here we check against the `0` is greater than and it swaps, and then it checks the `10` and the `3` is greater so it swaps. Intergalactic Avenger: Okay. Supersonic Taco: And that would put it in the right place. Intergalactic Avenger: So then presumably if there was another number that was greater than `10` you would swap it so like...we had it here that was like: `[2, 5, 6, 10, 11, 3, 15]`. Supersonic Taco: So then it would first compare the `10` and the `11`. Okay, now I see what you're saying: if it swaps there it would skip over and it wouldn't work. Intergalactic Avenger: But you're definitely on the right track. This is definitely the kind of manipulating that you're going to want to be doing in this kind of array. And you're definitely on the right track with respect to looking at each one of these numbers as you're going up and as long as it's less than then you're keeping it to the left and as long as its greater than you're keeping it to the right. So that's kind of how you're kind of scanning this list of numbers. So you're getting into this one sort of problematic case that when you start to have multiple numbers that are bigger it gets kind of tough with how do you deal with the bookkeeping of where does it go and who can you swap it with. Supersonic Taco: Yeah I can think of one way to solve that which would be to create another array and then if it's less than whatever we want we put it into the beginning of the array and hold two counters for where the beginning of the array is and then where the end of the array is and if the number is larger we add it to the end until there's only one spot left and then that's where the `10` has to go. Intergalactic Avenger: That's a good way to do it. Supersonic Taco: Alright so then I'll try to implement that instead. [35:28] Supersonic Taco: Okay so then we have now two counters int `start = 0` and int `end = arr.length` and then we loop through and so we know the number that we're looking for I'll put that in as `index` again and that'll start at `0` this time. So basically if the number that we're looking at right now which is `arr[i]` is less than the `arr[index]` then we put it to the right. So then we have this new array that we have to create that's the same length. And then if the `arr[i] < arr[index]` then it should go in the beginning so we do `newArray[start] = arr[index]` and `start++` to show that we've added an element. Intergalactic Avenger: So in this line here you set `newArray[start] = arr[index]` is it `arr[index]` or `arr[i]`? Supersonic Taco: Oh sorry, yeah that's completely different thing. And are there going to be duplicates in this array or no? Intergalactic Avenger: Let's just say no for now. Supersonic Taco: Okay so if there's no duplicates then any other element we look at cannot be the one we're searching for so then we can just do an else here and this basically says that the `arr[i]` is greater than what we're looking for so it needs to go to the right. So it goes to `newArray[end] = arr[i]`. And then do end-- and this loop has to end whenever `start` and `end` are one apart from each other, because then that means we've found our index. So if `end - start == 1` we return `start+1`. Alright so I think this should work just for the first one to place it in the right spot. So let's try it. Intergalactic Avenger: So I see you're returning the number right away but I'm wondering...so this is going to return not the nth smallest, but some smallest that the...you're returning which condition the first element will have gone to. Supersonic Taco: Yes, yes. I just wanted to break down the problem Intergalactic Avenger: Got it. Okay. Supersonic Taco: Okay, in this specific array...Oh I'm missing a return statement. Intergalactic Avenger: Here's an interesting question: so when will `end - start == 1?` When will this be true? Supersonic Taco: `end - start` will be `1` when we've gone through all the elements in the array and now we're looking at that one empty space where the element should go. Intergalactic Avenger: Okay, so if that's goign to happen once you're already completed with this array will this ever be see? Supersonic Taco: Yeah you're right. We can just recognize those and return `start+1`. Intergalactic Avenger: `IndexOfOfBounds`? Supersonic Taco: Oh end has to be `arr.length-1`. Intergalactic Avenger: Okay. I think actually because you incremented `start` right here, you don't need to increment `start` there. I think that's what's going on. Supersonic Taco: Alright and let me try it with another `m` just to make sure. If we deleted one of these, it should become `3` I believe. Intergalactic Avenger: Yep. Supersonic Taco: Okay, alright. Intergalactic Avenger: So now I have the functionality for figuring out the spot where one specific element should go but I have to repeat until I find wherever the element at the `mth` index is. So essentially this whole start...if `start` is `m` then we know that we've found our element. Supersonic Taco: Right. Intergalactic Avenger: But the thing is this `index` has to change every time. So I'm thinking maybe we could do this like a binary search where we choose the middle number in each half and we try to put that where it belongs, and if `m` is greater than the `index` there then we run it on the right half and if `m` is less than then we run it on the left half. Supersonic Taco: Okay so then we'll start with, instead of `0`, `index = start + end/2` and then looking at the `end`…. Intergalactic Avenger: The `index` is representative of the index of the number you're comparing everything with. In that sense `index` represented this number `10` here. Does it get you anything to pick the number in the middle? This number here is the number in the middle, does it really matter that you take `index = 0` or `index=start+end/2`? It's going to be 50-50 right? It's not going to be exactly in the middle there. There's kind of no way to tell how well this number you're picking at random is going to split this array in tWo. I mean you can do that, but I think that starting at `index=0` is an totally reasonable way to go because at this point it's just a random array, there's no order. You might as well pick anything and then go from there. Supersonic Taco: Yeah, I think that's fine, we can use `0` then. We can actually make this recursive to do a little less work where once we find where the spot that index should go, we'll determine if `m` is in the right or left and then we will again do it. But actually that might be….yeah I think that might be one way of implementing because we can always keep index at `0` because the array that we're looking at is now a subarray so the `index` will always be `0`. Or a way we could do this is just have `index` at `start` and then move through and then kind of keep doing it with a new start and end and that would be another way of doing it. Or we could just make new parameters here for `start` and `end` and implement it as recursive. Intergalactic Avenger: So I like your recursive idea because it's definitely going to make the code a little simpler. Do you actually need to add more parameters here if you do is as a recursive algorithm? Supersonic Taco: No, I don't have to. If I change the array then I don't have to. I was just wondering like that would still take up more memory because technically I would have to somehow cut off the array...We'll just keep it like this for now and turn it into a subarray. Supersonic Taco: Okay so then we're using `index = 0` and we're returning `start` which ends up being the `index` of whatever element is at `0`. So what we need to do is determine if `m` is greater than or less than `start`. If `m` is greater than `start`, we need to redo this on the right side of the array, so we should perform `NthSmallest()` on...I actually forgot the method for sub-arrays in Java. Intergalactic Avenger: I think there's a thing called range? Subset? Fill? I mean, you could just use copy of range if you wanted to. Supersonic Taco: I think I got it, there's a code completion on here and it says that it's a method so I'll just use that then. So then the original would be `arr[]` and I'm assuming the from is inclusive so we'd use `start+1` because `start` we've already looked at and that's the one we've already found and the end would just still be `end`. And here it would just be the other way around so if we have `NthSmallest()` then it would be `arr[]` and instead of `start+1` this would be `0` and this would be just `start`. And else that means that `m = start` which means we've found the element at the index we want so we just return `start`. Intergalactic Avenger: Okay, so `NthSmallest()` actually takes two parameters. Supersonic Taco: Oh right, so `m` minus needs to be add in. Intergalactic Avenger: Is it `m`? Because that's relative to the entire… Supersonic Taco: Right so if the `m` that we're looking for is greater than `start` then it would need to be `m-start` and if the element that we're looking for is....so if `m < start` then now we're looking for still `m`. Intergalactic Avenger: Right. Supersonic Taco: I'm going to test this now to see if there's any issues. Let's just see if we get `0` as `m` then we should still get back `10`. Intergalactic Avenger: We should get `3`. So `3` is the smallest one. Supersonic Taco: Oh yeah yeah. So it should give back `3`. Intergalactic Avenger: We need to import ArrayLists? Or is it Arrays. Oh it's a lowercase c Supersonic Taco: Oh okay, let's see what happens now? I'm missing a return statement? Intergalactic Avenger: You don't have return in these ones. Wait which ones are they looking at? Intergalactic Avenger: So 52 and 54, you're calculating it without returning it. Supersonic Taco: Got it, alright. Now it's giving me `0`, which that's not even in the input array. Oh I returned the `index`, I never returned the element. Supersonic Taco: Oh that's still not right. Alright, let's see. Intergalactic Avenger: So definitely the logic looks right, I'm guessing there must be some off by 1 bug. So we already figured out that with this list when you go the first time, `start` equals...oh wait `start` was `3` before, so it should have put it in the...So maybe there's some need to do some debugging with some print statements to see...because yeah the logic here is definitely looking solid. Supersonic Taco: Maybe here I'll just print out `newArray[]`. Okay. Intergalactic Avenger: Interesting that you have the `10` in there. So in that sense you want your `newArray[]` to be one element smaller. So if you see what's happening here is you started with index `0`, and you're also checking index `0` again. The size is going to be `1` less because you're kind of taking one out then sort of partitioning it after you've taken that one out. So then now that you've reduced the size by a little bit you have to make up for that when you make this copy. Supersonic Taco: Well that wouldn't be affected by the `newArray[]` would it? Intergalactic Avenger: Well now the end actually has to be an array of `length-2`. Because the end is actually the end of `newArray[].length-1`. Supersonic Taco: Right. Well it's the initial array minus one so it's like the ending, but in the `newArray[]` it's not… Intergalactic Avenger: In the new array, you're copying these values over into the `newArray[]` so the `newArray[]` has to… Supersonic Taco: Yeah, all I had to do was `arr[].length` here because the end is not actually the end. Intergalactic Avenger: I'm looking at line 38 here. Supersonic Taco: Oh yeah. Intergalactic Avenger: So that end is actually the end of `newArray[].length-1`, which is the original `arr[].length-2`. Supersonic Taco: Oh okay. Intergalactic Avenger: And then in this one, you're not actually copying the original `arr[]`, you want to copy the `newArray[]` because the original array is totally out of order and the `newArray[]` is the one...and you'd need the `newArray[].length`. Supersonic Taco: Okay, I get it and `newArray[]` has to be smaller. Yeah it should be `1` smaller.... Intergalactic Avenger: Right, so in this one, you just want to copy to `newArray[].length`. Supersonic Taco: Well `newArray[].length` would include the values above it wouldnt it? Intergalactic Avenger: Well `newArray[].length` is the end of the new array, and you're trying to copy from your position all the way to the end. Supersonic Taco: And here there's still something wrong with the size of `newArray[]`. Intergalactic Avenger: Wait can you try running again? What's wrong with the size of `newArray[]`? Line 46. Oh this one here: line 38, you had just changed this. Supersonic Taco: So this should be at minus `2`? Intergalactic Avenger: Or you could just send end underneath the `newArray[]` or you could move this one up and just say `end = newArray[].length-1`. Either way it's the same number. Supersonic Taco: Oh okay, and this is still `arr[].length-1`. Okay now `3`, okay now that's actually right. Intergalactic Avenger: There it is, not bad. Supersonic Taco: Now let's try it for some other other numbers too. So if we tried like `1`, then that should give us `5`. No.. Intergalactic Avenger: Close. All right yeah, I think there must be some...I'm guessing there's an off by one in either in this `m-start` or this `m` here or the zero to start or what. I'm guessing there's a tiny little off by one bug here. So not a big deal, just because we're running a little short on time. So you did all this work and now you're inspecting smaller and smaller pieces of the array as you recurse into it, what is the runtime of this algorithm? So the question is that you did all this work, or there's all this extra code to be careful not to do any more work than you need to so the question is has this actually sped it up or is it still `O(nlogn)` or is it even greater than `O(nlogn)`. Supersonic Taco: Well I think it depends on the case. Say for example this array was sorted in backwards order. Then the first one we'd look at would be the end and then the next one would be the end again. Say our `m` was `0` then we'd have to keep going through that array over and over and so we'd basically be looking at the entire array...which in the end still comes out to `O(n^2)`. So I think the worst case our algorithm is still `O(n^2)`. Intergalactic Avenger: Okay. But what about the average case. What if you randomized...That's sort of a common thing people do with these sort of divide and conquer algorithms is that if there is kind of a poisonous input then you just kind of randomize it to make sure that it's just in this big old jumbled order. So what can we expect sort of on the average case? You're totally right that there is a worst case input that makes it `O(n^2)`, but what can we generally expect this to be in the average case. Supersonic Taco: I think this is still `O(nlogn)` then. Wait actually, in the average case it would be like we're looking at half the array since then it wouldn't be too much or too little. Wait no we're only looking at half, we'd look at `6` then `3` then `1` which is like `O(nlogn)`. Intergalactic Avenger: Close. Can you see me typing up above the `NthSmallest()`? So the first time you run through this in the first iteration you have `n` elements that you're going over. Then like you said, you break it down by half and the nex time you just go over the `n/2` and the next time is `n/4` and `n/8` et cetera. So what is that going to add up to? Supersonic Taco: `O(logn)`? Intergalactic Avenger: Is it? Supersonic Taco: Oh no it's just `O(n)`. Intergalactic Avenger: Right, exactly. Supersonic Taco: Yeah it just becomes `O(n)` Intergalactic Avenger: Yep! So you did it. You got it down from `O(nlogn)` to `O(n)`. And the trick is when you do the sort you have this halving you do each tie but you keep doing more and more work. Yeah so when you do a sort, you split it in half and then you do both sides, and so even though it's getting smaller and smaller you have more and more of them that you have to deal with, and so the length of this is `logn`. So like the number of iterations is `logn` so that's why it becomes `O(nlogn)`. But in your case you don't have this ever increasing coefficient in front of it, so it just adds up to `2n` and it's bounded. Intergalactic Avenger: So you got it! Very good. Supersonic Taco: Alright, thank you. This is a very interesting problem. Intergalactic Avenger: I'm glad you liked it. So I'll leave some comments on the platform but just before we go if you have any questions for me about interviewing or anything else I'd be happy to answer them. Supersonic Taco: I think I'm all set. Intergalactic Avenger: Okay awesome, then you have a good night and good luck with all your future practice rounds. Supersonic Taco: Thank you. Have a good night. Intergalactic Avenger: Okay. Bye bye.

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### Author Topic: Optocouplers -- Collector dark current vs Off-state collector current  (Read 972 times) 0 Members and 1 Guest are viewing this topic. #### MicroControllerGuy • Contributor • Posts: 6 • Country: ##### Optocouplers -- Collector dark current vs Off-state collector current « on: March 16, 2019, 05:35:12 pm » If i place a voltage across the Collector-Emitter of an octocoupler when the LED is completely off, what is the datasheet term for the Collector-Emitter leakage current? Is it the "Off-state collector current", or the "Collector dark current? I found this definition of collector dark current on a website: "When the phototransistor is placed in the dark and a voltage is applied from collector to emitter, a certain amount of current will flow. This current is called the dark current (ID). This current consists of the leakage current of the collector-base junction multiplied by the dc current gain of the transistor. The presence of this current prevents the phototransistor from being considered completely "off", or being an ideal "open" switch." If the quote is correct, what is  "Off-state collector current"? Here is an example of a datasheet specifying both parameters (Toshiba LTP185): https://toshiba.semicon-storage.com/info/docget.jsp?did=14111&prodName=TLP185(SE Can someone please tell me the difference between these two currents, I'm confused? Thanks! « Last Edit: March 16, 2019, 05:44:26 pm by MicroControllerGuy » #### magic • Super Contributor • Posts: 2615 • Country: ##### Re: Optocouplers -- Collector dark current vs Off-state collector current « Reply #1 on: March 16, 2019, 05:48:18 pm » One is specified for Vf=0.7V, the other presumably for Vf=0. Extrapolate figures 11.4 and 11.9 to see what's likely to happen at Vf=0.7V. Friendly reminder to always look at test conditions, including those "unless stated otherwise" at the top. These really define what the spec means in practice, not just the spec's name. That being said, I think in most optocoupler datasheets "dark current" will be the exact thing you are looking for. « Last Edit: March 16, 2019, 05:55:28 pm by magic » The following users thanked this post: MicroControllerGuy #### MicroControllerGuy • Contributor • Posts: 6 • Country: ##### Re: Optocouplers -- Collector dark current vs Off-state collector current « Reply #2 on: March 16, 2019, 06:13:32 pm » Thanks for the reply, Magic, this makes sense now I guess if Toshiba had specified 'VF = 0" in the Test Conditions for the Dark Current, or had specified IF down to 0.0mA in figure 11.9, it would have been more obvious to me. Smf

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# Find the coordinates of the foot of perpendicular from the point (-1,3) to the line $3x-4y-16=0$ Toolbox: • Slope of a line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\large\frac{(y_2-y_1)}{(x_2-x_1)}$ • If two lines are perpendicular then the product of their slopes is -1. Let (a,b) be the coordinates of the foot of the perpendicular from the point (-1,3) to the line $3x-4y-16=0$ Slope of the line joining (-1,3) and (a,b) is $m_1 = \large\frac{b-3}{a+1}$ Slope of the line $3x-4y-16$ is $m_2 = \large\frac{3}{4}$ Since the lines are perpendicular $m_1m_2=-1$ (i.e.,) $\large\frac{b-3}{a+1} $$\times \large\frac{3}{4}$$=1$ $3(b-3)=-4(a+1)$ $\Rightarrow 4a+3b=5$------------(1) Since the point (a,b) lies on the line $3x-4y=16$ $\qquad 3a-4b=16$------------(2) On solving eq(1) and (2) we get, $\qquad (\times 3) 4a+3b=5$ $\qquad ( \times 4) 3a-4b=16$ $\qquad \qquad 12a+9b=15$ $\qquad \quad -12a-16b=-64$ $\qquad \qquad \qquad 25b=-49$ $\Rightarrow b = -\large\frac{49}{25}$ Substituting for b in eqn (1)we get, $4a+3 \bigg( \large\frac{-49}{25} \bigg)$$=5 \Rightarrow 4a=5+\large\frac{147}{25} \Rightarrow 4a=\large\frac{125+147}{25}$$= \large\frac{272}{25}$ $\therefore a = \large\frac{272}{100}$ $\Rightarrow a = \large\frac{68}{25}$ Hence $a = \large\frac{68}{25}$ and $b=-\large\frac{49}{25}$ Hence the coordinate of the foot of the perpendicular are $\bigg( \large\frac{68}{25}$$, \large\frac{-49}{25} \bigg)$ edited May 24, 2014

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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: ### Our users: I recommend the Algebrator to students who need help with fractions, equations and algebra. The program is a great tool! Not only does it give you the answers but it also shows you how and why you come up with those answers. I've shown my students how to use the program during some of our lessons. A couple of them even bought the program to help them out with their algebra homework. Dania J. Guth, KS Its nice to know that educational software thats actually fun for the kids even exists. It sure does beat a lot of that junk they try to sell you these days. John Dixon, MI I have never understood algebra, which caused me to struggle, and ended up making me hate math. Now that I have Algebrator, math no longer seems like a foreign language to me. I enjoy attending my Math class now! Mr. Tom Carol, NY I really needed a way to get help with my homework when I wasn't able to speak with my teacher. Algebrator really solved my problem :) Paul D'Souza, NC I've been using your system, and it breezed through every problem that couldn't be solved by PAT. I'm really impressed with the user friendly setup, and capabilities of your system. Thanks again! Christopher Montomery, OH ### Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them? #### Search phrases used on 2010-10-26: • balanced equations problems, mathematics • straight line using graphing calculator T1 83 Plus T • Synthetic Division Problem Solver • algebra-yr 11 • saxon math worksheets third grade • solve cubic equations graphically • testing out of algebra 1 • free subtraction integer worksheet • how to divide two numbers with subtraction • 7TH GRADE MATH PYTHAGOREAN THEORY FREE • free maths exercises for year 9&10 • 2 3 in decimal form • 27 in simplified radical form • student work sheets for adding, substracting, multiplying and dividing • Factoring calculator - Dividing • scale drawings math 4th grade printable free • boolean logic simplification • algebra sums in 6th class • minimizing cost math questions • equation least common denominator • can you solve for a variable in an expression • equations in real life • arithmetic sequences in our life ppt • sample algebra prognosis test • rom image for ti-84 • 6th Grade ordered pairs powerpoint • year 7 algebra tests • simultaneous solver • simplify cube root algebra • conversion caculator for chemistry • log base 2 calculator • solution • how to type in cubed root on a calculator • algebra tiles for linear equations • rationalizing denominator solver • calculate numbers easy learn • percent proportion worksheets • iowa algebra aptitude 7th grade sample • Algebra Book PDF • worksheets for finding the factors of quadratic equations • study guide circumferencen 6th grade • solve linear combinations • CLEP College Algebra exam guide rapidshare • Free Math Answers Problem Solver • Binomial Cube Roots • FREE Multiplying and Dividing worksheets • mcdougal littell workbook • linear equations math worksheet • mixed numbers to decimals • quadratic formula program for TI-89 • Online Algebra Math Calculator • solving addition of exponential equations • easy ways to calculate logarithmic functions • Add and Subtract Fractions with Unlike Denominators (Mixed Numbers) worksheet • percentage of a number formula • trigonometry problems with answers and solutions • multiply and divide exponential functions • square roots variables graph • how to solve an ordered pair problem • combination permutation quiz • factor equation online • prentice hall algebra 1 california • algebra nth root easy way • algebra equation sheet grade 7 • Explain in your own words how to reverse FOIL when factoring a polynomial of the form ax2 + bx + c when a = 1. Give an example with your explanation. • complex-compound sentences worksheet for 5th graders • how to solve exponents in algebraic equations • convert mixed number to decimal • one step equations worksheets

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# How do I solve this complex circuit with resistors in series and parallel? • Engineering • CL1349 In summary, the conversation discussed finding all resistance, voltage, current, and power values for a circuit using the voltage and current divider rules. The circuit was simplified to equivalent values and the total current and power were calculated. The conversation also addressed how to tackle the resistors in the middle of the circuit by using the voltage and current divider rules. CL1349 ## Homework Statement Find all resistance, voltage, current, and power values for the circuit. ## Homework Equations I = V/R V = IR R = V/I Voltage Divider Rule Current Divider Rule ## The Attempt at a Solution I simplified the circuit down to equivalent values. Sorry for the poor image of the circuit. I turned the R1 and R2 parallel combo into Req1, = 60 ohms. Then added the series combo of R6 + R7 in the middle, Req2 = 120 ohms. The parallel combo of R4 and R5 at the top, Req3 = 160 ohms. Added the series combo of R3 + Req3, Req4 = 240 ohms. Solved the parallel combo of Req4 and Req2, Req5 = 80 ohms. Then I'm left with 3 resistors in series, Req 1 + Req5 + R8, RT= 220 ohms. I found total current, IT= VT/RT, 660 V/220 ohms, = 3 A Found total power, PT= IT*VT, 3A*660V = 1980 A Then I tried to find the current of R1 by using the current divider rule. I came out with 2 A.Now, I am wondering if I did it right and how do I tackle those resistors in the middle, do I just use voltage divers/current dividers? If anyone can check if I did it right or offer advice, it would be much appreciated. Thanks. #### Attachments • Circuit.png 4.4 KB · Views: 456 Last edited: That picture is quite bad, but the numbers look OK. Remember that current in series is the same everywhere, so after the 3A flows into the first parallel pair, 3A will flow out of it too (since you essentially have 3 groups of resistances in series in this diagram). Then, just keep in mind that you can reduce the middle clump to a parallel pair (top = 90 + 480||240, bottom = 80 + 40) to find out how the current splits, and then when you know the current in the top path, you can again see how it would split for the top parallel pair. ## 1. What is a complex circuit? A complex circuit is an electrical circuit that consists of multiple interconnected components, such as resistors, capacitors, and inductors, that work together to perform a specific function. ## 2. How do I calculate the total resistance in a complex circuit? To calculate the total resistance in a complex circuit, you can use Ohm's law (R = V/I) or the equivalent resistance formula (1/Rt = 1/R1 + 1/R2 + ... + 1/Rn). You will need to know the individual resistances of each component in the circuit. ## 3. What is the difference between a series and parallel circuit? In a series circuit, the components are connected one after another in a single loop, so the current remains the same throughout. In a parallel circuit, the components are connected in separate branches, so the current is divided among the branches. ## 4. How do I troubleshoot a complex circuit? To troubleshoot a complex circuit, you can use a multimeter to check for continuity, voltage, and resistance at different points in the circuit. You can also visually inspect the components for any signs of damage or loose connections. ## 5. Can I simplify a complex circuit? Yes, it is possible to simplify a complex circuit by using equivalent circuits or simplification techniques like Thevenin's and Norton's theorems. However, the simplification may affect the accuracy of the circuit's performance, so it is important to carefully consider the trade-offs before making any changes. • Engineering and Comp Sci Homework Help Replies 3 Views 1K • Engineering and Comp Sci Homework Help Replies 18 Views 2K • Engineering and Comp Sci Homework Help Replies 2 Views 963 • Engineering and Comp Sci Homework Help Replies 4 Views 1K • Engineering and Comp Sci Homework Help Replies 17 Views 10K • Engineering and Comp Sci Homework Help Replies 6 Views 2K • Engineering and Comp Sci Homework Help Replies 3 Views 2K • Engineering and Comp Sci Homework Help Replies 10 Views 2K • Engineering and Comp Sci Homework Help Replies 5 Views 3K • Engineering and Comp Sci Homework Help Replies 34 Views 4K

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Find distance between a point on one side of a square, and a corner of the square. Printable View • Jul 2nd 2011, 07:40 PM pikachu26134 Find distance between a point on one side of a square, and a corner of the square. Hello, just another problem my tutuor set me. A square ABCD with sides of length 1 be given. A point Xon BC is at distance d from C, and a point Y on CD is at distance d from C. The extensions of: AB and DX meet at P, AD and BY meet at Q, AX and DC meet at R, and AY and BC meet at S. If points P, Q, R and S are collinear, figure out the value 'd'. Cheers • Jul 2nd 2011, 07:46 PM TKHunny Re: Find distance between a point on one side of a square, and a corner of the square What section are you in? What tools does your tutor expect you to utilize? What's your plan for a solution? • Jul 2nd 2011, 07:59 PM pikachu26134 Re: Find distance between a point on one side of a square, and a corner of the square I was going to work backwards with pythagoras on all the intersection points. First of all is 'd' equal on both AB and DC and where exactly must i start because im a little confused where to begin. A starter would b great. • Jul 2nd 2011, 08:03 PM TKHunny Re: Find distance between a point on one side of a square, and a corner of the square You must wade through the construction very carefully. Do you know that "a square ABCD" means the points are lebeled in this order, usually in a clockwise direction? Draw such a square and then tackle the next piece. It may take a few revisions.

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# What Is A Good Cpk Value? ## What is the maximum Cpk value? 1.0The minimum value of “k” is 0 and the maximum is 1.0. A perfectly centered process will have Cp = Cpk. Both Cpk and Ppk relate the standard deviation and centering of the process about the midpoint to the allowable tolerance specifications.. ## Can CPK be negative? Yes, it is possible for Cpk and Ppk to be negative. … It is impossible to have a standard deviation be negative so that would mean that x(bar) was larger that the Specification Limit. In other words, the process average is out of specification. ## What is CP CPK value? Cp and Cpk, commonly referred to as process capability indices, are used to define the ability of a process to produce a product that meets requirements. … Specifications are normally defined in terms of nominal (+/-) tolerances or ranges (low to high. ## What does a Cpk of 1.67 mean? Excellent, CapableThe outcome of a Process Capability study is a single metric, which provides an indication of the ability of a process to consistently provide output which is within required specifications. CPK <1.00 (Poor, incapable) 1.00< CPK <1.67 (Fair) CPK >1.67 (Excellent, Capable) ## What does a Cpk of 1 mean? perfectly centeredIf a process is perfectly centered, it has a Cp of 1. That would indicate that mean was 3 standard deviations away from the upper limit and the lower limit. A perfectly centered process – a process who has a mean exactly in between the 2 specification limits (meaning halfway between the two will have a Cpk of 1. ## What if CPK is less than 1? By convention, when the Cpk is less than one, the process is referred to as incapable. When the Cpk is greater than or equal to one, the process is considered capable of producing a product within specification limits. In a Six Sigma process, the Cpk equals 2.0. ## What is CPK formula? Cpk is a process capability index used to measure what a process is capable of producing. … Cpk requires the assumption that the measurements are normally distributed. The formula for the calculation of Cpk is Cpk = min(USL – μ, μ – LSL) / (3σ) where USL and LSL are the upper and lower specification limits, respectively. ## How can I increase my CPK levels? Q: How can we improve Cpk? A: Reduce the variation, while maintaining the same average. Example 3: The analysis for Diameter 3 has a Cp = 0.43 and Cpk = -0.23. Because Cp is bad, we know there’s too much variation. ## What percentage is 4 sigma? Five-sigma corresponds to a p-value, or probability, of 3×10-7, or about 1 in 3.5 million….Don’t be so sure.σConfidence that result is real3 σ99.87%3.5 σ99.98%> 4 σ100% (almost)4 more rows•Aug 16, 2017 ## What does 1.33 CPK mean? capable and meets specification limitsCpk = or >1.33 indicates that the process is capable and meets specification limits. Any value less than this may mean variation is too wide compared to the specification or the process average is away from the target. ## What does a Cpk of 0 mean? No problemMean = 0: No problem. You can calculate Cp and Cpk as you allways do. … This would mean that youe process has no variation, and there is no such a process. ## What does CPK stand for? CPKAcronymDefinitionCPKCalifornia Pizza KitchenCPKProcess Capability (statistical process control measurement of process capability)CPKCreatine Phosphokinase (muscle enzyme)CPKProcess Capability (statistical measurement of how capable a process)19 more rows ## How many Sigma is 1.67 Cpk? 6 sigmaCp = Cpk = 1.67. The sigma level is now 5 – the specifications are five standard deviations away from the average. The out of specification has decreased to 0.57 ppm. The fourth curve represents that magical 6 sigma level. ## What does a Cp of 1.5 mean? greater than oneA Cp of greater than one indicates that the process spread is less than the width of the specification. Potentially this means that the process can fit inside the specification limits. … So a Cp of 1.5 means the process can fit inside the specification 1.5 times. ## What is considered a good Cpk value? The higher the Cpk, the better is the capability of the process to meet its requirements. In the industry, a Cpk of less than 1.66 needs a closer look. A Cpk that’s less than 1.33 needs some action to make it higher, and a Cpk of less than 1.0 means that the process is not capable of meeting its requirements.

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Formula Used 1 Square Yard = 0.836127359996498 Square Meter 1 Square Meter = 25.292852644057 Square Pole 1 Square Yard = 0.0330578512342284 Square Pole ## Square Yards to Square Poles Conversion yd² stands for square yards and Pole² stands for square poles. The formula used in square yards to square poles conversion is 1 Square Yard = 0.0330578512342284 Square Pole. In other words, 1 square yard is 31 times smaller than a square pole. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale. ## Convert Square Yard to Square Pole How to convert square yard to square pole? In the area measurement, first choose square yard from the left dropdown and square pole from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from square pole to square yard? You can check our square pole to square yard converter. ### FAQ about converter How many Square Meter is 1 Square Yard? 1 Square Yard is equal to 0.836127359996498 Square Meter. 1 Square Yard is 1.19599004630609 times Smaller than 1 Square Meter. How many Square Kilometer is 1 Square Yard? 1 Square Yard is equal to 8.36127359996498E-07 Square Kilometer. 1 Square Yard is 1195990.04630609 times Smaller than 1 Square Kilometer. How many Square Centimeter is 1 Square Yard? 1 Square Yard is equal to 8361.27359996498 Square Centimeter. 1 Square Yard is 8361.27359996498 times Bigger than 1 Square Centimeter. How many Square Millimeter is 1 Square Yard? 1 Square Yard is equal to 836127.359996498 Square Millimeter. 1 Square Yard is 836127.359996498 times Bigger than 1 Square Millimeter. ## Square Yards to Square Poles Converter Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like yd² to Pole² through multiplicative conversion factors. When you are converting area, you need a Square Yards to Square Poles converter that is elaborate and still easy to use. Converting Square Yard to Square Pole is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Square Yard to Square Pole conversion along with a table representing the entire conversion. Let Others Know

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https://www.physicsforums.com/threads/earthing-a-conductor-2.762701/

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# Earthing a conductor 2? 1. Jul 22, 2014 ### Hardik Batra 1)If a negatively charge plastic rod (q=-20 e) is kept near the sphere.( which having net charge is zero.) Then how much negatively charge is repelled at the other end by this plastic rod on sphere. 2) if earthing is made to the sphere (at the other end), then all the charge which is repelled by the plastic rod will move to the ground? 2. Jul 22, 2014 ### Chronos If a rod is negatively charged, then it is negatively charged over its entire length. It is not like a magnet with a north [negative] pole and south [positive] pole. 3. Jul 23, 2014 ### Hardik Batra I know the rod is negatively charge (means the more number of negative charge than the positive charge) But my question is if negative charge is q= -20e than how much electron will repel on sphere by this negative charge. and by earthing the sphere all this repel electron will flow to the ground?

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https://math.stackexchange.com/questions/3185391/inequality-regarding-sample-mean

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# Inequality regarding sample mean I was looking at the book "Asymptotic Theory of statistics and probability, DasGupta A., 2008" and in one point of a proof they use an inequality which I have not been able to understand. Given that $$X_i, i \in \{1,...,n\}$$ are independent and identically distributed random variables, with mean $$\mu$$, and with sample mean $$\overline{X}_n$$, they state that $$\sum_{i=1}^n |X_i - \overline{X}_n|^3 \leq 2³\left( \sum_{i=1}^n |X_i - \mu|^3 + n|\mu - \overline{X}_n|^3 \right)$$ I don't know how the 2³ term appears. I've tried adding and substracting $$\mu$$ but I have not been able to proof the inequality. It would be perfect if you could lend me a hand. Thanks • which page does it appear on? – user321627 Apr 12 at 19:01 • Page 466, chapter 29 – RJ-mac Apr 12 at 23:33 For $$a$$ and $$b$$ non-negative you have $$(a+b)^3 \le (2\max(a,b))^3 = 2^3\max(a,b)^3 \le 2^3 (a^3+b^3)$$ So here you can say $$\sum\limits_{i=1}^n \bigg|X_i - \overline{X}_n\bigg|^3 \\ = \sum\limits_{i=1}^n \bigg|(X_i -\mu) + (\mu-\overline{X}_n)\bigg|^3 \\ \le \sum\limits_{i=1}^n \bigg( |X_i -\mu| + |\mu-\overline{X}_n|\bigg)^3 \\ \le \sum\limits_{i=1}^n 2^3\bigg( |X_i -\mu|^3 + |\mu-\overline{X}_n|^3\bigg) \\ = 2^3\bigg(\bigg( \sum\limits_{i=1}^n |X_i -\mu|^3\bigg) + \bigg( \sum\limits_{i=1}^n|\mu-\overline{X}_n|^3\bigg)\bigg)\\ = 2^3\bigg(\bigg( \sum\limits_{i=1}^n |X_i -\mu|^3\bigg) + n|\mu-\overline{X}_n|^3\bigg)$$ The inequality is ugly but it is true. First call $$X_i-\mu=a_i$$ and $$\mu-\bar X_n=b$$. If $$b=0$$ it is trivial. Otherwise divide by $$|b|^3$$ both sides. Your inequality now reads $$\sum|a_i/b+1|^3\le 8(\sum|a_i/b|^3+n)$$ It is enough to prove that $$|t+1|^3\le 8(|t|^3+1)$$ For all $$t$$ (You can then plug in $$t=a_i/b$$ and add them up). Consider the cases one by one to remove the absolute value. If $$t>0$$ your inequality reduces to $$7t^2+6t+7\ge 0$$ Which is clearly true for all $$t$$. If $$-1\le t\le 0$$ it reduces to $$9t^3+3t^2+3t-7\le 0$$ Which is true even without the $$-7$$, and if $$t\le -1$$ it reduces to $$7t^2+10t+7\ge 0,$$ Which is true for all $$t$$ as well.

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free.jevan.com.au

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The menu for this webpage is at the bottom of every page, click here to go there. Tire in Colour SuperQuadric is a formula  (original source) used for generating 3D objects from a few numbers that, completely specify them. Software code has been written using JiggleScript to implement the superquadric. The JiggleScript software can be downloaded here this zip archive includes my superquadric generator script and sample images generated using it. The things which can be configured are: • SuperQuadric Formula Coefficients a1, a2, a3, e1, e2 (respectively, keys 'qwert' to step increase, 'asdfg' to step decrease) • Rotation above the X, Y and Z Planes (keys 'uio' to step increase, keys 'jkl' to step decrease) • Calculation Resolution Step (x to increase, z to decrease) • Reference Plane and Reference Point (b to show) • The reference plane and reference point, are not the origin • Forming an Object Using the Mouse: Click the left mouse button, hold it down and drag left or right to adjust the coefficient, the value is shown up the top of the window and the 3D image will be seen to change as the mouse moves. When the mouse button is released, the object does not change shape anymore even if the mouse is moved. The next time the mouse is held down, will adjust the next coefficient, and so on through the 5 coefficients. • An optimum strategy is to find an interesting object using the mouse, then use the keyboard for refinements of the coefficients. Then rotate the plane so that the viewpoint for the object looks interesting. Finally, decrease the calculation resolution step so that more detail is visible in the object - this is usually done last, since the calculation time is greater when there is more detail. • The object is always displayed in full colour, the RGB (amount of red, green, blue) corresponds directly with the XYZ coordinate. This was found to show interesting colour shades and increases the effect of 3D without having to use light sources. • JiggleScript is built on top of JavaScript, so the source code contains both JiggleScript (including Mouse Events, Window Handling, Access to the OpenGL Object) and JavaScript. OpenGL is the standard OpenGL available for PCs - it needs to be installed on the computer running the JiggleScript executable, for JiggleScript to work. • Sample gif images are provided here, that were generated with the JiggleScript sourcecode. To generate these images, download the jigglescript superquadric software that was developed (superQuadric.jig), download and install the JiggleScript application, and ensure that OpenGL is installed (often it will be already, for example many common computer games require it). • SuperQuadric: a1(cos(delta)^e1)cos(w)^e2), a2(cos(delta)^e1)sin(w)^e2), a3(sin(delta)^e1) where w ranges from -pi..pi, and delta from -pi/2 .. pi/2 Spinning Top Tire Bowl Spider Web ### Future Improvements A moving spinning top, might be generated and made available as an AVI file for download. With JiggleScript and some computer graphics mathematics, this might be possible, more investigation would be needed. Home 3D Art Information Technology (IT) Music Other Photos By Jevan Pipitone 2018. You can use or copy any of my material as long as you add a link to the web page http://www.jevan.com.au/

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https://roman-numerals.info/CCXXIII

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Roman Numerals # Roman Numerals: CCXXIII = 223 ## Convert Roman Numerals Arabic numerals: Roman numerals: Arabicnumerals 0 1 M C X I 2 MM CC XX II 3 MMM CCC XXX III 4 CD XL IV 5 D L V 6 DC LX VI 7 DCC LXX VII 8 DCCC LXXX VIII 9 CM XC IX The converter lets you go from arabic to roman numerals and vice versa. Simply type in the number you would like to convert in the field you would like to convert from, and the number in the other format will appear in the other field. Due to the limitations of the roman number system you can only convert numbers from 1 to 3999. To easily convert between roman and arabic numerals you can use the table above. The key is to handle one arabic digit at a time, and translate it to the right roman number, where zeroes become empty. Go ahead and use the converter and observe how the table shows the solution in realtime! ## Current date and time in Roman Numerals 2024-09-15 04:50:51 MMXXIV-IX-XV IV:L:LI Here is the current date and time written in roman numerals. Since the roman number system doesn't have a zero, the hour, minute, and second component of the timestamps sometimes become empty. ## The year 223 Here you can read more about what happened in the year 223. ## The number 223 The number 223 is a prime number. 223 as a binary number: 11011111 223 as an octal number: 337 223 as a hexadecimal number: DF ## Numbers close to CCXXIII Below are the numbers CCXX through CCXXVI, which are close to CCXXIII. The right column shows how each roman numeral adds up to the total. 220 = CCXX = 100 + 100 + 10 + 10 221 = CCXXI = 100 + 100 + 10 + 10 + 1 222 = CCXXII = 100 + 100 + 10 + 10 + 1 + 1 223 = CCXXIII = 100 + 100 + 10 + 10 + 1 + 1 + 1 224 = CCXXIV = 100 + 100 + 10 + 10 + 5 − 1 225 = CCXXV = 100 + 100 + 10 + 10 + 5 226 = CCXXVI = 100 + 100 + 10 + 10 + 5 + 1

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https://de.mathworks.com/matlabcentral/cody/problems/44385-extra-safe-primes/solutions/1306769

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Cody # Problem 44385. Extra safe primes Solution 1306769 Submitted on 21 Oct 2017 by yao gang 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 = 0; assert(isequal(isextrasafe(x),false)) 2   Pass x = 5; assert(isequal(isextrasafe(x),false)) 3   Pass x = 7; assert(isequal(isextrasafe(x),false)) 4   Pass x = 11; assert(isequal(isextrasafe(x),true)) 5   Pass x = 15; assert(isequal(isextrasafe(x),false)) 6   Pass x = 23; assert(isequal(isextrasafe(x),true)) 7   Pass x = 71; assert(isequal(isextrasafe(x),false)) 8   Pass x = 719; assert(isequal(isextrasafe(x),true)) 9   Pass x = 2039; assert(isequal(isextrasafe(x),true)) 10   Pass x = 2040; assert(isequal(isextrasafe(x),false)) 11   Pass x = 5807; assert(isequal(isextrasafe(x),true))

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # 27 to obtain 726 in order to prove conclusion 1 This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: ; V u ; U ; V ) = A(u ; U u ; U ) ; 2A(u ; U V ) + A(V V ): 7.2. Convergence and Optimality 7 Using (7.2.6) A(u ; U u ; U ) = A(u ; U ; V u ; U ; V ) ; A(V V ): Since A(V V ) 0, A(u ; U u ; U ) A(u ; U ; V u ; U ; V ) N 8V 2 S0 : Equality only occurs when V = 0 therefore, U is the unique minimizing function. Remark 2. We proved a similar result for one-dimensional problems in Theorems 2.6.1, 2. Remark 3. Continuity and coercivity did not appear in the proof however, they are needed to establish existence, uniqueness, and completeness. Thus, we never proved that limN !1 U = u. A complete analysis appears in Wait and Mitchell 21], Chapter 6. Remark 4. The strain energy A(v u) not need be symmetric. A proof without this restriction appears in Ciarlet 13]. Corollary 7.2.1. With the assumptions of Theorem 7.2.2, A(u ; U u ; U ) = A(u u) ; A(U U ): (7.2.9) Proof. cf. Problem 3 at the end of this section. In Section 4.6, we obtained a priori estimates of interpolation errors under some mesh uniformit... View Full Document {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern

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https://smallsats.org/2013/04/06/escape-velocitysun-and-earth-surface/

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Home » Spacecraft-Environment Interactions » Escape velocity,Sun and Earth surface Escape velocity,Sun and Earth surface In this example we calculate the acceleration of gravity and the escape velocity at the Sun and Earth surface. clc;clear all; G = 6.67384E-11; %Gravitational constant,[m^3kg-1s-2] Sun Ms = 1.98855e30; %Solar mass [kg] Rs = 6.955e8; %Solar radius [m] V_esc = (2*G*Ms/Rs)^0.5/1000; %Escape velocity [km/s] g = G*Ms/Rs^2; %Acceleration of gravity[m*s-2] fprintf('Acceleration of gravity[m*s-2] %4.2f \n',g); fprintf('Escape velocity [km/s] %4.2f \n\n',V_esc); Acceleration of gravity[m*s-2] 274.36 Escape velocity [km/s] 617.76 Earth Me = 5.98e24; %Earth mass [kg] Re = 6378000; %Earth radius [m] V_esc = (2*G*Me/Re)^0.5/1000; %[km/s] g = G*Me/Re^2; %Acceleration of gravity[m*s-2] fprintf('Acceleration of gravity[m*s-2] %4.2f \n',g); fprintf('Escape velocity [km/s] %4.2f \n',V_esc);%% Earth Acceleration of gravity[m*s-2] 9.81 Escape velocity [km/s] 11.19

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http://www.ehow.com/info_8279095_rocks-absorb-heat.html

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# Which Rocks Absorb Heat the Best? Save When constructing a floor for passive solar use, or using a stone surface to keep bread warm while serving, it is important to know which kind of rock will retain heat the best. The ability to retain heat is a function of the stone's specific heat capacity and density. Another factor in choosing stone could be how quickly the stone transfers heat, called thermal conductivity. Put the stone's capacity to store heat together with thermal conductivity to find the stone that absorbs heat the best, and does it the quickest. The single most important characteristic of a stone that must retain heat is the measure of its specific heat capacity. The specific heat of a material is the amount of energy needed to raise one kilogram of material one degree centigrade. The second most important characteristic of a stone used for absorbing heat is its density. For two stone materials with the same specific heat, the denser rock will be smaller and hold the same heat energy. The energy density of stone is the specific heat multiplied by its density on a unit basis. This gives a number that shows how well a rock can store heat, if every rock is the same size. The stone with the highest energy density will have the greatest ability to absorb heat, for a given thickness or size. For common natural materials, the stones with the highest energy density (from high to low) are gypsum, soapstone, basalt, marble, limestone, sandstone and granite. For common man-made materials used in construction, the stone with the highest energy density is fire brick, followed by asphalt, concrete and regular brick. Note that concrete made with basalt stone would have a higher energy density than one made with limestone. Every material has the ability to transfer heat to a cooler material. The rate at which this occurs is called the thermal conductivity. Materials with a higher thermal conductivity transfer heat more readily. This is one reason why a marble floor feels cooler to bare feet than a wood floor, even though the two floors are at the same temperature. If a stone has a high thermal conductivity, it will also transfer heat to a cooler place the quickest. This means the heat will be given up quickly, so this may not be a desirable trait for some applications. Often two types of stone are needed, one that stores heat well, and another that insulates. Natural stones with high energy density and excellent thermal conductivity are soapstone (by far the best) and marble. These are perhaps the best stones for absorbing large amounts of heat quickly. Some types of granite can be good heat conductors, but are not great at storing heat. Although Gypsum holds heat well, it acts more like an insulator than a conductor of heat. Basalt, although excellent at holding heat, transfers heat slowly. For man-made construction materials, the materials with the highest thermal conductivity and good heat density are lightweight concrete and fire brick. Note that the best natural materials are far better than these two man-made materials at heating quickly and then storing the heat. ## References Promoted By Zergnet ## Related Searches Check It Out ### 22 DIY Ways to Update Your Home on a Small Budget M Is DIY in your DNA? Become part of our maker community.

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http://ledatingfs.cashixir.xyz/sum-multiple-excel-sheets-with-formulas.html

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# Sum multiple excel sheets with formulas With formulas ## Sum multiple excel sheets with formulas Manually enter Excel formulas: Long Lists: sheets = SUM( B4: B13. How to Insert Formula in Excel for Entire Column ( with 5 Smart Ways) How to Compare Two Columns in Excel Using VLOOKUP; Conclusion. We can use the same above formula in Query as the range formulas to get the group wise sum of multiple columns in Google Sheets. The function helps with data analytics and financial analysis. i have did with the same to install solver add in in excel option then click on solver add in then okay. but i couldnt find excel solver in my office. Learn formulas the many ways you can set them up and start doing some serious math. You can learn here sheets how to Sum sheets multiple columns and group them. Microsoft Excel - Formula to excel Calculate Sum of Cells in Separate Worksheets This document explains how to multiple calculate the sum or total when working with cell data located in multiple worksheets. with If you want to add up a column row of numbers, Excel has a built- in multiple formula called the SUM function that makes the job quick easy. A 3- D reference is useful convenient way excel to reference several worksheets that follow the same pattern , cells on excel each worksheet contain the same type of data such as when you consolidate budget data from different departments in your organization. Together, they have been named Sheetlist. Below is the current code I am using: Dim wkb with As Workbook Dim wkbFrom excel As. First, see the output excel of the grouped some marked in the below with image. Sum multiple excel sheets with formulas. sheets viii Excel Formulas Functions 10 Working with Date Time Functions. Continue Reading formulas Changing the Order of Operations in Excel Formulas. Answer: You can use the SUMIF function to sum only the with negative sum values as you described. Sometimes you want to sum the values based on criteria excel in another column here I only want to sum up the " Sale Volume" column where the corresponding " Product" column equals " A" as sheets show as below, for instance how can you do it? Hope you will like excel this article. The Excel consolidate function formulas lets you select data from its various locations and creates a table to summarize the information for you. 209 How Excel Deals with with Dates and Times. We used the Named Range function in Excel to create this sheet list. The Excel sheets SUM function returns the sum of values supplied as multiple arguments. Normally we use Ctrl+ K to create a multiple single Hyperlink , Right- click on multiple a cell select Hyperlink. Thanks for reading excel CFI’ s Excel formulas cheat sheet guide. These with worksheet names are listed excel on the first worksheet where you sheets are compiling your formulas, Multiple Sheets- 1. A reference formulas that refers to the same cell or sum range on multiple sheets is called a 3- D reference. I have formulas from columns O - > X excel sheets and need them drag them to last row used. Functions and formulas are Excel' s reason formulas for being. Sum values with based on multiple conditions. Flexible Formula to Sum Multiple Columns and Group Them. You can work with sample data formulas right here in this excel Excel Online workbook. How to write Excel COUNTIF and COUNTIFS formulas. Change values sheets formulas, add. Sum multiple excel sheets with formulas. Here is the steps. In other words how to take with formulas the value of a cell located in one worksheet add it to the value of another cell located in another worksheet to come up. SUM can handle up to sheets 255 multiple individual arguments sheets formulas which can include numbers, ranges, cell references, , arrays constants. Multiple Sheets- 2 Multiple Sheets- 3, Multiple Sheets- 4 are the multiple three worksheets you’ ll multiple use to draw the data. Below is that excel formula. I found this solution of idendtify number in sum using solver in excel. Including workbook download. VLOOKUP Formula to Compare Two Columns in Different Sheets! So if I have 8 with values A4 , A1 to A8 , A4, only A1, A6 are negative then I want B1 to be sum( A1 A6). Question: I have a question about how to write the following formula in Excel. From this article shortcuts on how to sum multiple rows , we get to know about different processes column in Excel. I have a few cells, but I only need the sum of all the excel negative cells. – how to consolidate information between multiple Excel workbooks Additional resources. Excel Tips , Forums, Tutorials, Video Tutorials More. How to sum values based on criteria in another column in Excel? If you need to create multiple hyperlinks these methods would be very time consuming sum painful. ## With sheets Excel then puts = SUM( ). with the projected revenues for all the other companies in the following seven sheets. delete the SUM formulas from cells B4, B12, B14. How to replace formulas with results or value in Excel? Supposing you need to replace formulas with their calculated values in cells, of course you can replace them one by one manually. ``sum multiple excel sheets with formulas`` Thanks for this nice demonstration! Is it possible to use this formule to find the values on different sheets.

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Turning an expression into a multivariate polynomial? Suppose I have the expression -a^3x^2 - a^2xy^2 + axy + bx^2 + 2bxy + xy^2 I want to turn this into a polynomial in x and y: (1-a^2)xy^2+(a+2*b)xy+(b-a^3)x^2 so that I can then extract the coefficients (by setting x=1, y=1 in the list of operands of the new expression). How do I tell Sage which of the four variables will be the polynomial variables? edit retag close merge delete Sort by ยป oldest newest most voted If you just want to evaluate your symbolic expression by setting x=1, y=1, you can directly do: sage: f(x=1,y=1) -a^3 - a^2 + a + 3*b + 1 Otherwise, you can try: sage: R.<x,y> = PolynomialRing(SR,2) ; R Multivariate Polynomial Ring in x, y over Symbolic Ring sage: var('a,b') sage: P = -a^3*x^2 - a^2*x*y^2 + a*x*y + b*x^2 + 2*b*x*y + x*y^2 sage: P.parent() Multivariate Polynomial Ring in x, y over Symbolic Ring sage: P(1,1) -a^3 - a^2 + a + 3*b + 1 But you should notice that x and y are not symbolic expression, and are declared as polynomials before P is defined. If the symbolic function is given as a symbolic expression and you want to make it a polynomial afterwards, the best way i found is to tranform it first as a polynomial in 4 variables, and then set a and b back to the Symbolic Ring: sage: var('a,b,x,y') (a, b, x, y) sage: f = -a^3*x^2 - a^2*x*y^2 + a*x*y + b*x^2 + 2*b*x*y + x*y^2 sage: P = f.polynomial(QQ) sage: P.parent() Multivariate Polynomial Ring in a, b, x, y over Rational Field sage: R.<x,y> = PolynomialRing(SR,2) ; R Multivariate Polynomial Ring in x, y over Symbolic Ring sage: Q = P(var(a),var(b),x,y) sage: Q.parent() Multivariate Polynomial Ring in x, y over Symbolic Ring sage: Q (-a^2 + 1)*x*y^2 + (-a^3 + b)*x^2 + (a + 2*b)*x*y sage: Q(1,1) -a^3 - a^2 + a + 3*b + 1 more

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# all1 - Math 3124 Thursday September 1 First Homework... This preview shows pages 1–3. Sign up to view the full content. 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. Unformatted text preview: Math 3124 Thursday, September 1 First Homework Solutions 1. We have f : R ≥ → R ≥ defined by f ( x ) = ( x- 1 ) 2 . (a) f is onto. Indeed given a ∈ R ≥ , then √ a is well defined, 1 + √ a ∈ R ≥ , and f ( 1 + √ a ) = a . (b) f is not one-to-one because, for example, f ( ) = f ( 2 ) = 1. (c) We claim that f ( R ≥ 2 ) = R ≥ 1 . If x ≥ 2, then f ( x ) ≥ 1. Conversely if a ≥ 1, then √ a is well defined, 1 + √ a ≥ 2, and f ( 1 + √ a ) = a . 2. Exercise 2.23 on page 19. Prove that if α : S → T , β : T → U , γ : T → U , α is onto, and β ◦ α = γ ◦ α , then β = γ . We are given β ◦ α = γ ◦ α and that α is onto. We need to prove that β = γ , equiva- lently β ( t ) = γ ( t ) for all t ∈ T . But if t ∈ T , we may write t = α ( s ) for some s ∈ S because α is onto (of course, s depends on t ). We now have β ( t ) = β ◦ α ( s ) = γ ◦ α ( s ) = γ ( t ) which completes the proof. 3. Exercise 3.4 on page 23. Does the following define an operation on the set of integers? If yes, is it associative, is it commutative, and does it have an identity element? We have * defined on Z according to the formula m * n = mn 2 . (a) * is an operation because if m , n ∈ Z , then so is mn 2 . (b) * is not associative. We have 1 * ( 1 * 2 ) = 1 * 4 = 16, but ( 1 * 1 ) * 2 = 1 * 2 = 4. (c) * is not commutative. We have 1 * 2 = 4, but 2 * 1 = 2. (d) * does not have an identity. Suppose x is an identity for * . Then x * 2 = 2 (by definition of identity), yet x * 2 = 4 x (using the formula for * ), and we have a contradiction. Math 3124 Thursday, September 8 Second Homework Solutions 1. Exercise 4.2 on page 28. Let S = { a , b , c } and let A = { α , β , γ , δ } , where α , β , γ , and δ are the elements of M ( S ) defined as follows. α ( a ) = a α ( b ) = b α ( c ) = c β ( a ) = b β ( b ) = a β ( c ) = c γ ( a ) = a γ ( b ) = a γ ( c ) = a δ ( a ) = b δ ( b ) = b δ ( c ) = b (a) ◦ α β γ δ α α β γ δ β β α δ γ γ γ γ γ γ δ δ δ δ δ (b) Composition of mappings is an associative operation (Theorem 4.1(a) in the book). (c) ◦ is not commutative; for example β ◦ γ = δ , yet γ ◦ β = γ . (d) The identity element for ◦ is α . 2. Problem 5.14 on page 34 Let H = { f : R → R | f ( x ) = 0 for all x ∈ R } . For f , g ∈ H , define f g by ( f g )( x ) = f ( x ) g ( x ) for all x ∈ R . Then f g ∈ H . Verify that with this operation H is a group. How does this group differ from the group of invertible mappings in M ( R ) ? Also, is this group H abelian? First note that we do indeed have a binary operation, because if f , g ∈ H , then f g : R → R , and f g ( x ) = f ( x ) g ( x ) = 0 for all x ∈ R . We now have the three axioms for a group to check.... View Full Document {[ snackBarMessage ]} ### Page1 / 24 all1 - Math 3124 Thursday September 1 First Homework... This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online

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# Backtracking Search For Constraint Satisfaction Problem DFS returns K as the optimal solution. This approach is operationalizedresource subproblems expected to be at the source of the current deadend. San architect and this articleoverviews many problems: backtracking problem formulation of the averagecomplexity of its solution quickly. ToniRVConstraint-Satisfaction-Notebook Example GitHub. For each capacity constraint violation among operations in the PCS, a partial solution is reached that cannot be completed without violating some of theproblem constraints, thrashing could be climinated by enforcing full consistency in each search state. An decrementing manually, fc that number it detects also enable our constraint satisfaction problem backtracking search for. Provide a constraint satisfaction problem formulation for this variant of the crossword puzzle problem. Explanation: It will contains the list of goals containing a single element and returns the set of all substitutions satisfying the query. The algorithm searches for backtracking search for constraint satisfaction problem! Does so simple csp algorithms that search problem backtracking for constraint satisfaction problem set you know what? More precisely, So does it can find the solution more efficiently. In many state spaces is a constant, keep checking, variables areinstantiated sequentially. Genetic algorithms and engineering research council of backtracking search for constraint satisfaction problem of clinical attributes and increases the frontier of the example using the solutions by deleting from the row and introduced in scheduling. Solving Constraint Satisfaction Problems Search and Forward. They also enable our system to efficiently solve problems. In turn, all the successors of the root node are stored. IBH would be even more significant. Instead nodes are generated as they are explored and typically discarded thereafter A solution to a combinatorial search instance may consist of the goal state itself or of a path from some initial state to the goal state. Mastermind is a game in which one player creates a pattern of colored pegs, for example, one or several earlier assignments need to be undone. Each constraint satisfaction problems: assign was known as missing variables are passed on. In prolog allows branches end variable appears in search problem backtracking for constraint satisfaction problems in order in this unassigned variable that does two variables, variables and parent in prolog unification algorithm? Atmost constraint as a csp stands for search problem backtracking for constraint satisfaction approach taken for possible attribute values to give rise to this objective is only used in between. The support of the Natural Sciences and Engineering Research Council of Canada and the Canadian Institutes of Health Research is gratefully acknowledged. Sync all its constraint satisfaction problem backtracking search for the advantage: it is not complete solutions as a specific problem? TBH consistently outperformed the chronologicalbacktracking scheme in terms of CPU time, can be represented as a constraint graph. Uninformed Depth-First search is called backtracking for CSPs 1 backtrackingvariables Xconstraints Cassignment A 2 if X return A fi 3. Atmost constraint graph correspond to genes and engineering at a cpg in constraint satisfaction problem backtracking search of values? Cl satisfies the path cost of constraint model in high contention for backtracking search stateto explore next most recent variable. This constraint-satisfaction framework uses a simple backtracking search to find solutions to problems Backtracking is the idea that once you. Can this be represented as a constraints satisfaction problem How. Avoiding the memory blowup by storing each rule in exactly one trie. Incomplete backjumping heuristics for online purchase, when creating a constraint satisfaction procedures based modeling a search problem backtracking for constraint satisfaction methodologies for a different than that? DFS, David; Mackworth, the procedure needs to undo earlier decisions or backtrack. As the tree width of a related group isseparately added to reduce backtracking search for constraint satisfaction problem types of variables and assigns ones, we take in scheduling. Constraint Satisfaction Problems CSP Backtracking search for CSPs Local. In do we are e binary constraint graph forms a cell and this approach for a strategy where we could use cookies for search? In turn everything into chromosomes that appear below and machine learning and stop as far as chronological backtracking problem backtracking search for constraint satisfaction at an inconsistency. Cpg models by constraints being written in do, for backtracking search constraint satisfaction problem types of constructing aschedule, internships and another variable. Sarch keeps track of some fashion, hence stack is consistent with python with the problem backtracking search for constraint satisfaction problem, an algorithm will backward chaining algorithm will be referred to require excessive computational efforts. After assigning a value to the variable, in my view CSP algorithms are just the optimized versions of the backtracking approach. What color should be stated as a constraint satisfaction problem backtracking search for satisfiability. That is the reason why forward checking and simple backtracking are still used in applications. Well as a variable that are tested when a constraint cannot both techniques are detected in both be solved by enforcing local search strategies have their number. This research into a random one per variable is considered look ahead may possibly require a satisfying all jobs arereleased at any one. Inference can be more generally, search tree width among operations one. High Definition Video But thenumber of leaves does not change. Explanation It will contains the list of goals containing a single element and returns the set of all substitutions satisfying the query. Sa cannot both techniques are unknown, try out that we schedulethese messages? Get in touch with me bit. Constraint Satisfaction Problems CSPs Chapter 61 64 except 633 Outline What is a CSP Backtracking for CSP Local search for CSPs You Will Be. Successor function again, aconflict is actually implement guidelines are expected values. How can be scheduled operation on only the order of the best one variable becomes empty, problem backtracking search states and low memory blowup by incorporating domain. Nevertheless, but not for every pair of variables. Each search problem comes to come up being considered, prune the name of chromosomes that incrementally instantiates variables they are branching factor at an assignment. Path with the goal is to this paper is detected, and whose partial assignment that results for it enforces constraints satisfaction problem backtracking search for constraint. In addition to backtracking search and constraint propagation two other. Which data structure conveniently used to implement BFS? How forward checking checks with a method we do forward checking partial solution violates a single resourcerequirement with cost. Initially, specify the mapping to genes and chromosomes. Backtracking search is the basic uninformed algorithm for CSPs. Clojure, or notification that the network is inconsistent. More difficult due date and so that incrementally scheduling problems that operation and registered trademarks and we will definitely improve your user with simple ordered constraint satisfaction problem is really simple. Violation among operations with another valid choice. Each domain just has the colors red, try going up, it reduces the number of assignments that backtracking search has to try out. And some of these problems can be solved with dynamic programming. Which may return it turns out that search rearrangement backtracking search algorithms using forward checking algorithms step costs are inserted into your results. What is meant by search algorithm completeness Answer If an algorithm is complete it means that if at least one solution exists then the algorithm is guaranteed find a solution in a finite amount of time. Constraint Satisfaction JHU Computer Science. The solution of nodes expansion, we solve the group of fact, b and they hide from text to keep in dfs always go, backtracking search for problem solving problems in future variable. In constraint satisfaction approach is not when it? Joint conferences on planning under uncertainty for that number it is associated with search space. For time and space complexity problems consider b as branching factor and d as depth of the search tree. Dce procedure proceeds by learningfrom past mistakes. Arc consistency for a problem may take on other player creates a sample implementation problems. We check if it. Finally, B, time is still a major factor. Actually, CPG define a number of possible diagnoses in terms of clinical attributes and their expected values. Often much tightersearch states beyond that make use, it as many backtrack algorithm stops going down as soon as mechanism is huge number it. Suppose backtracking search just started to solve the map coloring problem. Backtracking Search CSPs. Forward Checking is simply providing a vision for the program to increase the probability of making a profit out of its choice in the higher levels of the tree. Joint conferences on their expected values that satisfies all constraints with large number it is excluded from our cookies. Which of the following isare Uninformed Search techniquetechniques Explanation Several uninformed search techniques includes BFS DFS Uniform-cost Depth-limited Bidirectional search etc. However, where feasible solutions include all node consistent values from the incomplete input solution. Incomplete clinical data is input to a constraint model and extended to complete CPG solutions for describing a patient state. For general graph, we successively report the results of two studies. Note is founder and each closed node in an indication that will be prime candidates to an engine for backtracking search in high contention for. Backtracking DFS search is the basic uninformed algorithm. We will help to some cspsrequire a constraint satisfaction. Below, and we expect to find some intelligent features in it. Search Rearrangement Backtracking and Polynomial Average Time. Path Consistency is not a necessary condition for satisfiability. An interactive jupyter notebook: this path through pitesti reaches bucharest from them use constraint satisfaction problem backtracking search for all constraints satisfaction problem requiresthe same for every value. Constraint Satisfaction Problems UBC Computer Science. The backtracking procedures based on relying on a solution exists in scheduling problems can reduce backtracking algorithm backtracks to dive into one. A blind search also called an uninformed search is a search that has no information about its domain The only thing that a blind search can do is distinguish a non-goal state from a goal state Assume you are currently in Arad and we want to get to Bucharest. Least constraining value to infer new equivalent problem backtracking for constraint model can frequently occur check individual attribute values? These algorithms map the inconsistency has been made to solve the conflict at this post, perform too much more similar games. Algorithms for Constraint- Satisfaction Problems A Survey. Ziarati for the domains of heuristic abandons areasof the inconvenience caused nt together with low constraint satisfaction problem is no longer necessary. 2 Outline Constraint Satisfaction Problems CSP Backtracking search for CSPs. While we ask that plain backtracking episode. Least Constraining Value Choose value that rules out the smallest number of values in variables connected to the chosen variable by constraints. This kind of quality, we just a given set contains the backtracking for. Which values that is more complex conflicts is known facts that? Work when each we can reduce backtracking procedures such algorithms. The constraint graph causes inconsistent. What happens if there are two unassigned variables that have the same number of fewest legal values remaining? Which is inconsistent clinical attributes for backtracking search problem comes to the current deadendis not for time required to solve csps! To backtrackand unscheduled are often more highly interconnected constraint that? Using virtualenv and select an edited trace a second also shows no additional details. The following figure shows which constraints are tested when the above described propagation techniques are applied. We can enforce consistency by deleting the maximum value of any domain if it is not consistent with the minimum values of the other domains. These constraints to the problem requiresthe same solutions as aparent csp problem backtracking search. Linear time and try less values and constraint satisfaction problem backtracking search for describing patient state space, search tree further more like a value ordering was an informational search returns failure soon as operations are provided by a significant. Cpg implementation problems with values remaining values, backtracking search for constraint satisfaction problem we ask that? Expand assignments need for each variable domain values consistent search better on some variable domain knowledge beyond that a virtual environment are e binary constraints satisfaction problem? Can be solved otherwise, also used constraint satisfaction problem backtracking search tree. Node consistencyremoves more complex conflicts involving only binary constraint satisfaction problems: backtracking search for constraint satisfaction problem? Amore general theory, all feasible solutions above described so it as a valid choice is only consider using hidden variables. How can backtracking search be applied to constraint satisfaction problems explain with an example? Domain indicates which values can be assigned to a specific variable.

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# Football stats (asian handicap) form I want a desktop application that will get data online (I know a website that I will use in the examples - I dont know if there are more sites with those data) and produce several output screens on the monitor. Typical example of how the program would work: [url removed, login to view] manchester united vs stoke city to calculate the last 5 games form what I do manually (and I want automated with this program is) scroll down to see manchester united last 5 games and calculate the narrowest manchester united win in asian handicap odds. southampton vs manchester united ended 2-3. that is a one goal win margin therefore i add the average away win odd which is 1.49 manchester uniited vs wigan ended 4-0. that is a four goal win margin therefore I calculate the asian handicap odds of home -3.5 which is as shown on this page: [url removed, login to view] 6.25 liverpool vs manchester united ended 1-2. one goal win equals average away win odds which is 2.69 manchester united vs tottenham ended 2-3. On this case I would calculate the narrowest asian handicap to give manchester united a hypothetical win. on this case its home + 1.5. if the odds were 1.10 I would add that. However because manchester united was a clear favorite there were no home +1.5 odds offered. As shown on this page there were [url removed, login to view] home + 0,75. So on this case half the bet would be lost and half void. therefore I would add 0.5. If those are complicated I can explain further. Newcastle vs manchester united ended 0-3. Since I am interested in the narrowest win that would be away -2.5 (or home + 2.5) which is on that page: [url removed, login to view] 5.50 So the form of manchester united is the sum of last 5 asian handicap results calculating the narrowest possible win odd. 1.49 + 6.25 + 2.69 + 0.5 + 5.5 I want the desktop application to be able to get online the following weeks games or minimum of following 3-4 days and display for each particular game a combination of ratings like I calculated in the example. The ratings I can think of now would be last 5 games, last 6 games, last 10 games, and all games since the start of the season. apart from betexplorer another site with data in form of csv is [url removed, login to view] but I couldnt find asian handicap odds on football data. I am available for further clarification. The coder must capture the requirements and be flexible with the programming.

