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http://en.wikipedia.org/wiki/Particle_size | # Particle size
Particle size is a notion introduced for comparing dimensions of solid particles (flecks), liquid particles (droplets), or gaseous particles (bubbles).
The notion of particle size applies to
The particle size of a spherical object can be unambiguously and quantitatively defined by its diameter.
However, a typical material object is likely to be irregular in shape and non-spherical. The above quantitative definition of particle size cannot be applied to non-spherical particles. There are several ways of extending the above quantitative definition, so that a definition is obtained that also applies to non-spherical particles. Existing definitions are based on replacing a given particle with an imaginary sphere that has one of the properties identical with the particle.
• Volume based particle size equals the diameter of the sphere that has same volume as a given particle.
$D = 2 \sqrt[3] {\frac{3V}{4\pi}}$
where
$D$: diameter of representative sphere
$V$: volume of particle
• Weight based particle size equals the diameter of the sphere that has same weight as a given particle.
$D = 2 \sqrt[3] {\frac{3W}{4\pi dg}}$
where
$D$: diameter of representative sphere
$W$: weight of particle
$d$: density of particle
$g$: gravitational constant
• Area based particle size equals the diameter of the sphere that has the same surface area as a given particle.
$D = 2 \sqrt[2] {\frac{A}{4\pi}}$
where
$D$: diameter of representative sphere
$A$: surface area of particle
Another complexity in defining particle size appears for particles with sizes below a micrometre. When particle becomes that small, thickness of interface layer becomes comparable with the particle size. As a result, position of the particle surface becomes uncertain. There is convention for placing this imaginary surface at certain position suggested by Gibbs and presented in many books on Interface and Colloid Science.[1][2][3][4][5][6]
Definition of the particle size for an ensemble (collection) of particles presents another problem. Real systems are practically always polydisperse, which means that the particles in an ensemble have different sizes. The notion of particle size distribution reflects this polydispersity. There is often a need of a certain average particle size for the ensemble of particles. There are several different ways of defining such a particle size.
• There is an International Standard on presenting various characteristic particle sizes.[7] This set of various average sizes includes median size, geometric mean size, average size.
There are several methods for measuring particle size. Some of them are based on light, other on ultrasound, or electric field, or gravity, or centrifugation. They are briefly described in the section particle size distribution. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8864879012107849, "perplexity": 1008.5890378058072}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997888283.14/warc/CC-MAIN-20140722025808-00237-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://spmchemistry.blog.onlinetuition.com.my/2012/07/conversion-of-the-unit-of-concentration.html | # Conversion of the Unit of Concentration
## Conversion of the Unit of Concentration
1. The chart above shows how to convert the units of concentration from g dm-3 to mol dm-3 and vice versa.
2. The molar mass of the solute is equal to the relative molecular mass of the solute.
Example 1:
The concentration of a Potassium chloride solution is 14.9 g dm-3. What is the molarity ( mol dm-3) of the solution? [ Relative Atomic Mass: Cl = 35.5; K = 39 ]
Relative Formula Mass of Potassium Chloride (KCl)
= 39 + 35.5 = 74.5
Molar Mass of Potassium Chloride = 74.5 g/mol
Molarity of Potassium Chloride
Molarity = Concentration Molar Mass = 14.9gd m −3 74.5gmol−1 =0.2mol/dm3
Example 2
A solution of barium hydrokxide have molarity 0.1 mol dm-3. What is the concentration of the solution in g dm-3? [Relative Atomic Mass: Ba = 137; O = 16; H = 1 ] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9127719402313232, "perplexity": 4996.73930768253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00219.warc.gz"} |
https://www.physicsforums.com/threads/dervative-of-volume-of-sphere.649429/ | # Dervative of volume of sphere?
1. Nov 4, 2012
### Psyguy22
If you take the derivative of the area of a circle, you get the formula for circumference. When you take the derivative of the volume of the sphere, you do not get the formula for the area of a circle. Why not?
d/dr (4/3pi r^3) =4pi r^2
d/dr (pi r^2)= 2pi r
2. Nov 4, 2012
### Mute
What you actually get when you take the derivative of the volume with respect to radius is the surface area of the ball. Note that there is a technical difference between a ball and a "sphere": a sphere is, strictly speaker, the surface of a ball. It does not include the volume it encloses, whereas a "ball" includes the surface and the volume contained within. Occasionally the forum gets questions about the "volume of a sphere" and someone will answer that it is zero, which is technically correct because the volume of a "sphere", interpreted literally, refers to the volume the surface itself, which is zero, and not the volume contained by the surface; the OP in these cases pretty much always means the volume of the ball and was just unaware of the precise distinction in the terminology. So, just pointing that out.
Anyways, a way to see why it works like this is to consider the following: to build a circle of area $\pi R^2$, you can think of the process of building "shells" of circles of increasing radius, where each shell has an infinitesimal thickness $dr$. By adding more and more shells you are increasing the area of the circle you are building. If you have a circle of radius r, the infinitesimal change in area you get when adding another shell is $dA = 2\pi r dr$ - the circumference of the shell times the thickness.
When you then go and start building a ball in a similar manner, you are not adding shells of circles. Rather, you are adding shells of spheres of thickness $dr$ and surface area $4\pi r^2$. So, the infinitesimal change in volume as you add a shell to a ball of radius r is $dV = 4\pi r^2 dr$ - the surface area times the thickness.
Does that make sense?
3. Nov 4, 2012
### lurflurf
area of circle:circumference::volume of sphere:surface area of sphere
pi r^2:2pi r::4pi r^3/2:4pi r^2
A:A'::V=V'
you get surface area of the sphere
This follows from Stokes theorem
$$\int_\Omega \mathrm {d}\omega = \int_ {\partial \Omega} \omega$$
Last edited: Nov 4, 2012
4. Nov 4, 2012
### Psyguy22
So my old geometry book lied? Isn't a sphere a 3-D shape? Meaning it would have volume? What exactly does the equation 4/3pi r^3 represent then?
5. Nov 4, 2012
### pwsnafu
Sort of. In natural English sphere is a 3D object. In mathematics it's a 2D object. From Wikipedia:
Your geometry textbook wasn't being careful.
Mute already told you: the term is ball.
6. Nov 4, 2012
### Psyguy22
Ok. Thank you. And sorry for the misuse of words. So just to make sure I understand
4/3 pi r^3 is the volume of a 'ball' and 4 pi r^2 is the surface area of a 'sphere'?
7. Nov 4, 2012
### arildno
It is, of course, the surface area of the ball itself, the sphere being its SURROUNDING BOUNDARY.
(Just as the circumference of the DISK constitutes the surrounding circle bounding the disk)
For nice geometrical objects, the surrounding boundary of the object is of 1 dimension lower than the object itself. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.907912015914917, "perplexity": 615.1197623507875}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267861899.65/warc/CC-MAIN-20180619041206-20180619061206-00234.warc.gz"} |
https://perminc.com/resources/publications/role-of-medium-heterogeneity-and-viscosity-contrast-in-miscible-flow-regimes-and-mixing-zone-growth-a-computational-pore-scale-approach/ | ## Role of Medium Heterogeneity and Viscosity Contrast in Miscible Flow Regimes and Mixing Zone Growth: A Computational Pore-Scale Approach
Afshari, S., Kantzas, A., Hejazi, H.
DOI: 10.1103/PhysRevFluids.3.054501
Physical Review Fluids, 3(5), May 2018.
## ABSTRACT
Miscible displacement of fluids in porous media is often characterized by the scaling of the mixing zone length with displacement time. Depending on the viscosity contrast of fluids, the scaling law varies between the square root relationship, a sign for dispersive transport regime during stable displacement, and the linear relationship, which represents the viscous fingering regime during an unstable displacement. The presence of heterogeneities in a porous medium significantly affects the scaling behavior of the mixing length as it interacts with the viscosity contrast to control the mixing of fluids in the pore space. In this study, the dynamics of the flow and transport during both unit and adverse viscosity ratio miscible displacements are investigated in heterogeneous packings of circular grains using pore-scale numerical simulations. The pore-scale heterogeneity level is characterized by the variations of the grain diameter and velocity field. The growth of mixing length is employed to identify the nature of the miscible transport regime at different viscosity ratios and heterogeneity levels. It is shown that as the viscosity ratio increases to higher adverse values, the scaling law of mixing length gradually shifts from dispersive to fingering nature up to a certain viscosity ratio and remains almost the same afterwards. In heterogeneous media, the mixing length scaling law is observed to be generally governed by the variations of the velocity field rather than the grain size. Furthermore, the normalization of mixing length temporal plots with respect to the governing parameters of viscosity ratio, heterogeneity, medium length, and medium aspect ratio is performed. The results indicate that mixing length scales exponentially with log-viscosity ratio and grain size standard deviation while the impact of aspect ratio is insignificant. For stable flows, mixing length scales with the square root of medium length, whereas it changes linearly with length during unstable flows. This scaling procedure allows us to describe the temporal variation of mixing length using a generalized curve for various combinations of the flow conditions and porous medium properties.
A full version of this paper is available on ResearchGate Online. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212046265602112, "perplexity": 1161.1990471668341}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337731.82/warc/CC-MAIN-20221006061224-20221006091224-00030.warc.gz"} |
https://www.hackmath.net/en/math-problem/935 | # Fraction
Fraction
$\dfrac{0,\overline{ 40 }}{0,40 }$
write as fraction a/b, a, b is integers numerator/denominator.
Result
a = 100
b = 99
#### Solution:
$x = \dfrac{0,\overline{ 40 }}{0,40 } = \dfrac{1 }{0,40} \dfrac{ a_1}{1-q} = \dfrac{1 }{0,40} \dfrac{0,40}{1-\dfrac{1}{100}} \ \\ x = \dfrac{1}{1-\dfrac{1}{100}} = \dfrac{100}{99} \ \\$
$b=99$
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
Stella
i like this
Bob
good
#### Following knowledge from mathematics are needed to solve this word math problem:
Need help calculate sum, simplify or multiply fractions? Try our fraction calculator.
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Find two imaginary numbers whose sum is a real number. How are the two imaginary numbers related? What is its sum? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8983897566795349, "perplexity": 1951.3269561847808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540488620.24/warc/CC-MAIN-20191206122529-20191206150529-00396.warc.gz"} |
http://diposit.ub.edu/dspace/handle/2445/33855 | Please use this identifier to cite or link to this item: http://hdl.handle.net/2445/33855
Title: Sherali-Adams Relaxations and Indistinguishability in Counting Logics Author: Atserias, AlbertManeva, Elitza Keywords: Lògica de primer ordreProgramació linealTeoria de grafsFirst-order logicLinear programmingGraph theory Issue Date: 17-Jan-2013 Publisher: Society for Industrial and Applied Mathematics Abstract: Two graphs with adjacency matrices $\mathbf{A}$ and $\mathbf{B}$ are isomorphic if there exists a permutation matrix $\mathbf{P}$ for which the identity $\mathbf{P}^{\mathrm{T}} \mathbf{A} \mathbf{P} = \mathbf{B}$ holds. Multiplying through by $\mathbf{P}$ and relaxing the permutation matrix to a doubly stochastic matrix leads to the linear programming relaxation known as fractional isomorphism. We show that the levels of the Sherali--Adams (SA) hierarchy of linear programming relaxations applied to fractional isomorphism interleave in power with the levels of a well-known color-refinement heuristic for graph isomorphism called the Weisfeiler--Lehman algorithm, or, equivalently, with the levels of indistinguishability in a logic with counting quantifiers and a bounded number of variables. This tight connection has quite striking consequences. For example, it follows immediately from a deep result of Grohe in the context of logics with counting quantifiers that a fixed number of levels of SA suffice to determine isomorphism of planar and minor-free graphs. We also offer applications in both finite model theory and polyhedral combinatorics. First, we show that certain properties of graphs, such as that of having a flow circulation of a prescribed value, are definable in the infinitary logic with counting with a bounded number of variables. Second, we exploit a lower bound construction due to Cai, Fürer, and Immerman in the context of counting logics to give simple explicit instances that show that the SA relaxations of the vertex-cover and cut polytopes do not reach their integer hulls for up to $\Omega(n)$ levels, where $n$ is the number of vertices in the graph. Note: Reproducció del document publicat a: http://dx.doi.org/10.1137/120867834 It is part of: SIAM Journal on Computing, 2013, vol. 42, num. 1, p. 112-137 Related resource: http://dx.doi.org/10.1137/120867834 URI: http://hdl.handle.net/2445/33855 ISSN: 0097-5397 Appears in Collections: Articles publicats en revistes (Matemàtiques i Informàtica)
Files in This Item:
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http://mathhelpforum.com/advanced-algebra/342-need-help.html | # Math Help - need help
1. ## need help
Is there a difference between solving a system of equations by the algebraic method and the graphical method? Why?
2. ## Differences between the two methods
Yes and no. It's more a matter of precision than substantially different answers.
If you solve graphically, then you can estimate solutions by simply reading the graph. However, the solutions precision is limited by the size of the graph, how good your eyes are, the width of your pencil lines, etc.
An algebraic solution yields exact answers.
For example, if a solution was $\sqrt 2$, then that's what the algebraic method would give you, but graphically, you would get 1.4142136...
woot! it's been years since I used LaTeX, and I formatted the sqrt(2) above right on the first try! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9132897853851318, "perplexity": 771.4762522924043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657127222.34/warc/CC-MAIN-20140914011207-00150-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://astronomy.stackexchange.com/questions/19150/is-the-winds-intensity-on-mars-similar-to-earth/40155 | # Is the wind's intensity on Mars similar to Earth?
I've read that in Mars' poles, the winds can be as fast as 400 km/h, when the poles are exposed to sunlight because the frozen $CO_2$ sublimes. I know that the Martian atmosphere is much thinner than Earth's atmosphere.
So, by knowing the wind speeds on Mars, is there any way to get an idea of its intensity, or in other words, the intensity of a wind of x speed in Mars, to which speed of wind of Earth is comparable, for them to have the same intensity?
• Related questions here: space.stackexchange.com/questions/9301/… and space.stackexchange.com/questions/2621/… The first link has math where the wind-force can be calculated. – userLTK Nov 24 '16 at 22:29
• ok, so the pressure of the wind it would be 61,25 times lower? nice answer – Pablo Nov 24 '16 at 22:46
• do you want to post the answer here so I mark it as accepted? – Pablo Nov 24 '16 at 23:42
• I think your math is right at least, that's what I get too, but as for an answer, I didn't want to post or copy someone else's answer as my own. – userLTK Nov 25 '16 at 2:38
• @com.prehensible The atmospheric pressure on top of Olympus Mons is 0.0007 x the normal pressure at sea level on Earth, or 0.7 millibar. For comparison, a vacuum pump that you could buy online for 125 USD makes 0.1 millibar, only 7 times better; a pump that costs 50 USD makes 0.2 millibar, or 3.5 times better. Colloquially, I would describe the pressure on top of Olympus Mons as "pretty lousy vacuum". Seems like there's room for a lot of wind speed there before it really becomes threatening. – Florin Andrei Dec 14 '17 at 0:28
Credit to this question for inspiration, though my calculation methods differ.
The dynamic pressure equation is $$q=0.5\rho v^2$$ where $$q$$ is the pressure, $$\rho$$ is the atmospheric density, and $$v$$ is the wind speed. If we want to know what wind speeds give us equivalent pressures on Earth and Mars, we simply generate dynamic pressure equations for each of them: $$q=0.5\rho_e v_e^2$$ and $$q=0.5\rho_m v_m^2$$, set them equal $$q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$$, and solve for $$v_e$$ to get $$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$ where $$\rho_m=0.020 \space kg/m^3$$ is the atmospheric density for Mars, $$\rho_e=1.225 \space kg/m^3$$ is the atmospheric density on Earth, $$v_m$$ is the wind speed on Mars, and $$v_e$$ is the equivalent wind speed on Earth.
With a velocity ratio of about 7.826 we can plug in a few values for wind speed in kilometers per hour for Mars to get:
v_mars v_earth equivalent
10 1.28
50 6.39
100 12.8
200 25.6
400 51.1
These could be kph, or in fact, any units of velocity. screeenshot
and here's what hat looks like in a plot:
So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8155836462974548, "perplexity": 684.6338802505043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00053.warc.gz"} |
https://www.physicsforums.com/threads/trying-to-understand-absolute-uncertainties-in-geometric-shapes.188297/ | Trying to understand absolute uncertainties in geometric shapes
1. Oct 1, 2007
memsces
I've been studying absolute uncertainties and do not understand any of it. If someone can explain it will really help. Especially with uncertainties including diameters and area.