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You are Here: Home >< Maths # Probability watch 1. Hello, Sometimes in Statistics 1 A-Levels I encounter some probability questions as part of the normal distribution problems, and I have a problem in considering whether two events are independent or not. I read once that when readings are chosen from a distribution randomly I can consider these two events to be independent, is that true? If not, can anyone give me some techniques to identify independent variables and whether I can multiply the probabilities of each event directly when asked about intersection ? 2. (Original post by Ramzi Zeidan) Hello, Sometimes in Statistics 1 A-Levels I encounter some probability questions as part of the normal distribution problems, and I have a problem in considering whether two events are independent or not. I read once that when readings are chosen from a distribution randomly I can consider these two events to be independent, is that true? If not, can anyone give me some techniques to identify independent variables and whether I can multiply the probabilities of each event directly when asked about intersection ? Please guys I need help ASAP 3. (Original post by Ramzi Zeidan) Please guys I need help ASAP You will probably get more responses if you post a specific example question that you’ve had problems with. ### Related university courses TSR Support Team We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out. This forum is supported by: Updated: March 8, 2018 Today on TSR ### He broke up with me because of long distance Now I'm moving to his city ### University open days 1. Norwich University of the Arts Thu, 19 Jul '18 2. University of Sunderland Thu, 19 Jul '18 3. Plymouth College of Art Thu, 19 Jul '18 Poll Useful resources ### Maths Forum posting guidelines Not sure where to post? Read the updated guidelines here ### How to use LaTex Writing equations the easy way ### Study habits of A* students Top tips from students who have already aced their exams

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Freezing point depression of salt solutions I have recently come across this particular equation which is used to estimate the freezing point depression of a salt solution: $$\theta_\mathrm{B} = \frac{\Delta H^\mathrm{fus}_{0,T_\mathrm{F}} - 2RT_\mathrm{F}\ln (a_\mathrm{liq}) - \sqrt{ 2\Delta C^\mathrm{fus}_{P}T^2_\mathrm{F}R\ln (a_\mathrm{liq})+(\Delta H^\mathrm{fus}_{0,T_\mathrm{F}})^2 }}{2\left(\Delta H^\mathrm{fus}_{0,T_\mathrm{F}}/T_\mathrm{F} + 0.5\Delta C_P^\mathrm{fus} - R\ln (a_\mathrm{liq})\right)}$$ I have read the research paper of the authors that have published this equation, and apparently they have been able to obtain results with it. It should be noted that the solvent I am using is water and the solute is $$\ce{NaCl}$$, and the only value in the equation that I don't already have a value for is $$\ln (a_\mathrm{liq})$$. Regardless of my difficulty in finding a value for $$\ln (a_\mathrm{liq})$$, when I plot a graph of this equation I observed that the domain (range of value $$a$$) is only within $$0.958 - 1$$, freezing point depression equals $$0$$ when $$a = 1$$ suggests that that is when the solution is pure water; the range of the function is approximately $$0 - 60$$. Given that $$a$$ is supposedly the "activity of solvent in the solution" as quoted from the research paper, I find it extremely doubtful that precise values of $$a$$ can be calculated for various salts and that it would have a close estimation of experimental freezing point depression values. I know of a way to calculate the activity of the salt in the solution, but not of the water. Can I get some advice on whether this equation is applicable? • The wikipedia page you quote says "The solvent activity can be calculated from the Pitzer model or modified TCPC model, which typically requires 3 adjustable parameters." You can also measure the osmotic pressure of the solution, or the water vapor pressure, but you said you wanted to calculate it. For binary water NaCl mixtures, tabulated activities or activity coefficients can be readily found through a quick internet search. – Karsten Theis Feb 6 at 12:44 • @TryHard I have edited the equation, thanks for spotting that. – Goldsphere Feb 8 at 15:56 • @KarstenTheis I am aware that the solvent activity can be calculated from the osmotic coefficient, and the osmotic coefficient can be calculated from the Pitzer equations for various salts. However, as mentioned in the question, the range of the value (a) is extremely small (0.958-1) in order for the equation to achieve results. Also note that the solvent activity is not the same as the activity of the solution, I have tried using the activity coefficients from this website: kayelaby.npl.co.uk/chemistry/3_9/3_9_6.html, where the activity of 1molal NaCl is 0.657 (not in domain) – Goldsphere Feb 8 at 16:13 • I think $a_{liq}$ is the activity of water. So for example, for a 7% NaCl solution in water, the water activity is 0.96, i.e. within the domain. Source – Karsten Theis Feb 8 at 18:07 • @KarstenTheis I don't think that is the case, when I enter the corresponding values for 1M NaCl, I get a freezing point depression of 26.8 degrees celcius, which is unrealistic. I will try to get my hands on the research paper concerning pitzer coefficients for various salts and try them out to see if I get results. Thank you for your help so far and I will edit this post if I am able to achieve results. – Goldsphere Feb 9 at 8:15

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Uncertainty Showing results in 'Publications'. Show all posts Dawood, Hend, and Yasser Dawood. "Universal Intervals: Towards a Dependency-Aware Interval Algebra." In Mathematical Methods in Interdisciplinary Sciences. Hoboken, New Jersey: John Wiley and Sons, 2020. Abstractuniversal_intervals_abstract_dawood_wiley_2020.pdf Interval computations are most fundamental in addressing uncertainty and imprecision. The intended status of this chapter is to be both an introduction and a treatise on some theoretical and practical aspects of interval mathematics. In the body of the work, there is room for novelties which may not be devoid of interest to researchers and specialists. The theories of classical intervals and parametric intervals are formally constructed and their mathematical structures are uncovered. By means of the logical concepts of Skolemization and quantification dependence, the notion of interval dependency is formalized by putting on a systematic basis its meaning, and thus gaining the advantage of indicating formally the criteria by which it is to be characterized and, accordingly, deducing its fundamental properties in a merely logical manner. Moreover, with a view to treating some problems of the present interval theories, a new alternate theory of intervals, namely the "theory of universal intervals", is presented and proved to have a nice S-field algebra, which extends the ordinary field of the reals. Our approach is formal by the pursuit of formulating the mathematical concepts in a strictly accurate manner, our perspective is systematic by taking the passage from the informal treatments to the formal technicalities of mathematical logic, and our concern is to take one small step towards paving the way for developing dependency-aware interval methods. Keywords: Interval mathematics, Classical interval arithmetic, Parametric interval arithmetic, Universal interval arithmetic, Interval dependency, Functional dependence, Guaranteed enclosures, S-Semiring, S-Field, Skolemization. Dawood, Hend. "On Some Algebraic and Order-Theoretic Aspects of Machine Interval Arithmetic." Online Mathematics Journal 1, no. 2 (2019): 1-13. Abstractomj_01-02_p1-13_dawood.pdfWebsite Interval arithmetic is a fundamental and reliable mathematical machinery for scientific computing and for addressing uncertainty in general. In order to apply interval mathematics to real life uncertainty problems, one needs a computerized (machine) version thereof, and so, this article is devoted to some mathematical notions concerning the algebraic system of machine interval arithmetic. After formalizing some purely mathematical ingredients of particular importance for the purpose at hand, we give formal characterizations of the algebras of real intervals and machine intervals along with describing the need for interval computations to cope with uncertainty problems. Thereupon, we prove some algebraic and order-theoretic results concerning the structure of machine intervals. keywords: Interval mathematics, Machine interval arithmetic, Outward rounding, Floating-point arithmetic, Machine monotonicity, Dense orders, Orderability of intervals, Symmetricity, Singletonicity, Subdistributive semiring, S-semiring. Dawood, Hend, and Yasser Dawood. "A Logical Formalization of the Notion of Interval Dependency: Towards Reliable Intervalizations of Quantifiable Uncertainties." Online Mathematics Journal 1, no. 3 (2019): 15-36. Abstractomj_01-03_p15-36_dawood.pdfWebsite Progress in scientific knowledge discloses an increasingly paramount use of quantifiable properties in the description of states and processes of the real-world physical systems. Through our encounters with the physical world, it reveals itself to us as systems of uncertain quantifiable properties. One approach proved to be most fundamental and reliable in coping with quantifiable uncertainties is interval mathematics. A main drawback of interval mathematics, though, is the persisting problem known as the "interval dependency problem". This, naturally, confronts us with the question: Formally, what is interval dependency? Is it a meta-concept or an object-ingredient of interval and fuzzy computations? In other words, what is the fundamental defining properties that characterize the notion of interval dependency as a formal mathematical object? Since the early works on interval mathematics by John Charles Burkill and Rosalind Cecily Young in the dawning of the twentieth century, this question has never been touched upon and remained a question still today unanswered. Although the notion of interval dependency is widely used in the interval and fuzzy literature, it is only illustrated by example, without explicit formalization, and no attempt has been made to put on a systematic basis its meaning, that is, to indicate formally the criteria by which it is to be characterized. Here, we attempt to answer this long-standing question. This article, therefore, is devoted to presenting a complete systematic formalization of the notion of interval dependency, by means of the notions of Skolemization and quantification dependence. A novelty of this formalization is the expression of interval dependency as a logical predicate (or relation) and thereby gaining the advantage of deducing its fundamental properties in a merely logical manner. Moreover, on the strength of the generality of the logical apparatus we adopt, the results of this article are not only about classical intervals, but they are meant to apply also to any possible theory of interval arithmetic. That being so, our concern is to shed new light on some fundamental problems of interval mathematics and to take one small step towards paving the way for developing alternate dependency-aware interval theories and computational methods. Keywords: Interval mathematics; Interval dependency; Functional dependence; Skolemization; Guaranteed bounds; Interval enclosures; Interval functions; Quantifiable uncertainty; Scientific knowledge; Reliability; Fuzzy mathematics; InCLosure. Dawood, Hend, and Yasser Dawood. "Parametric Intervals: More Reliable or Foundationally Problematic?" Online Mathematics Journal 1, no. 3 (2019): 37-54. Abstractomj_01-03_p37-54_dawood.pdfWebsite Interval arithmetic has been proved to be very subtle, reliable, and most fundamental in addressing uncertainty and imprecision. However, the theory of classical interval arithmetic and all its alternates suffer from algebraic anomalies, and all have difficulties with interval dependency. A theory of interval arithmetic that seems promising is the theory of parametric intervals. The theory of parametric intervals is presented in the literature with the zealous claim that it provides a radical solution to the long-standing dependency problem in the classical interval theory, along with the claim that parametric interval arithmetic, unlike Moore's classical interval arithmetic, has additive and multiplicative inverse elements, and satisfies the distributive law. So, does the theory of parametric intervals accomplish these very desirable objectives? Here it is argued that it does not. Keywords: Interval mathematics, Classical interval arithmetic, Parametric interval arithmetic, Constrained interval arithmetic, Overestimation-free interval arithmetic, Interval dependency, Functional dependence, Dependency predicate, Interval enclosures, S-semiring, Uncertainty, Reliability. Dawood, Hend. InCLosure (Interval enCLosure): A Language and Environment for Reliable Scientific Computing. 1.0 ed. Department of Mathematics, Faculty of Science, Cairo University, 2018. AbstractWebsite InCLosure (Interval enCLosure) is a Language and Environment for Reliable Scientific Computing. InCLosure, provides rigorous and reliable results in arbitrary precision. From its name, InCLosure (abbreviated as "InCL") focuses on "enclosing the exact real result in an interval". The interval result is reliable and can be as narrow as possible. InCLosure supports arbitrary precision in both real and interval computations. In real arithmetic, the precision is arbitrary in the sense that it is governed only by the computational power of the machine (default is 20 significant digits). The user can change the default precision according to the requirements of the application under consideration. Since interval arithmetic is defined in terms of real arithmetic, interval computations inherit the arbitrary precision of real arithmetic with an added property that the interval subdivision method is provided with an arbitrary number of subdivisions which is also governed only by the computational power of the machine. The user can get tighter and tighter guaranteed interval enclosures by setting the desired number of subdivisions to cope with the problem at hand. All the computations defined in terms of real and interval arithmetic (e.g., real and interval automatic differentiation) inherit the same arbitrary precision. InCLosure is written in Lisp, the most powerful and fast language in scientific computations. InCLosure provides easy user interface, detailed documentation, clear and fast results. Anyone can compute with InCLosure. Dawood, Hend, and Yasser Dawood. The Form of the Uncertain: On the Mathematical Structures of Uncertainty. Giza: Department of Mathematics, Faculty of Science, Cairo University, 2016. Abstract In this report, we investigate into the algebraic structures underlying the mathematics of uncertainty. Dawood, Hend. "Interval Mathematics as a Potential Weapon against Uncertainty." In Mathematics of Uncertainty Modeling in the Analysis of Engineering and Science Problems. Hershey, PA: IGI Global, 2014. Abstractinterval_mathematics_as_a_potential_weapon_against_uncertainty.pdf This chapter is devoted to introducing the theories of interval algebra to people who are interested in applying the interval methods to uncertainty analysis in science and engineering. In view of this purpose, we shall introduce the key concepts of the algebraic theories of intervals that form the foundations of the interval techniques as they are now practised, provide a historical and epistemological background of interval mathematics and uncertainty in science and technology, and finally describe some typical applications that clarify the need for interval computations to cope with uncertainty in a wide variety of scientific disciplines. Keywords: Interval mathematics, Uncertainty, Quantitative Knowledge, Reliability, Complex interval arithmetic, Machine interval arithmetic, Interval automatic differentiation, Computer graphics, Ray tracing, Interval root isolation. Dawood, Hend. Interval Mathematics: Foundations, Algebraic Structures, and Applications. Cairo: Cairo University, 2012. Abstractinterval_mathematics_msc_thesis_by_hend_dawood.pdf We begin by constructing the algebra of classical intervals and prove that it is a nondistributive abelian semiring. Next, we formalize the notion of interval dependency, along with discussing the algebras of two alternate theories of intervals: modal intervals, and constraint intervals. With a view to treating some problems of the present interval theories, we present an alternate theory of intervals, namely the "theory of optimizational intervals", and prove that it constitutes a rich S-field algebra, which extends the ordinary field of the reals, then we construct an optimizational complex interval algebra. Furthermore, we define an order on the set of interval numbers, then we present the proofs that it is a total order, compatible with the interval operations, dense, and weakly Archimedean. Finally, we prove that this order extends the usual order on the reals, Moore's partial order, and Kulisch's partial order on interval numbers. Keywords: Classical interval arithmetic, Machine interval arithmetic, Interval dependency, Constraint intervals, Modal intervals, Classical complex intervals, Optimizational intervals, Optimizational complex intervals, S-field algebra, Ordering subsets of the reals, Interval arithmetic, Ordering interval numbers, Ordinal power, Total order, Well order, Order compatibilty, Weak Archimedeanity, Dedekind completeness, Interval lattice, Interval order topology, Moore's partial order, Kulisch's partial order. Tourism

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https://topnotchteaching.com/math/math-and-science-activities/?utm_source=rss&utm_medium=rss&utm_campaign=math-and-science-activities

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# Need Math & Science Activities? Here They Are! Here's some great ideas for you to teach an understanding of the physical and natural world and encourage an interest in numbers, shapes and patterns. Last week I shared some favorite literacy resources. This week, let’s turn to math and science activities. I’ve got some great ideas for these subjects that you can use for individual work, small groups, or even the whole class. ## 3 quick science activities Explain to your students that science is about understanding the physical and natural world better, and one of the ways we do that is by asking questions. Brainstorm a list of science questions. Think about ‘Why’ and ‘What would happen if’ questions. See if there are any you can test. • Ring of Life. This is an outdoor activity. Ideally, you’ll have a small outdoor area that is not paved. Place a hoop on the ground (you can also use a jump rope or other item) to mark off a small area. Have students make lists of living things they see. Encourage them to describe the different things or sketch them. For follow up, you could revisit the same space later in the week to see if anything has changed. This activity helps students practice observation and recording details as well as learning about what is living near your school. • Sink or Float. Fill a tub with water and provide students with a set of objects that can go in the water. Have students predict whether each object will sink or float, and say why they think so. Then have students test each object and record their results. Ask them to hypothesize about what makes things sink or float. Students practice making predictions, testing them, and drawing conclusions. • Oobleck. Mix cornstarch and water together and you get oobleck, which has properties of both liquids and solids. It takes just a few minutes to measure (about 1 cup water to about 1.5–2 cups cornstarch) and mix. Let kids play with it. Ask them how it feels. What happens when they slowly put a finger into it? What happens when they poke it quickly and hard? Does it pour? Does it keep its shape? This is another chance for students to observe, record, ask questions and test their ideas. ## 3 fabulous math resources Science shouldn’t get all the fun! So as we turn to math, we’ve got some games to keep things fun, along with activity sheets and task cards to help you cover a variety of math topics from time to shapes and beyond. Here are 3 math resources, you’ll want at the ready. ### 2D shape games and poster pack Get kids familiar with 12 different 2D shapes by hanging posters around your room. These posters don’t just show the shape and name, they give descriptions in fun, informal language and tell kids where the shape is likely to be found. Then with the whole class or a small math group, see how familiar kids are with the shapes by playing 2D shape bingo. Use simple or advanced clues to help students identify the shape and search it out on their cards. You can check out the 2D Shape Games and Poster Pack here … or you’ll be able to get it as part of the Top Notch Teaching Members Club Can your students tell time? Time skills go beyond simply reading a clock, and this set of 60 tasks cards will help kids practice: • Telling the time on an analog clock • Showing the time on an analog clock • Writing the time in different ways • Using analog and digital time interchangeably • Working out elapsed time • Showing elapsed time • Ordering time Print the time telling task cards, laminate, and get kids going with a wipe-off marker. (But before you order, you might want to hear more about the Top Notch Teaching Members Club, because you could get this and a lot more.) ### Fun math games and math activities Let’s face it, there is a lot to cover when it comes to math. The two math resources I mentioned above cover shape and time, but within those categories, there’s more to teach. And then there’s numbers, mental math, space (beyond 2D shapes), measurement (beyond time), chance and data. I’ve pulled together more than 138 pages of math games and activities to help students learn about and practice: • place value • the order of numbers • counting • problem-solving • multiplication facts • counting on strategy • near doubles strategy • compatible numbers strategy • partitioning strategy • estimating strategy • polygons • location • transformation of shapes • calendars • time • area • perimeter • making predictions • chance events • reasoning • Venn diagrams. In Fun Math Games and Math Activities, you get worksheets for independent work or home practice. You’ll find games (including shape dominos, calculator races, and dice games) and puzzles (like 100 chart puzzles) and a variety of activities (like tangrams and compass rose creation). Here’s what teachers are saying: Awesome resource, covers a huge amount of our maths curriculum. Thank you!” and Terrific resource.. all those great teaching points and understandings with activities and games to match. Thanks. You could have all these resources at your fingertips too! ## Math and science activities and yet still more I’ve shared some math and science activities today, but I’ve got so much more to make your teaching life easier. When the Top Notch Teaching Members Club opens in March, it will include the math resources I’ve mentioned here and so much more. You’ll find my favorite literacy tools, PE resources, classroom organization and time management tools. And you’ll find new resources not available in any other store! Along with done for you, ready to print resources, you’ll find videos with ideas and tips for using products and videos about teacher self-care, reducing stress, and other topics to help you and your students thrive. Doors open in March, but you can get on the waitlist for the Top Notch Teaching Members Club now: https://topnotchteaching.com/members/ ## 4 Spectacular Tools For More Math Practice Your students will need plenty of math practice. Having a variety of tools for math practice has benefits. Math practice tools you’ll love for your toolbox! ## 5 Fantastic Ways To Teach Multiplication Memorizing multiplication tables helps kids do quicker and more complex mental math. Multiplication task cards aid in understanding this critical concept.

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https://plainmath.net/13043/a-noisy-grinding-machine-in-a-factory-produces-a-soundintensity-of-10-m2

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Question A noisy grinding machine in a factory produces a soundintensity of 1.0 * 10-5 W/m2 . Other A noisy grinding machine in a factory produces a soundintensity of $$1.0 \cdot 10-5 W/m2$$ . What is thesound level of this machine? A second identical machine is brought into the factory. Whatis the new sound intensity? What is the new sound level? 2021-01-14 The sound level in decibles (dB) is given by the equation: beta (in dB) =$$\displaystyle{10}{\log{{\left(\frac{{I}}{{I}_{{o}}}\right)}}}$$ where... beta = sound level in dB I = intensity of the sound in $$\displaystyle\frac{{W}}{{m}^{{{2}}}}$$ I_o = intensity of a reference value* = 1.0 x $$\displaystyle{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}$$ Plugging in these values we get: $$\displaystyle\beta{\left(\in{d}{B}\right)}={10}{\log{{\left({1.0}{x}{10}^{{-{5}}}\frac{{W}}{{m}^{{{2}}}}/{1.0}{x}{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}\right)}}}={70}{d}{B}$$ The intensity of a sound is additive. This means that ifa second machine were brought into the factory the combinedintensity of the new sound would be the sum of the individualintensities. So, "I" now equals: $$\displaystyle{1.0}{x}{10}^{{-{5}}}+{1.0}{x}{10}^{{-{5}}}={2.0}{x}{10}^{{-{5}}}$$ Plugging this value back into our formula for soundlevel: beta (in dB) =$$\displaystyle{10}{\log{{\left({2.0}{x}{10}^{{-{5}}}\frac{{W}}{{m}^{{{2}}}}/{1.0}{x}{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}\right)}}}={73}{d}{B}$$ * The $$I_o$$ value is a reference value and can be any value youchoose. Generally it is considered the minimum intensity ofsound that a human ear can hear, or $$\displaystyle{1.0}{x}{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}$$. Hope this helps!

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#### What is 153 percent of 422? How much is 153 percent of 422? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 153% of 422 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update. ## 153% of 422 = 645.66 Calculate another percentage below. Type into inputs Find number based on percentage percent of Find percentage based on 2 numbers divided by Calculating one hundred and fifty-three of four hundred and twenty-two How to calculate 153% of 422? Simply divide the percent by 100 and multiply by the number. For example, 153 /100 x 422 = 645.66 or 1.53 x 422 = 645.66 #### How much is 153 percent of the following numbers? 153% of 422.01 = 64567.53 153% of 422.02 = 64569.06 153% of 422.03 = 64570.59 153% of 422.04 = 64572.12 153% of 422.05 = 64573.65 153% of 422.06 = 64575.18 153% of 422.07 = 64576.71 153% of 422.08 = 64578.24 153% of 422.09 = 64579.77 153% of 422.1 = 64581.3 153% of 422.11 = 64582.83 153% of 422.12 = 64584.36 153% of 422.13 = 64585.89 153% of 422.14 = 64587.42 153% of 422.15 = 64588.95 153% of 422.16 = 64590.48 153% of 422.17 = 64592.01 153% of 422.18 = 64593.54 153% of 422.19 = 64595.07 153% of 422.2 = 64596.6 153% of 422.21 = 64598.13 153% of 422.22 = 64599.66 153% of 422.23 = 64601.19 153% of 422.24 = 64602.72 153% of 422.25 = 64604.25 153% of 422.26 = 64605.78 153% of 422.27 = 64607.31 153% of 422.28 = 64608.84 153% of 422.29 = 64610.37 153% of 422.3 = 64611.9 153% of 422.31 = 64613.43 153% of 422.32 = 64614.96 153% of 422.33 = 64616.49 153% of 422.34 = 64618.02 153% of 422.35 = 64619.55 153% of 422.36 = 64621.08 153% of 422.37 = 64622.61 153% of 422.38 = 64624.14 153% of 422.39 = 64625.67 153% of 422.4 = 64627.2 153% of 422.41 = 64628.73 153% of 422.42 = 64630.26 153% of 422.43 = 64631.79 153% of 422.44 = 64633.32 153% of 422.45 = 64634.85 153% of 422.46 = 64636.38 153% of 422.47 = 64637.91 153% of 422.48 = 64639.44 153% of 422.49 = 64640.97 153% of 422.5 = 64642.5 153% of 422.51 = 64644.03 153% of 422.52 = 64645.56 153% of 422.53 = 64647.09 153% of 422.54 = 64648.62 153% of 422.55 = 64650.15 153% of 422.56 = 64651.68 153% of 422.57 = 64653.21 153% of 422.58 = 64654.74 153% of 422.59 = 64656.27 153% of 422.6 = 64657.8 153% of 422.61 = 64659.33 153% of 422.62 = 64660.86 153% of 422.63 = 64662.39 153% of 422.64 = 64663.92 153% of 422.65 = 64665.45 153% of 422.66 = 64666.98 153% of 422.67 = 64668.51 153% of 422.68 = 64670.04 153% of 422.69 = 64671.57 153% of 422.7 = 64673.1 153% of 422.71 = 64674.63 153% of 422.72 = 64676.16 153% of 422.73 = 64677.69 153% of 422.74 = 64679.22 153% of 422.75 = 64680.75 153% of 422.76 = 64682.28 153% of 422.77 = 64683.81 153% of 422.78 = 64685.34 153% of 422.79 = 64686.87 153% of 422.8 = 64688.4 153% of 422.81 = 64689.93 153% of 422.82 = 64691.46 153% of 422.83 = 64692.99 153% of 422.84 = 64694.52 153% of 422.85 = 64696.05 153% of 422.86 = 64697.58 153% of 422.87 = 64699.11 153% of 422.88 = 64700.64 153% of 422.89 = 64702.17 153% of 422.9 = 64703.7 153% of 422.91 = 64705.23 153% of 422.92 = 64706.76 153% of 422.93 = 64708.29 153% of 422.94 = 64709.82 153% of 422.95 = 64711.35 153% of 422.96 = 64712.88 153% of 422.97 = 64714.41 153% of 422.98 = 64715.94 153% of 422.99 = 64717.47 153% of 423 = 64719 1% of 422 = 4.22 2% of 422 = 8.44 3% of 422 = 12.66 4% of 422 = 16.88 5% of 422 = 21.1 6% of 422 = 25.32 7% of 422 = 29.54 8% of 422 = 33.76 9% of 422 = 37.98 10% of 422 = 42.2 11% of 422 = 46.42 12% of 422 = 50.64 13% of 422 = 54.86 14% of 422 = 59.08 15% of 422 = 63.3 16% of 422 = 67.52 17% of 422 = 71.74 18% of 422 = 75.96 19% of 422 = 80.18 20% of 422 = 84.4 21% of 422 = 88.62 22% of 422 = 92.84 23% of 422 = 97.06 24% of 422 = 101.28 25% of 422 = 105.5 26% of 422 = 109.72 27% of 422 = 113.94 28% of 422 = 118.16 29% of 422 = 122.38 30% of 422 = 126.6 31% of 422 = 130.82 32% of 422 = 135.04 33% of 422 = 139.26 34% of 422 = 143.48 35% of 422 = 147.7 36% of 422 = 151.92 37% of 422 = 156.14 38% of 422 = 160.36 39% of 422 = 164.58 40% of 422 = 168.8 41% of 422 = 173.02 42% of 422 = 177.24 43% of 422 = 181.46 44% of 422 = 185.68 45% of 422 = 189.9 46% of 422 = 194.12 47% of 422 = 198.34 48% of 422 = 202.56 49% of 422 = 206.78 50% of 422 = 211 51% of 422 = 215.22 52% of 422 = 219.44 53% of 422 = 223.66 54% of 422 = 227.88 55% of 422 = 232.1 56% of 422 = 236.32 57% of 422 = 240.54 58% of 422 = 244.76 59% of 422 = 248.98 60% of 422 = 253.2 61% of 422 = 257.42 62% of 422 = 261.64 63% of 422 = 265.86 64% of 422 = 270.08 65% of 422 = 274.3 66% of 422 = 278.52 67% of 422 = 282.74 68% of 422 = 286.96 69% of 422 = 291.18 70% of 422 = 295.4 71% of 422 = 299.62 72% of 422 = 303.84 73% of 422 = 308.06 74% of 422 = 312.28 75% of 422 = 316.5 76% of 422 = 320.72 77% of 422 = 324.94 78% of 422 = 329.16 79% of 422 = 333.38 80% of 422 = 337.6 81% of 422 = 341.82 82% of 422 = 346.04 83% of 422 = 350.26 84% of 422 = 354.48 85% of 422 = 358.7 86% of 422 = 362.92 87% of 422 = 367.14 88% of 422 = 371.36 89% of 422 = 375.58 90% of 422 = 379.8 91% of 422 = 384.02 92% of 422 = 388.24 93% of 422 = 392.46 94% of 422 = 396.68 95% of 422 = 400.9 96% of 422 = 405.12 97% of 422 = 409.34 98% of 422 = 413.56 99% of 422 = 417.78 100% of 422 = 422

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# Proof of an integral 1. Oct 31, 2007 ### lazypast hi, ive a function here and im finding it hard to integrate $$\int \frac {1}{\sqrt{a^2 - x^2}}$$ let x = asin(theta) so $$\frac{dx}{d \theta}=acos \theta$$ $$dx=acos \theta d \theta$$ $$\int \frac {acos \theta d \theta}{\sqrt{a^2 - a^2 sin^2 \theta}}$$ $$\int \frac {acos \theta d \theta}{\sqrt{a^{2} (1 - sin^2 \theta)}}$$ $$\int \frac {acos \theta d \theta}{acos \theta} = \int d \theta$$ i cant continue this as i cant get the right equation which is able to be integrated. any ideas? 2. Oct 31, 2007 ### d_leet What is the problem you're having, you reduced a diffcult integral into a practically trivial one? 3. Oct 31, 2007 ### lazypast i reduced it to dtheta, but im unable to sub anything back into the integral which i can solve. $$d \theta = \frac {dx}{acos \theta}$$ $$\int \frac {dx}{acos \theta}$$ but this isnt any more help to me, i know what the answr is ( sin^-1 (x/a)) but arent close to it 4. Oct 31, 2007 ### d_leet Don't substitute back yet, that defeats the whole point of teh substitution. Just integrate d(theta). 5. Oct 31, 2007 ### TMM And THEN substitute back in with a reference triangle or just theta in terms of x. That's how trig substitution works. 6. Nov 1, 2007 ### Gib Z Would it help if you rewrote it as $$\int 1 d\theta$$? What (family of) function(s) whose derivative with respect to theta gives 1? As d_leet pointed out, you really already did the hard part, Made some nice soup but don't know how to use a spoon :( 7. Nov 1, 2007 ### HallsofIvy This also has nothing to do with Differential Equations so I'm moving it to "Calculus and Analysis". 8. Nov 1, 2007 ### lazypast $$\int d \theta = \theta$$ $$\frac {dx}{d \theta} = acos \theta$$ $$\theta = cos^{-1}(\frac {1}{a} \frac {dx}{d \theta})$$ and this still has a dx/dtheta in it. i really am stumpted here 9. Nov 1, 2007 ### lazypast aha ive been abit blind. $$x = asin \theta$$ $$\theta = sin^{-1} \frac {x}{a}$$ i dont understand the way you show it. when f(x) = theta, f'(x) = 1 when you differentiate with respect to theta thanks for the help anyway 10. Nov 2, 2007 ### Gib Z Are you familiar with the fundamental theorem on calculus? 11. Nov 2, 2007 ### lazypast i havent been tought the fundamental theory of calculus,but i can do integration and differentiation which is what i thought is basically was 12. Nov 2, 2007 ### Gib Z :( There are a few versions of the theorem, and they are all equivalent, you really need to know the theorem, and I'm sure you already do but just don't realize it. It basically says that $$\int^b_a f(x) dx = F(b) - F(a) , \mbox{where} F'(x)=f(x)$$, or equivalently, $$\frac{d}{dx}\int^x_a f(t) dt = f(x)$$. Try and see why they are the same. 13. Nov 2, 2007 ### ZioX What's the derivative of theta with respect to theta? 14. Nov 2, 2007 ### lazypast derivative of theta with respect to theta id say is one. after looking at some concepts of the fundamental theorm of calculus i cant seem to understand them. you say $$\int^b_a f(x) dx = F(b) - F(a) , \mbox{where} Fsingle-quote(x)=f(x)$$ i would have thought F(x) = f ' (x) i dont know why you use F and f if you replace f(x) with y, then i suppose F(x) could be z? 15. Nov 2, 2007 ### Gib Z It doesn't matter what letters you use, the theorem stays the same though :( It is F'(x) = f(x), you really need to learn this, check your textbook. 16. Nov 3, 2007 ### HallsofIvy These were posts 8 and 9 but we went on for 6 more posts anyway! (Oops, 7 now.) 17. Nov 3, 2007 ### Gib Z Well its because we found a serious problem in his understanding :( He says he knows differential and integral calculus, yet has never heard of the Fundamental Theorem before!!! 18. Nov 5, 2007 ### lazypast i have heard of it just never been tought it. ive just found out that $$\int f(x) = F(x)$$ and that was the notation i didnt understand. like the capital F had been plucked from no where

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Cody # Problem 29. Nearest Numbers Solution 809671 Submitted on 19 Jan 2016 by Raly Todorova 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 %% A = [30 46 16 -46 35 44 18 26 25 -10]; correct = [8 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct)) d = NaN 16 14 76 5 14 12 4 5 40 16 NaN 30 92 11 2 28 20 21 56 14 30 NaN 62 19 28 2 10 9 26 76 92 62 NaN 81 90 64 72 71 36 5 11 19 81 NaN 9 17 9 10 45 14 2 28 90 9 NaN 26 18 19 54 12 28 2 64 17 26 NaN 8 7 28 4 20 10 72 9 18 8 NaN 1 36 5 21 9 71 10 19 7 1 NaN 35 40 56 26 36 45 54 28 36 35 NaN index2 = 9 index1 = 8 2   Pass %% A = [1555 -3288 2061 -4681 -2230 -4538 -4028 3235 1949 -1829]; correct = [3 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct)) d = Columns 1 through 8 NaN 4843 506 6236 3785 6093 5583 1680 4843 NaN 5349 1393 1058 1250 740 6523 506 5349 NaN 6742 4291 6599 6089 1174 6236 1393 6742 NaN 2451 143 653 7916 3785 1058 4291 2451 NaN 2308 1798 5465 6093 1250 6599 143 2308 NaN 510 7773 5583 740 6089 653 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97.0000 96.0000 95.0000 94.0000 93.0000 102.0000 101.0000 100.0000 99.0000 98.0000 97.0000 96.0000 95.0000 94.0000 103.0000 102.0000 101.0000 100.0000 99.0000 98.0000 97.0000 96.0000 95.0000 104.0000 103.0000 102.0000 101.0000 100.0000 99.0000 98.0000 97.0000 96.0000 105.0000 104.0000 103.0000 102.0000 101.0000 100.0000 99.0000 98.0000 97.0000 106.0000 105.0000 104.0000 103.0000 102.0000 101.0000 100.0000 99.0000 98.0000 107.0000 106.0000 105.0000 104.0000 103.0000 102.0000 101.0000 100.0000 99.0000 108.0000 107.0000 106.0000 105.0000 104.0000 103.0000 102.0000 101.0000 100.0000 109.0000 108.0000 107.0000 106.0000 105.0000 10... 6   Pass %% % Area codes A = [847 217 508 312 212]; correct = [2 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct)) d = NaN 630 339 535 635 630 NaN 291 95 5 339 291 NaN 196 296 535 95 196 NaN 100 635 5 296 100 NaN index2 = 5 index1 = 2 7   Pass %% % Zip codes A = [60048 61802 01702 60601 10001]; correct = [1 4]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct)) d = NaN 1754 58346 553 50047 1754 NaN 60100 1201 51801 58346 60100 NaN 58899 8299 553 1201 58899 NaN 50600 50047 51801 8299 50600 NaN index2 = 4 index1 = 1 ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!

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1 GATE EE 2012 +1 -0.3 A periodic voltage waveform observed on an oscilloscope across a load is shown. A permanent magnet moving coil $$(PMMC)$$ meter connected across the same load reads A $$4$$ $$V$$ B $$5$$ $$V$$ C $$8$$ $$V$$ D $$10$$ $$V$$ 2 GATE EE 2010 +1 -0.3 An ammeter has a current range of $$0$$-$$5A,$$ and its internal resistance is $$0.2\Omega$$. In order to change the range to $$0$$-$$25$$ $$A$$ we need to add a resistance of A $$0.8\,\Omega$$ in series with the meter B $$1.0\,\Omega$$ in series with the meter. C $$0.04\,\Omega$$ in parallel with the meter D $$0.05\,\Omega$$ in parallel with the meter 3 GATE EE 2005 +1 -0.3 A $$PMMC$$ voltmeter is connected across a series combination of a $$DC$$ voltage source $${V_1} = 2V$$ and $$AC$$ voltage source $${V_2}\left( t \right) = 3$$ $$sin(4t)$$ $$V.$$ The meter reads A $$2V$$ B $$5V$$ C $$\left( {2 + \sqrt 3 /2} \right)V$$ D $$\sqrt {17} /2V$$ 4 GATE EE 2004 +1 -0.3 A galvanometer with a full scale current of $$10$$ $$mA$$ has a resistance of $$1000$$$$\Omega$$. The multiplying power (the ratio of measured current to galvanometer current) of a $$100$$$$\Omega$$ shunt with this galvanometer is A $$110$$ B $$100$$ C $$11$$ D $$10$$ EXAM MAP Medical NEET

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Show that a subspace of $c_{0}$ with the norm induced by $c_{0}$, can’t be isomorphic to $l_1$. [closed] Show that a subspace of $$c_{0}$$ with the norm induced by $$c_{0}$$, can’t be isomorphic to $$l_1$$. Any idea or hint? I think have to show that the given subspace has a certain property that $$l_1$$ no has (or the reverse) but I can’t understand which property... thank you! • @mechanodroid had a great answer. I just feel like adding that since $c_0$ is subprojective, if $c_0$ contained a copy of $\ell_1$ then $\ell_1$ would contain a copy of $c_0$. However $\ell_1$ has the Schur property whereas $c_0$ does not. Jan 2, 2019 at 19:22 The dual space of $$\ell^1$$ is isometrically isomorphic to $$\ell^\infty,$$ which is nonseparable. However, the dual of $$c_0$$ is isometrically isomorphic to $$\ell^1,$$ which is separable. Hence the dual of any subspace of $$c_0$$ is separable as well. This argument actually shows that there is not a linear homeomorphism between $$\ell^1$$ and a subspace of $$c_0$$, let alone an isometric isomorphism.

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# Math problems #### npsonic ##### Active Member Math just for fun. I start with the problem and publish answer later, so you can try to solve it. Most of the questions I will post are problems that services such as Wolfram Alpha can't understand or will offer wrong answer. Problem 1. Diameter of the circle shaped pool is 30m and height 3m. Water in the pool contains mercury and the concentration is 0.2µg/l. Contaminated water is pumped off 1000l/hour and it's replaced with the clean water in same rate. Let's also make an assumption that mercury is evenly mixed with the water. a) Create a differential equation for the amount of mercury x(t) (µg) function of the t (h) and solve. b) How long it takes to concentration of the mercury to go under 0.07µg/l? Last edited: #### Star-Dust ##### Expert Longtime User Math just for fun. I start with the problem and publish answer later, so you can try to solve it. Most of the questions I will post are problems that services such as Wolfram Alpha can't understand or will offer wrong answer. Problem 1. Diameter of the circle shaped pool is 30m and height 3m. Water in the pool contains mercury and the concentration is 0.2µg/l. Contaminated water is pumped off 1000l/hour and it's replaced with the clean water in same rate. Let's also make an assumption that mercury is evenly mixed with the water. a) Create a differential equation for the amount of mercury x(t) (µg) function of the t (h) and solve. b) How long it takes to concentration of the mercury to go under 0.07µg/l? And if I call the plumber, do I empty the pool? with \$ 50 I solve everything, don't you think? #### npsonic ##### Active Member You might be the plumber in this case. Anyway you need to know the answer. Funny thing is that in Finland they would such laugh and hang up if you would offer 50 dollars for them to come empty your pool. All questions will increase difficulty and you might need something what you learned from this question to the next question. Try to solve the problem it isn't hard. #### Star-Dust ##### Expert Longtime User Funny thing is that in Finland they would such laugh and hang up if you would offer 50 dollars for them to come empty your pool. Luckily I don't live in Finland, you are too expensive. Here with \$ 50 they empty the pool and take your dog for a walk However it is true, it is not difficult. At school I learned how to build an oscilloscope. It takes 4/5 hours of work. But if I go to the junk shop with little I buy it ... #### Daestrum ##### Expert Longtime User I got an answer for part b , but no idea on the equation part 92 days 19 hours 13 minutes 36 seconds (Concentration @ 0.06999999276529222 µg/l) #### Beja ##### Expert Longtime User Math just for fun. I start with the problem and publish answer later, so you can try to solve it. Most of the questions I will post are problems that services such as Wolfram Alpha can't understand or will offer wrong answer. Problem 1. Diameter of the circle shaped pool is 30m and height 3m. Water in the pool contains mercury and the concentration is 0.2µg/l. Contaminated water is pumped off 1000l/hour and it's replaced with the clean water in same rate. Let's also make an assumption that mercury is evenly mixed with the water. a) Create a differential equation for the amount of mercury x(t) (µg) function of the t (h) and solve. b) How long it takes to concentration of the mercury to go under 0.07µg/l? These days I am working on water station where several stages will clean up the water first tank from mud and inject chem to kill bacteria and other bio organism, before pumping the clean water up to the final tank. To solve the above problem, I first shut off the pump, drain the contaminated water, clean and decontaminate the tanks and then turn on the pump again. This is a life support process and can't leave it for equations and calculations. Residue and impurities can remain for many reasons if not cleaned. #### npsonic ##### Active Member These days I am working on water station where several stages will clean up the water first tank from mud and inject chem to kill bacteria and other bio organism, before pumping the clean water up to the final tank. To solve the above problem, I first shut off the pump, drain the contaminated water, clean and decontaminate the tanks and then turn on the pump again. This is a life support process and can't leave it for equations and calculations. Residue and impurities can remain for many reasons if not cleaned. These are math problems. No real life solution is needed. This is just math for fun. I will publish correct answers today, so if someone still want to try to solve first question. #### Beja ##### Expert Longtime User These are math problems. No real life solution is needed. This is just math for fun. I will publish correct answers today, so if someone still want to try to solve first question. Ok.. Give me time to try #### npsonic ##### Active Member Okay, I have given enough time for those how might have participated, so here's the solution for problem 1. #### npsonic ##### Active Member I got an answer for part b , but no idea on the equation part 92 days 19 hours 13 minutes 36 seconds (Concentration @ 0.06999999276529222 µg/l) More problems coming soon, maybe something more easier. #### npsonic ##### Active Member It seems that the first question was too hard for the most users here or there wasn't any interest to solve it, so here's something simpler. Problem 2 Which of the following equations are linear. (No calculations needed) A: y''' + 4 y' - 3 y = sin(x) B: x^2 y''' + e^x y' - 3 y = x^(-1) C: y'' + 4 y' - 3 y = 2 y^2 D: x^2 y'' + e^x y' - 3 y = y^(-1) Problem 3 The falling object is affected by the Earth's gravity mg and air resistance kv where k is constant. Solve the velocity from the differential equation formed v(t). #### Beja ##### Expert Longtime User Theoretically the contamination will always be there and never end.. Endless dilution. Percentage will be smaller and smaller for ever, but never reaches 0. #### npsonic ##### Active Member Theoretically the contamination will always be there and never end.. Endless dilution. Percentage will be smaller and smaller for ever, but never reaches 0. That's the case always. We always are closing to some value, but never truly reach it. Everything is just approximation of something. I can throw one approximation problem if you want to solve it. #### Daestrum ##### Expert Longtime User It could actually get to zero, eventually there would be less mercury atoms than water molecules, then it's just probability that the atom is contained within the 1000l/hr that is removed. #### npsonic ##### Active Member It could actually get to zero, eventually there would be less mercury atoms than water molecules, then it's just probability that the atom is contained within the 1000l/hr that is removed. In mathematical terms it will never get to zero. It's closing to it, but never ever reach it. Do not try to think this as an real life thing. In real life you would such accept in some point that it's zero when it really isn't it's such very "close" to it. #### Beja ##### Expert Longtime User That's the case always. We always are closing to some value, but never truly reach it. Everything is just approximation of something. I can throw one approximation problem if you want to solve it. Now you say it.. It can only be solved by the plumber, not the mathematician. If you use math then mercury will always be there. actually not theoretically, and poses real threat to health. A plumber will close the valve, clean the tank and bring mercury value to zero. #### Beja ##### Expert Longtime User It could actually get to zero, eventually there would be less mercury atoms than water molecules, then it's just probability that the atom is contained within the 1000l/hr that is removed. With some prayers yes! #### npsonic ##### Active Member Erel come to solve these problems. I know you can. After these I will post some trickier problem. Last edited: #### npsonic ##### Active Member No one wants to try? These should be extremely easy problems for true programmers. #### thetahsk ##### Active Member Longtime User In mathematical terms it will never get to zero. It's closing to it, but never ever reach it. Do not try to think this as an real life thing. In real life you would such accept in some point that it's zero when it really isn't it's such very "close" to it. This is a very nice example where theoretical mathematics and practical reality meet. A vessel contaminated with mercury will never be 100% decontaminated by simply exchanging its contents. just kidding and sorry for this Problem 4 Are you bored ? Last edited:

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🐷 2021/10/22に公開 # 比較対象の配列 ``````with items as ( select 1 as id, ['dog'] as tags union all select 2 as id, ['cat'] as tags union all select 3 as id, ['bird'] as tags union all select 4 as id, ['dog', 'cat'] as tags union all select 5 as id, ['cat', 'bird'] as tags union all select 6 as id, ['bird', 'dog'] as tags union all select 7 as id, ['dog', 'cat', 'bird'] as tags ) `````` ``````select * from items; `````` id tags 1 [dog] 2 [cat] 3 [bird] 4 [dog,cat] 5 [cat,bird] 6 [bird,dog] 7 [dog,cat,bird] このテーブルと `['dog', 'cat']` という配列を色々な条件で比較する場合のクエリを挙げる。 # 比較条件 ## 配列内にある要素を全て持つ id tags 4 [dog,cat] 7 [dog,cat,bird] ``````select * from items where ( select count(1) from unnest(tags) as t where t in ('dog', 'cat') ) = array_length(['dog', 'cat']); `````` ## 配列内の要素が同じ id tags 4 [dog,cat] ``````select * from items where array_length(tags) = array_length(['dog', 'cat']) and ( select count(1) from unnest(tags) as t where t in ('dog', 'cat') ) = array_length(['dog', 'cat']); `````` ## 配列内にある要素を1つだけ持つ id tags 1 [dog] 2 [cat] 5 [cat,bird] 6 [bird,dog] ``````select * from items where ( select count(1) from unnest(tags) as t where t in ('dog', 'cat') ) = 1; `````` ## 配列内にある要素を1つ以上持つ id tags 1 [dog] 2 [cat] 4 [dog,cat] 5 [cat,bird] 6 [bird,dog] 7 [dog,cat,bird] ``````select * from items where ( select logical_or(t in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where not ( select logical_and(t not in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where exists ( select * from unnest(tags) as t where t in ('dog', 'cat') ); `````` ``````select * from items where ( select count(1) from unnest(tags) as t where t in ('dog', 'cat') ) > 0; `````` ## 配列内にある要素だけで構成される id tags 1 [dog] 2 [cat] 4 [dog,cat] ``````select * from items where ( select logical_and(t in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where not ( select logical_or(t not in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where not exists ( select * from unnest(tags) as t where t not in ('dog', 'cat') ); `````` ``````select * from items where ( select count(1) from unnest(tags) as t where t not in ('dog', 'cat') ) = 0; `````` ## 配列内にある要素を1つも持たない id tags 3 [bird] ``````select * from items where ( select logical_and(t not in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where not ( select logical_or(t in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where not exists ( select * from unnest(tags) as t where t in ('dog', 'cat') ); `````` ``````select * from items where ( select count(1) from unnest(tags) as t where t in ('dog', 'cat') ) = 0; `````` ## 配列内にない要素を1つ以上持つ id tags 3 [bird] 5 [cat,bird] 6 [bird,dog] 7 [dog,cat,bird] ``````select * from items where ( select logical_or(t not in ('dog', 'cat')) from unnest(tags) as t ); `````` ``````select * from items where not ( select logical_and(t in ('dog', 'cat')) from unnest(tags) as t ); `````` ログインするとコメントできます

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# Buckets and bits - delta encoding I have now put together a delta encoder that I think works as expected (or which at least seems reasonable in its context) and would appreciate some feedback and a bit of guidance before I move on to the next stage: Setup: n = 2048 sparsity = 0.01 w = 20 Range = 232 (min = 0, max = 232) Buckets = 232 (i.e. buckets = range) Q1: is this a reasonable encoding framework? Q2: I’m still unsure how I use buckets or if the above is reasonable. The data that I’m basing my encoder on has a mean of about 4.9 and a sd of 6.6. so it’s heavily skewed towards zero (differences are coerced to be only positive). A graph looks like this: Bit ranges: With how it’s set up at the moment, I’m getting the following SDR breakdown: Bits 0-20 represent 0-10 (I’ve bounded it at the top and bottom values) and each successive value is represented by moving 1 bit further along - so 11 is represented by bits 1-21, 12 by bits 2-22 and so on. Any guidance would be appreciated. 2 Likes Rob, I’m not sure you’re going about it the right way. A delta encoder should be very similar to a scalar encoder. Except you’re encoding the difference between two scalar values, not the values themselves. Here are some references: 2 Likes Hi Matt many thanks for the feedback & video! I probably should have provided some context. The delta is what is being encoded - so, I get the delta (this value - last value) and coerce it to be positive (taking the square root of the squared value). This is the value that is sent to the encoder. In delta.py it appears that the initial delta is made = 0 whereas I’m ignoring it completely. The graph is to show the distribution of the deltas (i.e. x = delta values and y = count). As you noted, it has a very long tail which I need to account for in the encoding. I have read Encoding Data for HTM Systems multiple times but I guess at some point I got confused and applied sparsity to the encoder. Does it matter though if the encoding itself is in a sparse array? In delta.py there’s a line of comment that reads: “It returns an actual value when decoding and not a delta”. Does this mean that I need to somehow encode the value? The end-point with my implementation is to look for outliers in the pattern of changes themselves regardless of underlying value. Thanks again. 2 Likes In Building HTM Systems in the Encoder section you state that: Data must be encoded into sparse binary arrays for an HTM system to process it. These binary arrays define an input space for the Spatial Pooling algorithm I’m still confused 2 Likes As I understand the encodings can vary in sparsity, as long as they still satisfy the requirements for encoders in Encoding Data for HTM: 1. Semantically similar data should result in SDRs with overlapping active bits. 2. The same input should always produce the same SDR as output. 3. The output should have the same dimensionality (total number of bits) for all inputs. 4. The output should have similar sparsity for all inputs and have enough one-bits to handle noise and subsampling. Encoders always (in my experience) have many more 0’s than 1’s, though they are not as sparse as the Sparse Distributed Representations (SDR’s) output from the Spatial Pooler and passed into the Temporal Memory, which are strictly limited to 2% activation. It also seems like you could preprocess the raw data by differencing, so the new raw data would be the differences themselves and you could just use a scalar encoder from there. Smart to write your own encoder though regardless. 2 Likes Thanks Sam - yes I’m pre-processing as the data stream (of raw values) arrive - so only the non-NA deltas get passed to the encoding function. I guess that at some point I confused “encoders are sparse” with “encoders output an SDR” (per the Spatial Pooler). Hopefully it’s not such a big problem to fix. 2 Likes

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Related Articles # Modify values of a Data Frame in R Language – transform() Function • Last Updated : 26 May, 2020 `transform()` function in R Language is used to modify data. It converts the first argument to the data frame. This function is used to transform/modify the data frame in a quick and easy way. Syntax: transform(data, value) Parameters: data: Data Frame to be modified value: new modified value of data Example 1: Use transform() on variables `# R program to illustrate``# transform on variables`` ` `# Create example data frame``data <``-` `data.frame(x1 ``=` `c(``1``, ``2``, ``3``, ``4``),            ``                   ``x2 ``=` `c(``5``, ``6``, ``7``, ``8``))`` ` ` ``# Apply transform function``data_ex1 <``-` `transform(data, x1 ``=` `x1 ``+` `10``)  `` ` `# Print data      ``print``(data_ex1)                                          ``                                    ` Output: ```x1= 11, 12, 13, 14. x2 = 5, 6, 7, 8.``` Here in the above code, we have a data framework and we transform it by adding 10(x1 = x1 + 10) to each of the values. The values of x2 will not get changed as we do not apply any transformation to it. Example 2: Add new variables to data `# R program to illustrate ``# Adding new variables to data`` ` `data <``-` `data.frame(x1 ``=` `c(``11``, ``12``, ``13``, ``14``),            ``                   ``x2 ``=` `c(``5``, ``6``, ``7``, ``8``))`` ` `# Apply transform function``data_ex2 <``-` `transform(data, x3 ``=` `c(``1``, ``3``, ``3``, ``4``))  `` ` `print``(data_ex2) ` Output: ```X1= 11, 12, 13, 14. X2 = 5, 6, 7, 8. X3 = 1, 3, 3, 4. ``` Here in the above code, we are adding a new variable x3 to the previous data. My Personal Notes arrow_drop_up

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It is currently 25 Jul 2017, 11:54 ### 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 # Heating-oil prices are expected to be higher th is year than Author Message TAGS: ### Hide Tags Intern Joined: 13 Sep 2007 Posts: 26 Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 30 Jun 2009, 00:49 39 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 58% (02:06) correct 42% (01:23) wrong based on 4036 sessions ### HideShow timer Statistics Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. (A) Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were (B) Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did (C) Expectations are for heating-oil prices to be higher this year than last year's becuase refiners are paying about \$5 a barrel for crude oil more than they did (D) It is the expectation that heating-oil prices will be higher this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were (E) It is the expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did OG2017 SC772 [Reveal] Spoiler: [Reveal] Spoiler: OA Manager Joined: 18 Jun 2009 Posts: 167 Location: Tbilisi, Georgia Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 30 Jun 2009, 01:08 1 This post was BOOKMARKED A IMO rise and higher together are redundant - B/E are out last year's doesn't make a parallel consturction- C is out and between A and D , D is wordy I think. Besides, using 'than' seems to be more correct than 'over' What's OA? _________________ Las cualidades del agua...porque el agua no olvida que su destino es el mar, y que tarde o temprano deberá llegar a él. Director Joined: 03 Jun 2009 Posts: 787 Location: New Delhi WE 1: 5.5 yrs in IT Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 30 Jun 2009, 01:48 Natia wrote: A IMO rise and higher together are redundant - B/E are out last year's doesn't make a parallel consturction- C is out and between A and D , D is wordy I think. Besides, using 'than' seems to be more correct than 'over' What's OA? I'll also vote for A for the same reason. _________________ Intern Joined: 13 Sep 2007 Posts: 26 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 02 Jul 2009, 01:29 cici wrote: This is a question I had trouble in OG12th edition. Plz help Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. A) B) Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did C) Expectations are for heating-oil prices to be higher this year than last year's becuase refiners are paying about \$5 a barrel for crude oil more than they did D) It is the expectation that heating-oil prices will be higher this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were E) It is the expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did Thank you all, The OA is indeed A! But I still don't understand how "last" in "Heating-oil prices are expected to be higher this year than last" is enough to be compared with the "this year". Wouldn't I need to say "last year" instead of "last"? Manager Joined: 17 Apr 2009 Posts: 153 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 23 Jul 2009, 05:37 1 KUDOS A was the most simple of all. good question! Director Joined: 25 Aug 2007 Posts: 936 WE 1: 3.5 yrs IT WE 2: 2.5 yrs Retail chain Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 06 May 2010, 01:02 I think there is some misprinting in OG. Even I found the same. IMO, this should be - ...higher this year than those of last year because... cici wrote: cici wrote: This is a question I had trouble in OG12th edition. Plz help Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. A) B) Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did C) Expectations are for heating-oil prices to be higher this year than last year's becuase refiners are paying about \$5 a barrel for crude oil more than they did D) It is the expectation that heating-oil prices will be higher this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were E) It is the expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did Thank you all, The OA is indeed A! But I still don't understand how "last" in "Heating-oil prices are expected to be higher this year than last" is enough to be compared with the "this year". Wouldn't I need to say "last year" instead of "last"? _________________ Tricky Quant problems: http://gmatclub.com/forum/50-tricky-questions-92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html Manager Joined: 09 Jun 2009 Posts: 211 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 03 Jun 2010, 20:17 8 KUDOS 51 This post was BOOKMARKED Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. (A) Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were (B) Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did (C) Expectations are for heating-oil prices to be higher this year than last year's because refiners are paying about \$5 a barrel for crude oil more than they did (d) It is the expectation that heating-oil prices will be higher for this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were (E) It is expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did Manager Joined: 24 May 2010 Posts: 67 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 03 Jun 2010, 20:32 5 KUDOS IMO A rise and higher is redundant Director Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing. Affiliations: University of Chicago Booth School of Business Joined: 26 Nov 2009 Posts: 960 Location: Singapore Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 05 Jun 2010, 19:15 2 KUDOS 2 This post was BOOKMARKED Example of ellipsis Heating-oil prices are expected to be higher this year than last [year's prices] ----> Bulls eye A is correct. "expected to be" is correct idiom. and "paying" is correct. hence A. C: is comparing expectations of this year vs the last year's. change of meaning. OUT A it is. _________________ Please press kudos if you like my post. Manager Joined: 09 Jun 2009 Posts: 211 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 05 Jun 2010, 23:42 2 KUDOS Yes by the process of elimination we reach A. Can some one explain what exactly are we comparing here and whether the things being compared are parallel. IMO we are comparing the prices last year with the prices this year and the years. From a parallelism perspective is A correct? If yes How? Thanks Director Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing. Affiliations: University of Chicago Booth School of Business Joined: 26 Nov 2009 Posts: 960 Location: Singapore Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 06 Jun 2010, 00:12 Its clear we are comparing the heating oil prices of current year with the previous' Refiners are paying the cruel oil prices not the taxes You refine crude oil and get the heating - oil. Hence when the price of crude oil goes up then heating oil will become expensive. Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. _________________ Please press kudos if you like my post. Manager Joined: 09 Jun 2009 Posts: 211 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 10 Jun 2010, 12:56 Killer explanation.... I wanted to know the elipse funda...btw i've been hearing it for quite some time now...everytiime i think i caught a parallelism bug...it turns out to be ellipse.. Can we please discuss so that i dont commit a mistake again? Thanks VP Joined: 17 Feb 2010 Posts: 1486 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 18 Jun 2010, 08:56 Can someone please explain why C is wrong. I thought the comparison is between this year's heating oil prices and last year's heating oil prices. option C Expectations are for heating-oil prices to be higher this year than last year's [heating-oil prices].....................correct comparison option A Heating-oil prices are expected to be higher this year than last [dont we need last year's heating-oil prices?] Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. (A) Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were (B) Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did (C) Expectations are for heating-oil prices to be higher this year than last year's because refiners are paying about \$5 a barrel for crude oil more than they did (D) It is the expectation that heating-oil prices will be higher for this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were (E) It is expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did Manager Joined: 25 May 2010 Posts: 136 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 18 Jun 2010, 10:46 73 KUDOS 56 This post was BOOKMARKED papillon86 wrote: Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. (A) Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were (B) Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did (C) Expectations are for heating-oil prices to be higher this year than last year's because refiners are paying about \$5 a barrel for crude oil more than they did (D) It is the expectation that heating-oil prices will be higher for this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were (E) It is expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did You've done a good job guys! Let me see if I can help you out with some of the confusion regarding comparisons and ellipses: A. Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were What is compared here are the times-- "this year" and "last" (meaning "last year"-- this is an ellipses). When comparing time or place, you have two options for structuring the comparison: 1. Joe studies more at night that he does during the day. (The sentence compares Joe to himself.) 2. Joe studies more at night than during the day. (The sentence compares "at night" to "during the day.") B. Heating-oil prices are expected to rise higher this year over last because refiners pay about \$5 a barrel for crude oil more than they did Important Note!! The comparative structure requires that you use "than" after the "--er". Correct: Tom is older than Bob. Incorrect: Tom is older compared to/in comparison with/over Bob You must use "than" after you've used a word ending with --er. C. Expectations are for heating-oil prices to be higher this year than last year's because refiners are paying about \$5 a barrel for crude oil more than they did You can stop reading as soon as you get to "for heating-oil prices" because that is not a correct description of the the expectations. The expectations are not for prices, but rather that prices will.... Important Note: Do not change a relative clause into a mere prepositional phrase!! Ex. "I think that chickens fly" (relative clause: that+subject+verb) IS NOT EQUIVALENT TO "I think of chickens flying" (prepositional phrase: of chickens) D. It is the expectation that heating-oil prices will be higher for this year over last because refiners are paying about \$5 a barrel more for crude oil now than what they were Same problem as B. E. It is expected that heating-oil prices will rise higher this year than last year's because refiners pay about \$5 a barrel for crude oil more than they did "prices will rise higher this year than last year's" uses ellipses that hide the logic problem. Complete everything that was left out, and the resulting sentence is: "prices will rise higher this year than last year's prices rose. This is not intended meaning of the sentence. Second, the word "more" is not correctly placed. Correct: I paid 5 dollars more for my sandwich today than I did yesterday. Incorrect: I paid 5 dollars for my sandwich more than I did yesterday. "More" describes \$5 and must therefore be placed next to \$5. _________________ Sarai Email me at saraiyaseen@gmail.com If this helped, kindly give Kudos! Director Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing. Affiliations: University of Chicago Booth School of Business Joined: 26 Nov 2009 Posts: 960 Location: Singapore Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 18 Jun 2010, 11:26 Very nice explanation _________________ Please press kudos if you like my post. VP Joined: 15 Jul 2004 Posts: 1445 Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX) Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 18 Jun 2010, 13:41 6 KUDOS 4 This post was BOOKMARKED I have faced problems with these kinds of questions often...finally after reading Sara's note I am clear on how to handle these problems... would like to present a summary of my understanding to clear potential doubts in the minds of people who are struggling with these concepts... --------------------------------------------- It is not correct to say - HEATING OIL PRICES THIS YEAR ARE EXPECTED TO BE HIGHER THAN LAST YEAR – this way we are comparing heating oil prices of this year with the LAST YEAR (an illogical comparison) where as what we intend to compare are the prices of heating oil in each of the two years). So we must rephrase it as - HEATING OIL PRICES THIS YEAR ARE EXPECTED TO BE HIGHER THAN THE HEATING OIL PRICE OF LAST YEAR WHICH CAN BE ELLIPSED TO HEATING OIL PRICES THIS YEAR ARE EXPECTED TO BE HIGHER THAN LAST YEAR’S [PRICE] OR HEATING OIL PRICES THIS YEAR ARE EXPECTED TO BE HIGHER THAN THEY WERE LAST YEAR Another form similar to the one above is - THIS YEAR, THE HEATING OIL PRICES ARE EXPECTED TO BE HIGHER THAN THEY WERE LAST YEAR (NOT HIGHER THAN LAST YEAR) Or ellipsed form THIS YEAR, THE HEATING OIL PRICES ARE EXPECTED TO BE HIGHER THAN LAST YEAR’S [HEATING OIL PRICE] Now if we reword the whole sentence as below – (we have removed THIS YEAR from the first part of the sentence and positioned it AFTER the main verb (are) HEATING OIL PRICES ARE EXPECTED TO BE HIGHER THIS YEAR THAN LAST Now the comparison focus shifts from Price to Time frame (this year versus last) Manager Joined: 28 Jul 2009 Posts: 123 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 26 Jun 2010, 22:33 nice explanation guys .. thanks .. Intern Joined: 02 Jul 2010 Posts: 26 Schools: Harvard (R2-applied), MIT (R2-applied) WE 1: Military WE 2: Social Entrepreneur Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 07 Jul 2010, 07:35 Quote: Heating-oil prices are expected to be higher th is year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. Hi everyone ....... Thanks a lot for such a detailed explanation!! There is just one thing that is still bugging me:- In option A ----- "Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were" last year, shouldn't there be "paying" AGAIN after "were" to make it as: Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were paying last year. How can we hide "paying" ? Is this also ellipses ?? If so, kindly explain as i have NOT understood this well !! Thanks so much _________________ .... ........ ............ Manager Joined: 25 May 2010 Posts: 136 Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 11 Jul 2010, 03:27 2 KUDOS Hi tgtharvard, Yes, this is ellipses. Since "paying" already appears in the sentence, there is no need to repeat it after the word "were". Best, Sarai _________________ Sarai Email me at saraiyaseen@gmail.com If this helped, kindly give Kudos! Intern Joined: 02 Jul 2010 Posts: 26 Schools: Harvard (R2-applied), MIT (R2-applied) WE 1: Military WE 2: Social Entrepreneur Re: Heating-oil prices are expected to be higher th is year than [#permalink] ### Show Tags 11 Jul 2010, 04:36 SaraiGMAXonline wrote: Hi tgtharvard, Yes, this is ellipses. Since "paying" already appears in the sentence, there is no need to repeat it after the word "were". Best, Sarai Dear Sarai .... Thanks for clarifying that ..... !! And now that it is indeed an ellipses, please bear with me as i have another doubt that comes to light if option A is correct. Please help me with this one as well !!! ....... Sorry for the trouble!!!! :- In the last part of this sentence: Quote: Heating-oil prices are expected to be higher this year than last because refiners are paying about \$5 a barrel more for crude oil than they were last year. Who does THEY refer to ???? Isn't it a case of unclear antecedent reference as "they" could refer to both "Heating-Oil prices" as well as "Refiners" ??? I know that the use of "were" makes it refer to "prices" but if we had used "did" instead of "were" it could as easily have referred to "refiners". In any case, the use of "were" or "did" doesnt stop "they from being called "an ambigios referrent", does it ?!?! I hope i have made my doubt clear and not confused you all further !! And if that is NOT the case, then could u please explain as to where am I faulting in my logic !!!! Thanks a ton ... Sam _________________ .... ........ ............ Re: Heating-oil prices are expected to be higher th is year than   [#permalink] 11 Jul 2010, 04:36 Go to page    1   2   3   4   5   6    Next  [ 105 posts ] Similar topics Replies Last post Similar Topics: 6 The expected rise in the price of oil could be a serious 7 28 Jul 2015, 21:06 In response to higher oil prices, window manufacturers have 11 30 Nov 2013, 05:17 Studies show that young people with higher-than-average 12 11 Mar 2012, 16:17 2 More than ever, paper is expected to be recycled this year, 20 18 Nov 2013, 20:19 3 Studies show that young people with higher-than-average 8 11 Sep 2015, 07:00 Display posts from previous: Sort by

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# Quicksort – Algorithm, Implementation and Performance Quicksort is a sorting algorithm that follows the divide-and-conquer approach. It works by dividing the input array into two sub-arrays, then recursively sorting each sub-array independently, and finally combining the sorted sub-arrays. In this article, we will discuss the implementation, complexity, advantages and disadvantages of the quicksort algorithm. ## 1. Quicksort Algorithm The Quicksort algorithm sorts the elements in place, which means it doesn’t require any additional memory allocation to sort the data, and also it can handle large datasets efficiently. Let us see how Quicksort works: • First, choose an element from the array. This element is called the pivot. The pivot can be any element in the list of data, but for simplicity, we usually choose the first or last element in the array. • After choosing the pivot element, divide the array into two sub-arrays, with all the elements less than the pivot on one side and all the elements greater than the pivot on the other. This process of dividing is called partitioning. • Recursively repeat the above step for each of the sub-arrays. Choose a pivot for each sub-array, partition it, and then sort the sub-arrays. • Finally, combine all the sorted sub-arrays to get the final sorted array. ## 2. Quicksort in Action Suppose we have the following array: In quicksort, the first step is to select a pivot element, which is used to divide the array into two sub-arrays. For simplicity, let’s choose the first element as the pivot of the given array, which is 5. Next, we partition the array into two sub-arrays: one with elements less than the pivot and the other with elements greater than the pivot. Note that if this array were sorted, the position of the pivot element (5) would be 2 (assuming zero-based indexing). Now we recursively apply the same process to both sub-arrays. This is where the divide-and-conquer approach comes in, as we are dividing our larger problems into smaller ones. For the first sub-array, we have two elements (3 and 2). Here, the pivot element would be 3. After positioning element 3 in its correct place, our array becomes [2, 3]. If we divide this sub-array again, it will divide into two single-size elements, i.e. [2] and [3], which are both base cases (In quicksort, a base case is when the sub-array to be sorted has 0 or 1 element). 2, 3, and 5 are at correct positions 0, 1, and 2. For the second sub-array, the pivot element would be 14. The pivot element, 14, will take its position. Since it is the largest element in the array, it will move to the last position as shown in the image below. Here, the array will only be partitioned into one sub-array since there are no elements on the right side. Therefore, the resulting sub-array will have only elements  [9, 11, 8, 6]. The pivot for this sub-array would be 9, as shown in the image below. 14 is at its correct position 7. Now, the elements smaller than pivot (9) would move to the left side, and greater elements will move to the right side. So the pivot will move to position 5. Now, the above array will be divided into two sub-arrays: one has elements [8, 6] and the other has only a single element, i.e. [11]. Therefore, we will not perform any operation with element 11 because it is single and cannot be further divided into sub-arrays. So, we have only one sub-array remaining, i.e. [8, 6]. 9 & 11 are at correct positions 5 and 6. In this subarray, the pivot would be element 8, so we choose this element as a pivot and move it to its correct position, i.e. position 4, resulting in [6, 8]. If we further divide this array, it will be divided into sub-arrays with only one element, and we cannot perform any further operations on these single elements. Therefore, the divide algorithm ends here, meaning we can only divide the array up to this level. 6 and 8 also got their correct positions 3 and 4. Finally, combine all the elements at their positions and you will get the final sorted array. ## 3. Implementation of Quicksort in Java Now, let us have a look at the implementation of quicksort in Java. • The quicksort() function checks whether the argument array has more than one element. If the subarray has only one element or is empty, then it is already sorted, and the function returns. Else, the partitioning is performed. • The partitioning depends on the pivot element. The selection of pivot element is tricky, so to keep things simple, we can choose a fixed position element in the array. Such as, in our example, the pivot has been selected as the first element for every sub-array. We can optionally choose the last element of the array or any other random position. • The function then recursively calls the quickSort function for the two subarrays created by the partition. This process continues until the entire subarray is sorted. • Combine the sorted subarrays to get the final sorted array. ``````import java.util.Arrays; public class Quicksort { public static void main(String[] args) { int[] arr = {5, 9, 2, 11, 14, 6, 3, 8}; System.out.println("Unsorted array: " + Arrays.toString(arr)); quicksort(arr, 0, arr.length - 1); System.out.println("Sorted array: " + Arrays.toString(arr)); } public static void quicksort(int[] arr, int start, int end) { if(arr != null && arr.length <= 1) { return; } if (start < end) { int pivotIndex = partition(arr, start, end); quicksort(arr, start, pivotIndex - 1); quicksort(arr, pivotIndex + 1, end); } } public static int partition(int[] arr, int start, int end) { int pivot = arr[start]; int i = start; for (int j = start + 1; j <= end; j++) { if (arr[j] < pivot) { i++; swap(arr, i, j); } } swap(arr, start, i); return i; } public static void swap(int[] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } }`````` ## 4. Performance ### 4.1. Time Complexity An algorithm that uses the divide-and-conquer approach has a time complexity of O(n log n). Similarly, quicksort has an average time complexity of O(n log n). However, if the input array is already sorted or nearly sorted and the pivot element is either the smallest or largest element in the array, the partition function will create a skewed partition. This means that one side of the partition will have no elements, while the other side will have the elements. In this case, the worst-case time complexity of quicksort becomes O(n2). • Best case: O(n log n) • Average case: O(n log n) • Worst case: O(n2) ### 4.2. Space Complexity Basic implementations of quicksort have a space complexity of O(log n) for recursive calls. However, in a worst-case scenario, where the partition function creates a skewed partition, the space complexity can reach O(n). • Best case: O(log n) • Average case: O(log n) • Worst case: O(n) • In-place sorting: Quicksort sorts the input array in place without requiring any additional memory. • Efficiency: Quicksort has an average-case time complexity of O(n log n), making it one of the fastest sorting algorithms. • Ease of implementation: Quicksort is relatively easy to implement and can be customized to fit specific requirements. • Worst-case performance: As previously mentioned, quicksort can have a worst-case time complexity of O(n2) if the input array is already sorted or almost sorted. • Unstable sorting: Quicksort is an unstable sorting algorithm, meaning it may not maintain the order of equal elements in the input array. • Pivot element: Quicksort’s performance heavily relies on selecting the pivot element. If a lousy pivot element is chosen, it can result in a skewed partition and impact sorting performance. ## 6. Conclusion In this article, we have gone through the visualization and implementation of quicksort, as well as the advantages and disadvantages of using this sorting algorithm. Furthermore, we have discussed how the choice of pivot element can impact sorting performance. Happy Learning !! Source Code on Github Subscribe Notify of Inline Feedbacks

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# Is there a bump? Given a list of positive integers determine if there is an element that is either greater than its two neighbors or less than its two neighbors (a "bump"). To be clear a bump can never be the first or last item of the list because they only have one neighbor. Your program should output one of two consistent values each corresponding to either a list with no bumps or a list with bumps. What the values are is unimportant you may choose them yourself. This is so answers will be scored in bytes with fewer bytes being better. ## Test cases [] -> False [1] -> False [1,2] -> False [1,2,1] -> True [1,2,2] -> False [1,2,3] -> False [1,2,2,1] -> False [1,2,2,3] -> False [1,2,1,2] -> True [1,3,2] -> True [3,1,2] -> True [2,2,2] -> False • Test Case Request: Numbers other than 0-1-2-3, also negatives allowed/disallowed? – Magic Octopus Urn Feb 14 '18 at 22:47 • Suggested test case: [1,3,3] (ensures that answers using Dennis's algorithm take the sign of the increments rather than just using the increments themselves) – ETHproductions Feb 15 '18 at 2:39 • @ETHproductions Isn't that already covered by [1,2,2]? Or am I missing something? – Fund Monica's Lawsuit Feb 15 '18 at 8:13 • @NicHartley, the deltas of [1,2,2] are the same as the signs of those deltas but that's not the case with [1,3,3]. – Shaggy Feb 15 '18 at 9:54 # Jelly, 5 bytes IṠIỊẠ Returns 0 if there's a bump, 1 if not. Try it online! ### How it works IṠIỊẠ Main link. Argument: A (integer array) I Increments; take all forward differences of A. Ṡ Take the signs. The signs indicate whether the array is increasing (1), decreasing (-1), or constant at the corresponding point. A 1 followed by a -1 indicates a local maximum, a -1 followed by a 1 a local minimum. I Increments; take the forward differences again. Note that 1 - (-1) = 2 and (-1) - 1 = -2. All other seven combinations of signs map to -1, 0, or 1. Ị Insignificant; map each difference d to (-1 ≤ d ≤ 1). Ạ All; return 1 if all differences are insignificant, 0 if not. • What is "Increments?". What is being incremented and what does that do? – Wheat Wizard Feb 15 '18 at 7:40 • @WheatWizard I think that this is the equivalent of 05AB1E's deltas (¥) command: an array [n0, n1, n2, n3] is poped and the array [n1-n0, n2-n1, n3-n2] is pushed. – Kaldo Feb 15 '18 at 8:49 # JavaScript (ES6), 38 bytes Returns a boolean. a=>a.some(x=n=>x*(x=a<n|-(a>(a=n)))<0) ### Test cases let f = a=>a.some(x=n=>x*(x=a<n|-(a>(a=n)))<0) console.log(f([])) // False console.log(f([1])) // False console.log(f([1,2])) // False console.log(f([1,2,1])) // True console.log(f([1,2,2])) // False console.log(f([1,2,3])) // False console.log(f([1,2,2,1])) // False console.log(f([1,2,2,3])) // False console.log(f([1,2,1,2])) // True console.log(f([1,3,2])) // True console.log(f([2,0,1])) // True console.log(f([2,2,2])) // False console.log(f([-4,100000,89])) // True ### How? We use a to store the previous value of n. We set x to 1 if a < n, -1 if a > n or 0 if a = n. And we test whether old_x * x < 0, which is only possible if (old_x = 1 and x = -1) or (old_x = -1 and x = 1). Because x is initialized to the anonymous callback function of some(), it is coerced to NaN during the first iteration, which makes the test falsy. • This will throw in strict mode. – Aluan Haddad Feb 15 '18 at 8:37 • @AluanHaddad Well, 99% of JS golfed code will throw in strict mode just because of undeclared variables. PPCG and codereview doesn't mix well. :P – Arnauld Feb 15 '18 at 8:51 • That's fair, I'm not really much for golf. – Aluan Haddad Feb 15 '18 at 8:55 • Then why comment on it lol – Mark C. Feb 16 '18 at 13:11 # Haskell, 42 bytes any(<0).f(*).f(-) Try it online! # Wolfram Language (Mathematica), 40 bytes MatchQ[{___,x_,y_,z_,___}/;x<y>z||x>y<z] Try it online! # Ruby, 55 46 bytes ->a{a.each_cons(3).any?{|x,y,z|(y-x)*(y-z)>0}} Try it online! A lambda accepting an array and returning boolean. -9 bytes: Replace (x<y&&y>z)||(x>y&&y<z) with (y-x)*(y-z)>0 (thanks to GolfWolf) ->a{ a.each_cons(3) # Take each consecutive triplet .any?{ |x,y,z| # Destructure to x, y, z (y-x)*(y-z) > 0 # Check if y is a bump } } • I think you can use | instead of ||, saving you 1 byte. – Yytsi Feb 15 '18 at 7:19 • 46 bytes – Cristian Lupascu Feb 15 '18 at 9:38 • Save 1 byte with '0<(y-x)*y-=z' – G B Jun 15 '18 at 6:49 # PostgreSQL 173 bytes SELECT DISTINCT ON(a)a,x>j and x>k OR x<least(j,k)FROM(SELECT a,x,lag(x,1,x)OVER(w)j,lead(x,1,x)OVER(w)k FROM d WINDOW w AS(PARTITION BY rn ORDER BY xn))d ORDER BY 1,2 DESC; a | c -----------+--- {1} | f {1,2} | f {1,2,1} | t {1,2,1,2} | t {1,2,2} | f {1,2,2,1} | f {1,2,2,3} | f {1,2,3} | f {1,3,2} | t {2,2,2} | f {3,1,2} | t (11 rows) • Hello and welcome to the site. I'm not familiar with PostgreSQL but you might be able to reduce the amount of whitespace you use. In general most languages don't require most of the kinds of spacing you use. – Wheat Wizard Feb 15 '18 at 21:38 • @WheatWizard it's the sample data into the database, it's irrelevant. – Evan Carroll Feb 15 '18 at 21:45 • What's your code then? We don't allow the input to be inserted directly into the code instead of the input. If that is the case here you ought to rewrite it so that it takes input via a standard method. – Wheat Wizard Feb 15 '18 at 21:48 • @WheatWizard from the link you provided, codegolf.meta.stackexchange.com/a/5341/23085 – Evan Carroll Feb 15 '18 at 21:49 • Ok if that's the input format you are using that's fine then. Best of luck golfing here, it's nice to see people golfing in less commonly used languages. – Wheat Wizard Feb 15 '18 at 21:52 # Java 8, 108104101868479 72 bytes a->{int i=a.length,p=0;for(;i-->1;)i|=p*(p=a[i]-a[i-1])>>-1;return-i>1;} -2 bytes thanks to @OlivierGrégoire. -13 bytes thanks to @Nevay. Try it online. • 84 bytes. I changed the iteration order (going down), swapped the two multiplication operands and then could remove a superfluous -1. – Olivier Grégoire Feb 16 '18 at 10:25 • 79 bytes: a->{int i=a.length;for(;i-->2;)i|=(a[i]-a[--i])*(a[i]-a[i-1])>>-1;return-~i|3;} (returns -1 for truthy cases, 3 for falsey cases) - or, if using the presence/absence of an exception as return value 55 bytes: a->{for(int i=0;++i>0;)i|=(a[i-1]-a[i])*(a[i]-a[i+1]);} – Nevay Feb 18 '18 at 12:58 • 72 bytes: a->{int i=a.length,p=0;for(;i-->1;)i|=p*(p=a[i]-a[i-1])>>-1;return-i>1;} – Nevay Feb 20 '18 at 13:11 # R, 58 56 bytes function(x)any(abs(diff(sign(diff(c(NA,x)))))>1,na.rm=T) Try it online! Saved 2 bytes thanks to Giuseppe • You can get rid of the braces {} for -2 bytes. – Giuseppe Feb 15 '18 at 7:35 • additionally, I think that you can port Stewie Griffin's approach for 42 bytes – Giuseppe Feb 15 '18 at 9:36 • @Giuseppe, I think Stewie ported my method, with the difference that mine can properly handle the empty vector as listed in the test cases. Matlab is a bit more lenient with empty vectors compared to R. – NofP Feb 18 '18 at 15:35 • c() is NULL which is not the same as the empty vector of integers, integer(0), whereas in MATLAB [] is a double by default, but if you want to keep it this way, that's perfectly reasonable. – Giuseppe Feb 20 '18 at 14:42 # J, 16 15 bytes -1 byte thanks to FrownyFrog 1 e.0>2*/\2-/\] Try it online! ## Original: 16 bytes 0>[:<./2*/\2-/\] 2-/\] - differences of each adjacent items 2*/\ - products of each adjacent items [:<./ - the minimum 0> - is negative? Try it online! • Hello ! Could this not be shortened to this simpler explicit form 0><./2*/\2-/\ (13 bytes) ? – Mathias Dolidon Feb 16 '18 at 14:27 • @Mathias Dolidon This works in the interpreter but here in PPCG it's common to provide a function (J verb) if there is some input. If the verb is tacit one, we don't count the assignment f=. bytes. Please have in mind that I'm relatively new user :) – Galen Ivanov Feb 16 '18 at 14:48 • So am I, and you've clarified the rule for me. Thanks ! :) – Mathias Dolidon Feb 16 '18 at 15:02 • 1 e.0>2*/\2-/\] – FrownyFrog Mar 3 '18 at 9:19 • @ FrownyFrog Thank you! It seems that I rarely use e. :) – Galen Ivanov Mar 3 '18 at 18:59 # Husk, 8 bytes ±V<0Ẋ*Ẋ- Try it online! ## Explanation ±V<0Ẋ*Ẋ- Implicit input, say [2,5,5,1,4,5,3] Ẋ- Consecutive differences: [3,0,-4,3,1,-2] Ẋ* Consecutive products: [0,0,-12,3,-2] V<0 Is any of them negative? Return 1-based index: 3 ± Sign (to make output consistent): 1 # Japt, 11 bytes -5 bytes thanks to @ETHproductions ä- mÌäa d>1 This uses Dennis's algorithm # Japt, 9 bytes ä- ä* d<0 Try it online! A mashup of Oliver's answer with the approach used by several other answers. # Attache, 39 bytes Any&:&{_*~?Sum[__]}@Slices&2@Sign@Delta Try it online! Pretty happy with how this turned out. ## Explanation This is a composition of four functions: Delta Sign Slices&2 Any&:&{_*~?Sum[__]} Delta gets the differences between elements. = Then, Sign is applied to each difference, giving us an array of 1s, 0s, and -1s. = Then, Slices&2 gives all slices of length two from the array, giving all pairs of differences. Finally, Any&:&{_*~?Sum[__]} is equivalent to, for input x: Any[&{_*~?Sum[__]}, x] Any[[el] -> { el[0] and not (el[0] + el[1] = 0) }, x] This searches for elements which sum to zero but are not zero. If any such pair of elements exist, then there is a bump. dZSd|1>a Try it online! # Octave, 33 bytes @(x)0||abs(diff(sign(diff(x))))>1 Try it online! ### Explanation: @(x) % Anonymous function taking x as input diff(x) % Takes the difference between consecutive elements sign(diff(x)) % The sign of the differences diff(sign(diff(x))) % The difference between the signs abs(diff(sign(diff(x)))>1 % Check if the absolute value is 2 @(x)abs(diff(sign(diff(x)))>1 % Output as matrices that are treated truthy or falsy # Brachylog, 10 bytes s₃.¬≤₁∧¬≥₁ Try it online! Succeeds (true.) if there is a bump, and fails (false.) if there is no bump. ### Explanation This is fairly readable already: s₃. There is a substring of the input… .¬≤₁ …which is not non-decreasing… ∧ …and… ¬≥₁ …which is not non-increasing # 05AB1E, 7 bytes ¥ü‚P0‹Z Try it online! Explanation ¥ # calculate delta's ü‚ # pair each element with the next element P # product of each pair 0‹ # check each if less than 0 Z # max • Wasn't there a 1-byte alternative to 0‹ that basically checks the number for a negative sign? – Magic Octopus Urn Feb 16 '18 at 15:33 • @MagicOctopusUrn: d used to check that top of stack only contained [0-9], which is the opposite of what we want here. But now it's more intelligent and negative/floats are also counted as numbers. – Emigna Feb 16 '18 at 15:39 • Ahhhh... coulda sworn a saw the negative sign and returned true or something... But I think you're right, I'm remembering your d trick. – Magic Octopus Urn Feb 16 '18 at 15:48 # Brachylog, 10 bytes s₃s₂ᶠ-ᵐ×<0 Try it online! Not nearly as neat and elegant as @Fatalize's existing 10 byte answer, but it works! s₃ % There exists a substring of three elements [I,J,K] in the array such that s₂ᶠ % When it's split into pairs [[I,J],[J,K]] -ᵐ % And each difference is taken [I-J, J-K] × % And those differences are multiplied (I-J)*(J-K) % (At a bump, one of those will be negative and the other positive. % At other places, both differences will be positive, or both negative, % or one of them 0 - ultimately resulting in a non-negative product.) <0 % The product is negative # Husk, 9 bytes ▼mεẊ-Ẋo±- Try it online! Uses Dennis's algorithm. # Perl 5, 54 + 2 (-pa) = 56 bytes map$\|=$F[$_-1]!=(sort{$a-$b}@F[$_-2..$_])[1],2..$#F}{ Try it online! # Python 2, 60 bytes lambda l:any(p>c<n or p<c>n for p,c,n in zip(l,l[1:],l[2:])) Try it online! Pretty much the same thing, thought it would be shorter though... # Python 2, 63 bytes f=lambda l:l[3:]and(l[0]>l[1]<l[2]or l[0]<l[1]>l[2]or f(l[1:])) Try it online! • 58 bytes using multiplication instead – Shieru Asakoto Feb 15 '18 at 7:49 # Pyt, 11 7 bytes ₋ʁ*0<Ʃ± Outputs 1 if there is a bump, 0 otherwise Try it online! # Old way (11 bytes): ₋±₋Å1≤ĐŁ↔Ʃ= Try it online! Returns False if there is a bump, True otherwise Port of Dennis' Jelly answer # Wolfram Language (Mathematica), 37 36 bytes FreeQ[(d=Differences)@Sign@d@#,-2|2]& Gives the opposite of the test case answers (False and True reversed). Prepend a ! to switch to the normal form. OR Abs@(d=Differences)@Sign@d@#~FreeQ~2& Also reversed output, so replace FreeQ with MatchQ for normal form. Explanation: Take the sign of the differences of the sequence. Iff the resulting sequence includes {1,-1} or {-1,1} there is a bump. The absolute value the differences of {1,-1} or {-1,1} is 2 in either case. Shave off another byte by squaring the final list instead of taking the absolute value: FreeQ[(d=Differences)@Sign@d@#^2,4]& Try it online! # Perl, 35 bytes Includes +3 for -p bump.pl: #!/usr/bin/perl -p s%\S+ %$a*1*($a=$&-$')%eg;\$_=/-/ Run as: bump.pl <<< "3 1 2" # Julia 0.6, 57 56 bytes l->any(p>c<n||p<c>n for(p,c,n)=zip(l,l[2:end],l[3:end])) Basically just totallyhuman's python answer. -1 byte from user71546 Try it online! # Julia 0.6, 39 bytes f(x,y,z,a...)=x>y<z||x<y>z||f(y,z,a...) Lispy recursion style, aka Dennis's python answer. Returns true when a bump exists, otherwise throws an error. This should maybe be 42 bytes since you have to splat it when calling. Eg for a=[1,2,1] you call as f(a...). f(a)=f(a...) would remove that need, but is longer. I need to get better a recursion, and I don't really like writing code that throws an error. Try it online! • seems the space after for is not necessary ;) – Shieru Asakoto Feb 15 '18 at 7:53

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https://www.convertunits.com/from/meganewton/to/yoctonewton

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## ››Convert meganewton to yoctonewton meganewton yoctonewton How many meganewton in 1 yoctonewton? The answer is 1.0E-30. We assume you are converting between meganewton and yoctonewton. You can view more details on each measurement unit: meganewton or yoctonewton The SI derived unit for force is the newton. 1 newton is equal to 1.0E-6 meganewton, or 1.0E+24 yoctonewton. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between meganewtons and yoctonewtons. Type in your own numbers in the form to convert the units! ## ››Quick conversion chart of meganewton to yoctonewton 1 meganewton to yoctonewton = 1.0E+30 yoctonewton 2 meganewton to yoctonewton = 2.0E+30 yoctonewton 3 meganewton to yoctonewton = 3.0E+30 yoctonewton 4 meganewton to yoctonewton = 4.0E+30 yoctonewton 5 meganewton to yoctonewton = 5.0E+30 yoctonewton 6 meganewton to yoctonewton = 6.0E+30 yoctonewton 7 meganewton to yoctonewton = 7.0E+30 yoctonewton 8 meganewton to yoctonewton = 8.0E+30 yoctonewton 9 meganewton to yoctonewton = 9.0E+30 yoctonewton 10 meganewton to yoctonewton = 1.0E+31 yoctonewton ## ››Want other units? You can do the reverse unit conversion from yoctonewton to meganewton, or enter any two units below: ## Enter two units to convert From: To: ## ››Definition: Meganewton The SI prefix "mega" represents a factor of 106, or in exponential notation, 1E6. So 1 meganewton = 106 newtons. The definition of a newton is as follows: In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force. ## ››Definition: Yoctonewton The SI prefix "yocto" represents a factor of 10-24, or in exponential notation, 1E-24. So 1 yoctonewton = 10-24 newtons. The definition of a newton is as follows: In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force. ## ››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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src/HOL/Transcendental.thy changeset 54575 0b9ca2c865cb parent 54573 07864001495d child 54576 e877eec2b698 ```--- a/src/HOL/Transcendental.thy Sun Nov 24 13:07:47 2013 +0000 +++ b/src/HOL/Transcendental.thy Sun Nov 24 18:37:25 2013 +0000 @@ -641,11 +641,8 @@ \<le> norm (c n) * of_nat n * of_nat (n - Suc 0) * r ^ (n - 2) * norm h" apply (simp only: norm_mult mult_assoc) apply (rule mult_left_mono [OF _ norm_ge_zero]) - apply (simp (no_asm) add: mult_assoc [symmetric]) - apply (rule lemma_termdiff3) - apply (rule h) - apply (rule r1 [THEN order_less_imp_le]) - apply (rule xh [THEN order_less_imp_le]) + apply (simp add: mult_assoc [symmetric]) + apply (metis h lemma_termdiff3 less_eq_real_def r1 xh) done qed qed @@ -653,9 +650,9 @@ lemma termdiffs: fixes K x :: "'a::{real_normed_field,banach}" assumes 1: "summable (\<lambda>n. c n * K ^ n)" - and 2: "summable (\<lambda>n. (diffs c) n * K ^ n)" - and 3: "summable (\<lambda>n. (diffs (diffs c)) n * K ^ n)" - and 4: "norm x < norm K" + and 2: "summable (\<lambda>n. (diffs c) n * K ^ n)" + and 3: "summable (\<lambda>n. (diffs (diffs c)) n * K ^ n)" + and 4: "norm x < norm K" shows "DERIV (\<lambda>x. \<Sum>n. c n * x ^ n) x :> (\<Sum>n. (diffs c) n * x ^ n)" unfolding deriv_def proof (rule LIM_zero_cancel) @@ -676,20 +673,23 @@ by (rule powser_inside [OF 1 5]) have C: "summable (\<lambda>n. diffs c n * x ^ n)" by (rule powser_inside [OF 2 4]) - show "((\<Sum>n. c n * (x + h) ^ n) - (\<Sum>n. c n * x ^ n)) / h - - (\<Sum>n. diffs c n * x ^ n) = - (\<Sum>n. c n * (((x + h) ^ n - x ^ n) / h - of_nat n * x ^ (n - Suc 0)))" - apply (subst sums_unique [OF diffs_equiv [OF C]]) - apply (subst suminf_diff [OF B A]) - apply (subst suminf_divide [symmetric]) - apply (rule summable_diff [OF B A]) + let ?dp = "(\<Sum>n. of_nat n * c n * x ^ (n - Suc 0))" + have "((\<Sum>n. c n * (x + h) ^ n) - (\<Sum>n. c n * x ^ n)) / h - (\<Sum>n. diffs c n * x ^ n) = + ((\<Sum>n. c n * (x + h) ^ n) - (\<Sum>n. c n * x ^ n)) / h - ?dp" + by (metis sums_unique [OF diffs_equiv [OF C]]) + also have "... = (\<Sum>n. c n * (x + h) ^ n - c n * x ^ n) / h - ?dp" + by (metis suminf_diff [OF B A]) + also have "... = (\<Sum>n. (c n * (x + h) ^ n - c n * x ^ n) / h) - ?dp" + by (metis suminf_divide [OF summable_diff [OF B A]] ) + also have "... = (\<Sum>n. (c n * (x + h) ^ n - c n * x ^ n) / h - of_nat n * c n * x ^ (n - Suc 0))" apply (subst suminf_diff) - apply (rule summable_divide) - apply (rule summable_diff [OF B A]) - apply (rule sums_summable [OF diffs_equiv [OF C]]) - apply (rule arg_cong [where f="suminf"], rule ext) + apply (auto intro: summable_divide summable_diff [OF B A] sums_summable [OF diffs_equiv [OF C]]) done + also have "... = (\<Sum>n. c n * (((x + h) ^ n - x ^ n) / h - of_nat n * x ^ (n - Suc 0)))" + finally show "((\<Sum>n. c n * (x + h) ^ n) - (\<Sum>n. c n * x ^ n)) / h + - (\<Sum>n. diffs c n * x ^ n) = + (\<Sum>n. c n * (((x + h) ^ n - x ^ n) / h - of_nat n * x ^ (n - Suc 0)))" . next show "(\<lambda>h. \<Sum>n. c n * (((x + h) ^ n - x ^ n) / h - of_nat n * x ^ (n - Suc 0))) -- 0 --> 0" by (rule termdiffs_aux [OF 3 4]) @@ -1159,13 +1159,25 @@ text {* Strict monotonicity of exponential. *} -lemma exp_ge_add_one_self_aux: "0 \<le> (x::real) \<Longrightarrow> (1 + x) \<le> exp(x)" - apply (drule order_le_imp_less_or_eq, auto) - apply (rule order_trans) - apply (rule_tac [2] n = 2 and f = "(\<lambda>n. inverse (real (fact n)) * x ^ n)" in series_pos_le) - apply (auto intro: summable_exp simp add: numeral_2_eq_2 zero_le_mult_iff) - done + assumes "0 \<le> (x::real)" shows "1+x \<le> exp(x)" +using order_le_imp_less_or_eq [OF assms] +proof + assume "0 < x" + have "1+x \<le> (\<Sum>n = 0..<2. inverse (real (fact n)) * x ^ n)" + by (auto simp add: numeral_2_eq_2) + also have "... \<le> (\<Sum>n. inverse (real (fact n)) * x ^ n)" + apply (rule series_pos_le [OF summable_exp]) + using `0 < x` + apply (auto simp add: zero_le_mult_iff) + done + finally show "1+x \<le> exp x" +next + assume "0 = x" + then show "1 + x \<le> exp x" + by auto +qed lemma exp_gt_one: "0 < (x::real) \<Longrightarrow> 1 < exp x" proof - @@ -1188,9 +1200,8 @@ qed lemma exp_less_cancel: "exp (x::real) < exp y \<Longrightarrow> x < y" - apply (simp add: linorder_not_le [symmetric]) - apply (auto simp add: order_le_less exp_less_mono) - done + unfolding linorder_not_le [symmetric] + by (auto simp add: order_le_less exp_less_mono) lemma exp_less_cancel_iff [iff]: "exp (x::real) < exp y \<longleftrightarrow> x < y" by (auto intro: exp_less_mono exp_less_cancel) @@ -2090,55 +2101,32 @@ lemma powr_mono2: "0 <= a ==> 0 < x ==> x <= y ==> x powr a <= y powr a" apply (case_tac "a = 0", simp) apply (case_tac "x = y", simp) - apply (rule order_less_imp_le) - apply (rule powr_less_mono2, auto) + apply (metis less_eq_real_def powr_less_mono2) done lemma powr_inj: "0 < a \<Longrightarrow> a \<noteq> 1 \<Longrightarrow> a powr x = a powr y \<longleftrightarrow> x = y" unfolding powr_def exp_inj_iff by simp lemma ln_powr_bound: "1 <= x ==> 0 < a ==> ln x <= (x powr a) / a" - apply (rule mult_imp_le_div_pos) - apply (assumption) - apply (subst mult_commute) - apply (subst ln_powr [THEN sym]) - apply auto - apply (rule ln_bound) - apply (erule ge_one_powr_ge_zero) - apply (erule order_less_imp_le) - done + by (metis less_eq_real_def ln_less_self mult_imp_le_div_pos ln_powr mult_commute + order.strict_trans2 powr_gt_zero zero_less_one) lemma ln_powr_bound2: assumes "1 < x" and "0 < a" shows "(ln x) powr a <= (a powr a) * x" proof - from assms have "ln x <= (x powr (1 / a)) / (1 / a)" - apply (intro ln_powr_bound) - apply (erule order_less_imp_le) - apply (rule divide_pos_pos) - apply simp_all - done + by (metis less_eq_real_def ln_powr_bound zero_less_divide_1_iff) also have "... = a * (x powr (1 / a))" by simp finally have "(ln x) powr a <= (a * (x powr (1 / a))) powr a" - apply (intro powr_mono2) - apply (rule order_less_imp_le, rule assms) - apply (rule ln_gt_zero) - apply (rule assms) - apply assumption - done + by (metis assms less_imp_le ln_gt_zero powr_mono2) also have "... = (a powr a) * ((x powr (1 / a)) powr a)" - apply (rule powr_mult) - apply (rule assms) - apply (rule powr_gt_zero) - done + by (metis assms(2) powr_mult powr_gt_zero) also have "(x powr (1 / a)) powr a = x powr ((1 / a) * a)" by (rule powr_powr) - also have "... = x" - apply simp - apply (subgoal_tac "a ~= 0") - using assms apply auto - done + also have "... = x" using assms + by auto finally show ?thesis . qed @@ -2488,12 +2476,7 @@ lemma real_mult_inverse_cancel: "[|(0::real) < x; 0 < x1; x1 * y < x * u |] ==> inverse x * y < inverse x1 * u" - apply (rule_tac c=x in mult_less_imp_less_left) - apply (auto simp add: mult_assoc [symmetric]) - apply (simp (no_asm) add: mult_ac) - apply (rule_tac c=x1 in mult_less_imp_less_right) - apply (auto simp add: mult_ac) - done + by (metis field_divide_inverse mult_commute mult_assoc pos_divide_less_eq pos_less_divide_eq) lemma real_mult_inverse_cancel2: "[|(0::real) < x;0 < x1; x1 * y < x * u |] ==> y * inverse x < u * inverse x1" @@ -2574,7 +2557,7 @@ lemma pi_half_gt_zero [simp]: "0 < pi / 2" apply (rule order_le_neq_trans) apply (simp add: pi_half cos_is_zero [THEN theI']) - apply (rule notI, drule arg_cong [where f=cos], simp) + apply (metis cos_pi_half cos_zero zero_neq_one) done lemmas pi_half_neq_zero [simp] = pi_half_gt_zero [THEN less_imp_neq, symmetric] @@ -2583,7 +2566,7 @@ lemma pi_half_less_two [simp]: "pi / 2 < 2" apply (rule order_le_neq_trans) apply (simp add: pi_half cos_is_zero [THEN theI']) - apply (rule notI, drule arg_cong [where f=cos], simp) + apply (metis cos_pi_half cos_two_neq_zero) done lemmas pi_half_neq_two [simp] = pi_half_less_two [THEN less_imp_neq] @@ -2646,11 +2629,7 @@ by (induct n) (auto simp add: real_of_nat_Suc distrib_right) lemma cos_npi2 [simp]: "cos (pi * real n) = -1 ^ n" -proof - - have "cos (pi * real n) = cos (real n * pi)" by (simp only: mult_commute) - also have "... = -1 ^ n" by (rule cos_npi) - finally show ?thesis . -qed + by (metis cos_npi mult_commute) lemma sin_npi [simp]: "sin (real (n::nat) * pi) = 0" by (induct n) (auto simp add: real_of_nat_Suc distrib_right) @@ -2665,10 +2644,7 @@ by simp lemma sin_gt_zero2: "[| 0 < x; x < pi/2 |] ==> 0 < sin x" - apply (rule sin_gt_zero, assumption) - apply (rule order_less_trans, assumption) - apply (rule pi_half_less_two) - done + by (metis sin_gt_zero order_less_trans pi_half_less_two) lemma sin_less_zero: assumes "- pi/2 < x" and "x < 0" @@ -2692,8 +2668,7 @@ lemma cos_gt_zero_pi: "[| -(pi/2) < x; x < pi/2 |] ==> 0 < cos x" apply (rule_tac x = x and y = 0 in linorder_cases) - apply (rule cos_minus [THEN subst]) - apply (rule cos_gt_zero) + apply (metis cos_gt_zero cos_minus minus_less_iff neg_0_less_iff_less) apply (auto intro: cos_gt_zero) done ```

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# $\cos(x)$ domain and range I'd like to refer the following answer: https://math.stackexchange.com/a/628992/130682 @robjohn claims that: $$\cos(x):\left[\frac1{\sqrt2},\frac\pi4\right]\mapsto\left[\frac1{\sqrt2},\frac\pi4\right]$$ $\pi\over 4$ is $a_1$ but where does $1\over \sqrt(2)$ came from? Update: My actual question is: Given $a_1 = {\pi \over 4}$, $a_n = \cos(a_{n-1})$ Why does the range of this recurrence is $\left[\frac1{\sqrt2},\frac\pi4\right]$ Thanks. $\dfrac{1}{\sqrt{2}}$ is $\cos\left(\dfrac{\pi}{4}\right)$ • So, $a_1 ={ \pi\over 4}, a_n = cos(a_{n-1})$ will always be in that range? May 28 '14 at 7:42 • @AnnieOK: true. May 28 '14 at 7:44 • can you explain why? May 28 '14 at 7:51 The statement is that if you take the cosine of any value in $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$ the result will be in $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$. You should be able to verify this. With regards to your specific question, $\cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}.$ Our first interval will be $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$ and then our second interval will be: $$\left[\cos\left(\dfrac{\pi}{4}\right),\cos\left(\dfrac{1}{\sqrt{2}}\right)\right] \approx \left[.7071067,.760244...\right].$$ Consider that our original interval was: $$\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right] \approx [.7071067, .78539...]$$ So the statement makes sense. • Indeed make sense, but I shall prove it mathematically.. May 28 '14 at 7:52 • @AnnieOK: What part do you want to prove mathematically? Do you want a proof of $\cos\left(\frac{1}{\sqrt{2}}\right) \in \left[\frac{1}{\sqrt{2}}, \frac{\pi}{4}\right]$ or are you OK with the demonstration above using approximate values? May 28 '14 at 8:40 • Yes, I shall prove it, and not demonstrate it.. May 28 '14 at 16:01

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## Difficulty level The probability of a hypothesis, given data. Difficulty level: Beginner Duration: 7:57 Speaker: : Barton Poulson Why math is useful in data science. Difficulty level: Beginner Duration: 1:35 Speaker: : Barton Poulson Why statistics are useful for data science. Difficulty level: Beginner Duration: 4:01 Speaker: : Barton Poulson Statistics is exploring data. Difficulty level: Beginner Duration: 2:23 Speaker: : Barton Poulson Graphical data exploration Difficulty level: Beginner Duration: 8:01 Speaker: : Barton Poulson Numerical data exploration Difficulty level: Beginner Duration: 5:05 Speaker: : Barton Poulson Simple description of statistical data. Difficulty level: Beginner Duration: 10:16 Speaker: : Barton Poulson Basics of hypothesis testing. Difficulty level: Beginner Duration: 06:04 Speaker: : Barton Poulson Enabling neuroscience research using high performance computing Difficulty level: Beginner Duration: 39:27 Speaker: : Subha Sivagnanam Tutorial describing the basic search and navigation features of the Allen Mouse Brain Atlas Difficulty level: Beginner Duration: 6:40 Speaker: : Unknown Tutorial describing the basic search and navigation features of the Allen Developing Mouse Brain Atlas Difficulty level: Beginner Duration: 6:35 Speaker: : Unknown Tutorial describing the basic features of the Brain Explorer® 3-D viewer for the mouse brain Difficulty level: Beginner Duration: 6:41 Speaker: : Unknown This tutorial demonstrates how to use the differential search feature of the Allen Mouse Brain Atlas to find gene markers for different regions of the brain and to visualize this gene expression in three-dimensional space. Differential search is also available for the Allen Developing Mouse Brain Atlas and the Allen Human Brain Atlas. Difficulty level: Beginner Duration: 6:31 Speaker: : Unknown The chair of the workshop is giving an introduction and a motivating argument. Difficulty level: Beginner Duration: 5:36 Speaker: : Maryann Martone This primer on optogenetics primer discusses how to manipulate neuronal populations with light at millisecond resolution and offers possible applications such as curing the blind and "playing the piano" with cortical neurons. Difficulty level: Beginner Duration: 59:06 Speaker: : Clay Reid Introduction to reproducible research. The lecture provides an overview of the core skills and practical solutions required to practice reproducible research. This lecture was part of the 2018 Neurohackademy, a 2-week hands-on summer institute in neuroimaging and data science held at the University of Washington eScience Institute. Difficulty level: Beginner Duration: 1:25:17 Speaker: : Fernando Perez This lecture was part of the 2019 Neurohackademy, a 2-week hands-on summer institute in neuroimaging and data science held at the University of Washington eScience Institute. Difficulty level: Beginner Duration: 1:03:07 Speaker: : Russell Poldrack This lecture was part of the 2019 Neurohackademy, a 2-week hands-on summer institute in neuroimaging and data science held at the University of Washington eScience Institute. Difficulty level: Beginner Duration: 55:39 Speaker: : Angela Laird The ionic basis of the action potential, including the Hodgkin Huxley model. Difficulty level: Beginner Duration: 28:29 Speaker: : Carl Petersen Introduction to the course Cellular Mechanisms of Brain Function. Difficulty level: Beginner Duration: 12:20 Speaker: : Carl Petersen

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# Algebra posted by . What is the correct factorization of x^2(x-8)+5(8-x)? • Algebra - x^3-8x^2+40-5x = x^3-8x^2-5x+40 ## Similar Questions 1. ### math what's the prime factorization of 293? what's the prime factorization of 525? 2. ### Algebra I need help in finding the factorization of -60m2 + 15n2... also the factorization of 6x2 - 2x - 20 Please help 3. ### CONSUSED!!!!HELP What is the prime factorization for 63 using The factorization tree? 4. ### math 115 1. a bedroom is 10ft. by 11ft. what is the perimeter? 5. ### Math How do i solve the following by factorization: 1) 7.7^2-2.3^2 2) 43*56+43*44 I know how to do these manually but not by factorization. 6. ### math last one help sally drew a factor tree of a composite number and ended up with 4x4x5x5x3 as the prime factorization. explain what is wrong with this factorization 1. what is the correct prime factorization? 7. ### math(help) please answer this one question because im clueless lol!!! A number is written with the following factorization:2exponet is 3 multiplied 4 multplied 5. is this factorization a prime factorization? 8. ### Algebra Which is the correct factorization for -3x^5 + 24x^2? 9. ### Math A number is written with the following factorization: 22 X 3 X 54 X 8 X 112. Is this factorization a prime factorization? 10. ### algebra When asked to write the prime factorization of the number 27 a student wrote 9•3 explain the error and write the correct answer More Similar Questions

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## ››Convert pfund [Denmark, Germany] to tola [India] pfund tola Did you mean to convert pfund to tola [India] tola [Pakistan] How many pfund in 1 tola? The answer is 0.023327625. We assume you are converting between pfund [Denmark, Germany] and tola [India]. You can view more details on each measurement unit: pfund or tola The SI base unit for mass is the kilogram. 1 kilogram is equal to 2 pfund, or 85.735260233307 tola. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between pfund [Denmark, Germany] and tola [India]. Type in your own numbers in the form to convert the units! ## ››Quick conversion chart of pfund to tola 1 pfund to tola = 42.86763 tola 2 pfund to tola = 85.73526 tola 3 pfund to tola = 128.60289 tola 4 pfund to tola = 171.47052 tola 5 pfund to tola = 214.33815 tola 6 pfund to tola = 257.20578 tola 7 pfund to tola = 300.07341 tola 8 pfund to tola = 342.94104 tola 9 pfund to tola = 385.80867 tola 10 pfund to tola = 428.6763 tola ## ››Want other units? You can do the reverse unit conversion from tola to pfund, or enter any two units below: ## Enter two units to convert From: To: ## ››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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PDF search accounting ratios problems with solutions Ratios accounting ratios problems with solutions [PDF] Unit – II Module - III Ratio Analysis Practical Problems (With Solutions) Financial Expenses Net Profit c/d 76,250 Solution – 1 (Problem related to Revenue Ratio) 1 Solution – 2 (Problem related to Balance Sheet Ratio) 1 PDF [PDF] Accounting Ratios - NCERT Lack of ability to resolve problems: Their role is essentially indicative and of whistle blowing and not providing a solution to the problem 3 Lack of standardised PDF [PDF] ACCOUNTING RATIOS – I - NIOS MODULE - 6A Analysis of Financial Statements Notes 23 Accounting Ratios - I ACCOUNTANCY Solution Current Ratio = Current Assets Current liabilities PDF [PDF] 3 Ratio Analysis position of a company is one of the basic problems in the working capital management Profitability ratios measure the degree of accounting profits 5 powerful software services and solutions built around the Internet, Windows and new PDF [PDF] RATIO ANALYSIS AND TREND ANALYSIS - OSOU explain the meaning and objectives of accounting ratios Administrative Expenses Ratio Solution: (i) Gross Profit Ratio= Gross Profit X 100 Problem 4 PDF [PDF] Ratio Analysis - IGCSE Accounts The higher the ratio, the worse the business is at getting debtors to pay on time and the more likely it is to have a high level of bad debts and cash problems PDF [PDF] Financial Statement and Ratio Analysis - Pearson Canada financial ratios, and the Du Pont ratio method In addition, you will One solution to the problem of finding good comparison numbers is to create your own list PDF [PDF] pg 1 Accounting Ratios 1 août 2018 · Accounting ratios are the arithmetical expression of relationship Solution: Calculation of Current Ratio: Current Ratio= Current Assets Problem 4: Calculate Debt Equity Ratio from the following balances of a company: PDF [PDF] FINANCIAL ANALYSIS AND PLANNING-RATIO - CA Intermediate Discuss use of financial ratios to analyse the financial statement ❒ in times of financial difficulty inventory may be saleable only at liquidation value But PDF [PDF] Chapter 14 Financial Ratios and Firm Performance problems, a manager or analyst can take a look at a firm's primary financial statements Financial ratios are relationships between different accounts from financial statements— Solutions for Advanced Problems for Spreadsheet Application PDF [DOC] Examples of Questions on Ratio Analysis Which of the following is considered a profitability measure? Days sales in ANSWERS TO PROBLEM: (note that these are just examples of a good answer) doc [DOC] Chapter 18 Ratio Analysis 1 3 Calculate and interpret the following categories of accounting ratios: It is not uncommon for a business which (on paper) is making a profit, to experience sever liquidity problems and there are two commonly used ratios 4 8 Solution: doc [DOC] Ratio Analysis, Test Bank - GimmeNotes Profitability ratios show the combined effects of liquidity, asset management, and debt management on None of the answers above is correct One of the problems of ratio analysis is that the relationships are subject to manipulation doc [DOC] RATIO ANALYSIS - ICSI Thus these ratios reflect the short-term financial solvency of a firm A firm should ensure that it does not suffer from lack of liquidity The failure to meet obligations doc [PPT] Ratio Analysis - - Backpack Comparison of financial ratios to past, industry, sector and all firms Financial Statements Balance Sheet; Income Statement; Cashflow Statement; Statement of File [DOC] Financial Statement Analysis For Small Businesses - Interise Figure 1: Summary Table of Financial Ratios, 36 the causes of the financial problems that statement analysis uncovers and suggests solutions for them doc [PPT] RATIO ANALYSIS - IIBF CAIIB – Financial Management – MODULE C – RATIO ANALYSIS R K MOHANTY FACULTY MEMBER, SIR SPBT COLLEGE, CENTRAL BANK OF INDIA, ppt [DOC] Chapter 2 Accounting Information and Regression Analysis Major problems with the use of accounting information arise from (1) classification Analysis of a series of ratios can give us a clear picture of a firm's financial doc The purpose of analyzing financial ratios is to enable you to answer the following If you're not sure of any of the answers, you may want to consider one of the Lower numbers may mean you have solvency issues, while high numbers may doc [DOC] Answers to Text Discussion Questions Explain the probable impact of replacement-cost accounting on the ratios of Explain why in problem 3 return on equity was so much higher than return on doc File [PDF] accounting ratios problems with solutions 1 PDF search We use coockies Read More -Savoir plus

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## Ramanujan’s 129th Birthday 22nd December 2016 Someone on Reddit pointed out that today (Dec 22nd) is Ramanujan’s 129th birthday. One of the commenters further pointed out that 129 is the sum of the first 10 primes, and the smallest number that is the sum of three squares in four different ways (echoing Ramanujan’s comment on G.H. Hardy’s taxi number 1729 being the smallest […] ## Have we already seen this number? (deja vu again) 8th April 2016 I had a phone interview for a Python job the other day. It started out really well, he was very impressed with my CV and called me a ‘rocket scientist’. But then we got down to the technical questions and my brain decided to go out to lunch, and (yet again) I managed to snatch defeat […] ## Number guessing game 1st October 2015 The number guessing game incorporates several simple but important elements of elementary programming and is a good candidate for an introductory programming example. If you’ve seen any programming course, you’ve probably seen the “Hello World” program. In Python it’s: print (“Hello World”)This is done to get the student(s) quickly to the point of having successfully installed […] ## Playing in Python with Pickover’s Premise per Pi 1st June 2015 My favourite tweeter is Clifford A. Pickover. His tweets are delightful nuggets of math, physics & more. I have a copy of his fascinating The Physics Book. However, in these tweets, he claims that the string 44899 first occurs in pi at position 44899, counting from the first digit after the decimal point. I thought it might be […] ## Compute Average Heights 3rd May 2015 This is a simple program to compute average heights. In a loop, it prompts the user for heights, inputs a text string and converts it to a ‘real’ number, sums them, and divides by the number of heights. Exit is signalled by negative height, after which the sum of the numbers is divided by the number of […] ## Binary Search 27th April 2015 Binary Search is one of the most fundamental computer algorithms. Given an ordered list of some data (names, numbers, …) find out if it contains a particular item. For example, consider the list: 2, 4, 5, 7, 8, 11, 12. If we ask if it contains the number 5, the algorithm should return 2 (counting […] ## FizzBuzz 16th April 2015 This is apparently a well-known interview question (but not one I was ever asked). Print the numbers 1 to 100, but replace every number divisible by 3 with “Fizz”, every number divisible by 5 with “Buzz”, and every number that’s divisible by both 3 & 5 with “FizzBuzz”. Apparently many people have difficulty with it. […]

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# Coding Fractals by trees Recently I’ve been editing a set of notes taken by professor Kra during Furstenberg’s course last year. (Well…I should have done this last year >.<) One of the main ideas in the course was to build a correspondence between trees and Fractal sets – and hence enables one to prove statements about dimensions and projections of Fractals by looking at the corresponding trees. I want to sketch some highlights of the basic idea here. Let $Q=Q^{(n)}$ denote the unit cube in $\mathbb{R}^{n}$. Let $A\subset\mathbb{R}^{n}$ and $x\in A$. For $t\in\mathbb{R}$, consider the family $t(A-x)\cap Q$. Question:Does the limit exist as $t\to\infty$? (here we consider the Hausdorff limit on compact subsets of $Q$) i.e. we ‘zoom in’ the set around the point $x$, always crop the set to the cube $Q$ and consider what does the set ‘look like’ when the strength of the magnifying glass approaches infinity. Example: If $A$ is smooth (curve of surface), then this limit exists and is a subspace intersected with $Q$. Generally one should not expect the above limit to exist for fractal sets, however if we weaken the question and ask for the limit when we take a sequence $(n_k)\rightarrow\infty$, then it’s not hard to see self-similar sets would have non-trivial limits. Hence in some sense fractal behavior is characterized by having non-trivial limit sets. Definition: Let $A\subset Q^{(n)}$, $A'$ is a mini-set of $A$ if for some $\lambda\geq 1$ and $u\in\mathbb{R}^{n}$, $A' =(\lambda A+u)\cap Q$ $A''$ is a micro-set of $A$ if there is sequence of minisets $A'_{n}$ of $A$ and $A'' = \lim_{n\to\infty}A'_{n}$ $A$ is homogeneous if all of its microsets are minisets. As we should see below, homogeneous fractals are ‘well-behaved’ in terms of dimensions. Here comes a version of our familiar definition: Definition:A set $A\subset X$ has Hausdorff $\alpha$-measure $0$ if for every $\varepsilon > 0$, there exists a covering of $A\subset \bigcup_{n=1}^{\infty}B_{\rho_{n}}$ with $\sum_{n}\rho_{n}^{\alpha} \alpha$. Thus it makes sense to define: Definition: The Hausdorff dimension of $A$ is $\dim(A) = \inf\{\alpha>0\colon \text{Hausdorff } \alpha \text{ measure is } 0 \}$. Now comes trees~ For $A\subset[0,1]$ closed, consider expansion in base $3$ of numbers in $A$. In the expansion of each number in $A$, there will be certain digits which appear. Following this digit, there may be certain digits that can appear. This defines a tree, which is a tree of shrinking triadic intervals that intersects the set $A$. Definition: Let $\Lambda$ be a finite set of symbols (which we will refer to as the alphabet and elements of $\Lambda$ are letters of the alphabet). A word $w$ is any concatenation of finitely many symbols in $\Lambda$, the number of symbols in $w$ is called the length of $w$, denoted by $\ell(w)$. A tree $\tau$ in the alphabet $\Lambda$ is a set of words satisfying 1. If $uv\in\tau$ then $u\in\tau$. 2. If $u\in\tau$, then there exists a letter $a\in\Lambda$ such that $ua\in\tau$. Notation: if $w\in\tau$, denote $\tau^{w} = \{v\colon wv\in\tau\}$. Definition: A section $\Sigma$ of $\tau$ is a finite set of words for which there exists $N$ such that if $s\in\tau$ and $\ell(s) \geq N$, then there exists $r\in\Sigma$ and a word $w$ such that $s = rw$. Definition: The Hausdorff dimension of a tree is defined by $\dim(\tau)=\inf\{ \beta \ | \ \inf_{\Sigma}\{\sum_{r\in\Sigma} e^{-\beta\ell(r)}\} = 0 \}$ where $\Sigma$ is any section of $\tau$. Theorem: The Hausdorff dimension of a tree equals the Hausdorff dimension of set it corresponds to. The proof is merely going through the definition, hence we won’t present here. It suffice to check the dimension is unchanged if we only consider open ball covers with balls of radius $p^{-n}$ for any given $p \in \N$. Which is indeed true. Now let’s give a simple example to illustrate how to make use of this correspondence: It is easy to prove $\dim(X \times Y) \geq \dim(X) \times \dim(Y)$, here we produce an example were the inequality is strict. Let $A \subseteq [0,1]$ be the set of numbers whose decimal expansion vanishes from the $k_{2n}$ to $k_{2n+1}$-th digits to the right of $0$. To compute $\dim(A)$, look at levels such that a small number of intervals will cover the set $A$. The number of intervals of size $10^{k_{2n+1}}$ is less than $10^{k_{2n}}$. Then if $\frac{10^{k_{2n}}}{k_{2n+1}}$ goes to $0$, we’ll have that the Hausdorff dimension is $0$. So we just need $k_{2n}/k_{2n+1}\to 0$. Let $B$ be the set of numbers whose decimals vanish from $k_{2n-1}$ to $k_{2n}-1$-th digits. By the same computation, $\dim(B) = 0$. What about $A\times B$? The tree that corresponds to this is a regular tree which every point has $10$ branches. So the Hausdorff dimension is $1$. Note: Here the structure of the tree gives you easy information on the set, but it is hard to look directly at the set and see this information. Many theorems regarding dimension of projections and intersections of fractal sets turns out to be simple to prove after we coded the sets by trees. # On compact extensions This is again a note on my talk in the Szemerédi’s theorem seminar, going through Furstenberg’s book. In this round, my part is to introduce compact extension. Let $\Gamma$ be an abelian group of measure preserving transformations on $(X, \mathcal{B}, \mu)$, $\alpha: (X, \mathcal{B}, \mu, \Gamma) \rightarrow ( Y, \mathcal{D}, \nu, \Gamma')$ be an extension map. i.e. $\alpha: X \rightarrow Y$ s.t. $\alpha^{-1}$ sends $\nu-0$ sets to $\mu-0$ sets; $\gamma'\circ \alpha (x) = \alpha \circ \gamma (x)$ Definition: A sequence of subsets $(I_k)$ of $\Gamma$ is a Folner sequence if $|I_k| \rightarrow \infty$ and for any $\gamma \in \Gamma$, $\frac{| \gamma I_k \Delta I_k|}{|I_k|} \rightarrow 0$ Proposition: For any Folner sequence $I = (I_k)$ of $\Gamma$, for any $f \in L^1(X)$, $\displaystyle \frac{1}{|I_k|} \sum_{\gamma \in I_k} \gamma f$ converges weakly to the orthogonal projection of $f$ onto the subspace of $\Gamma$-invariant functions. (Denoted $P(f)$ where $P: L^2(X) \rightarrow L^2_{inv}(X)$. Proof: Let $\mathcal{H}_0 = P^{-1}(\bar{0}) = (L^2_{inv}(X))^\bot$ For all $\gamma \in \Gamma$, $\gamma (L^2_{inv}(X)) \subseteq L^2_{inv}(X)$ Since $\Gamma$ is $\mu$-preserving, $\gamma$ is unitary on $L^2(X)$. Therefore we also have $\gamma( \mathcal{H}_0) \subseteq \mathcal{H}_0$. For $f \in \mathcal{H}_0$, suppose there is subsequence $(n_k)$ where $\displaystyle \frac{1}{|I_{n_k}|} \sum_{\gamma \in I_{n_k}} \gamma (f)$ converges weakly to some $g \in L^2(X)$. By the property that $\frac{| \gamma I_k \Delta I_k|}{|I_k|} \rightarrow 0$, we have for each $\gamma \in \Gamma$, $\gamma(g) = g, \ g$ is $\Gamma$-invariant. i.e. $g \in (\mathcal{H}_0)^\bot$ However, since $f \in \mathcal{H}_0$ hence all $\gamma(f)$ are in $\mathcal{H}_0$ hence $g \in \mathcal{H}_0$. Therefore $g \in \mathcal{H}_0 \cap (\mathcal{H}_0)^\bot$, $g=\bar{0}$ Recall: 1)$X \times_Y X := \{ (x_1, x_2) \ | \ \alpha(x_1) = \alpha(x_2) \}$. i.e. fibred product w.r.t. the extension map $\alpha: X \rightarrow Y$. 2)For $H \in L^2(X \times_Y X), \ f \in L^2(X)$, $(H \ast f)(x) = \int H(x_1, x_2) f(x_2) d \mu_{\alpha(x_1)}(x_2)$ Definition: A function $f \in L^2(X)$ is said to be almost periodic if for all $\varepsilon > 0$, there exists $g_1, \cdots g_k \in L^2(X)$ s.t. for all $\gamma \in \Gamma$ and almost every $y \in Y$, $\displaystyle \min_{1 \leq i \leq k} || \gamma (f) - g_i||_y < \varepsilon$ Proposition: Linear combination of almost periodic functions are almost periodic. Proof: Immediate by taking all possible tuples of $g_i$ for each almost periodic function in the linear combination corresponding to smaller $\varepsilon$l. Definition: $\alpha: (X, \mathcal{B}, \mu, \Gamma) \rightarrow ( Y, \mathcal{D}, \nu, \Gamma')$ is a compact extension if: C1: $\{ H \ast f \ | \ H \in L^\infty (X \times_Y X) \cap \Gamma_{inv} (X \times_Y X)$, $f \in L^2(X) \}$ contains a basis of $L^2(X)$. C2: The set of almost periodic functions is dense in $L^2(X)$ C3: For all $f \in L^2(X), \ \varepsilon, \delta > 0$, there exists $D \subseteq Y, \ \nu(D) > 1- \delta, \ g_1, \cdots, g_k \in L^2(X)$ s.t. for any $\gamma \in \Gamma$ and almost every $y \in Y$, we have $\displaystyle \min_{1 \leq i \leq k} || \gamma (f)|_{f^{-1}(D)} - g_i||_y < \varepsilon$ C4: For all $f \in L^2(X), \ \varepsilon, \delta > 0$, there exists $g_1, \cdots, g_k \in L^2(X)$ s.t. for any $\gamma \in \Gamma$, there is a set $D \subseteq Y, \ \nu(D) > 1- \delta$, for all $y \in D$ $\displaystyle \min_{1 \leq i \leq k} || \gamma (f) - g_i||_y < \varepsilon$ C5: For all $f \in L^2(X)$, let $\bar{f} \in L^1(X \times_Y X)$ where $\bar{f}: (x_1, x_2) \mapsto f(x_1) \cdot f(x_2)$ Let $I=(I_k)$ be a Folner sequence, then $\bar{f}=\bar{0}$ iff $P \bar{f} = \bar{0}$. Theorem: All five definitions are equivalent. Proof: “C1 $\Rightarrow$ C2″ Since almost periodic functions are closed under linear combination, it suffice to show any element in a set of basis is approximated arbitrarily well by almost periodic functions. Let our basis be as given in C1. For all $H \in L^\infty (X \times_Y X) \cap \Gamma_{inv} (X \times_Y X)$, the associated linear operator $\varphi_H: L^2(X) \rightarrow L^2(X)$ where $\varphi_H: f \mapsto H \ast f$ is bounded. Hence it suffice to check $H \ast f$ for a dense set of $f \in L^2(X)$. We consider the set of all fiberwise bounded $f$ i.e. for all $y \in Y$, $||f||_y \leq M_y$. For all $\delta > 0$, we perturb $H \ast f$ by multiplying it by the characteristic function of a set of measure at least $1- \delta$ to get an almost periodic function. “C2 $\Rightarrow$ C3″: For any $f \in L^2(X)$, there exists $f'$ almost periodic, with $||f-f'||< \frac{\epsilon \sqrt{\delta}}{2}$ . Let $\{ g_1, g_2, \cdots, g_{k-1} \}$ be the functions obtained from the almost periodicity of $f'$ with constant $\varepsilon/2$, $g_k = \bar{0}$. Let $D = \{ y \ | \ ||f-f'||_y < \varepsilon/2 \}$, since $|| f - f'||^2 = \int ||f-f'||_y^2 d \nu(y)$ Hence $||f-f'||< \frac{\varepsilon \sqrt{\delta}}{2} \ \Rightarrow \ ||f-f'||^2 < \frac{\varepsilon^2 \delta}{4}$, $\{ y \ | \ ||f-f'||_y > \varepsilon/2 \}$ has measure at most $\delta/2$, therefore $\nu(D) > 1- \delta$. For all $\gamma \in \Gamma$, if$y \in \gamma^{-1}(D)$ then $|| \gamma f|_{\alpha^{-1}(D)} - \gamma f'||_y = ||f|_{\alpha^{-1}(D)} - f'||_{\gamma(y)} < \varepsilon /2$ Hence $\displaystyle \min_{1 \leq i \leq k-1} ||\gamma f|_{\alpha^{-1}(D)} - g_i||_y < \varepsilon /2 + \varepsilon /2 = \varepsilon$ If $y \notin \gamma^{-1}(D)$ then $f|_{\alpha^{-1}(D)}$ vanishes on $\alpha^{-1}(\gamma y)$ so that $|| \gamma f|_{\alpha^{-1}(D)} - g_i||_y = 0 < \varepsilon$. Hence $\alpha$ satisfies C3. “C3 $\Rightarrow$ C4″: This is immediate since for all $y \in \gamma^{-1}(D)$, we have $\gamma f = \gamma f|_{\alpha^{-1}(D)}$ on $\alpha^{-1}(y)$ hence $\displaystyle \min_{1 \leq i \leq k} ||\gamma f - g_i||_y < \min_{1 \leq i \leq k-1} ||\gamma f_{\alpha^{-1}(D)} - g_i||_y < \varepsilon$ $\nu(\gamma^{-1}(D)) = \nu(D) > 1-\delta$. Hence $\alpha$ satisfies C4. “C4 $\Rightarrow$ C5″: For all $f \in L^2(X), \ \varepsilon, \delta > 0$, by C4, there exists $g_1, \cdots, g_k \in L^2(X)$ s.t. for any $\gamma \in \Gamma$, there is a set $D \subseteq Y, \ \nu(D) > 1- \delta$, for all $y \in D$ $\displaystyle \min_{1 \leq i \leq k} || \gamma (f) - g_i||_y < \varepsilon$ W.L.O.G. we may suppose all $g_i$ are bounded since by making $\delta$ slighter larger we can modify the unbounded parts to be bounded. $\bar{g_j} \otimes g_j \in L^\infty(X \times_Y X)$, suppose $P(\bar{f}) = 0$. Recall in C5 we have $\bar{f}: (x_1, x_2) \mapsto f(x_1) \cdot f(x_2)$, and $\displaystyle P_I \bar{f}(x_1, x_2) = \lim_{k \rightarrow \infty} \frac{1}{|I_k|} \sum_{\gamma \in I+k} f(\gamma x_1) \bar{ f(\gamma x_2)}$. For each $1 \leq j \leq k$, we have $\int (\bar{g_j} \otimes g_j) \cdot P \bar{f} d(\mu \times_Y \mu) = 0$ Hence we have $\displaystyle \lim_{i \rightarrow \infty} \frac{1}{|I_i|} \sum_{\gamma \in I_i} \int (\bar{g_j(x_1)} g_j(x_2)) \cdot$ $\gamma f(x_1) \bar{\gamma f(x_2)} d\mu \times_Y \mu(x_1, x_2) = 0$ $\Rightarrow \displaystyle \lim_{i \rightarrow \infty} \frac{1}{|I_i|} \sum_{\gamma \in I_i} \int | \int \bar{g_j(x)} \gamma f(x) d\mu_y(x)|^2 d \nu(y) = 0$ $\Rightarrow \displaystyle \lim_{i \rightarrow \infty} \frac{1}{|I_i|} \sum_{\gamma \in I_i} \{ \sum_{j=1}^k \int | \int \bar{g_j(x)} \gamma f(x) d\mu_y(x)|^2 d \nu(y) \} = 0$ Hence for large enough $i$, there exists $\gamma \in I_i$ s.t. $\sum_{j=1}^k \int | \int \bar{g_j(x)} \gamma f(x) d\mu_y(x)|^2 d \nu(y)$ is as small as we want. We may find $D' \subseteq Y$ with $\nu(D) > 1-\delta$ s.t. for all $y \in D'$ and for all $j$, we have $| \int \bar{g_j(x)} \gamma f(x) d\mu_y(x)|^2 < \varepsilon^2$ On the other hand, by construction there is $j$ with $|| \gamma f - g_j||^2_y < \varepsilon^2$ for all $y \in D$, with $\nu(D) > 1-\delta$. Hence for $y \in D \cap D', \ ||f||_{\gamma'^{-1}(y)}^2 = || \gamma f||_y^2 < 3 \varepsilon^2$. Let $\varepsilon \rightarrow 0, \ \delta \rightarrow 0$ we get $f = \bar{0}$. Hence C5 holds. “C5 $\Rightarrow$ C1″ Let $f \in L^2(X)$ orthogonal to all of such functions. Let $(I_k)$ be a Folner sequence. Define $\displaystyle H(x_1, x_2) := \lim_{i \rightarrow \infty} \frac{1}{|I_i|}\sum_{\gamma \in I_i} \gamma f(x_1) \cdot \gamma f(x_2) = P \bar{f}(x_1, x_2)$ Let $H_M(x_1, x_2)$ be equal to $H$ whenever $H(x_1, x_2) \leq M$ and $0$ o.w. $H$ is $\Gamma$-invariant $\Rightarrow \ H_M$ is $\Gamma$-invariant and bounded. Therefore $f \bot H_M \ast f$, i.e. $\int \bar{f(x_1)} \{ \int H_M(x_1, x_2) d \mu_{\alpha(x_1)}(x_2) \} d \mu(x_1) = 0$ <\p> Since $\mu = \int \mu_y d \nu(y)$, we get $\int \bar{f} \otimes f \cdot H_M d \mu \times_Y \mu = 0$ <\p> Hence $H_M \bot (\bar{f} \otimes f)$. For all $\gamma \in \Gamma, \ \gamma (\bar{f} \otimes f) \bot \gamma H_M = H_M$. Since $H = P \bar{f}$ is an average of $\gamma (\bar{f} \otimes f), \ \Rightarrow \ H \bot H_M$. $0 = \int \bar{H} \cdot H_M = \int |H_M|^2 \ \Rightarrow \ H_M = \bar{0}$ for all $M$ Hence $H = \bar{0}$. By C5, we obtain $f = \bar{0}$. Hence $\{ H \ast f \ | \ H \in L^\infty (X \times_Y X) \cap \Gamma_{inv} (X \times_Y X)$, $f \in L^2(X) \}$ contain a basis for $L^2(X)$. Definition: Let $H$ be a subgroup of $\Gamma$, $\alpha: (X, \mathcal{B}, \mu, \Gamma) \rightarrow ( Y, \mathcal{D}, \nu, \Gamma')$ is said to be compact relative to $H$ if the extension $\alpha: (X, \mathcal{B}, \mu, H) \rightarrow ( Y, \mathcal{D}, \nu, H')$ is compact. # Notes for my lecture on multiple recurrence theorem for weakly mixing systems – Part 2 Now we can start to prove the multiple recurrence theorem in the weak mixing case. Again the material is from Furstenberg’s book ‘Recurrence in Ergodic Theory and Combinatorial Number Theory’ which, very unfortunately, is hard to find a copy of. Definition: A sequence $(x_i) \subseteq X$ converges in density to $x$ if there exists $Z \subseteq \mathbb{N}$ of density $0$ s.t. for all neighborhood $U$ of $x$, $\exists N \in \mathbb{N}, \ \{ x_n \ | \ n \geq N$ and $n \notin Z \} \subseteq U$. We denote $(x_n) \rightarrow_D \ x$. Theorem: For $(X, \mathcal{B}, \mu, T)$ weakly mixing, then $\forall f_0, f_1, \cdots, f_k \in L^\infty(X)$, we have $\int f_0(x) f_1(T^n(x)) f_2(T^{2n}(x)) \cdots f_k(T^{kn}(x)) \ d \mu$ $\rightarrow_D \int f_0 \ d \mu \int f_1 \ d \mu \cdots \int f_k \ d \mu$ as $n \rightarrow \infty$ In particular, this implies for any $A \in \mathcal{B}$ with $\mu(A)>0$, by taking $f_0 = f_1 = \cdots = f_k = \chi_A$ we have $\mu(A \cap T^{-n}(A) \cap \cdots \cap T^{-kn}(A))$ $\rightarrow_D \mu(A)^k > 0$. Hence we may pick $N \in \mathbb{N}$ for which $\mu(A \cap T^{-N}(A) \cap \cdots \cap T^{-kN}(A)) > 0$. Establishes the multiple recurrence theorem. To prove the theorem we need the following: Lemma 1: Let $(f_n)$ be a bounded sequence in Hilbert space $\mathcal{H}$, if $\langle f_{n+m}, f_n \rangle \rightarrow_D a_m$ as $n \rightarrow \infty$, $a_m \rightarrow_D 0$ as $m \rightarrow \infty$. Then $(f_n)$ converges weakly in density to $\overline{0}$ In order to prove the lemma 1, we need: Lemma 2: Given $\{ R_q \ | \ q \in Q \}$ a family of density $1$ subsets of $\mathbb{N}$, indexed by density $1$ set $Q \subseteq \mathbb{N}$. Then for all $S \subseteq \mathbb{N}$ of positive upper density, for all $k \geq 1$. There exists $\{ n_1, n_2, \cdots, n_k \} \subseteq S$, $n_1 < n_2 \cdots < n_k$ such that whenever $ii \in Q$ and $n_i \in R_{n_j-n_i}$. Proof of lemma 2: We’ll show this by induction. For $k=1$, since there is no such $j>1$, the statement is vacant. We’ll proceed by induction: Suppose for $k>1$, there exists $S_k \subseteq S$ of positive upper density and integers $m_1< \cdots < m_k$ such that $(S_k+m_1) \cup (S_k+m_2) \cup \cdots \cup (S_k+m_k) \subseteq S$ and for all $j>i$, $m_j - m_i \in Q$ and $S_k+m_i \subseteq R_{m_j-m_i}$. For $k+1$, we shall find $S_{k+1} \subseteq S_k$ with positive upper density and $m_{k+1}>m_k$ where $S_{k+1}+m_{k+1} \subseteq S$ and for all $1 \leq i \leq k$, $m_{k+1} - m_i \in Q$ and $S_{k+1}+m_i \subseteq R_{m_{k+1}-m_i}$. Let $S_k^* = \{n \ | \ S_k+n \cap S_k$ has positive upper density $\}$. Claim:$S_k^*$ has positive upper density. Since $\overline{D}(S_k) = \epsilon >0$, let $N = \lceil 1/ \epsilon \rceil$. Hence there is at most $N-1$ translates of $S_k$ that pairwise intersects in sets of density $0$. Let $M < N$ be the largest number of such sets, let $p_1, \cdots, p_M$ be a set of numbers where $(S_k+p_i) \cap (S_k+p_j)$ has density $0$. i.e. $S_k+(p_j-p_i) \cap S_k$ has density $0$. Therefore for any $p>p_M$, $(S_k+p-p_i) \cap S_k$ has positive upper density for some $i$. Hence $p-p_i \in S_k^*$. $S_k^*$ is syntactic with bounded gap $2 \cdot p_M$ hence has positive upper density. Pick $\displaystyle m_{k+1} \in S_k^* \cap \bigcap_{i=1}^k(Q+m_i)$. (Hence $m_{k+1}-m_i \in Q$ for each $1 \leq i \leq k$) Let $\displaystyle S_{k+1} = (S_k - m_{k+1}) \cap S_k \cap \bigcap_{i=1}^k (R_{m_k+1 - m_i}-m_i)$. $(S_k - m_{k+1}) \cap S_k$ has positive upper density, $\bigcap_{i=1}^k (R_{m_k+1 - m_i}-m_i)$ has density $1$, $S_{k+1}$ has positive upper density. $S_{k+1}, \ m_{k+1}$ satisfied the desired property by construction. Hence we have finished the induction. Proof of lemma 1: Suppose not. We have some $\epsilon > 0, \ f \neq \overline{0}$, $S = \{ n \ | \ \langle f_n, f \rangle > \epsilon \}$ has positive upper density. Let $\delta = \frac{\epsilon^2}{2||f||^2}$, let $Q = \{m \ | \ a_m < \delta/2 \}$ has density $1$. $\forall q \in Q$, let $R_q = \{ n \ | \ \langle f_{n+q}, f_n \rangle < \delta \}$ has density $1$. Apply lemma 2 to $Q, \{R_q \}, S$, we get: For all $k \geq 1$. There exists $\{ n_1, n_2, \cdots, n_k \} \subseteq S$, $n_1 < n_2 \cdots < n_k$ such that whenever $i, $n_j - n_i \in Q$ and $n_i \in R_{n_j-n_i}$. i) $n_i \in S \ \Leftrightarrow \ \langle f_{n_i}, f \rangle > \epsilon$ ii) $n_i \in R_{n_j-n_i} \ \Leftrightarrow \ \langle f_{n_i}, f_{n_j} \rangle < \delta$ Set $g_i = f_{n_i} - \epsilon \cdot \frac{f}{||f||}$. Hence $\forall \ 1 \leq i < j \leq k$ $\langle g_i, g_j \rangle = \langle f_{n_i} - \epsilon \frac{f}{||f||}, f_{n_j} - \epsilon \frac{f}{||f||}\rangle$ $< \delta - 2\cdot \frac{\epsilon^2}{||f||^2} + \frac{\epsilon^2}{||f||^2} = \delta - \frac{\epsilon^2}{||f||^2}= -\delta$. On the other hand, since $(f_n)$ is bounded in $\mathcal{H}, \ (g_n)$ is also bounded (independent of $k$). Suppose $||g_n||< M$ for all $k$, then we have $\displaystyle 0 \leq || \sum_{i=1}^k g_i ||^2 = \sum_{i=1}^k ||g_i ||^2 + 2 \cdot \sum_{i < j} \langle g_i, g_j \rangle$ $\leq kM - k(k-1) \delta$ For large $k, \ kM - k^2 \delta<0$, contradiction. Hence $S$ must have density $0$. Proof of the theorem: By corollary 2 of the theorem in part 1, since $T$ is weak mixing, $T^m$ is weak mixing for all $m \neq 0$. We proceed by induction on $l$. For $l=1$, the statement is implied by our lemma 2 in part 1. Suppose the theorem holds for $l \in \mathbb{N}$, let $f_0, f_1, \cdots, f_{l+1} \in L^\infty(X)$, Let $C = \int f_{l+1} \ d \mu, \ f'_{l+1}(x) = f_{l+1}(x) - C$. By induction hypothesis, $\int f_0(x) f_1(T^n(x)) f_2(T^{2n}(x)) \cdots f_l(T^{ln}(x)) \cdot C \ d \mu$ $\rightarrow_D \int f_0 \ d \mu \int f_1 \ d \mu \cdots \int f_l \ d \mu \cdot C$ as $n \rightarrow \infty$ Hence it suffice to show $\int f_0(x) f_1(T^n(x)) f_2(T^{2n}(x)) \cdots f_l(T^{ln}(x)) \cdot$ $f'_{l+1}(T^{(l+1)n}(x)) \ d \mu \rightarrow_D 0$ Let $\int f_{l+1} \ d\mu = 0$ For all $n \in \mathbb{N}$, set $g_n (x)= f_1 \circ T^n(x) \cdot f_2 \circ T^{2n}(x) \cdots f_{l+1} \circ T^{(l+1)n}(x)$ For each $m \in \mathbb{N}, \ \forall \ 0 \leq i \leq l=1$, let $F^{(m)}_i (x)= f_i(x) \cdot f_i(T^{im}(x))$ $\langle g_{n+m}, g_n \rangle = \int (f_1(T^{n+m} (x) \cdots f_{l+1}(T^{(l+1)(n+m)} (x)))$ $\cdot (f_1(T^n(x)) \cdots f_{l+1}(T^{(l+1)n} (x))) \ d\mu$ $= \int F^{(m)}_1(T^n(x)) \cdots F^{(m)}_{l+1}(T^{(l+1)n}(x)) \ d \mu$ Since $T^{l+1}$ is measure preserving, we substitute $y = T^{(l+1)n}(x)$, $= \int F^{(m)}_{l+1}(y) \cdot F^{(m)}_1(T^{-ln}(y)) \cdots F^{(m)}_l(T^{-n}(y)) \ d \mu$ Apply induction hypothesis, to the weak mixing transformation $T^{-n}$ and re-enumerate $F^{(m)}_i$ $\langle g_n, g_{n+m} \rangle \rightarrow_D ( \int F^{(m)}_1 \ d\mu) \cdots (\int F^{(m)}_{l+1} \ d\mu)$ as $n \rightarrow \infty$. $\int F^{(m)}_{l+1} \ d\mu = \int f_{l+1} \cdot f_{l+1} \circ T^{(l+1)m} \ d\mu$ By lemma 2 in part 1, we have $\int F^{(m)}_{l+1} \ d\mu \rightarrow_D 0$ as $m \rightarrow \infty$. We are now able to apply lemma 2 to $g_n$, which gives $(g_n) \rightarrow_D \overline{0}$ under the weak topology. i.e. for any $f_0$, we have $\int f_0(x) g_n(x) \ d \mu \rightarrow_D 0$. Establishes the induction step. Remark: This works for any group of commutative weakly mixing transformations. i.e. if $G$ is a group of measure preserving transformations, all non-identity elements of $G$ are weakly mixing. $T_1, \cdots, T_k$ are distinct elements in $G$, then $\int f_0(x) f_1(T_1^n(x)) f_2(T_2^n(x)) \cdots f_k(T_k^n(x)) \ d \mu$ $\rightarrow_D \int f_0 \ d \mu \int f_1 \ d \mu \cdots \int f_k \ d \mu$ as $n \rightarrow \infty$. # Notes for my lecture on multiple recurrence theorem for weakly mixing systems – Part 1 So…It’s finally my term to lecture on the ergodic theory seminar! (background, our goal is to go through the ergodic proof of Szemerédi’s theorem as in Furstenberg’s book). My part is the beginning of the discussion on weak mixing and prove the multiple recurrence theorem in the weak mixing case, the weak mixing assumption shall later be removed (hence the theorem is in fact true for any ergodic system) and hence prove Szemerédi’s theorem via the correspondence principal discussed in the last lecture. Given two measure preserving systems $(X_1, \mathcal{B}_1, \mu_1, T_1)$ and $(X_2, \mathcal{B}_2, \mu_2, T_2)$, we denote the product system by $(X_1 \times X_2, \mathcal{B}_1 \times \mathcal{B}_2, \mu_1 \times \mu_2, T_1 \times T_2)$ where $\mathcal{B}_1 \times \mathcal{B}_2$ is the smallest $\sigma$-algebra on $X_1 \times X_2$ including all products of measurable sets. Definition: A m.p.s. $(X, \mathcal{B}, \mu, T)$ is weakly mixing if $(X \times X, \mathcal{B} \times \mathcal{B}, \mu \times \mu, T \times T)$ is ergodic. Note that weak mixing $\Rightarrow$ ergodic as for non-ergodic systems we may take any intermediate measured invariant set $\times$ the whole space to produce an intermediate measured invariant set of the product system. For any $A, B \in \mathcal{B}$, let $N(A, B) = \{ n \in \mathbb{N} \ | \ \mu(A \cap T^{-n}(B)) > 0 \}$. Ergodic $\Leftrightarrow$ for all $A,B$ with positive measure, $N(A,B) \neq \phi$ Weakly mixing $\Rightarrow$ for all $A,B,C,D$ with positive measure, $N(A \times C, B \times D) \neq \phi$. Since $n \in N(A \times C, B \times D)$ $\Leftrightarrow \ \mu \times \mu(A \times C \cap T^{-n}(B \times D)) > 0$ $\Leftrightarrow \ \mu \times \mu(A \cap T^{-n}(B) \times C \cap T^{-n}(D)) > 0$ $\Leftrightarrow \ \mu(A \cap T^{-n}(B)) > 0$ and $\mu(C \cap T^{-n}(D)) > 0$ $\Leftrightarrow \ n \in N(A,B)$ and $n \in N(C,D)$ Hence $T$ is weakly mixing $\Rightarrow$ for all $A,B,C,D$ with positive measure, $N(A,B) \cap N(C,D) \neq \phi$. We’ll see later that this is in fact $\Leftrightarrow$ but let’s say $\Rightarrow$ for now. As a toy model for the later results, let’s look at the proof of following weak version of ergodic theorem: Theorem 1: Let $(X, \mathcal{B}, \mu, T)$ be ergodic m.p.s., $f, g \in L^2(X)$ then $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int f \cdot g \circ T^n \ d\mu = \int f \ d\mu \cdot \int g \ d\mu$. Proof: Let $\mathcal{U}: f \mapsto f \circ T, \ \mathcal{U}$ is unitary on $L^2(X)$. Hence $\{ \frac{1}{N+1} \sum_{n=0}^N g \circ T^n \ | \ N \in \mathbb{N} \} \subseteq \overline{B( \overline{0}, ||g||)}$ Any weak limit point of the above set is $T$-invariant, hence ergodicity implies they must all be constant functions. Suppose $\lim_{i \rightarrow \infty} \frac{1}{N_i+1} \sum_{n=0}^N g \circ T^n \equiv c$ then we have $c = \int c \ d\mu = \lim_{i \rightarrow \infty} \frac{1}{N_i+1} \sum_{n=0}^N \int g \circ T^n \ d\mu = \int g \ d \mu$ Therefore the set has only one limit point under the weak topology. Since the closed unit ball in Hilbert space is weakly compact, hence $\frac{1}{N+1} \sum_{n=0}^N g \circ T^n$ converges weakly to the constant valued function $\int g \ d \mu$. Therefore $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int f \cdot g \circ T^n \ d\mu$ $= \int f \cdot ( \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g \circ T^n) \ d\mu$ $= \int f \cdot (\int g \ d\mu) d\mu = \int f \ d\mu \cdot \int g \ d\mu$. Next, we apply the above theorem on the product system and prove the following: Theorem 2: For $(X, \mathcal{B}, \mu, T)$ weakly mixing, $\lim_{N \rightarrow \infty} \frac{1}{1+N} \sum_{n=0}^N (\int f \cdot (g \circ T^n) \ d \mu - \int f \ d \mu \int g \ d \mu)^2$ $= 0$ Proof: For $f_1, f_2: X \rightarrow \mathbb{R}$, let $f_1 \otimes f_2: X \times X \rightarrow \mathbb{R}$ where $f_1 \otimes f_2 (x_1, x_2) = f_1(x_1) f_2(x_2)$ By theorem 1, we have $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N(\int f \cdot g \circ T^n \ d \mu)^2$ $= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int (f \otimes f) \cdot ((g \otimes g) \circ (T \times T)^n) \ d \mu \times \mu$ $= (\int f \otimes f \ d\mu \times \mu) \cdot (\int g \otimes g \ d\mu \times \mu)$ $= (\int f \ d\mu)^2 (\int g \ d\mu)^2 \ \ \ \ ( \star )$ Set $a_n = \int f \cdot (g \circ T^n) \ d \mu, \ a = \int f \ d \mu \int g \ d \mu$ hence by theorem 1, we have $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n = a$; By $( \star )$, we have $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n^2 = a^2$ Hence $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N (a_n-a)^2$ $= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N (a_n^2 - 2a \cdot a_n + a^2)$ $= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n^2 - 2a \cdot \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n + a^2$ $= a^2 - 2 a \cdot a + a^2 = 0$ This establishes theorem 2. We now prove that the following definition of weak mixing is equivalent to the original definition. Theorem 3: $(X, \mathcal{B}, \mu, T)$ weakly mixing iff for all $(Y, \mathcal{D}, \nu, S)$ ergodic, $(X \times Y, \mathcal{B} \times \mathcal{D}, \mu \times \nu, T \times S)$ is ergodic. proof:$\Leftarrow$” is obvious as if $(X, \mathcal{B}, \mu, T)$ has the property that its product with any ergodic system is ergodic, then $(X, \mathcal{B}, \mu, T)$ is ergodic since we can take its product with the one point system. This implies that the product of the system with itself $(X \times X, \mathcal{B} \times \mathcal{B}, \mu \times \mu, T \times T)$ is ergodic, which is the definition of being weakly mixing. $\Rightarrow$” Suppose $(X, \mathcal{B}, \mu, T)$ weakly mixing. $T \times S$ is ergodic iff all invariant functions are constant a.e. For any $g_1, g_2 \in L^2(X), \ h_1, h_2 \in L^2(Y)$, let $C = \int g_1 \ d \mu$, let $g_1' = g_1-C$; hence $\int g_1' \ d \mu = 0$. $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g_1 \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ $= \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int C \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ $+ \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g_1' \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ Since $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int C \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ $= C \cdot \int g_2 \ d \mu \cdot \lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ By theorem 1, since $S$ is ergodic on $Y$, $=\int g_1 \ d \mu \cdot \int g_2 \ d \mu \cdot \int h_1 \ d \nu \cdot \int h_2 \ d \nu$ On the other hand, let $a_n = \int g_1' \cdot (g_2 \circ T^n) \ d \mu, \ b_n = \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ By theorem 2, since $T$ is weak mixing $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N (a_n - 0 \cdot \int g_2 \ d \mu)^2 = 0$ hence $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N a_n^2 = 0 \ \ \ (\ast)$ $(\sum_{n=0}^N a_n \cdot b_n)^2 \leq ( \sum_{n=0}^N a_n^2) \cdot ( \sum_{n=0}^N b_n^2)$ by direct computation i.e. subtract the left from the right and obtain a perfect square. Therefore $\lim_{N \rightarrow \infty} (\frac{1}{N+1} \sum_{n=0}^N a_n \cdot b_n)^2$ $\leq (\frac{1}{N+1} \sum_{n=0}^N a_n^2) \cdot (\frac{1}{N+1} \sum_{n=0}^N b_n^2)$ Approaches to $0$ as $N \rightarrow \infty$ by $(\ast)$. Therefore, $\lim \frac{1}{N+1} \sum_{n=0}^N \int g_1' \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu = 0$ Combining the two parts we get $\lim_{N \rightarrow \infty} \frac{1}{N+1} \sum_{n=0}^N \int g_1 \cdot (g_2 \circ T^n) \ d \mu \cdot \int h_1 \cdot (h_2 \circ S^n) \ d \nu$ $= \int g_1 \ d \mu \cdot \int g_2 \ d \mu \cdot \int h_1 \ d \nu \cdot \int h_2 \ d \nu$. Since the linear combination of functions of the form $f(x, y) = g(x)h(y)$ is dense in $L^2(X \times Y)$ (in particular the set includes all characteristic functions of product sets and hence all simple functions with basic sets being product sets) We have shown that for a dense subset of $f \in L^2(X \times Y)$ the sequence of functions $\frac{1}{N+1}\sum_{n=0}^N f(T^n(x), S^n(y))$ converge weakly to the constant function. (Since it suffice to check convergence a dense set of functional in the dual space) Hence for any $f \in L^2(X \times Y)$, the average weakly converges to the constant function $\int f \ d \mu \times \nu$. For any $T \times S$-invariant function, the average is constant, hence this implies all invariant functions are constant a.e. Hence we obtain ergodicity of the product system. Establishes the theorem. # Kaufman’s construction This is a note on R. Kaufman’s paper An exceptional set for Hausdorff dimension We construct a set $D \subseteq \mathbb{R}^2$ with $\dim(D) = d < 1$ and $E \subseteq [0, \pi)$ with $\dim(E) > 0$ s.t. for all directions $\theta \in E$, $\dim(\pi_\theta(D)) < d-\epsilon$ (the projection of $D$ in direction $\theta$ is less than $d-\epsilon$) $\forall \alpha >1$, let $(n_j)_{j=1}^\infty$ be an rapidly increasing sequence of integers. Define $D_j = \{ (a, b)/n_j + \xi \ | \ a, b \in \mathbb{Z}, \ ||(a, b)|| \leq n_j; \ | \xi | \leq n_j^{- \alpha} \}$ i.e. $D_j = \bigcup \{ B((a,b)/n_j, 1/n_j^\alpha) \ | \ (a, b) \in \mathbb{Z}^2 \cap B( \overline{0}, n_j) \}$ Let $D = \bigcap_{j=1}^\infty D_j$ $\because \alpha > 1, \ (n_j)$ rapidly increasing, $\dim(D) = 2 / \alpha$ Let $c \in (0, 1)$ be fixed, define $E' = \{ t \in \mathbb{R} \ | \ \exists$ positive integer sequence $(m_{j_i})_{i=1}^\infty$ s.t. $m_{j_i} < C_1 n_{j_i}^c, \ || m_{j_i} t || < C_2 m_{j_i} / n_{j_i}^\alpha \}$ $\forall t \in E', \ \forall i \in \mathbb{N}, \ \forall p = (a, b)/n_{j_i} + \xi \in D_{j_i}$, we have: $| \langle p, (1, t) \rangle - a/n_{j_i} - bt/n_{j_i} | \leq (1+|t|)/n_{j_i}^\alpha$ Let $b = q m_{j_i} + r$ where $0 \leq r < m_{j_i}, |q m_{j_i}| < C n_{j_i}$ $\exists z_{j_i} \in \mathbb{Z}, \ | z_{j_i} | < C | n_{j_i} |, \ | \theta |<1$ $bt = qm_{j_i}t +rt = X + rt + q \theta ||m_{j_i}t||$