2. Oct 5, 2007 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9100137948989868, "perplexity": 1487.2673037175946}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501174135.70/warc/CC-MAIN-20170219104614-00312-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/273164/whats-the-automorphism-group-of-this-covering | What's the automorphism group of this covering?
What's the automorphism group of this covering?
I know why this is a covering, but I don't know how to find the automorphism group of this covering.
I need help, thanks
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Two automorphisms of a path-connected covering coincide iff they coincide at one point. 6 points lie above the node, thus the automorphism group can be identified as a subgroup of the permutation group on a 6 point set. Call the six points $\lbrace i_1,i_2,i_3,o_1,o_2,o_3\rbrace$ : the nodes labeled $i$ are those on the inner circle and one labeled $o$ lie on the outer circle. The lift of $b$ produces the permutation $(i_1i_2i_3)(o_1o_2o_3)$, while the lift of $a$ gives the permutation $(i_1o_1)(i_2o_2)(i_3o_3)$, so it seems that the group $G$ of the covering is isomorphic to the subgroup of $S_6$ generated by these two permutations. They commute, and so we should have $G\simeq\Bbb Z/6\Bbb Z$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9401921629905701, "perplexity": 78.4463966872758}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00549-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/186780/show-different-limits-under-different-mode-of-convergence-equals-almost-everywhe | Show different limits under different mode of convergence equals almost everywhere
Suppose that a sequence of bounded and continuous functions $f_n$ converges uniformly to $f_1$ and $f_n$ converges to $f_2$ in $L^2$ sense, then how to show $f_1= f_2$ a.e.?
I tried the following: let $A_\epsilon = \{x:|f_1(x)-f_2(x)|>\epsilon\}$, then $m(A_\epsilon) < m(|f_n - f_1|>\epsilon) + m(|f_n - f_2|>\epsilon)$. Let $n$ go to infinity, then the first part of RHS goes to zero by uniform convergence, but I cannot do anything to $L^2$-convergence.
Can anyone show me how to solve this question? Thanks in advance .
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You are on the right track: use $m(|f_n-f_2|\gt\epsilon)\leqslant\epsilon^{-2}\|f_n-f\|_2^2$. – Did Aug 25 '12 at 15:17
@Norbert Asked 5 hours ago is a bit soon for a question to be declared unanswered, don't you think? – Did Aug 25 '12 at 20:10
I think people shy to post answer that you have already gave in first comment. I really don't like common practice of posting answers as comments – Norbert Aug 25 '12 at 22:24
@Norbert What you say does not correspond to my experience, as I have seen countless examples of the opposite happening on this site. Anyway, your second comment forces me to interpret your first one quite differently than I first did, and in a way which I really don't like. – Did Aug 25 '12 at 22:48
@did So what you don't like? – Norbert Aug 25 '12 at 23:49
Markov's inequality does the job: we get that for each $\varepsilon>0$, $$\lambda\{x,|f_n(x)-f_2(x)|>\delta\}\leqslant \frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ hence following the notations in the OP, we get $$\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant \lambda(\{x,|f_n(x)-f_1(x)|>\varepsilon\}\cap [-N,N])+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ hence $$\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant 2N\cdot \left[\sup_{[-N,N]}|f_n-f_1|>\varepsilon\right]+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$$ where $[P]$ is when when $P$ is satisfied and $0$ otherwise.
When $N$ and $\varepsilon$ are fixed, the RHS goes to $0$ as $n$ goes to infinity. Hence, $\lambda(A_{2\varepsilon}\cap [-N,N])=0$ for all $N$ and $\varepsilon$, giving the wanted conclusion. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9677801132202148, "perplexity": 190.46564785453023}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049279525.46/warc/CC-MAIN-20160524002119-00086-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://psjd.icm.edu.pl/psjd/element/bwmeta1.element.bwnjournal-article-appv81z302kz | PL EN
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## Acta Physica Polonica A
1992 | 81 | 3 | 353-360
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### Regular and Chaotic Behaviour of a Kicked Damped Spin
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The dynamics of a kicked, anisotropic, damped spin is reduced to a two-dimensional map. This map exhibits such features as bifurcation diagrams, regular or chaotic attractors/repellors and intermittent-like transitions between two strange attractors. With increase of damping a transition from chaos to the fixed point attractor occurs. On the contrary to the Hamiltonian case the type of magnetic anisotropy plays a crucial role for damped models.
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1992-03 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8312307596206665, "perplexity": 3731.3004720368085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00448.warc.gz"} |
http://twittercensus.se/british-national-xccek/hydrogen-energy-levels-diagram-bf6149 | Energy Level Diagram for Hydrogen Atom: Energy level diagrams indicate us the different series of lines observed in a spectrum of the hydrogen atom. The formula defining the energy levels of a Hydrogen atom are given by the equation: E = -E0/n2, where E0 = 13.6 eV (1 eV = 1.602×10-19 Joules) and n = 1,2,3… and so on. This is the energy carried away by the photon. For hydrogen-like atoms (ions) only, the Rydberg levels depend only on the principal quantum number n. ... Energy level diagrams. This effect is now called Lamb shift. Note that the difference in energy between orbitals decreases rapidly with increasing values of n. In general, both energy and radius decrease as the nuclear charge increases. Author: Barb Newitt. The energy levels are shown as numbers on the left side with the lowest energy level at the bottom. Post was not sent - check your email addresses! Home A Level Quantum Physics & Lasers (A Level) Energy Level Diagram For Hydrogen. The electron in a hydrogen atom is in the n = 2 state. n represents the principle quantum number and only takes integral values from 1 to infinity. 100 or more) are so weakly bound that … This lecture would give you an idea about the energy levels of hydrogen atoms. The electron energy level diagram for the hydrogen atom. The diffusion velocity is proportional to the diffusion coefficient and varies with temperature according to T n with n in the range of 1.72-1.8. Energy levels. (transition from ground state n = 0 to infinity n =. However, the fundamental difference between the two is that, while the planetary system is held in place by the gravitational force, the nucl… Each box corresponds to one orbital. 2 Hydrogen gas is highly diffusive and highly buoyant; it rapidly mixes with the ambient air upon release. The energy levels of hydrogen, including fine structure (excluding Lamb shift and hyperfine structure), are ... (which is able to deal with these vacuum fluctuations and employs the famous Feynman diagrams for approximations using perturbation theory). The arrows represent transitions from one energy level to another (in this case they are all emissions). The energy difference between any two adjacent levels gets smaller as n increases, which results in the higher energy levels getting very close and crowded together just below n =, The ionization energy of an atom is the energy required to remove the electron completely from the atom. Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. It's often helpful to draw a diagram showing the energy levels for the particular element you're interested in. The energy level diagram for the H atom. Hydrogen Spectrum introduction. Justify your answer. energy level diagram for hydrogen. The figure below is an energy level diagram for a hydrogen atom. (i) Find out the transition which results in the emission of a photon of wavelength 496 nm. Thus, the lower energy states correspond to more stable states. For hydrogen, the ionization energy = 13.6eV When an excited electron returns to a lower level, it loses an exact amount of energy by emitting a photon. Which photon has the longer wavelength? The electron normally occupies this level unless given sufficient energy to move up to a higher level. Energy Level Diagrams - Hydrogen. Is this in the visible spectrum? This is called the Balmer series. Figure 7. If you spot any errors or want to suggest improvements, please contact us. These are obtained by substituting all possible values of n into equation (1). Fig. The Lyman(ultraviolet) series of spectral lines corresponds to electron transitions from higher energy levels to level n = 1. The ground state is the lowest energy equilibrium state measured for hydrogen molecules. 3.3.1a - Bohr Diagram. Bohr explained the hydrogen spectrum in terms of electrons absorbing and emitting photons to change energy levels, where the photon energy is h\nu =\Delta E = \left (\dfrac {1} { {n_ {low}}^2}-\dfrac {1} { {n_ {high}}^2}\right) \cdot 13.6\,\text {eV} hν = ΔE = (nlow 21 Now the hydrogen atom, there is a classical description, a diagram, for the hydrogen atom and in fact, for any central potential. The greatest possible fall in energy will therefore produce the highest frequency line in the spectrum. Energy Level Diagrams. We all know that electrons in an atom or a molecule absorb energy and get excited, they jump from a lower energy level to a higher energy level, and they emit radiation when they come back to their original states. 0 votes . An atom is said to be in an excited state when its electrons are found in the higher energy levels. Let's take a look at how to draw Bohr diagrams: For a hydrogen atom, H, the one electron goes into the first energy level. Examples Molecular orbital diagrams, Jablonski diagrams, and Franck–Condon diagrams. (ii) Which transition corresponds to the emission of radiation of maximum wavelength ? This is called the Balmer series. Transitions to n = 2 and n = 3are called the Balmer(visible) and Paschen(Infra Red) series, respectively. What is the wavelength of the photon? The diagram for hydrogen is shown above. Negative value of energy indicates that the electron is bound to the nucleus and there exists an attractive force between the electron and the nucleus. The energy levels agree with the earlier Bohr model, and agree with experiment within a small fraction of an electron volt. The horizontal lines of the diagram indicate different energy levels. Figure 2. It's a negative energy. Well, just to straighten things out first, the term “multielectron” is a tiny misuse here. The last equation can therefore be re-written as a measure of the energy gap between two electron levels. It really isn’t the number of electrons that matter, it is the potential energy of the electrons and type of orbital that matter. The energy level diagram of a hypothetical atom is shown below. 5 9 Sample Calculation nCalculate the wavelength at which the least energetic emission spectral line of the Lyman Series(nf = 1) is observed. That energy must be exactly the same as the energy gap between the 3-level and the 2-level in the hydrogen atom. Administrator of Mini Physics. On this diagram, the n = 0 energy level was represented. Mini Physics is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.sg. What is the energy of the n- o energy level? If a photon with an energy equal to the energy difference between two levels is incident on an atom, the photon can be absorbed, raising the electron up to the higher level. If the electron in the atom makes a transition from a particular state to a lower state, it is losing energy. The 2p level is split into a … Label the arc 1e -to represent that there is one electron in this energy level. The lower the energy level, the more negative the energy value associated with that level. The smaller the energy the longer the wavelength. The vertical lines indicate the transition of an electron from a higher energy level to a lower energy level. Consider the photon emitted when an electron drops from the n=4 to the n=2 state to the photon emitted when an electron drops from n=3 to n=2. The energy level diagram gives us a way to show what energy the electron has without having to draw an atom with a bunch of circles all the time. So if you're looking at bound states, the way we do bound states and represent them for central potentials is by a diagram in which you put the energy on the vertical line. To conserve energy, a photon with an energy equal to the energy difference between the states will be emitted by the atom. Converting this to joules gives E = 10.2 * 1.60 x 10-19 J/eV = 1.632 x 10-18 J, λ = hc/E = 6.63 x 10-34 * 3 x 108 / 1.632 x 10-18. Area Approximations; Slope Exploration 1; segments in a triangle Discover Resources. These spectra can be used as analytical tools to assess composition of matter. The photon emitted in the n=4 to n=2 transition, The photon emitted in the n=3 to n=2 transition. Draw an arc to represent the first energy level. The molecular orbital energy level diagram of H 2 molecule is given in Fig.. Energy level diagram The energy of the electron in the nth orbit of the hydrogen atom is given by, En = -13.6 /n2 eV Energy associated with the first orbit of the hydrogen atom is, Energy Level Diagram for Hydrogen Atom: Energy level diagrams indicate us the different series of lines observed in a spectrum of the hydrogen atom. Sorry, your blog cannot share posts by email. When you learned about the energy levels of hydrogen, an energy level diagram was introduced. Bond order = (N b -N a) /2 = 2-0/2 = 2 i. There are various types of energy level diagrams for bonds between atoms in a molecule. 3-2. Let's say our pretend atom has electron energy levels of zero eV, four eV, six eV, and seven eV. answered Oct 5, 2018 by Supria (63.9k points) selected … The figure shows energy level diagram of hydrogen atom. Why the energy levels have negative values? Energy level transitions. This phenomenon accounts for the emission spectrum through hydrogen too, … The energy is expressed as a negative number because it takes that much energy to unbind (ionize) the electron from the nucleus. Fig. Thus the most stable orbitals (those with the lowest energy) are those closest to the nucleus. The horizontal lines of the diagram indicate different energy levels. In tables of atomic energy levels, however, it is more usual to take the energy of the ground state ($$n=1$$) to be zero. If the electron in the atom makes a transition from a particular state to a lower state, it is losing energy. The n = 1 state is known as the ground state, while higher n states are known as excited states. 1-1: Phase diagram of hydrogen . “d” represents the distance between adjacent scratches on the diffraction grating. For instance, our knowledge of the atomic composition of the sun was in part aided by considering the spectra of the radiation from the sun. a. Nature of bond: This means that the two hydrogen atoms in a molecule of hydrogen are bonded by a single covalent bond. The absorption of what frequency photon would result in a ground state electron transitioning to its first excited state? According to Rutherford’s model, an atom has a central nucleus and electron/s revolve around it like the sun-planet system. The ground state refers to the lowest energy level n=1 in which the atom is the most stable. Draw a circle and label it with the symbol of the nucleus, H. Write the number of protons for the nucleus, 1p +. Go to the Hydrogen Atom simulation (Unit D), and complete the n=1 to n transition Assignment Booklet 10 Observe the energy state data. Figure 7 shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). One way to do this is to first calculate the energy of the electron in the initial and final states using the equation: In dropping from the n = 2 state to the ground state the electron loses 10.2 eV worth of energy. In that case the energy levels are given by The Lyman (ultraviolet) series of spectral lines corresponds to electron transitions from higher energy levels to level n = 1. Made with | 2010 - 2020 | Mini Physics |. The photon has a smaller energy for the n=3 to n=2 transition. He then mathematically showed which energy level transitions corresponded to the spectral lines in the atomic emission spectrum ( Figure 2). b. The diagram for hydrogen is shown above. atoms; cbse; class-12; Share It On Facebook Twitter Email. Back To Quantum Physics And Lasers (A Level). Each group of transitions is given the name of the scientist who identified their origin. Topic: Diagrams He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). Notify me of follow-up comments by email. Hydrogen Energy Level Diagram. If you look at the hydrogen energy levels at extremely high resolution, you do find evidence of some other small effects on the energy. The electron energy level diagram for the hydrogen atom. Figure $$\PageIndex{7}$$: Orbital Energy Level Diagram for the Hydrogen Atom. When an atom is excited from the ground state to a higher energy, it becomes unstable and falls back to one of the lower energy levels by emitting photon(s)/electromagnetic radiation. The bond order of H 2 molecule can be calculated as follows. In the hydrogen atom, with Z = 1, the energy of the emitted photon can be found using: Atoms can also absorb photons. Energy level diagrams are a means of assessing the energies electrons may take and release as the transition occurs, from one accepted orbital to another one. Bohr model of the hydrogen atom attempts to plug in certain gaps as suggested by Rutherford’s model by including ideas from the newly developing Quantum hypothesis. Hydrogen’s Energy Level Diagram When nf = 2: Balmer Series-visibleemission When nf = 3: Paschen Series-infraredemission. 1 Answer. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Reddit (Opens in new window), Click to share on Telegram (Opens in new window), Click to share on WhatsApp (Opens in new window), Click to email this to a friend (Opens in new window), Click to share on LinkedIn (Opens in new window), Click to share on Tumblr (Opens in new window), Click to share on Pinterest (Opens in new window), Click to share on Pocket (Opens in new window), Click to share on Skype (Opens in new window), The Schrodinger Equation And Wave Function, Case Study 2: Energy Conversion for A Bouncing Ball, Case Study 1: Energy Conversion for An Oscillating Ideal Pendulum, UY1: Electric Field Of Uniformly Charged Disk, Induced Magnetism & Electrical Method Of Magnetisation. This is in the ultraviolet part of the spectrum, so it would not be visible to us. Each line dentoes an allowed energy for the atom. Also, since the potential at infinity is defined as zero, energy levels at a distance below infinity are negative. The vertical lines indicate the transition of an electron from a higher energy level to a lower energy level. The n = 1 state is known as the ground state, while higher n states are known as excited states. We can again construct an energy level diagram listing the allowed energy values . Hence the energy of all bound orbits is negative. In practice, electrons with high n (e.g. The different energy levels of Hydrogen are denoted by the quantum number n where n varies from 1 for the ground state (the lowest energy level) to ∞, corresponding to unbound electrons. Transitions between the energy states (levels) of individual atoms give rise to characteristic atomic spectra. Here, N b = 2 and N a = 0. The orbital energies are calculated using the above equation, first derived by Bohr. When an excited electron returns to a lower level, it loses an exact amount of energy by emitting a photon. Hydrogen Spectrum - Wavelength, Diagram, Hydrogen Emission Spectrum . ii. When it drops to the ground state a photon is emitted. Diffraction Grating You will use the diffraction grating relation, which may be written as m is the angle at which the m th order maximum occurs for light of wavelength . Printer Friendly Version: Refer to the following information for the next four questions. That is, the energy level we have calculated for a bound orbit is expressed relative to the energy of ionized hydrogen. Diffusion in multi-component mixtures is usually described by the Stefan-Maxwell equation. The name of the energy levels possible fall in energy will therefore produce highest! Associated with that level at the bottom fraction of an electron from higher! By a single covalent bond a hypothetical atom is shown below can again construct an equal! Accounts for the next four questions out the transition of an electron from a particular state to a state... Level was represented and varies with temperature according to Rutherford ’ s energy level diagram of,! Printer Friendly Version: Refer to the nucleus types of energy level at bottom! Electron levels agree with experiment within a small fraction of an electron from the nucleus a covalent. The ambient air upon release associated with that level to the energy carried away by the Stefan-Maxwell equation it! Your blog can not Share posts by email to suggest improvements, please contact us sun-planet system scientist identified... “ multielectron ” is a tiny misuse here Lasers ( a level quantum Physics & (... The 2p level is split into a … hydrogen spectrum - wavelength, hydrogen energy levels diagram. ” represents the principle quantum number and only takes integral values from 1 to infinity n = 1 state known. Series of spectral lines corresponds to the energy level diagram for the hydrogen atom Version... The n- o energy level diagrams for bonds between atoms in a molecule of hydrogen are bonded by a covalent... Jablonski diagrams, Jablonski diagrams, Jablonski diagrams, and agree with the earlier Bohr model, and eV! A lower energy level diagrams of energy level diagram of H 2 molecule can be as! = 1 energy carried away by the atom energy will therefore produce the highest frequency line in higher... Only, the lower the energy of ionized hydrogen horizontal lines of the spectrum order (... Of 1.72-1.8 information for the hydrogen atom first energy level diagram of hydrogen atoms in a ground electron... The sun-planet system number n.... energy level two electron levels diagram indicate different energy levels n! Are shown as numbers on the principal quantum number n.... energy level diagram for the next four.. Orbital energy level diagrams hydrogen energy levels diagram bonds between atoms in a hydrogen atom with in... N with n in the range of 1.72-1.8 the Rydberg levels depend only on the principal quantum and. The first energy level diffusion coefficient and varies with temperature according to Rutherford ’ energy. 'S say our pretend atom has electron energy levels this level unless given sufficient to! Smaller energy for the hydrogen atom | Mini Physics | orbital diagrams, Jablonski,. The atom makes a transition from a higher energy levels to level n = 0 energy level diagram for hydrogen! Obtained by substituting all possible values of n into equation ( 1 ) T n n. Was introduced | 2010 - 2020 | Mini Physics | hydrogen atoms not visible... All bound orbits is negative you an idea about the energy value associated that. Exact amount of energy by emitting a photon with an energy level was. Hence the energy of all bound orbits is negative carried away by the atom makes transition! First energy level can again construct an energy level diagrams for bonds between in! Orbital energy level n=1 in which the atom makes hydrogen energy levels diagram transition from ground state a is. ) are so weakly bound that … we can again construct an level! Improvements, please contact us states are known as excited states Mini Physics.... Shown below orbital energy level diagram of hydrogen atoms in a hydrogen atom this energy level to another ( this... That … we can again construct an energy level to another ( in energy. Improvements, please contact us sorry, your blog can not Share posts by email types energy! Not sent - check your email addresses a ground state n = to! Of matter wavelength 496 nm bonded by a single covalent bond element you 're interested in is... Blog can not Share posts by email lower energy level diagram for the of. Is a tiny misuse here the 2p level is split into a … hydrogen spectrum -,. The spectrum Lyman ( ultraviolet ) series, respectively this diagram, hydrogen emission spectrum hydrogen... Well, just to straighten things out first, the Rydberg levels depend only the! Single covalent bond energy for the atom when an excited state when electrons. Spot any errors or want to suggest improvements, please contact us you any!, first derived by Bohr measured for hydrogen showing the energy levels agree with the ambient air release... = 3are called the Balmer ( visible ) and Paschen series of spectral lines to! It like the sun-planet system an electron volt what is the most stable Twitter email model, and Franck–Condon.. It drops to the following information for the n=3 to n=2 transition,,. Highest frequency line in the atom makes a transition from a higher energy level diagram for hydrogen )! From one energy level was represented Lyman, Balmer, and agree with lowest! Revolve around it like the sun-planet system lower level, it is losing.! 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Is given the name of the scientist who identified their origin states are known as excited states,... 2020 | Mini Physics | is split into a … hydrogen spectrum - wavelength, diagram, Rydberg... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8668105006217957, "perplexity": 633.7677276470627}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00632.warc.gz"} |
http://www.pearltrees.com/gnatalia/physics/id13263770 | # Physics
Frontier of Physics: Interactive Map. “Ever since the dawn of civilization,” Stephen Hawking wrote in his international bestseller A Brief History of Time, “people have not been content to see events as unconnected and inexplicable.
They have craved an understanding of the underlying order in the world.” In the quest for a unified, coherent description of all of nature — a “theory of everything” — physicists have unearthed the taproots linking ever more disparate phenomena. With the law of universal gravitation, Isaac Newton wedded the fall of an apple to the orbits of the planets. Albert Einstein, in his theory of relativity, wove space and time into a single fabric, and showed how apples and planets fall along the fabric’s curves. And today, all known elementary particles plug neatly into a mathematical structure called the Standard Model. Our map of the frontier of fundamental physics, built by the interactive developer Emily Fuhrman, weights questions roughly according to their importance in advancing the field. Vector Model of Angular Momentum. Once you have combined orbital and spin angular momenta according to the vector model, the resulting total angular momentum can be visuallized as precessing about any externally applied magnetic field.
This is a useful model for dealing with interactions such as the Zeeman effect in sodium. The magnetic energy contribution is proportional to the component of total angular momentum along the direction of the magnetic field, which is usually defined as the z-direction. The z-component of angular momentum is quantized in values one unit apart, so for the upper level of the sodium doublet with j=3/2, the vector model gives the splitting shown. Even with the vector model, the determination of the magnitude of the Zeeman spliting is not trivial since the directions of S and L ar constantly changing as they precess about J. This problem is handled with the Lande' g-factor.
Quantized Angular Momentum. Lagrangian formalism - Intuition Behind Conservation of Angular Momentum. Threshold size for quantum effects. Angular Momentum. Angular momentum. This gyroscope remains upright while spinning due to the conservation of its angular momentum.
In physics, angular momentum, (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the angular momentum of a system remains constant unless acted on by an external torque. Angular momentum in classical mechanics Definition First principle. A first principle is a basic proposition or assumption that cannot be deduced from any other proposition or assumption.
In philosophy, first principles are from First Cause[1] attitudes and taught by Aristotelians, and nuanced versions of first principles are referred to as postulates by Kantians.[2] In mathematics, first principles are referred to as axioms or postulates. In physics and other sciences, theoretical work is said to be from first principles, or ab initio, if it starts directly at the level of established science and does not make assumptions such as empirical model and parameter fitting.
In formal logic In a formal logical system, that is, a set of propositions that are consistent with one another, it is possible that some of the statements can be deduced from other statements. For example, in the syllogism, "All men are mortal; Socrates is a man; Socrates is mortal" the last claim can be deduced from the first two. Philosophy in general Terence Irwin writes:
## Mètodes numèrics
Particle and nuclear physics. Advanced mathematical methods. Quantum physics. Pràctiques externes. Thermodynamics and statistical mechanics. Optics. Numerical methods. Symmetries, conservation laws and Noether's Theorem. Electro. Chemistry. Differential equations. Mechanics LAB. Classical Mechanics. Multivariable Calculus. The Speed Of Light Can Vary In A Vacuum.
Quantum physics just got less complicated. Here's a nice surprise: quantum physics is less complicated than we thought.
An international team of researchers has proved that two peculiar features of the quantum world previously considered distinct are different manifestations of the same thing. The result is published 19 December in Nature Communications. Patrick Coles, Jedrzej Kaniewski, and Stephanie Wehner made the breakthrough while at the Centre for Quantum Technologies at the National University of Singapore. They found that 'wave-particle duality' is simply the quantum 'uncertainty principle' in disguise, reducing two mysteries to one. "The connection between uncertainty and wave-particle duality comes out very naturally when you consider them as questions about what information you can gain about a system.
The discovery deepens our understanding of quantum physics and could prompt ideas for new applications of wave-particle duality. Explore further: A new 'lens' for looking at quantum behavior. Gauge esto, Gauge lo otro… ¿Qué es una teoría gauge? La palabra gauge la encontramos por doquier en los escritos sobre física.
Aparecen expresiones como simetría gauge, invariancia gauge, bosones gauge, teorías gauge, etc. Sin embargo, pocas veces se explica con propiedad qué es esta teoría, por qué es tan fundamental y cómo la entienden y por qué la veneran tanto los físicos. ¿Por qué La Tierra está achatada por los polos? La densidad de La Tierra. En su elaboración de la Teoría de Gravitación Universal, Newton ya se dio cuenta de que en La Tierra, a consecuencia de su movimiento de rotación y según su Ley de atracción, cada partícula de masa m a diferente distancia del eje, estaría expuesta a una diferente Fuerza Centrípeta, ya que describe un movimiento circular uniforme de diferente radio alrededor del eje de rotación de La Tierra.
Según la segunda ley de Newton, para que se produzca una aceleración debe actuar una fuerza en la dirección de esa aceleración. Así, si consideramos una partícula de masa m en movimiento circular uniforme, estará sometida a una fuerza centrípeta dada por: F=-m · w^2 · r Esta fuerza es precisamente la que deforma La Tierra, que deja de ser una perfecta esfera para convertirse en un elipsoide, o geoide si se prefiere. El radio de rotación de la particula de masa m irá desde cero en el eje hasta el Radio de La Tierra en la superficie. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9155085682868958, "perplexity": 3391.5162067510637}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00499.warc.gz"} |
https://groupprops.subwiki.org/wiki/Linear_representation_theory_of_dihedral_groups | # Linear representation theory of dihedral groups
This article gives specific information, namely, linear representation theory, about a family of groups, namely: dihedral group.
View linear representation theory of group families | View other specific information about dihedral group
Note first that all dihedral groups are ambivalent groups -- every element is conjugate to its inverse. Thus, all the irreducible representations of a dihedral group over the complex numbers can be realized over the real numbers.
## Summary
We consider here the dihedral group $D_{2n}$ of degree $n$ and order $2n$. So, for instance, for $n = 4$, the corresponding group is dihedral group:D8.
Item Value
degrees of irreducible representations over a splitting field Case $n$ odd: 1 (2 times), 2 ($(n - 1)/2$ times)
Case $n$ even: 1 (4 times), 2 ($(n - 2)/2$ times)
maximum: 2 (if $n \ge 3$), lcm: 2 (if $n \ge 3$), number: $(n + 3)/2$ for $n$ odd, $(n + 6)/2$ for $n$ even, sum of squares: $2n$
Schur index values of irreducible representations over a splitting field 1 (all of them)
condition for a field to be a splitting field First, the field should have characteristic not equal to 2 or any prime divisor of $n$. Also, take the cyclotomic polynomial $\Phi_n(x)$. Let $\zeta$ be a root of the polynomial. Then, the field should contain the element $\zeta + \zeta^{-1}$, i.e., the minimal polynomial for $\zeta + \zeta^{-1}$ should split completely.
smallest ring of realization (characteristic zero) $\mathbb{Z}[2\cos(2\pi/n)]$
smallest field of realization (characteristic zero) $\mathbb{Q}(\cos(2\pi/n))$. Note that a degree two extension of this gives the cyclotomic extension of $\mathbb{Q}$ for primitive $n^{th}$ roots of unity. The given field can be thought of as the intersection of the cyclotomic extension and the real numbers.
smallest size splitting field unclear. Definitely, for a field of odd size $q$, $n$ dividing $q - 1$ is sufficient, but not necessary.
degrees of irreducible representations over rational numbers PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
YOU MAY ALSO BE INTERESTED IN: linear representation theory of generalized dihedral groups, linear representation theory of dicyclic groups
## Particular cases
Note that that cases $n = 1$ and $n = 2$ are atypical.
Degree $n$ Order $2n$ $n$ odd or even? Dihedral group Degrees of irreducible representations Number of irreducible representations ($(n + 3)/2$ if $n$ odd, $(n + 6)/2$ if $n$ even) Smallest splitting field (characteristic zero) Linear representation theory page
1 2 odd cyclic group:Z2 1,1 2 $\mathbb{Q}$ linear representation theory of cyclic group:Z2
2 4 even Klein four-group 1,1,1,1 4 $\mathbb{Q}$ linear representation theory of Klein four-group
3 6 odd symmetric group:S3 1,1,2 3 $\mathbb{Q}$ linear representation theory of symmetric group:S3
4 8 even dihedral group:D8 1,1,1,1,2 5 $\mathbb{Q}$ linear representation theory of dihedral group:D8
5 10 odd dihedral group:D10 1,1,2,2 4 $\mathbb{Q}(\sqrt{5})$ linear representation theory of dihedral group:D10
6 12 even dihedral group:D12 1,1,1,1,2,2 6 $\mathbb{Q}$ linear representation theory of dihedral group:D12
7 14 odd dihedral group:D14 1,1,2,2,2 5 $\mathbb{Q}(\cos(2\pi/7))$ linear representation theory of dihedral group:D14
8 16 even dihedral group:D16 1,1,1,1,2,2,2 7 $\mathbb{Q}(\sqrt{2})$ linear representation theory of dihedral group:D16
## The linear representation theory of dihedral groups of odd degree
Consider the dihedral group $D_{2n}$, where $n$ is odd:
$D_{2n} := \langle a,x \mid a^n = x^2 = e, xax = a^{-1} \rangle$.
The group $D_{2n}$ has a total of $(n+3)/2$ conjugacy classes: the identity element, $(n-1)/2$ other conjugacy classes in $\langle a \rangle$, and the conjugacy class of $x$. Thus, there are $(n+3)/2$ irreducible representations. We discuss these representations.
### The two one-dimensional representations
The derived subgroup is $\langle a \rangle$, and hence the abelianization of the group is cyclic of order two. Thus, there are two one-dimensional representations:
• The trivial representation, sending all elements to the $1 \times 1$ matrix $(1)$.
• The representation sending all elements in $\langle a \rangle$ to $(1)$ and all elements outside $\langle a \rangle$ to $(-1)$.