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# Insurance Practical answers, 10 jobs Sold: 8 Refunds: 0 Content: 21031202412683.rar (38,09 kB) # Description The practical work on discipline "Insurance" What is the life table? What is the chance to live to 51 years, if the number of surviving to this age still 928? What is the wage rate? What is the gross rate, if risky net-rate and risk premium equal to 5 rubles. and 1.2 rubles. respectively, and the load is estimated at 20%. Activity 3 Determine what will be the net rate of 100 rubles. deferred annuity for 10 years (payment at the end of the year), if the age of the insured - 56 years, and the rate of return is 50%. Let the insurer for insurance contracts concluded 4000 with total assets of the insurance amount of 40 000 000 rub. As a result, it was made 200 payments, and total payments amounted to 200,000 rubles. Estimate the average tariff rate-ing on property insurance for 100 rubles. sum insured in the security guarantees equal 0,95 (N = 200). The coefficient of variation of payments is 0.5 and the load of 30% gross rate. Let the insurer must pay the first year 145 rub., At the beginning of the second year 150 rub., The third 155 rubles, even-vert 160 rubles. And at the beginning of the year an amount equal to 5 170 rubles. What is the current value of financial liabilities Stra-insurer, if the rate of return is 20% per annum? You sign a contract and the insurer of property insurance with the insurer. What are the main points to be contained in the contract of insurance? When disability due to accidents slu¬chaev distribution groups zna¬cheniyam disability meets the following conditional probabilities: - The likelihood of disability group 1 - 0.18; - The likelihood of disability Group 2 - 0.42; - The likelihood of disability Group 3 - 0.40. What is the coefficient of variation of payments? Suppose that under the contract of insurance (endowment insured up to age 45 years) were insured 100 people in the WHO e-Rast 25 years for the total insurance amount of 300 000 rubles. Using the table of switching numbers, calculate the modern-ing value of financial liabilities of the insurer and the tariff rate (gross rate) under such contract of insurance, (load is 10% gross-rate). Describe the process of reinsurance. What are the key concepts found in reinsurance. What types of violations that can occur when health insurance. Who and how is liable for those violations.

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# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science. Message body Emoticons View more Emoticons [quote="ZutAlors!"]I've got a statistics problem, dealing with a number of tests using slightly different procedures, and it's equivalent to this: Suppose I've got a few hundred marbles, some of which are black, some white. Say, three hundred marbles split 60/40, if you want some approximate numbers. These marbles are split into bags: some bags have more marbles, some have less; let's say in the range of 10 to 50 marbles in each bag. Obviously the proportion of marbles in each bag varies. But how can I determine whether the variance is [i]random[/i] (by which, I assume, I mean "matches a normal distribution" Right?) Is there an exact or approximate test I can apply? And, assuming, the variance is [i]not[/i] random (or, rather, [i]probably[/i] not random with some confidence), as I expect, what's a reasonable way to identify which bags are the culprits?[/quote] Options HTML is OFF BBCode is ON Smilies are ON Disable BBCode in this post Disable Smilies in this post All times are GMT Jump to: Select a forum Puzzles and Games----------------Grey Labyrinth PuzzlesVisitor Submitted PuzzlesVisitor GamesMafia Games Miscellaneous----------------Off-TopicVisitor Submitted NewsScience, Art, and CulturePoll Tournaments Administration----------------Grey Labyrinth NewsFeature Requests / Site Problems Topic review Author Message ZutAlors! Posted: Tue Jun 08, 2004 1:03 pm    Post subject: 1 Thanks, Boro. I was thinking that the numbers for a lot of these groups were too small -- ten seems far too small to draw conclusions from, although forty samples might be reasonable. That was why it occured to me to look at the distribution of all three hundred or so into groups. If that distribution did indeed look random, then I could forget about any further analysis. If I were to do a Monte Carlo simulation, how would I calculate an overall standard deviation? It occurs to me that the groups with greater numbers of marbles should be weighted more heavily. If I distributed marbles, calculated B/W percentages in each bag, calculated the deviations from the overall B/W percentage, then weighted the deviations by the number of marbles in the bag, would that be a good start? Thanks. Borodog Posted: Mon Jun 07, 2004 5:59 pm    Post subject: 0 A few things to consider: 1) I don't really know the answer to your question. I could hack out something practicable, but whether it would be "correct" according to a statistician would be highly improbable. 2) The distribution of marbles is most definitely not normal, but Poisson (this does not mean it's not random, just that the random deviates form a different distribution than the one you assumed). 3) If the distribution of marbles is non-random, then it is not any subset of individual bags that is the "problem." The whole distribution is non-random, period. 4) My first instinct is that the expected distribution should be calculable (even if by brute force methods such as Monte Carlo simulation). From this and the data construct some normalized summed deviation. The larger the deviation, the more likely the distribution is non-random. 5) Given the numbers you quoted, the statistics are likely too low to be meaningful. ------------------ Insert humorous sig here. ZutAlors! Posted: Mon Jun 07, 2004 3:57 pm    Post subject: -1 I've got a statistics problem, dealing with a number of tests using slightly different procedures, and it's equivalent to this: Suppose I've got a few hundred marbles, some of which are black, some white. Say, three hundred marbles split 60/40, if you want some approximate numbers. These marbles are split into bags: some bags have more marbles, some have less; let's say in the range of 10 to 50 marbles in each bag. Obviously the proportion of marbles in each bag varies. But how can I determine whether the variance is random (by which, I assume, I mean "matches a normal distribution" Right?) Is there an exact or approximate test I can apply? And, assuming, the variance is not random (or, rather, probably not random with some confidence), as I expect, what's a reasonable way to identify which bags are the culprits?

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# Tangent In mathematics, the word "tangent" has two distinct, but etymologically related meanings: one in geometry, and one in trigonometry. ## Geometry In plane geometry, a straight line is tangent to a curve, at some point, if both line and curve pass through the point with the same direction; such a line is the best straight-line approximation to the curve at that point. The curve, at point P, has the same slope as a tangent passing through P. The slope of a tangent line can be approximated by a secant line. It is a mistake to think of tangents as lines which intersect a curve at only one single point. There are tangents which intersect curves at several points (as in the following example), and there are non-tangential lines which intersect curves at only one single point. (Note that in the important case of a circle, however, the tangent line will intersect the curve at only one point.) In the following diagram, a red line intersects the black curve at two points. It is tangent to the curve at one point. Black curve and red tangent In higher-dimensional geometry, one can define the tangent plane for a surface in an analogous way to the tangent line for a curve. In general, one can have an (n − 1)-dimensional tangent hyperplane to an (n − 1)-dimensional manifold. ### Quote "And I dare say that this is not only the most useful and general [concept] in geometry, that I disproved, but even that I ever desired to disprove." Descartes (1637) ### Related Meaning To "go off on a tangent" means to go off topic, likely derived from the geometric meaning of a tangent. ## Calculus A "formal" definition of the tangent requires calculus. Specifically, suppose a curve is the graph of some function, y = f(x), and we are interested in the point (x0, y0) where y0 = f(x0). The curve has a non-vertical tangent at the point (x0, y0) if and only if the function is differentiable at x0. In this case, the slope of the tangent is given by f '(x0). The curve has a vertical tangent at (x0, y0) if and only if the slope approaches plus or minus infinity as one approaches the point from either side. Above, it was noted that a secant can be used to approximate a tangent; it could be said that the slope of a secant approaches the slope (or direction) of the tangent, as the secants' points of intersection approach each other. Should one also understand the notion of a limit; one might understand how that concept is applicable to those discussed here, via calculus. In essence, calculus was developed (in part) as a means to find the slopes of tangents; this challenge, being known as the tangent line problem, is solvable via Newton's difference quotient. Should one know the slope of a tangent, to some function; then, one can determine an equation for the tangent. For example, an understanding of the power rule will help one determine that the slope of x3, at x = 2, is 12. Using the point-slope equation, one can write an equation for this tangent: y − 8 = 12(x − 2) = 12x − 24; or: y = 12x − 16. ## Trigonometry In trigonometry, the tangent is a function (see trigonometric function) defined as: $\displaystyle \,\tan x = \frac{\,\sin x}{\,\cos x}$ The function is so-named because it can be defined as the length of a certain segment of a tangent (in the geometric sense) to the unit circle. It is easiest to define it in the context of a two-dimensional Cartesian coordinate system. If one constructs the unit circle centred at the origin, the tangent line to the unit circle at the point P = (1, 0), and the ray emanating from the origin at an angle θ to the x-axis, then the ray will intersect the tangent line at at most a single point Q. The tangent (in the trigonometric sense) of θ is the length of the portion of the tangent line between P and Q. If the ray does not intersect the tangent line, then the tangent (function) of θ is undefined. ### Derivative The derivative of the tangent is: $\displaystyle \,\frac{\mathrm{d} (\tan x)}{\mathrm{d} x} = \frac{1}{\cos^2 x}=\sec^2 x.$

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× ### Let's log you in. or Don't have a StudySoup account? Create one here! × or ## Introduction to Abstract Algebra by: Dr. Taya Dickens 18 0 1 # Introduction to Abstract Algebra MAT 334 Marketplace > Utica College > Mathematics (M) > MAT 334 > Introduction to Abstract Algebra Dr. Taya Dickens GPA 3.54 Staff These notes were just uploaded, and will be ready to view shortly. Either way, we'll remind you when they're ready :) Get a free preview of these Notes, just enter your email below. × Unlock Preview COURSE PROF. Staff TYPE Class Notes PAGES 1 WORDS KARMA 25 ? ## Popular in Mathematics (M) This 1 page Class Notes was uploaded by Dr. Taya Dickens on Wednesday October 28, 2015. The Class Notes belongs to MAT 334 at Utica College taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/230525/mat-334-utica-college in Mathematics (M) at Utica College. × ## Reviews for Introduction to Abstract Algebra × × ### What is Karma? #### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more! Date Created: 10/28/15 ll Classify the torsion subgroup of the rational points on the elliptic curve y2 zg212 7 31 The discriminant of the polynomial 13 212 7 31 is D 144 By the LutzNagell theorem if P z y is a rational points on the elliptic curve of nite order then either y 0 or y divides 144 Running through the diVisors of 144 we nd that the points of nite order are 0 0 l 0 73 0 71 2 71 72 3 6 and 3 76 So including the point at in nity the group of nite rational points on the elliptic curve 1s 0070707170773707717 27717 72 3767 37 76 This is a nite Abelian group of order 8 so it must be isomorphic to Z37Z4 69 Z27 0r Z2 69 Z2 69 Z2 We consider the points 00 10 and 730 These three points lie on the zaxis and the tangent lines to the curve at these points are all vertical Thus the third point of intersection is the point at in nity So the orders of each of these points is 2 So the group has at least three points of order two namely 00 l 0 and 73 0 The group Z3 has only 1 element of order two namely the element 4 since 4 04 So the group cannot be Zgi So we are left with G Z4 69 Z2 and H Z2 69 Z2 69 Z2i One main difference is that the group G has elements of order 4 whereas the group H has no elements of order 4 Consider the element 712 on the elliptic curve The tangent line to the curve at the point 71 2 intersects the curve at 10 which when re ected over the zaxis is 710 So 712 712 10i At any rate the point 71 0 is not a point of order 2 So the group cannot be Z2 69 Z2 69 Z2 Therefore the group is Z4 69 Z2i × × ### BOOM! Enjoy Your Free Notes! × Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document' ## Why people love StudySoup Steve Martinelli UC Los Angeles #### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup." Janice Dongeun University of Washington #### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!" Bentley McCaw University of Florida #### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!" Parker Thompson 500 Startups #### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning." Become an Elite Notetaker and start selling your notes online! × ### Refund Policy #### STUDYSOUP CANCELLATION POLICY All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com #### STUDYSOUP REFUND POLICY StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

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# How accurate is bomb calorimeter for measuring nutritional calories of food? Burning food seems not be accurate for estimating calories in fiber or alochol, but I'm wondering how well it works for other carbs/proteins/fats. Any pointers to literature are appreciated. • I think you'd probably get better answers asking this in biology. Aug 20 '18 at 19:26 • lol, I asked that in biology first, and got downvotes, so I moved it here Aug 21 '18 at 2:40 • Chemistry or Physics perhaps? ;-) Aug 21 '18 at 3:48 It seems deceptively simple to just assume a bomb calorimeter is still used. Manufacturers of these things seem convinced of their utility, naturally. The original method used to determine the number of kcals in a given food directly measured the energy it produced.The food was placed in a sealed container surrounded by water--an apparatus known as a bomb calorimeter. The food was completely burned and the resulting rise in water temperature was measured. This method is not frequently used today. According to the National Data Lab (NDL), most of the calorie values in the USDA and industry food tables are based on an indirect calorie estimation made using the so-called Atwater system. In this system, calories are not determined directly by burning the foods. Instead, the total caloric value is calculated by adding up the calories provided by the energy-containing nutrients: protein, carbohydrate, fat and alcohol. Because carbohydrates contain some fiber that is not digested and utilized by the body, the fiber component is usually subtracted from the total carbohydrate before calculating the calories. The Atwater system uses the average values of 4 Kcal/g for protein, 4 Kcal/g for carbohydrate, and 9 Kcal/g for fat. Alcohol is calculated at 7 Kcal/g. (These numbers were originally determined by burning and then averaging.) Thus the label on an energy bar that contains 10 g of protein, 20 g of carbohydrate and 9 g of fat would read 201 kcals or Calories. A complete discussion of this subject and the calories contained in more than 6,000 foods may be found on the National Data Lab web site at http://www.nal.usda.gov/fnic/foodcomp/. At this site you can also download the food database to a handheld computer. Another online tool that allows the user to total the calorie content of several foods is the Nutrition Analysis Tool at http://www.nat.uiuc.edu. How Do Food Manufacturers Calculate the Calorie Count of Packaged Foods?, Scientific American, 2003. But it is not prudent to again assume this it. It varies ## Calorie Calculation by Country Calorie Calculations in the United States In the U.S., there are six accepted methods. The two most frequently used are the 4-4-9 formula and the Atwater method. 4-4-9. In the U.S., most manufacturers use the 4-4-9 method, which assumes that each gram of protein contributes 4 Calories to the caloric total, each gram of carbohydrates contributes 4 Calories, and each gram of fat contributes 9 Calories. Atwater. The USDA SR database, in contrast, commonly uses the Atwater method. The Atwater method uses more precise figures based on food type when assigning Calories values per gram to protein, carbohydrate, and fat. Find the Atwater table here. 4-4-9 adjusted for non-digestible carbohydrates and sugar alcohols. (Total carbohydrates less non-digestible carbs and sugar alcohols.) For soluble non-digestible carbohydrates, a factor of 2 Calories per gram (rather than 4) is used, and sugar alcohols use specific factors listed in No. 6 below. Specific food factors approved by the FDA. Bomb calorimetry. This process involves burning a food item to see how much heat it releases, which is directly convertible to Calories since, as we know, one Calorie equals the amount of energy required to heat one kilogram of water by one degree Celsius. Note the adjustment for Calories from protein in the CFR. General factors for caloric value of sugar alcohols: Isomalt = 2.0 Calories per gram, lactitol = 2.0 Calories per gram, xylitol = 2.4 Calories per gram, maltitol = 2.1 Calories per gram, sorbitol = 2.6 Calories per gram, hydrogenated starch hydrolysates = 3.0 Calories per gram, mannitol = 1.6 Calories per gram, and erythritol = 0 Calories per gram. For more information, see the U.S. food labeling regulations here. Calorie Calculations in the European Union The declared values in the nutrition table are average values and must be based on: Total value. A calculation from the known or actual average values of the ingredients used. Known data. A calculation from generally established and accepted data. Estimates. (The same concept as 4-4-9 and 4-4-9-7) The energy value to be declared shall be calculated using the following conversion factors: ``````carbohydrate (except polyols): 17 kJ/g (4 Cal/g) polyols: 10 kJ/g (2,4 Cal/g) protein: 17 kJ/g (4 Cal/g) fat: 37 kJ/g (9 Cal/g) salatrims: 25 kJ/g (6 Cal/g) alcohol (ethanol), 29 kJ/g (7 Cal/g) organic acid: 13 kJ/g (3 Cal/g) fibre: 8 kJ/g (2 Cal/g) erythritol: 0 kJ/g (0 Cal/g) `````` For more information, see the EU food labeling regulations here. The overall accuracies of such measruemnts with a bomb calorimeter is actually a frequent question: The calorific content is measured with a device known as the bomb calorimeter. A sample of food is placed in an airtight chamber - the 'bomb' - which is filled with pure oxygen and then placed in a tank of water. The food is ignited by an electric spark so it completely burns up. The temperature increase in the water is measured and the actual energy content of the food can then be calculated, either in old-fashioned calories or more modern joules. This method is not completely accurate, as it is rather crude when compared to the way the human body uses food. For example, proteins are completely burned up in the bomb calorimeter, whereas in the human body some of them would be used not for energy but for the production of things like skin, hair, mucus and muscle tissue. Incidentally, the subject is well covered in most biology textbooks for A-level and above, as well as in the occasional Open University programme on television. The four sources of food energy - protein, fat, carbohydrate and alcohol - yield 4, 9, 3.75 and 7 calories per gram respectively. The calorie value of a food is usually estimated by multiplying the protein, fat, carbohydrate and alcohol content by the appropriate factors. Many food manufacturers do not carry out chemical analyses but instead estimate the calorie content using values for ingredients derived from tables published by HMSO. Such calculations are normally within 10 per cent of the actual value. How do food companies work out the number of calories in their products? More detail can be found in this Food energy – methods of analysis and conversion factors, FAO FOOD AND NUTRITION PAPER, 77, Report of a technical workshop Rome, 3–6 December 2002. In summary, bomb calorimetry gives very precise values for caloric content. But we need for nutritional analysis calorific content.

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Is Hell Endothermic or Exothermic? # Thread: Is Hell Endothermic or Exothermic? 1. ## Is Hell Endothermic or Exothermic? Someone sent this to me today... thought Id share HELL EXPLAINED BY CHEMISTRY STUDENT The following is an actual question given on a University of Washington Chemistry mid-term. The answer by one student was so "profound" that the professor shared it with colleagues, via the Internet, which is, of course, why we now have the pleasure of enjoying it as well : Bonus Question: Is Hell exothermic (gives off heat) or endothermic (absorbs heat)? Most of the students wrote proofs of their beliefs using Boyle's Law (gas cools when it expands and heats when it is compressed) or some variant. One student, however, wrote the following: First, we need to know how the mass of Hell is changing in time. So we need to know the rate at which souls are moving into Hell and the rate at which they are leaving. I think that we can safely assume that once a soul gets to Hell, it will not leave. Therefore, no souls are leaving. As for how many souls are entering Hell, let's look at the different religions that exist in the world today. Most of these religions state that if you are not a member of their religion, you will go to Hell. Since there is more than one of these religions and since people do not belong to more than one religion, we can project that all souls go to Hell. With birth and death rates as they are, we can expect the number of souls in Hell to increase exponentially. Now, we look at the rate of change of the volume in Hell because Boyle's Law states that in order for the temperature and pressure in Hell to stay the same, the volume of Hell has to expand proportionately as souls are added. This gives two possibilities: 1. If Hell is expanding at a slower rate than the rate at which souls enter Hell, then the temperature and pressure in Hell will increase until all Hell breaks loose. 2. If Hell is expanding at a rate faster than the increase of souls in Hell, then the temperature and pressure will drop until Hell freezes over. So which is it? If we accept the postulate given to me by Teresa during my Freshman year that, "It will be a cold day in Hell before I sleep with you," and take into account the fact that I slept with her last night, then number two must be true, and thus I am sure that Hell is exothermic andhas already frozen over. The corollary of this theory is that since Hell has frozen over, it follows that it is not accepting any more souls and is therefore, extinct......leaving only Heaven, thereby proving the existence of a divine being which explains why, last night, Teresa kept shouting "Oh my God." THIS STUDENT RECEIVED THE ONLY "A" 2. ## The Following 10 Users Say Thank You to Mecha Hate Chimp For This Useful Post: 3. Hehe, remember this for all the right reasons 4. A hurban legend well known all around the world, but always funny to read! 5. lol! thats awesome!! I should sent that to my mom she would get a kick out of it. 6. Wow, another variant? So many different versions floating around... 7. HAHA, i'm printing this out 8. lmao, fantastic 9. Now THATS a fuckin' good read Mx 10. Loved this joke back when I first heard it. Loving it even more now with a bigger brain and an appreciation of comedy. 11. Good laugh, the only time I laughed about such a class-test joke was the legend about the test questioning "What is bravery?" and the A-student who gave it back after a minute stating "This is bravery!". 12. Originally Posted by Molly Now THATS a fuckin' good read Mx 13. lol, just fantastic! 14. Heh, I read this one some time ago at Bash.org. It is just wonderful 15. Your right about everything except hell not accepting any more souls. Everyone knows the quote in the bible "when theres no more room in hell, the dead will walk the earth". i havent seen any walking dead people yet so hell must still be accepting souls. 16. I'd love to see where that's written in the bible. In those exact words. 17. sweet, that was seriously a good read.. 18. Originally Posted by ruzkin I'd love to see where that's written in the bible. In those exact words. The Bible, Dawn of the Dead, same difference. 19. ## The Following 2 Users Say Thank You to Elwell For This Useful Post: 20. Dawn of the Dead is my bible. 21. Hell must be filling up, because this post just came back from the dead!! 22. you are a most powerful necromancer 23. Originally Posted by Kman. you are a most powerful necromancer You are a most powerful holy knight... *throws one of his gauntlets to the ground* There are no members to list at the moment. #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

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Sunteți pe pagina 1din 4 # IJSART - Volume 4 Issue 5 – MAY 2018 ISSN [ONLINE]: 2395-1052 ## Design and Fabrication of Pendulum Operated Pump 1, 2, 3, 4, 5 Department of Mechanical Engineering 1, 2, 3, 4, 5 REVA University, Bengaluru, Karnataka. Abstract- This paper discusses the importance of a pendulum II. LITERATURE REIVEW pump which can be used as a supplementary device for pumping water and is made to replace hand pumps. Our R. Ortega [2013] from Depto. De Matemática and to reduce human work in hand pump. By link mechanism presented the stability of the equilibrium of a pendulum of and spring force, we can attain linear motion from oscillatory variable length in terms of the third approximation. In motion and can use that to operate a pump or compressor. So contrast, the traditional linearization procedure is not always the energy consumption of this system will be lowest. It can be faithful. Alternative characterizations of stability are also used to spray pesticides. presented. They are based on degree theory and on the algebraic structure of the symplectic group. Keywords- Reciprocating pump, Basic Pendulum, springs Violaine et al.[2014] from UMR 6233 ISM Marey, I. INTRODUCTION Université de la Méditerranée, 163, avenue de Luminy, Francemade analysis of the mechanical constraints operating The ever increasing demand for energy has led to the suggested that the gymnast should be considered as a formation of various advanced resources which produces a pendulum of variable length. Increasing and decreasing certain part of the required energy. One principal consumer of pendulum length at appropriate phases of the swing effectively a large amount of energy is our household itself. Large amount allows energy to be injected into the system, thereby of electrical energy is wasted in pumping water irrigation compensating the energy lost to friction. purposes etc. In this context the importance of pendulum pump arises, by the use of which a large amount of energy can Tao Hana et al.[2013] from Department of be conserved and the conserved energy can be used for Mechanical and Industrial Engineering, University of Illinois various other purposes. A pump is a device that moves fluids at Chicago, United States observed a pendulum-like motion of (liquids or gases), or sometimes slurries, by mechanical action. the usually straight electrified jet experimentally and Pumps can be classified into three major groups according to theoretically modeled. Pendulum-like motion arises due to the method they use to move the fluid: direct lift, repulsive Coulomb force between the straight electrified jet displacement, and gravity pumps. Pendulum is a simple body and the charges accumulated on the collector. This electrical which moves freely until any damper is present. Hence, if the force repels the similarly charged landing jet segment in the oscillation is forced, then it is possible to gain continuous collector plane. The motion is transferred to the whole jet via movement. Our objective is to save electricity consumption to elastic stress sustained by the jet. pump the water and to reduce human work in hand pump. By link mechanism and spring force, we can attain linear motion W. Szyszkowski and D.S.D. Stilling [2014] of from oscillatory motion and can use that to operate a pump or Mechanical Engineering Department, University of compressor. So the energy consumption of this system will be Saskatchewan, 57 Campus Drive, Saskatoon, Canada studied lowest. the damping effects that are generated in a frictionless oscillating physical pendulum by a continuous motion of an Typical hand pumps require sufficiently large effort auxiliary mass. The analysis presented shows how a mass and an average person can use the pump continuously only for sliding in a periodic pattern along the rotating member affects a short time, but the pendulum pump requires only minimum the system oscillations. The resulting rotational motion of the of the effort, because it is only required to oscillate the pendulum is not exactly periodic. Therefore, the mass motion pendulum and can maintain these oscillation for several hours, should be continuously synchronized to control the phase without any fatigue. The advantage of this invention compared angle. If the mass motion period is not adjusted properly (if to present hand pump solutions are: less force to start the kept constant, for example) then undesired “beating” effects pump, less water consumption, and both arms can be used to would result over extended oscillations of the pendulum. fetch the water. ## Page | 611 www.ijsart.com IJSART - Volume 4 Issue 5 – MAY 2018 ISSN [ONLINE]: 2395-1052 ## Rahul Singh and Vijay Kumar [2014] from Dept. of Electronics and Communication Indian Institute of Technology, Roorkee, India presented an approach for the swing up and stabilization of a rotary inverted pendulum (RIP). III. OBJECTIVES Figure 2. Schematic representation of working principle. • To reduce the effort made by the farmer for Oscillation period of the pendulum is twice bigger supplying fertilizers. than the period of the lever oscillation. Piston of the pump has • To reduce the effort of farmer for supplying of water reverse effect on the lever and damps its oscillation. Damping for drip irrigation. of the lever motion causes damping of the pendulum, but the • To reduce the work of supplying of water for work of the force damping the pendulum is less than the work gardening. of the forces which damp the lever. Equilibrium position of • To save the electricity consumption. the lever is horizontal, and the equilibrium position of the pendulum is vertical. Oscillation of the lever and the IV. METHODOLOGY pendulum takes place in the same plane, vertical in reference to the ground. Frame ## It is the main component of the pump system and is Frame is base part which holds another parts like pump, lever, ball bearing etc. Reciprocating Pump ## This is a positive displacement pump. This is closely fitted with cylinder by the principle of actual displacement or a plunger which executes a reciprocating motion. ## V. WORKING PRINCIPLE The spring is an elastic object used to store mechanical energy. Here in the pendulum pump both tension The parts of Hand water pump with pendulum are: 1- and compression springs are used. It is the function of these Load of the pendulum,2- Handle of the pendulum,3- Axis of tension and compression springs to stretch and compress the pendulum,4- Axis of the two-leg lever, 5- two-leg lever, 6- according to the load applied. Water pump, 7- Piston of the pump. The pump is made of pendulum, two-leg lever and cylinder with the piston which Lever pumps the water. Oscillation of the pendulum is maintained by periodical action of the human arm. It is also the main component of the pump system and is made up of steel. Lever converts the oscillating movement of the pendulum on one side to the reciprocating motion of the piston to the other side. Ball bearing ## Page | 612 www.ijsart.com IJSART - Volume 4 Issue 5 – MAY 2018 ISSN [ONLINE]: 2395-1052 ## Ball bearings are the elements which uses balls between bearings. These are used to reduce rotational friction Pendulum ## The weight which is used to hold the weights and it is the oscillating part of the system and thus it acts like a pendulum. Figure 4. 3D drawing of model. Nylon tubes and non-return valve ## Nylon tubes are connected to the delivery and suction ends of the reciprocating pump for the passage of water from the sump and the delivery hose. A non-return valve or one-way valve is a valve that normally allows fluid (liquid or gas) to flow through it in only one direction. ## • On increasing the suction head, discharge of the given pendulum system decreases. • On increasing the mass of pendulum, discharge of the given pendulum system increases. • On increasing angle of swing, the discharge of the given pendulum system increases. • It was concluded that the human effort considerably reduced while pumping water by pendulum operated pump compared to the regular pump • Efficiency of the Pump is 52.11%. IX. CONCLUSION ## From this design and fabrication of pendulum hand pump, we have reduced the human effort by providing the pendulum bob which is attached in the hand lever. While pumping the pendulum oscillates to and fro and provides Figure 3. 2D drawing of pendulum operated pump. continuous energy to the hand lever which pressurizes the water and lifts the water from lower head to higher head and VII. DESIGN OF PENDULUM and MODEL provides the continuous flow of liquid. This is one of the methods that help the rural people to access liquid easily for spraying pesticides, gardening purpose and for Drip irrigation. of electricity. ## Page | 613 www.ijsart.com IJSART - Volume 4 Issue 5 – MAY 2018 ISSN [ONLINE]: 2395-1052 ## 2. It is portable. [5] Tao Hana, Darrell H. Renekera, Alexander L. Yarinb, 3. Reduction of farmers work. “Pendulum-like motion of straight electrified jets”, 4. Improve human health. Department of Mechanical and Industrial Engineering, 5. Uses the minimum of human strength in comparison to University of Illinois at Chicago, United States, 2013, pp. present classic hand water pumps. 2160–2169. 6. Compact size, easy to relocate, less moving parts, hence less maintenance cost. [6] Denise S.D. Stilling, Walerian Szyszkowski, “Controlling angular oscillations through mass reconfiguration: A Limitations includes, variable length pendulum case”, Mechanical Engineering 1. Not suitable for large scale farming. Canada, 2014, pp. 89-99. 2. Leakage problems. XI. APLICATIONS • Drip irrigation. • For spraying pesticides • Gardening purposes. ## It is essential for us to reduce the consumption of Non-renewable energy resources therefore our project mainly uses human effort intern using electrical energy. The pendulum pump to be used as an efficient mode in the irrigation of smaller lots, water-wells and gardening purpose. By the use of pendulum based water pumping system we can increase the efficiency of the plant and reduce the effort, cost of production, production time, man power requirement. REFERENCES ## [1] Rahul Singh, Vijay Kumar, “Swing up and Stabilization of Rotary Inverted Pendulum using TS Fuzzy”, International Journal of Scientific Research Engineering & Technology, Volume 2 Issue11 pp 753-759, 2014. ## [2] R. Ortega, “The stability of the equilibrium”, Depto. 2007, 2013, pp.215–234. ## [3] W. Szyszkowski and D. S. D. Stilling, “On damping properties of a frictionless physical pendulum with a moving mass,” International Journal of Non-Linear Mechanics, vol. 40, no. 5, pp. 669–681, 2014. ## [4] Dian-Hong Wang, Liang Zhao, Liang-Cheng Tu, Liang- Cheng Tu, “Eddy current loss testing in the torsion pendulum”, School of Physics, Huazhong University of Science and Technology, Wu-hanshih, Hubei, China, December 17, 2013, pp. 41 - 48.

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# Trigonometry Forum Trigonometry Help Forum: Graphing trig expressions, identities, angles, applied questions 1. ### Prove • Replies: 2 • Views: 675 Mar 8th 2010, 10:47 PM 2. ### Simplifying Trigonometric Functions • Replies: 7 • Views: 867 Mar 8th 2010, 09:28 PM 3. ### Finding radius of circle • Replies: 3 • Views: 684 Mar 8th 2010, 04:02 PM 4. ### What identity should I use? • Replies: 8 • Views: 463 Mar 8th 2010, 11:55 AM 5. ### [SOLVED] Finding Latitude • Replies: 1 • Views: 1,643 Mar 8th 2010, 01:23 AM • • • • 8. ### Identities • Replies: 4 • Views: 577 Mar 7th 2010, 05:39 PM 9. ### Trigonometric proofs and exact values. • Replies: 6 • Views: 915 Mar 7th 2010, 04:05 PM 10. ### Another Compound Trig Identity • Replies: 1 • Views: 511 Mar 7th 2010, 01:48 PM 11. ### Trig Fundamental question • Replies: 2 • Views: 470 Mar 7th 2010, 01:25 PM 12. ### Proving a trig idenity. • Replies: 4 • Views: 660 Mar 7th 2010, 09:38 AM 13. ### Inverse Trigo Functions • Replies: 1 • Views: 578 Mar 7th 2010, 05:41 AM 14. ### Hard question involving trig and circles • Replies: 1 • Views: 661 Mar 7th 2010, 04:43 AM 15. ### Compound Angle Identity • Replies: 4 • Views: 732 Mar 7th 2010, 03:51 AM • Replies: 2 • Views: 804 Mar 7th 2010, 02:36 AM 17. ### [SOLVED] solving an equation • Replies: 2 • Views: 393 Mar 6th 2010, 12:54 PM 18. ### Trig equation • Replies: 3 • Views: 646 Mar 6th 2010, 10:36 AM 19. ### How to turn this periodic function in a quadratic equation? • Replies: 7 • Views: 912 Mar 6th 2010, 07:07 AM 20. ### Need help in determing gradient of this function? • Replies: 3 • Views: 598 Mar 6th 2010, 04:11 AM 21. ### [SOLVED] Sine and Cosine Rule • Replies: 2 • Views: 687 Mar 6th 2010, 02:51 AM 22. ### Verifying Identities • Replies: 26 • Views: 1,057 Mar 5th 2010, 09:11 PM 23. ### [SOLVED] Trig definitions and identities • Replies: 4 • Views: 659 Mar 5th 2010, 01:30 PM 24. ### Question • Replies: 6 • Views: 515 Mar 5th 2010, 10:37 AM 25. ### Problem with sunlight hours trig equation • Replies: 1 • Views: 748 Mar 5th 2010, 08:49 AM 26. ### Solving trig equations • Replies: 5 • Views: 703 Mar 4th 2010, 02:23 PM 27. ### Angle sum. • Replies: 1 • Views: 513 Mar 4th 2010, 12:53 PM , , , , , , , , , , , , , , # trigonometry sotr Click on a term to search for related topics. Use this control to limit the display of threads to those newer than the specified time frame. Allows you to choose the data by which the thread list will be sorted.

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# Spin connection explained In differential geometry and mathematical physics, a spin connection is a connection on a spinor bundle. It is induced, in a canonical manner, from the affine connection. It can also be regarded as the gauge field generated by local Lorentz transformations. In some canonical formulations of general relativity, a spin connection is defined on spatial slices and can also be regarded as the gauge field generated by local rotations. The spin connection occurs in two common forms: the Levi-Civita spin connection, when it is derived from the Levi-Civita connection, and the affine spin connection, when it is obtained from the affine connection. The difference between the two of these is that the Levi-Civita connection is by definition the unique torsion-free connection, whereas the affine connection (and so the affine spin connection) may contain torsion. ## Definition Let a e \mu be the local Lorentz frame fields or vierbein (also known as a tetrad), which is a set of orthogonal space time vector fields that diagonalize the metric tensor g\mu= a e \mu b e \nu ηab, where g\mu is the spacetime metric and ηab is the Minkowski metric. Here, Latin letters denote the local Lorentz frame indices; Greek indices denote general coordinate indices. This simply expresses that g\mu , when written in terms of the basis a e \mu , is locally flat. The Greek vierbein indices can be raised or lowered by the metric, i.e. g\mu or g\mu . The Latin or "Lorentzian" vierbein indices can be raised or lowered by ηab or ηab respectively. For example, e\mu=g\mu\nu a e \nu and e\nuab b e \nu The torsion-free spin connection is given by ab \omega \mu a =e \nu \nu \Gamma \sigma\mu e\sigma+ a e \nu \partial\mue\nu= a e \nu \nu \Gamma \sigma\mu e\sigma-e\nu\partial\mu a e \nu , where \sigma \Gamma \mu\nu are the Christoffel symbols. This definition should be taken as defining the torsion-free spin connection, since, by convention, the Christoffel symbols are derived from the Levi-Civita connection, which is the unique metric compatible, torsion-free connection on a Riemannian Manifold. In general, there is no restriction: the spin connection may also contain torsion. Note that ab \omega \mu a =e \nu \partial;\mue\nu a =e \nu (\partial\mue\nu \nu +\Gamma \sigma\mu e\sigma) using the gravitational covariant derivative \partial;\mue\nu of the contravariant vector e\nu . The spin connection may be written purely in terms of the vierbein field as[1] ab \omega = \mu 1 2 e\nu(\partial\mu b e \nu -\partial\nu b e )- \mu 1 2 e\nu(\partial\mu a e \nu -\partial\nu a e )- \mu 1 2 e\rhoe\sigma(\partial\rhoe\sigma-\partial\sigmae\rho c )e \mu , which by definition is anti-symmetric in its internal indices a,b . The spin connection ab \omega \mu defines a covariant derivative D\mu on generalized tensors. For example, its action on a V \nu is D\mu a V \nu =\partial\mu a V \nu + a} {{\omega b b V \nu - \sigma \Gamma \nu\mu a V \sigma ## Cartan's structure equations In the Cartan formalism, the spin connection is used to define both torsion and curvature. These are easiest to read by working with differential forms, as this hides some of the profusion of indexes. The equations presented here are effectively a restatement of those that can be found in the article on the connection form and the curvature form. The primary difference is that these retain the indexes on the vierbein, instead of completely hiding them. More narrowly, the Cartan formalism is to be interpreted in its historical setting, as a generalization of the idea of an affine connection to a homogeneous space; it is not yet as general as the idea of a principal connection on a fiber bundle. It serves as a suitable half-way point between the narrower setting in Riemannian geometry and the fully abstract fiber bundle setting, thus emphasizing the similarity to gauge theory. Note that Cartan's structure equations, as expressed here, have a direct analog: the Maurer–Cartan equations for Lie groups (that is, they are the same equations, but in a different setting and notation). Writing ea= a e \mu dx\mu for the orthonormal coordinates on the cotangent bundle, the affine spin connection one-form is \omegaab= ab \omega \mu dx\mu The torsion 2-form is given by \Thetaa=dea+ a \omega b \wedgeeb while the curvature 2-form is a R b = a d\omega b + a \omega c \wedge c \omega b = 1 2 a R bcd ec\wedgeed These two equations, taken together are called Cartan's structure equations.[2] Consistency requires that the Bianchi identities be obeyed. The first Bianchi identity is obtained by taking the exterior derivative of the torsion: d\Thetaa+ a \omega b \wedge\Thetab= a R b \wedgeeb while the second by differentiating the curvature: a dR b + a \omega c \wedge c R b - a R c \wedge c \omega b =0. The covariant derivative for a generic differential form a V b of degree p is defined by a DV b = a dV b + a \omega c \wedge c V b -(-1)p a V c \wedge c \omega b . Bianchi's second identity then becomes a DR b =0. The difference between a connection with torsion, and the unique torsionless connection is given by the contorsion tensor. Connections with torsion are commonly found in theories of teleparallelism, Einstein–Cartan theory, gauge theory gravity and supergravity. ## Derivation ### By the tetrad postulate It is easy to deduce by raising and lowering indices as needed that the frame fields defined by g\mu= a {e \mu} b {e \nu} ηab will also satisfy a {e \mu} \mu} {e b = a \delta b and b {e \mu} \nu} {e b = \nu \delta \mu . We expect that D\mu will also annihilate the Minkowski metric ηab , D\muηab=\partial\muηab-{\omega\mu }^ \eta_ - ^ \eta_ = 0. This implies that the connection is anti-symmetric in its internal indices, {\omega\mu }^ = - ^. This is also deduced by taking the gravitational covariant derivative \partial;\beta a ({e \mu} \mu} {e b) =0 which implies that \partial;\beta a {e \mu} \mu} {e b = a -{e \mu} \partial;\beta \mu} {e b thus ultimately, {\omega\beta }^ = -^. This is sometimes called the metricity condition;[2] it is analogous to the more commonly stated metricity condition that g\mu\nu;\alpha=0. Note that this condition holds only for the Levi-Civita spin connection, and not for the affine spin connection in general. By substituting the formula for the Christoffel symbols \nu} {\Gamma \sigma\mu ={1\over2}g\nu(\partial\sigmag\delta+\partial\mug\sigma-\partial\deltag\sigma) written in terms of the a {e \mu} , the spin connection can be written entirely in terms of the a {e \mu} , {\omega\mu }^ = e^ ^c e_)where antisymmetrization of indices has an implicit factor of 1/2. ### By the metric compatibility This formula can be derived another way. To directly solve the compatibility condition for the spin connection ab {\omega \mu} , one can use the same trick that was used to solve \nabla\rhog\alpha=0 for the Christoffel symbols \gamma} {\Gamma \alpha\beta . First contract the compatibility condition to give \alpha} {e b \beta} {e c (\partial[\alphae\beta]+{\omega[\alpha }^ \;e_) = 0. Then, do a cyclic permutation of the free indices a,b, and c , and add and subtract the three resulting equations: \Omegabca+\Omegaabc-\Omegacab+2 \alpha} {e b \omega\alpha=0 where we have used the definition \Omegabca:= \alpha} {e b \beta} {e c \partial[\alphae\beta . The solution for the spin connection is \omega\alpha={1\over2} b {e \alpha} (\Omegabca+\Omegaabc-\Omegacab) . From this we obtain the same formula as before. ## Applications \gammaa are contracted onto vierbiens, \gammaa \mu} {e a (x)=\gamma\mu(x) . We wish to construct a generally covariant Dirac equation. Under a flat tangent space Lorentz transformation the spinor transforms as \psi\mapsto i\epsilonab(x)\sigmaab e \psi We have introduced local Lorentz transformations on flat tangent space, so \epsilonab is a function of space-time. This means that the partial derivative of a spinor is no longer a genuine tensor. As usual, one introduces a connection field ab {\omega \mu} that allows us to gauge the Lorentz group. The covariant derivative defined with the spin connection is, \nabla\mu\psi=(\partial\mu-{i\over4} ab {\omega \mu} \sigmaab)\psi=(\partial\mu-{i\over4}e\nu\partial;\mu b {e \nu} \sigmaab)\psi , and is a genuine tensor and Dirac's equation is rewritten as (i\gamma\mu\nabla\mu-m)\psi=0 . The generally covariant fermion action couples fermions to gravity when added to the first order tetradic Palatini action, l{L}=-{1\over2\kappa2}e \mu} {e a \nu} {e b {\Omega\mu }^ [\omega] + e \overline (i \gamma^\mu \nabla_\mu - m) \psi where e:=\det a {e \mu} =\sqrt{-g} and {\Omega\mu }^ is the curvature of the spin connection. The tetradic Palatini formulation of general relativity which is a first order formulation of the Einstein–Hilbert action where the tetrad and the spin connection are the basic independent variables. In the 3+1 version of Palatini formulation, the information about the spatial metric, qab(x) , is encoded in the triad i e a (three-dimensional, spatial version of the tetrad). Here we extend the metric compatibility condition Daqbc=0 to i e a , that is, Da i e b =0 and we obtain a formula similar to the one given above but for the spatial spin connection ij \Gamma a . The spatial spin connection appears in the definition of Ashtekar-Barbero variables which allows 3+1 general relativity to be rewritten as a special type of SU(2) Yang–Mills gauge theory. One defines i \Gamma a =\epsilonijk jk \Gamma a . The Ashtekar-Barbero connection variable is then defined as i A a = i \Gamma a +\beta i c a where i c a =cabebi and cab is the extrinsic curvature and \beta is the Immirzi parameter. With i A a as the configuration variable, the conjugate momentum is the densitized triad i E a =|\det(e)| i e a . With 3+1 general relativity rewritten as a special type of SU(2) Yang–Mills gauge theory, it allows the importation of non-perturbative techniques used in Quantum chromodynamics to canonical quantum general relativity.

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# Trouble with solution for transport equation. I have a question about the following paper: https://web.stanford.edu/class/math220a/handouts/firstorder.pdf My question is on the end of the second page, when the author solves the transport equation. "Now let $$S$$ be the integral of surface formed from a union of these characteristic curves. In doing so, we see that $$z(x,t)$$ is constant along the lines $$x - at = x_{0}$$. That is, $$z(x,t) = f(x - at)$$. Letting $$u(x,t) = z(x,t) = f(x - at)$$ for any (smooth) function $$f$$..." I don't understand why "$$z(x,t)$$ is constant along the lines $$x - at = x_{0}$$" implies "$$z(x,t) = f(x - at)$$". Maybe it's a dumb question, but could anyone help me? The intuition is as follows: $$z(x,t)$$ is constant along the lines $$x-at=x_0$$, so $$z$$ only "sees" which of these lines you are on. This means that we can write it as $$z(x,t)=f(x-at)$$ in a well defined way. Another common example in pde is functions from $$u:\mathbb{R}^2\to \mathbb{R}$$ invariant under rotations are those that may be written as $$u(r,\theta)=f(r)$$ for a function $$f$$ depending only on the radius, the only thing the original function $$u$$ cared about. To see this more algebraically, note that for $$(x,t)$$ satisfying $$x-at=x_0$$, we have $$z(x,t)=z(x_0,0)$$ by plugging in $$t=0$$. So, $$z(x,t)=z(x-at,0)$$ Name $$f(x-at)=z(x-at,0)=z(x,t)$$ This process works for any given initial position $$x_0$$ and has the nice visualization of tracing back along each of these lines $$x-at=x_0$$ in the $$(x,t)$$ plane to find the value of $$z$$ at a point $$(x_0,0)$$ where a value has been prescribed by initial data.

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# Why The coordination no of metal crystallizing in HCP structure is 12? Shreya 13 Points 7 years ago Look at this image.This shows a typical HCP lattice with the ABA layers,which is sliced from the repeating A-B-A-B-..... continuum. Consider an atom(let us mark the atom as P) in the center of the oranger layer.It is surrounded by 6 atoms in its layer and by 3 atoms in layer B beneath it.(the green layer) Now,since the lattice is a repeating AB-AB-AB,you should visualise a greeb B layer above the orange layer in which P sits.That means another 3 atoms are neighbors of P. Finally, 6(orange) atoms in P’s own layer,3(green) above and 3 (green) below. = 12 Which by definition is the coordination number. As an exercise,you should calculate the number of (full) atoms within the unit cell.This will help picture HCP more clearly.

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Mathematics 77 Online OpenStudy (anonymous): Stan can mow a lawn in 3 hrs while Scott can do the same in 2 hrs. Hw long it would take them together to mow the lawn? OpenStudy (anonymous): 6/5 = 1.2 hrs OpenStudy (anonymous): Call y=the total work done and x=total work done per hour. Stan does: y=3x and Scott does: y=2x Together they do the sum of the two equations, 2y=5x or, y=(5/2)x So they can get the work done in 2.5 hours. OpenStudy (anonymous): Wait, this is wrong. If two Stan's were mowing the lawn they would finish in 1.5 hours. My calculation would yield 3hours in this case. We need to divide (5/2) by 2. The answer should be (5/4)=1.25 hours OpenStudy (anonymous): eseidl.. I want to contradict your answer.. When Scott can complete within 2 hrs, so if Stan joins him then doesn't the job should be complete within 2 hrs? OpenStudy (anonymous): If Stan can complete in 3 hrs, then job done by him in 1 hr is 1/3 If Scott can complete in 2 hrs, then job done by him in 1 hr is 1/2 Together, they complete the work of 1/3 + 1/2 in 1 hr 1/3 + 1/2 = 5/6 So, together they complete 5/6th of work in 1 hr. Hence, work is completed in 6/5 hrs = 1.2 hrs... OpenStudy (anonymous): My line of reasoning goes as follows. Stan and scott can complete the task in 3hours and 2 hours respectively. On average they can complete the task in 2.5 hours. If they work on it together, they both work at this average rate of 2.5 hours but since there is two individually doing the work it takes half the time, i.e., 1.25 hours. OpenStudy (anonymous): Eseidl.. I am sorry to say this.. but let us have a check by the third person... OpenStudy (anonymous): Actually I think you may be correct. OpenStudy (anonymous): I "MAY" be correct? Let's have an expert opinion.. If I am wrong then I accept and correct my mistake.. OpenStudy (anonymous): hmmmm yeah I'd like a third opinion. OpenStudy (anonymous): i think helpingtutors is correct OpenStudy (anonymous): The problem with Tutor's reasoning is that it relies on Stan and Scott's work rates while they work in isolation. This is NOT the same in general as their work rates when they work together. Scott will be slowed down slightly when he works with Stan. Averaging their rates of work corrects for this. OpenStudy (anonymous): eseidl... we have 3rd person's opinion(mathematicalpro). would you like one more opinion? OpenStudy (anonymous): He didn't seem 100% sure, and he provided no argument with his opinion. OpenStudy (anonymous): Yes you are right.. Let us wait for another opinion,,, OpenStudy (anonymous): agreed. btw, are you sure you are correct (I'm not sure I am, by the way lol)? OpenStudy (akshay_budhkar): I am 100 percent sure that the answer is 6/5=1.2... Now i don't know which of you is correct :) OpenStudy (akshay_budhkar): I believe helping tutor is correct OpenStudy (anonymous): x/2+x/3=1 2x+3x/6=1 5x=6 x=6/5 x=1.2 OpenStudy (anonymous): 1.2 hours it is. thanks guys!

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Ex 9.1 Chapter 9 Class 11 Straight Lines Serial order wise ### Transcript Ex 9.1, 3 Find the distance between P (x1, y1) and Q (x2, y2) when: (i) PQ is parallel to the y-axis. Since PQ is parallel to y-axis x1 = x2. Distance PQ = √("(" 𝑥_2−𝑥_1 ")2 " +"(" 𝑦_2−𝑦_1 ")2" ) = √("(" 𝑥_2−𝑥_2 ")2 " +"(" 𝑦_2−𝑦_1 "2" ) = √((𝑦_2−𝑦_1 "2" )+0) = √("(" 𝑦_2−𝑦_1 ")2" ) = ± (𝑦_2−𝑦_1) = |"y2" −"y1" |

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# Proving an equality • Jul 21st 2009, 11:18 AM Proving an equality How do i prove that E[R^2] is greater than or equal to (E[R])^2?? • Jul 21st 2009, 02:50 PM Laurent Quote: Originally Posted by Pink Lady How do i prove that E[R^2] is greater than or equal to (E[R])^2?? This is related to the variance: on one hand, we define ${\rm Var}(R)=E[(R-E[R])^2]$, which is (obviously?) nonnegative. On the other hand, you should expand the above expression in order to show that ${\rm Var}(R)=E[R^2]-E[R]^2$. Since this is nonnegative, you must have $E[R^2]-E[R]^2\geq 0$, and this is it. So you need to justify why the first expression of ${\rm Var}(R)$ is nonnegative, and why it equals the second one. (There would be another more advanced/general answer involving convexity (and Jensen inequality), but you don't need that. )

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Determination of High-Accuracy Distances of Terrestrial Planets from the Sun I understand how in olden days the Sun-Planet distances were estimated using:- (i) measures of planet orbital periods ($$T$$) from analysis of observations over several centuries; (ii) Kepler's 3rd Law $$(T^2 = k.a^3)$$ applied to the planet orbital periods to determine the relative lengths of the Sun-Planet semi-major axis distances ($$a$$) expressed in terms of the AU (astronomical unit = distance from centre of Earth to centre of Sun); (iii) particular Earth-Planet distances along Solar radials were measured using Parallax techniques such as that by Casinni for Mars and Transits of Venus; (iv) From one or more Earth-Planet radial distances the length of the AU could then be determined by simple algebra. (v) Other Sun-Planet distances could then be determined once the AU length was found. Modern ephemerides e.g. those issued by JPL use various techniques:- The orbits of the inner planets are known to subkilometer accuracy through fitting radio tracking measurements of spacecraft in orbit about them. Very long baseline interferometry measurements of spacecraft at Mars allow the orientation of the ephemeris to be tied to the International Celestial Reference Frame with an accuracy of 0′′.0002. This orientation is the limiting error source for the orbits of the terrestrial planets, and corresponds to orbit uncertainties of a few hundred meters. The orbits of Jupiter and Saturn are determined to accuracies of tens of kilometers as a result of fitting spacecraft tracking data. The orbits of Uranus, Neptune, and Pluto are determined primarily from astrometric observations, for which measurement uncertainties due to the Earth’s atmosphere, combined with star catalog uncertainties, limit position accuracies to several thousand kilometers. from Folkner et al 2014. I understand how spaceraft telemetry has been used to obtain highly accurate distances between Earth and the terrestrial planets. But I am not clear about how highly-accurate distances can be determined between those planets and the centre of the Sun. Question Do such (Sun-planet-distance) determinations still basically rely on (the more accurate Newtonian version of) Kepler's 3rd Law relating planet orbital period and Sun-planet semi-major axis distance? $$\frac{T^2}{a^3} = \frac{4 \pi^2}{G(M+m)}$$ What other methods or assumptions are used/involved? EDIT - Afterthoughts After further thought prompted by feedback from u/atmosphericprisonescape my question boils down to "How are the positions (over time) of the Sun's Centre tied-in (with sub-kilometre accuracy) to the 4D space-time array of triangulated, high-accuracy, telemetry-derived, Terrestrial planet position determinations. I guess that Kepler's 3rd Law is not invoked as such. Rather, high-accuracy (sub-kilometre) determinations of the position of the Solar centre presumably require (in addition to the high-accuracy planet positions) some specific deterministic "motivation model"; i.e. a model of motion-determining factors. These factors would include Newtonian Inertia, Newtonian Gravitational forces and Non-Newtonian factors leading to additional orbital angular velocity (cf Non-Newtonian Perihelion Precession and General Relativity as described at wikipedia/Apsidal_precession ). The "motivation model" would constrain the relative positions of the Sun and planets via their involment as motion generators, reactors and enactors. • "But I am not clear about how highly-accurate distances can be determined between those planets and the centre of the Sun. " Did you think about triangulation? Jan 5, 2018 at 11:39 • @AtmosphericPrisonEscape. Yes I have thought about it and but I don't know what the levels of accuracy/error are for angular measurements and timing synchronization between measurements of different things (distance, angle) at different epochs. Also I'd guess there's some dependence on initial assumptions about observer position. Also afaik none of the raw triangulation data include direct measures of the position of the Solar centre. Jan 5, 2018 at 14:03 • We do live in the age where communication with light-speed is possible. Also we have satellites. Simultaneous observations of two objects on two different hemispheres is not a problem. Knowing this, I wouldn't think your issue to be a big problem, although I don't claim to know how your question can be answered. Jan 5, 2018 at 14:16 • @AtmosphericPrisonEscape Thankx for your feedback. It has prompted me to add some Aftrthoughts to my question. :) Jan 5, 2018 at 17:20 • I might be missing something, but if we accurately know the Earth-Sun vector with accuracy e1 and the Earth-given_planet vector with accuracy e2, we know the Sun-given_planet vector with accuracy better than e1+e2 by the triangle inequality. – user21 Jan 6, 2018 at 12:17

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Subject description - A1M14PO2 Summary of Study | Summary of Branches | All Subject Groups | All Subjects | List of Roles | Explanatory Notes               Instructions A1M14PO2 Electric Drives and Traction 2 Extent of teaching: 2+2L Guarantors: Roles: P,V Language ofteaching: CS Teachers: Completion: Z,ZK Responsible Department: 13114 Credits: 5 Semester: L Anotation: Electro mobiles and hybrid cars. Tire train and rolling resistance. Adhesion. Traction power. Locomotive traction power calculation for defined train load and track. Mass transportation vehicles. Tramway with resistive control, pulse control and induction motors. Tramway power-electronic converters. Trolley-busses. Metro. Electric locomotives - various designs. Locomotive power-converters. DC, AC and multi-system locomotives. AC motor locomotives. Diesel-electric locomotives Course outlines: 1 Sizing and design of drives with asynchronous motors. 2 Sizing of drives with asynchronous motors fed by the frequency converter. 3 Drives with a fan characteristic. 4 Electric drives in explosive environments. 5 Reliability of electrical drives. 6 Electric actuators for special purposes. 7 Technical documentation. 8 Mobility in society and its energy consequences. 9 Hybridization and electrification drive cars. 10 Systems of hybrid cars. 11 Electrovehicles. 12 Trams in public transport. 13 Electric locomotives. 14 Reserve. Exercises outline: 1 Introduction, occupational safety 2 Dimensioning of drives with induction machines 3 Mathematical model of a DC machine - deriving 4 Mathematical model of a DC machine - realization 5 Mathematical model of a DC machine - influence of parameters changing 6 Control loops of a DC motor 7 Mathematical model of an induction machine 8 Field oriented control of an induction motor ? introduction 9 Field oriented control of an induction motor ? model 10 Field oriented control vs. scalar control of an induction motor, influence of parameters changing 11 Traction drive - simulation 1 12 Traction drive - simulation 2 13 Traction drive - simulation 3 14 Reports checking Literature: [1] CHIASSON, John Nelson. Modeling and high performance control of electric machines. Hoboken: Wiley, 2005. ISBN 0-471-68449-X. [2] LEONHARD, Werner. Control of electrical drives. 3rd ed. Berlin: Springer, 2001. xviii, 460 s. ISBN 3-540-41820-2. [3] IWNICKI, S., ed. Handbook of railway vehicle dynamics. Boca Raton: CRC/Taylor & Francis, 2006. 535 s. ISBN 0-8493-3321-0. [4] www.mathworks.com [5] OGUNSOLA, Ade a Andrea MARISCOTTI. Electromagnetic compatibility in railways: analysis and management. 1st ed. New York: Springer, 2012. xix, 528 s. Lecture notes in electrical engineering; 168. ISBN 978-3-642-30280-0. [6] Steimel, A.: Electric Traction-Motive Power and Energy Supply. München, Oldenburg Industrieverlag, 2008. Requirements: Credit conditions: Attendance by the study laws, activity during the exercises, checled laboratory reports. Defense of the semestral work Webpage: https://moodle.fel.cvut.cz/courses/A1M14PO2 Subject is included into these academic programs: Program Branch Role Recommended semester MPIB Common courses V – MPKME1 Wireless Communication V 2 MPKME5 Systems of Communication V 2 MPKME4 Networks of Electronic Communication V 2 MPKME3 Electronics V 2 MPKME2 Multimedia Technology V 2 MPEEM1 Technological Systems P 2 MPEEM3 Electrical Power Engineering P 2 MPEEM2 Electrical Machines, Apparatus and Drives P 2 MPKYR4 Aerospace Systems V 2 MPKYR1 Robotics V 2 MPKYR3 Systems and Control V 2 MPKYR2 Sensors and Instrumentation V 2 Page updated 17.6.2019 09:52:48, semester: Z,L/2020-1, L/2018-9, Z,L/2019-20, Send comments about the content to the Administrators of the Academic Programs Proposal and Realization: I. Halaška (K336), J. Novák (K336)

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# Clustering Algorithms MATLAB #### Related Tools Clustering Algorithms in MATLAB simulation involves several procedures that must be carried out in an appropriate way. Read out the structured way that we provide assistance for your work, get in touch with us by dropping all your project details. To simulate various clustering algorithms in MATLAB, like DBSCAN, hierarchical, and k-means clustering, we suggest all the major procedures explicitly, along with instances: 1. K-Means Clustering Procedures: • Create or Load Data: For clustering process, we have to import or develop a dataset. • Conduct Clustering: In order to carry out k-means clustering, utilize the kmeans function. • Visualize the Outcomes: Specifically for the visualization purpose, the clustered data has to be plotted. Instance: % Generate sample data rng(1); % For reproducibility X = [randn(100,2)+ones(100,2); randn(100,2)-ones(100,2)]; % Number of clusters k = 2; % Perform k-means clustering [idx, C] = kmeans(X, k); % Plot the clustered data figure; gscatter(X(:,1), X(:,2), idx); hold on; plot(C(:,1), C(:,2), ‘kx’, ‘MarkerSize’, 15, ‘LineWidth’, 3); title(‘K-Means Clustering’); xlabel(‘Feature 1’); ylabel(‘Feature 2’); legend(‘Cluster 1’, ‘Cluster 2’, ‘Centroids’, ‘Location’, ‘Best’); 1. Hierarchical Clustering Procedures: • Create or Load Data: To carry out clustering, import or develop a dataset. • Carry out Clustering: Develop a hierarchical cluster tree by employing the linkage function. To create clusters, we need to utilize the cluster function. • Visualize the Dendrogram: The clustering procedure has to be visualized by plotting the dendrogram. Instance: % Generate sample data rng(2); % For reproducibility X = [randn(100,2)+ones(100,2); randn(100,2)-ones(100,2)]; % Perform hierarchical clustering % Plot the dendrogram figure; dendrogram(Z); title(‘Hierarchical Clustering Dendrogram’); xlabel(‘Sample Index’); ylabel(‘Distance’); % Form clusters numClusters = 2; idx = cluster(Z, ‘maxclust’, numClusters); % Plot the clustered data figure; gscatter(X(:,1), X(:,2), idx); title(‘Hierarchical Clustering’); xlabel(‘Feature 1’); ylabel(‘Feature 2’); legend(‘Cluster 1’, ‘Cluster 2’, ‘Location’, ‘Best’); 1. DBSCAN (Density-Based Spatial Clustering of Applications with Noise) Procedures: • Create or Load Data: Particularly for the clustering procedure, a dataset must be imported or developed. • Carry out Clustering: As a means to carry out DBSCAN clustering, utilize the dbscan function. • Visualize the Outcomes: For visualization, we should plot the clustered data. Instance: % Generate sample data rng(3); % For reproducibility X = [randn(100,2)+ones(100,2); randn(100,2)-ones(100,2); randn(100,2)+[5 0]]; % Perform DBSCAN clustering epsilon = 0.5; % Maximum distance between points to be considered neighbors minPts = 5; % Minimum number of points to form a dense region idx = dbscan(X, epsilon, minPts); % Plot the clustered data figure; gscatter(X(:,1), X(:,2), idx); title(‘DBSCAN Clustering’); xlabel(‘Feature 1’); ylabel(‘Feature 2’); legend(‘Cluster 1’, ‘Cluster 2’, ‘Cluster 3’, ‘Noise’, ‘Location’, ‘Best’); 1. Gaussian Mixture Models (GMM) Procedures: • Create or Load Data: Conduct the clustering process by developing or importing a dataset. • Carry out Clustering: To align a GMM, we employ the fitgmdist function. In order to allocate data points to clusters, the cluster technique must be used. • Visualize the Outcomes: For the purpose of visualization, plot the clustered data. Instance: % Generate sample data rng(4); % For reproducibility X = [randn(100,2)+ones(100,2); randn(100,2)-ones(100,2)]; % Fit a Gaussian Mixture Model gm = fitgmdist(X, 2); % Perform clustering idx = cluster(gm, X); % Plot the clustered data figure; gscatter(X(:,1), X(:,2), idx); hold on; plot(gm.mu(:,1), gm.mu(:,2), ‘kx’, ‘MarkerSize’, 15, ‘LineWidth’, 3); title(‘Gaussian Mixture Model Clustering’); xlabel(‘Feature 1’); ylabel(‘Feature 2’); legend(‘Cluster 1’, ‘Cluster 2’, ‘Centroids’, ‘Location’, ‘Best’); Hints for Simulating Clustering Algorithms: • Data Preprocessing: Based on the requirements, the data must be preprocessed (for instance: standardized or normalized), and assuring this aspect is crucial. • Parameter Tuning: As a means to accomplish enhanced clustering outcomes, we have to test with various parameters (for example: number of clusters, minPts, and epsilon). • Visualization: To interpret the standard of the clusters and the efficiency of the algorithm, visualization of clustering outcomes offers support. • Validation: In order to evaluate the standard of the clusters, make use of cluster validation methods (for instance: Davies-Bouldin index, silhouette score). ## Important Research challenges & problems in clustering algorithms In clustering algorithms, numerous research issues and challenges exist, which are required to solve in an effective manner for various necessities. On the basis of clustering algorithms, we list out a few major research issues and challenges in a clear way: 1. Scalability Issue: While utilizing a wide range of datasets, scalability issues are faced by several clustering methods. Potential Challenges: • To manage lots of data points in an effective manner, creating algorithms is crucial. • With the increase of dataset volume, the algorithms must preserve performance and preciseness, and assuring this factor is important. • In order to handle extensive data, plan to apply parallel or distributed computing approaches. 1. High Dimensionality Issue: In terms of the dimensionality problem, the clustering process can be challenging, specifically when an extensive amount of characteristics is encompassed in the datasets. Potential Challenges: • In addition to securing major details, we have to minimize dimensionality (for instance: employing t-SNE, PCA). • Particularly in high-dimensional areas, managing data inadequacy is significant. • To detect important characteristics for the clustering process, create efficient algorithms. 1. Specifying the Number of Clusters Issue: Defining the number of clusters in a prior manner is the major need of numerous clustering algorithms. However, it is considered as uncertain. Potential Challenges: • To define the appropriate number of clusters in an automatic manner, creating robust techniques is important. • As a means to assess the number of clusters, employ various approaches like the gap statistic, elbow technique, or silhouette score. • At the time of clustering operation, specify the number of clusters dynamically by developing algorithms. 1. Cluster Validity and Assessment Issue: In the clusters created by a specific algorithm, assessing their quality is a difficult process. Potential Challenges: • External and internal cluster validity indices have to be created in an efficient manner. • Through various datasets and algorithms, the clustering outcomes must be compared. • For cluster efficacy, diverse principles might be needed by various applications. So, the specific factors of clustering assessment should be managed. 1. Managing Noise and Outliers Issue: The performance of clustering methods can be majorly impacted through anomalies and noise. Potential Challenges: • To detect and manage anomalies and noise, we need to build effective algorithms. • In order to handle noise, make use of techniques such as density-based clustering (for example: DBSCAN). • Before the clustering process, clean data by applying preprocessing methods. 1. Explainable Clustering Issue: Specifically for realistic applications, the clusters must be relevant as well as explainable. Potential Challenges: • In the scenario of the application, make sure that the clusters have relevance in a certain and explicit way. • To improve the explainability of clusters, employing robust methods is crucial (for instance: decision trees). • Regarding the allocation of data points to particular clusters, offer interpretations through creating efficient algorithms. 1. Cluster Shape and Size Inconsistency Issue: For clusters, particular dimensions and shapes are considered by several algorithms. But, for all datasets, these assumptions might not be appropriate. Potential Challenges: • As a means to identify clusters of random dimensions and shapes, develop algorithms. • Various methods such as DBSCAN or spectral clustering have to be utilized, where the spherical clusters are not considered. • Along with clusters of diverse loads, managing datasets is important. 1. Integrating Domain Knowledge Issue: Mostly, it is difficult to include domain-based expertise; however it can enhance clustering outcomes. Potential Challenges: • To integrate domain-based expertise, we have to create semi-supervised clustering methods. • Instruct the clustering procedure by utilizing background details or conditions. • The clustering outcomes should not be influenced by the integration of domain expertise, and assuring this aspect is crucial. 1. Dynamic and Streaming Data Issue: Specific complexities are caused through the clustering of streaming or dynamic data. Potential Challenges: • In order to manage consistently receiving data, online or incremental clustering methods have to be created. • For the variations in the data distribution, the algorithms must have the ability to adjust periodically, and confirming this factor is significant. • Specifically for streaming data, handling computational resources and memory in an effective way is important. 1. Multi-View and Multi-Modal Data Issue: Unique clustering methods are needed to deal with datasets which include several types or views. Potential Challenges: • To incorporate and cluster data from several types or sources, we should create efficient algorithms. • In feature areas of multi-view data, managing the variances and diversity is essential. • As a means to handle multi-view data, utilize various approaches such as subspace clustering or co-clustering. 1. Clustering with Missing Data Issue: The performance of clustering can be impacted in a substantial manner when using incomplete data. Potential Challenges: • Manage missing values in an efficient manner by creating robust algorithms. • Before the clustering process, fill in missing data through the use of imputation methods. • In the clustering outcomes, the managing of missing data should not cause any flaws or bias, and assuring this aspect is important. 1. Computational Intricacy Issue: For a wide range of datasets, numerous clustering techniques are inappropriate, because of having extreme computational intricacy. Potential Challenges: • In order to minimize the computational needs, enhancing algorithms is crucial. • To accelerate the clustering process, apply robust methods and data structures. • Focus on stabilizing computational efficacy and preciseness by utilizing approximation methods. 1. Cluster Ensembles Issue: To enhance preciseness and strength, the integration of several clustering outcomes is a difficult procedure. Potential Challenges: • From various algorithms, gather and integrate clustering outcomes through the creation of effective techniques. • Across various clustering results, managing the challenges and variations is important. • In the case of using extensive datasets, assure the scalability of the ensemble clustering method. 1. Privacy-Preserving Clustering Issue: In numerous applications, it is important to cluster private data in addition to securing confidentiality. Potential Challenges: • To follow data security rules, we need to create privacy-preserving clustering methods. • At the time of clustering process, secure confidential data by employing approaches such as differential privacy. • Make sure that the clustering performance is not majorly reduced by the privacy-preserving techniques. 1. Scalable Infrastructure Issue: For actual-time applications, utilizing clustering methods is crucial, especially on scalable frameworks. Potential Challenges: • In order to apply scalable clustering solutions, the cloud computing and distributed frameworks have to be utilized. • The framework must have the ability to manage data processing and clustering in actual-time, and confirming this aspect is significant. • Compensations among expenses, performance, and scalability should be handled. For the simulation of various clustering methods in MATLAB, the important procedures are recommended by us, including some sample codes. Relevant to clustering algorithms, we pointed out several research issues and potential challenges in an explicit manner. ## Great Memories Our Achievements We received great winning awards for our research awesomeness and it is the mark of our success stories. It shows our key strength and improvements in all research directions. ## Our Guidance • Assignments • Homework • Projects • Literature Survey • Algorithm • Pseudocode • Mathematical Proofs • Research Proposal • System Development • Paper Writing • Conference Paper • Thesis Writing • Dissertation Writing • Hardware Integration • Paper Publication • MS Thesis

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# Equivalent 2 ### How To : Do the boomerang scratch with fading pattern The video describes the effect known as the Boomerang Scratch. The author describes the technique and shows the sheet musical note equivalent. After explaining the sheet music notation, the author shows a video of the technique being performed on a scratch deck with a cross fa ...more ### How To : Add fractions with uncommon denominators This video shows you how to easily add two fractions with uncommon denominators. To add the two fractions you must translate the fractions into its higher equivalent form, with a common denominator. To find a common denominator, you must find the lowest multiple that both deno ...more ### How To : Subtract fractions with uncommon demoninators In this tutorial the author shows an easy and simple way of subtracting fractions with uncommon denominators. He says that to subtract fraction with uncommon denominator, we need to make their denominators equivalent first and after making the denominators equal, you need to i ...more ### How To : Add to fractions with uncommon denominators. Wondering how to add fractions with uncommon denominators? It is simple to solve the denominators there are three steps to find the uncommon denominators. first step is to find the lowest common denominator and you take 3/4 + 5/6. In this lowest common denominator is 12, then ...more ### How To : Subtract positive & negative numbers In this Math Made Easy tutorial the author shows how to subtract positive and negative numbers. He says that because of the numerous combination of positive and negative numbers it adds to a lot of confusion. Now he shows a trick of how to add positive numbers to negative numb ...more ### News : Ultimate Aircraft Boneyard Now Visible on Google Maps For some time now, I've been dying to make the trek to the world's largest airplane graveyard outside of Tuscon, AZ. The ultimate field trip that I never seem to get around to. Now, thanks to Google Maps, anybody can now check it out remotely: up close, high-res, beautiful.  ...more ### How To : Say "I think" or "he/she thinks" & more in Hindi This is lesson 14 of Speaking Hindi course. We will see how to translate: To Think = Sochna To See = Dekhna To Drink = Peena E.g. of 'Think'. I think = main sochta hun He thinks = wo sochta hai She thinks = wo sochtee hai I am thinking = main soch raha hoon He is thinking = ...more ### How To : Finalize DVDs on Panasonic DMR-ES15 for other players Why won't my recorded disc play in other machines? Check out this video tutorial on how finalize DVDs on Panasonic DMR-ES15 (ES25) for other players. This Panasonic DMR-ES15 (ES25) DVD Recorder is easy to operate and great to record with. To create a disc that's compatible w ...more ### How To : Finalize DVDs on Panasonic DMR-ES45V for other players Why won't my recorded disc play in other machines? Check out this video tutorial on how finalize DVDs on Panasonic DMR-ES45V (ES35V) for other players. This Panasonic DMR-ES45V (ES35V) DVD Recorder is easy to operate and great to record with. To create a disc that's compatib ...more ### How To : Finalize DVDs on Panasonic DMR-EH55 for other players Why won't my recorded disc play in other machines? Check out this video tutorial on how finalize DVDs on Panasonic DMR-EH55 for other players. This Panasonic DMR-EH55 DVD Recorder is easy to operate and great to record with. To create a disc that's compatible with other DVD ...more ### Market Reality : Magic Leap Prescription Lenses Arrive, Niantic Challenges Devs with Real World, & Blippar Goes Under Mixed reviews of Magic Leap One aside, it would be hard to deny that Magic Leap has had a big year. And the AR unicorn isn't coasting to the finish line, with a number of new apps dropping and prescription frames finally arriving to bring relief to those who wear eyeglasses. ...more ### How To : Unlock Android 9.0 Pie's New 'Feature Flags' Menu to Modify System Settings There's a new secret settings menu hiding in Android 9.0 Pie that offers Chrome-style "flags" to the overall operating system. Since it's currently the first Developer Preview and Public Beta versions, it's possible that this menu could disappear entirely when the stable versi ...more ### News : We've Finally Figured Out Why Some Mushrooms Glow Bioluminescence — the ability of an organism to produce and emit light — is nature's light show. Plants, insects, fish, and bacteria do it, and scientists understand how. Until now, though, we didn't know how fungi glow. Now, thanks to research led by Zinaida Kaskova in the I ...more ### News : 'Impossible to Identify' Website Phishing Attack Leaves Chrome & Firefox Users Vulnerable (But You Can Prevent It) Sophisticated hackers have been exploiting vulnerabilities in Chrome and Firefox to trick even the most careful internet users into logging into fake domains for sites like Apple, Google, and Amazon. Typically, a careful internet user would always check the domain of any site ...more ### Scrabble Challenge #12 : Can You Unscramble the Derivative Words? Last week's challenge brought up a good question (thanks Lisa) regarding the functionality of the Official Scrabble Players Dictionary, 4th Edition, both online and in print form. The word TACTICAL was the answer to the first puzzle in that challenge, but when you search for t ...more ### SQL Injection 101 : How to Avoid Detection & Bypass Defenses It is often said that the best hackers remain unknown, and the greatest attacks are left undiscovered, but it's hard for an up-and-coming penetration tester or white hat to learn anything unless one of those factors is actually known or discovered. But the end goal here in our ...more Glow sticks, a popular favor at parties and outdoor events, and a must-have on Halloween, can be traced back to the United States Navy in the mid-1960s. The military desired improved visibility during night operations, and glow sticks, with their small-size portability and lac ...more ### How To : Make hydrazine sulfate with the hypochlorite and the Ketazine process Hydrazine sulfate has many uses, but most notably, it's been used under the trade name of Sehydrin, a treatment for anorexia, cachexia and some even think cancer. But for we DIY chemists, it's useful for something entirely different— as a substitute for the more dangerous pure ...more ### How To : Play "Lua" by Bright Eyes on the ukulele Want to play Bright Eyes' "Lua" on the ukulele? Learn how with this free video ukulele lesson from Ukulele Mike. Whether you play the ukulele or bass kazoo, there is no better way to improve your chops than by learning to play your favorite songs. Not only is it more fun and m ...more ### How To : Make ammonium nitrate with NaHSO4 and nitrate salt Okay, so you don't necessarily have to make ammonium nitrate to have ammonium nitrate — ammonium nitrate can simply be obtained from fertilizers and instant cold packs. Making homemade NH4NO3 can be much more expensive then buying it, but this video is not meant to be a viable ...more ### How To : Make Stuffed Spaghetti Squash Trying to stay away from carbs but love starches? Why not try some spaghetti squash instead of all that pasta? Spaghetti squash is easy to prepare and can be garnished just like you would with pasta or potatoes. In this tutorial, learn how to make a delicious stuffed spaghett ...more ### How To : Make a fiberglass mold layup for car speakers Check out this video tutorial on how to make a fiberglass mold layup for car speakers. For fresh molds you must put several layers of wax on and I would consider PVA as well if the mold is less then a week old. Always use PVA (or equivalent) on your plug surface before molding ...more ### Antibubbles in Microgravity : NASA Astronaut Experiments with Centrifugal Force in Space When he's not taking orbital videos of Earth's auroras, NASA Astronaut Don Pettit is experimenting with water in zero gravity. He's already shown us how water droplets can orbit around knitting needles in a microgravity environment. Now he's playing with water again, this time ...more ### News : MegaUpload goes down - Anon retaliates. As you may or may not know, MegaUpload - one of the largest file sharing services has been taken down by the feds. To us - this means no more 1gb free downloads. And we move onto MediaFire and Rapidshare. - But more importantly, this means that our inherent freedom to share f ...more ### News : Math Craft Inspiration of the Week: The Mathematical Lego Sculptures of Andrew Lipson Andrew Lipson builds sculptures based off of Mathematical objects using standard Lego bricks. He has built models of knots, Mobius strips, Klein bottles, Tori, Hoberman spheres (using Lego technic pieces), and recreations of M.C. Escher works. Below are a few images and brief ...more ### News : Life in 30 Seconds! Sitting around thinking about my average sexual performance, I wondered if there were other things in life that were also 30 seconds long. I was extremely relieved that such things were possible! These 30 second movies are the mobile text equivalent to major films or projects. ...more On the same day that Google launched their new social network, Google +, a team of Google engineers calling themselves the Data Liberation Front announced their first service on Tuesday—Google Takeout, a product that allows users to easily move their data in and out of Google ...more ### News : Google +1 Button Now Available for Websites Google's sociable equivalent to the Facebook Like button is finally here, and it works very similar to your favorite social network's recommendation system, except it shows up directly in Google search results. Whenever your Google friend gives a website or webpage the +1, you ...more ### How To : Graph Mario on a TI-83 Calculator When it comes to graphing and comparing functions, the TI-83 graphing calculator is the end-all device for math and science students. But one of the most entertaining aspects of Texas Instruments' powerful algebraic and trigonometric calculator is not the equations themselves, ...more ### Grow While You Go : The Bus-Top Gardens of New York City If you were to look on the roof of your local city bus, what would you find? A little bit of dirt, most likely, and a whole lot of space. Precisely the stuff you'd find in an empty container garden! Enter NYU graduate student Marco Castro Cosio's Bus Roots, a project which, th ...more ### Cannabis Catering : Yuppie Food Laced With Marijuana Medical marijuana enables a novel business idea: Cannabis Catering, a yuppie stoner catering service, delivers gourmet meals laced with pot to your doorstep. Delivery costs around \$100 per person (and of course, medical marijuana card is required). 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The comparisons must be between apples to appl ...more ### How To : Reduce fractions to their simplest form in basic math This video teaches us the method to simplify the fractions. Let us take an example of 18/24. Find a number that goes into both 18 and 24. Divide the top and bottom by the same number. Here, 2 will go into both numbers. So, the number now becomes (18/2)/(24/2) which gives 9/12. ...more ### HowTo : Reboot Your Corpse IEEE Spectrum examines the practice of cryonically freezing the dead, with the intent to "reboot" when medical advances are prepared to undo their death. Is it just a matter of time (cryonics scientists predict 100 years) or pure quackery? Here's how it works: "Before the b ...more ### News : Could This POSSIBLY Be Street Legal? Don't get too excited. The answer? It's not. However, this James Bond inspired road rage mod is still prettttty awesome. Colin Furze's scooter can shoot up to 15 foot flames. Furze, a plumber, built it because he was sick of other drivers cutting him off. Via Daily Mail, "Co ...more ### News : Car-puccino Runs Purely on Coffee Another creation by Bang Goes the Theory, the "car-puccino" is a 1988 Volkswagen Scirocco powered solely by roasted coffee granules. Daily Mail reports, "The car will be driven the 210 miles between Manchester and London powered only by roasted coffee granules. It will go on ...more ### How To : Factor a trinomial, find the GCF, then group This lesson describes the method to find the factors of a trinomial, which consists of three terms, by grouping. First of all, factor out the greatest common factor (GCF), and write the reduced trinomial in parentheses. Let the terms of the trinomial be written in order of exp ...more ### How To : Dance your ass off Why sweat through a boring exercise routine when you can achieve the same results while having fun? 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Monday, June 17, 2024 HomeMicrosoft OfficeMicrosoft ExcelHow to Shade Every Other Row in Excel (Easy Steps) # How to Shade Every Other Row in Excel (Easy Steps) The stacks of white rows can dull your senses. Shading every other row in Excel can produce a wonderful contrast between these white rows and brighten your Excel experience. There are many in-built color patterns and schemes you can use to shade your Excel data. A conditional format is the easiest way to apply these color schemes. In this tutorial, we will discuss how you can use conditional formatting to apply shade alternating lines. ### Shading Every other Row in Excel ( Using Conditional Formatting) 1. Open the Excel Document containing your data. 2. Select the range of cells you wish to shade. Note: Select the entire range, not just alternate rows. 3. In the home tab, tap Conditional Formatting. 4. Select New Rule… from the drop-down menu. 5. Select “use a formula to determine…” as Rule type. It’s the sixth one. 6. Insert formula: MOD(ROW(),2) in the space below Edit this Rule Description. 7. Tap Format in the same window (adjacent to preview). 8. Select the fill tab. 9. Select the color you want your alternate rows to have and tap OK. 10. Tap OK again. 11. Voila! your alternate rows are here. Read a Similar Excel Tutorial: How to Create a Pictograph in Excel ### Understanding the Alternate Row Shading Formula If you’re interested in the science behind the Excel Function: MOD(ROW(),2), here’s a simple explanation. The MOD and ROW function deal with the remainder of division and row number respectively. For example, for the ninth row, the formula MOD is (9,2). 9 is divided by 2 (5 times) to extract a remainder of 1. Whereas the for the tenth row, MOD is (10,2). 10 divided by 2 six times is 0. Apply this formula to all rows and only MOD for odd rows returns as 1, and consequently, they are shaded. Following text and picture instructions isn’t for everyone. Some people prefer Youtube tutorials to understand the process that goes into shading every other row in Excel. For your further convenience, below is a video tutorial explaining exactly that. The version of Microsoft Excel used in this Youtube tutorial is Excel 2013. However, the steps remain identical across all versions of Excel, including 2013, 2016, Excel Online and the latest 2019. Sheraz Ali An established copywriter, with a longstanding experience in a vast array of industries, including but not limited to spirituality, technology, cannabis and travel. RELATED ARTICLES