### The two-dimensional representations
There are $(n-1)/2$ irreducible two-dimensional representations. The $k^{th}$ representation is given in the following equivalent forms:
Group element Matrix as real orthogonal Matrix as complex unitary Matrix as real, non-orthogonal, in $\mathbb{Q}(\cos(2\pi/n))$ Character (trace of any of the matrices) Minimal polynomial
$a$ $\begin{pmatrix} \cos(2\pi k/n) & -\sin (2\pi k/n) \\ \sin (2\pi k/n) & \cos (2\pi k/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ik/n} & 0 \\ 0 & e^{-2\pi ik/n}\end{pmatrix}$ $\begin{pmatrix} 0 & -1 \\ 1 & 2 \cos(2 \pi k/n)\end{pmatrix}$ $2\cos(2\pi k/n)$ $t^2 - 2\cos(2\pi k/n)t + 1$
$a^l$ $\begin{pmatrix} \cos(2\pi kl/n) & -\sin(2\pi kl/n) \\ \sin(2\pi kl/n) & \cos(2 \pi kl/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ikl/n} & 0 \\ 0 & e^{-2\pi ikl/n}\end{pmatrix}$ ? $2 \cos(2 \pi kl/n)$ $t^2 - 2\cos(2 \pi kl/n)t + 1$
$x$ $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $0$ $t^2 - 1$
$a^l x$ $\begin{pmatrix} \cos(2 \pi kl/n) & \sin(2\pi kl/n) \\ \sin(2 \pi kl/n) & -\cos(2\pi kl/n)\end{pmatrix}$ $\begin{pmatrix} 0 & e^{2\pi ikl/n} \\ e^{-2\pi ikl/n} & 0 \\\end{pmatrix}$ ? $0$ $t^2 - 1$
## The linear representation theory of dihedral groups of even degree
Consider the dihedral group $D_{2n}$, where $n$ is even:
$D_{2n} := \langle a,x \mid a^n = x^2 = e, xax = a^{-1} \rangle$.
This group has $(n+6)/2$ conjugacy classes: the identity element, the element $a^{n/2}$, $(n-2)/2$ other conjugacy classes in $\langle a \rangle$, and two conjugacy classes outside $\langle a \rangle$, with representatives $x$ and $ax$.
### The four one-dimensional representations
The commutator subgroup is $\langle a^2 \rangle$, which has index four, and the quotient group (the [[abelianization]) is a Klein four-group. There are thus four one-dimensional representations:
• The trivial representation, sending all elements to the $1 \times 1$ matrix $(1)$.
• The representation sending all elements in $\langle a \rangle$ to $(1)$ and all elements outside $\langle a \rangle$ to $(-1)$.
• The representation sending all elements in $\langle a^2, x \rangle$ to $(1)$ and $a$ to $-1$.
• The representation sending all elements in $\langle a^2, ax \rangle$ to $(1)$ and $a$ to $-1$.
### The two-dimensional representations
There are $(n-2)/2$ irreducible two-dimensional representations. All of these can be realized over $\mathbb{Q}(\cos(2\pi/n))$. The representations can be described in a number of different ways. The description of the $k^{th}$ representation is given below:
Group element Matrix as real orthogonal Matrix as complex unitary Matrix as real, non-orthogonal, in $\mathbb{Q}(\cos(2\pi/n))$ Character (trace of any of the matrices) Minimal polynomial
$a$ $\begin{pmatrix} \cos(2\pi k/n) & -\sin (2\pi k/n) \\ \sin (2\pi k/n) & \cos (2\pi k/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ik/n} & 0 \\ 0 & e^{-2\pi ik/n}\end{pmatrix}$ $\begin{pmatrix} 0 & -1 \\ 1 & 2 \cos(2 \pi k/n)\end{pmatrix}$ $2\cos(2\pi k/n)$ $t^2 - 2\cos(2\pi k/n)t + 1$
$a^l$ $\begin{pmatrix} \cos(2\pi kl/n) & -\sin(2\pi kl/n) \\ \sin(2\pi kl/n) & \cos(2 \pi kl/n) \\\end{pmatrix}$ $\begin{pmatrix} e^{2\pi ikl/n} & 0 \\ 0 & e^{-2\pi ikl/n}\end{pmatrix}$ ? $2 \cos(2 \pi kl/n)$ $t^2 - 2\cos(2 \pi kl/n)t + 1$
$x$ $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $\begin{pmatrix}0 & 1 \\ 1 & 0 \\\end{pmatrix}$ $0$ $t^2 - 1$
$a^l x$ $\begin{pmatrix} \cos(2 \pi kl/n) & \sin(2\pi kl/n) \\ \sin(2 \pi kl/n) & -\cos(2\pi kl/n)\end{pmatrix}$ $\begin{pmatrix} 0 & e^{2\pi ikl/n} \\ e^{-2\pi ikl/n} & 0 \\\end{pmatrix}$ ? $0$ $t^2 - 1$
Note that for the representations for $k$ and $n - k$ are equivalent, hence we get distinct representations only for $k = 1,2, \dots, (n-2)/2$. (The representations for $k = 0$ and $k = n/2$ are not irreducible and they break up into one-dimensional representations already discussed).
## Degrees of irreducible representations
FACTS TO CHECK AGAINST FOR DEGREES OF IRREDUCIBLE REPRESENTATIONS OVER SPLITTING FIELD:
Divisibility facts: degree of irreducible representation divides group order | degree of irreducible representation divides index of abelian normal subgroup
Size bounds: order of inner automorphism group bounds square of degree of irreducible representation| degree of irreducible representation is bounded by index of abelian subgroup| maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup
Cumulative facts: sum of squares of degrees of irreducible representations equals order of group | number of irreducible representations equals number of conjugacy classes | number of one-dimensional representations equals order of abelianization
The summary (based on the detailed description above):
Parity of $n$ Number of degree 1 representations Number of degree 2 representations Total number
odd 2 $(n - 1)/2$ $(n + 3)/2$
even 4 $(n - 2)/2$ $(n + 6)/2$
## Realizability information
### Quick summary for realization of characters
Case Character ring (ring generated by characters) Degree of field extension of $\mathbb{Q}$ containing character ring
$n$ odd $\mathbb{Z}[2\cos(2\pi/n)] = \mathbb{Z}[2\cos(\pi/n)]$ $(1/2)\varphi(n)$
$n = 2m$, $m$ odd, $\mathbb{Z}[2\cos(2\pi/n)] = \mathbb{Z}[2\cos(2\pi/m)]$ $(1/2)\varphi(n) = (1/2)\varphi(m)$
$n = 4m$, $m$ odd or even $\mathbb{Z}[2\cos(2\pi/n)] = \mathbb{Z}[2\sin(2\pi/n)]$ $(1/2)\varphi(n)$
### Schur index and realization of representations
It turns out that for all irreducible representations of the dihedral group, the Schur index equals one. This means that every irreducible representation of the dihedral group can be realized over its field of character values. In particular, all the irreducible representations of the dihedral group can be realized over the field $\mathbb{Q}(\cos(2\pi/n))$.
### Quick summary for orthogonal representations
Case Smallest ring over which all irreducible representations written as orthogonal representations are realized Degree of field extension of $\mathbb{Q}$ containing it
$n$ odd $\mathbb{Z}[\cos(2\pi/n),\sin(2\pi/n)]$ $\varphi(n)$
$n = 2m$, $m$ odd $\mathbb{Z}[\cos(2\pi/n), \sin(2\pi/n)]$ $\varphi(n) = \varphi(m)$
$n = 4m$, $m$ odd or even $\mathbb{Z}[\cos(2\pi/n)]$ $(1/2)\varphi(n)$
### Smaller splitting fields: some specific examples
The symmetric group of degree three, which is also the dihedral group of order six (and degree three) is an example of a group for which the splitting field is $\mathbb{Q}(\cos (2\pi/3))$, which is equal to $\mathbb{Q}$ itself. Note, however, that the degree two representation we obtain is not in terms of orthogonal matrices. Further information: Linear representation theory of symmetric group:S3
The Klein four-group (which is a dihedral group of order four and degree two) and dihedral group:D8 (which has order eight and degree four) are the only examples where the representations described above are naturally over $\mathbb{Q}$.
### Representations in prime characteristic
If $p$ is a prime number not dividing the order of the dihedral group, we can discuss the linear representation theory in characteristic $p$. The representations remain the same; however, we need to replace $\cos (2\pi k/n)$ with the element $\zeta_n^k + \zeta_n^{-k}$, where $\zeta_n$ is a primitive $n^{th}$ root of unity. All the irreducible characters take values in the field $\mathbb{F}_p(\zeta_n + \zeta_n^{-1})$, while all the irreducible representations are realized over the field $\mathbb{F}_p(\zeta)$.
The groups $D_4$ (Klein four-group, order four, degree two), $D_6$ (also symmetric group of degree three, order six, degree three), and $D_8$ (dihedral group of order eight, degree four) have representations that can be realized over a prime field of any characteristic relatively prime to their respective orders.
## Orthogonality relations and numerical checks
• The degrees of irreducible representations are all $1$ or $2$. This confirms the fact that degree of irreducible representation divides index of abelian normal subgroup. In this case, the abelian normal subgroup is the cyclic subgroup $\langle a \rangle$ and it is a subgroup of index two.
• The number of irreducible representations is $(n+3)/2$ for odd $n$ and $(n+6)/2$ for even $n$, which is equal to the number of conjugacy classes in either case.
• The number of one-dimensional representations is $2$ for odd $n$ and $4$ for even $n$, which is equal to the order of the abelianization in either case.
• For odd $n$, the sum of squares of degrees of irreducible representations is $2(1)^2 + ((n-1)/2)(2)^2 = 2n$, which is equal to the order of the group. For even $n$, the sum of squares of degrees of irreducible reprensetations is $4(1)^2 + ((n-2)/2)(2)^2 = 2n$, which is equal to the order of the group. This confirms the fact that sum of squares of degrees of irreducible representations equals order of group.
• The character table satisfies the orthogonality relations: in particular, the row orthogonality theorem and the column orthogonality theorem.
## Action of automorphism group
### The special case $n = 2$
For $n = 2$, the automorphism group permutes the three nontrivial one-dimensional representations. This anomalous behavior is explained by the fact that in the $n = 2$ case, $a$ and $x$ are related by an automorphism.
### The general case of odd $n$
We have the following:
• Both one-dimensional representations are preserved by the action of the automorphism group.
• For the two-dimensional representations, there are $\tau(n) - 1$ equivalence classes under the action of the automorphism group, where $\tau(n)$ is the number of divisors of $n$. Specifically, for every divisor $d > 1$ of $n$, there is an equivalence class of irreducible two-dimensional representations of size $\varphi(d)$ (where $\varphi$ is the Euler totient function) comprising the $k^{th}$ irreducible representation for all $k$ with $\operatorname{gcd}(n,k) = n/d$.
### The general case of even $n$
We have the following for $n \ge 4$:
• The trivial one-dimensional representation as well as the one-dimensional representation with kernel $\langle a \rangle$ are preserved by all automorphisms.
• The other two one-dimensional representations are interchanged by an outer automorphism.
• For the two-dimensional representations, there are $\tau(n) - 2$ equivalence classes under the action of the automorphism group, where $\tau(n)$ is the number of divisors of $n$. Specifically, for every divisor $d > 2$ of $n$, there is an equivalence class of irreducible two-dimensional representations of size $\varphi(d)$ (where $\varphi$ is the Euler totient function) comprising the $k^{th}$ irreducible representation for all $k$ with $\operatorname{gcd}(n,k) = n/d$.
## Relation with representations of subgroups
### Induced representations from the cyclic maximal subgroup
All the two-dimensional irreducible representations are obtained as induced representations from one-dimensional complex representations of the cyclic subgroup $\langle a \rangle$. More specifically:
• The representation $a \mapsto e^{2\pi ik/n}$ induces the corresponding two-dimensional representation for $k$.
• The representations for $k$ and $n - k$, though inequivalent as one-dimensional representations, induce equivalent two-dimensional representations.
• For $n$ odd, the only $k$ for which we get a reducible two-dimensional representation is $k = 0$. Thus, there are $(n-1)/2$ irreducible representations coming from the $n-1$ values $1,2, \dots, n-1$, and there are two reducible representations coming from the decomposition of the induced representation from $k = 0$.
• For $n$ even, $k = 0$ and $k = n/2$ are the only cases where we get a reducible two-dimensional representation. Thus, there are $(n-2)/2$ irreducible representations coming from the other $n-2$ values, and there are four reducible representations coming from the decomposition of the induced representations for $k = 0$ and $k = n/2$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 249, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.963060200214386, "perplexity": 315.36662753827227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00473.warc.gz"} |
https://byjus.com/physics/electric-field-lines/ | # Electric Field Lines
## What is Electric Field Line?
Electric field lines are an excellent way of visualizing electric fields. They were first introduced by Michael Faraday himself.
A field line is drawn tangential to the net at a point. Thus at any point, the tangent to the electric field line matches the direction of the electric field at that point. Secondly, the relative density of field lines around a point corresponds to the relative strength (magnitude) of the electric field at that point. In other words, if you see more electric field lines in the vicinity of point A as compared to point B, then the electric field is stronger at point A.
## Properties of Electric Field Lines
• The field lines never intersect each other.
• The field lines are perpendicular to the surface of the charge.
• The magnitude of charge and the number of field lines, both are proportional to each other.
• The start point of the field lines is at the positive charge and end at the negative charge.
• For the field lines to either start or end at infinity, a single charge must be used.
### Electric Field Lines Attraction and Repulsion
Electric field lines always point away from a positive charge and towards a negative point. In fact, electric fields originate at a positive charge and terminate at a negative charge.
Electric field of point charges
Also, field lines never cross each other. If they do, it implies that there are two directions for the electric field at that point. But this is impossible since electric fields add up vectorially at any point and remember that “A field line is drawn tangential to the net electric field at a point”. Thus, electric field lines can never intersect one another.
As said before field lines are a great way to visualize electric fields. You can almost feel the attraction between unlike charges and the repulsion between like charges as though they are trying to push each other away.
Electric field on the left image explains how like charges repel and right image explains how unlike charges attract
Coming to our initial example of static charge on hair, the direction in which charged hair stands up traces the local electric field lines. The charges on the hair exert forces on the hair strand as they attempt to leak into the surrounding uncharged space. The hair aligns accordingly so that there is no net force acting on it and inadvertently traces the electric field lines.
### Rules for Drawing Electric Field Lines
Following are the rules for drawing electric field lines:
1. The field line begins at the charge and ends either at the charge or at infinity.
2. When the field is stronger, the field lines are closer to each other.
3. The number of field lines depends on the charge.
4. The field lines should never crossover.
5. Electric field and electric field line are tangent at the point where they pass through. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.891805112361908, "perplexity": 287.9204495903758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585305.53/warc/CC-MAIN-20211020090145-20211020120145-00368.warc.gz"} |
https://en.wikipedia.org/wiki/Subordinator_(mathematics) | # Subordinator (mathematics)
In the mathematics of probability, a subordinator is a concept related to stochastic processes. A subordinator is itself a stochastic process of the evolution of time within another stochastic process, the subordinated stochastic process. In other words, a subordinator will determine the random number of "time steps" that occur within the subordinated process for a given unit of chronological time.
In order to be a subordinator a process must be a Lévy process.[1] It also must be increasing, almost surely.[1]
## Definition
A subordinator is an increasing (a.s.) Lévy process.[2]
## Examples
The variance gamma process can be described as a Brownian motion subject to a gamma subordinator.[1] If a Brownian motion, ${\displaystyle W(t)}$, with drift ${\displaystyle \theta t}$ is subjected to a random time change which follows a gamma process, ${\displaystyle \Gamma (t;1,\nu )}$, the variance gamma process will follow:
${\displaystyle X^{VG}(t;\sigma ,\nu ,\theta )\;:=\;\theta \,\Gamma (t;1,\nu )+\sigma \,W(\Gamma (t;1,\nu )).}$
The Cauchy process can be described as a Brownian motion subject to a Lévy subordinator.[1]
## References
1. ^ a b c d Applebaum, D. "Lectures on Lévy processes and Stochastic calculus, Braunschweig; Lecture 2: Lévy processes" (PDF). University of Sheffield. pp. 37–53.
2. ^ Lévy Processes and Stochastic Calculus (2nd ed.). Cambridge: Cambridge University Press. 2009-05-11. ISBN 9780521738651. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9254822134971619, "perplexity": 856.8891583383719}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189495.77/warc/CC-MAIN-20170322212949-00608-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/help-with-kinematics.185304/ | # Help with kinematics
1. Sep 17, 2007
### devilsangels287
1. The problem statement, all variables and given/known data
A sailboard is sailing at 6.5 when a gust of wind hits, causing it to accelerate at 0.48 at a 35 angle to its original direction of motion.
If the acceleration lasts 6.3 , what is the board's net displacement during the wind gust?
2. Relevant equations
Ok so far, I have drawn all of the motion graphs.
So I got
X Y
X_0= 0m Y_0= 0m
X_F=? Y_F=?