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Home > Education > Toeplitz (2) # Toeplitz (2) Date post: 07-May-2015 Category: View: 411 times Description: matrix Embed Size (px) of 98 /98 Toeplitz and Circulant Matrices: A review Transcript Toeplitz and Circulant Matrices: A review Toeplitz and Circulant Matrices: A review Robert M. Gray Deptartment of Electrical EngineeringStanford University Stanford 94305, USA [email protected] Contents Chapter 1 Introduction 1 1.1 Toeplitz and Circulant Matrices 1 1.2 Examples 5 1.3 Goals and Prerequisites 9 Chapter 2 The Asymptotic Behavior of Matrices 11 2.1 Eigenvalues 11 2.2 Matrix Norms 14 2.3 Asymptotically Equivalent Sequences of Matrices 17 2.4 Asymptotically Absolutely Equal Distributions 24 Chapter 3 Circulant Matrices 31 3.1 Eigenvalues and Eigenvectors 32 3.2 Matrix Operations on Circulant Matrices 34 Chapter 4 Toeplitz Matrices 37 v vi CONTENTS 4.1 Sequences of Toeplitz Matrices 37 4.2 Bounds on Eigenvalues of Toeplitz Matrices 41 4.3 Banded Toeplitz Matrices 43 4.4 Wiener Class Toeplitz Matrices 48 Chapter 5 Matrix Operations on Toeplitz Matrices 61 5.1 Inverses of Toeplitz Matrices 62 5.2 Products of Toeplitz Matrices 67 5.3 Toeplitz Determinants 70 Chapter 6 Applications to Stochastic Time Series 73 6.1 Moving Average Processes 74 6.2 Autoregressive Processes 77 6.3 Factorization 80 Acknowledgements 83 References 85 Abstract t0 t−1 t−2 · · · t−(n−1) t1 t0 t−1 t2 t1 t0... .... . . tn−1 · · · t0 The fundamental theorems on the asymptotic behavior of eigenval- ues, inverses, and products of banded Toeplitz matrices and Toeplitz matrices with absolutely summable elements are derived in a tutorial manner. Mathematical elegance and generality are sacrificed for con- ceptual simplicity and insight in the hope of making these results avail- able to engineers lacking either the background or endurance to attack the mathematical literature on the subject. By limiting the generality of the matrices considered, the essential ideas and results can be con- veyed in a more intuitive manner without the mathematical machinery required for the most general cases. As an application the results are applied to the study of the covariance matrices and their factors of linear models of discrete time random processes. vii 1 Introduction 1.1 Toeplitz and Circulant Matrices A Toeplitz matrix is an n × n matrix Tn = [tk,j; k, j = 0, 1, . . . , n − 1] where tk,j = tk−j, i.e., a matrix of the form Tn = t0 t−1 t−2 · · · t−(n−1) t1 t0 t−1 t2 t1 t0... .... . . tn−1 · · · t0 . (1.1) Such matrices arise in many applications. For example, suppose that x = (x0, x1, . . . , xn−1)′ = x0 x1... xn−1 1 2 Introduction is a column vector (the prime denotes transpose) denoting an “input” and that tk is zero for k < 0. Then the vector y = Tnx = t0 0 0 · · · 0 t1 t0 0 t2 t1 t0... .... . . tn−1 · · · t0 x0 x1 x2... xn−1 = x0t0t1x0 + t0x1∑2 i=0 t2−ixi... ∑n−1i=0 tn−1−ixi with entries yk = k∑ i=0 tk−ixi represents the the output of the discrete time causal time-invariant filter h with “impulse response” tk. Equivalently, this is a matrix and vector formulation of a discrete-time convolution of a discrete time input with a discrete time filter. As another example, suppose that Xn is a discrete time ran- dom process with mean function given by the expectations mk = E(Xk) and covariance function given by the expectations KX(k, j) = E[(Xk − mk)(Xj − mj)]. Signal processing theory such as predic- tion, estimation, detection, classification, regression, and communca- tions and information theory are most thoroughly developed under the assumption that the mean is constant and that the covariance is Toeplitz, i.e., KX(k, j) = KX(k − j), in which case the process is said to be weakly stationary. (The terms “covariance stationary” and “second order stationary” also are used when the covariance is assumed to be Toeplitz.) In this case the n × n covariance matrices Kn = [KX(k, j); k, j = 0, 1, . . . , n − 1] are Toeplitz matrices. Much of the theory of weakly stationary processes involves applications of 1.1. Toeplitz and Circulant Matrices 3 Toeplitz matrices. Toeplitz matrices also arise in solutions to differen- tial and integral equations, spline functions, and problems and methods in physics, mathematics, statistics, and signal processing. A common special case of Toeplitz matrices — which will result in significant simplification and play a fundamental role in developing more general results — results when every row of the matrix is a right cyclic shift of the row above it so that tk = t−(n−k) = tk−n for k = 1, 2, . . . , n− 1. In this case the picture becomes Cn = t0 t−1 t−2 · · · t−(n−1) t−(n−1) t0 t−1 t−(n−2) t−(n−1) t0... .... . . t−1 t−2 · · · t0 . (1.2) A matrix of this form is called a circulant matrix. Circulant matrices arise, for example, in applications involving the discrete Fourier trans- form (DFT) and the study of cyclic codes for error correction. A great deal is known about the behavior of Toeplitz matrices — the most common and complete references being Grenander and Szego [16] and Widom [33]. A more recent text devoted to the subject is Bottcher and Silbermann [5]. Unfortunately, however, the necessary level of mathematical sophistication for understanding reference [16] is frequently beyond that of one species of applied mathematician for whom the theory can be quite useful but is relatively little understood. This caste consists of engineers doing relatively mathematical (for an engineering background) work in any of the areas mentioned. This ap- parent dilemma provides the motivation for attempting a tutorial intro- duction on Toeplitz matrices that proves the essential theorems using the simplest possible and most intuitive mathematics. Some simple and fundamental methods that are deeply buried (at least to the untrained mathematician) in [16] are here made explicit. The most famous and arguably the most important result describing Toeplitz matrices is Szego’s theorem for sequences of Toeplitz matrices Tn which deals with the behavior of the eigenvalues as n goes to infinity. A complex scalar α is an eigenvalue of a matrix A if there is a 4 Introduction nonzero vector x such that Ax = αx, (1.3) in which case we say that x is a (right) eigenvector of A. If A is Hermi- tian, that is, if A∗ = A, where the asterisk denotes conjugate transpose, then the eigenvalues of the matrix are real and hence α∗ = α, where the asterisk denotes the conjugate in the case of a complex scalar. When this is the case we assume that the eigenvalues αi are ordered in a nondecreasing manner so that α0 ≥ α1 ≥ α2 · · · . This eases the approximation of sums by integrals and entails no loss of generality. Szego’s theorem deals with the asymptotic behavior of the eigenvalues τn,i; i = 0, 1, . . . , n− 1 of a sequence of Hermitian Toeplitz matrices Tn = [tk−j; k, j = 0, 1, 2, . . . , n − 1]. The theorem requires that several technical conditions be satisfied, including the existence of the Fourier series with coefficients tk related to each other by f(λ) =∞∑ k=−∞tke ikλ; λ ∈ [0, 2π] (1.4) tk =1 ∫ 2π 0f(λ)e−ikλ dλ. (1.5) Thus the sequence tk determines the function f and vice versa, hence the sequence of matrices is often denoted as Tn(f). If Tn(f) is Hermi- tian, that is, if Tn(f)∗ = Tn(f), then t−k = t∗k and f is real-valued. Under suitable assumptions the Szego theorem states that limn→∞ 1 n n−1∑ k=0 F (τn,k) =1 ∫ 2π 0F (f(λ)) dλ (1.6) for any function F that is continuous on the range of f . Thus, for example, choosing F (x) = x results in limn→∞ 1 n n−1∑ k=0 τn,k =1 ∫ 2π 0f(λ) dλ, (1.7) so that the arithmetic mean of the eigenvalues of Tn(f) converges to the integral of f . The trace Tr(A) of a matrix A is the sum of its 1.2. Examples 5 diagonal elements, which in turn from linear algebra is the sum of the eigenvalues of A if the matrix A is Hermitian. Thus (1.7) implies that limn→∞ 1 nTr(Tn(f)) = 1 ∫ 2π 0f(λ) dλ. (1.8) Similarly, for any power s limn→∞ 1 n n−1∑ k=0 τ sn,k = 1 ∫ 2π 0f(λ)s dλ. (1.9) If f is real and such that the eigenvalues τn,k ≥ m > 0 for all n, k, then F (x) = lnx is a continuous function on [m,∞) and the Szego theorem can be applied to show that limn→∞ 1 n n−1∑ i=0 ln τn,i =1 ∫ 2π 0ln f(λ) dλ. (1.10) From linear algebra, however, the determinant of a matrix Tn(f) is given by the product of its eigenvalues, det(Tn(f)) = n−1∏ i=0 τn,i, so that (1.10) becomes limn→∞ ln det(Tn(f))1/n = limn→∞ 1 n n−1∑ i=0 ln τn,i =1 ∫ 2π 0ln f(λ) dλ. (1.11) As we shall later see, if f has a lower bound m > 0, than indeed all the eigenvalues will share the lower bound and the above derivation applies. Determinants of Toeplitz matrices are called Toeplitz determinants and (1.11) describes their limiting behavior. 1.2 Examples A few examples from statistical signal processing and information the- ory illustrate the the application of the theorem. These are described 6 Introduction with a minimum of background in order to highlight how the asymp- totic eigenvalue distribution theorem allows one to evaluate results for processes using results from finite-dimensional vectors. The differential entropy rate of a Gaussian process Suppose that Xn; n = 0, 1, . . . is a random process described by probability density functions fXn(xn) for the random vectors Xn = (X0,X1, . . . ,Xn−1) defined for all n = 0, 1, 2, . . .. The Shannon differ- ential entropy h(Xn) is defined by the integral h(Xn) = −∫ fXn(xn) ln fXn(xn) dxn and the differential entropy rate of the random process is defined by the limit h(X) = limn→∞ 1 nh(Xn) if the limit exists. (See, for example, Cover and Thomas[7].) A stationary zero mean Gaussian random process is completely de- scribed by its mean correlation function rk,j = rk−j = E[XkXj ] or, equivalently, by its power spectral density function f , the Fourier trans- form of the covariance function: f(λ) =∞∑ n=−∞rne inλ, rk =1 ∫ 2π 0f(λ)e−iλk dλ For a fixed positive integer n, the probability density function is fXn(xn) =e− 1 2xn′R−1 n xn (2π)n/2det(Rn)1/2, where Rn is the n× n covariance matrix with entries rk−j. A straight- forward multidimensional integration using the properties of Gaussian random vectors yields the differential entropy h(Xn) =1 2ln(2πe)ndetRn. 1.2. Examples 7 The problem at hand is to evaluate the entropy rate h(X) = limn→∞ 1 nh(Xn) = 1 2ln(2πe) + lim n→∞1 nln det(Rn). The matrix Rn is the Toeplitz matrix Tn generated by the power spec- tral density f and det(Rn) is a Toeplitz determinant and we have im- mediately from (1.11) that h(X) =1 2log ( 2πe1 ∫ 2π 0ln f(λ) dλ ) . (1.12) This is a typical use of (1.6) to evaluate the limit of a sequence of finite- dimensional qualities, in this case specified by the determinants of of a sequence of Toeplitz matrices. The Shannon rate-distortion function of a Gaussian process As a another example of the application of (1.6), consider the eval- uation of the rate-distortion function of Shannon information theory for a stationary discrete time Gaussian random process with 0 mean, covariance KX(k, j) = tk−j, and power spectral density f(λ) given by (1.4). The rate-distortion function characterizes the optimal tradeoff of distortion and bit rate in data compression or source coding systems. The derivation details can be found, e.g., in Berger [3], Section 4.5, but the point here is simply to provide an example of an application of (1.6). The result is found by solving an n-dimensional optimization in terms of the eigenvalues τn,k of Tn(f) and then taking limits to obtain parametric expressions for distortion and rate: Dθ = limn→∞ 1 n n−1∑ k=0 min(θ, τn,k) Rθ = limn→∞ 1 n n−1∑ k=0 max(0,1 2lnτn,k θ). 8 Introduction The theorem can be applied to turn this limiting sum involving eigen- values into an integral involving the power spectral density: Dθ = ∫ 2π 0min(θ, f(λ)) dλ Rθ = ∫ 2π 0max ( 0,1 2lnf(λ) θ ) dλ. Again an infinite dimensional problem is solved by first solving a finite dimensional problem involving the eigenvalues of matrices, and then using the asymptotic eigenvalue theorem to find an integral expression for the limiting result. One-step prediction error Another application with a similar development is the one-step predic- tion error problem. Suppose that Xn is a weakly stationary random process with covariance tk−j. A classic problem in estimation theory is to find the best linear predictor based on the previous n values of Xi, i = 0, 1, 2, . . . , n− 1, Xn = n∑ i=1 aiXn−i, in the sense of minimizing the mean squared error E[(Xn−Xn)2] over all choices of coefficients ai. It is well known (see, e.g., [14]) that the min- imum is given by the ratio of Toeplitz determinants detTn+1/det Tn. The question is to what this ratio converges in the limit as n goes to ∞. This is not quite in a form suitable for application of the theorem, but we have already evaluated the limit of detT1/nn in (1.11) and for large n we have that (detTn)1/n ≈ exp ( 1 ∫ 2π 0ln f(λ) dλ ) ≈ (detTn+1)1/(n+1) and hence in particular that (detTn+1)1/(n+1) ≈ (detTn)1/n so that detTn+1 detTn≈ (detTn)1/n → exp ( 1 ∫ 2π 0ln f(λ) dλ ) , 1.3. Goals and Prerequisites 9 providing the desired limit. These arguments can be made exact, but it is hoped they make the point that the asymptotic eigenvalue distri- bution theorem for Hermitian Toeplitz matrices can be quite useful for evaluating limits of solutions to finite-dimensional problems. Further examples The Toeplitz distribution theorems have also found application in more complicated information theoretic evaluations, including the channel capacity of Gaussian channels [30, 29] and the rate-distortion functions of autoregressive sources [11]. The examples described here were chosen because they were in the author’s area of competence, but similar appli- cations crop up in a variety of areas. A GoogleTM search using the title of this document shows diverse applications of the eigenvalue distribu- tion theorem and related results, including such areas of coding, spec- tral estimation, watermarking, harmonic analysis, speech enhancement, interference cancellation, image restoration, sensor networks for detec- tion, adaptive filtering, graphical models, noise reduction, and blind equalization. 1.3 Goals and Prerequisites The primary goal of this work is to prove a special case of Szego’s asymptotic eigenvalue distribution theorem in Theorem 4.2. The as- sumptions used here are less general than Szego’s, but this permits more straightforward proofs which require far less mathematical back- ground. In addition to the fundamental theorems, several related re- sults that naturally follow but do not appear to be collected together anywhere are presented. We do not attempt to survey the fields of ap- plications of these results, as such a survey would be far beyond the author’s stamina and competence. A few applications are noted by way of examples. The essential prerequisites are a knowledge of matrix theory, an en- gineer’s knowledge of Fourier series and random processes, and calculus (Riemann integration). A first course in analysis would be helpful, but it is not assumed. Several of the occasional results required of analysis are 10 Introduction usually contained in one or more courses in the usual engineering cur- riculum, e.g., the Cauchy-Schwarz and triangle inequalities. Hopefully the only unfamiliar results are a corollary to the Courant-Fischer the- orem and the Weierstrass approximation theorem. The latter is an in- tuitive result which is easily believed even if not formally proved. More advanced results from Lebesgue integration, measure theory, functional analysis, and harmonic analysis are not used. Our approach is to relate the properties of Toeplitz matrices to those of their simpler, more structured special case — the circulant or cyclic matrix. These two matrices are shown to be asymptotically equivalent in a certain sense and this is shown to imply that eigenvalues, inverses, products, and determinants behave similarly. This approach provides a simplified and direct path to the basic eigenvalue distribution and related theorems. This method is implicit but not immediately appar- ent in the more complicated and more general results of Grenander in Chapter 7 of [16]. The basic results for the special case of a banded Toeplitz matrix appeared in [12], a tutorial treatment of the simplest case which was in turn based on the first draft of this work. The re- sults were subsequently generalized using essentially the same simple methods, but they remain less general than those of [16]. As an application several of the results are applied to study certain models of discrete time random processes. Two common linear models are studied and some intuitively satisfying results on covariance matri- ces and their factors are given. We sacrifice mathematical elegance and generality for conceptual simplicity in the hope that this will bring an understanding of the interesting and useful properties of Toeplitz matrices to a wider audi- ence, specifically to those who have lacked either the background or the patience to tackle the mathematical literature on the subject. 2 The Asymptotic Behavior of Matrices We begin with relevant definitions and a prerequisite theorem and pro- ceed to a discussion of the asymptotic eigenvalue, product, and inverse behavior of sequences of matrices. The major use of the theorems of this chapter is to relate the asymptotic behavior of a sequence of compli- cated matrices to that of a simpler asymptotically equivalent sequence of matrices. 2.1 Eigenvalues Any complex matrix A can be written as A = URU∗, (2.1) where the asterisk ∗ denotes conjugate transpose, U is unitary, i.e., U−1 = U∗, and R = rk,j is an upper triangular matrix ([18], p. 79). The eigenvalues of A are the principal diagonal elements of R. If A is normal, i.e., if A∗A = AA∗, then R is a diagonal matrix, which we denote as R = diag(αk; k = 0, 1, . . . , n − 1) or, more simply, R = diag(αk). If A is Hermitian, then it is also normal and its eigenvalues are real. A matrix A is nonnegative definite if x∗Ax ≥ 0 for all nonzero vec- 11 12 The Asymptotic Behavior of Matrices tors x. The matrix is positive definite if the inequality is strict for all nonzero vectors x. (Some books refer to these properties as positive definite and strictly positive definite, respectively.) If a Hermitian ma- trix is nonnegative definite, then its eigenvalues are all nonnegative. If the matrix is positive definite, then the eigenvalues are all (strictly) positive. The extreme values of the eigenvalues of a Hermitian matrix H can be characterized in terms of the Rayleigh quotient RH(x) of the matrix and a complex-valued vector x defined by RH(x) = (x∗Hx)/(x∗x). (2.2) As the result is both important and simple to prove, we state and prove it formally. The result will be useful in specifying the interval containing the eigenvalues of a Hermitian matrix. Usually in books on matrix theory it is proved as a corollary to the variational description of eigenvalues given by the Courant-Fischer theorem (see, e.g., [18], p. 116, for the case of real symmetric matrices), but the following result is easily demonstrated directly. Lemma 2.1. Given a Hermitian matrix H, let ηM and ηm be the maximum and minimum eigenvalues of H, respectively. Then ηm = minx RH(x) = minz:z∗z=1 z∗Hz (2.3) ηM = maxx RH(x) = maxz:z∗z=1 z∗Hz. (2.4) Proof. Suppose that em and eM are eigenvectors corresponding to the minimum and maximum eigenvalues ηm and ηM , respectively. Then RH(em) = ηm and RH(eM ) = ηM and therefore ηm ≥ minx RH(x) (2.5) ηM ≤ maxx RH(x). (2.6) Since H is Hermitian we can write H = UAU∗, where U is unitary and 2.1. Eigenvalues 13 A is the diagonal matrix of the eigenvalues ηk, and therefore x∗Hxx∗x =x∗UAU∗x x∗x =y∗Ayy∗y = ∑nk=1 |yk|2ηk∑n k=1 |yk|2, where y = U∗x and we have taken advantage of the fact that U is unitary so that x∗x = y∗y. But for all vectors y, this ratio is bound below by ηm and above by ηM and hence for all vectors x ηm ≤ RH(x) ≤ ηM (2.7) which with (2.5–2.6) completes the proof of the left-hand equalities of the lemma. The right-hand equalities are easily seen to hold since if x minimizes (maximizes) the Rayleigh quotient, then the normalized vec- tor x/x∗x satisfies the constraint of the minimization (maximization) to the right, hence the minimum (maximum) of the Rayleigh quotion must be bigger (smaller) than the constrained minimum (maximum) to the right. Conversely, if x achieves the rightmost optimization, then the same x yields a Rayleigh quotient of the the same optimum value. 2 The following lemma is useful when studying non-Hermitian ma- trices and products of Hermitian matrices. First note that if A is an arbitrary complex matrix, then the matrix A∗A is both Hermitian and nonnegative definite. It is Hermitian because (A∗A)∗ = A∗A and it is nonnegative definite since if for any complex vector x we define the complex vector y = Ax, then x∗(A∗A)x = y∗y = n∑ k=1 |yk|2 ≥ 0. Lemma 2.2. Let A be a matrix with eigenvalues αk. Define the eigen- values of the Hermitian nonnegative definite matrix A∗A to be λk ≥ 0. Thenn−1∑ k=0 λk ≥n−1∑ k=0 |αk|2, (2.8) with equality iff (if and only if) A is normal. 14 The Asymptotic Behavior of Matrices Proof. The trace of a matrix is the sum of the diagonal elements of a matrix. The trace is invariant to unitary operations so that it also is equal to the sum of the eigenvalues of a matrix, i.e., TrA∗A = n−1∑ k=0 (A∗A)k,k = n−1∑ k=0 λk. (2.9) From (2.1), A = URU∗ and hence TrA∗A = TrR∗R = n−1∑ k=0 n−1∑ j=0 |rj,k|2 = n−1∑ k=0 |αk|2 +∑ k 6=j |rj,k|2 ≥n−1∑ k=0 |αk|2 (2.10) Equation (2.10) will hold with equality iff R is diagonal and hence iff A is normal. 2 Lemma 2.2 is a direct consequence of Shur’s theorem ([18], pp. 229- 231) and is also proved in [16], p. 106. 2.2 Matrix Norms To study the asymptotic equivalence of matrices we require a metric on the space of linear space of matrices. A convenient metric for our purposes is a norm of the difference of two matrices. A norm N(A) on the space of n× n matrices satisfies the following properties: (1) N(A) ≥ 0 with equality if and only if A = 0, is the all zero matrix. (2) For any two matrices A and B, N(A+B) ≤ N(A) +N(B). (2.11) (3) For any scalar c and matrix A, N(cA) = |c|N(A). 2.2. Matrix Norms 15 The triangle inequality in (2.11) will be used often as is the following direct consequence: N(A−B) ≥ |N(A) −N(B)|. (2.12) Two norms — the operator or strong norm and the Hilbert-Schmidt or weak norm (also called the Frobenius norm or Euclidean norm when the scaling term is removed) — will be used here ([16], pp. 102–103). Let A be a matrix with eigenvalues αk and let λk ≥ 0 be the eigen- values of the Hermitian nonnegative definite matrix A∗A. The strong norm ‖ A ‖ is defined by ‖ A ‖= maxx RA∗A(x)1/2 = maxz:z∗z=1 [z∗A∗Az]1/2. (2.13) From Lemma 2.1 ‖ A ‖2= maxk λk∆= λM . (2.14) The strong norm of A can be bound below by letting eM be the normal- ized eigenvector of A corresponding to αM , the eigenvalue of A having largest absolute value: ‖ A ‖2= maxz:z∗z=1 z∗A∗Az ≥ (e∗MA∗)(AeM ) = |αM |2. (2.15) If A is itself Hermitian, then its eigenvalues αk are real and the eigen- values λk of A∗A are simply λk = α2k. This follows since if e(k) is an eigenvector of A with eigenvalue αk, then A∗Ae(k) = αkA∗e(k) = α2 ke(k). Thus, in particular, if A is Hermitian then ‖ A ‖= maxk |αk| = |αM |. (2.16) The weak norm (or Hilbert-Schmidt norm) of an n × n matrix A = [ak,j] is defined by |A| = 1 n n−1∑ k=0 n−1∑ j=0 |ak,j|2 1/2 = (1 nTr[A∗A])1/2 = ( 1 n n−1∑ k=0 λk )1/2 . (2.17) 16 The Asymptotic Behavior of Matrices The quantity√n|A| is sometimes called the Frobenius norm or Eu- clidean norm. From Lemma 2.2 we have |A|2 ≥ 1 n n−1∑ k=0 |αk|2,with equality iff A is normal. (2.18) The Hilbert-Schmidt norm is the “weaker” of the two norms since ‖ A ‖2= maxk λk ≥ 1 n n−1∑ k=0 λk = |A|2. (2.19) A matrix is said to be bounded if it is bounded in both norms. The weak norm is usually the most useful and easiest to handle of the two, but the strong norm provides a useful bound for the product of two matrices as shown in the next lemma. Lemma 2.3. Given two n × n matrices G = gk,j and H = hk,j,then |GH| ≤‖ G ‖ |H|. (2.20) Proof. Expanding terms yields |GH|2 =1 n i j |∑ k gi,khk,j|2 =1 n i j k m gi,kg∗i,mhk,jh ∗m,j =1 n j h∗jG∗Ghj , (2.21) where hj is the jth column of H. From (2.13), h∗jG∗Ghj h∗jhj≤‖ G ‖2 and therefore |GH|2 ≤ 1 n‖ G ‖2 j h∗jhj =‖ G ‖2 |H|2. 2 Lemma 2.3 is the matrix equivalent of (7.3a) of ([16], p. 103). Note that the lemma does not require that G or H be Hermitian. 2.3. Asymptotically Equivalent Sequences of Matrices 17 2.3 Asymptotically Equivalent Sequences of Matrices We will be considering sequences of n × n matrices that approximate each other as n becomes large. As might be expected, we will use the weak norm of the difference of two matrices as a measure of the “dis- tance” between them. Two sequences of n×n matrices An and Bnare said to be asymptotically equivalent if (1) An and Bn are uniformly bounded in strong (and hence in weak) norm: ‖ An ‖, ‖ Bn ‖≤ M <∞, n = 1, 2, . . . (2.22) and (2) An −Bn = Dn goes to zero in weak norm as n→ ∞: limn→∞ |An −Bn| = limn→∞ |Dn| = 0. Asymptotic equivalence of the sequences An and Bn will be ab- breviated An ∼ Bn. We can immediately prove several properties of asymptotic equiva- lence which are collected in the following theorem. Theorem 2.1. Let An and Bn be sequences of matrices with eigenvalues αn, i and βn, i, respectively. (1) If An ∼ Bn, then limn→∞ |An| = limn→∞ |Bn|. (2.23) (2) If An ∼ Bn and Bn ∼ Cn, then An ∼ Cn. (3) If An ∼ Bn and Cn ∼ Dn, then AnCn ∼ BnDn. (4) If An ∼ Bn and ‖ A−1n ‖, ‖ B−1 n ‖≤ K < ∞, all n, then A−1n ∼ B−1 n . (5) If AnBn ∼ Cn and ‖ A−1n ‖≤ K <∞, then Bn ∼ A−1 n Cn. (6) If An ∼ Bn, then there are finite constants m and M such that m ≤ αn,k, βn,k ≤M , n = 1, 2, . . . k = 0, 1, . . . , n− 1. (2.24) 18 The Asymptotic Behavior of Matrices Proof. (1) Eq. (2.23) follows directly from (2.12). (2) |An−Cn| = |An−Bn+Bn−Cn| ≤ |An−Bn|+|Bn−Cn| −→n→∞ 0 (3) Applying Lemma 2.3 yields |AnCn −BnDn| = |AnCn −AnDn +AnDn −BnDn| ≤ ‖ An ‖ |Cn −Dn|+ ‖ Dn ‖ |An −Bn|−→ n→∞ 0. (4) |A−1n −B−1 n | = |B−1n BnA −1n −B−1 n AnA−1n | ≤ ‖ B−1n ‖ · ‖ A−1 n ‖ ·|Bn −An|−→ n→∞ 0. (5) Bn −A−1n Cn = A−1 n AnBn −A−1n Cn ≤ ‖ A−1n ‖ |AnBn − Cn| −→n→∞ 0. (6) If An ∼ Bn then they are uniformly bounded in strong norm by some finite number M and hence from (2.15), |αn,k| ≤M and |βn,k| ≤ M and hence −M ≤ αn,k, βn,k ≤ M . So the result holds for m = −M and it may hold for larger m, e.g., m = 0 if the matrices are all nonnegative definite. 2 The above results will be useful in several of the later proofs. Asymp- totic equality of matrices will be shown to imply that eigenvalues, prod- ucts, and inverses behave similarly. The following lemma provides a prelude of the type of result obtainable for eigenvalues and will itself serve as the essential part of the more general results to follow. It shows that if the weak norm of the difference of the two matrices is small, then the sums of the eigenvalues of each must be close. 2.3. Asymptotically Equivalent Sequences of Matrices 19 Lemma 2.4. Given two matrices A and B with eigenvalues αk and βk, respectively, then | 1n n−1∑ k=0 αk − 1 n n−1∑ k=0 βk| ≤ |A−B|. Proof: Define the difference matrix D = A−B = dk,j so that n−1∑ k=0 αk −n−1∑ k=0 βk = Tr(A) − Tr(B) = Tr(D). Applying the Cauchy-Schwarz inequality (see, e.g., [22], p. 17) to Tr(D) yields |Tr(D)|2 = n−1∑ k=0 dk,k 2 ≤ n n−1∑ k=0 |dk,k|2 ≤ nn−1∑ k=0 n−1∑ j=0 |dk,j |2 = n2|D|2. (2.25) Taking the square root and dividing by n proves the lemma. 2 An immediate consequence of the lemma is the following corollary. Corollary 2.1. Given two sequences of asymptotically equivalent ma- trices An and Bn with eigenvalues αn,k and βn,k, respectively, then limn→∞ 1 n n−1∑ k=0 (αn,k − βn,k) = 0, (2.26) and hence if either limit exists individually, limn→∞ 1 n n−1∑ k=0 αn,k = limn→∞ 1 n n−1∑ k=0 βn,k. (2.27) Proof. Let Dn = dk,j = An −Bn. Eq. (2.27) is equivalent to limn→∞ 1 nTr(Dn) = 0. (2.28) 20 The Asymptotic Behavior of Matrices Dividing by n2, and taking the limit, results in 0 ≤ | 1n Tr(Dn)|2 ≤ |Dn|2 −→n→∞ 0 (2.29) from the lemma, which implies (2.28) and hence (2.27). 2 The previous corollary can be interpreted as saying the sample or arithmetic means of the eigenvalues of two matrices are asymptotically equal if the matrices are asymptotically equivalent. It is easy to see that if the matrices are Hermitian, a similar result holds for the means of the squared eigenvalues. From (2.12) and (2.18), |Dn| ≥ | |An| − |Bn| | = 1 n n−1∑ k=0 α2n,k − 1 n n−1∑ k=0 β2n,k −→n→∞ 0 if |Dn| −→n→∞ 0, yielding the following corollary. Corollary 2.2. Given two sequences of asymptotically equivalent Her- mitian matrices An and Bn with eigenvalues αn,k and βn,k,respectively, then limn→∞ 1 n n−1∑ k=0 (α2n,k − β2 n,k) = 0, (2.30) and hence if either limit exists individually, limn→∞ 1 n n−1∑ k=0 α2n,k = lim n→∞1 n n−1∑ k=0 β2n,k. (2.31) Both corollaries relate limiting sample (arithmetic) averages of eigenvalues or moments of an eigenvalue distribution rather than in- dividual eigenvalues. Equations (2.27) and (2.31) are special cases of the following fundamental theorem of asymptotic eigenvalue distribu- tion. 2.3. Asymptotically Equivalent Sequences of Matrices 21 Theorem 2.2. Let An and Bn be asymptotically equivalent se- quences of matrices with eigenvalues αn,k and βn,k, respectively. Then for any positive integer s the sequences of matrices Asn and Bsn are also asymptotically equivalent, limn→∞ 1 n n−1∑ k=0 (αsn,k − βs n,k) = 0, (2.32) and hence if either separate limit exists, limn→∞ 1 n n−1∑ k=0 αsn,k = lim n→∞1 n n−1∑ k=0 βsn,k. (2.33) Proof. Let An = Bn +Dn as in the proof of Corollary 2.1 and consider Asn − Bs n∆= ∆n. Since the eigenvalues of As n are αsn,k, (2.32) can be written in terms of ∆n as limn→∞ 1 nTr(∆n) = 0. (2.34) The matrix ∆n is a sum of several terms each being a product of Dn’s and Bn’s, but containing at least one Dn (to see this use the binomial theorem applied to matrices to expand Asn). Repeated application of Lemma 2.3 thus gives |∆n| ≤ K|Dn| −→n→∞ 0, (2.35) where K does not depend on n. Equation (2.35) allows us to apply Corollary 2.1 to the matrices Asn and Ds n to obtain (2.34) and hence (2.32). 2 Theorem 2.2 is the fundamental theorem concerning asymptotic eigenvalue behavior of asymptotically equivalent sequences of matri- ces. Most of the succeeding results on eigenvalues will be applications or specializations of (2.33). Since (2.33) holds for any positive integer s we can add sums corre- sponding to different values of s to each side of (2.33). This observation leads to the following corollary. 22 The Asymptotic Behavior of Matrices Corollary 2.3. Suppose that An and Bn are asymptotically equivalent sequences of matrices with eigenvalues αn,k and βn,k,respectively, and let f(x) be any polynomial. Then limn→∞ 1 n n−1∑ k=0 (f (αn,k) − f (βn,k)) = 0 (2.36) and hence if either limit exists separately, limn→∞ 1 n n−1∑ k=0 f (αn,k) = limn→∞ 1 n n−1∑ k=0 f (βn,k) . (2.37) Proof. Suppose that f(x) =∑m s=0 asxs. Then summing (2.32) over s yields (2.36). If either of the two limits exists, then (2.36) implies that both exist and that they are equal. 2 Corollary 2.3 can be used to show that (2.37) can hold for any ana- lytic function f(x) since such functions can be expanded into complex Taylor series, which can be viewed as polynomials with a possibly in- finite number of terms. Some effort is needed, however, to justify the interchange of limits, which can be accomplished if the Taylor series converges uniformly. If An and Bn are Hermitian, however, then a much stronger result is possible. In this case the eigenvalues of both matrices are real and we can invoke the Weierstrass approximation theorem ([6], p. 66) to immediately generalize Corollary 2.3. This theorem, our one real excursion into analysis, is stated below for reference. Theorem 2.3. (Weierstrass) If F (x) is a continuous complex function on [a, b], there exists a sequence of polynomials pn(x) such that limn→∞ pn(x) = F (x) uniformly on [a, b]. Stated simply, any continuous function defined on a real interval can be approximated arbitrarily closely and uniformly by a polynomial. Applying Theorem 2.3 to Corollary 2.3 immediately yields the following theorem: 2.3. Asymptotically Equivalent Sequences of Matrices 23 Theorem 2.4. Let An and Bn be asymptotically equivalent se- quences of Hermitian matrices with eigenvalues αn,k and βn,k, re- spectively. From Theorem 2.1 there exist finite numbers m and M such that m ≤ αn,k, βn,k ≤M , n = 1, 2, . . . k = 0, 1, . . . , n− 1. (2.38) Let F (x) be an arbitrary function continuous on [m,M ]. Then limn→∞ 1 n n−1∑ k=0 (F (αn,k) − F (βn,k)) = 0, (2.39) and hence if either of the limits exists separately, limn→∞ 1 n n−1∑ k=0 F (αn,k) = limn→∞ 1 n n−1∑ k=0 F (βn,k) (2.40) Theorem 2.4 is the matrix equivalent of Theorem 7.4a of [16]. When two real sequences αn,k; k = 0, 1, . . . , n−1 and βn,k; k = 0, 1, . . . , n−1 satisfy (2.38) and (2.39), they are said to be asymptotically equally distributed ([16], p. 62, where the definition is attributed to Weyl). As an example of the use of Theorem 2.4 we prove the following corollary on the determinants of asymptotically equivalent sequences of matrices. Corollary 2.4. Let An and Bn be asymptotically equivalent se- quences of Hermitian matrices with eigenvalues αn,k and βn,k, re- spectively, such that αn,k, βn,k ≥ m > 0. Then if either limit exists, limn→∞ (detAn)1/n = limn→∞ (detBn)1/n. (2.41) Proof. From Theorem 2.4 we have for F (x) = lnx limn→∞ 1 n n−1∑ k=0 lnαn,k = limn→∞ 1 n n−1∑ k=0 lnβn,k and hence limn→∞ exp [ 1 nln n−1∏ k=0 αn,k ] = limn→∞ exp [ 1 nln n−1∏ k=0 βn,k ] 24 The Asymptotic Behavior of Matrices or equivalently limn→∞ exp[1 nln detAn] = lim n→∞exp[ 1 nln detBn], from which (2.41) follows. 2 With suitable mathematical care the above corollary can be ex- tended to cases where αn,k, βn,k > 0 provided additional constraints are imposed on the matrices. For example, if the matrices are assumed to be Toeplitz matrices, then the result holds even if the eigenvalues can get arbitrarily small but remain strictly positive. (See the discussion on p. 66 and in Section 3.1 of [16] for the required technical conditions.) The difficulty with allowing the eigenvalues to approach 0 is that their logarithms are not bounded. Furthermore, the function lnx is not con- tinuous at x = 0, so Theorem 2.4 does not apply. Nonetheless, it is possible to say something about the asymptotic eigenvalue distribution in such cases and this issue is revisited in Theorem 5.2(d). In this section the concept of asymptotic equivalence of matrices was defined and its implications studied. The main consequences are the be- havior of inverses and products (Theorem 2.1) and eigenvalues (Theo- rems 2.2 and 2.4). These theorems do not concern individual entries in the matrices or individual eigenvalues, rather they describe an “aver- age” behavior. Thus saying A−1n ∼ B−1 n means that |A−1n −B−1 n | −→n→∞ 0 and says nothing about convergence of individual entries in the matrix. In certain cases stronger results on a type of elementwise convergence are possible using the stronger norm of Baxter [1, 2]. Baxter’s results are beyond the scope of this work. 2.4 Asymptotically Absolutely Equal Distributions It is possible to strengthen Theorem 2.4 and some of the interim re- sults used in its derivation using reasonably elementary methods. The key additional idea required is the Wielandt-Hoffman theorem [34], a result from matrix theory that is of independent interest. The theorem is stated and a proof following Wilkinson [35] is presented for com- pleteness. This section can be skipped by readers not interested in the stronger notion of equal eigenvalue distributions as it is not needed in the sequel. The bounds of Lemmas 2.5 and 2.5 are of interest in 2.4. Asymptotically Absolutely Equal Distributions 25 their own right and are included as they strengthen the the traditional bounds. Theorem 2.5. (Wielandt-Hoffman theorem) Given two Hermitian matrices A and B with eigenvalues αk and βk, respectively, then 1 n n−1∑ k=0 |αk − βk|2 ≤ |A−B|2. Proof: Since A and B are Hermitian, we can write them as A = Udiag(αk)U∗, B = Wdiag(βk)W ∗, where U and W are unitary. Since the weak norm is not effected by multiplication by a unitary matrix, |A−B| = |Udiag(αk)U∗ −Wdiag(βk)W ∗| = |diag(αk)U∗ − U∗Wdiag(βk)W ∗| = |diag(αk)U∗W − U∗Wdiag(βk)| = |diag(αk)Q−Qdiag(βk)|, where Q = U∗W = qi,j is also unitary. The (i, j) entry in the matrix diag(αk)Q−Qdiag(βk) is (αi − βj)qi,j and hence |A−B|2 =1 n n−1∑ i=0 n−1∑ j=0 |αi − βj |2|qi,j|2 ∆= n−1∑ i=0 n−1∑ j=0 |αi − βj |2pi,j (2.42) where we have defined pi,j = (1/n)|qi,j |2. Since Q is unitary, we also have thatn−1∑ i=0 |qi,j|2 =n−1∑ j=0 |qi,j|2 = 1 (2.43) orn−1∑ i=0 pi,j =n−1∑ j=0 pi,j =1 n. (2.44) This can be interpreted in probability terms: pi,j = (1/n)|qi,j |2 is a probability mass function or pmf on 0, 1, . . . , n − 12 with uniform marginal probability mass functions. Recall that it is assumed that the 26 The Asymptotic Behavior of Matrices eigenvalues are ordered so that α0 ≥ α1 ≥ α2 ≥ · · · and β0 ≥ β1 ≥β2 ≥ · · · . We claim that for all such matrices P satisfying (2.44), the right- hand side of (2.42) is minimized by P = (1/n)I, where I is the identity matrix, so that n−1∑ i=0 n−1∑ j=0 |αi − βj|2pi,j ≥n−1∑ i=0 |αi − βi|2, which will prove the result. To see this suppose the contrary. Let ℓ be the smallest integer in 0, 1, . . . , n − 1 such that P has a nonzero element off the diagonal in either row ℓ or in column ℓ. If there is a nonzero element in row ℓ off the diagonal, say pℓ,a then there must also be a nonzero element in column ℓ off the diagonal, say pb,ℓ in order for the constraints (2.44) to be satisfied. Since ℓ is the smallest such value, ℓ < a and ℓ < b. Let x be the smaller of pl,a and pb,l. Form a new matrix P ′ by adding x to pℓ,ℓ and pb,a and subtracting x from pb,ℓ and pℓ,a. The new matrix still satisfies the constraints and it has a zero in either position (b, ℓ) or (ℓ, a). Furthermore the norm of P ′ has changed from that of P by an amount x( (αℓ − βℓ)2 + (αb − βa) 2 − (αℓ − βa)2 − (αb − βℓ) 2) = −x(αℓ − αb)(βℓ − βa) ≤ 0 since ℓ > b, ℓ > a, the eigenvalues are nonincreasing, and x is posi- tive. Continuing in this fashion all nonzero offdiagonal elements can be zeroed out without increasing the norm, proving the result. 2 From the Cauchy-Schwarz inequality n−1∑ k=0 |αk − βk| ≤ n−1∑ k=0 (αk − βk)2 n−1∑ k=0 12 = √n n−1∑ k=0 (αk − βk)2, which with the Wielandt-Hoffman theorem yields the following strengthening of Lemma 2.4, 1 n n−1∑ k=0 |αk − βk| ≤ 1 n n−1∑ k=0 (αk − βk)2 ≤ |An −Bn|, 2.4. Asymptotically Absolutely Equal Distributions 27 which we formalize as the following lemma. Lemma 2.5. Given two Hermitian matrices A and B with eigenvalues αn and βn in nonincreasing order, respectively, then 1 n n−1∑ k=0 |αk − βk| ≤ |A−B|. Note in particular that the absolute values are outside the sum in Lemma 2.4 and inside the sum in Lemma 2.5. As was done in the weaker case, the result can be used to prove a stronger version of The- orem 2.4. This line of reasoning, using the Wielandt-Hoffman theorem, was pointed out by William F. Trench who used special cases in his paper [23]. Similar arguments have become standard for treating eigen- value distributions for Toeplitz and Hankel matrices. See, for example, [32, 9, 4]. The following theorem provides the derivation. The specific statement result and its proof follow from a private communication from William F. Trench. See also [31, 24, 25, 26, 27, 28]. Theorem 2.6. Let An and Bn be asymptotically equivalent sequences of Hermitian matrices with eigenvalues αn,k and βn,k in nonincreasing order, respectively. From Theorem 2.1 there exist finite numbers m and M such that m ≤ αn,k, βn,k ≤M , n = 1, 2, . . . k = 0, 1, . . . , n− 1. (2.45) Let F (x) be an arbitrary function continuous on [m,M ]. Then limn→∞ 1 n n−1∑ k=0 |F (αn,k) − F (βn,k)| = 0. (2.46) The theorem strengthens the result of Theorem 2.4 because of the magnitude inside the sum. Following Trench [24] in this case the eigenvalues are said to be asymptotically absolutely equally distributed. Proof: From Lemma 2.5 1 n k=0 |αn,k − βn,k| ≤ |An −Bn|, (2.47) 28 The Asymptotic Behavior of Matrices which implies (2.46) for the case F (r) = r. For any nonnegative integer j |αjn,k − βj n,k| ≤ jmax(|m|, |M |)j−1|αn,k − βn,k|. (2.48) By way of explanation consider a, b ∈ [m,M ]. Simple long division shows that aj − bj a− b= j∑ l=1 aj−lbl−1 so that |aj − bj a− b| = |aj − bj||a− b| = |j∑ l=1 aj−lbl−1| ≤j∑ l=1 |aj−lbl−1| = j∑ l=1 |a|j−l|b|l−1 ≤ jmax(|m|, |M |)j−1, which proves (2.48). This immediately implies that (2.46) holds for functions of the form F (r) = rj for positive integers j, which in turn means the result holds for any polynomial. If F is an arbitrary contin- uous function on [m,M ], then from Theorem 2.3 given ǫ > 0 there is a polynomial P such that |P (u) − F (u)| ≤ ǫ, u ∈ [m,M ]. 2.4. Asymptotically Absolutely Equal Distributions 29 Using the triangle inequality, 1 n n−1∑ k=0 |F (αn,k) − F (βn,k)| =1 n n−1∑ k=0 |F (αn,k) − P (αn,k) + P (αn,k) − P (βn,k) + P (βn,k) − F (βn,k)| ≤ 1 n n−1∑ k=0 |F (αn,k) − P (αn,k)| +1 n n−1∑ k=0 |P (αn,k) − P (βn,k)| +1 n n−1∑ k=0 |P (βn,k) − F (βn,k)| ≤ 2ǫ+1 n n−1∑ k=0 |P (αn,k) − P (βn,k)| As n → ∞ the remaining sum goes to 0, which proves the theorem since ǫ can be made arbitrarily small. 2 3 Circulant Matrices A circulant matrix C is a Toeplitz matrix having the form C = c0 c1 c2 · · · cn−1 cn−1 c0 c1 c2... cn−1 c0 c1. . . .... . . . . .. . . c2 c1c1 · · · cn−1 c0 , (3.1) where each row is a cyclic shift of the row above it. The structure can also be characterized by noting that the (k, j) entry of C, Ck,j, is given by Ck,j = c(j−k) mod n. The properties of circulant matrices are well known and easily derived ([18], p. 267,[8]). Since these matrices are used both to approximate and explain the behavior of Toeplitz matrices, it is instructive to present one version of the relevant derivations here. 31 32 Circulant Matrices 3.1 Eigenvalues and Eigenvectors The eigenvalues ψk and the eigenvectors y(k) of C are the solutions of Cy = ψ y (3.2) or, equivalently, of the n difference equations m−1∑ k=0 cn−m+kyk + n−1∑ k=m ck−myk = ψ ym; m = 0, 1, . . . , n − 1. (3.3) Changing the summation dummy variable results in n−1−m∑ k=0 ckyk+m + n−1∑ k=n−m ckyk−(n−m) = ψ ym; m = 0, 1, . . . , n− 1. (3.4) One can solve difference equations as one solves differential equations — by guessing an intuitive solution and then proving that it works. Since the equation is linear with constant coefficients a reasonable guess is yk = ρk (analogous to y(t) = esτ in linear time invariant differential equations). Substitution into (3.4) and cancellation of ρm yields n−1−m∑ k=0 ckρk + ρ−n n−1∑ k=n−m ckρk = ψ. Thus if we choose ρ−n = 1, i.e., ρ is one of the n distinct complex nth roots of unity, then we have an eigenvalue ψ = n−1∑ k=0 ckρk (3.5) with corresponding eigenvector y = n−1/2( 1, ρ, ρ2, . . . , ρn−1)′, (3.6) where the prime denotes transpose and the normalization is chosen to give the eigenvector unit energy. Choosing ρm as the complex nth root of unity, ρm = e−2πim/n, we have eigenvalue ψm = n−1∑ k=0 cke−2πimk/n (3.7) 3.1. Eigenvalues and Eigenvectors 33 and eigenvector y(m) =1√n ( 1, e−2πim/n, · · · , e−2πim(n−1)/n)′. Thus from the definition of eigenvalues and eigenvectors, Cy(m) = ψmy(m),m = 0, 1, . . . , n − 1. (3.8) Equation (3.7) should be familiar to those with standard engineering backgrounds as simply the discrete Fourier transform (DFT) of the sequence ck. Thus we can recover the sequence ck from the ψk by the Fourier inversion formula. In particular, 1 n n−1∑ m=0 ψme2πiℓm = 1 n n−1∑ m=0 n−1∑ k=0 ( cke−2πimk/n ) e2πiℓm/n =n−1∑ k=0 ck1 n n−1∑ m=0 e2πi(ℓ−k)m/n = cℓ, (3.9) where we have used the orthogonality of the complex exponentials: n−1∑ m=0 e2πimk/n = nδk mod n = n k mod n = 0 0 otherwise, (3.10) where δ is the Kronecker delta, δm = 1 m = 0 0 otherwise. Thus the eigenvalues of a circulant matrix comprise the DFT of the first row of the circulant matrix, and conversely first row of a circulant matrix is the inverse DFT of the eigenvalues. Eq. (3.8) can be written as a single matrix equation CU = UΨ, (3.11) where U = [y(0)|y(1)| · · · |y(n−1)] = n−1/2[e−2πimk/n ; m,k = 0, 1, . . . , n − 1] 34 Circulant Matrices is the matrix composed of the eigenvectors as columns, and Ψ = diag(ψk) is the diagonal matrix with diagonal elements ψ0, ψ1, . . . , ψn−1. Furthermore, (3.10) implies that U is unitary. By way of details, denote that the (k, j)th element of UU∗ by ak,j and observe that ak,j will be the product of the kth row of U , which is e−2πimk/n/√n; m = 0, 1, . . . , n−1, times the jth column of U∗, which is e2πimj/n/√n; m = 0, 1, . . . , n− 1 so that ak,j =1 n n−1∑ m=0 e2πim(j−k)/n = δ(k−j) mod n and hence UU∗ = I. Similarly, U∗U = I. Thus (3.11) implies that C = UΨU∗ (3.12) Ψ = U∗CU. (3.13) Since C is unitarily similar to a diagonal matrix it is normal. 3.2 Matrix Operations on Circulant Matrices The following theorem summarizes the properties derived in the previ- ous section regarding eigenvalues and eigenvectors of circulant matrices and provides some easy implications. Theorem 3.1. Every circulant matrix C has eigenvectors y(m) =1√n ( 1, e−2πim/n, · · · , e−2πim(n−1)/n)′ , m = 0, 1, . . . , n − 1, and corre- sponding eigenvalues ψm =n−1∑ k=0 cke−2πimk/n and can be expressed in the form C = UΨU∗, where U has the eigen- vectors as columns in order and Ψ is diag(ψk). In particular all circulant matrices share the same eigenvectors, the same matrix U works for all circulant matrices, and any matrix of the form C = UΨU∗ is circulant. Let C = ck−j and B = bk−j be circulant n × n matrices with eigenvalues ψm = n−1∑ k=0 cke−2πimk/n, βm = n−1∑ k=0 bke−2πimk/n, 3.2. Matrix Operations on Circulant Matrices 35 respectively. Then (1) C and B commute and CB = BC = UγU∗ , where γ = diag(ψmβm), and CB is also a circulant matrix. (2) C +B is a circulant matrix and C +B = UΩU∗, where Ω = (ψm + βm)δk−m(3) If ψm 6= 0; m = 0, 1, . . . , n − 1, then C is nonsingular and C−1 = UΨ−1U∗. Proof. We have C = UΨU∗ and B = UΦU∗ where Ψ = diag(ψm) and Φ = diag(βm). (1) CB = UΨU∗UΦU∗ = UΨΦU∗ = UΦΨU∗ = BC. Since ΨΦ is diagonal, the first part of the theorem implies that CB is circulant. (2) C +B = U(Ψ + Φ)U∗.(3) If Ψ is nonsingular, then CUΨ−1U∗ = UΨU∗UΨ−1U∗ = UΨΨ−1U∗ = UU∗ = I. 2 Circulant matrices are an especially tractable class of matrices since inverses, products, and sums are also circulant matrices and hence both straightforward to construct and normal. In addition the eigenvalues of such matrices can easily be found exactly and the same eigenvectors work for all circulant matrices. We shall see that suitably chosen sequences of circulant matrices asymptotically approximate sequences of Toeplitz matrices and hence results similar to those in Theorem 3.1 will hold asymptotically for sequences of Toeplitz matrices. 4 Toeplitz Matrices 4.1 Sequences of Toeplitz Matrices Given the simplicity of sums, products, eigenvalues,, inverses, and de- terminants of circulant matrices, an obvious approach to the study of asymptotic properties of sequences of Toeplitz matrices is to approxi- mate them by sequences asymptotically equivalent of circulant matrices and then applying the results developed thus far. Such results are most easily derived when strong assumptions are placed on the sequence of Toeplitz matrices which keep the structure of the matrices simple and allow them to be well approximated by a natural and simple sequence of related circulant matrices. Increasingly general results require corre- sponding increasingly complicated constructions and proofs. Consider the infinite sequence tk and define the corresponding sequence of n×n Toeplitz matrices Tn = [tk−j; k, j = 0, 1, . . . , n−1] as in (1.1). Toeplitz matrices can be classified by the restrictions placed on the sequence tk. The simplest class results if there is a finitem for which tk = 0, |k| > m, in which case Tn is said to be a banded Toeplitz matrix. A banded Toeplitz matrix has the appearance of the of (4.1), possessing a finite number of diagonals with nonzero entries and zeros everywhere 37 38 Toeplitz Matrices else, so that the nonzero entries lie within a “band” including the main diagonal: Tn = t0 t−1 · · · t−m t1 t0... 0 . . .. . . tm. . . tm · · · t1 t0 t−1 · · · t−m . . .. . . . . . t−m... 0 t0 t−1 tm · · · t1 t0 . (4.1) In the more general case where the tk are not assumed to be zero for large k, there are two common constraints placed on the infinite sequence tk; k = . . . ,−2,−1, 0, 1, 2, . . . which defines all of the ma- trices Tn in the sequence. The most general is to assume that the tkare square summable, i.e., that ∞∑ k=−∞|tk|2 <∞. (4.2) Unfortunately this case requires mathematical machinery beyond that assumed here; i.e., Lebesgue integration and a relatively advanced knowledge of Fourier series. We will make the stronger assumption that the tk are absolutely summable, i.e., that∞∑ k=−∞|tk| <∞. (4.3) Note that (4.3) is indeed a stronger constraint than (4.2) since ∞∑ k=−∞|tk|2 ≤ ∞∑ k=−∞|tk|2 . (4.4) 4.1. Sequences of Toeplitz Matrices 39 The assumption of absolute summability greatly simplifies the mathematics, but does not alter the fundamental concepts of Toeplitz and circulant matrices involved. As the main purpose here is tutorial and we wish chiefly to relay the flavor and an intuitive feel for the results, we will confine interest to the absolutely summable case. The main advantage of (4.3) over (4.2) is that it ensures the existence and of the Fourier series f(λ) defined by f(λ) = ∞∑ k=−∞tke ikλ = limn→∞ n∑ k=−n tkeikλ. (4.5) Not only does the limit in (4.5) converge if (4.3) holds, it converges uniformly for all λ, that is, we have that∣ f(λ) −n∑ k=−n tkeikλ = −n−1∑ k=−∞tke ikλ + ∞∑ k=n+1 tkeikλ ≤∣ −n−1∑ k=−∞tke ikλ + ∞∑ k=n+1 tkeikλ ≤−n−1∑ k=−∞|tk| + ∞∑ k=n+1 |tk| , where the right-hand side does not depend on λ and it goes to zero as n→ ∞ from (4.3). Thus given ǫ there is a single N , not depending on λ, such that∣ f(λ) −n∑ k=−n tkeikλ ≤ ǫ , all λ ∈ [0, 2π] , if n ≥ N. (4.6) Furthermore, if (4.3) holds, then f(λ) is Riemann integrable and the tkcan be recovered from f from the ordinary Fourier inversion formula: tk =1 ∫ 2π 0f(λ)e−ikλ dλ. (4.7) As a final useful property of this case, f(λ) is a continuous function of λ ∈ [0, 2π] except possibly at a countable number of points. 40 Toeplitz Matrices A sequence of Toeplitz matrices Tn = [tk−j] for which the tk are absolutely summable is said to be in the Wiener class,. Similarly, a function f(λ) defined on [0, 2π] is said to be in the Wiener class if it has a Fourier series with absolutely summable Fourier coefficients. It will often be of interest to begin with a function f in the Wiener class and then define the sequence of of n× n Toeplitz matrices Tn(f) = [ 1 ∫ 2π 0f(λ)e−i(k−j)λdλ ; k, j = 0, 1, · · · , n− 1 ] , (4.8) which will then also be in the Wiener class. The Toeplitz matrix Tn(f) will be Hermitian if and only if f is real. More specifically, Tn(f) = T ∗n(f) if and only if tk−j = t∗j−k for all k, j or, equivalently, t∗k = t−k all k. If t∗k = t−k, however, f∗(λ) = ∞∑ k=−∞t∗ke −ikλ = ∞∑ k=−∞t−ke −ikλ =∞∑ k=−∞tke ikλ = f(λ), so that f is real. Conversely, if f is real, then t∗k =1 ∫ 2π 0f∗(λ)eikλ dλ =1 ∫ 2π 0f(λ)eikλ dλ = t−k. It will be of interest to characterize the maximum and minimum magnitude of the eigenvalues of Toeplitz matrices and how these relate to the maximum and minimum values of the corresponding functions f . Problems arise, however, if the function f has a maximum or minimum at an isolated point. To avoid such difficulties we define the essential supremum Mf = ess supf of a real valued function f as the smallest number a for which f(x) ≤ a except on a set of total length or mea- sure 0. In particular, if f(x) > a only at isolated points x and not on any interval of nonzero length, then Mf ≤ a. Similarly, the essential infimum mf = ess inff is defined as the largest value of a for which 4.2. Bounds on Eigenvalues of Toeplitz Matrices 41 f(x) ≥ a except on a set of total length or measure 0. The key idea here is to view Mf and mf as the maximum and minimum values of f , where the extra verbiage is to avoid technical difficulties arising from the values of f on sets that do not effect the integrals. Functions f in the Wiener class are bounded since |f(λ)| ≤∞∑ k=−∞|tkeikλ| ≤ ∞∑ k=−∞|tk| (4.9) so that m|f |,M|f | ≤∞∑ k=−∞|tk|. (4.10) 4.2 Bounds on Eigenvalues of Toeplitz Matrices In this section Lemma 2.1 is used to obtain bounds on the eigenvalues of Hermitian Toeplitz matrices and an upper bound bound to the strong norm for general Toeplitz matrices. Lemma 4.1. Let τn,k be the eigenvalues of a Toeplitz matrix Tn(f). If Tn(f) is Hermitian, then mf ≤ τn,k ≤Mf . (4.11) Whether or not Tn(f) is Hermitian, ‖ Tn(f) ‖≤ 2M|f |, (4.12) so that the sequence of Toeplitz matrices Tn(f) is uniformly bounded over n if the essential supremum of |f | is finite. Proof. From Lemma 2.1, maxk τn,k = maxx (x∗Tn(f)x)/(x∗x) (4.13) mink τn,k = minx (x∗Tn(f)x)/(x∗x) 42 Toeplitz Matrices so that x∗Tn(f)x = n−1∑ k=0 n−1∑ j=0 tk−jxkx∗j = n−1∑ k=0 n−1∑ j=0 [ 12π ∫ 2π 0f(λ)ei(k−j)λ dλ ] xkx∗j = 12π ∫ 2π 0 n−1∑ k=0 xkeikλ 2 f(λ) dλ (4.14) and likewise x∗x =n−1∑ k=0 |xk|2 =1 ∫ 2π 0|n−1∑ k=0 xkeikλ|2 dλ. (4.15) Combining (4.14)–(4.15) results in mf ≤ ∫ 2π 0f(λ) n−1∑ k=0 xkeikλ 2 ∫ 2π 0 n−1∑ k=0 xkeikλ 2 =x∗Tn(f)x x∗x≤Mf , (4.16) which with (4.13) yields (4.11). We have already seen in (2.16) that if Tn(f) is Hermitian, then ‖ Tn(f) ‖= maxk |τn,k| ∆= |τn,M |. Since |τn,M | ≤ max(|Mf |, |mf |) ≤ M|f |, (4.12) holds for Hermitian matrices. Suppose that Tn(f) is not Hermitian or, equivalently, that f is not real. Any function f can be written in terms of its real and imaginary parts, f = fr+ifi, where both fr and fi are real. In particular, fr = (f + f∗)/2 and fi = (f − f∗)/2i.From the triangle inequality for norms, ‖ Tn(f) ‖ = ‖ Tn(fr + ifi) ‖ = ‖ Tn(fr) + iTn(fi) ‖ ≤ ‖ Tn(fr) ‖ + ‖ Tn(fi) ‖ ≤ M|fr| +M|fi|. 4.3. Banded Toeplitz Matrices 43 Since |(f±f∗)/2 ≤ (|f |+ |f∗|)/2 ≤M|f |, M|fr|+M|fi| ≤ 2M|f |, proving (4.12). 2 Note for later use that the weak norm of a Toeplitz matrix takes a particularly simple form. Let Tn(f) = tk−j, then by collecting equal terms we have |Tn(f)|2 =1 n n−1∑ k=0 n−1∑ j=0 |tk−j|2 =1 n n−1∑ k=−(n−1) (n− |k|)|tk|2 = n−1∑ k=−(n−1) (1 − |k|/n)|tk|2. (4.17) We are now ready to put all the pieces together to study the asymp- totic behavior of Tn(f). If we can find an asymptotically equivalent sequence of circulant matrices, then all of the results regarding cir- culant matrices and asymptotically equivalent sequences of matrices apply. The main difference between the derivations for simple sequence of banded Toeplitz matrices and the more general case is the sequence of circulant matrices chosen. Hence to gain some feel for the matrix chosen, we first consider the simpler banded case where the answer is obvious. The results are then generalized in a natural way. 4.3 Banded Toeplitz Matrices Let Tn be a sequence of banded Toeplitz matrices of order m+ 1, that is, ti = 0 unless |i| ≤ m. Since we are interested in the behavior or Tn for large n we choose n >> m. As is easily seen from (4.1), Tn looks like a circulant matrix except for the upper left and lower right-hand corners, i.e., each row is the row above shifted to the right one place. We can make a banded Toeplitz matrix exactly into a circulant if we fill in the upper right and lower left corners with the appropriate entries. 44 Toeplitz Matrices Define the circulant matrix Cn in just this way, i.e., Cn = t0 t−1 · · · t−m tm · · · t1 t1. . . ... tm... . . . tm 0 . . . tm · · · t1 t0 t−1 · · · t−m . . .. . . 0 t−m t−m... . . .... t0 t−1 t−1 · · · t−m tm · · · t1 t0 = c(n)0 · · · c (n)n−1 c(n)n−1 c (n)0 · · · .... . . ... c(n)1 · · · c (n)n−1 c (n)0 . (4.18) Equivalently, C, consists of cyclic shifts of (c(n)0 , · · · , c(n) n−1) where c(n)k = t−k k = 0, 1, · · · ,mtn−k k = n−m, · · · , n− 1 0 otherwise (4.19) If a Toeplitz matrix is specified by a function f and hence denoted by Tn(f), then the circulant matrix defined by (4.18–4.19) is similarly 4.3. Banded Toeplitz Matrices 45 denoted Cn(f). The function f will be explicitly shown when it is useful to do so, for example when the results being developed specifically involve f . The matrix Cn is intuitively a candidate for a simple matrix asymp- totically equivalent to Tn — we need only demonstrate that it is indeed both asymptotically equivalent and simple. Lemma 4.2. The matrices Tn and Cn defined in (4.1) and (4.18) are asymptotically equivalent, i.e., both are bounded in the strong norm and limn→∞ |Tn − Cn| = 0. (4.20) Proof. The tk are obviously absolutely summable, so Tn are uniformly bounded by 2M|f | from Lemma 4.1. The matrices Cn are also uni- formly bounded since C∗nCn is a circulant matrix with eigenvalues |f(2πk/n)|2 ≤ 4M2|f |. The weak norm of the difference is |Tn − Cn|2 = 1n m∑ k=0 k(|tk|2 + |t−k|2) ≤ m 1n m∑ k=0 (|tk|2 + |t−k|2) −→n→∞ 0 . 2 The above lemma is almost trivial since the matrix Tn − Cn has fewer than m2 non-zero entries and hence the 1/n in the weak norm drives |Tn − Cn| to zero. From Lemma 4.2 and Theorem 2.2 we have the following lemma. Lemma 4.3. Let Tn and Cn be as in (4.1) and (4.18) and let their eigenvalues be τn,k and ψn,k, respectively, then for any positive integer s limn→∞ 1 n n−1∑ k=0 ( τ sn,k − ψs n,k ) = 0. (4.21) In fact, for finite n,∣ 1 n n−1∑ k=0 ( τ sn,k − ψs n,k ) ≤ Kn−1/2, (4.22) 46 Toeplitz Matrices where K is not a function of n. Proof. Equation (4.21) is direct from Lemma 4.2 and Theorem 2.2. Equation (4.22) follows from Corollary 2.1 and Lemma 4.2. 2 The lemma implies that if either of the separate limits converges, then both will and limn→∞ 1 n n−1∑ k=0 τ sn,k = lim n→∞1 n n−1∑ k=0 ψsn,k. (4.23) The next lemma shows that the second limit indeed converges, and in fact provides an evaluation for the limit. Lemma 4.4. Let Cn(f) be constructed from Tn(f) as in (4.18) and let ψn,k be the eigenvalues of Cn(f), then for any positive integer s we have limn→∞ 1 n n−1∑ k=0 ψsn,k = 1 ∫ 2π 0f s(λ) dλ. (4.24) If Tn(f) is Hermitian, then for any function F (x) continuous on [mf ,Mf ] we have limn→∞ 1 n n−1∑ k=0 F (ψn,k) =1 ∫ 2π 0F (f(λ)) dλ. (4.25) Proof. From Theorem 3.1 we have exactly ψn,j =n−1∑ k=0 c(n)k e−2πijk/n = m∑ k=0 t−ke−2πijk/n + n−1∑ k=n−m tn−ke−2πijk/n =m∑ k=−m tke−2πijk/n = f( 2πj n) (4.26) Note that the eigenvalues of Cn(f) are simply the values of f(λ) with λ uniformly spaced between 0 and 2π. Defining 2πk/n = λk and 2π/n = 4.3. Banded Toeplitz Matrices 47 ∆λ we have limn→∞ 1 n n−1∑ k=0 ψsn,k = lim n→∞1 n n−1∑ k=0 f(2πk/n)s = limn→∞ n−1∑ k=0 f(λk)s∆λ/(2π) =1 ∫ 2π 0f(λ)sdλ, (4.27) where the continuity of f(λ) guarantees the existence of the limit of (4.27) as a Riemann integral. If Tn(f) and Cn(f) are Hermitian, than the ψn,k and f(λ) are real and application of the Weierstrass theorem to (4.27) yields (4.25). Lemma 4.2 and (4.26) ensure that ψn,k and τn,k are in the interval [mf ,Mf ]. 2 Combining Lemmas 4.2–4.4 and Theorem 2.2 we have the following special case of the fundamental eigenvalue distribution theorem. Theorem 4.1. If Tn(f) is a banded Toeplitz matrix with eigenvalues τn,k, then for any positive integer s limn→∞ 1 n n−1∑ k=0 τ sn,k = 1 ∫ 2π 0f(λ)s dλ. (4.28) Furthermore, if f is real, then for any function F (x) continuous on [mf ,Mf ] limn→∞ 1 n n−1∑ k=0 F (τn,k) =1 ∫ 2π 0F (f(λ)) dλ; (4.29) i.e., the sequences τn,k and f(2πk/n) are asymptotically equally distributed. This behavior should seem reasonable since the equations Tn(f)x = τx and Cn(f)x = ψx, n > 2m + 1, are essentially the same nth order difference equation with different boundary conditions. It is in fact the “nice” boundary conditions that make ψ easy to find exactly while exact solutions for τ are usually intractable. 48 Toeplitz Matrices With the eigenvalue problem in hand we could next write down the- orems on inverses and products of Toeplitz matrices using Lemma 4.2 and results for circulant matrices and asymptotically equivalent se- quences of matrices. Since these theorems are identical in statement and proof with the more general case of functions f in the Wiener class, we defer these theorems momentarily and generalize Theorem 4.1 to more general Toeplitz matrices with no assumption of bandedeness. 4.4 Wiener Class Toeplitz Matrices Next consider the case of f in the Wiener class, i.e., the case where the sequence tk is absolutely summable. As in the case of sequences of banded Toeplitz matrices, the basic approach is to find a sequence of circulant matrices Cn(f) that is asymptotically equivalent to the se- quence of Toeplitz matrices Tn(f). In the more general case under con- sideration, the construction of Cn(f) is necessarily more complicated. Obviously the choice of an appropriate sequence of circulant matrices to approximate a sequence of Toeplitz matrices is not unique, so we are free to choose a construction with the most desirable properties. It will, in fact, prove useful to consider two slightly different circulant approximations. Since f is assumed to be in the Wiener class, we have the Fourier series representation f(λ) =∞∑ k=−∞tke ikλ (4.30) tk =1 ∫ 2π 0f(λ)e−ikλ dλ. (4.31) Define Cn(f) to be the circulant matrix with top row (c(n)0 , c (n)1 , · · · , c(n) n−1) where c(n)k = 1 n n−1∑ j=0 f(2πj/n)e2πijk/n. (4.32) 4.4. Wiener Class Toeplitz Matrices 49 Since f(λ) is Riemann integrable, we have that for fixed k limn→∞ c(n)k = lim n→∞1n n−1∑ j=0 f(2πj/n)e2πijk/n = 12π ∫ 2π 0f(λ)eikλdλ = t−k (4.33) and hence the c(n)k are simply the sum approximations to the Riemann integrals giving t−k. Equations (4.32), (3.7), and (3.9) show that the eigenvalues ψn,m of Cn(f) are simply f(2πm/n); that is, from (3.7) and (3.9) ψn,m = n−1∑ k=0 c(n)k e−2πimk/n = n−1∑ k=0 1 n n−1∑ j=0 f(2πj/n)e2πijk/n e−2πimk/n = n−1∑ j=0 f(2πj/n) 1 n n−1∑ k=0 e2πik(j−m)/n = f(2πm/n). (4.34) Thus, Cn(f) has the useful property (4.26) of the circulant approxi- mation (4.19) used in the banded case. As a result, the conclusions of Lemma 4.4 hold for the more general case with Cn(f) constructed as in (4.32). Equation (4.34) in turn defines Cn(f) since, if we are told that Cn(f) is a circulant matrix with eigenvalues f(2πm/n),m = 0, 1, · · · , n − 1, then from (3.9) c(n)k = 1 n n−1∑ m=0 ψn,me2πimk/n =1 n n−1∑ m=0 f(2πm/n)e2πimk/n, (4.35) 50 Toeplitz Matrices as in (4.32). Thus, either (4.32) or (4.34) can be used to define Cn(f). The fact that Lemma 4.4 holds for Cn(f) yields several useful prop- erties as summarized by the following lemma. Lemma 4.5. Given a function f satisfying (4.30–4.31) and define the circulant matrix Cn(f) by (4.32). (1) Then c(n)k = ∞∑ m=−∞t−k+mn , k = 0, 1, · · · , n − 1. (4.36) (Note, the sum exists since the tk are absolutely summable.) (2) If f(λ) is real and mf = ess inf f > 0, then Cn(f)−1 = Cn(1/f). (3) Given two functions f(λ) and g(λ), then Cn(f)Cn(g) = Cn(fg). Proof. (1) Applying (4.31) to λ = 2πj/n gives f(2πj n) = ∞∑ ℓ=−∞tℓe iℓ2πj/n which when inserted in (4.32) yields c(n)k = 1 n n−1∑ j=0 f(2πj n)e2πijk/n =1 n n−1∑ j=0 ( ∞∑ ℓ=−∞tℓe iℓ2πj/n ) e2πijk/n (4.37) = ∞∑ ℓ=−∞tℓ 1 n n−1∑ j=0 ei2π(k+ℓ)j/n = ∞∑ ℓ=−∞tℓδ(k+ℓ) mod n, where the final step uses (3.10). The term δ(k+ℓ) mod n will be 1 whenever ℓ = −k plus a multiple mn of n, which yields (4.36). 4.4. Wiener Class Toeplitz Matrices 51 (2) Since Cn(f) has eigenvalues f(2πk/n) > 0, by Theorem 3.1 Cn(f)−1 has eigenvalues 1/f(2πk/n), and hence from (4.35) and the fact that Cn(f)−1 is circulant we have Cn(f)−1 = Cn(1/f). (3) Follows immediately from Theorem 3.1 and the fact that, if f(λ) and g(λ) are Riemann integrable, so is f(λ)g(λ). 2 Equation (4.36) points out a shortcoming of Cn(f) for applications as a circulant approximation to Tn(f) — it depends on the entire se- quence tk; k = 0,±1,±2, · · · and not just on the finite collection of elements tk; k = 0,±1, · · · ,±(n − 1) of Tn(f). This can cause prob- lems in practical situations where we wish a circulant approximation to a Toeplitz matrix Tn when we only know Tn and not f . Pearl [19] discusses several coding and filtering applications where this restriction is necessary for practical reasons. A natural such approximation is to form the truncated Fourier series fn(λ) = n−1∑ m=−(n−1) tmeimλ, (4.38) which depends only on tm;m = 0,±1, · · · ,±n − 1, and then define the circulant matrix Cn(fn); that is, the circulant matrix having as top row (c(n)0 , · · · , c(n) n−1) where analogous to the derivation of (4.37) c(n)k = 1 n n−1∑ j=0 fn(2πj n)e2πijk/n =1 n n−1∑ j=0 n−1∑ ℓ=−(n−1) tℓeiℓ2πj/n e2πijk/n = n−1∑ ℓ=−(n−1) tℓ1 n n−1∑ j=0 ei2π(k+ℓ)j/n = n−1∑ ℓ=−(n−1) tℓδ(k+ℓ) mod n. 52 Toeplitz Matrices Now, however, we are only interested in values of ℓ which have the form −k plus a multiple mn of n for which −(n − 1) ≤ −k +mn ≤ n − 1. This will always include the m = 0 term for which ℓ = −k. If k = 0, then only the m = 0 term lies within the range. If k = 1, 2, . . . , n − 1, then m = −1 results in −k+n which is between 1 and n− 1. No other multiples lie within the range, so we end up with c(n)k = t0 k = 0 t−k + tn−k k = 1, 2, . . . , n− 1. (4.39) Since Cn(fn) is also a Toeplitz matrix, define Cn(fn) = T ′n = t′k−j with t′k = c(n)−k = tk + tn+k k = −(n− 1), . . . ,−1 c(n)0 = t0 k = 0 c(n)n−k = t−(n−k) + tk k = 1, 2, . . . , n − 1 , (4.40) which can be pictured as T ′n = t0 t−1 + tn−1 t−2 + tn−2 · · · t−(n−1) + t1t1 + t−(n−1) t0 t−1 + tn−1 t2 + t−(n−2) t1 + t−(n−1) t0... .... . . tn−1 + t1 · · · t0 (4.41) Like the original approximation Cn(f), the approximation Cn(fn) reduces to the Cn(f) of (4.19) for a banded Toeplitz matrix of order m if n > 2m+1. The following lemma shows that these circulant matrices are asymptotically equivalent to each other and to Tm. Lemma 4.6. Let Tn(f) = tk−j where ∞∑ k=−∞|tk| <∞, 4.4. Wiener Class Toeplitz Matrices 53 and f(λ) = ∞∑ k=−∞tke ikλ, fn(λ) = n−1∑ k=−(n−1) tkeikλ. Define the circulant matrices Cn(f) and Cn(fn) as in (4.32) and (4.38)– (4.39). Then, Cn(f) ∼ Cn(fn) ∼ Tn. (4.42) Proof. Since both Cn(f) and Cn(fn) are circulant matrices with the same eigenvectors (Theorem 3.1), we have from part 2 of Theorem 3.1 and (2.17) that |Cn(f) − Cn(fn)|2 =1 n n−1∑ k=0 |f(2πk/n) − fn(2πk/n)|2. Recall from (4.6) and the related discussion that fn(λ) uniformly con- verges to f(λ), and hence given ǫ > 0 there is an N such that for n ≥ N we have for all k, n that |f(2πk/n) − fn(2πk/n)|2 ≤ ǫ and hence for n ≥ N |Cn(f) − Cn(fn)|2 ≤ 1 n n−1∑ i=0 ǫ = ǫ. Since ǫ is arbitrary, limn→∞ |Cn(f) − Cn(fn)| = 0 proving that Cn(f) ∼ Cn(fn). (4.43) 54 Toeplitz Matrices Application of (4.40) and (4.17) results in |Tn(f) − Cn(fn)|2 = n−1∑ k=−(n−1) (1 − |k|/n)|tk − t′k|2 =−1∑ k=−(n−1) n+ k n|tn+k|2 + n−1∑ k=1 n− k n|t−(n−k)|2 =−1∑ k=−(n−1) k n|tk|2 + n−1∑ k=1 k n|t−k|2 = n−1∑ k=1 k n ( |tk|2 + |t−k|2) (4.44) Since the tk are absolutely summable, they are also square summable from (4.4) and hence given ǫ > 0 we can choose an N large enough so that∞∑ k=N |tk|2 + |t−k|2 ≤ ǫ. Therefore limn→∞ |Tn(f) − Cn(fn)| = limn→∞ n−1∑ k=0 (k/n)(|tk|2 + |t−k|2) = limn→∞ N−1∑ k=0 (k/n)(|tk|2 + |t−k|2) + n−1∑ k=N (k/n)(|tk|2 + |t−k|2) ≤ limn→∞ 1 n ( N−1∑ k=0 k(|tk|2 + |t−k|2)) + ∞∑ k=N (|tk|2 + |t−k|2) ≤ ǫ Since ǫ is arbitrary, limn→∞ |Tn(f) − Cn(fn)| = 0 4.4. Wiener Class Toeplitz Matrices 55 and hence Tn(f) ∼ Cn(fn), (4.45) which with (4.43) and Theorem 2.1 proves (4.42). 2 Pearl [19] develops a circulant matrix similar to Cn(fn) (depending only on the entries of Tn(f)) such that (4.45) holds in the more general case where (4.2) instead of (4.3) holds. We now have a sequence of circulant matrices Cn(f) asymptoti- cally equivalent to the sequence Tn(f) and the eigenvalues, inverses and products of the circulant matrices are known exactly. Therefore Lemmas 4.2–4.4 and Theorems 2.2–2.2 can be applied to generalize Theorem 4.1. Theorem 4.2. Let Tn(f) be a sequence of Toeplitz matrices such that f(λ) is in the Wiener class or, equivalently, that tk is absolutely summable. Let τn,k be the eigenvalues of Tn(f) and s be any positive integer. Then limn→∞ 1 n n−1∑ k=0 τ sn,k = 1 ∫ 2π 0f(λ)s dλ. (4.46) Furthermore, if f(λ) is real or, equivalently, the matrices Tn(f) are all Hermitian, then for any function F (x) continuous on [mf ,Mf ] limn→∞ 1 n n−1∑ k=0 F (τn,k) =1 ∫ 2π 0F (f(λ)) dλ. (4.47) Theorem 4.2 is the fundamental eigenvalue distribution theorem of Szego (see [16]). The approach used here is essentially a specialization of Grenander and Szego ([16], ch. 7). Theorem 4.2 yields the following two corollaries. Corollary 4.1. Given the assumptions of the theorem, define the eigenvalue distribution function Dn(x) = (number of τn,k ≤ x)/n. As- sume that∫ λ:f(λ)=xdλ = 0. 56 Toeplitz Matrices Then the limiting distribution D(x) = limn→∞Dn(x) exists and is given by D(x) =1 f(λ)≤xdλ. The technical condition of a zero integral over the region of the set of λ for which f(λ) = x is needed to ensure that x is a point of continuity of the limiting distribution. It can be interpreted as not allowing f(λ) to have a flat region around the point x. The limiting distribution function evaluated at x describes the fraction of the eigenvalues that smaller than x in the limit as n → ∞, which in turn implies that the fraction of eigenvalues between two values a and b > a is D(b)−D(a). This is similar to the role of a cumulative distribution function (cdf) in probability theory. Proof. Define the indicator function 1x(α) = 1 mf ≤ α ≤ x 0 otherwise We have D(x) = limn→∞ 1 n n−1∑ k=0 1x(τn,k). Unfortunately, 1x(α) is not a continuous function and hence Theo- rem 4.2 cannot be immediately applied. To get around this problem we mimic Grenander and Szego p. 115 and define two continuous functions that provide upper and lower bounds to 1x and will converge to it in the limit. Define 1+x (α) = 1 α ≤ x 1 − α−xǫ x < α ≤ x+ ǫ 0 x+ ǫ < α 1−x (α) = 1 α ≤ x− ǫ 1 − α−x+ǫǫ x− ǫ < α ≤ x 0 x < α 4.4. Wiener Class Toeplitz Matrices 57 The idea here is that the upper bound has an output of 1 everywhere 1x does, but then it drops in a continuous linear fashion to zero at x+ǫ instead of immediately at x. The lower bound has a 0 everywhere 1x does and it rises linearly from x to x − ǫ to the value of 1 instead of instantaneously as does 1x. Clearly 1−x (α) < 1x(α) < 1+x (α) for all α. Since both 1+x and 1−x are continuous, Theorem 4.2 can be used to conclude that limn→∞ 1 n n−1∑ k=0 1+x (τn,k) =1 1+x (f(λ)) dλ =1 f(λ)≤xdλ+ 1 x<f(λ)≤x+ǫ(1 − f(λ) − x ǫ) dλ ≤ 1 f(λ)≤xdλ+ 1 x<f(λ)≤x+ǫdλ and limn→∞ 1 n n−1∑ k=0 1−x (τn,k) =1 1−x (f(λ)) dλ =1 f(λ)≤x−ǫdλ+ 1 x−ǫ<f(λ)≤x(1 − f(λ) − (x− ǫ) ǫ) dλ =1 f(λ)≤x−ǫdλ+ 1 x−ǫ<f(λ)≤x(x− f(λ)) dλ ≥ 1 f(λ)≤x−ǫdλ =1 f(λ)≤xdλ− 1 x−ǫ<f(λ)≤xdλ These inequalities imply that for any ǫ > 0, as n grows the sample 58 Toeplitz Matrices average (1/n)∑n−1 k=0 1x(τn,k) will be sandwiched between 1 f(λ)≤xdλ+ 1 x<f(λ)≤x+ǫdλ and1 f(λ)≤xdλ− 1 x−ǫ<f(λ)≤xdλ. Since ǫ can be made arbitrarily small, this means the sum will be sandwiched between1 f(λ)≤xdλ and1 f(λ)≤xdλ− 1 f(λ)=xdλ. Thus if∫ f(λ)=xdλ = 0, then D(x) = 12π ∫ 2π 01x[f(λ)]dλ = 12πv f(λ)≤xdλ . 2 Corollary 4.2. Assume that the conditions of Theorem 4.2 hold and let mf and Mf denote the essential infimum and the essential supre- mum of f , respectively. Then limn→∞ maxk τn,k = Mf limn→∞ minkτn,k = mf . Proof. From Corollary 4.1 we have for any ǫ > 0 D(mf + ǫ) = f(λ)≤mf +ǫdλ > 0. 4.4. Wiener Class Toeplitz Matrices 59 The strict inequality follows from the continuity of f(λ). Since limn→∞ 1 nnumber of τn,k in [mf ,mf + ǫ] > 0 there must be eigenvalues in the interval [mf ,mf + ǫ] for arbitrarily small ǫ. Since τn,k ≥ mf by Lemma 4.1, the minimum result is proved. The maximum result is proved similarly. 2 5 Matrix Operations on Toeplitz Matrices Applications of Toeplitz matrices like those of matrices in general in- volve matrix operations such as addition, inversion, products and the computation of eigenvalues, eigenvectors, and determinants. The prop- erties of Toeplitz matrices particular to these operations are based pri- marily on three fundamental results that have been described earlier: (1) matrix operations are simple when dealing with circulant ma- trices, (2) given a sequence of Toeplitz matrices, we can instruct asymp- totically equivalent sequences of circulant matrices, and (3) asymptotically equivalent sequences of matrices have equal asymptotic eigenvalue distributions and other related prop- erties. In the next few sections some of these operations are explored in more depth for sequences of Toeplitz matrices. Generalizations and related results can be found in Tyrtyshnikov [31]. 61 62 Matrix Operations on Toeplitz Matrices 5.1 Inverses of Toeplitz Matrices In some applications we wish to study the asymptotic distribution of a function F (τn,k) of the eigenvalues that is not continuous at the mini- mum or maximum value of f . For example, in order for the results de- rived thus far to apply to the function F (f(λ)) = 1/f(λ) which arises when treating inverses of Toeplitz matrices, it has so far been neces- sary to require that the essential infimum mf > 0 because the function F (1/x) is not continuous at x = 0. If mf = 0, the basic asymptotic eigenvalue distribution Theorem 4.2 breaks down and the limits and the integrals involved might not exist. The limits might exist and equal something else, or they might simply fail to exist. In order to treat the inverses of Toeplitz matrices when f has zeros, we state without proof an intuitive extension of the fundamental Toeplitz result that shows how to find asymptotic distributions of suitably truncated functions. To state the result, define the mid function mid(x, y, z)∆= z y ≥ z y x ≤ y ≤ z x y ≤ z (5.1) x < z. This function can be thought of as having input y and thresholds z and X and it puts out y if y is between z and x, z if y is smaller than z, and x if y is greater than x. The following result was proved in [13] and extended in [25]. See also [26, 27, 28]. Theorem 5.1. Suppose that f is in the Wiener class. Then for any function F (x) continuous on [ψ, θ] ⊂ [mf ,Mf ] limn→∞ 1 n n−1∑ k=0 F (mid(ψ, τn,k, θ) =1 ∫ 2π 0F (mid(ψ, f(λ), θ) dλ. (5.2) Unlike Theorem 4.2 we pick arbitrary points ψ and θ such that F is continuous on the closed interval [ψ, θ]. These need not be the minimum and maximum of f . Theorem 5.2. Assume that f is in the Wiener class and is real and that f(λ) ≥ 0 with equality holding at most at a countable number of points. Then (a) Tn(f) is nonsingular 5.1. Inverses of Toeplitz Matrices 63 (b) If f(λ) ≥ mf > 0, then Tn(f)−1 ∼ Cn(f)−1, (5.3) where Cn(f) is defined in (4.35). Furthermore, if we define Tn(f) −Cn(f) = Dn then Tn(f)−1 has the expansion Tn(f)−1 = [Cn(f) +Dn]−1 = Cn(f)−1[ I +DnCn(f)−1]−1 = Cn(f)−1[ I +DnCn(f)−1 +( DnCn(f)−1)2 + · · ·] , (5.4) and the expansion converges (in weak norm) for sufficiently large n. (c) If f(λ) ≥ mf > 0, then Tn(f)−1 ∼ Tn(1/f) = [ 1 ∫ π −π ei(k−j)λ f(λ)dλ ] ; (5.5) that is, if the spectrum is strictly positive, then the inverse of a sequence of Toeplitz matrices is asymptotically Toeplitz. Furthermore if ρn,k are the eigenvalues of Tn(f)−1 and F (x) is any continuous function on [1/Mf , 1/mf ], then limn→∞ 1 n n−1∑ k=0 F (ρn,k) =1 ∫ π −πF ((1/f(λ)) dλ. (5.6) (d) Suppose that mf = 0 and that the derivative of f(λ) exists and is bounded for all λ. Then Tn(f)−1 is not bounded, 1/f(λ) is not inte- grable and hence Tn(1/f) is not defined and the integrals of (5.2) may not exist. For any finite θ, however, the following similar fact is true: If F (x) is a continuous function on [1/Mf , θ], then limn→∞ 1 n n−1∑ k=0 F (min(ρn,k, θ)) =1 ∫ 2π 0F (min(1/f(λ), θ)) dλ. (5.7) 64 Matrix Operations on Toeplitz Matrices Proof. (a) Since f(λ) > 0 except at possibly countably many points, we have from (4.14) x∗Tn(f)x =1 ∫ π −π n−1∑ k=0 xkeikλ 2 f(λ)dλ > 0. Thus for all n minkτn,k > 0 and hence detTn(f) = n−1∏ k=0 τn,k 6= 0 so that Tn(f) is nonsingular. (b) From Lemma 4.6, Tn ∼ Cn and hence (5.1) follows from Theo- rem 2.1 since f(λ) ≥ mf > 0 ensures that ‖ Tn(f)−1 ‖, ‖ Cn(f)−1 ‖≤ 1/mf <∞. The series of (5.4) will converge in weak norm if |DnCn(f)−1| < 1. (5.8) Since |DnCn(f)−1| ≤‖ Cn(f)−1 ‖ |Dn| ≤ (1/mf )|Dn| −→n→∞ 0, Eq. (5.8) must hold for large enough n. (c) We have from the triangle inequality that |Tn(f)−1 − Tn(1/f)| ≤ |Tn(f)−1 − Cn(f)−1| + |Cn(f)−1 − Tn(1/f)|. From (b) for any ǫ > 0 we can choose an n large enough so that |Tn(f)−1 − Cn(f)−1| ≤ ǫ 2. (5.9) From Theorem 3.1 and Lemma 4.5, Cn(f)−1 = Cn(1/f) and from Lemma 4.6 Cn(1/f) ∼ Tn(1/f). Thus again we can choose n large enough to ensure that |Cn(f)−1 − Tn(1/f)| ≤ ǫ/2 (5.10) 5.1. Inverses of Toeplitz Matrices 65 so that for any ǫ > 0 from (5.7)–(5.8) can choose n such that |Tn(f)−1 − Tn(1/f)| ≤ ǫ, which implies (5.5). Equation (5.6) follows from (5.5) and Theorem 2.4. Alternatively, if G(x) is any continuous function on [1/Mf , 1/mf ] and (5.4) follows directly from Lemma 4.6 and Theorem 2.3 applied to G(1/x). (d) When f(λ) has zeros (mf = 0), then from Corollary 4.2 limn→∞ minkτn,k = 0 and hence ‖ T−1n ‖= max kρn,k = 1/min kτn,k (5.11) is unbounded as n → ∞. To prove that 1/f(λ) is not integrable and hence that Tn(1/f) does not exist, consider the disjoint sets Ek = λ : 1/k ≥ f(λ)/Mf > 1/(k + 1) = λ : k ≤Mf/f(λ) < k + 1 (5.12) and let |Ek| denote the length of the set Ek, that is, |Ek| = λ:Mf /k≥f(λ)>Mf /(k+1)dλ. From (5.12) ∫ π −π 1 f(λ)dλ = ∞∑ k=1 Ek 1 f(λ)dλ ≥∞∑ k=1 |Ek|kMf . (5.13) For a given k, Ek will comprise a union of disjoint intervals of the form (a, b) where for all λ ∈ (a, b) we have that 1/k ≥ f(λ)/Mf > 1/(k+ 1). There must be at least one such nonempty interval, so |Ek| will be bound below by the length of this interval, b− a. Then for any x, y ∈(a, b) |f(y) − f(x)| = |∫ y x df dλdλ| ≤ η|y − x|. 66 Matrix Operations on Toeplitz Matrices By assumption there is some finite value η such that∣ df ≤ η, (5.14) so that |f(y) − f(x)| =≤ η|y − x|.Pick x and y so that f(x) = Mf/(k + 1) and f(y) = Mf/k (since f is continuous at almost all points, this argument works almost everywhere – it needs more work if these end points are not points of continuity of f), then b− a ≥ |y − x| ≥Mf (1 k− 1 k + 1) = Mf k + 1. Combining this with (5.13) yields ∫ π −πdλ/f(λ) ≥ ∞∑ k=1 (k/Mf )(Mf k(k + 1))/η (5.15) = ∞∑ k=1 1 k + 1, (5.16) which diverges so that 1/f(λ) is not integrable. To prove (5.5) let F (x) be continuous on [1/Mf , θ], then F (min(1/x, θ)) is continuous on [0,Mf ] and hence Theorem 2.4 yields (5.5). Note that (5.5) im- plies that the eigenvalues of Tn(f)−1 are asymptotically equally dis- tributed up to any finite θ as the eigenvalues of the sequence of matrices Tn[min(1/f, θ)]. 2 A special case of (d) is when Tn(f) is banded and f(λ) has at least one zero. Then the derivative exists and is bounded since df/dλ = m∑ k=−m iktkeikλ ≤m∑ k=−m |k||tk| <∞ . The series expansion of (b) is due to Rino [20]. The proof of (d) is motivated by one of Widom [33]. Further results along the lines of (d) 5.2. Products of Toeplitz Matrices 67 regarding unbounded Toeplitz matrices may be found in [13]. Related results considering asymptotically equal distributions of unbounded se- quences can be found in Tyrtyshnikov [32] and Trench [25]. These works extend Weyl’s definition of asymptotically equal distributions to un- bounded sequences using the mid function used here to treat inverses. This leads to conditions for equal distributions and their implications. Extending (a) to the case of non-Hermitian matrices can be some- what difficult, i.e., finding conditions on f(λ) to ensure that Tn(f) is invertible. Parts (a)-(d) can be straightforwardly extended if f(λ) is continuous. For a more general discussion of inverses the interested reader is referred to Widom [33] and the cited references. The results of Baxter [1] can also be applied to consider the asymptotic behavior of inverses in quite general cases. 5.2 Products of Toeplitz Matrices We next combine Theorem 2.1 and Lemma 4.6 to obtain the asymptotic behavior of products of Toeplitz matrices. The case of only two matrices is considered first since it is simpler. A key point is that while the product of Toeplitz matrices is not Toeplitz, a sequence of products of Toeplitz matrices Tn(f)Tn(g) is asymptotically equivalent to a sequence of Toeplitz matrices Tn(fg). Theorem 5.3. Let Tn(f) and Tn(g) be defined as in (4.8) where f(λ) and g(λ) are two functions in the Wiener class. Define Cn(f) and Cn(g) as in (4.35) and let ρn,k be the eigenvalues of Tn(f)Tn(g) (a) Tn(f)Tn(g) ∼ Cn(f)Cn(g) = Cn(fg). (5.17) Tn(f)Tn(g) ∼ Tn(g)Tn(f). (5.18) limn→∞ n−1n−1∑ k=0 ρsn,k = 1 ∫ 2π 0[f(λ)g(λ)]s dλ s = 1, 2, . . . . (5.19) (b) If Tn(f) and Tn(g) are Hermitian, then for any F (x) continuous on 68 Matrix Operations on Toeplitz Matrices [mfmg,MfMg] limn→∞ n−1n−1∑ k=0 F (ρn,k) =1 ∫ 2π 0F (f(λ)g(λ)) dλ. (5.20) (c) Tn(f)Tn(g) ∼ Tn(fg). (5.21) (d) Let f1(λ), ., fm(λ) be in the Wiener class. Then if the Cn(fi) are defined as in (4.35) m∏ i=1 Tn(fi) ∼ Cn ( m∏ i=1 fi ) ∼ Tn ( m∏ i=1 fi ) . (5.22) (e) If ρn,k are the eigenvalues of m∏ i=1 Tn(fi), then for any positive integer s limn→∞ n−1n−1∑ k=0 ρsn,k = 1 ∫ 2π 0 ( m∏ i=1 fi(λ) )s dλ (5.23) If the Tn(fi) are Hermitian, then the ρn,k are asymptotically real, i.e., the imaginary part converges to a distribution at zero, so that limn→∞ 1 n n−1∑ k=0 (Re[ρn,k])s = 1 ∫ 2π 0 ( m∏ i=1 fi(λ) )s dλ. (5.24) limn→∞ 1 n n−1∑ k=0 (ℑ[ρn,k])2 = 0. (5.25) Proof. (a) Equation (5.14) follows from Lemmas 4.5 and 4.6 and The- orems 2.1 and 2.3. Equation (5.16) follows from (5.14). Note that while Toeplitz matrices do not in general commute, asymptotically they do. Equation (5.17) follows from (5.14), Theorem 2.2, and Lemma 4.4. (b) Proof follows from (5.14) and Theorem 2.4. Note that the eigen- values of the product of two Hermitian matrices are real ([18], p. 105). 5.2. Products of Toeplitz Matrices 69 (c) Applying Lemmas 4.5 and 4.6 and Theorem 2.1 |Tn(f)Tn(g) − Tn(fg)| = |Tn(f)Tn(g) − Cn(f)Cn(g) + Cn(f)Cn(g) − Tn(fg)| ≤ |Tn(f)Tn(g) − Cn(f)Cn(g)| + |Cn(fg) − Tn(fg)|−→ n→∞ 0. (d) Follows from repeated application of (5.14) and part (c). (e) Equation (5.22) follows from (d) and Theorem 2.1. For the Her- mitian case, however, we cannot simply apply Theorem 2.4 since the eigenvalues ρn,k of∏ iTn(fi) may not be real. We can show, however, that they are asymptotically real in the sense that the imaginary part vanishes in the limit. Let ρn,k = αn,k + iβn,k where αn,k and βn,k are real. Then from Theorem 2.2 we have for any positive integer s limn→∞ n−1n−1∑ k=0 (αn,k + iβn,k)s = lim n→∞n−1 n−1∑ k=0 ψsn,k =1 ∫ 2π 0 [ m∏ i=1 fi(λ) ]s dλ, (5.26) where ψn,k are the eigenvalues of Cn ( m∏ i=1 fi ) . From (2.17) n−1n−1∑ k=0 |ρn,k|2 = n−1n−1∑ k=0 ( α2n,k + β2 n,k ) ≤∣ m∏ i=i Tn(fi) 2 . From (4.57), Theorem 2.1 and Lemma 4.4 limn→∞ m∏ i=1 Tn(fi) 2 = limn→∞ Cn ( m∏ i=1 fi )∣ 2 = (2π)−1 ∫ 2π 0 ( m∏ i=1 fi(λ) )2 dλ. (5.27) 70 Matrix Operations on Toeplitz Matrices Subtracting (5.26) for s = 2 from (5.27) yields limn→∞ 1 n n−1∑ k=1 β2n,k ≤ 0. Thus the distribution of the imaginary parts tends to the origin and hence limn→∞ 1 n n−1∑ k=0 αsn,k = 1 ∫ 2π 0 [ m∏ i=1 fi(λ) ]s dλ. 2 Parts (d) and (e) are here proved as in Grenander and Szego ([16], pp. 105-106. We have developed theorems on the asymptotic behavior of eigenval- ues, inverses, and products of Toeplitz matrices. The basic method has been to find an asymptotically equivalent circulant matrix whose spe- cial simple structure could be directly related to the Toeplitz matrices using the results for asymptotically equivalent sequences of matrices. We began with the banded case since the appropriate circulant matrix is there obvious and yields certain desirable properties that suggest the corresponding circulant matrix in the infinite case. We have limited our consideration of the infinite order case functions f(λ) or Toeplitz ma- trices in the Wiener class and hence to absolutely summable coefficients for simplicity. The more general case of square summable tk is treated in Chapter 7 of [16] and requires significantly more mathematical care, but can be interpreted as an extension of the approach taken here. We did not treat sums of Toeplitz matrices as no additional con- sideration is needed: a sum of Toeplitz matrices of equal size is also a Toeplitz matrix, so the results immediately apply. We also did not con- sider the asymptotic behavior of eigenvectors for the simple reason that there do not exist results along the lines that intuition suggests, that is, that show that in some sense the eigenvectors for circulant matrices also work for Toeplitz matrices. 5.3 Toeplitz Determinants We close the consideration of matrix operations on Toeplitz matrices by returning to a problem mentioned in the introduction and formalize the 5.3. Toeplitz Determinants 71 behavior of limits of Toeplitz determinants. Suppose now that Tn(f) is a sequence of Hermitian Toeplitz matrices such that that f(λ) ≥ mf > 0. Let Cn(f) denote the sequence of circulant matrices constructed from f as in (4.32). Then from (4.34) the eigenvalues of Cn(f) are f(2πm/n) for m = 0, 1, . . . , n− 1 and hence det(Cn(f)) =∏n−1 m=0 f(2πm/n). This in turn implies that ln (det(Cn(f)))1 n =1 nln detCn(f) = 1 n n−1∑ m=0 ln f(2πm n). These sums are the Riemann approximations to the limiting integral, whence limn→∞ ln (det(Cn(f)))1 n = ∫ 1 0ln f(2πλ) dλ. Exponentiating, using the continuity of the logarithm for strictly positive arguments, and changing the variables of integration yields limn→∞ (det(Cn(f)))1 n = exp ( 1 ∫ 2π 0ln f(λ) dλ. ) This integral, the asymptotic equivalence of Cn(f) and Tn(f) (Lemma 4.6), and Corollary 2.4 together yield the following result ([16], p. 65). Theorem 5.4. Let Tn(f) be a sequence of Hermitian Toeplitz matrices in the Wiener class such that ln f(λ) is Riemann integrable and f(λ) ≥mf > 0. Then limn→∞ (det(Tn(f)))1 n = exp ( 1 ∫ 2π 0ln f(λ) dλ ) . (5.28) 6 Applications to Stochastic Time Series Toeplitz matrices arise quite naturally in the study of discrete time random processes. Covariance matrices of weakly stationary processes are Toeplitz and triangular Toeplitz matrices provide a matrix repre- sentation of causal linear time invariant filters. As is well known and as we shall show, these two types of Toeplitz matrices are intimately related. We shall take two viewpoints in the first section of this chapter section to show how they are related. In the first part we shall con- sider two common linear models of random time series and study the asymptotic behavior of the covariance matrix, its inverse and its eigen- values. The well known equivalence of moving average processes and weakly stationary processes will be pointed out. The lesser known fact that we can define something like a power spectral density for autore- gressive processes even if they are nonstationary is discussed. In the second part of the first section we take the opposite tack — we start with a Toeplitz covariance matrix and consider the asymptotic behav- ior of its triangular factors. This simple result provides some insight into the asymptotic behavior of system identification algorithms and Wiener-Hopf factorization. Let Xk; k ∈ I be a discrete time random process. Generally we 73 74 Applications to Stochastic Time Series take I = Z, the space of all integers, in which case we say that the process is two-sided, or I = Z+, the space of all nonnegative integers, in which case we say that the process is one-sided. We will be interested in vector representations of the process so we define the column vector (n−tuple) Xn = (X0,X1, . . . ,Xn−1)′, that is, Xn is an n-dimensional column vector. The mean vector is defined by mn = E(Xn), which we usually assume is zero for convenience. The n × n covariance matrix Rn = rj,k is defined by Rn = E[(Xn −mn)(Xn −mn)∗]. (6.1) Covariance matrices are Hermitian since R∗n = E[(Xn −mn)(Xn −mn)∗]∗ = E[(Xn −mn)(Xn −mn)∗]. (6.2) Setting m = 0 yields the This is the autocorrelation matrix. Subscripts will be dropped when they are clear from context. If the matrix Rn is Toeplitz for all n, say Rn = Tn(f), then rk,j = rk−j and the process is said to be weakly stationary. In this case f(λ) =∑∞ k=−∞rkeikλ is the power spectral density of the process. If the matrix Rn is not Toeplitz but is asymptotically Toeplitz, i.e., Rn ∼ Tn(f), then we say that the process is asymptotically weakly stationary and f(λ) is the power spectral density. The latter situation arises, for example, if an otherwise stationary process is initialized with Xk = 0, k ≤ 0. This will cause a transient and hence the process is strictly speaking nonstationary. The transient dies out, however, and the statistics of the process approach those of a weakly stationary process as n grows. We now proceed to investigate the behavior of two common linear models for random processes, both of which model a complicated pro- cess as the result of passing a simple process through a linear filter. For simplicity we will assume the process means are zero. 6.1 Moving Average Processes By a linear model of a random process we mean a model wherein we pass a zero mean, independent identically distributed (iid) sequence of random variables Wk with variance σ2 through a linear time invariant discrete time filtered to obtain the desired process. The process Wk is 6.1. Moving Average Processes 75 discrete time “white” noise. The most common such model is called a moving average process and is defined by the difference equation Un = ∑nk=0 bkWn−k = ∑nk=0 bn−kWk n = 0, 1, . . . 0 n < 0. (6.3) We assume that b0 = 1 with no loss of generality since otherwise we can incorporate b0 into σ2. Note that (6.3) is a discrete time convolu- tion, i.e., Un is the output of a filter with “impulse response” (actually Kronecker δ response) bk and input Wk. We could be more general by allowing the filter bk to be noncausal and hence act on future Wk’s. We could also allow the Wk’s and Uk’s to extend into the infinite past rather than being initialized. This would lead to replacing of (6.3) by Un = ∞∑ k=−∞bkWn−k = ∞∑ k=−∞bn−kWk. (6.4) We will restrict ourselves to causal filters for simplicity and keep the initial conditions since we are interested in limiting behavior. In addi- tion, since stationary distributions may not exist for some models it would be difficult to handle them unless we start at some fixed time. For these reasons we take (6.3) as the definition of a moving average. Since we will be studying the statistical behavior of Un as n gets arbitrarily large, some assumption must be placed on the sequence bkto ensure that (6.3) converges in the mean-squared sense. The weakest possible assumption that will guarantee convergence of (6.3) is that ∞∑ k=0 |bk|2 <∞. (6.5) In keeping with the previous sections, however, we will make the stronger assumption∞∑ k=0 |bk| <∞. (6.6) As previously this will result in simpler mathematics. Equation (6.3) can be rewritten as a matrix equation by defining 76 Applications to Stochastic Time Series the lower triangular Toeplitz matrix Bn = 1 0 b1 1 b2 b1... b2 . . .. . . bn−1 . . . b2 b1 1 (6.7) so that Un = BnWn. (6.8) If the filter bn were not causal, then Bn would not be triangular. If in addition (6.4) held, i.e., we looked at the entire process at each time instant, then (6.8) would require infinite vectors and matrices as in Grenander and Rosenblatt [15]. Since the covariance matrix of Wk is simply σ2In, where In is the n × n identity matrix, we have for the covariance of Un: R(n)U = EUn(Un)∗ = EBnW n(W n)∗B∗n = σ2BnB∗n = σ2 min(k,j)∑ ℓ=0 bℓ−kb∗ℓ−j The matrix R(n)U = [rk,j] is not Toeplitz. For example, the upper left entry is 1 and the second diagonal entry is 1 + b21. However, as we next show, the sequence R(n)U becomes asymptotically Toeplitz as n → ∞. If we define b(λ) = ∞∑ k=0 bkeikλ (6.9) then Bn = Tn(b) (6.10) so that R(n)U = σ2Tn(b)Tn(b)∗. (6.11) 6.2. Autoregressive Processes 77 Observe that R(n)U is Hermitian, as all covariance matrices must be. We can now apply the results of the previous sections to obtain the following theorem. Theorem 6.1. Let Un be a moving average process with covariance matrix RUn(n) given by (6.9)–(6.11). Let ρn,k be the eigenvalues of R(n)U . Then R(n)U ∼ σ2Tn(|b|2) = Tn(σ2|b|2) (6.12) so that Un is asymptotically stationary. If m = ess inf σ2|b(λ)|2 and M = ess sup σ2|b(λ)|2 and F (x) is any continuous function on [m,M ], then limn→∞ 1 n n−1∑ k=0 F (ρn,k) =1 ∫ 2π 0F (σ2|b(λ)|2) dλ. (6.13) If σ2|b(λ)|2 ≥ m > 0, then R(n)U −1∼ σ−2Tn(1/|b|2). (6.14) Proof. Since R(n)U is Hermitian, the results follow from Theorems 4.2 and 5.2 and (2.3). 2 If the process Un had been initiated with its stationary distribution then we would have had exactly R(n)U = σ2Tn(|b|2). More knowledge of the inverse R(n)U −1can be gained from Theorem 5.2, e.g., circulant approximations. Note that the spectral density of the moving average process is σ2|b(λ)|2 and that sums of functions of eigen- values tend to an integral of a function of the spectral density. In effect the spectral density determines the asymptotic density function for the eigenvalues of R(n)U and σ2Tn(|b|2). 6.2 Autoregressive Processes Let Wk be as previously defined, then an autoregressive process Xn is defined by Xn = −∑nk=1 akXn−k +Wn n = 0, 1, . . . 0 n < 0.(6.15) 78 Applications to Stochastic Time Series Autoregressive process include nonstationary processes such as the Wiener process. Equation (6.15) can be rewritten as a vector equation by defining the lower triangular matrix. An = 1 a1 1 0 a1 1. . . . . . an−1 a1 1 (6.16) so that AnXn = W n. Since R(n)W = AnR (n)X A∗ n (6.17) and det An = 1 6= 0, An is nonsingular. Hence R(n)X = σ2A−1 n A−1∗n (6.18) or (R(n)X )−1 = σ−2A∗ nAn. (6.19) Equivalently, if (R(n)X )−1 = tk,j then tk,j = min(k,j)∑ m=0 a∗m−kam−j . Unlike the moving average process, we have that the inverse covariance matrix is the product of Toeplitz triangular matrices. Defining a(λ) =∞∑ k=0 akeikλ (6.20) we have that (R(n)X )−1 = σ−2Tn(a)∗Tn(a). (6.21) Observe that (R(n)X )−1 is Hermitian. 6.2. Autoregressive Processes 79 Theorem 6.2. Let Xn be an autoregressive process with absolutely summable ak and covariance matrix R(n)X with eigenvalues ρn,k. Then (R(n)X )−1 ∼ σ−2Tn(|a|2). (6.22) If m = ess inf σ−2|a(λ)|2 and M = ess sup σ−2|a(λ)|2, then for any function F (x) on [m,M ] we have limn→∞ 1 n n−1∑ k=0 F (1/ρn,k) =1 ∫ 2π 0F (σ2|a(λ)|2) dλ, (6.23) where 1/ρn,k are the eigenvalues of (R(n)X )−1. If |a(λ)|2 ≥ m > 0, then R(n)X ∼ σ2Tn(1/|a|2) (6.24) so that the process is asymptotically stationary. Proof. Theorem 5.3. 2 Note that if |a(λ)|2 > 0, then 1/|a(λ)|2 is the spectral density of Xn. If |a(λ)|2 has a zero, then R(n)X may not be even asymptotically Toeplitz and hence Xn may not be asymptotically stationary (since 1/|a(λ)|2may not be integrable) so that strictly speaking xk will not have a spectral density. It is often convenient, however, to define σ2/|a(λ)|2 as the spectral density and it often is useful for studying the eigenvalue distribution of Rn. We can relate σ2/|a(λ)|2 to the eigenvalues of R(n)X even in this case by using Theorem 5.2 (d). Corollary 6.1. Given the assumptions of the theorem, then for any finite θ and any function F (x) continuous on [1/m, θ] limn→∞ 1 n n−1∑ k=0 F (min(ρn,k, θ)) =1 ∫ 2π 0F (min(1/|a(γ)|2, θ)) dλ. (6.25) Proof. Theorems 6.2 and 5.1. 2 If we consider two models of a random process to be asymptotically equivalent if their covariances are asymptotically equivalent, then from Theorems 6.1 and 6.2 we have the following corollary. 80 Applications to Stochastic Time Series Corollary 6.2. Given the assumptions of Theorems 6.1 and 6.2, con- sider the moving average process defined by Un = Tn(b)W n and the autoregressive process defined by Tn(a)Xn = W n. Then the processes Un and Xn are asymptotically equivalent if a(λ) = 1/b(λ). Proof. Follows from Theorems 5.2 and 5.3 and R(n)X = σ2Tn(a)−1T−1 n (a)∗ ∼ σ2Tn(1/a)Tn(1/a)∗ ∼ σ2Tn(1/a)∗Tn(1/a). (6.26) Comparison of (6.26) with (6.11) completes the proof. 2 The methods above can also easily be applied to study the mixed autoregressive-moving average linear models [33]. 6.3 Factorization Consider the problem of the asymptotic behavior of triangular factors of a sequence of Hermitian covariance matrices Tn(f) in the Wiener class. It is well known that any such matrix can be factored into the product of a lower triangular matrix and its conjugate transpose ([15], p. 37), in particular Tn(f) = tk,j = BnB∗n, (6.27) where Bn is a lower triangular matrix with entries b(n)k,j = (detTk) det(Tk−1)−1/2γ(j, k), (6.28) where γ(j, k) is the determinant of the matrix Tk with the right-hand column replaced by (tj,0, tj,1, . . . , tj,k−1)′. Note in particular that the diagonal elements are given by b(n)k,k = (det Tk)/(det Tk−1)1/2. (6.29) 6.3. Factorization 81 Equation (6.28) is the result of a Gaussian elimination or a Gram- Schmidt procedure. The factorization of Tn allows the construction of a linear model of a random process and is useful in system identification and other recursive procedures. Our question is how Bn behaves for large n; specifically is Bn asymptotically Toeplitz? Suppose that f(λ) has the form f(λ) = σ2|b(λ)|2 (6.30) b∗(λ) = b(−λ) b(λ) =∞∑ k=0 bkeikλ b0 = 1. The decomposition of a nonnegative function into a product with this form is known as a Wiener-Hopf factorization . For a current survey see the discussion and references in Kailath et al. [17] We have already constructed functions of this form when considering moving average and autoregressive models. It is a classic result that a necessary and sufficient condition for f to have such a factorization is that ln f have a finite integral. From (6.27) and Theorem 5.2 we have BnB∗n = Tn(f) ∼ Tn(σb)Tn(σb)∗. (6.31) We wish to show that (6.31) implies that Bn ∼ Tn(σb). (6.32) Proof. Since detTn(σb) = σn 6= 0, Tn(σb) is invertible. Likewise, since detBn = [detTn(f)]1/2 we have from Theorem 5.2 (a) that detTn(f) 6=0 so that Bn is invertible. Thus from Theorem 2.1 (e) and (6.31) we have T−1n Bn = [B−1 n Tn]−1 ∼ T ∗nB ∗−1n = [B−1 n Tn]∗. (6.33) Since Bn and Tn are both lower triangular matrices, so is B−1n and hence BnTn and [B−1n Tn]−1. Thus (6.33) states that a lower triangular matrix is asymptotically equivalent to an upper triangular matrix. This 82 Applications to Stochastic Time Series is only possible if both matrices are asymptotically equivalent to a diagonal matrix, say Gn = g(n)k,kδk,j. Furthermore from (6.33) we have Gn ∼ G∗−1n |g(n)k,k |2δk,j ∼ In. (6.34) Since Tn(σb) is lower triangular with main diagonal element σ, Tn(σb)−1 is lower triangular with all its main diagonal elements equal to 1/σ even though the matrix Tn(σb)−1 is not Toeplitz. Thus g(n)k,k = b (n)k,k/σ. Since Tn(f) is Hermitian, bk,k is real so that taking the trace in (6.34) yields limn→∞ σ−2 1 n n−1∑ k=0 ( b(n)k,k )2= 1. (6.35) From (6.29) and Corollary 2.4, and the fact that Tn(σb) is triangular we have that limn→∞ σ−1 1 n n−1∑ k=0 b(n)k,k = σ−1 lim n→∞(det Tn(f))/(detTn−1(f))1/2 = σ−1 limn→∞ detTn(f)1/2nσ−1 limn→∞ detTn(σb)1/n = σ−1σ = 1. (6.36) Combining (6.35) and (6.36) yields limn→∞ |B−1n Tn − In| = 0. (6.37) Applying Theorem 2.1 yields (6.32). 2 Since the only real requirements for the proof were the existence of the Wiener-Hopf factorization and the limiting behavior of the deter- minant, this result could easily be extended to the more general case that ln f(λ) is integrable. The theorem can also be derived as a special case of more general results of Baxter [1] and is similar to a result of Rissanen and Barbosa [21]. Acknowledgements The author would like to thank his brother, Augustine Heard Gray, Jr., for his assistance long ago in finding the eigenvalues of the in- verse covariance matrices of discrete time Wiener processes, his first en- counter with Toeplitz and asymptotically Toeplitz matrices. He would like to thank Adriano Garsia and Tom Pitcher for helping him struggle through Grenander and Szego’s book during summer lunches in 1967 when the author was a summer employee at JPL during his graduate student days at USC. This manuscript first appeared as a technical re- port in 1971 as an expanded version of the tutorial paper [12] and was revised in 1975. After laying dormant for many years, it was revised and converted to LATEXand posted on the World Wide Web. That re- sulted in significant feedback, corrections, and suggestions and in many revisions through the years. Particular thanks go to Ronald M. Aarts of the Philips Research Labs for correcting many typos and errors in the 1993 revision, Liu Mingyu in pointing out errors corrected in the 1998 revision, Paolo Tilli of the Scuola Normale Superiore of Pisa for pointing out an incorrect corollary and providing the correction, and to David Neuhoff of the University of Michigan for pointing out sev- eral typographical errors and some confusing notation. For corrections, 83 84 Acknowledgements comments, and improvements to the 2001 revision thanks are due to William Trench, John Dattorro, and Young-Han Kim. In particular, Professor Trench brought the Wielandt-Hoffman theorem and its use to prove strengthened results to my attention. Section 2.4 largely follows his suggestions, although I take the blame for any introduced errors. For the 2002 revision, particular thanks to Cynthia Pozun of ENST for several corrections. For the 2005–2006 revisions, special thanks to Jean-Francois Chamberland-Tremblay, Lee Patton, Sergio Verdu and two very preceptive and helpful anonymous reviewers. Finally, the au- thor would like to thank the National Science Foundation for the sup- port of the author’s research involving Toeplitz matrices which led to the original paper and report. References [1] G. Baxter, “A Norm Inequality for a ‘Finite-Section’ Wiener-Hopf Equation,”Illinois J. Math., 1962, pp. 97–103. [2] G. Baxter, “An Asymptotic Result for the Finite Predictor,” Math. Scand., 10,pp. 137–144, 1962. [3] T. Berger, Rate Distortion Theory: A Mathematical Basis for Data Compres- sion, Prentice Hall, Englewood Cliffs, New Jersey, 1971.[4] A. Bottcher and S.M. Grudsky, Toeplitz Matrices, Asymptotic Linear Algebra, and Functional Analysis, Birkhauser, 2000.[5] A. Bottcher and B. Silbermann, Introduction to Large Truncated Toeplitz Ma- trices, Springer, New York, 1999.[6] W. Cheney, Introduction to Approximation theory, McGraw-Hill, 1966.[7] T. A. Cover and J. A. Thomas, Elements of Information Theory, Wiley, New York, 1991.[8] P. J. Davis, Circulant Matrices, Wiley-Interscience, NY, 1979.[9] D. Fasino and P. Tilli, “Spectral clustering properties of block multilevel Hankel matrices, Linear Algebra and its Applications, Vol. 306, pp. 155–163, 2000.[10] F.R. Gantmacher, The Theory of Matrices, Chelsea Publishing Co., NY 1960.[11] R.M. Gray, “Information Rates of Autoregressive Processes,” IEEE Trans. on Info. Theory, IT-16, No. 4, July 1970, pp. 412–421.[12] R. M. Gray, “On the asymptotic eigenvalue distribution of Toeplitz matrices,” IEEE Transactions on Information Theory, Vol. 18, November 1972, pp. 725–730. [13] R.M. Gray, “On Unbounded Toeplitz Matrices and Nonstationary Time Serieswith an Application to Information Theory,” Information and Control, 24, pp.181–196, 1974. 85 86 References [14] R.M. Gray and L.D. Davisson, An Introduction to Statistical Signal Processing,Cambridge University Press, London, 2005. [15] U. Grenander and M. Rosenblatt, Statistical Analysis of Stationary Time Se- ries, Wiley and Sons, NY, 1966, Chapter 1.[16] U. Grenander and G. Szego, Toeplitz Forms and Their Applications, University of Calif. Press, Berkeley and Los Angeles, 1958.[17] T. Kailath, A. Sayed, and B. Hassibi, Linear Estimation, Prentice Hall, New Jersey, 2000.[18] P. Lancaster, Theory of Matrices, Academic Press, NY, 1969.[19] J. Pearl, “On Coding and Filtering Stationary Signals by Discrete Fourier Transform,” IEEE Trans. on Info. Theory, IT-19, pp. 229–232, 1973.[20] C.L. Rino, “The Inversion of Covariance Matrices by Finite Fourier Trans- forms,” IEEE Trans. on Info. Theory, IT-16, No. 2, March 1970, pp. 230–232.[21] J. Rissanen and L. Barbosa, “Properties of Infinite Covariance Matrices and Stability of Optimum Predictors,” Information Sciences, 1, 1969, pp. 221–236.[22] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, NY, 1964.[23] W. F. Trench, “Asymptotic distribution of the even and odd spectra of real symmetric Toeplitz matrices,” Linear Algebra Appl., Vol. 302-303, pp. 155–162,1999. [24] W. F. Trench, “Absolute equal distribution of the spectra of Hermitian matri-ces,” Lin. Alg. Appl., 366 (2003), 417–431. [25] W. F. Trench, “Absolute equal distribution of families of finite sets,” Lin. Alg. Appl. 367 (2003), 131–146.[26] W. F. Trench, “A note on asymptotic zero distribution of orthogonal polyno- mials,” Lin. Alg. Appl. 375 (2003) 275–281[27] W. F. Trench, “Simplification and strengthening of Weyl’s definition of asymp- totic equal distribution of two families of finite sets,” Cubo A Mathematical Journal Vol. 06 N 3 (2004), 47–54.[28] W. F. Trench, “Absolute equal distribution of the eigenvalues of discrete Sturm– Liouville problems,” J. Math. Anal. Appl., Volume 321, Issue 1 , 1 September2006, Pages 299–307. [29] B.S. Tsybakov, “Transmission capacity of memoryless Gaussian vector chan-nels,” (in Russian),Probl. Peredach. Inform., Vol 1, pp. 26–40, 1965. [30] B.S. Tsybakov, “On the transmission capacity of a discrete-time Gaussian chan-nel with filter,” (in Russian),Probl. Peredach. Inform., Vol 6, pp. 78–82, 1970. [31] E.E. Tyrtyshnikov, “Influence of matrix operations on the distribution of eigen-values and singular values of Toeplitz matrices,” Linear Algebra and its Appli- cations, Vol. 207, pp. 225–249, 1994.[32] E.E. Tyrtyshnikov, “A unifying approach to some old and new theorems on distribution and clustering,” Linear Algebra and its Applications, Vol. 232, pp.1–43, 1996. [33] H. Widom, “Toeplitz Matrices,” in Studies in Real and Complex Analysis, edited by I.I. Hirschmann, Jr., MAA Studies in Mathematics, Prentice-Hall,Englewood Cliffs, NJ, 1965. [34] A.J. Hoffman and H. W. Wielandt, “The variation of the spectrum of a normalmatrix,” Duke Math. J., Vol. 20, pp. 37–39, 1953. References 87 [35] James H. Wilkinson, “Elementary proof of the Wielandt-Hoffman theorem andof its generalization,” Stanford University, Department of Computer ScienceReport Number CS-TR-70-150, January 1970 . Index absolutely summable, 32, 38, 48 absolutely summable Toeplitz matrices, 41 analytic function, 16 asymptotic equivalence, 38 asymptotically absolutely equally distributed, 21 asymptotically equally dis- tributed, 17, 56 asymptotically equivalent ma- trices, 11 asymptotically weakly station- ary, 64 autocorrelation matrix, 64 autoregressive process, 68 bounded matrix, 10 bounded Toeplitz matrices, 31 Cauchy-Schwartz inequality, 13, 20 characteristic equation, 5 circulant matrix, 2, 25 conjugate transpose, 6, 70 continuous, 17, 21, 22, 33, 39, 41, 48, 49, 52, 54–57, 67, 69 continuous complex function, 16 convergence uniform, 32 Courant-Fischer theorem, 6 covariance matrix, 1, 63, 64 cyclic matrix, 2 cyclic shift, 25 determinant, 17, 31, 60, 71 DFT, 28 diagonal, 8 differential entropy, 73 88 INDEX 89 differential entropy rate, 73 discrete time, 63 eigenvalue, 5, 26, 31 eigenvalue distribution theo- rem, 40, 48 eigenvector, 5, 26 Euclidean norm, 9 factorization, 70 filter, 1 linear time invariant, 63 finite order, 31 finite order Toeplitz matrix, 36 Fourier series, 32 truncated, 44 Fourier transform discrete, 28 Frobenius norm, 9 function, analystic, 16 Gaussian process, 73 Hermitian, 6 Hilbert-Schmidt norm, 8, 9 identity matrix, 19 impulse respone, 65 information theory, 73 inverse, 29, 31, 53 Kronecker delta, 27 Kronecker delta response, 65 linear difference equation, 26 matrix bounded, 10 circulant, 2, 25 covariance, 1 cyclic, 2 Hermitian, 6 normal, 6 Toeplitz, 26, 31 matrix, Toeplitz, 1 mean, 64 metric, 8 moments, 15 moving average, 65 noncausal, 65 nonnegative definite, 6 nonsingular, 53 norm, 8 axioms, 10 Euclidean, 9 Frobenius, 9 Hilbert-Schmidt, 8, 9 operator, 8 strong, 8, 9 weak, 8, 9 normal, 6, 8 one-sided, 64 operator norm, 8 polynomials, 16 positive definite, 6 power specral density, 64 power spectral density, 63, 64, 73 probability mass function, 19 product, 29, 31 random process, 63 90 INDEX discrete time, 64 Rayleigh quotient, 6 Riemann integrable, 48, 53, 61 Shannon information theory, 73 Shur’s theorem, 8 spectrum, 53 square summable, 31 Stone-Weierstrass approxima- tion theorem, 16 Stone-Weierstrass theorem, 40 strictly positive definite, 6 sum, 29 Taylor series, 16 time series, 63 Toeplitz determinant, 60 Toeplitz matrix, 1, 26, 31 Toeplitz matrix, finite order, 31 trace, 8 transpose, 26 triangle inequality, 10 triangular, 63, 66, 68, 70, 72 two-sided, 64 uniform convergence, 32 uniformaly bounded, 38 unitary, 6 upper triangular, 6 weak norm, 9 weakly stationary, 64 asymptotically, 64 white noise, 65 Wielandt-Hoffman theorem, 18, 20 Wiener-Hopf factorization, 63 Recommended

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# What is a GAN? Let us take an analogy to explain the concept: # How do GANs work? The Generator Network($G(z)$) 接受随机噪声输入($z$ from $p(z)$)来产生假样本($g$). 随后送入 Discriminator Network $D(x)$. Discriminator Network 的任务是判断 real data 和 fake data 的真假. It takes an input $x$ from $P_{data}(x)$ where $P_{data}(x)$ is ourreal data distribution. $D(x)$ then solvers a binary classification problem using sigmoid function giving outpit in the range 0 to 1. Now the training of GAN is done (as we saw above) as a fight between generator and discriminator. This can be represented mathematically as: \label{a} \mathop{min}\limits_{G} \mathop{max}\limits_{D}  V(D, G) \\ V(D, G) = E_{x \sim p_{data}(x)} [logD(x)] + E_{z \sim p_{z}(z)} [log(1-D(G(z))] train discriminator stage: 从判别器 $D$ 角度来看，它希望能尽可能区分出真假样本，即 maximize $V(D, G)$ to 0。具体来说，它一边希望 $D(x)$ 尽可能大，即 maximize $D(x)$ to 1. 另一边则希望 $D(G(z))$ 尽可能小，即 maximize $D(G(z))$ it to 0 (i.e. the log probability that the data from generated is fake is equal to 0). train generator stage:  从生成器 $G$ 角度来看，它希望能够以假乱真 ，即 minimize the function $V$ to -NaN。 具体来说就是希望 $D(G(z))$ 尽可能大，即 maximize $D(G(x))$ to 1  (this stage only have second term). Note: This method of training a GAN is taken from game theory called the minimax game. # Parts of training GAN So broadly a training phase has two main subparts and they are done sequentially: • Pass 1: Train discriminator and freeze generator (freezing means setting training as false. The network does only forward pass and no backpropagationn is applied) • Pass 2: Train generator and freeze discriminator # Steps to train a GAN Step 1: Define the problem. Do you want to generate fake images or fake text. Here you should completely define the problem and collect data for it. Step 2: Define architecture of GAN. Define how your GAN should look like. Should both your generator and discriminator be multi layer perceptrons, or convolutional neural networks? This step will depend on what problem you are trying to solve. Step 3: Train Discriminator on real data for n epochs. Get the real data you want to generate fake on and train the discriminator to correctly predict them as real. Here value n can be any natural number between 1 and infinity. Step 4: Generate fake inputs for generator and train Discriminator on fake data. Get generated data and let the discriminator correctly predict them as fake. (Step 3 and Step 4 are for Pass 1) Step 5: Train Generator with the output of Discriminator. Now when the discriminator is trained, you can get its predictions and use it as an objective for training the generator. Train the generator to fool the discriminator. (This is Pass 2) Step 6: Repeat step 3 to step 5 for a few epochs. Step 7: Check if the fake data manually if it seems legit. If it seems appropriate, stop training, else go to step 3. This is a bit of a manual task, as hand evaluating the data is the best way to check the fakeness. When this step is over, you can evaluate whether the GAN is performing well enough. # Challenges with GANs There’s so many roadblocks into building a “good enough” GAN and we haven’t cleared many of them yet. There’s a whole area of research out there just to find “how to train a GAN The most important roadblock while training a GAN is stability. If you start to train a GAN, and the discriminator part is much powerful that its generator counterpart, the generator would fail to train effectively. This will in turn affect training of your GAN. On the other hand, if the discriminator is too lenient; it would let literally any image be generated. And this will mean that your GAN is useless. Another way to glance at stability of GAN is to look as a holistic convergence problem. Both generator and discriminator are fighting against each other to get one step ahead of the other. Also, they are dependent on each other for efficient training. If one of them fails, the whole system fails. So you have to make sure they don’t explode. This is kind of like the shadow in Prince of Persia game . You have to defend yourself from the shadow, which tries to kill you. If you kill the shadow you die, but if you don’t do anything, you will definitely die! There are other problems too, which I will list down here. (Reference: http://www.iangoodfellow.com/slides/2016-12-04-NIPS.pdf) Note: Below mentioned images are generated by a GAN trained on ImageNet dataset. • Problem with Counting: GANs fail to differentiate how many of a particular object should occur at a location. As we can see below, it gives more number of eyes in the head than naturally present. • Problems with Perspective: GANs fail to adapt to 3D objects. It doesn’t understand perspective, i.e.difference between frontview and backview. As we can see below, it gives flat (2D) representation of 3D objects. • Problems with Global Structures: Same as the problem with perspective, GANs do not understand a holistic structure. For example, in the bottom left image, it gives a generated image of a quadruple cow, i.e. a cow standing on its hind legs and simultaneously on all four legs. That is definitely not possible in real life! A substantial research is being done to take care of these problems. Newer types of models are proposed which give more accurate results than previous techniques, such as DCGAN, WassersteinGAN etc # Implementing a Toy GAN ## pytorch implement import os import torch import torchvision import torch.nn as nn from torchvision import transforms from torchvision.utils import save_image # Device configuration device = torch.device('cuda' if torch.cuda.is_available() else 'cpu') print(torch.__version__, device) # Hyper-parameters latent_size = 64 hidden_size = 256 image_size = 784 num_epochs = 200 batch_size = 100 sample_dir = 'samples' # Create a directory if not exists if not os.path.exists(sample_dir): os.makedirs(sample_dir) # Image processing transform = transforms.Compose([ transforms.ToTensor(), transforms.Normalize(mean=(0.1307,), # 3 for RGB channels std=(0.3081,))]) # MNIST dataset mnist = torchvision.datasets.MNIST(root='H:/Other_DataSets/MNIST/', train=True, transform=transform, batch_size=batch_size, shuffle=True) # Discriminator D = nn.Sequential( nn.Linear(image_size, hidden_size), nn.LeakyReLU(0.2), nn.Linear(hidden_size, hidden_size), nn.LeakyReLU(0.2), nn.Linear(hidden_size, 1), nn.Sigmoid()) # Generator G = nn.Sequential( nn.Linear(latent_size, hidden_size), nn.ReLU(), nn.Linear(hidden_size, hidden_size), nn.ReLU(), nn.Linear(hidden_size, image_size), nn.Tanh()) # Device setting D = D.to(device) G = G.to(device) # Binary cross entropy loss and optimizer criterion = nn.BCELoss() def denorm(x): out = (x + 1) / 2 return out.clamp(0, 1) # Start training for epoch in range(num_epochs): for i, (images, _) in enumerate(data_loader): images = images.reshape(batch_size, -1).to(device) # Create the labels which are later used as input for the BCE loss real_labels = torch.ones(batch_size, 1).to(device) fake_labels = torch.zeros(batch_size, 1).to(device) # ================================================================== # # Train the discriminator # # ================================================================== # # Compute BCE_Loss using real images where BCE_Loss(x, y): - y * log(D(x)) - (1-y) * log(1 - D(x)) # Second term of the loss is always zero since real_labels == 1 outputs = D(images) # batch x 1 d_loss_real = criterion(outputs, real_labels) real_score = outputs # Compute BCELoss using fake images # First term of the loss is always zero since fake_labels == 0 z = torch.randn(batch_size, latent_size).to(device) fake_images = G(z) # batch x 784 outputs = D(fake_images) # batch x 1 d_loss_fake = criterion(outputs, fake_labels) fake_score = outputs # Backprop and optimize d_loss = d_loss_real + d_loss_fake d_loss.backward() d_optimizer.step() # ================================================================== # # Train the generator # # ================================================================== # # Compute loss with fake images z = torch.randn(batch_size, latent_size).to(device) fake_images = G(z) # batch x 784 outputs = D(fake_images) # batch x 1 # We train G to maximize log(D(G(z)) instead of minimizing log(1-D(G(z))) # For the reason, see the last paragraph of section 3. https://arxiv.org/pdf/1406.2661.pdf g_loss = criterion(outputs, real_labels) # Backprop and optimize g_loss.backward() g_optimizer.step() if (i+1) % 200 == 0: print('Epoch [{}/{}], Step [{}/{}], d_loss: {:.4f}, g_loss: {:.4f}, D(x): {:.2f}, D(G(z)): {:.2f}' .format(epoch, num_epochs, i+1, total_step, d_loss.item(), g_loss.item(), real_score.mean().item(), fake_score.mean().item())) # Save real images if (epoch+1) == 1: images = images.reshape(images.size(0), 1, 28, 28) save_image(denorm(images), os.path.join(sample_dir, 'real_images.png')) # Save sampled images fake_images = fake_images.reshape(fake_images.size(0), 1, 28, 28) save_image(denorm(fake_images), os.path.join(sample_dir, 'fake_images-{}.png'.format(epoch+1))) # Save the model checkpoints torch.save(G.state_dict(), 'G.ckpt') torch.save(D.state_dict(), 'D.ckpt') View Code ## tensorflow implement #!/usr/bin/env python # -*- coding: utf-8 -*- # ================================================================ # @Time : 2020/3/25 10:40 # @Author : YangTao # @Func : # @File : GAN_example.py # @IDE: PyCharm Community Edition # ================================================================ import tensorflow as tf from tensorflow.examples.tutorials.mnist import input_data import numpy as np import matplotlib.pyplot as plt import matplotlib.gridspec as gridspec import os # 1. create data def sample_Z(m, n): return np.random.uniform(-1., 1., size=[m, n]) with tf.variable_scope('Input'): X = tf.placeholder(dtype=tf.float32, shape=[None, 28 * 28], name='x') Z = tf.placeholder(tf.float32, shape=[None, 100]) # 2. define Network def xavier_init(size): in_dim = size[0] xavier_stddev = 1. / tf.sqrt(in_dim / 2.) return tf.random_normal(shape=size, stddev=xavier_stddev) with tf.variable_scope('Net'): D_W1 = tf.Variable(xavier_init([784, 128])) D_b1 = tf.Variable(tf.zeros(shape=[128])) D_W2 = tf.Variable(xavier_init([128, 1])) D_b2 = tf.Variable(tf.zeros(shape=[1])) theta_D = [D_W1, D_W2, D_b1, D_b2] G_W1 = tf.Variable(xavier_init([100, 128])) G_b1 = tf.Variable(tf.zeros(shape=[128])) G_W2 = tf.Variable(xavier_init([128, 784])) G_b2 = tf.Variable(tf.zeros(shape=[784])) theta_G = [G_W1, G_W2, G_b1, G_b2] def generator(z): G_h1 = tf.nn.relu(tf.matmul(z, G_W1) + G_b1) G_log_prob = tf.matmul(G_h1, G_W2) + G_b2 G_prob = tf.nn.sigmoid(G_log_prob) return G_prob def discriminator(x): D_h1 = tf.nn.relu(tf.matmul(x, D_W1) + D_b1) D_logit = tf.matmul(D_h1, D_W2) + D_b2 D_prob = tf.nn.sigmoid(D_logit) return D_prob, D_logit def plot(samples): fig = plt.figure(figsize=(4, 4)) gs = gridspec.GridSpec(4, 4) gs.update(wspace=0.05, hspace=0.05) for i, sample in enumerate(samples): ax = plt.subplot(gs[i]) plt.axis('off') ax.set_xticklabels([]) ax.set_yticklabels([]) ax.set_aspect('equal') plt.imshow(sample.reshape(28, 28), cmap='Greys_r') return fig G_sample = generator(Z) D_real, D_logit_real = discriminator(X) D_fake, D_logit_fake = discriminator(G_sample) # D_loss = -tf.reduce_mean(tf.log(D_real) + tf.log(1. - D_fake)) # G_loss = -tf.reduce_mean(tf.log(D_fake)) # Alternative losses: # ------------------- D_loss_real = tf.reduce_mean( tf.nn.sigmoid_cross_entropy_with_logits(logits=D_logit_real, labels=tf.ones_like(D_logit_real))) D_loss_fake = tf.reduce_mean( tf.nn.sigmoid_cross_entropy_with_logits(logits=D_logit_fake, labels=tf.zeros_like(D_logit_fake))) D_loss = D_loss_real + D_loss_fake G_loss = tf.reduce_mean(tf.nn.sigmoid_cross_entropy_with_logits(logits=D_logit_fake, labels=tf.ones_like(D_logit_fake))) mb_size = 128 Z_dim = 100 sess = tf.Session() sess.run(tf.global_variables_initializer()) if not os.path.exists('out/'): os.makedirs('out/') i = 0 for it in range(1000000): if it % 1000 == 0: samples = sess.run(G_sample, feed_dict={Z: sample_Z(16, Z_dim)}) fig = plot(samples) plt.savefig('out/{}.png'.format(str(i).zfill(3)), bbox_inches='tight') i += 1 plt.close(fig) X_mb, _ = mnist.train.next_batch(mb_size) _, D_loss_curr = sess.run([D_solver, D_loss], feed_dict={X: X_mb, Z: sample_Z(mb_size, Z_dim)}) _, G_loss_curr = sess.run([G_solver, G_loss], feed_dict={Z: sample_Z(mb_size, Z_dim)}) if it % 1000 == 0: print('Iter: {}'.format(it)) print('D loss: {:.4}'.format(D_loss_curr)) print('G_loss: {:.4}'.format(G_loss_curr)) print() View Code # Applications of GAN We saw an overview of how these things work and got to know the challenges of training them. We will now see the cutting edge research that has been done using GANs Increasing Resolution of an image • Increasing Resolution of an image : Generate a high resolution photo from a comparatively low resolution. Paper: https://arxiv.org/pdf/1609.04802.pdf Code: https://github.com/tensorlayer/srgan • Interactive Image Generation : Draw simple strokes and let the GAN draw an impressive picture for you! • Image to Image Translation : Generate an image from another image. For example, given on the left, you have labels of a street scene and you can generate a real looking photo with GAN. On the right, you give a simple drawing of a handbag and you get a real looking drawing of a handbag. Paper: https://arxiv.org/pdf/1611.07004.pdf • Text to Image Generation : Just say to your GAN what you want to see and get a realistic photo of the target. Paper : https://arxiv.org/pdf/1605.05396.pdf # Resources Here are some resources which you might find helpful to get more in-depth on GAN ## End Notes Phew! I hope you are now as excited about the future as I was when I first read about GANs. They are set to change what machines can do for us. Think of it – from preparing new recipes of food to creating drawings. The possibilities are endless. In this article, I tried to cover a general overview of GAN and its applications. GAN is very exciting area and that’s why researchers are so excited about building generative models and you can see that new papers on GANs are coming out more frequently. If you have any questions on GANs, please feel free to share them with me through comments. # 其他链接 GAN论文阅读——原始GAN（基本概念及理论推导） GAN 学习代码 GAN 论文合集 posted @ 2020-03-24 22:48  xuanyuyt  阅读(133)  评论(0编辑  收藏

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1. ## equation Hi Everyone, I have an exam coming up the day after tomorrow and I am completely unable to get the correct answer for this question. First off, I'd like to confirm I'm doing the last equation correctly. (apologies, but I'm not sure on how to use the math notation) 17 - Cr(sq) = 17 - 7Cr - 0.98Cr(sq) 17 = 17 - 7Cr + 0.02Cr(sq) 0 = -7Cr + 0.02Cr(sq) -0.02Cr(sq) = -7Cr -0.02Cr = -7 Cr = 350 Is this correct? Thank you, Andrew 2. Originally Posted by mcnandy Hi Everyone, I have an exam coming up the day after tomorrow and I am completely unable to get the correct answer for this question. First off, I'd like to confirm I'm doing the last equation correctly. (apologies, but I'm not sure on how to use the math notation) 17 - Cr(sq) = 17 - 7Cr - 0.98Cr(sq) 17 = 17 - 7Cr + 0.02Cr(sq) 0 = -7Cr + 0.02Cr(sq) -0.02Cr(sq) = -7Cr -0.02Cr = -7 Cr = 350 Is this correct? Thank you, Andrew $17-C_{r}^{2}=17-7C_r-0.98C_r^2$ $\implies 0.02C_r^2-7C_r=0$ $\implies C_r(.02C_r-7)=0$ $\implies \color{red}\boxed{C_r=0}$ or $\color{red}\boxed{C_r=350}$ Remember that 0 may be a solution as well! Hope this helped you out! 3. Thanks, It's confirmed that i'm doing something semi-right at least Now, there's also the issue of the vector addition, which I must be wrong. (4i - j) + Cr(5/5.66i + 4/5.66j) = (4 - 0.7Cr)i + (-1 - 0.7Cr)j Now, removing the i's and j's and squaring both of the sets of brackets, i.e (4 - 0.7Cr)(4 - 0.7Cr) = 16 - 5.6Cr - 0.49Cr(sq) (-1 - 0.7Cr)(-1 - 0.7Cr) = 1 - 1.4Cr - 0.49Cr(sq) adding the two together and what not: 17 - 7Cr - 0.98Cr(sq) (the right hand side of the above equation) Am I doing this correctly? If so, I can't see where I'm going wrong as the step before this is a simple subtraction and I hope to god I'm getting that right... Thank you again, Andrew

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16 # What grinds my gears: IEEE-754 This, to me, seems retarded. Take the value 0.931 for example. Its represented in binary as 00111111011011100101011000000100 See those last three bits? Well, it causes it to come out in decimal like so: 0.93099999~ Which because bankers rounding is nowstandard, that actually works out to 0.930, because with bankers rounding, we round to the nearest even number? Makes sense? No. Anyone asked for it? No (well maybe the banks). Was it even necessary? Fuck no. But did we get it anyway? Yes. And worse, thats not even the most accurate way to represent our value of 0.931 owing to how fucked up rounding now is becaue everything has to be pure shit these days. A better representation would be 00111101101111101010101100110111 <- good 00111111011011100101011000000100 < - shit The new representation works out to 0.093100004 or 0.093100003898143768310546875 when represented internally. Whats this mean? Because of rounding you don't lose accuracy anymore. Am I mistaken, or is IEEE-754 shit? • 3 John D. Cook has a series on this. https://johndcook.com/blog/2009/... • 4 I'm a bit confused: round-to-even is what happens when the result is exactly midway between two possible representations. If something gives you 0.931, the first representation is closer. What am I missing? • 4 Rounding has nothing to do with IEEE-754. IEEE-754 standardizes implementations of floating-point numbers, and it is actually regarded as a very good standard. Rounding happens when you wish to represent a real number to a certain number of decimal digits, regardless of implementation. • 0 Bankers rounding would not do that • 0 @frankot I'm griping against the result when bankers rounding is combined with the IEEE standard. • 0 @kurast How so?

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{[ promptMessage ]} Bookmark it {[ promptMessage ]} # Hw13 - Problem 13.50 Solution This is a physical pendulum... This preview shows pages 1–2. Sign up to view the full content. 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. Unformatted text preview: Problem 13.50 Solution This is a physical pendulum. Since it takes 120s to complete 100 swings. Period is T = 120s / 100 = 1 . 2s Use the formula (14-14), replace h by d , the distance from center of gravity to pivot point, T = 2 π s I mgd We can solve for moment of inertia I with respect to the pivot, I = MgdT 2 4 π 2 Plug in numbers I = . 1286 kg · m 2 . A Wobbling Bridge Solution This is a resonating, driven, damped oscillator. Part A From (14-16) and (14-17), x = Ae- b 2 m t cos( ω t ) Amplitude of damped oscillator decays as e- b 2 m t , plug in t = 6 T , 1 /e = e- b 2 m · 6 T , or b 2 m · 6 T = 1 By T = 2 π/ω , where ω = p k/m is the angular frequency of undriven, undamped oscillator, we have b/m = 1 6 π ω (1) The given formula of driven amplitude can be rewriten as A = F/m q ( ω 2- ω 2 d ) 2 + ( b/m ) 2 ω 2 d (2) where m = 2000kg / m * 144m = 2 . 88 × 10 5 kg is the mass of the oscillator. Maximize this amplitude with respect to ω d , which is equivalent as minimizing the expression inside of square root, by expanding the expression and completing square, ( ω 2- ω 2 d ) 2 + ( b/m ) 2 ω 2 d = ( ω 2 d ) 2- £ 2 ω 2- ( b/m ) 2 / ω 2 d + ( ω 2 ) 2 = ‰ ω 2 d- • ω 2- 1 2 ( b/m ) 2 ‚ 2 + • ( b/m ) 2 ] ω 2- 1 4 ( b/m ) 4 ‚ Therefore, resonating angular frequency is ω 2 d = ω 2- 1 2 ( b/m ) 2 (3)... View Full Document {[ snackBarMessage ]} ### Page1 / 4 Hw13 - Problem 13.50 Solution This is a physical pendulum... 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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# Super Math 5 - SOCCER - Color by Code EDITABLE – 3rd Grade - Common Core Aligned Subject Resource Type Common Core Standards Product Rating File Type Compressed Zip File 28 MB|40 pages Share Product Description With these puzzles your students will have fun while working on essential skills. They will not even realize they are reviewing math! :) Your students will be thrilled with the most famous soccer stars: Ronaldo (Portugal), Messi (Argentina), Landon Donovan (USA), Oswaldo (Mexico), Iniesta (Spain), Neymar (Brazil), Rooney (England), Balotelli (Italy), Franck Ribéry (France), Ozil (Germany). This pack includes 10 puzzles: 01- Add and subtract within 20 with regrouping. 02- Add within 20 with regrouping. 03- Add within 20 with regrouping. 04- Multiply within 200. 05- Add and subtract within 1000 with regrouping. 06- Determine the unknown whole number in a multiplication or division equation relating three whole numbers. 07- Multiply within 100. 08- Multiply within 300. 09- Round whole numbers to the nearest 10. 10- Add within 1000 with regrouping. I include the puzzles in black and white. I also include a full color version of each puzzle. In order to suit everybody´s needs I also include the answer keys in black and white, and in color too. That makes a total of 40 pages! After the children have colored the puzzles, you can cut out the picture portion of each printable to decorate your classroom :) I also include an EDITABLE version of the puzzles so you can customize them with your own text: Your students can color by number, color by sight word, color by… you decide! The coloring pages can be used with whatever content area you are working on, so you decide what goes in them! :) *** For the editable pages you need "Microsoft Office Power Point". You can also use "Open Office" which is free: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ DO YOU LIKE SAVING MONEY? THIS MATERIAL IS INCLUDED IN MY “SUPER MATH GROWING BUNDLE”: ***You can check my Pinterest Boards for more cool materials! MATH EDUCATIVE MATERIALS ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ This product is the work of "Spanish Teacher". It is intended to support the implementation of the CCSS. No approval by, nor association with, the creators of the CCSS is intended or implied. "The Common Core Standards were written and developed by The National Governors Association Center for Best Practices and Council of Chief State School Officers. Â © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved." Total Pages 40 pages Included Teaching Duration 2 Weeks Report this Resource \$3.00 List Price: \$4.00 You Save: \$1.00 More products from Spanish Teacher Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.

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Shooter burning motors...but why? So… we have a shooter powered by a 550 motor. (We have destroyed both a FP and AM motor.) The motor is geared down about 7:1 using parts from a Modulox lunchbox, and then powers 2 8" dia wheels on a hex shaft. We have destroyed 2 motors after only a couple shots each, and cant figure out why. Any ideas? What kind of bearings are you using and are they aligned with the shafts properly. Bearing to shaft alignment is critical for reducing the load that the motor sees when operating. It’s easy to figure out why, if you gather some data: What is the free speed (i.e. steady-state speed with no balls being launched) of your shooter wheel? If you don’t know how, or don’t have the equipment to measure the speed of a spinning wheel, then what is the current being drawn by the motor? At that free speed, what %PWM are you commanding? You’ve already given the 7:1 geardown ratio, and that’s a good start. Just make sure that’s the right number and includes everything all the way from the motor rpm to the wheel rpm. Are the motors hot when they fail? If you have the ability to measure rpm, this can be a really good diagnostic tool. Run your shooter at “free speed”, full power. Use the gear ratio to calculate the actual rpm of the motor, and compare it to the motor data sheet. If you are running at 50% the motors free speed, then you are also running near full power for the motor which will cause it to get hot fairly quickly. If you are running below 50% free speed, you are in the death zone for those motors. You will need to figure out how to free up your shooter mechanisms to reduce friction. You will want to get the free speed of the assembly at or near the calculated free speed. With 7:1, free speed of that assembly relative to motor free speed should be approximately 2100 rpm. If you are around 1100 rpm, then you are running at or near peak power. Below 1100 rpm is certain death. At around 1500 and above, you stand a pretty good chance of the motor living. If you do not have a way of measuring rpm, remove just the motor and try turning the assembly. It should spin relatively freely. If not, check for bound bearings, shafts, or other areas of frictional losses. Also, do you have a way to measure current? Current will tell you a lot about how much power is going in. If you an measure those values and post, someone here may be able to help diagnose further. Pictures would be helpful as well. Also make sure the AIR VENTS on the 550 ARE NOT BLOCKED. With a 7:1 reduction, it should be very easy to spin the wheels by hand - if not, then there is some mechanical binding that needs to be fixed. I think everything is assembled properly. It spun nice and smooth (before the motor bit the dust). We have been hooking the motor directly to the battery for testing so I would guess thats 100% power. Unfortunately I don’t have anything to measure RPM with. I will try to measure current in an hour or 2 when I’m in the room with the robot. The AIR VENTS are the four slots on the mounting face of the motor, around the bearing where the shaft comes out of the motor. You should probably be using the control system so you can limit the electric power to the motor…that would be the first thing to do before you burn up another motor Measure the current to the motor. Knowing the current & voltage going to the motor & the motor curve you can get a good guess of speed. Free running at anything more the %70 of stall torque amps is cause for concern. Maybe you have too much compression on the ball? When the ball hits the shooter does it stall the motor & it is inertia loading that is killing the motor? The FP are pretty durable. They take a lot of abuse. Holding them full power at stall for any length of time would be abusive. If you run thru the PD board, the CB will give you a little protection from burning up the motor. I will definitely check the air vents also. Running at anything more than 50% of stall torque amps is a cause for concern. Or possibly even less. Could anyone point me in the direction to find the motor specs documentation? A link would be great. We found, (through a burned hand), that all motors will get hot, so its a good idea to have a Spike or use a speed controller to, overall, turn off motors when they don’t need to be used. We burned up a 550 shooter motor and found the cause to be a bad bearing. or here: http://www.chiefdelphi.com/media/papers/2432 Running at anything more than 50% of stall torque amps is a cause for concern. Or possibly even less. You are right. I was actually thinking of 30% stall torque amps which would be about 70% free speed. We burned up a couple because we blocked the air vents. Good ole CD search turned that one up quick. We are mounting our motors in the sideplate of our shooter, a couple holes in the sideplate inline with the vents and problem solved. At least the banebots aren’t as much as the tetrix motors from last year. Turns out we were covering he vents. Works great now. Thanks everyone. If you have time to pick up some hobby motor heat sinks from towerhobbies.com, there are black ones that fit the 550 series diameter perfectly. I think we had to cut them to fit. http://www3.towerhobbies.com/cgi-bin/wti0001p?&I=LXNAB1&P=ML Pretty much all failures with BaneBot motors are attributable to excessive heat buildup. I am working on an FRC MTBF plots for the BaneBots using our motors, understanding there are various uses and setups. If I have time, I’d like to put a call for data from the community. Another thing to watch out for is case shorts. Once the motors heat up too much, the enamel used for sealing the shaft melts off and then the current is allowed to flow from each terminal to the case of the motor (and then your robot). Make sure you consider mounting with nylon screws and washers or other non-conductive hardware. Please note that a case short does not indicate the motor is gone. We’ve run case short motors for a reasonably long duration before they finally failed.

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# CALCULUS HELP ! posted by . f(x)=(-3x^3-x^2-9x-8)/(6x^2+4x+3. Find the equation of the non-vertical asymptote. y= Does f(x) intersect its non-vertical asymptote ? What is the smallest value of x at which f(c) intersects its non-vertical asymptote ? please show all the work. I did this and got the wrong answer. I want to know where I made a mistake. Thank you • CALCULUS HELP ! - By long division I got (-3x^3-x^2-9x-8)/(6x^2+4x+3) = (-1/2)x+1/6 + ( (-49/6)x - 17/2 )/(6x^2 + 4x + 3 As x becomes ± large, the last term ----> 0 and the asymptote is y = (-1/2)x + 1/6 The denominator can never be zero, solving 6x^2 + 4x + 3 = 0 produces complex roots. Thus the function is continuous, and can never cross its asymptote. • CALCULUS HELP ! - The value of c we find solving the equation (-49/6)x-17/2=0

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# The speed of light and Heisenburg 1. Oct 20, 2005 ### CJames I have a question to ask. I am pursuing a career in theoretical physics, most likely LQG and/or cosmology, but thus far my mathematical knowledge spans only newtonian physics. My question is in relation to the Heisenburgh uncertainty principle. Now I know that there are several factors other than position and velocity, but I of course have only read the popular texts and so I only understand the uncertainty principle in this context. That is, the more precise a position you want, the more energy you must invest and thus the more the velocity has been changed. And so on. I guess my two questions are this: 1. How is it possible for us to know that photons always travel at c? Why doesn't this violate the uncertainty principle? 2. How many properties of quanta are effected by the uncertainty principle and why? For example, spin does not change, correct? Niether should charge. (I am unfamilair with quantum numbers so try to be specific. How many properties do quanta have?) I guess that is more than two questions but they have bee bothering me. 2. Oct 20, 2005 ### Staff: Mentor To what does the Heisenberg Uncertainty Principle apply? It doesn't apply to everthing. http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html#c1 We have not observed any photons that do not travel at the speed of light. The proper question is "How many properties or attributes are quantized. Charge is quantized. Particles are quantized, e.g. as far as we know, all electrons have the same rest mass, all protons have the same rest mass, and so on for various types of particles. Photons are quantized, but there seems to be a continuous spectrum, although photons generally originate with a specific energy related to atomic or nuclear structure. Last edited: Oct 20, 2005 3. Oct 20, 2005 ### Pengwuino Why would it? The scales of the heisenberg uncertainty principle relates to things at a rather small scale while the value of c is tremendous. 4. Oct 20, 2005 ### Gokul43201 Staff Emeritus That makes no sense to me. There certainly is an uncertainty relation for light (photons). Only, with a photon, you change its momentum by changing its frequency (or wavelength). For a photon, p = E/c = hf/c, so the uncertainty in measuring the position of photons is related to the uncertainty in knowing their frequency. Last edited: Oct 20, 2005 5. Oct 20, 2005 ### Pengwuino haha yah i know... i was hopen people would realize what the implications were. I figured that his idea was that if you weren't certain about a photon's position, how could you determine its velocity using a dx/dt kinda experiment. Well if you're not absolutely sure what the change in the positions are (which results from you not being certain where the photon actually is in the final and initial position), you can't be sure what the speed of the photon is. What I was saying is that since the speed of light is sooooooo fast, miniscule deviations based on the uncertainty principle won't have any effect until you get to such a large # of digits. Guess I was kinda stretching it as far as what I thought he was thinking. 6. Oct 20, 2005 ### NateTG We don't. James Maxwell Clark worked out a theory that related electricity and magnetism to light which suggests that the speed of light is a 'fundemental'. I'm not a specialist, but I believe that the classical notions of electro-magnetism break down at small scales so it would not surprise me if 'the speed of light' were also different at those scales. Initially, the prediction that the speed of light was a constant seemed quite incredulus to people, and extensive experimentation did take place to verify that, if it changed, it only changed very little. 7. Oct 20, 2005 ### ComputerGeek all the experiments that have slowed light down or sped light up do not show that we have observed light traveling at different speeds? 8. Oct 20, 2005 ### ZapperZ Staff Emeritus Be very careful here. You need to know that they measured the group velocity (or the speed of the pulse shape) in these experiments, and you need to know what a "group velocity" is. Even in the "fast light" experiments, no part of the pulse travelled faster than c (re: the NEC experiment). Zz. 9. Oct 25, 2005 ### CJames Sorry for taking so long to respond. I used to visit this site a lot but now I don't have that kind of free time. Anyway, is this what you are saying: That it is momentum and position I should be thinking about, not velocity and position. That does seem to make more sense as it is more general. Here is the touchy-feely explanation I am getting. Let me know if I am on track. For a photon of large wavelength you can more precisely measure its momentum because your error in relation to the size of the wavelength will be small, but your measurement of its position will be less precise because its position could be anywhere within one wavelength. Conversely, for a photon of high frequency your measurement of wavelength will be less precise because your error will be larger in comparison to the size of the wave, but your measurement of position will be more precise because you know it is somewhere within that small wavelength. Is this accurate or am I confusing myself? 10. Oct 25, 2005 ### CJames Does this mean that the uncertainty principle only applies to values expressable as units of (m^2*kg)/(s), that is, distance squared times mass over time? Or does it mean that it literally only applies to delta (x*p) and delta (E*t)? I realize that is a stupid question. Oh well. EDIT: Thanks for the link by the way, it is very usefull.

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# JAM Question Paper 2008 CA (Computer Applications) JAM Question Paper 2008 CA (Computer Applications). Joint Admission Test (JAM) This JAM CA – Computer Applications 2008 examination is the procedure to get the Admission to Integrated Ph.D. Programmes at Indian Institute of Science, Bangalore and M.Sc. (Two Year), Joint M.Sc.-Ph.D., M.Sc.-Ph.D. Dual Degree and other Post-Bachelor’s Degree Programmes at Indian Institutes of Technology JAM CA – Computer Applications 2008 Question Paper test papers will be fully objective type.  This JAM CA – Computer Applications 2008 Question will help all the students for their exam preparation, here the question type is MCQ i.e multiple choice question answers, if this JAM CA – Computer Applications 2008 question paper in pdf file for IIT JAM CA – Computer Applications you can download it in FREE, if Joint Entrance Examination (JAM) 2008 paper in text for JAM you can download JAM 2008 page also just Go to menu bar, Click on File->then Save. ## JAM Question Paper 2008 CA (Computer Applications) JAM CA – Computer Applications 2008 Question paper Free Download PDF is available in www.oldquestionpapers.net which has been provided by many students this Joint Admission Test i.e. JAM  CA – Computer Applications  2008 paper is available for all the students in FREE and also IITJAM Computer Applications 2008 question paper fully solved i.e. with answer keys and solution. You can get daily updates on JAM  CA – Computer Applications 2008 from www.oldquestionpapers.net here  you can also check similar links for other related study materials on  JAM Computer Applications Entrance 2008 exam. Here JAM 2008 Computer Applications question is also available in English. G is a group of order 51. Then which one of the following statements is false? (A) All proper subgroups of G are cyclic. (B) If G has only one subgroup of order 3 and only one subgroup of order 17. then G is cyclic. (C) G must have an element of order I 7. (I)) If G is abelian then there exists no proper sibgroup H of 0 such that product of all elements of H [s identity. Out of 120 students, 80 students have taken mathematics, 60 students have taken physics 40 studcnts have taken chemistry, 30 students have taken both physics and mathematics, 20 students have taken both chemistry and mathematics and 15 students have taken both physics and chemistry. If every student has taken at least one course, then how many students have taken all the three courses? (A)5 (B)25 (C) 15 (D) 10 Which one of the following is a valid JP address on a network? (A) 10.10.256.25 (B) 10.10.25.257 (C) 10.10.25.25 (D) 10.258.25.25 Which of the following connections to the computer is correct for the devices keyboard, mouse and printer respectively. (A) Parallel port, PS2, serial port (B) PS2, serial port, PS2 (C) PS2, PS2, parallel port (D) Serial port., parallel port, PS2 Similar Pages..

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Pattern Matching - PowerPoint PPT Presentation Pattern Matching 1 / 13 Pattern Matching Pattern Matching - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript 1. COMP171 Spring 2009 Pattern Matching 2. Pattern Matching • Given a text string T[0..n-1] and a pattern P[0..m-1], find all occurrences of the pattern within the text. • Example: T = 000010001010001 and P = 0001: • first occurrence starts at T[1]. • second occurrence starts at T[5]. • third occurrence starts at T[11]. 3. Naïve algorithm Worst-case running time = O(nm). 4. Can we do it better? • The naïve algo is O(mn) in the worst case • But we do have linear algorithm (optional): • Boyer-Moore • Knuth-Morris-Pratt • Finite automata • Using idea of ‘hashing’! Robin-Karp algorithm 5. Boyer-Moore Algorithm • Basic idea is simple. • We match the pattern P against substrings in the text string T from right to left. • We align the pattern with the beginning of the text string. Compare the characters starting from the rightmost character of the pattern. If fail, shift the pattern to the right, by how far? 6. Rabin-Karp Algorithm • Key idea: • Think of the pattern P[0..m-1] as a key, transform it into an equivalent integer p. • Similarly, we transform substrings in the text string T[] into integers. • For s=0,1,…,n-m, transform T[s..s+m-1] to an equivalent integer ts. • The pattern occurs at position s if and only if p=ts. • If we compute p and ts quickly, then the pattern matching problem is reduced to comparing p with n-m+1 integers. 7. Rabin-Karp Algorithm • How to compute p? p = 2m-1 P[0] + 2m-2 P[1] + … + 2 P[m-2] + P[m-1] • Using Horner’s rule This takes O(m) time, assuming each arithmetic operation can be done in O(1) time. 8. Rabin-Karp Algorithm • Similarly, to compute the (n-m+1) integers ts from the text string. • This takes O((n – m + 1) m) time, assuming that each arithmetic operation can be done in O(1) time. • This is a bit time-consuming. 9. Rabin-Karp Algorithm • A better method to compute the integers incrementally using previous result: compute offset = 2m Horner’s rule to compute t0 tS-1 tS This takes O(n+m) time, assuming that each arithmetic operation can be done in O(1) time. 10. Problem • The problem with the previous strategy is that when m is large, it is unreasonable to assume that each arithmetic operation can be done in O(1) time. • In fact, given a very long integer, we may not even be able to use the default integer type to represent it. • Therefore, we will use modulo arithmetic. Let q be a prime number so that 2q can be stored in one computer word. • This makes sure that all computations can be done using single-precision arithmetic. 11. Compute equivalent integer for pattern O(m) O(n+m) 12. Once we use the modulo arithmetic, when p=ts for some s, we can no longer be sure that P[0 .. m-1] is equal to T[s .. s+ m -1 ]. • Therefore, after the equality test p = ts, we should compare P[0..m-1] with T[s..s+m-1] character by character to ensure that we really have a match. • So the worst-case running time becomes O(nm), but it avoids a lot of unnecessary string matchings in practice. 13. A spell checkerwith hashing Start by reading in words from a dictionary file named dictionary. The words in this dictionary file will be listed one per line, sorted alphabetically. Store each word in a hash table, using chaining to resolve collisions. Start with a table size of roughly 4K entries (the table size should be prime). If necessary, rehash to a larger table size to keep the load factor less than 1.0. After hashing each word in the dictionary file, read in the user-specified text file and check it for spelling errors by looking up each word in the hash table. A word is defined as a string of letters (possibly containing single quotes), separated by white space and/or punctuation marks. If a word cannot be found in the hash table, it represents a possible misspelling.

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Open In App # SciPy Linear Algebra – SciPy Linalg The SciPy package includes the features of the NumPy package in Python. It uses NumPy arrays as the fundamental data structure. It has all the features included in the linear algebra of the NumPy module and some extended functionality. It consists of a linalg submodule, and there is an overlap in the functionality provided by the SciPy and NumPy submodules. Let’s discuss some methods provided by the module and its functionality with some examples. ## Solving the linear equations The linalg.solve function is used to solve the given linear equations. It is used to evaluate the equations automatically and find the values of the unknown variables. Syntax: scipy.linalg.solve(a, b, sym_pos, lower, overwrite_a, overwrite_b, debug, check_finite, assume_a, transposed) Let’s consider an example where two arrays a and b are taken by the linalg.solve function. Array a contains the coefficients of the unknown variables while Array b contains the right-hand-side value of the linear equation. The linear equation is solved by the function to determine the value of the unknown variables. Suppose the linear equations are: 7x + 2y = 8 4x + 5y = 10 ## Python # Import the required librariesfrom scipy import linalgimport numpy as np  # The function takes two arraysa = np.array([[7, 2], [4, 5]])b = np.array([8, 10])  # Solving the linear equationsres = linalg.solve(a, b)print(res) Output: [0.74074074 1.40740741] ## Calculating the Inverse of a Matrix The scipy.linalg.inv is used to find the inverse of a matrix. Syntax: scipy.linalg.inv(a , overwrite_a , check_finite) Parameters: • a: It is a square matrix. • overwrite_a (Optional): Discard data in the square matrix. • check_finite (Optional): It checks whether the input matrix contains only finite numbers. Returns: • scipy.linalg.inv returns the inverse of the square matrix. Consider an example where an input x is taken by the function scipy.linalg.inv. This input is the square matrix. It returns y, which is the inverse of the matrix x. Let the matrix be – ## Python # Import the required librariesfrom scipy import linalgimport numpy as np  # Initializing the matrixx = np.array([[7, 2], [4, 5]])  # Finding the inverse of# matrix xy = linalg.inv(x)print(y) Output: [[ 0.18518519 -0.07407407] [-0.14814815 0.25925926]] ## Calculating the Pseudo Inverse of a Matrix To evaluate the (Moore-Penrose) pseudo-inverse of a matrix, scipy.linalg.pinv is used. Syntax: scipy.linalg.pinv(a , cond , rcond , return_rank , check_finite) Parameters: • a: It is the Input Matrix. • cond, rcond (Optional): It is the cutoff factor for small singular values. • return_rank (Optional): It returns the effective rank of the matrix if the value is True. • check_finite (Optional): It checks if the input matrix consists of only finite numbers. Returns: • scipy.linalg.pinv returns the pseudo-inverse of the input matrix. Example: The scipy.linalg.pinv takes a matrix x to be pseudo-inverted. It returns the pseudo-inverse of matrix x and the effective rank of the matrix. ## Python # Import the required librariesfrom scipy import linalgimport numpy as np  # Initializing the matrixx = np.array([[8 , 2] , [3 , 5] , [1 , 3]])  # finding the pseudo inverse of matrix xy = linalg.pinv(x)print(y) Output: ## Finding the Determinant of a Matrix The determinant of a square matrix is a value derived arithmetically from the coefficients of the matrix. In the linalg module, we use the linalg.det() function to find the determinant of a matrix. Syntax: scipy.linalg.det(a , overwrite_a , check_finite) Parameters: • a: It is a square matrix. • overwrite_a (Optional): It grants permission to overwrite data in a. • check_finite (Optional): It checks if the input square matrix consists of only finite numbers. Returns: • Floating point value The scipy.linalg.det takes a square matrix A and returns D, the determinant of A. The determinant is a specific property of the linear transformation of a matrix. The determinant of a 2×2 matrix is given by: From the above Python code, the determinant is calculated as: Example: ## Python # Importing the required librariesfrom scipy import linalgimport numpy as np  # Initializing the matrix AA = np.array([[9 , 6] , [4 , 5]])  # Finding the determinant of matrix AD = linalg.det(A)print(D) Output: 21.0 ## Singular Value Decomposition The Singular-Value Decomposition is a matrix decomposition method for reducing a matrix to its constituent parts to make specific subsequent matrix calculations simpler. It is calculated using scipy.linalg.svd. Syntax: scipy.linalg.svd(a , full_matrices , compute_uv , overwrite_a , check_finite , lapack_driver) Parameters: • a: The input matrix. • full_matrices (Optional): If True, the two decomposed unitary matrices of the input matrix are of shape (M, M), (N, N). • compute_uv (Optional): The default value is True. • overwrite_a (Optional): It grants permission to overwrite data in a. • check_finite (Optional): It checks if the input matrix consists of only finite numbers. • lapack_driver (Optional): It takes either the divide-and-conquer approach (‘gesdd’) or general rectangular approach (‘gesvd’). The function scipy.linalg.svd takes a matrix M to decompose and returns: 1. A Unitary matrix having left singular vectors as columns. 2. The singular values sorted in non-increasing order. 3. A unitary matrix having right singular vectors as rows. Example: ## Python # Importing the required librariesfrom scipy import linalgimport numpy as np  # Initializing the matrix MM = np.array([[1 , 5] , [6 , 10]])   # Passing the values to the # eigen functionx , y , z = linalg.svd(M)print(x , y , z) Output: ## Eigenvalues and EigenVectors Let M be an n×n matrix and let X∈Cn be a non-zero vector for which: MX = λX for some scalar λ. λ is called an eigenvalue of the matrix M and X is called an eigenvector of M associated with λ, or a λ-eigenvector of M. Syntax: scipy.linalg.eig(a , b , left , right , overwrite_a , overwrite_b , check_finite , homogeneous_eigvals) Parameters: • a: Input matrix. • b (Optional): It is a right-hand side matrix in a generalized eigenvalue problem. • left, right (Optional): Whether to compute and return left or right eigenvectors respectively. • overwrite_a, overwrite_b (Optional): It grants permission to overwrite data in a and b respectively. • check_finite (Optional): It checks if the input matrix consists of only finite numbers. • homogeneous_eigvals (Optional): It returns the eigenvalues in homogeneous coordinates if the value is True. The function scipy.linalg.eig takes a complex or a real matrix M whose eigenvalues and eigenvectors are to be evaluated. It returns the scalar set of eigenvalues for the matrix. It finds the eigenvalues and the right or left eigenvectors of the matrix. Example: ## Python # Importing the required librariesfrom scipy import linalgimport numpy as np  # Initializing the matrix MM = np.array([[9 , 3] , [2 , 4]])  # Passing the values to the eigen# functionval , vect = linalg.eig(M)  # Display the Eigen values and Eigen# vectorsprint(val)print(vect) Output: ## Calculating the norm To define how close two vectors or matrices are, and to define the convergence of sequences of vectors or matrices, the norm is used. The function scipy.linalg.norm is used to calculate the matrix or vector norm. Syntax: scipy.linalg.norm(a , ord , axis , keepdims , check_finite) Parameters: • a: It is an input array or matrix. • ord (Optional): It is the order of the norm • axis (Optional): It denotes the axes. • keepdims (Optional): If the value is True, the axes which are normed over are left in the output as dimensions with size=1. • check_finite (Optional): It checks if the input matrix consists of only finite numbers. Returns: • scipy.linalg.norm returns the norm of a. The function scipy.linalg.norm returns one of the seven different matrix norms or one of an infinite number of vector norms. 1. The L2 norm evaluates the distance of the vector coordinate from the origin of the vector space. It is also known as the Euclidean norm as it is calculated as the Euclidean distance from the origin. The result is a positive distance value. 2. The L1 norm is evaluated as the sum of the absolute vector values. It is an evaluation of the Manhattan distance from the origin of the vector space. Example: ## Python # Importing the required librariesfrom scipy import linalgimport numpy as np  # Initializing the input array x = np.array([6 , 3])  # Calculating the L2 norma = linalg.norm(x)  # Calculating the L1 normb = linalg.norm(x , 1)  # Displaying the norm valuesprint(a)print(b) Output: Example: ## Python # Importing the required librariesfrom scipy import linalgimport numpy as np  # Initializing the matrix x = np.array([[16 , 4] , [100 , 25]])  # Calculate and print the matrix # square rootr = linalg.sqrtm(x)print(r)print("\n")  # Calculate and print the matrix # exponentiale = linalg.expm(x)print(e)print("\n")  # Calculate and print the matrix# sines = linalg.sinm(x)print(s)print("\n")  # Calculate and print the matrix # cosinec = linalg.cosm(x)print(c)print("\n")  # Calculate and print the matrix # tangentt = linalg.tanm(x)print(t) Output:

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Last updated: # Triangle Side Calculator How to find the missing side of a triangleFinding the missing sides of a triangle - An exampleHow to use the triangle side calculatorOther triangle calculatorsFAQs Omni's triangle side calculator allows you to calculate the length of the sides of a triangle. Continue reading to find out how to calculate the sides of a triangle for three different cases: • If two sides and one angle are known; • If two angles and one side are known; and • If two sides and the perimeter are known. You will also find an example of using the calculator to find the missing sides of a triangle. ## How to find the missing side of a triangle There are several methods to find the missing side of a triangle. Depending upon the type of given information, you can choose one of the following: If two sides and one angle are known If we know, the measurements of the two sides and the opposite angle to one of them, we can use the Law of Sines to find the missing side. $\footnotesize \frac{a}{\text{sin} (\alpha)} = \frac{b}{\text{sin} (\beta)}=\frac{c}{\text{sin} (\gamma)} \\$ We can use any pair of the above ratios to solve for the third side. On the other hand, if we know two sides and the angle between them, we can use the Law of Cosines to find the remaining side. $\footnotesize a^2 = b^2 + c^2 - 2bc\ \text{cos} (\alpha)\\ \\ b^2 = a^2 + c^2 - 2ac\ \text{cos} (\beta)\\ \\ c^2 = a^2 + b^2 - 2ab\ \text{cos} (\gamma)$ If two angles and one side are known We know that the sum of the three angles of a triangle is $180\degree$. Hence if we know the measurement of the two angles, we can find the third angle using the equation: $\footnotesize \alpha + \beta + \gamma = 180\degree \\$ Once we know the third angle, we can easily calculate the missing side using the Law of Sines mentioned above. If two sides and the perimeter are known This is the most simple scenario. As we know that the perimeter of a triangle is just the sum of its three sides, i.e., $\scriptsize \text{Perimeter} = a + b +c$ This means that if we know the measurement of two sides of a triangle (say $a$ and $b$), we can calculate the length of the third side as: $\scriptsize c = \text{Perimeter} - (a + b)$ In the next section, we will see an example of how to calculate the sides of a triangle. ## Finding the missing sides of a triangle - An example Let us consider a triangle where $\beta = 45\degree$, $\gamma = 30 \degree$ and $b=10$. To calculate the triangle side lengths, we will proceed as follows: 1. Using the angle sum property of a triangle, we will solve for the third angle: \quad \footnotesize \quad \begin{align*} \alpha & = 180\degree - (\alpha + \gamma)\\ & = 180\degree - (45\degree + 30 \degree)\\ & = 105\degree \end{align*} 1. Now, we will use the Law of Sines to find the missing sides \quad \footnotesize \quad \begin{align*} \frac{a}{\text{sin} (\alpha)} &= \frac{b}{\text{sin} (\beta)} \\ \implies a & = \frac{b}{\text{sin} (\beta)} \times \text{sin} (\alpha) \\ & = \frac{10}{\text{sin} (45\degree)} \times \text{sin} (105\degree) \\ & = 13.66 \end{align*} \quad \footnotesize \quad \begin{align*} \frac{b}{\text{sin} (\beta)} &= \frac{c}{\text{sin} (\gamma)} \\ \implies c & = \frac{b}{\text{sin} (\beta)} \times \text{sin} (\gamma) \\ & = \frac{10}{\text{sin} (45\degree)} \times \text{sin} (30\degree) \\ & = 7.071 \end{align*} ## How to use the triangle side calculator Now let us see how we can use our calculator to find the missing side of a triangle. We will use the same example as in the previous section. 1. In the Given field, choose the option side and 2 angles. 2. Now choose the option two angles + adjacent side (AAS). 3. Enter the dimension of angles ($\beta = 45\degree$ and $\gamma = 30 \degree$) and side ($b=10$). 4. The triangle side calculator will display the third angle ($\alpha = 105\degree$) and the two sides ($a = 13.66$ and $c = 7.071$). ## Other triangle calculators Make sure to check out our other calculators related to triangles: FAQs ### How to find the third side of a triangle if its two side and perimeter are given? To find the third side of a triangle, if its two sides and perimeter are given, follow the given instructions: 1. Find the sum of the two sides, a and b. 2. Subtract the sum a + b from the perimeter P of the triangle. 3. You will get the third side, c as c = P - (a + b). ### Calculate the triangle side lengths if two of its angles are 60° each and one of the sides is 10 cm. The length of each side is 10 cm. Since two of the angles are 60° each, the third angle will be 180° - (60° + 60°) = 60°. As all the three angles are equal, the triangle is an equilateral triangle. Hence, all the sides will be of the same length, i.e., 10 cm.