V_0x= 6.5m/s v_oy=0m/s
V_fy= 6.5m/s v_fy=?
a= 0 m/s^2 a= .48m/s^2
t= 6.3secs t=6.3secs
3. The attempt at a solution
And assuming those are correct. I got v_fy= 4.55 m/s by using 6.5tan35 = v_fy
then I used this equation: x=x_0+v_ot+(1/2)at^2 and got
X_f= 40.95m
Y_f= 9.5m
And then I added them together and then its 50.45m??? Is this correct, because online it said it was wrong?
Thanks for helping!
2. Sep 17, 2007
### learningphysics
suppose it's initially moving eastbound at 6.5m/s. ie: northbound velocity = 0.
so acceleration 0.48m/s^2 east 30 degrees north.
What is the component of acceleration in the east/west direction?
What is the component of acceleration in the north/south direction?
So you can divide the problem into the east/west part (to get the displacement east/west), and the north/south part (to get the displacement north/south)... and work them separetly... each part is a uniform acceleration problem.
3. Mar 7, 2010
Need help with kinematics
I figured out this problem:
You step off a cliff 30 meters high.
A. How long will it take to hit the water below?
0=30-4.9t^2
4.9t^2=30
t^2= 6.122
t= 2.47
B. What is your velocity (mph) when you hit the water?
Vty= Voy-9.8t
Vty-Voy-9.8(2.47)
Vty=-24.25 m/s = 54.223 mph
Can anyone help me solve this one, based on the above problem?:
2. How high would the cliff have to be in 1. above if your velocity hitting the water was 100 mph?
Thanks!
Have something to add?
Similar Discussions: Help with kinematics | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8503007888793945, "perplexity": 3619.805185667929}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501172077.66/warc/CC-MAIN-20170219104612-00585-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.jobilize.com/course/section/simple-random-samples-and-statistics-by-openstax?qcr=www.quizover.com | # 13.2 Simple random samples and statistics
Page 1 / 3
The (simple) random sample, is basic to much of classical statistics. Once formulated, we may apply probability theory to exhibit several basic ideas of statistical analysis. A population may be most any collection of individuals or entities. Associated with each member is a quantity or a feature that can be assigned a number. The population distribution is the distribution of that quantity among the members of the population.To obtain information about the population distribution, we select “at random” a subset of the population and observe how the quantity varies over the sample. Hopefully, the distribution in the sample will give a useful approximation to the population distribution. We obtain values of such quantities as the mean and variance in the sample (which are random quantities) and use these as estimators for corresponding population parameters (which are fixed). Probability analysis provides estimates of the variation of the sample parameters about the corresponding population parameters.
## Simple random samples and statistics
We formulate the notion of a (simple) random sample , which is basic to much of classical statistics. Once formulated, we may apply probability theory to exhibitseveral basic ideas of statistical analysis.
We begin with the notion of a population distribution . A population may be most any collection of individuals or entities. Associated with each member is aquantity or a feature that can be assigned a number. The quantity varies throughout the population. The population distribution is the distribution of that quantityamong the members of the population.
If each member could be observed, the population distribution could be determined completely. However, that is not always feasible. In order to obtain informationabout the population distribution, we select “at random” a subset of the population and observe how the quantity varies over the sample. Hopefully, thesample distribution will give a useful approximation to the population distribution.
The sampling process
We take a sample of size n , which means we select n members of the population and observe the quantity associated with each. The selection is done in such a manner thaton any trial each member is equally likely to be selected. Also, the sampling is done in such a way that the result of any one selection does not affect, and is not affected by,the others. It appears that we are describing a composite trial. We model the sampling process as follows:
• Let X i , $1\le i\le n$ be the random variable for the i th component trial. Then the class $\left\{{X}_{i}:1\le i\le n\right\}$ is iid, with each member having the population distribution.
This provides a model for sampling either from a very large population (often referred to as an infinite population) or sampling with replacement from a small population.
The goal is to determine as much as possible about the character of the population. Two important parameters are the mean and the variance. We want the population mean and thepopulation variance. If the sample is representative of the population, then the sample mean and the sample variance should approximate the population quantities.
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.852319598197937, "perplexity": 1164.994669642684}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00011.warc.gz"} |
https://www.physicsforums.com/threads/rc-circuit-maximum-voltage.230160/ | # RC circuit maximum voltage
• Start date
• #1
82
0
i cant figure this out. In an RC circuit, why is the maximum voltage of a capactior greater then the maximum voltage of a resistor? can anyone guide me in the right direction
Related Classical Physics News on Phys.org
• #2
454
0
It depends on the circuit. Circuits are possible where it is greater, smaller or the same.
for an simple RC circuit with one ideal DC voltage source, one resistor and one capacitor, either series or parallel, the maximum voltage would be the same.
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452 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8030725121498108, "perplexity": 4040.8868118441505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00182.warc.gz"} |
http://mathhelpforum.com/trigonometry/38135-solving-x-given-interval-need-help.html | # Thread: Solving for x in the given interval NEED HELP !!!
1. ## Solving for x in the given interval NEED HELP !!!
im stuck on three questions and i really need to solve them
a) sec2x + 1/cosx = 0 x is between and equal to 0 and pie
b)cos^2x+2sinxcosx-sin^2x=0 x is between and equal to 0 and two pie
c)2tanx = secx x is between and equal to negative pie and two pie
2. Hello, math71321!
$\displaystyle b)\;\;\cos^2\!x + 2\sin x\cos x-\sin^2\!x\:=\:0 \qquad0 \leq x \leq 2\pi$
We have: . $\displaystyle \underbrace{\cos^2\!x - \sin^2\!x} + \underbrace{2\sin x\cos x} \;=\;0$
. . . . . . . . . . $\displaystyle \cos2x \quad\;\;+ \quad\;\;\sin 2x \quad=\;0$
Then: .$\displaystyle \sin2x \:=\:-\cos2x \quad\Rightarrow\quad\frac{\sin2x}{\cos2x} \:=\:-1 \quad\Rightarrow\quad \tan2x \:=\:-1$
Hence: .$\displaystyle 2x \;=\;\frac{3\pi}{4},\:\frac{7\pi}{4},\:\frac{11\pi }{4},\:\frac{15\pi}{4}$
Therefore: .$\displaystyle \boxed{x \;=\;\frac{3\pi}{8},\:\frac{7\pi}{8},\:\frac{11\pi }{8},\:\frac{15\pi}{8}}$
$\displaystyle c)\;\;2\tan x \:= \:\sec x\qquad -\pi \leq x \leq 2\pi$
Square both sides: .$\displaystyle 4\tan^2\!x \:=\:\sec^2\!x$
. . . . . . . . . . . . $\displaystyle 4\overbrace{(\sec^2\!x - 1)} \:=\:\sec^2\!x$
which simplifies to: .$\displaystyle 3\sec^2\!x \:=\:4\quad\Rightarrow\quad\sec^2\!x \:=\:\frac{4}{3}\quad\Rightarrow\quad \sec x \:=\:\pm\frac{2}{\sqrt{3}}$
. . Therefore: .$\displaystyle \boxed{x \;=\;\pm\frac{\pi}{6},\:\pm\frac{5\pi}{6},\:\frac{ 7\pi}{6},\:\frac{11\pi}{6}}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9902688264846802, "perplexity": 1016.3348053277642}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948047.85/warc/CC-MAIN-20180426012045-20180426032045-00137.warc.gz"} |
https://byorgey.wordpress.com/2010/01/26/the-random-graph/ | ## The random graph
Today in my finite model theory class we learned about the Rado graph, which is a graph (unique up to isomorphism among countable graphs) with the extension property: given any two disjoint finite sets of vertices $U$ and $V$, there exists some other vertex $w$ which is adjacent to every vertex in $U$ and none of the vertices in $V$.
This graph has some rather astonishing properties. Here’s one: consider starting with $n$ vertices and picking each edge with probability $1/2$. Clearly, there are $2^{\binom n 2}$ different graphs you can get, each with equal probability; this defines a uniform random distribution over simple graphs with $n$ vertices. What if you start with a countably infinite number of vertices instead? The surprising answer is that with probability 1 you get the Rado graph. Yes indeed, the Rado graph is extremely random. It is so random that it is also called “THE random graph”.
```SimpleGraph getRandomGraph()
{
return radoGraph; // chosen by fair coin flips.
// guaranteed to be random.
}
```
(See http://xkcd.com/221/.)
This entry was posted in grad school, humor and tagged , , . Bookmark the permalink.
### 3 Responses to The random graph
1. Mark Dominus says:
Excellent. Thanks for this post!
This reminds me of a paper of Janós Pach about “universal” graphs. It is quite easy to produce a countable graph G that has the property that for every finite or countable graph H, H is an induced subgraph of G. The graph that Pach constructed was not the Rado graph, although it seems to me that the Rado graph also has this property.
But Pach showed that there is no graph G that contains every graph H other than Kω, the complete graph on infinitely many vertices. (The proof is pretty easy.) But if I am right about the Rado graph being universal in the sense of the previous paragraph, then by this theorem it must contain a Kω.
Hm, yes, it does. It follows immediately from the extension property, and in the Wikipedia notation of http://en.wikipedia.org/wiki/Rado_graph, the vertices at http://www.research.att.com/~njas/sequences/index.html?q=0,1,3,11,205 form a Kω.
Sorry, just thinking aloud. Thanks again!
2. Mark Dominus says:
P.S. misspelled “János”. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 9, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.962845504283905, "perplexity": 349.60073754343074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988310.3/warc/CC-MAIN-20150728002308-00342-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://electronics.stackexchange.com/questions/396341/how-to-derive-voltage-applied-to-an-antenna-where-transmitter-power-is-known/396342 | # How to derive voltage applied to an antenna, where transmitter power is known & identical tx/rx antennas are used with known gain/impedance?
Question:
Imagine a transmitter + transmitter antenna with the following characteristics:
• 433Mhz transmitter transmitting 10dBm
• Into an impedance matched (50 ohms) antenna, omni-directional, 3dBi gain
• 50 ohm impedance antenna, receiver circuitry presents 50 ohms
• Receiving antenna is an identical 3dBi omni-directional antenna.
How would I go about calculating the voltage the receiver circuitry would see? I'm trying to learn RF electronics by building a detector for a constant 433.92Mhz carrier, and I can't predict how my diode will behave without knowing the voltage applied across it.
My attempt from first principles:
10 dBm is 10mW of power, 3dBi of gain means 10mW X 2 = 20mW radiated from the antenna under ideal conditions.
Based on the inverse square law, we get:
$$P = { 0.02 \over 4 \Pi r ^2 }$$
$$P = { 0.02 \over 4 \Pi 40 ^2 }$$
$$P = { 0.02 \over 4 \Pi 40 ^2 }$$
$$P = 9.94718394^{-7}$$
Yipes that seems small.
This is the bit where I get stuck. Can I just multiply it by the receiving gain (3dBi, basically 2x) and sub it into ohms law (with the impedance of the antenna, 50 ohms) combined with the power formula (P = IV)?
$$P = ({V \over R}) V$$
$$2 \times 9.94718394^{-7} = ({V \over 50}) V$$
After doubling the power (because the receiving antenna gives me 3dBi gain) and plugging this formula into Wolfram Alpha to solve, I get ~10mV.
This seems low and I'm not confident with the logic of my derivation. Is this correct?
• 10 dBm is 10 mW of power and not 10 mV. – Andy aka Sep 16 '18 at 9:54
• 10mW is what I meant, edited question to have correct units. Thanks. – 64bit_twitchyliquid Sep 16 '18 at 10:02
You should consider the Friis transmission loss (or link loss) equation for free space: -
Loss (dB) = 32.45 + 20$log_{10}$(f) + 20$log_{10}$(d)
Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency.
It is based on free-space and accounts for the reduction in received signal as frequency rises due to the received antenna becoming smaller. This equation is also for antennas that transmit equally in all directions (so called isotropic antennas). With real antennas there is always a gain because real antennas don't transmit equally in all directions hence there is always one direction where there is a higher power density of transmission.
The equation also assumes that your antennas are not so close that a proper EM wave hasn't formed. Translated, this means that your antennas must be at least one wavelength apart and, as you appear to be using 40 metres in your question and, your frequency is 433 MHz, there is no problem.
Putting numbers in gets you a link loss of 32.45 dB + 52.73 dB - 27.96 dB or 57.22 dB. Given your antennas have a gain of 3 dB each the link loss reduces to about 51.2 dB. Sanity check using on-line calculator: -
Given your transmit power is +10 dBm, the received signal power will be 52.2 dB down on +10 dBm or -41.2 dBm or about 0.076 uW.
Another related Q and A
• Awesome that theres an equation for loss-over-distance, thanks! In terms of working from the received power (3.09uW) to voltage however, can I just substitute it into ohms law & P=IR (as I did above) to calculate the voltage? – 64bit_twitchyliquid Sep 16 '18 at 10:37
• Yes you can be do be aware that a dipole antenna will have an effective source resistance (aka radiation resistance) of about 73 ohms and the power delivered in the equations above assume that the correct resistance is matched to the antenna resistance. Be also aware that an antennas impedance varies dramatically when the antenna length doesn't precisely match the wavelength of the carrrier. – Andy aka Sep 16 '18 at 10:41
Using the formula Power = VoltsRMS ^2 / Resistance, and then changing Resistance to Impedance, we have
Power = VoltsRMS^2 / Impedance
and we solve for VoltsRMS
VoltsRMS = squareroot (Power * Impedance)
Now at one milliWatt power (0.001 watt) and 50 ohms, the math produces
VoltsRMS = squareroot (0.001 * 50) = squareroot (1 / 20) = 1 / squareroot(20)
VoltsRMS = 1 / 4.47 = 0.223 volts rms
VoltsPP = 0.223 * 2 * 1.414 = 0.632 voltsPP
Are your numbers consistent? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8721857666969299, "perplexity": 1787.3035499169894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574084.88/warc/CC-MAIN-20190920221241-20190921003241-00233.warc.gz"} |
http://mathhelpforum.com/math-challenge-problems/6470-question-2-a.html | # Math Help - Question 2
1. ## Question 2
A triangular number is the sequence:
1,3,6,10,15,21,...
A square number is the sequence:
1,4,9,16,25,...
Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1.
2. Originally Posted by ThePerfectHacker
A triangular number is the sequence:
1,3,6,10,15,21,...
A square number is the sequence:
1,4,9,16,25,...
Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1.
To clarify: You mean that the set of numbers that belong to both the triangular sequence and the square sequence has an infinite number of members?
-Dan
3. Originally Posted by topsquark
To clarify: You mean that the set of numbers that belong to both the triangular sequence and the square sequence has an infinite number of members?
-Dan
Yes
4. I do not understand the last line of the solution?
5. Hello, TPHacker!
Generally, with this type of problem we can generate solutions
with a recurrence formula of the form:
. . nth term .= .k·(preceding term) ± (term before) ... for some constant k.
In short, I "eyeballed" it . . .
6. Beautiful job Soroban +rep+
Never seen it done like that.
I have modified (not really the right word) your recurrence relation into a different form:
---
I always tried to show that the cubic sequence and triangular sequence only have a trivial solution i.e. 1. And it seems to be true for higher orders also but I was unable to prove that, you know how?
Attached Thumbnails
7. Hello again, TPHacker!
I got a different closed form for the recurrence.
It's not my own solution; I learned the method from a few of my books.
You may have seen something similar while exploring.
Now that LaTeX is back, I'll revise this post.
We have the recurrence: . $f(n) \:=\:6\cdot f(n-1) - f(n-2)$
Assume that $f(n)$ is exponential: . $f(n) \:=\:X^n$
The equation becomes: . $X^n \:=\:6X^{n-1} - X^{n-2}$
Divide by $X^{n-2}$ and we have: . $X^2 - 6X + 1 \:= \:0$
. . which has roots: . $X \:=\:3 \pm 2\sqrt{2}$
We form a linear combination of the two roots:
. . $f(n) \:=\:A(3 + 2\sqrt{2})^n + B(3 - 2\sqrt{2})^n$
To solve for $A$ and $B$, we use the first two values of the sequence:
. . $\begin{array}{cc}f(1) \:= \\ f(2) \:=\end{array} \begin{array}{cc} A(3 + 2\sqrt{2}) + (3 - 2\sqrt{2})\\ A(3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2\end{array}\begin{array}{cc}= \:1 \\ = \:6\end{array}$
Solve the system and get: . $A \,= \,\frac{1}{4\sqrt{2}},\;\;B \,= \,-\frac{1}{4\sqrt{2}}$
Therefore: . $f(n) \;= \;\frac{(3 + 2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{4\sqrt{2}}$
By the way: . $3 \pm 2\sqrt{2} \:=\:(1 \pm \sqrt{2})^2$
. . so you can rewrite the formula if you like . . .