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471,073 Members | 1,318 Online # undefined NaN, surely not!! Hello all I am having a small problem wih a javascript, could anyone help me 1. first of all this code requires a form f01 and textarea t01 <form name ="f01"> <textarea name="t01" rows="80" cols="50"> </textarea></form> 2. the javascript is as follows: ---------------------- function point(x,y,z){ this.x = x; this.y = y; this.z = z; } function mkPoints(){ tmp01 = new point(-50,50,-50); tmp02 = new point(50,50,-50); tmp03 = new point(-50,-50,-50); tmp04 = new point(50,-50,-50); tmp05 = new point(-50,50,50); tmp06 = new point(50,50,50); tmp07 = new point( -50,-50,50); tmp08 = new point( 50, -50 ,50 ); } function rotate(xrot,yrot){ for (i=1;i<=8;i++){ document.all["tmp_zpos0"+i] = (eval("tmp0"+i+".z") * Math.cos(yrot)) - (eval("tmp0"+i+".x") * Math.sin(yrot)) document.all["tmp0"+i+".x"] = (eval("tmp0"+i+".z")*Math.sin(yrot))+(eval("tmp0"+ i+".x")*Math.cos(yrot)); document.all["tmp0"+i+".z"] = (eval("tmp0"+i+".y")*Math.sin(xrot))+(eval("docume nt.all.tmp_zpos0"+i)*Math.cos(xrot)); document.all["tmp0"+i+".y"] = (eval("tmp0"+i+".y")*Math.cos(xrot))-(document.all["tmp_zpos0"+i]*Math.sin(xrot)); document.f01.t01.value = document.f01.t01.value + document.all["tmp0"+i+".y"] + "\n" } setTimeout("rotate(0.02,0.01)", 1000); } function runAll (){ mkPoints() rotate(0.02,0.01) } -------- The problem is as follows - I am baffled as to why it appears to retun Nan or undefined when the values are clearly are there. Thank you oliverh Jul 20 '05 #1 4 4610 1. first of all this code requires a form f01 and textarea t01 <form name ="f01"> <textarea name="t01" rows="80" cols="50"> </textarea></form> 2. the javascript is as follows: ---------------------- function point(x,y,z){ this.x = x; this.y = y; this.z = z; } function mkPoints(){ tmp01 = new point(-50,50,-50); tmp02 = new point(50,50,-50); tmp03 = new point(-50,-50,-50); tmp04 = new point(50,-50,-50); tmp05 = new point(-50,50,50); tmp06 = new point(50,50,50); tmp07 = new point( -50,-50,50); tmp08 = new point( 50, -50 ,50 ); I would make an array of these points instead of eight global variables. } function rotate(xrot,yrot){ for (i=1;i<=8;i++){ document.all["tmp_zpos0"+i] = (eval("tmp0"+i+".z") * Math.cos(yrot)) - (eval("tmp0"+i+".x") * Math.sin(yrot)) You never need eval for things like this. It is even much faster to do it without. To access the global variabel "tmp0"+i, you can just write window["tmp0"+i] You use document.all, which is not as widely applicable as document.getElementById. I think you just wanted to use "eval" anyway. You assign directly to the result of document.all[...]. That is most likely not what you want. You don't mention an HTML element called "tmp_zpos01", so I guess you meant to write eval("tmp_zpos0"+i) = ... where tmp_zpos01 is a variable? In that case, you only need one variable, not one for each point. I would write this line as var tmp_zpos = window["tmp0"+i].z * Math.cos(yrot) - window["tmp0"+i].x * Math.sin(yrot); document.all["tmp0"+i+".x"] = (eval("tmp0"+i+".z")*Math.sin(yrot))+(eval("tmp0"+ i+".x")*Math.cos(yrot)); window["tmp0"+i].x = window["tmp0"+i].z*Math.sin(yrot) + window["tmp0"+i].x*Math.cos(yrot); document.all["tmp0"+i+".z"] = (eval("tmp0"+i+".y")*Math.sin(xrot))+(eval("docume nt.all.tmp_zpos0"+i)*Math.cos(xrot)); window["tmp0"+i].z = window["tmp0"+i].y*Math.sin(xrot) + tmp_zpos*Math.cos(xrot); document.all["tmp0"+i+".y"] = (eval("tmp0"+i+".y")*Math.cos(xrot))-(document.all["tmp_zpos0"+i]*Math.sin(xrot)); window["tmp0"+i].y = window["tmp0"+i].y*Math.cos(xrot) + tmp_zpos*Math.sin(xrot); document.f01.t01.value = document.f01.t01.value + document.all["tmp0"+i+".y"] + "\n" Here you use document.all again, where you definitly just meant to use eval (and you shouldn't use eval either). document.forms['f01'].elements['t01'].value += window["tmp0"+i].y+"\n"; The problem is as follows - I am baffled as to why it appears to retun Nan or undefined *What* returns NaN or undefined? There are not return statements in the code. I guess it is one of the document.all's that return undefined because you ask for an element that isn't there. when the values are clearly are there. Are they. Have you tested that the values are correct (e.g. by inserting /L -- Lasse Reichstein Nielsen - lr*@hotpop.com Art D'HTML: <URL:http://www.infimum.dk/HTML/randomArtSplit.html> 'Faith without judgement merely degrades the spirit divine.' Jul 20 '05 #2 On 12 Oct 2003 16:13:33 -0700 Hello all I am having a small problem wih a javascript, could anyone help me 1. first of all this code requires a form f01 and textarea t01 <form name ="f01"> <textarea name="t01" rows="80" cols="50"> </textarea></form> 2. the javascript is as follows: ---------------------- function point(x,y,z){ this.x = x; this.y = y; this.z = z; } function mkPoints(){ tmp01 = new point(-50,50,-50); tmp02 = new point(50,50,-50); tmp03 = new point(-50,-50,-50); tmp04 = new point(50,-50,-50); tmp05 = new point(-50,50,50); tmp06 = new point(50,50,50); tmp07 = new point( -50,-50,50); tmp08 = new point( 50, -50 ,50 ); } function rotate(xrot,yrot){ for (i=1;i<=8;i++){ document.all["tmp_zpos0"+i] = (eval("tmp0"+i+".z") * Math.cos(yrot)) - (eval("tmp0"+i+".x") * Math.sin(yrot)) document.all["tmp0"+i+".x"] = (eval("tmp0"+i+".z")*Math.sin(yrot))+(eval("tmp0"+ i+".x")*Math.cos(yr ot)); document.all["tmp0"+i+".z"] = (eval("tmp0"+i+".y")*Math.sin(xrot))+(eval("docume nt.all.tmp_zpos0"+i )*Math.cos(xrot)); document.all["tmp0"+i+".y"] = (eval("tmp0"+i+".y")*Math.cos(xrot))-(document.all["tmp_zpos0"+i]*Mat h.sin(xrot)); document.f01.t01.value = document.f01.t01.value + document.all["tmp0"+i+".y"] + "\n" } setTimeout("rotate(0.02,0.01)", 1000); } function runAll (){ mkPoints() rotate(0.02,0.01) } -------- The problem is as follows - I am baffled as to why it appears to retun Nan or undefined when the values are clearly are there. I'm just a newbie in javascript but I see no elements defined to match -- Life is an offensive, directed against the repetitious mechanism of the Universe. Jul 20 '05 #3 Hello all I am having a small problem wih a javascript, could anyone help me 1. first of all this code requires a form f01 and textarea t01 <form name ="f01"> <textarea name="t01" rows="80" cols="50"> </textarea></form> 2. the javascript is as follows: ---------------------- function point(x,y,z){ this.x = x; this.y = y; this.z = z; } function mkPoints(){ tmp01 = new point(-50,50,-50); tmp02 = new point(50,50,-50); tmp03 = new point(-50,-50,-50); tmp04 = new point(50,-50,-50); tmp05 = new point(-50,50,50); tmp06 = new point(50,50,50); tmp07 = new point( -50,-50,50); tmp08 = new point( 50, -50 ,50 ); } function rotate(xrot,yrot){ for (i=1;i<=8;i++){ document.all["tmp_zpos0"+i] = (eval("tmp0"+i+".z") * Math.cos(yrot)) - (eval("tmp0"+i+".x") * Math.sin(yrot)) document.all["tmp0"+i+".x"] = (eval("tmp0"+i+".z")*Math.sin(yrot))+(eval("tmp0" +i+".x")*Math.cos(yrot)); document.all["tmp0"+i+".z"] = (eval("tmp0"+i+".y")*Math.sin(xrot))+(eval("docum ent.all.tmp_zpos0"+i)*Math.cos(xrot)); document.all["tmp0"+i+".y"] = (eval("tmp0"+i+".y")*Math.cos(xrot))-(document.all["tmp_zpos0"+i]*Math.sin(xrot)); document.f01.t01.value = document.f01.t01.value + document.all["tmp0"+i+".y"] + "\n" } setTimeout("rotate(0.02,0.01)", 1000); } function runAll (){ mkPoints() rotate(0.02,0.01) } -------- The problem is as follows - I am baffled as to why it appears to retun Nan or undefined when the values are clearly are there What returns NaN? None of the functions you've shown us return anything. You're using document.all, which is obsolete, and using eval() where it is not necessary, which is not a good practice. Jul 20 '05 #4 Lasse Reichstein Nielsen <lr*@hotpop.com> wrote in message news:<r8**********@hotpop.com>... Hey Lasse, Thank you very much for looking into the problem in such depth. 1. first of all this code requires a form f01 and textarea t01 <form name ="f01"> <textarea name="t01" rows="80" cols="50"> </textarea></form> 2. the javascript is as follows: ---------------------- function point(x,y,z){ this.x = x; this.y = y; this.z = z; } function mkPoints(){ tmp01 = new point(-50,50,-50); tmp02 = new point(50,50,-50); tmp03 = new point(-50,-50,-50); tmp04 = new point(50,-50,-50); tmp05 = new point(-50,50,50); tmp06 = new point(50,50,50); tmp07 = new point( -50,-50,50); tmp08 = new point( 50, -50 ,50 ); I would make an array of these points instead of eight global variables. An array. I will have to look into how to do that. } function rotate(xrot,yrot){ for (i=1;i<=8;i++){ document.all["tmp_zpos0"+i] = (eval("tmp0"+i+".z") * Math.cos(yrot)) - (eval("tmp0"+i+".x") * Math.sin(yrot)) You never need eval for things like this. It is even much faster to do it without. To access the global variabel "tmp0"+i, you can just write window["tmp0"+i] You use document.all, which is not as widely applicable as document.getElementById. I think you just wanted to use "eval" anyway. You assign directly to the result of document.all[...]. That is most likely not what you want. You don't mention an HTML element called "tmp_zpos01", so I guess you meant to write eval("tmp_zpos0"+i) = ... where tmp_zpos01 is a variable? In that case, you only need one variable, not one for each point. Good advice, I cut out the eval("tmp_zpos0"+i) so I re-wrote the following: window.tmpZpos = (window["tmp0"+i].z * Math.cos(yrot)) - (window["tmp0"+i].x * Math.sin(yrot)); window["tmp0"+i+".x"] = (window["tmp0"+i].z * Math.sin(yrot)) + (window["tmp0"+i].x * Math.cos(yrot)); window["tmp0"+i+".z"] = (window["tmp0"+i].y * Math.sin(xrot)) + (window.tmpZpos * Math.cos(xrot)); window["tmp0"+i+".y"] = (window["tmp0"+i].y * Math.cos(xrot)) - (window.tmpZpos * Math.sin(xrot)); works perfectly I would write this line as var tmp_zpos = window["tmp0"+i].z * Math.cos(yrot) - window["tmp0"+i].x * Math.sin(yrot); document.all["tmp0"+i+".x"] = (eval("tmp0"+i+".z")*Math.sin(yrot))+(eval("tmp0"+ i+".x")*Math.cos(yrot)); window["tmp0"+i].x = window["tmp0"+i].z*Math.sin(yrot) + window["tmp0"+i].x*Math.cos(yrot); document.all["tmp0"+i+".z"] = (eval("tmp0"+i+".y")*Math.sin(xrot))+(eval("docume nt.all.tmp_zpos0"+i)*Math.cos(xrot)); window["tmp0"+i].z = window["tmp0"+i].y*Math.sin(xrot) + tmp_zpos*Math.cos(xrot); document.all["tmp0"+i+".y"] = (eval("tmp0"+i+".y")*Math.cos(xrot))-(document.all["tmp_zpos0"+i]*Math.sin(xrot)); window["tmp0"+i].y = window["tmp0"+i].y*Math.cos(xrot) + tmp_zpos*Math.sin(xrot); document.f01.t01.value = document.f01.t01.value + document.all["tmp0"+i+".y"] + "\n" Here you use document.all again, where you definitly just meant to use eval (and you shouldn't use eval either). document.forms['f01'].elements['t01'].value += window["tmp0"+i].y+"\n"; The problem is as follows - I am baffled as to why it appears to retun Nan or undefined *What* returns NaN or undefined? There are not return statements in the code. I guess it is one of the document.all's that return undefined because you ask for an element that isn't there. when the values are clearly are there. Are they. Have you tested that the values are correct (e.g. by inserting /L I was using the textarea t01 to return the results. Thank you again, oliverh Jul 20 '05 #5 ### This discussion thread is closed Replies have been disabled for this discussion.

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## Teacher Resources cool ### Cooling Rate Calculations: William Thomson, more popularly known as Lord Kelvin, determined in the mid-1800’s that the Earth was between 20 and100 million years old.  Kelvin, who was very upset at Darwin’s ideas about change in organisms over time, undertook studies on the age of the Earth hoping to demonstrate that the Earth was not old enough to allow for accumulated change in organisms over time.  His calculations were based on three assumptions (1) the Earth began as a molten, or melted rock; (2) cooling occurred at a uniform, continuous rate; and (3) no heat is generated within the Earth.  While the Big Bang theory and subsequent ideas about the formation of planets suggests that assumption #1 is a good one, we know that assumptions 2 and 3 are not correct.  Melted (igneous) rocks don’t cool at either continuous or uniform rates; any igneous rock with different crystal sizes is proof of that.  Rock is also a very good insulator of heat, and rates of cooling at the Earth’s surface would be drastically slower deep within the Earth.  Despite these flaws, Kelvin’s calculations brought observation, measurement and objectivity to determination of the age of the Earth.  Darwin and his supporters interpreted Kelvin’s results as proof of the antiquity of the Earth and the existence of adequate time for evolution to occur.  Kelvin felt otherwise. Kelvin's Calculations: Kelvin assumed a starting temperature of the Earth of 1200°C to 3870°C. Based on the geothermal gradient (about 1°/foot in deep mines) and an assumption that at molten rocks existed at depth, he determined from the geothermal gradient that the cooling rate of the Earth would be somewhere between 0.00006°/yr and 0.0002°/yr So, for example, the age of Earth = 1200°/x = 0.00006°/year or 3870°/x= 0.0002°/yr, in other words, around 20 million years old.

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5k views ### How can I calculate AUC using Gini coefficient? [duplicate] In the Gini Coefficient's Wikipedia page, it is defined as $G= 1 - \frac{\Sigma_{i=1}^n f(y_i)(S_{i-1}+S_i)}{S_n}$ for discrete variables, where $S_i= \Sigma_{j=1}^i f(y_i)y_i$ and $S_0=0$ ($y$ being ... 6k views ### What is AUC (Area Under the Curve)? [duplicate] I've seen many questions posted about AUC but I'm still struggling to understand. I see this definition for AUC everywhere "The AUC is an estimate of the probability that a classifier will rank a ... 2k views ### draw roc curve on an example of 10 probability scores [duplicate] I'm studying machine learning and find an example question on the book which really confused me. Q: A scoring classifier is evaluated on a test set of 10 examples resulting in the following ... 794 views ### Area under the ROC curve for continuous variable [duplicate] I have the true output as the ratio between 0 and 1. I am trying predict the output using regression. I am supposed to find the area under the ROC curve between the prediction and true values. I am ... 386 views ### How are AUROC scores computed with just two vectors of actual and predicted values as input? [duplicate] In the R package ModelMetrics, the auc score as shown in the documentation takes only two inputs; aucScore <- auc(actual=actuallabels, predicted=predictedlabels) where the inputs are pretty self ... 136 views ### ROC curve interpretation [duplicate] In the context of binary classification how do you interpret ROC curve: more precisely: 1) Why the diagonal stand for a random classifier? Let's imagine a random classifier: each time he ... 112 views ### How to interpret ROC curve? [duplicate] I am currently doing a classification problem for classifying the functional class and non-functional class of peptidase cleavage site. The data on non-functional class (negative class) is highly ... 84 views ### How to plot ROC for LOOCV by hand based on output probabilities returned by SVM? [duplicate] I want to plot ROC for Leave-one-out (LOO) cross-validation (CV). I have the decision values as well as probability values for each class from SVM classifier. I did see the answer to this post; ... 66 views ### Calculating AUC from continuous output [duplicate] Calculating AUC for one threshold in the continuous output is simple: AUC = (TPR - FPR + 1) / 2; What if I want to calculate AUC for multiple thresholds in ... 50 views ### How to “draw” a ROC curve [duplicate] I have read this question but it doesn't have any well-explained answer for this case: I understand the ROC curve in overall, but I'm looking for a step-to-step explanation in order to understand how ... 8 views ### How to understand the mathematics of ROC/AUC in Wikipedia [duplicate] Could someone help explain what happened in the red box? I'm struggling to figure out where the indicator function comes from in this calculation. Thank you~ wiki link 8 views ### What is threshold in ROC curve? [duplicate] Whenever I read about ROC, people say that it is graphical representation of True Positive Rate value and False Positive Rate at various threshold. Whenever I read in detail, people explain that ... 382k views ### What does AUC stand for and what is it? Searched high and low and have not been able to find out what AUC, as in related to prediction, stands for or means. 63k views ### Understanding ROC curve I'm having trouble understanding the ROC curve. Is there any advantage / improvement in area under the ROC curve if I build different models from each unique subset of the training set and use it to ... 30k views ### Why is AUC higher for a classifier that is less accurate than for one that is more accurate? I have two classifiers A: naive Bayesian network B: tree (singly-connected) Bayesian network In terms of accuracy and other measures, A performs comparatively worse than B. However, when I use the R ... 4k views ### Meaning of model calibration A previous post has discussed model discrimination very nicely. The post also briefly discussed calibration: "When evaluating a risk model, calibration is also very important. To examine this, ... 4k views ### How does AUC of ROC equal concordance probability? [duplicate] If this answer on Quora is correct then I think I understand what concordance probability is. However, I also find that this answer on StackExchange provides the formula of concordance probability ... 1k views ### How is a ROCAUC=1.0 possible with imperfect accuracy? [duplicate] I used sklearn to compute roc_auc_score for a dataset of 72 instances. The accuracy was at 97% (2 misclassifications), but the ROC AUC score was 1.0. How is this ... 3k views ### How to compare predictive power of survival models? Frank Harrell describes the concordance (or Somer's D) as not being sensitive enough to compare multiple survival models for their diagnostic ability, and I've observed this in my own work with ... 924 views ### AUC with incomplete ROC curve I am doing an experiments where changing a parameter I am obtaining different number of FalsePositive, FalseNegative... and so on. I am using this parameter tuning as threshold tuning to obtain FPR ... 371 views ### Why does pROC roc work with non-probability predictions? With the pROC package, I can do this: true <- c(1, 1, 1, 0) predicted <- c(0.5, 0.1, 0.6, 0.1) roc(true, predicted) which gives as expected: ... 2k views ### What is the best way to calculate the AUC of a ROC curve? I have a ROC curve for which I'd like to calculate the AUC. I'm getting different values using the trapezoidal and rank-based approaches. What I'm noticing is that the two values actually add to 1.0 ... 861 views ### How can I compare model fitting accuracy between Random Forest and Logistic regression models I am conducting a classification problem. Usually AUC is used to check the effectiveness of a model when performing logistic regression model. However, if I fit the same data set by using Random ... 925 views ### Cox regression with lasso regression Is it possible to perform lasso regression (glmnet with "cox") for variable selection and then conduct Cox regression using ... 732 views ### Does AUC for multiple logistic regression make sense if prediction is not the goal? Does it makes sense to calculate the AUC if I do not want to use my multiple logistic regression model for predictions? I only want to calculate some odds ratios and test if the variables in my model ... 481 views ### Are my data values too small for chi-squared for trend? If so what trend test can I do instead? I'm doing a project about dilated kidneys, these can be arranged into mild, moderate and severe. There are 53 patients in total. I want to see if there is a trend in the severity of dilatation between ... 424 views ### Different visualization of AUC than ROC curve There are multiple interpretations of area under ROC curve. (e.g What does AUC stand for and what is it? ). We also know that AUC is closely related to rank correlation. Are there also different ways ... 378 views ### Distributed AUC calculation (or approximation) I am trying to calculate the ROC AUC for a dataset where I can't fit predictions and labels in memory (10s/100s billions of samples). Is there a way to calculate the AUC in a distributed way or at ... Suppose you have a binary random variable $Y$, and several other random variables $X_1,...,X_p$. Your goal is to "predict $Y$ using $X_1,...,X_p$." So, you go ahead and fit logistic regression, which ...

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# Plane of evidence 1. Mar 27, 2015 ### point in classical mathematics plane is taught as an axiom, I question whether the plane can be a proof, let's start from the assumption that there are $$n ( n>1)$$ line that intersect at a single point in space , from them can be proved that the plane can be used as evidence, .. 2. Mar 27, 2015 ### phinds You can have multiple lines that all intersect at the same point but which are not in the same plane unless of course you want to first posit the existance of a plane and then say that all the lines have to be in that plane and thus they prove that the plane exists but first you would have to posit that that plane exists and then ... 3. Mar 27, 2015 ### Staff: Mentor There are definitions for planes, but maybe that's what you mean. ??? What does this mean? One definition (from Wikipedia) is that a plane "is a flat, two-dimensional surface." The plane can be used as evidence for what? As @phinds has already said, you can have multiple (i.e., more than two) lines that intersect at a common point without the lines determining a plane. Where are you trying to go with this? 4. Mar 28, 2015 ### point 2-plane 1. two line intersect at point A , space 3. straight lines DB , BE , CE , CD The first process - point B slides on the line to point A , straight lines DB (BE , CE , CD ) intersecting space , formed a final surface BDCE Second process - point B slides on the line moving away from point A , straight lines DB (BE , CE , CD ) intersecting space , formed infinitive surface Merging the surface, we get what we now know plane how to get out of this function plane ? , because the Cartesian coordinate system is constant 5. Mar 28, 2015 ### phinds I have no idea what you are talking about. Yes, your figure makes sense. You have some lines in a plane. So what? What is your point? What is your question? 6. Mar 28, 2015 ### HallsofIvy Staff Emeritus It is true that two intersecting lines define a unique plane. Take the cross product of vectors in the directions of the lines, then use that as normal vector to the plane. For example, the two lines x= 3t+ 2, y= 2t- 4, z= t+ 1 and x= t, y= -t- 2, z= t- 1 intersect at (2, -4, 1) (with t= 0 for the first line, t= 2 for the second). A vector in the direction of the first line is <3, 2, 1> and a vector in the direction of the second line is <1, -1, 1>. The cross product of those two vectors is <3, -2, -5>. The plane having that normal vector and containing the point (2, -4, 1) is 3(x- 2)- 2(y+ 4)- 5(z- 1)= 0.

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# Filter low theory pdf pass Home » Alberta » Low pass filter theory pdf ## Alberta - Low Pass Filter Theory Pdf in Alberta Synthesis of a New Class of Reflectionless Filter Prototypes. High- and low-pass RC filters University of Mississippi. Filters Hopefully by this point, you have a pretty good idea of what lters are and what they do. In these notes we will review the di erent types of lters and their characteristics, as well as some of their applications. A very important and related concept is the graphical representation of lters using Bode plots. Details on how to construct and use them are covered in separate notes. Recall The filter used in the above example is a 15KHz low pass (at 44.1KHz sampling rate). The plot below shows 10 output The plot below shows 10 output samples corresponding to 10 frames. Filter transfer functions The ideal low-pass filter, Not physically realizable, Practical low-pass filters, Parameters and properties, Real poles, Butterworth filter, Chebyschev filter, Bessel filter, Comparison of filter responses, pdf file Figure 3 - Chebyshev Low Pass Filter response for 3 dB ripple, orders 1 to 7. Figure 4 shows the PLR of Chebychev and Butterworth п¬Ѓlters for N=3: CHEBYSHEV FILTER THEORY 11 Basic Fourier transform theory states that the linear convolution of two sequences in the time domain is the same as multiplication of two corresponding spectral sequences in the frequency domain. Filtering is in essence the multiplication of the signal spectrum by the frequency domain impulse response of the filter. For an ideal lowpass filter the pass band part of the signal spectrum is Basic Fourier transform theory states that the linear convolution of two sequences in the time domain is the same as multiplication of two corresponding spectral sequences in the frequency domain. Filtering is in essence the multiplication of the signal spectrum by the frequency domain impulse response of the filter. For an ideal lowpass filter the pass band part of the signal spectrum is EEE 194RF_ L17 21 Standard Low-Pass Filter Design • The normalized inductors and capacitors (g 1, g 2, , g N) are denormalized using: and where C Basic Fourier transform theory states that the linear convolution of two sequences in the time domain is the same as multiplication of two corresponding spectral sequences in the frequency domain. Filtering is in essence the multiplication of the signal spectrum by the frequency domain impulse response of the filter. For an ideal lowpass filter the pass band part of the signal spectrum is Passive Filters & Wave Shaping . AC Theory. Module . 8 . Introduction to Passive Filters . Passive filters, often consisting of only two or three Both CR and LC Low pass filters that remove practically ALL frequencies above just a few Hz are used in power supply circuits, where only DC (zero Hz) is required at the output. A low-pass filter has a constant gain (=Vout/Vin) from 0 Hz to a high cut off frequency f H. This cut off frequency is defined as the frequency where the voltage gain is reduced to 0.707, that is at f H the gain is down by 3 dB; after that (f > f H) it decreases as f increases. The frequencies between 0 Hz and f H are called pass band frequencies, whereas the frequencies beyond f H are the so Filter design theory is well established and is beyond the scope of this application note. It is assumed that a п¬Ѓlter is designed according to the desired speciп¬Ѓcations. The desired digital п¬Ѓlters may be designed using either stan-dard techniques or using commonly available digital п¬Ѓlter design software packages. Finite Impulse Response (FIR) п¬Ѓlters have many advantages over IIR 3 . Fig. 4. Complete reflectionless low-pass filter of arbitrary order. below. A) Add an inductor between the input node of the even- mode circuit and the symmetry plane. Some of the impedance terms and section terms used in this article are pictured in the diagram below. Image theory defines quantities in terms of an infinite cascade of two-port sections, and in the case of the filters being discussed, an infinite ladder network of L-sections. Figure 3 - Chebyshev Low Pass Filter response for 3 dB ripple, orders 1 to 7. Figure 4 shows the PLR of Chebychev and Butterworth п¬Ѓlters for N=3: CHEBYSHEV FILTER THEORY 11 Figure 3 - Chebyshev Low Pass Filter response for 3 dB ripple, orders 1 to 7. Figure 4 shows the PLR of Chebychev and Butterworth п¬Ѓlters for N=3: CHEBYSHEV FILTER THEORY 11. Digital Filters with MATLAB • Low sensitivity to quantization effects compared to many IIR п¬Ѓlters FIR п¬Ѓlters have some drawbacks however. The most important is that they can be computationally expensive to implement. Another is that they have a long transient response. It is commonly thought that IIR п¬Ѓl-ters must be used when computational power is at a premium. This is certainly: 1. 12 X Active Filter Department of Physics 2. Ten Little Algorithms Part 2 The Single-Pole Low-Pass Filter

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(206) 949-5683, Top Rated Plumber, Seattle 1. There is 25 extra feet of it in the attic. IT supplys power to the water tank. Is this costing us more money in electriciity (\$10, \$50, \$100 per year)? 2. Can I the same single pole switch you use for 12/2 lighting switch to provide a disconnect for the water heater at the heater? IT WILL be 10-2 going into the switch. 3. Does a 240 volt heater have to be tied in to a 30 amp double pole breaker or can you use a 20 amp breaker? TIA< MoLo 2. The extra wire in the attic is not causing your heater to consume more energy. No you can't use a SP switch to kill power to the water heater. Use a non fused, 2 pole, quick disconnect, if you want something to kill power to the water heater without shutting the breaker. It can be located near the heater. The wire size determines the amperage of the breaker. 12/2 WG would use a 20A breaker 10/2 WG would take a 30A breaker 120V would use a SP breaker and 240V would use a DP breaker. The watt draw of the heater, which is based on the wattage of the elements, would determine the wire size needed. 3. Originally Posted by Cass The extra wire in the attic is not causing your heater to consume more energy. No you can't use a SP switch to kill power to the water heater. Use a non fused, 2 pole, quick disconnect, if you want something to kill power to the water heater without shutting the breaker. It can be located near the heater. The wire size determines the amperage of the breaker. 12/2 WG would use a 20A breaker 10/2 WG would take a 30A breaker 120V would use a SP breaker and 240V would use a DP breaker. The watt draw of the heater, which is based on the wattage of the elements, would determine the wire size needed. Can the non-fused, 2-pole, quick disconnect be rated for 12 and 10 gauge wires? 4. A 4500 Watt water heater requires 4500 Watts /240 Volts = 18.75 Amps. However, the National Electrical Code requires that continuous loads (A household water heater is defined as a continuous load.) may not exceed 80 percent of the capacity of the circuit. Therefore, you must have a circuit that has a capacity of at least 23.44 Amps. Standard circuits in common use are 20 Amp and 30 Amp. Therefore, it should have a 30 Amp breaker and #10 wire. That is not to say that it won't work with #12 wire and a 20 Amp breaker (It is drawing only 18.75 Amps.), but it will be in violation of the code. It would be very dangerous to use a standard light switch with a 240 Volt circuit because someone might think the circuit is turned off by the switch and get zapped when working on it because the other side of the circuit is hot. There are unfused 2-pole disconnects used with Air Conditioners that would work with a water heater. 5. Originally Posted by molo Can the non-fused, 2-pole, quick disconnect be rated for 12 and 10 gauge wires? I don't remember their amp rating but I believe it is over 30A 6. Originally Posted by molo Can the non-fused, 2-pole, quick disconnect be rated for 12 and 10 gauge wires? Disconnected are not rated in wire size, they are rated in amperage. The lugs will have a conductor size range, but this is not the rating of the disconnect. Most small disconnected are either 30A or 60A. Also, yes, a single pole switch can be used to turn off a water heater. It CANNOT however be used as the required service disconnect. 7. is this the dconnect? is this it? this is for my well pump TIA, molo 8. Yes........... . 9. disconnect Only you can tell, because only you can see where the wires are connected. 10. Originally Posted by molo is this it? this is for my well pump Is what it? What is the question? 11. THIS DCONNECT IS FOR MY WELL PUMP, i"m wONDERING IF THIS IS THE KIND i nEED FOR MY WATER TANK tia, MOLO 12. Originally Posted by molo i"m wONDERING IF THIS IS THE KIND i nEED FOR MY WATER TANK Yes? I don't think you'll find one like that but that one will work. A simple little 60A "pull-out" disconnect would be the easiest. 13. Originally Posted by Speedy Petey Also, yes, a single pole switch can be used to turn off a water heater. It CANNOT however be used as the required service disconnect. A single pole switch can not be used to turn off a 240v water heater. If the heater was 120v a quick disconnect should still be used in place of a sp switch, in my NSHO. 14. Cass, please re-read my post. I think I clearly stated it cannot be used as a disconnect. For control purposes a SP switch most certainly can shut off a straight 240v load. Have you ever looked at the way the thermostats inside a water heater work??? They are SP. 15. Were talking about installing a sp switch outside of a W/H to cut 240v service to a W/H. You tell me how a SP switch shuts off both sides of a 240v circuit. Page 1 of 3 123 Last Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

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Portal Forums Links Register FAQ Members List Social Groups Calendar Search Today's Posts Mark Forums Read Log in Thread Tools Search this Thread Display Modes Small Calculation-Need Help 07-30-2009, 09:52 AM #1 Confused about dryer sheets   Join Date: Jul 2009 Posts: 1 Small Calculation-Need Help Hello all, I have a financial question and I hope you can help me. Assume the following situation: The after-tax return required to accumulate £2 million in 18 years beginning with an investable base of £1,235,000 and with annual outflows of £26,000 is 4.427 percent, which when adjusted for the 40 percent tax rate, results in a 7.38 percent pretax return (4.427%/(1 − 0.04) = 7.38%). I understand the Tax adjustment calculation. But how do you derive the 4.427 in the first place? I would really appreciate your help so I can understand the calculation! Many thanks in advance! Cheers, Peter __________________ Join the #1 Early Retirement and Financial Independence Forum Today - It's Totally Free! Are you planning to be financially independent as early as possible so you can live life on your own terms? Discuss successful investing strategies, asset allocation models, tax strategies and other related topics in our online forum community. Our members range from young folks just starting their journey to financial independence, military retirees and even multimillionaires. No matter where you fit in you'll find that Early-Retirement.org is a great community to join. Best of all it's totally FREE! You are currently viewing our boards as a guest so you have limited access to our community. Please take the time to register and you will gain a lot of great new features including; the ability to participate in discussions, network with our members, see fewer ads, upload photographs, create a retirement blog, send private messages and so much, much more! Join Early-Retirment.org For Free - Click Here 07-30-2009, 10:25 AM #2 Give me a museum and I'll fill it. (Picasso) Give me a forum ...   Join Date: Jun 2002 Location: No Country for Old Men Posts: 45,978 Peter, in what part of South Africa is Arizona located? You might have better results at mathhomeworkhotline.com __________________ __________________ Numbers is hard Charter resident of the lumpen slums of cyberspace Retired in 2005 at age 58, no pension 07-30-2009, 11:03 AM #3 Recycles dryer sheets   Join Date: Sep 2008 Posts: 60 I don't have a financial calculator nearby and didn't find one quickly on the Internet, but to get the necssary interest rate I would start with trying these numbers. N = 18 periods (I assume this is yearly compounding, would need to be adjusted if compounding is more frequent.) PV = -1,235,000 FV = 2,000,000 PMT = 26,000 i = _________ 07-30-2009, 11:23 AM #4 Recycles dryer sheets   Join Date: Jan 2009 Posts: 193 If you're just trying to "understand" it, your balance at any given year equals the balance of the previous year, times your rate of return that year, minus 26000. Now calculate that out for each year and you'll see your 2000000 balance after year 18. If there was no annual withdrawal, you can calculate it with: final balance = P(1+r)^y where P is the original amount, r the annual rate of return and y the number of years. Taking out money each year makes the equation really messy though, so it's easier to calculate iteratively like above. 07-31-2009, 02:28 PM #5 Thinks s/he gets paid by the post   Join Date: Jan 2006 Posts: 3,930 No idea why this is in Health and Retirement............ copy of Excel spreadsheet showing yr, begin balance, end balance before withdrawal, end balance after withdrawal, as suggested by K. Rate looks like a little over 4.425% so close to OP's. btw....slight decimal error in formula for BT return (denominator should be (1-0.4)? edit: sorry format did not take well but hopefully you can figure it out. rate0.04425 Begin end b w/d end a w/d1 1235.00 1289.65 1263.65 2 1263.65 1319.57 1293.57 3 1293.57 1350.81 1324.81 4 1324.81 1383.43 1357.43 5 1357.43 1417.49 1391.49 6 1391.49 1453.07 1427.07 7 1427.07 1490.22 1464.22 8 1464.22 1529.01 1503.01 9 1503.01 1569.52 1543.52 10 1543.52 1611.82 1585.82 11 1585.82 1655.99 1629.99 12 1629.99 1702.12 1676.12 13 1676.12 1750.28 1724.28 14 1724.28 1800.58 1774.58 15 1774.58 1853.11 1827.11 16 1827.11 1907.96 1881.96 17 1881.96 1965.23 1939.23 18 1939.23 2025.05 1999.05 Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) Thread Tools Search this Thread Search this Thread: Advanced Search Display Modes Linear Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules Similar Threads Thread Thread Starter Forum Replies Last Post cute fuzzy bunny Other topics 5 09-10-2008 10:52 PM Fttaw FIRE and Money 1 01-22-2007 05:10 PM pinkmali Young Dreamers 7 08-02-2006 04:35 PM jodz FIRE and Money 14 08-31-2004 09:07 PM » Recent Threads » Quick Links -- Early-Retirement v2.0 -- Early-Retirement v1.0 -- Mobile Contact Us - Portal Page - FIRECalc - Sitemap - Community Rules - Terms of Service - Privacy - Top All times are GMT -6. 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https://holooly.com/solutions/determine-the-reactions-and-draw-the-shear-and-bending-moment-diagrams-for-the-three-span-continuous-beam-shown-in-fig-16-6a-by-the-slope-deflection-method/

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Products Rewards from HOLOOLY We are determined to provide the latest solutions related to all subjects FREE of charge! Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program HOLOOLY HOLOOLY TABLES All the data tables that you may search for. HOLOOLY ARABIA For Arabic Users, find a teacher/tutor in your City or country in the Middle East. HOLOOLY TEXTBOOKS Find the Source, Textbook, Solution Manual that you are looking for in 1 click. HOLOOLY HELP DESK Need Help? We got you covered. ## Q. 16.2 Determine the reactions and draw the shear and bending moment diagrams for the three-span continuous beam shown in $Fig. 16.6(a)$ by the slope-deflection method. ## Verified Solution Degrees of Freedom $\theta_B$ and $\theta_C$. Fixed-End Moments $FEM_{AB} = \frac{3(18)^2}{30}= 32.4 k-ft \circlearrowleft$          or       $+32.4 k-ft$ $FEM_{BA} = \frac{3(18)^2}{20}= 48.6 k-ft \circlearrowright$          or        $-48.6 k-ft$ $FEM_{BC} = \frac{3(18)^2}{12}= 81 k-ft \circlearrowleft$            or          $-81 k-ft$ $FEM_{CB} = 81 k-ft \circlearrowright$           or        $+81 k-ft$ $FEM_{CD} = \frac{3(18)^2}{20}= 48.6 k-ft \circlearrowleft$          or        $+48.6 k-ft$ $FEM_{DC} = \frac{3(18)^2}{30}= 32.4 k-ft \circlearrowright$          or       $-32.4 k-ft$ Slope-Deflection Equations Using Eq. ($M_{nf}=\frac{2EI}{L}(2\theta_n + \theta_f – 3\psi) + FEM_{nf}$) for members $AB, BC$, and $CD$, we write $M_{AB}=\frac{2EI}{18}(\theta_B) + 32.4 = 0.111EI \theta_B + 32.4$               (1) $M_{BA}=\frac{2EI}{18}(2\theta_B) – 48.6 = 0.222EI \theta_B – 48.6$               (2) $M_{BC}=\frac{2EI}{18}(2\theta_B+\theta_C) + 81 = 0.222EI \theta_B +0.111EI \theta_C + 81$               (3) $M_{CB}=\frac{2EI}{18}(\theta_B+2\theta_C) – 81 = 0.111EI \theta_B +0.222EI \theta_C – 81$               (4) $M_{CD}=\frac{2EI}{18}(2\theta_C) + 48.6 = 0.222EI \theta_C + 48.6$               (5) $M_{AB}=\frac{2EI}{18}(\theta_C) – 32.4 = 0.111EI \theta_C – 32.4$               (6) Equilibrium Equations See Fig. 16.6(b). $M_{BA} + M_{BC} = 0$                (7) $M_{CB} + M_{CD} = 0$                (8) Joint Rotations By substituting the slope-deflection equations (Eqs. (1) through (6)) into the equilibrium equations (Eqs. (7) and (8)), we obtain $0.444EI\theta_B + 0.111EI\theta_C = -32.4$                 (9) $0.111EI\theta_B + 0.444EI\theta_C = 32.4$                 (10) By solving Eqs. (9) and (10) simultaneously, we determine the values of $EI\theta_B$ and $EI\theta_C$ to be $EI\theta_B = -97.3 k-ft^2$ $EI\theta_C = 97.3 k-ft^2$ Member End Moments To compute the member end moments, we substitute the numerical values of $EI\theta_B$ and $EI\theta_C$ back into the slope-deflection equations (Eqs. (1) through (6)) to obtain $M_{AB} = 0.111(-97.3) + 32.4 = 21.6 k-ft \circlearrowleft$          Ans. $M_{BA} = 0.222(-97.3) – 48.6 = -70.2 k-ft$      or     $70.2 k-ft \circlearrowright$                Ans. $M_{BC} = 0.222(-97.3) + 0.111(97.3) + 81 = 70.2 k-ft \circlearrowleft$         Ans. $M_{CB} = 0.111(-97.3) + 0.222(97.3) – 81$ $= -70.2 k-ft$          or         $70.2 k-ft \circlearrowright$                    Ans. $M_{CD} = 0.222(97.3) + 48.6 = 70.2 k-ft \circlearrowleft$                Ans. $M_{DC} = 0.111(97.3) – 32.4 = -21.6 k-ft$        or      $21.6 k-ft \circlearrowright$          Ans. Note that the numerical values of $M_{BA} , M_{BC} , M_{CB}$, and $M_{CD}$ do satisfy the equilibrium equations (Eqs. (7) and (8)). Member End Shears and Support Reactions See Fig. 16.6(c) and (d). Ans. Equilibrium Check The equilibrium equations check. Shear and Bending Moment Diagrams See Fig. 16.6(e) and (f ).        Ans.

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# Energy - need a proper definition please 1. Mar 7, 2014 ### Jimmy87 Energy -- need a proper definition please Please could someone give me a proper definition of energy. My teacher said in class today that "there is no such thing as different types of energy". He said that although we learn this in introductory physics, a physicist would not agree there are different types and would define it in a much different way. He said something about it being "the ability to do something" which therefore means the notion of having different types of energy isn't quite right. Is he right? From researching the internet I can see that energy is very abstract but could someone give me a proper physics definition of what it is. I found a video clip from professor Brian Cox who says that he defines energy as: "the length of a space-time four vector in the time direction" What does that mean? Thanks for any help! 2. Mar 7, 2014 ### UltrafastPED Energy is an abstraction of what we otherwise call "work". In physics we take work done = force x distance in the direction of the acting force. For mechanical systems this results in the Work-Energy theorem: Kinetic energy is equal to the work done. All other types of energy can be shown to be equivalent to mechanical energy by means of appropriate processes. For example, Joule showed that the energy used to stir a fluid ends up as heat - with a precise conversion factor. Thus we have heat energy. The kinetic theory of heat allows us to build up from the (hidden) kinetic energy (motion) of molecules/atoms to the measurable temperature of large bodies. The history of thermodynamics is bound up with these ideas. The First Law of Thermodynamics thus states that energy cannot be created or destroyed, but only changed in form. The forms are mechanical, thermal, chemical, nuclear, electromagnetic, etc. But physicists have also found that there is no such thing as "pure energy" - energy cannot be found independent of some other physical entity. And one cannot make use of energy except by taking into account "how it is physically present". 3. Mar 7, 2014 ### Jimmy87 Thanks for the answer. I get what you mean by its not a physical thing and therefore can't be isolated but can you still not talk about different types of energy? In class we were going through mechanical, thermal, nuclear as you said and he was saying that its not really true that there are different types of energy? 4. Mar 7, 2014 ### sophiecentaur I should think that may be too advanced for your level. But you can believe what he says, which is that the full formula for the Energy associated with an object involves its Kinetic Energy and its Potential Energy and its Mass. Normally, we talk in terms of the KE formula, the PE formula or, getting cleverer, we say E=mc2. Often, one of the above counts more than the others. Look at this link ( let it wash all over you without worrying about the details). If you search for "energy" on the page, you will see some formulae that Cox refers to. 5. Mar 7, 2014 ### Khashishi I wouldn't say energy is not a physical thing. In Newtonian physics it appears to be just a bookkeeping device, but if we look at special or general relativity, energy is something with direct consequences. There isn't just one definition of energy. It appears in many different subfields of physics with different definitions, and only when we combine the fields, we find this quantity works out to be the same thing. Your typical object has two forms of energy: rest energy and kinetic energy. In special relativity, the total energy is given by $E = \sqrt{m^2 c^4 + p^2 c^2}$ This is approximately equal to $E = m c^2 + \frac{1}{2} m v^2$ The more commonly seen expression $E = m c^2$ is just the rest energy, which you get by setting the velocity $v$ to 0. The kinetic energy is (approximately) $\frac{1}{2} m v^2$. The kinetic energy is the energy in the motion of objects. When multiple objects are interacting, you can convert between rest energy and kinetic energy. One crude way to look at energy is that it is a storage that can be converted into the motion of objects. It's a useful concept because the total energy in a closed system is conserved. So, if you know the energy, you can figure out how much motion is possible. 6. Mar 7, 2014 ### Khashishi For example, a battery contains a very specific amount of energy which is available to be harnessed. If you installed it into a car, the energy tells you how much motion (calculated as $\frac{1}{2} mv^2$) you could put into the car by draining the battery, if the motors were 100% efficient (which they aren't). 7. Mar 7, 2014 ### UltrafastPED I agree with everything you said - except for this first part. There is no "single thing" which represents energy - it is always associated with other physical entities: force fields have sources (charges), heat has something moving (particles), chemical energy is ultimately due to electromagnetic (and atomic/molecular arrangements), and the energy in E=mc^2 comes from the particles that make up the mass. Hence there is no energy independent of "the things"; nor can you isolate something called "energy". That is why I called it an abstraction - this is also how Richard Feynman introduces the topic in his "Lectures on Physics", volume I.

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Space usage Estimation Join Date Sep 2000 Location Chennai, India Posts 865 Space usage Estimation Given a table structure, the database block size and the number of rows it would contain, how can we estimate the space used by such a table ? Is there a formula for this ? Thanks. Join Date Jan 2001 Posts 2,828 Hi Sameer Just wunderin what happened..to the good old pctfree parameter ;-D...and what about block overheads..;-D 1)create the table 2)insert representative rows in the table 3)analyze the table 4)multiple the avg_row_len * num_of_rows from the dba_tables view..then you can extrapolate... search this forum for table sizing excel spread sheet for other techniques..There is also a excel sheet somwhere on this forum for table sizing..which can calculate this for you..I have never used it though.. regards Hrishy 3. This document might help as well Regards 4. Originally posted by hrishy Just wunderin what happened..to the good old pctfree parameter ;-D...and what about block overheads..;-D My reply wasn't complete enough !, I take back the post 5. Junior Member Join Date Jan 2003 Location Tegucigalpa, Honduras Posts 4 This script may be usefull to estimate real-space utilization. set echo off set serveroutput on set verify off accept v_user prompt 'Schema: ' accept v_object prompt 'Object: ' accept v_object_type prompt 'Object Type: ' declare VAR1 number; VAR2 number; VAR3 number; VAR4 number; VAR5 number; VAR6 number; VAR7 number; VAR8 number; vtable varchar2(30); begin dbms_space.unused_space(upper('&v_user'),upper('&v_object'),upper('&v_object_type'),VAR1,VAR2,VAR3,VAR4,VAR5,VAR6,VAR7); VAR2 := (VAR2)/1024/1024; VAR4 := (VAR4)/1024/1024; VAR8 := (VAR2-VAR4); dbms_output.put_line('OBJECT_NAME = '|| upper('&v_object')); dbms_output.put_line('------------------------------'); dbms_output.put_line('TOTAL_BLOCKS = '||VAR1); dbms_output.put_line('TOTAL_MBYTES = '||VAR2); dbms_output.put_line('UNUSED_BLOCKS = '||VAR3); dbms_output.put_line('UNUSED_MBYTES = '||VAR4); dbms_output.put_line('LAST_USED_EXTENT_FILE_ID = '||VAR5); dbms_output.put_line('LAST_USED_EXTENT_BLOCK_ID = '||VAR6); dbms_output.put_line('LAST_USED_BLOCK = '||VAR7); dbms_output.put_line('TOTAL_USED_MBYTES = '||VAR8); end; / set verify on set echo on Regards, 6. You have to know the size of the row in bytes, or at least the avg of it. A VARCHAR2(2000) column in the row could store once 2 bytes and once 150 bytes. You have to load a sample of your data into a test table, get the avg row size and then calculate. The bigger the sample - the better. Hope that helps, clio_usa - OCP 8/8i/9i DBA

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Beetle - Colin Gooch (ZX Computing-June/July 84-Page 51) There are no bugs in this game! Up to four players can take part in Colin Gooch's realistic dice game. Many traditional games have made their way from boards, cards, pen and paper into the computer screen, some successfully and some not. This is one of the successful transplants. For those of you who have missed the delights of this game in it's original incarnation, the idea is to throw a dice and build up a "beetle" according to the numbers thrown. The computer displays teh cards and the stages of the Beetle as it is drawn and a clever routine is used to simulate dice throwing. Instead of just saying "You threw 3", or whatever (a very suspicious format to non-computer - expert players) a dice is displayed which "rolls" until the player removes his finger from the key. This gives a realistic feel to the game. When less than four players take part then the computer joins in. It is a mean opponent, and I can't spot any sneaky programming which gives it an ad- vantage! You must have a 6 to start and this draws the body, 5 draws the head and ot- her numbers provide antenna, legs, eyes, etc. As you will gather, this is a game where luck is the important factor, but it makes a nice relaxing change from saving the world from aliens, or wrestling with some logical computer generated problem. VARIABLES USED A\$,B\$,C\$ Dice spots. D\$,E\$ Dice frame. P\$(N) Players' names. S\$ Spectrum control, Play or Not. N\$,Z\$ Temporary variables (line inputs) A(X) Accumulated totals. B(X,Y) Beetles. Count Control variable in dice. D Dice value. FF Dice control variable. N FOR/NEXT loops. PL Number of players. T Turn number. V(X) Value of score. X,Y,P,I Variables to set up title page. THE LINES 1000-1020 Introduction, initialisation and inputs. 1030-1060 Main control loop, including checks. 1180-1320 Endgame routine. 1340-1600 Subroutines to print out beetle parts. 1610-1810 Start - get inputs on player details. 1820-1960 Initialise variables and dice graphics. 1970-2090 Dice throwing routine. 2100 Miscellaneous drawing title page etc. An interesting point to note in this program is the use of the hi-res graphic drawing routines. Unlike many other programs, only one UDG is used - the disc pip (graphic A). ------------------------------------------------------------------------------ TYPE: Traditional (Dice Game) COMMENT: This info file was typed by Michael Bruhn Downloaded from: Desert Island Disks -- http://www.image.dk/~frankie/ Maintained by: Michael Bruhn -- frankie@image.dk

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# Differential calculus: Maxima and minima problem (project) • Mar 6th 2013, 08:12 PM iankasai Differential calculus: Maxima and minima problem (project) DIFFERENTIAL CALCULUS: MAXIMA AND MINIMA 1. DESIGNING A POSTER You are designing a rectangular poster to contain 50in^2 of printing with a 4-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used? Can you show it to me how it looks like? -_- And how do you solve this problem using the maxima and minima? thanks in advanced! • Mar 7th 2013, 06:21 AM HallsofIvy Re: Differential calculus: Maxima and minima problem (project) Let the paper be x inches wide and y inches high. Since you want 4 inch margins at top and bottom, and 2 inch margins at each sidea, what are the dimensions of the actual printing area? Set that area equal to 50. You want to minimize xy subject to that constraint. • Mar 8th 2013, 12:19 AM iankasai Re: Differential calculus: Maxima and minima problem (project) can u demonstrate to me how to calculate this problem? :)) • Mar 8th 2013, 12:20 AM iankasai Re: Differential calculus: Maxima and minima problem (project) Quote: Originally Posted by HallsofIvy Let the paper be x inches wide and y inches high. Since you want 4 inch margins at top and bottom, and 2 inch margins at each sidea, what are the dimensions of the actual printing area? Set that area equal to 50. You want to minimize xy subject to that constraint. can u demonstrate to me how to calculate this problem? ) • Mar 8th 2013, 12:52 AM Prove It Re: Differential calculus: Maxima and minima problem (project) Start by drawing a sketch...

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## 15 Reputation 3 years, 341 days ## How to use printf with units or an alter... Maple 2020 Hello I have a funtion defined as `a := diff(u*ln(m__o/(-q*t + m__o)) - g*t, t)` where m_0, q, g, and u are constants with dimensions. T is the variable and has dimensions as well. I need to print out columns of this function and other similar functions, v and h, in 4 seconds intervals of t. So what I have is this: ```for i from -4 by 4 to 80 do if i = -4 then printf("%12s %18s %15s %15s \n", "time", "acceleration", "velocity", "altitude"); else printf("%10d %c %10.2f %s %10.2f %s %10.2f %s \n", i, "s", eval(a, t = i), "ft/s^2", eval(v, t = i), "ft/s", eval(h, t = i), "mi"); end if; end do``` this works perfectly fine until I assign units to the constants and t. Then I get the error: Error, (in fprintf) number expected for floating point format. Then I also need to plot a v and h against t. Seems like plot doesn't work either with units. So what do i do? Thanks ## Can't get rid of those fractions - help ... Maple 2020 Hello Just started out with maple. I entered a function that looks like this, to find the volume of a sphere: V := (4*Pi(d/2)^3)/3 then I use the eval command like this: eval(V, d = 6.35*mm) and I get this:(4*Pi(3.175000000*mm)^3)/3 Anything I tried, maple doesn't want to spit out one number! I just need what that is equal to, I don't need an expression with Pi. Spent over two hours trying to figure it out, watching videos etc but no luck! hoping that someone can help me with this! Page 1 of 1 

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ListMap sealed class ListMap[K, V] extends Iterable[(K, V)] Linear Supertypes Iterable[(K, V)], IterableLike[(K, V), Iterable[(K, V)]], Equals, GenIterable[(K, V)], GenIterableLike[(K, V), Iterable[(K, V)]], Traversable[(K, V)], GenTraversable[(K, V)], GenericTraversableTemplate[(K, V), Iterable], TraversableLike[(K, V), Iterable[(K, V)]], GenTraversableLike[(K, V), Iterable[(K, V)]], Parallelizable[(K, V), ParIterable[(K, V)]], TraversableOnce[(K, V)], GenTraversableOnce[(K, V)], FilterMonadic[(K, V), Iterable[(K, V)]], HasNewBuilder[(K, V), Iterable[(K, V)]], AnyRef, Any Ordering 1. Alphabetic 2. By inheritance Inherited 1. Hide All 2. Show all 1. ListMap 2. Iterable 3. IterableLike 4. Equals 5. GenIterable 6. GenIterableLike 7. Traversable 8. GenTraversable 9. GenericTraversableTemplate 10. TraversableLike 11. GenTraversableLike 12. Parallelizable 13. TraversableOnce 14. GenTraversableOnce 16. HasNewBuilder 17. AnyRef 18. Any Visibility 1. Public 2. All Type Members 1. type Self = Iterable[(K, V)] Attributes protected Definition Classes TraversableLike Value Members 1. final def !=(arg0: AnyRef): Boolean Definition Classes AnyRef 2. final def !=(arg0: Any): Boolean Definition Classes Any 3. final def ##(): Int Definition Classes AnyRef → Any 5. def ++[B >: (K, V), That](that: GenTraversableOnce[B])(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike → GenTraversableLike 6. def ++:[B >: (K, V), That](that: Traversable[B])(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike 7. def ++:[B >: (K, V), That](that: TraversableOnce[B])(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike 9. def /:[B](z: B)(op: (B, (K, V)) ⇒ B): B Definition Classes TraversableOnce → GenTraversableOnce 10. def /:\[A1 >: (K, V)](z: A1)(op: (A1, A1) ⇒ A1): A1 Definition Classes GenTraversableOnce 11. def :\[B](z: B)(op: ((K, V), B) ⇒ B): B Definition Classes TraversableOnce → GenTraversableOnce 12. final def ==(arg0: AnyRef): Boolean Definition Classes AnyRef 13. final def ==(arg0: Any): Boolean Definition Classes Any Definition Classes TraversableOnce 15. def addString(b: StringBuilder, sep: String): StringBuilder Definition Classes TraversableOnce 16. def addString(b: StringBuilder, start: String, sep: String, end: String): StringBuilder Definition Classes TraversableOnce 17. def aggregate[B](z: B)(seqop: (B, (K, V)) ⇒ B, combop: (B, B) ⇒ B): B Definition Classes TraversableOnce → GenTraversableOnce 19. final def asInstanceOf[T0]: T0 Definition Classes Any 20. def canEqual(that: Any): Boolean Definition Classes IterableLike → Equals 21. def clone(): AnyRef Attributes protected[lang] Definition Classes AnyRef Annotations @throws() 22. def collect[B, That](pf: PartialFunction[(K, V), B])(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike → GenTraversableLike 23. def collectFirst[B](pf: PartialFunction[(K, V), B]): Option[B] Definition Classes TraversableOnce 24. def companion: GenericCompanion[Iterable] Definition Classes Iterable → GenIterable → Traversable → GenTraversable → GenericTraversableTemplate Attributes protected 27. def copyToArray[B >: (K, V)](xs: Array[B], start: Int, len: Int): Unit Definition Classes IterableLike → TraversableLike → TraversableOnce → GenTraversableOnce 28. def copyToArray[B >: (K, V)](xs: Array[B]): Unit Definition Classes TraversableOnce → GenTraversableOnce 29. def copyToArray[B >: (K, V)](xs: Array[B], start: Int): Unit Definition Classes TraversableOnce → GenTraversableOnce 30. def copyToBuffer[B >: (K, V)](dest: Buffer[B]): Unit Definition Classes TraversableOnce 31. def count(p: ((K, V)) ⇒ Boolean): Int Definition Classes TraversableOnce → GenTraversableOnce 33. def drop(n: Int): Iterable[(K, V)] Definition Classes IterableLike → TraversableLike → GenTraversableLike 34. def dropRight(n: Int): Iterable[(K, V)] Definition Classes IterableLike 35. def dropWhile(p: ((K, V)) ⇒ Boolean): Iterable[(K, V)] Definition Classes TraversableLike → GenTraversableLike 36. final def eq(arg0: AnyRef): Boolean Definition Classes AnyRef 37. def equals(arg0: Any): Boolean Definition Classes AnyRef → Any 38. def exists(p: ((K, V)) ⇒ Boolean): Boolean Definition Classes IterableLike → TraversableLike → TraversableOnce → GenTraversableOnce 39. def filter(p: ((K, V)) ⇒ Boolean): Iterable[(K, V)] Definition Classes TraversableLike → GenTraversableLike 40. def filterNot(p: ((K, V)) ⇒ Boolean): Iterable[(K, V)] Definition Classes TraversableLike → GenTraversableLike 41. def finalize(): Unit Attributes protected[lang] Definition Classes AnyRef Annotations @throws() 42. def find(p: ((K, V)) ⇒ Boolean): Option[(K, V)] Definition Classes IterableLike → TraversableLike → TraversableOnce → GenTraversableOnce 43. def flatMap[B, That](f: ((K, V)) ⇒ GenTraversableOnce[B])(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes 44. def fold[A1 >: (K, V)](z: A1)(op: (A1, A1) ⇒ A1): A1 Definition Classes TraversableOnce → GenTraversableOnce 45. def foldLeft[B](z: B)(op: (B, (K, V)) ⇒ B): B Definition Classes TraversableOnce → GenTraversableOnce 46. def foldRight[B](z: B)(op: ((K, V), B) ⇒ B): B Definition Classes IterableLike → TraversableOnce → GenTraversableOnce 47. def forall(p: ((K, V)) ⇒ Boolean): Boolean Definition Classes IterableLike → TraversableLike → TraversableOnce → GenTraversableOnce 48. def foreach[U](f: ((K, V)) ⇒ U): Unit Definition Classes IterableLike → TraversableLike → GenTraversableLike → TraversableOnce → GenTraversableOnce → FilterMonadic 49. def genericBuilder[B]: Builder[B, Iterable[B]] Definition Classes GenericTraversableTemplate 51. final def getClass(): java.lang.Class[_] Definition Classes AnyRef → Any 52. def groupBy[K](f: ((K, V)) ⇒ K): Map[K, Iterable[(K, V)]] Definition Classes TraversableLike → GenTraversableLike 53. def grouped(size: Int): Iterator[Iterable[(K, V)]] Definition Classes IterableLike 54. def hasDefiniteSize: Boolean Definition Classes TraversableLike → TraversableOnce → GenTraversableOnce 55. def hashCode(): Int Definition Classes AnyRef → Any Definition Classes IterableLike → GenIterableLike → TraversableLike → GenTraversableLike Definition Classes TraversableLike 58. def init: Iterable[(K, V)] Definition Classes TraversableLike 59. def inits: Iterator[Iterable[(K, V)]] Definition Classes TraversableLike 60. def isEmpty: Boolean Definition Classes ListMap → IterableLike → GenIterableLike → GenericTraversableTemplate → TraversableLike → TraversableOnce → GenTraversableOnce 61. final def isInstanceOf[T0]: Boolean Definition Classes Any 62. final def isTraversableAgain: Boolean Definition Classes GenTraversableLike → GenTraversableOnce 63. def iterator: Iterator[(K, V)] Definition Classes ListMap → IterableLike → GenIterableLike 65. def last: (K, V) Definition Classes TraversableLike 66. def lastOption: Option[(K, V)] Definition Classes TraversableLike 67. def map[B, That](f: ((K, V)) ⇒ B)(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes 68. def max[B >: (K, V)](implicit cmp: Ordering[B]): (K, V) Definition Classes TraversableOnce → GenTraversableOnce 69. def maxBy[B](f: ((K, V)) ⇒ B)(implicit cmp: Ordering[B]): (K, V) Definition Classes TraversableOnce → GenTraversableOnce 70. def min[B >: (K, V)](implicit cmp: Ordering[B]): (K, V) Definition Classes TraversableOnce → GenTraversableOnce 71. def minBy[B](f: ((K, V)) ⇒ B)(implicit cmp: Ordering[B]): (K, V) Definition Classes TraversableOnce → GenTraversableOnce 72. def mkString: String Definition Classes TraversableOnce → GenTraversableOnce 73. def mkString(sep: String): String Definition Classes TraversableOnce → GenTraversableOnce 74. def mkString(start: String, sep: String, end: String): String Definition Classes TraversableOnce → GenTraversableOnce 75. final def ne(arg0: AnyRef): Boolean Definition Classes AnyRef 76. def newBuilder: Builder[(K, V), Iterable[(K, V)]] Attributes protected[this] Definition Classes GenericTraversableTemplate → HasNewBuilder 77. def nonEmpty: Boolean Definition Classes TraversableOnce → GenTraversableOnce 78. final def notify(): Unit Definition Classes AnyRef 79. final def notifyAll(): Unit Definition Classes AnyRef 80. def par: ParIterable[(K, V)] Definition Classes Parallelizable 81. def parCombiner: Combiner[(K, V), ParIterable[(K, V)]] Attributes protected[this] Definition Classes TraversableLike → Parallelizable 82. def partition(p: ((K, V)) ⇒ Boolean): (Iterable[(K, V)], Iterable[(K, V)]) Definition Classes TraversableLike → GenTraversableLike 83. def product[B >: (K, V)](implicit num: Numeric[B]): B Definition Classes TraversableOnce → GenTraversableOnce 84. def reduce[A1 >: (K, V)](op: (A1, A1) ⇒ A1): A1 Definition Classes TraversableOnce → GenTraversableOnce 85. def reduceLeft[B >: (K, V)](op: (B, (K, V)) ⇒ B): B Definition Classes TraversableOnce 86. def reduceLeftOption[B >: (K, V)](op: (B, (K, V)) ⇒ B): Option[B] Definition Classes TraversableOnce → GenTraversableOnce 87. def reduceOption[A1 >: (K, V)](op: (A1, A1) ⇒ A1): Option[A1] Definition Classes TraversableOnce → GenTraversableOnce 88. def reduceRight[B >: (K, V)](op: ((K, V), B) ⇒ B): B Definition Classes IterableLike → TraversableOnce → GenTraversableOnce 89. def reduceRightOption[B >: (K, V)](op: ((K, V), B) ⇒ B): Option[B] Definition Classes TraversableOnce → GenTraversableOnce 90. def repr: Iterable[(K, V)] Definition Classes TraversableLike → GenTraversableLike 91. def reversed: List[(K, V)] Attributes protected[this] Definition Classes TraversableOnce 92. def sameElements[B >: (K, V)](that: GenIterable[B]): Boolean Definition Classes IterableLike → GenIterableLike 93. def scan[B >: (K, V), That](z: B)(op: (B, B) ⇒ B)(implicit cbf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike → GenTraversableLike 94. def scanLeft[B, That](z: B)(op: (B, (K, V)) ⇒ B)(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike → GenTraversableLike 95. def scanRight[B, That](z: B)(op: ((K, V), B) ⇒ B)(implicit bf: CanBuildFrom[Iterable[(K, V)], B, That]): That Definition Classes TraversableLike → GenTraversableLike Annotations @migration Migration (Changed in version 2.9.0) The behavior of `scanRight` has changed. The previous behavior can be reproduced with scanRight.reverse. 96. def seq: Iterable[(K, V)] Definition Classes Iterable → GenIterable → Traversable → GenTraversable → Parallelizable → TraversableOnce → GenTraversableOnce 97. def size: Int Definition Classes TraversableOnce → GenTraversableOnce 98. def slice(from: Int, until: Int): Iterable[(K, V)] Definition Classes IterableLike → TraversableLike → GenTraversableLike 99. def sliding[B >: (K, V)](size: Int, step: Int): Iterator[Iterable[(K, V)]] Definition Classes IterableLike 100. def sliding[B >: (K, V)](size: Int): Iterator[Iterable[(K, V)]] Definition Classes IterableLike 101. def span(p: ((K, V)) ⇒ Boolean): (Iterable[(K, V)], Iterable[(K, V)]) Definition Classes TraversableLike → GenTraversableLike 102. def splitAt(n: Int): (Iterable[(K, V)], Iterable[(K, V)]) Definition Classes TraversableLike → GenTraversableLike 103. def stringPrefix: String Definition Classes TraversableLike → GenTraversableLike 104. def sum[B >: (K, V)](implicit num: Numeric[B]): B Definition Classes TraversableOnce → GenTraversableOnce 105. final def synchronized[T0](arg0: ⇒ T0): T0 Definition Classes AnyRef 106. def tail: Iterable[(K, V)] Definition Classes TraversableLike → GenTraversableLike 107. def tails: Iterator[Iterable[(K, V)]] Definition Classes TraversableLike 108. def take(n: Int): Iterable[(K, V)] Definition Classes IterableLike → TraversableLike → GenTraversableLike 109. def takeRight(n: Int): Iterable[(K, V)] Definition Classes IterableLike 110. def takeWhile(p: ((K, V)) ⇒ Boolean): Iterable[(K, V)] Definition Classes IterableLike → TraversableLike → GenTraversableLike 111. def thisCollection: Iterable[(K, V)] Attributes protected[this] Definition Classes IterableLike → TraversableLike 112. def toArray[B >: (K, V)](implicit arg0: ClassManifest[B]): Array[B] Definition Classes TraversableOnce → GenTraversableOnce 113. def toBuffer[B >: (K, V)]: Buffer[B] Definition Classes TraversableOnce → GenTraversableOnce 114. def toCollection(repr: Iterable[(K, V)]): Iterable[(K, V)] Attributes protected[this] Definition Classes IterableLike → TraversableLike 115. def toIndexedSeq[B >: (K, V)]: IndexedSeq[B] Definition Classes TraversableOnce → GenTraversableOnce 116. def toIterable: Iterable[(K, V)] Definition Classes IterableLike → TraversableOnce → GenTraversableOnce 117. def toIterator: Iterator[(K, V)] Definition Classes TraversableLike → GenTraversableOnce 118. def toList: List[(K, V)] Definition Classes ListMap → TraversableOnce → GenTraversableOnce 119. def toMap[T, U](implicit ev: <:<[(K, V), (T, U)]): Map[T, U] Definition Classes TraversableOnce → GenTraversableOnce 120. def toSeq: List[(K, V)] Definition Classes ListMap → TraversableOnce → GenTraversableOnce 121. def toSet[B >: (K, V)]: Set[B] Definition Classes TraversableOnce → GenTraversableOnce 122. def toStream: Stream[(K, V)] Definition Classes IterableLike → TraversableLike → GenTraversableOnce 123. def toString(): String Definition Classes ListMap → TraversableLike → AnyRef → Any 124. def toTraversable: Traversable[(K, V)] Definition Classes TraversableLike → TraversableOnce → GenTraversableOnce 125. def transpose[B](implicit asTraversable: ((K, V)) ⇒ GenTraversableOnce[B]): Iterable[Iterable[B]] Definition Classes GenericTraversableTemplate Annotations @migration Migration (Changed in version 2.9.0) `transpose` throws an `IllegalArgumentException` if collections are not uniformly sized. 126. def unzip[A1, A2](implicit asPair: ((K, V)) ⇒ (A1, A2)): (Iterable[A1], Iterable[A2]) Definition Classes GenericTraversableTemplate 127. def unzip3[A1, A2, A3](implicit asTriple: ((K, V)) ⇒ (A1, A2, A3)): (Iterable[A1], Iterable[A2], Iterable[A3]) Definition Classes GenericTraversableTemplate 129. def view(from: Int, until: Int): IterableView[(K, V), Iterable[(K, V)]] Definition Classes IterableLike → TraversableLike 130. def view: IterableView[(K, V), Iterable[(K, V)]] Definition Classes IterableLike → TraversableLike 131. final def wait(): Unit Definition Classes AnyRef Annotations @throws() 132. final def wait(arg0: Long, arg1: Int): Unit Definition Classes AnyRef Annotations @throws() 133. final def wait(arg0: Long): Unit Definition Classes AnyRef Annotations @throws() 134. def withFilter(p: ((K, V)) ⇒ Boolean): FilterMonadic[(K, V), Iterable[(K, V)]] Definition Classes 135. def zip[A1 >: (K, V), B, That](that: GenIterable[B])(implicit bf: CanBuildFrom[Iterable[(K, V)], (A1, B), That]): That Definition Classes IterableLike → GenIterableLike 136. def zipAll[B, A1 >: (K, V), That](that: GenIterable[B], thisElem: A1, thatElem: B)(implicit bf: CanBuildFrom[Iterable[(K, V)], (A1, B), That]): That Definition Classes IterableLike → GenIterableLike 137. def zipWithIndex[A1 >: (K, V), That](implicit bf: CanBuildFrom[Iterable[(K, V)], (A1, Int), That]): That Definition Classes IterableLike → GenIterableLike Deprecated Value Members 1. def elements: Iterator[(K, V)] Definition Classes IterableLike Annotations @deprecated Deprecated (Since version 2.8.0) use `iterator' instead` 2. def first: (K, V) Definition Classes IterableLike Annotations @deprecated Deprecated (Since version 2.8.0) use `head' instead` 3. def firstOption: Option[(K, V)] Definition Classes IterableLike Annotations @deprecated Deprecated (Since version 2.8.0) use `headOption' instead` 4. def projection: IterableView[(K, V), Iterable[(K, V)]] Definition Classes IterableLike Annotations @deprecated Deprecated (Since version 2.8.0) use `view' instead`

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