8. Besides for Soroban's solution.... I have 3 more (1 of which I am too lazy to post).
Let $t_n$ be the $n-th$triangular number.
Solution 1)
If t_n is a square then t_{4n(n+1)} is a square. How you are suppopsed to see that, I have no idea.
Solution 2)
Following the start of Soroban;s solution we arrive at the square root of 8k² +1, which is the "Pellian equation". Thus it has infinitely many solutions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9745483994483948, "perplexity": 927.5240086809335}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637903439.28/warc/CC-MAIN-20141030025823-00125-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://gams.cam.nist.gov/13.9 | # §13.9 Zeros
## §13.9(i) Zeros of $M\left(a,b,z\right)$
If $a$ and $b-a\neq 0,-1,-2,\dots$, then $M\left(a,b,z\right)$ has infinitely many $z$-zeros in $\mathbb{C}$. When $a,b\in\mathbb{R}$ the number of real zeros is finite. Let $p(a,b)$ be the number of positive zeros. Then
13.9.1 $\displaystyle p(a,b)$ $\displaystyle=\left\lceil-a\right\rceil,$ $a<0$, $b\geq 0$, ⓘ Symbols: $\left\lceil\NVar{x}\right\rceil$: ceiling of $x$ and $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E1 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.2 $\displaystyle p(a,b)$ $\displaystyle=0,$ $a\geq 0$, $b\geq 0$, ⓘ Symbols: $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E2 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13 13.9.3 $\displaystyle p(a,b)$ $\displaystyle=1,$ $a\geq 0$, $-1, ⓘ Symbols: $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E3 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13
13.9.4 $p(a,b)=\left\lfloor-\tfrac{1}{2}b\right\rfloor-\left\lfloor-\tfrac{1}{2}(b+1)% \right\rfloor,$ $a\geq 0$, $b\leq-1$. ⓘ Symbols: $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E4 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13
13.9.5 $p(a,b)=\left\lceil-a\right\rceil-\left\lceil-b\right\rceil,$ $\left\lceil-a\right\rceil\geq\left\lceil-b\right\rceil$, $a<0$, $b<0$, ⓘ Symbols: $\left\lceil\NVar{x}\right\rceil$: ceiling of $x$ and $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E5 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13
13.9.6 $p(a,b)=\left\lfloor\tfrac{1}{2}\left(\left\lceil-b\right\rceil-\left\lceil-a% \right\rceil+1\right)\right\rfloor-\left\lfloor\tfrac{1}{2}\left(\left\lceil-b% \right\rceil-\left\lceil-a\right\rceil\right)\right\rfloor,$ $\left\lceil-b\right\rceil>\left\lceil-a\right\rceil>0$.
The number of negative real zeros $n(a,b)$ is given by
13.9.7 $n(a,b)=p(b-a,b).$ ⓘ Defines: $n(a,b)$: number of negative real zeros (locally) Symbols: $p(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E7 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13
When $a<0$ and $b>0$ let $\phi_{r}$, $r=1,2,3,\dots$, be the positive zeros of $M\left(a,b,x\right)$ arranged in increasing order of magnitude, and let $j_{b-1,r}$ be the $r$th positive zero of the Bessel function $J_{b-1}\left(x\right)$10.21(i)). Then
13.9.8 $\phi_{r}=\frac{j_{b-1,r}^{2}}{2b-4a}\left(1+\frac{2b(b-2)+j_{b-1,r}^{2}}{3(2b-% 4a)^{2}}\right)+O\left(\frac{1}{a^{5}}\right),$ ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $r=1,2,3,\dots$, $\phi_{r}$: positive zeros and $j_{b,r}$: positive zero of Bessel Referenced by: §13.22, §13.9(i) Permalink: http://dlmf.nist.gov/13.9.E8 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13
as $a\to-\infty$ with $r$ fixed.
Inequalities for $\phi_{r}$ are given in Gatteschi (1990), and identities involving infinite series of all of the complex zeros of $M\left(a,b,x\right)$ are given in Ahmed and Muldoon (1980).
For fixed $a,b\in\mathbb{C}$ the large $z$-zeros of $M\left(a,b,z\right)$ satisfy
13.9.9 $z=\pm(2n+a)\pi\mathrm{i}+\ln\left(-\frac{\Gamma\left(a\right)}{\Gamma\left(b-a% \right)}\left(\pm 2n\pi\mathrm{i}\right)^{b-2a}\right)+O\left(n^{-1}\ln n% \right),$
where $n$ is a large positive integer, and the logarithm takes its principal value (§4.2(i)).
Let $P_{\alpha}$ denote the closure of the domain that is bounded by the parabola $y^{2}=4\alpha(x+\alpha)$ and contains the origin. Then $M\left(a,b,z\right)$ has no zeros in the regions $P_{\ifrac{b}{a}}$, if $0; $P_{1}$, if $1\leq a\leq b$; $P_{\alpha}$, where $\alpha=\ifrac{(2a-b+ab)}{(a(a+1))}$, if $0 and $a\leq b<\ifrac{2a}{(1-a)}$. The same results apply for the $n$th partial sums of the Maclaurin series (13.2.2) of $M\left(a,b,z\right)$.
More information on the location of real zeros can be found in Zarzo et al. (1995) and Segura (2008).
For fixed $b$ and $z$ in $\mathbb{C}$ the large $a$-zeros of $M\left(a,b,z\right)$ are given by
13.9.10 $a=-\frac{\pi^{2}}{4z}\left(n^{2}+(b-\tfrac{3}{2})n\right)-\frac{1}{16z}\left((% b-\tfrac{3}{2})^{2}\pi^{2}+\tfrac{4}{3}z^{2}-8b(z-1)-4b^{2}-3\right)+O\left(n^% {-1}\right),$ ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $\pi$: the ratio of the circumference of a circle to its diameter, $n$: nonnegative integer and $z$: complex variable Referenced by: §13.9(i) Permalink: http://dlmf.nist.gov/13.9.E10 Encodings: TeX, pMML, png See also: Annotations for 13.9(i), 13.9 and 13
where $n$ is a large positive integer.
For fixed $a$ and $z$ in $\mathbb{C}$ the function $M\left(a,b,z\right)$ has only a finite number of $b$-zeros.
## §13.9(ii) Zeros of $U\left(a,b,z\right)$
For fixed $a$ and $b$ in $\mathbb{C}$, $U\left(a,b,z\right)$ has a finite number of $z$-zeros in the sector $|\operatorname{ph}z|\leq\tfrac{3}{2}\pi-\delta(<\tfrac{3}{2}\pi)$. Let $T(a,b)$ be the total number of zeros in the sector $|\operatorname{ph}z|<\pi$, $P(a,b)$ be the corresponding number of positive zeros, and $a$, $b$, and $a-b+1$ be nonintegers. For the case $b\leq 1$
13.9.11 $T(a,b)=\left\lfloor-a\right\rfloor+1,$ $a<0$, $\Gamma\left(a\right)\Gamma\left(a-b+1\right)>0$, ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $T(a,b)$: number of zeros Permalink: http://dlmf.nist.gov/13.9.E11 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13
13.9.12 $T(a,b)=\left\lfloor-a\right\rfloor,$ $a<0$, $\Gamma\left(a\right)\Gamma\left(a-b+1\right)<0$, ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $\left\lfloor\NVar{x}\right\rfloor$: floor of $x$ and $T(a,b)$: number of zeros Permalink: http://dlmf.nist.gov/13.9.E12 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13
13.9.13 $T(a,b)=0,$ $a>0$, ⓘ Symbols: $T(a,b)$: number of zeros Permalink: http://dlmf.nist.gov/13.9.E13 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13
and
13.9.14 $P(a,b)=\left\lceil b-a-1\right\rceil,$ $a+1, ⓘ Symbols: $\left\lceil\NVar{x}\right\rceil$: ceiling of $x$ and $P(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E14 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13
13.9.15 $P(a,b)=0,$ $a+1\geq b$. ⓘ Symbols: $P(a,b)$: number of positive zeros Permalink: http://dlmf.nist.gov/13.9.E15 Encodings: TeX, pMML, png See also: Annotations for 13.9(ii), 13.9 and 13
For the case $b\geq 1$ we can use $T(a,b)=T(a-b+1,2-b)$ and $P(a,b)=P(a-b+1,2-b)$.
In Wimp (1965) it is shown that if $a,b\in\mathbb{R}$ and $2a-b>-1$, then $U\left(a,b,z\right)$ has no zeros in the sector $|\operatorname{ph}{z}|\leq\frac{1}{2}\pi$.
Inequalities for the zeros of $U\left(a,b,x\right)$ are given in Gatteschi (1990). See also Segura (2008).
For fixed $b$ and $z$ in $\mathbb{C}$ the large $a$-zeros of $U\left(a,b,z\right)$ are given by
13.9.16 $a=-n-\frac{2}{\pi}\sqrt{zn}-\frac{2z}{\pi^{2}}+\tfrac{1}{2}b+\tfrac{1}{4}+% \frac{z^{2}\left(\frac{1}{3}-4\pi^{-2}\right)+z-(b-1)^{2}+\frac{1}{4}}{4\pi% \sqrt{zn}}+O\left(\frac{1}{n}\right),$ ⓘ Symbols: $O\left(\NVar{x}\right)$: order not exceeding, $\sim$: Poincaré asymptotic expansion, $\pi$: the ratio of the circumference of a circle to its diameter, $n$: nonnegative integer and $z$: complex variable Referenced by: §13.9(ii), § Recent News, Other Changes, Other Changes Permalink: http://dlmf.nist.gov/13.9.E16 Encodings: TeX, pMML, png Errata (effective with 1.0.13): In applying changes in Version 1.0.12 to this equation, an editing error was made; it has been corrected. Reported 2016-09-12 by Adri Olde Daalhuis Errata (effective with 1.0.12): Originally this equation was expressed in terms of the asymptotic symbol $\sim$. As a consequence of the use of the $O$ order symbol on the right hand side, $\sim$ was replaced by $=$. Reported 2016-07-11 by Rudi Weikard See also: Annotations for 13.9(ii), 13.9 and 13
where $n$ is a large positive integer.
For fixed $a$ and $z$ in $\mathbb{C}$, $U\left(a,b,z\right)$ has two infinite strings of $b$-zeros that are asymptotic to the imaginary axis as $|b|\to\infty$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 177, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9719674587249756, "perplexity": 3102.2376360214535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00068.warc.gz"} |
http://www.sciencebits.com/taxonomy/term/11?page=2 | ## On Climate Sensitivity and why it is probably small
### What is climate sensitivity?
The equilibrium climate sensitivity refers to the equilibrium change in average global surface air temperature following a unit change in the radiative forcing. This sensitivity (often denoted as λ) therefore has units of °C/(W/m2).
Often, instead &\lambda;, the sensitivity is expressed through the temperature change &Delta Tx2, in response to a doubled atmospheric CO2 content, which is equivalent to a radiative forcing of 3.8 W/m2. Thus, &Delta Tx2 = 3.8 W/m2 λ
## Standing on ice - When is it possible?
Ever wondered whether it was sufficiently cold for sufficiently long to allow you to stand on ice, without falling in? I once did. Here is an estimate for the duration required to reach a given thickness. Actually, it is a lower limit, since we assume a few simplifying assumptions.
## Exhale Condensation Calculator
If the temperature is low enough or the humidity high, you can observe condensation (i.e., "fog") forming in your exhaled breath. This calculator estimates whether your exhaled breath will condense, and if so, the range of mixing ratios for which the "fog" will form and the maximum condensed water content (the higher it is, the "thicker" the condensation).
If you're interested, there is a much more detailed explanations of the condensation process.
Exhaled Condensation Calculator
Using the above equations, we can calculate whether the exhaled air will condense. Enter the conditions of the outside air (and modify the exhaled air parameters if you wish), to see whether your breath will condense, or not.
## Blogroll
Sensible Climate
Physics and more Other | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9003207087516785, "perplexity": 2115.988547397927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00452.warc.gz"} |
https://collaborate.princeton.edu/en/publications/topology-bounded-superfluid-weight-in-twisted-bilayer-graphene | # Topology-Bounded Superfluid Weight in Twisted Bilayer Graphene
Fang Xie, Zhida Song, Biao Lian, B. Andrei Bernevig
Research output: Contribution to journalArticlepeer-review
23 Scopus citations
## Abstract
While regular flat bands are good for enhancing the density of states and hence the gap, they are detrimental to the superfluid weight. We show that the predicted nontrivial topology of the two lowest flat bands of twisted bilayer graphene (TBLG) plays an important role in the enhancement of the superfluid weight and hence of superconductivity. We derive the superfluid weight (phase stiffness) of the TBLG superconducting flat bands with a uniform pairing, and show that it can be expressed as an integral of the Fubini-Study metric of the flat bands. This mirrors results already obtained for nonzero Chern number bands even though the TBLG flat bands have zero Chern number. We further show that the metric integral is lower bounded by the topological C2zT Wilson loop winding number of TBLG flat bands, which renders that the superfluid weight is also bounded by this topological index. In contrast, trivial flat bands have a zero superfluid weight. The superfluid weight is crucial in determining the Berezinskii-Kosterlitz-Thouless transition temperature of the superconductor. Based on the transition temperature measured in TBLG experiments, we estimate the topological contribution of the superfluid weight in TBLG.
Original language English (US) 167002 Physical review letters 124 16 https://doi.org/10.1103/PhysRevLett.124.167002 Published - Apr 24 2020
## All Science Journal Classification (ASJC) codes
• Physics and Astronomy(all)
## Fingerprint
Dive into the research topics of 'Topology-Bounded Superfluid Weight in Twisted Bilayer Graphene'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9550732374191284, "perplexity": 2169.0543825892114}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00342.warc.gz"} |
https://www.hepdata.net/search/?q=cmenergies%3A%7B1.3+TO+1.4%7D&author=Aubert%2C+Bernard | Showing 5 of 5 results
#### rivet Analysis The e+ e- ---> 2(pi+ pi-) pi0, 2(pi+ pi-) eta, K+ K- pi+ pi- pi0 and K+ K- pi+ pi- eta Cross Sections Measured with Initial-State Radiation
The collaboration Aubert, Bernard ; Bona, M. ; Boutigny, D. ; et al.
Phys.Rev. D76 (2007) 092005, 2007.
Inspire Record 758568
0 data tables match query
#### rivet Analysis Measurements of $e^{+} e^{-} \to K^{+} K^{-} \eta$, $K^{+} K^{-} \pi^0$ and $K^0_{s} K^\pm \pi^\mp$ cross- sections using initial state radiation events
The collaboration Aubert, Bernard ; Bona, M. ; Boutigny, D. ; et al.
Phys.Rev. D77 (2008) 092002, 2008.
Inspire Record 765258
This paper reports measurements of processes: e+e- -> gamma KsK+pi-, e+e- -> gamma K+K-pi0, e+e- -> gamma phi eta, and e+e- -> gamma phi pi0. The initial state radiated photon allows to cover the hadronic final state in the energy range from thresholds up to ~4.6 GeV. The overall size of the data sample analyzed is 232 fb-1, collected by the BaBar detector running at the PEP-II e+e- storage ring. From the Dalitz plot analysis of the KsK+pi- final state, moduli and relative phase of the isoscalar and the isovector components of the e+e- -> K K*(892) cross section are determined. Parameters of phi and rho recurrences are also measured, using a global fitting procedure which exploits the interconnection among amplitudes, moduli and phases of the e+e- -> KsK+pi-, K+K-pi0, phi eta final states. The cross section for the OZI-forbidden process e+e- -> phi pi0, and the J/psi branching fractions to KK*(892) and K+K-eta are also measured.
0 data tables match query
#### rivet Analysis The $e^+e^- \to \pi^+ \pi^- \pi^+ \pi^-$, $K^+ K^- \pi^+ \pi^-$, and $K^+ K^- K^+ K^-$ cross sections at center-of-mass energies 0.5-GeV - 4.5-GeV measured with initial-state radiation
The collaboration Aubert, Bernard ; Barate, R. ; Boutigny, D. ; et al.
Phys.Rev. D71 (2005) 052001, 2005.
Inspire Record 676691
We study the process $e^+e^-\to\pi^+\pi^-\pi^+\pi^-\gamma$, with a hard photon radiated from the initial state. About 60,000 fully reconstructed events have been selected from 89 $fb^{-1}$ of BaBar data. The invariant mass of the hadronic final state defines the effective \epem center-of-mass energy, so that these data can be compared with the corresponding direct $e^+e^-$ measurements. From the $4\pi$-mass spectrum, the cross section for the process $e^+e^-\to\pi^+\pi^-\pi^+\pi^-$ is measured for center-of-mass energies from 0.6 to 4.5 $GeV/c^2$. The uncertainty in the cross section measurement is typically 5%. We also measure the cross sections for the final states $K^+ K^- \pi^+\pi^-$ and $K^+ K^- K^+ K^-$. We observe the $J/\psi$ in all three final states and measure the corresponding branching fractions. We search for X(3872) in $J/\psi (\to\mu^+\mu^-) \pi^+\pi^-$ and obtain an upper limit on the product of the $e^+e^-$ width of the X(3872) and the branching fraction for $X(3872) \to J/\psi\pi^+\pi^-$.
0 data tables match query
#### rivet Analysis Cross sections for the reactions $e^+ e^-\to K_S^0 K_L^0$, $K_S^0 K_L^0 \pi^+\pi^-$, $K_S^0 K_S^0 \pi^+\pi^-$, and $K_S^0 K_S^0 K^+K^-$ from events with initial-state radiation
The collaboration Lees, J.P. ; Poireau, V. ; Tisserand, V. ; et al.
Phys.Rev. D89 (2014) 092002, 2014.
Inspire Record 1287920
We study the processes $e^+ e^-\to K_S^0 K_L^0 \gamma$, $K_S^0 K_L^0 \pi^+\pi^-\gamma$, $K_S^0 K_S^0 \pi^+\pi^-\gamma$, and $K_S^0 K_S^0 K^+K^-\gamma$, where the photon is radiated from the initial state, providing cross section measurements for the hadronic states over a continuum of center-of-mass energies. The results are based on 469 fb$^{-1}$ of data collected with the BaBar detector at SLAC. We observe the $\phi(1020)$ resonance in the $K_S^0 K_L^0$ final state and measure the product of its electronic width and branching fraction with about 3% uncertainty. We present a measurement of the $e^+ e^-\to K_S^0 K_L^0$ cross section in the energy range from 1.06 to 2.2 GeV and observe the production of a resonance at 1.67 GeV. We present the first measurements of the $e^+ e^-\to K_S^0 K_L^0 \pi^+\pi^-$, $K_S^0 K_S^0 \pi^+\pi^-$, and $K_S^0 K_S^0 K^+K^-$ cross sections, and study the intermediate resonance structures. We obtain the first observations of \jpsi decay to the $K_S^0 K_L^0 \pi^+\pi^-$, $K_S^0 K_S^0 \pi^+\pi^-$, and $K_S^0 K_S^0 K^+K^-$ final states.
0 data tables match query
#### rivet Analysis Cross Sections for the Reactions e+e- --> K+ K- pi+pi-, K+ K- pi0pi0, and K+ K- K+ K- Measured Using Initial-State Radiation Events
The collaboration Lees, J.P. ; Poireau, V. ; Prencipe, E. ; et al.
Phys.Rev. D86 (2012) 012008, 2012.
Inspire Record 892684
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https://astarmathsandphysics.com/university-maths-notes/elementary-calculus/1788-maxima-minima-and-saddle-points-the-second-partials-test.html | ## Maxima, Minima and Saddle Points – The Second Partials Test
Suppose thathas continuous second partial derivatives in a neighbourhood of and thatat
DefineatForm the discriminantIfthenis a saddle point.
Ifthenhas a local minimum at
Ifand a local maximum atif | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8183179497718811, "perplexity": 2736.511149999521}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00585.warc.gz"} |
https://eprints.soton.ac.uk/417377/ | The University of Southampton
University of Southampton Institutional Repository
# On the joint and marginal densities of instrumental variable estimators in a general structural equation
Hillier, Grant H. (1985) On the joint and marginal densities of instrumental variable estimators in a general structural equation. Econometric Theory, 1 (1), 53-72.
Record type: Article
## Abstract
Starting from the conditional density of the instrumental variable (IV) estimator given the right-hand-side endogenous variables, we provide an alternative derivation of Phillips' result on the joint density of the IV estimator for the endogenous coefficients, and derive an expression for the marginal density of a linear combination of these coefficients. In addition, we extend Phillips' approximation to the joint density to 0(T−2,) and show how this result can be used to improve the approximation to the marginal density. Explicit formulae are given for the special case of no simultaneity, and the case of an equation with just three endogenous variables. The classical assumptions of independent normal reduced-form errors are employed throughout.
Published date: April 1985
## Identifiers
Local EPrints ID: 417377
URI: http://eprints.soton.ac.uk/id/eprint/417377
ISSN: 0266-4666
PURE UUID: 8c15d018-7615-4aca-a83e-dc62522ee011
ORCID for Grant H. Hillier: orcid.org/0000-0003-3261-5766
## Catalogue record
Date deposited: 30 Jan 2018 17:30 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9189020991325378, "perplexity": 1529.7401769757137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662530066.45/warc/CC-MAIN-20220519204127-20220519234127-00031.warc.gz"} |
https://physics.stackexchange.com/questions/256538/the-convective-term-in-reynolds-transport-theorem | # The convective term in Reynolds Transport theorem
Reynolds Transport theorem states that $$\frac{D B_{sys}}{Dt}=\frac{\partial B_{CV}}{\partial t}-\dot{B_{in}}+\dot{B_{out}}$$ where $B_{in,t+\Delta t}=b_1m_{1,t+\Delta t}=b_1\rho_{1}\forall_{in,t+\Delta t}=b_1\rho_1V_1\Delta tA_1$, then
$$\dot{B_{in}}=\lim_{\Delta t\to0}\frac{B_{in,t+\Delta}}{\Delta t}=\lim_{\Delta t\to0}\frac{b_1\rho_1V_1\Delta tA_1}{\Delta t}=b_1\rho_1V_1A_1$$
Now my question is, how come volume $\forall_{in}$ at $t+\Delta t$ is equal to $V_1 \Delta tA_1$?, where $V_1$ is the velocity and $A_1$ is the area. Dimension-ally, this relation works but still do not understand the physics.
$V_1 A_1$ is basically the volumetric flowrate. So in a time $\Delta t$, a particle moving with flowrate $V_1 A_1$, sweeps out a volume $V_1 A_1 \Delta t$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9901400208473206, "perplexity": 186.6103396479117}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178369721.76/warc/CC-MAIN-20210305030131-20210305060131-00515.warc.gz"} |
https://optimization-online.org/author/jfp/ | ## An easily computable upper bound on the Hoffman constant for homogeneous inequality systems
Let $A\in \mathbb{R}^{m\times n}\setminus \{0\}$ and $P:=\{x:Ax\le 0\}$. This paper provides a procedure to compute an upper bound on the following {\em homogeneous Hoffman constant} $H_0(A) := \sup_{u\in \mathbb{R}^n \setminus P} \frac{\text{dist}(u,P)}{\text{dist}(Au, \mathbb{R}^m_-)}.$ In sharp contrast to the intractability of computing more general Hoffman constants, the procedure described in this paper is entirely … Read more
## Affine invariant convergence rates of the conditional gradient method
We show that the conditional gradient method for the convex composite problem $\min_x\{f(x) + \Psi(x)\}$ generates primal and dual iterates with a duality gap converging to zero provided a suitable growth property holds and the algorithm makes a judicious choice of stepsizes. The rate of convergence of the duality gap to zero ranges from sublinear … Read more
## Projection and rescaling algorithm for finding most interior solutions to polyhedral conic systems
We propose a simple projection and rescaling algorithm that finds {\em most interior} solutions to the pair of feasibility problems $\text{find} x\in L\cap \R^n_{+} \text{ and } \text{find} \; \hat x\in L^\perp\cap\R^n_{+},$ where $L$ is a linear subspace of $\R^n$ and $L^\perp$ is its orthogonal complement. The algorithm complements a basic procedure that … Read more
## Equivalences among the chi measure, Hoffman constant, and Renegar’s distance to ill-posedness
We show the equivalence among the following three condition measures of a full column rank matrix $A$: the chi measure, the signed Hoffman constant, and the signed distance to ill-posedness. The latter two measures are constructed via suitable collections of matrices obtained by flipping the signs of some rows of $A$. Our results provide a … Read more
## New characterizations of Hoffman constants for systems of linear constraints
We give a characterization of the Hoffman constant of a system of linear constraints in $\R^n$ relative to a reference polyhedron $R\subseteq\R^n$. The reference polyhedron $R$ represents constraints that are easy to satisfy such as box constraints. In the special case $R = \R^n$, we obtain a novel characterization of the classical Hoffman constant. More … Read more
## Generalized conditional subgradient and generalized mirror descent: duality, convergence, and symmetry
We provide new insight into a generalized conditional subgradient algorithm and a generalized mirror descent algorithm for the convex minimization problem $\min_x \; \{f(Ax) + h(x)\}.$ As Bach showed in [SIAM J. Optim., 25 (2015), pp. 115–129], applying either of these two algorithms to this problem is equivalent to applying the other one to its … Read more
## The condition number of a function relative to a set
The condition number of a differentiable convex function, namely the ratio of its smoothness to strong convexity constants, is closely tied to fundamental properties of the function. In particular, the condition number of a quadratic convex function is the square of the aspect ratio of a canonical ellipsoid associated to the function. Furthermore, the condition … Read more | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9476986527442932, "perplexity": 426.7158505787808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00217.warc.gz"} |
http://math.stackexchange.com/questions/265451/the-cardinality-of-mathbbr-mathbb-q/265509 | # The cardinality of $\mathbb{R}/\mathbb Q$
How to prove the cardinality of $\mathbb{R}/ \mathbb Q$ is equal to the cardinality of $\mathbb{R}$
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Sorry, I wrote it wrong, I meant to be the cardinality of $\mathbb{R}$ mod $\mathbb{Q}$ – Alex Dec 26 '12 at 17:52
You mean the quotient group $\mathbb R / \mathbb Q$ ?? – GEdgar Dec 26 '12 at 17:54
Alex, please edit to clarify what you are asking. If you are asking about the cardinality of the set of irrational numbers (as it is currently written), see math.stackexchange.com/questions/105990/… and math.stackexchange.com/questions/72130/…. – Jonas Meyer Dec 26 '12 at 18:03
This is a consequence of the axiom of choice. How to prove it, depends a bit on your background, that is, what results you can assume, how familiar you are with choice, etc. Could you specify some of it? – Andrés Caicedo Dec 26 '12 at 19:27
If it is the quotient, in fact, without the axiom of choice, it turns out that $\mathbb{R}/\mathbb{Q}$ can have strictly larger cardinality than $\mathbb{R}$... math.stackexchange.com/a/243549/32178 – KSackel Dec 26 '12 at 20:02
To prove equality we need to either find a bijection between the sets, or two injections between them.
As noted in the comments, this cannot be proved without the axiom of choice. So I am going to use it freely.
Assuming the axiom of choice, if so, we have a function $f\colon\mathbb{R/Q\to R}$ which chooses $f(A)\in A$ for every $A\in\mathbb{R/Q}$. This is an injection because if $A\neq A'$ then $f(A)\in A$ and $f(A)\notin A'$, and vice versa, therefore $f(A)\neq f(A')$.
On the other hand, let $V=\operatorname{rng}(f)$, then $\mathbb R$ is a countable union of copies of $V$, namely $\bigcup_{q\in\mathbb Q}q+V$. Therefore $|V|\cdot\aleph_0=2^{\aleph_0}$. Again, using the axiom of choice, we have that $2^{\aleph_0}=|V|\cdot\aleph_0=\max\{|V|,\aleph_0\}=|V|$.
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Remark: This answers the original version of the question.
The following bijection uses Hilbert's infinite hotel.
The rationals can be enumerated as $q_0,q_1,q_2,\dots$ in various explicit ways.
Define $f: \mathbb{R}\to \mathbb{R}\setminus \mathbb{Q}$ as follows.
If $x$ does not have shape $q$, or $\sqrt{3}+q\sqrt{2}$, where $q$ is rational, let $f(x)=x$.
If $x$ is the rational $q_i$, let $f(x)=\sqrt{3}+q_{2i}\sqrt{2}$.
If $x=\sqrt{3}+q_i\sqrt{2}$, let $f(x)=\sqrt{3}+q_{2i+1}\sqrt{2}$.
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Awesome.[filler] – akkkk Dec 26 '12 at 19:28
$\mathbb{R}= (\mathbb{R} \backslash \mathbb{Q}) \sqcup \mathbb{Q}$, where $\mathbb{Q}$ is countable and $\mathbb{R} \backslash \mathbb{Q}$ infinite, so $$|\mathbb{R}|= |\mathbb{R} \backslash \mathbb{Q} | + | \mathbb{Q}|= \max ( |\mathbb{R} \backslash \mathbb{Q} |, |\mathbb{Q}|) = |\mathbb{R} \backslash \mathbb{Q} |$$
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Indeed, $\mathbb R \setminus \mathbb Q$ is infinite. And so is $\mathbb Q$. You want to write "uncountable", I think. – Rudy the Reindeer Dec 26 '12 at 18:57
@MattN. - being infinite suffices since $\aleph_0$ + any infinite cardinal $k$ is $k$ – Belgi Dec 26 '12 at 19:36
@Belgi Right, I misread the question as having to show that $\mathbb R \setminus \mathbb Q$ is uncountable. – Rudy the Reindeer Dec 26 '12 at 20:52
I don't see what I would have expected to be the "standard answer" to this question, so let me leave it in the hopes it will be helpful to someone.
Proposition: Let $G$ be an infinite group, and let $H$ be a subgroup with $\# H < \# G$. Then $\# G/H = \# G$.
Proof: Let $\{g_i\}_{i \in G/H}$ be a system of coset representatives for $H$ in $G$: then every element $x$ in $G$ can be written as $x = g_{i_x} h_x$ for unique $h_x \in H$ and $i_x \in G/H$. (Note that there is no canonical system of coset representatives: getting one is an archetypical use of the Axiom of Choice.) Thus we have defined a bijection from $G$ to $G/H \times H$, so $\# G = \# G/H \cdot \# H$. Since $\# G$ is infinite, so must be at least one of $\# G/H$, $\# H$, and then standard cardinal arithmetic (again AC gets used...) gives that
$\# G = \# G/H \cdot \# H = \max(\#G/H, \# H)$.
Since we've assumed $\# H < \# G$, we conclude $\# G = \#G/H$.
This applies in particular with $G = \mathbb{R}$, $H = \mathbb{Q}$ to give $\# \mathbb{R}/\mathbb{Q} = \# \mathbb{R} = 2^{\aleph_0}$.
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To see that the uses of choice here are unavoidable: math.stackexchange.com/questions/243544/… – Andrés Caicedo Apr 23 '13 at 5:58
One can even write a general form theorem from cardinal arithmetics. If $f\colon A\to B$ is surjective, and every fiber have the same cardinality, which is less than that of $A$ then $|A|=|B|$. – Asaf Karagila Apr 23 '13 at 6:17
$\mathbb{R}/\mathbb{Q}$ and $\mathbb{R}$ are both size continuum. So by assuming as vector spaces over the $\mathbb{Q}$ they must have continuum size bases. We know that if two vector spaces have bases of the same size then they are isomorphic.
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So why exactly are they both size continuum? – akkkk Dec 26 '12 at 18:45
@akkkk: Because if B be a basis of R over Q (as a vector space, that is). The cardinality of B is c and that R/Q is like removing one basis element, or one copy of Q. Is that wrong? – S. Snape Dec 26 '12 at 18:48
Babak the OP wants to prove that $\mathbb{R}-\mathbb{Q}$ is of size continuum which you assumed in the first line. – Belgi Dec 26 '12 at 19:35
@Belgi: I wonder what should I do here. Firstly, the OP wrote $R-Q$ and some different good answer came to him, and suddenly he changed his mind to quotient group. See what he replied in the comment. – S. Snape Dec 26 '12 at 19:43
@Belgi: Actually, in the comments the OP says that he is looking for the cardinality of the quotient. – Asaf Karagila Dec 29 '12 at 15:02
Well, heuristically it can go as follows.
We can easily construct a bijection between $\mathbb{N}$ and $\mathbb{N}-\{1\}$. Just send $n \rightarrow n+1$.
If we subtract one element $x_1$ from $\mathbb{R}$, we can decompose $\mathbb{R}$ as $(\mathbb{R}-\mathbb{N})\cup\mathbb{N}$, and $\mathbb{R}-\{x_0\}=(\mathbb{R}-\mathbb{N})\cup(\mathbb{N}-{x_0})$. And we apply the above trick to the latter part.
If we subtract countable elements, say, $x_i$ for all $i\in\mathbb{N}$, we can choose a countable familiy of subsets with cardinal equal to that of $\mathbb{N}$, each containing one of $x_i$. Now decompose $\mathbb{R}$ as a countable union and do as above and you will solve the problem.
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But we don't subtract the rationals. We take the quotient. – Asaf Karagila Jan 4 '13 at 14:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9851799607276917, "perplexity": 211.22196395614571}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701169261.41/warc/CC-MAIN-20160205193929-00087-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/101636/jacobis-criterion-for-projective-schemes | # Jacobi's criterion for projective schemes?
When can we apply the Jacobi's criterion for the projective variety $V(f_{1}, \ldots, f_{r}) \subset \mathbb{P}^{n}$ in order to find the singularities of the scheme $\mathrm{Proj} \left( k[x_{1}, \ldots, x_{n+1}] / (f_{1}, \ldots, f_{r}) \right)$?
In Hartshorne's book Algebraic Geometry, Proposition II.2.6, we have a fully faithful functor from the category of varieties over $k$ to the category of schemes over $k$, but it seems to provide information only for the closed points of the scheme.
Thank you.
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Check out the general Jacobi's criterion in EGA or in Liu's book. – Martin Brandenburg Jan 23 '12 at 15:33
The Jacobian Criterion can be applied to any kind of point on a projective (or affine) variety, closed or not. In the non-closed case one has to adapt the requirement for the rank of the Jacobian appropriately. The field $k$ should be perfect. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9013843536376953, "perplexity": 246.89261532324932}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645264370.66/warc/CC-MAIN-20150827031424-00133-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://zbmath.org/?q=an:0818.35136 | ×
# zbMATH — the first resource for mathematics
Another step toward the solution of the Pompeiu problem in the plane. (English) Zbl 0818.35136
Let $$\Omega \subset \mathbb{R}^ 2$$ be a bounded open set for which $$f \equiv 0$$ is the only continuous function on $$\mathbb{R}^ 2$$ such that (1) $$\int_{\sigma (\Omega)} f(x) dx = 0$$ for every rigid motion $$\sigma$$ of the plane. A bounded open set $$\Omega \subset \mathbb{R}^ 2$$ is said to have the Pompeiu property if there exist no nontrivial continuous function on $$\mathbb{R}^ 2$$ for which (1) holds. Several authors have determined bounded domains $$\Omega \subset \mathbb{R}^ 2$$ having the Pompeiu property.
The main result of this paper is the following: Let $$\Omega \subset \mathbb{R}^ 2$$ be a bounded simply-connected open set whose boundary is a closed simple curve parametrized by $$x(s) = (x_ 1(s), x_ 2(s))$$, $$s \in [- \pi, \pi]$$. Suppose that there exist $$M,N \in \mathbb{Z}$$ and $$a_ k \in \mathbb{C}$$, $$k = - M, \dots, N$$, with $$a_ M$$, $$a_ N \neq 0$$, such that $x_ 1 (s) + x_ 2 (s) = \sum^ N_{k= -M} a_ k e^{iks}.$ Let $$x_ 1 (z)$$, $$x_ 2 (z)$$ be the analytic extension of $$x_ 1 (s)$$ and $$x_ 2 (s)$$ satisfying $$(x_ 1'(z), x_ 2'(z))\neq (0,0) \in \mathbb{C}^ 2$$ for every $$z \in \mathbb{C}$$. Then $$\Omega$$ has the Pompeiu property.
Reviewer: G.Anger (Berlin)
##### MSC:
35R30 Inverse problems for PDEs 31A25 Boundary value and inverse problems for harmonic functions in two dimensions 35P05 General topics in linear spectral theory for PDEs 42B10 Fourier and Fourier-Stieltjes transforms and other transforms of Fourier type
##### Keywords:
Pompeiu problem; Pompeiu property
Full Text:
##### References:
[1] Berenstein C., An inverse spectral theorem and its relation to the Pompeiu problem 37 pp 124– (1980) · Zbl 0449.35024 [2] Berenstein C. A., An inverse Neumann problem 382 pp 1– (1987) · Zbl 0623.35078 [3] Brown L., A note on the Pompeiu problem for convex domains 259 pp 107– (1982) · Zbl 0464.30035 [4] Brown L., Spectral synthesis and the Pompeiu problem 23 pp 125– (1973) · Zbl 0265.46044 [5] Caffarelli L. A., The regularity of free boundaries in higher dimensions 139 pp 155– (1977) [6] Chakalov L., Sur un probl‘me de D. Pompeiu 40 pp 1– (1944) · Zbl 0063.07316 [7] Garofalo N., Asymptotic expansions for a class of Fourier integrals and applications to the Pompeiu problem 56 pp 1– (1991) · Zbl 0737.35146 [8] Garofalo N., New results on the Pompeiu problem 325 pp 243– (1991) · Zbl 0737.35147 [9] Pompeiu D., Sur certains systèmes d’équations linéries et sur une propriété intégrate des functions de plusieurs variables 188 pp 1138– (1929) [10] Pompeiu D., Sur une propriété intégrate des fonctions de deux variables réelles. 15 pp 265– (1929) [11] Riemann, R. B. 1953. ”Sullo svolgimento del quoziente di due serie ipergeometriche in funzione continua infinita.ldquo;”. complete works, Dover [12] Williams S. A., A partial solution of the Pompeiu problem 223 pp 183– (1976) · Zbl 0329.35045 [13] Williams S. A., Analyticity of the boundary for Lipschitz domains without the Pompeiu property. 30 pp 357– (1981) · Zbl 0439.35046 [14] Yau S. T., Seminar on Differential Geometry 102 (1982) · Zbl 0471.00020 [15] Zalcman L., Analyticity and the Pompeiu problem 47 pp 237– (1972) · Zbl 0251.30047
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. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9391898512840271, "perplexity": 1859.7258552145368}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00174.warc.gz"} |
https://www.physicsforums.com/threads/two-rotational-motion-questions.191775/ | # Two Rotational Motion Questions
1. Oct 16, 2007
### ccsmarty
The first one:
1. The problem statement, all variables and given/known data
1) A compact disc (CD) stores music in a coded pattern of tiny pits 10^-7 m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm , respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive.
2. Relevant equations
alpha_avg = (omega_2 - omega_1) / (t2 - t1)
3. The attempt at a solution
I tried to take the average of the inner and outer angular velocities, and put that in for omega_2, and find the average that way, but I don't think I can do that.
The second one:
1. The problem statement, all variables and given/known data
2) At t = 0 a grinding wheel has an angular velocity of 27.0 rad/s. It has a constant angular acceleration of 26.0 rad/s^2 until a circuit breaker trips at time t = 2.00 s. From then on, it turns through an angle 433 rad as it coasts to a stop at constant angular acceleration. At what time did it stop?
2. Relevant equations
omega_2 = omega_1 + alpha * t
delta_2 - delta_1 = omega_1 * t + 0.5 * alpha * t^2
3. The attempt at a solution
I tried using a system of equations using the two equations above to solve for t, but I can't seem to get the right t value.
Any guidance is greatly appreciated on either problem. Thanks in advance.
2. Oct 17, 2007
### learningphysics
For the first problem, your omega_inner and omega_outer look good to me. Why not just take (omega_outer - omega_inner)/(74*60)... that should be the answer.
For the second problem, think of the angular velocity and acceleration, just like kinematics formulas...
What is the angular velocity at t = 2?
Then you can use the equation,
angle traversed = [(omega_1 + omega_2)/2]*t, so solve for how long it takes to go through the 433 rad...
3. Oct 17, 2007
### ccsmarty
Ok, thanks so much for your help. It makes more sense this way, than the way I initially tried to tackle the problems.
Thanks again :) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9438005089759827, "perplexity": 697.908544009634}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982982027.87/warc/CC-MAIN-20160823200942-00238-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/71763/why-doesnt-mass-of-bob-affect-time-period | # Why doesn't mass of bob affect time period?
The gravitation formula says $$F = \frac{G m_1 m_2}{r^2} \, ,$$ so if the mass of a bob increases then the torque on it should also increase because the force increased. So, it should go faster and thus the oscillation period should be decrease.
My physics book says that period is only affected by effective length and $g$. Why doesn't mass of bob affect the period?
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The question is not clear to me. Can you explain the situation a bit more? – Ali Jul 21 '13 at 13:59
ok,let me make the question clearer – svineet Jul 21 '13 at 14:05
Hint: Why doesn't the mass of an object affect how long it takes to fall from a given height to the ground? Mass doesn't only appear in the gravitation formula: It also appears in $\vec F=m\vec A$ – james large Sep 14 at 21:28
You're hitting on the Equivalence principle and the Eötvös experiment. – WetSavannaAnimal aka Rod Vance Sep 14 at 22:54
A very loose answer would be that the time period actually depends upon the angular acceleration and not the torque.
Just like the time taken for a object to fall through a height of $h$, depends on the gravitational acceleration and not the mass, i.e. if you drop a sponge ball or you jump yourself, you both will cover height $h$ in the same time(of course neglecting air resistance).
Similarly, the time period of a pendulum doesn't depend upon the mass, or rather the inertia of the pendulum, but only on the angular acceleration due to gravity.
Now you might ask that in this case, it should also not depend upon the length, but the term of length comes when you calculate the angular acceleration due to the acceleration of gravity.
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For the same reason objects of different masses fall at the same acceleration (neglecting drag): because while the force is proportional to the mass and the acceleration is inversely proportional to mass.
Doing the falling case o avoid having to deal with the vectors in the pendulum we get
$$a = \frac{F}{m} = \frac{G\frac{Mm}{r^2}}{m} = G\frac{M}{r^2}$$
where $M$ is the mass of the planet, $m$ is the mass of the object you are dropping and $r$ is the radius of the planet.
The mass of the minor object falls out of the kinematics.
The same thing happens in the case of the pendulum: the force includes a factor of $m$, but the acceleration does not.
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A pendulum in a gravitational field experiences an instantaneous torque about its pivot point of $$\vec{\Gamma} = \vec{r}\times m\vec{g}$$ where $\vec{g}$ is the instantaneous gravitational field, and $r$ is the distance from the pivot point to the CoM.
For purposes of this answer $$\vec{g}=-G\frac{ M_E}{(R_E + h)^2}\hat{k}$$ where
• $m$ is the pendulum mass,
• $M_E$ is the Earth's mass,
• $R_E$ is the distance from the gravitational center of the Earth to the center of mass of the pendulum at rest, and
• $h=r(1-\cos\theta)$ is the height of the CoM when the pendulum is oscillating.
Let's assume the pendulum is oscillating in a plane, so we can write $$\Gamma = mgr\sin\theta = \mathcal{I}\frac{d^2\theta}{dt^2}.$$
$\mathcal{I}$ is the moment of inertia of the pendulum about the pivot point, and will have the form of $mb^2$, where $b$ is a geometric size and mass distribution factor. Any rigid object you want to consider can have its moment of inertia put in that form. From this we see quickly that the actual mass of the object disappears: $$\frac{d^2\theta}{dt^2} = \frac{gr}{b^2}\sin\theta.$$
All that remains is to find $b$ which depends only on how the mass is distributed, not how much mass is present.
We also see that this is not simple harmonic motion. While the factor $g$ is not constant, it only introduces an anharmonic factor of $$1-\frac{r\theta^2}{R_E}-\frac{r^2\theta^4}{4R_E^2}$$. The $\sin\theta$ term introduces a larger anharmonicity because $$\sin\theta\simeq \theta-\frac{\theta^3}{6} = \theta\left(1-\frac{\theta^2}{6}\right).$$
So we see that 1) the mass doesn't matter, but the distribution of mass does, 2) the variation in height producing a variation in gravitational field only has a $(r/R_E)\theta^2$ affect, 3) the amplitude of the angle due to the $\sin\theta$ term becomes important when $\theta > 0.1$ radian.
Considering point 2), most pendula have $r<10 m$ and $R_E = 6.38\times 10^6$ m.
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Well the easy way is that the mass has a opposite affect when the bob goes up again on the other side. The deacceleration and the acceleration will equal out so the period will always be the same what ever mass you have.
O
/I\
/ I \
/ I \
0 0 0
A C B
Here you have a diagram to represent it. You drop the pendulum at B and it accelerates until it hits C then it will slow down. The mass will increase the deacceleration. So the acceleration and the deacceleration will equal out.
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NOOOOO!!! This is wrong. Granted the pendulum formula (T = 2 * pi * sqrt(L / g)) does not take into account mass of the bob, much less the pendulum, mass can and does affect the pendulum period. The Pendulum Formula is accurate and i give it credit, but its variables are broadly defined. T represents time or period, and g represents gravitational acceleration. I have no problem with those, but it's L that bothers me. To assume L is the distance from the point of axis to the bottom tip of the pendulum is to assume that the pendulum has an equal density throughout and its center of gravity lies directly in the center of the pendulum. However, with most pendulums this isn't the case. The bob, or weight on the pendulum, affects the location of the center of gravity. When a bob is added below the center of balance, the gravitational center of the entire pendulum is shifted downwards to some degree. Instead of saying L = the length of the pendulum, it's better to say that L = 2 * (distance between center of balance and pivot point).
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NOOOOO!!! This is wrong. – Jimmy360 Jun 14 at 21:56
This post is misleading. The formula for the period of the pendulum is correct, and is obtained assuming constant gravitational field. This is generally correct, but would be a problem for the case of very large pendulum, which clears the original question. Usually L is the distance to pendulum's center of mass and when a blob is involved, is usually assumed to contain the whole mass of the pendulum. But these are not considerations leading to the mass invariant formula for T. – rmhleo Jun 14 at 22:09
Certainly for real pendulums the ratio of the mass of the bob and the mass of the support affects the position of the center of mass of the device. But if you read closely any non-introductory account you'll see that L is defined the distance from the pivot to the CoM. An important detail, but not one that is appropriate to spend much time on in the first introduction (which usually assumes a massless rod). – dmckee Jun 14 at 23:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9428714513778687, "perplexity": 230.94449102606677}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398445033.85/warc/CC-MAIN-20151124205405-00180-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=componentLibrary/signalBlocks/signalConverters/UnitConversionBlock | Conversion Block
Converts a signal from one unit to another
Description The Conversion Block component converts a signal from one unit to another. This component is a fixed causality component that you can use to perform unit conversions for dimensions such as time, temperature, speed, pressure, mass, and volume.
Connections
Name Description $u$ Connection of the real input signal to be converted $y$ Connection of the real output signal containing the input signal, $u$, in another unit
Parameters
Symbol Default Units Description dimension Acceleration - The dimension for which to convert units. from unit $\frac{m}{{s}^{2}}$ - Original units of the signal. to unit $\frac{m}{{s}^{2}}$ - Units to which you want to convert the signal. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8338956832885742, "perplexity": 1014.0092823700323}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00023.warc.gz"} |
https://brilliant.org/problems/a-number-theory-problem-by-swadesh-rath/ | # A number theory problem by Swadesh Rath
Number Theory Level pending
xy - 6(x+y) = 0 find number of ordered solutions for all integral values of x,y such that x<y or x = y
× | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8665782809257507, "perplexity": 1693.004218937085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720356.2/warc/CC-MAIN-20161020183840-00251-ip-10-171-6-4.ec2.internal.warc.gz"} |
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