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https://scribesoftimbuktu.com/convert-to-a-fraction-19-3-5/ | # Convert to a Fraction 19 3/5%
1935%
Convert the percentage to a fraction by placing the expression over 100. Percentage means ‘out of 100’.
1935100
Reduce the fraction.
A mixed number is an addition of its whole and fractional parts.
19+35100
Simplify the numerator.
To write 19 as a fraction with a common denominator, multiply by 55.
19⋅55+35100
Combine 19 and 55.
19⋅55+35100
Combine the numerators over the common denominator.
19⋅5+35100
Simplify the numerator.
Multiply 19 by 5.
95+35100
985100
985100
985100
Multiply the numerator by the reciprocal of the denominator.
985⋅1100
Cancel the common factor of 2.
Factor 2 out of 98.
2(49)5⋅1100
Factor 2 out of 100.
2⋅495⋅12⋅50
Cancel the common factor.
2⋅495⋅12⋅50
Rewrite the expression.
495⋅150
495⋅150
Multiply 495 and 150.
495⋅50
Multiply 5 by 50.
49250
49250
Convert to a Fraction 19 3/5%
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
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http://www.ck12.org/physics/Longitudinal-Wave/lesson/Longitudinal-Waves-Intermediate/r13/ | <meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.
Longitudinal Wave
Mechanical waves in which energy travels parallel to the wave's oscillation.
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Progress
Progress
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Longitudinal Waves
Playing with a Slinky is a childhood tradition, but few children realize they are actually playing with physics.
Longitudinal Waves
Like transverse waves, longitudinal waves are mechanical waves, which means they transfer energy through a medium. Unlike transverse waves, longitudinal waves cause the particles of medium to move parallel to the direction of the wave. They are most common in springs, where they are caused by the pushing an pulling of the spring. Although the surface waves on water are transverse waves, fluids (liquids, gases, and plasmas) usually transmit longitudinal waves.
As shown in the image below, longitudinal waves are a series of compressions and rarefactions, or expansions. The wavelength of longitudinal waves is measured by the distance separating the densest compressions. The amplitude of longitudinal waves is the difference in media density between the undisturbed density to the highest density in a compression.
Example Problem: A sonar signal (sonar is sound waves traveling through water) of 1.00×106\begin{align*}1.00 \times 10^6\end{align*} Hz frequency has a wavelength of 1.50 mm in water. What is the speed of sound in water?
Solution: v=λf=(0.00150 m)(1.00×106 s1)=1500 m/s\begin{align*}v=\lambda f =(0.00150 \ \text{m})(1.00 \times 10^6 \ \text{s}^{-1})=1500 \ \text{m/s}\end{align*}
Example Problem: A sound wave of wavelength 0.70 m and velocity 330 m/s is produced for 0.50 s.
1. What is the frequency of the wave?
2. How many complete waves are emitted in this time interval?
3. After 0.50 s, how far is the wave front from the source of the sound?
Solution:
1. f=vλ=330 m/s0.70 m=470 s1\begin{align*}f=\frac{v}{\lambda}=\frac{330 \ \text{m/s}}{0.70 \ \text{m}}=470 \ \text{s}^{-1}\end{align*}
2. complete waves=(470 cycles/s)(0.50 s)=235 cycles\begin{align*}\text{complete waves} = (470 \ \text{cycles/s})(0.50 \ \text{s}) = 235 \ \text{cycles}\end{align*}
3. distance=(330 m/s)(0.50 s)=115 m\begin{align*}\text{distance} = (330 \ \text{m/s})(0.50 \ \text{s}) = 115 \ \text{m}\end{align*}
Summary
• Longitudinal waves cause the particles of medium to move parallel to the direction of the wave.
Practice
Questions
The following video explains how a tuning fork creates sound. Use this resource to answer the questions that follow.
1. In your own words, how are compressions and rarefactions produced by the tuning fork?
2. Make a guess why sound can easily travel around corners (Hint: think of its medium).
Review
Questions
1. Some giant ocean waves have a wavelength of 25 m long, and travel at speeds of 6.5 m/s. Determine the frequency and period of such a wave.
2. Bats use sound echoes to navigate and hunt. They emit pulses of high frequency sound waves which reflect off obstacles in the surroundings. By detecting the time delay between the emission and return of a pulse, a bat can determine the location of the object. What is the time delay between the sending and return of a pulse from an object located 12.5 m away? The approximate speed of sound is 340 m/s.
3. Sachi is listening to her favorite radio station which broadcasts radio signals with a frequency of 1.023×108 Hz\begin{align*}1.023 \times 10^8 \ \text{Hz}\end{align*}. If the speed of the signals in air is 2.997×108 m/s\begin{align*}2.997 \times 10^8 \ \text{m/s}\end{align*}, what is the wavelength of these radio signals?
4. A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to move past a given point in 9.1 s. Determine the wavelength, frequency, and speed of this wave.
5. A sonar signal leaves a submarine, travels through the water to another submarine and reflects back to the original submarine in 4.00 s. If the frequency of the signal was 512 cycles per second and the wavelength of the signal was 2.93 m, how far away is the second submarine? | {"extraction_info": {"found_math": true, "script_math_tex": 7, "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": 0.9122370481491089, "perplexity": 1333.5587284639214}, "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-2015-40/segments/1443736678190.22/warc/CC-MAIN-20151001215758-00247-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://encyclopediaofmath.org/wiki/Normally-imbedded_subspace | # Normally-imbedded subspace
2010 Mathematics Subject Classification: Primary: 54-XX [MSN][ZBL]
A subspace \$A\$ of a space \$X\$ such that for every neighbourhood \$U\$ of \$A\$ in \$X\$ there is a set \$H\$ that is the union of a countable family of sets closed in \$X\$ and with \$A \subset H \subset U\$. If \$A\$ is normally imbedded in \$X\$ and \$X\$ is normally imbedded in \$Y\$, then \$A\$ is normally imbedded in \$Y\$. A normally-imbedded subspace of a normal space is itself normal in the induced topology, which explains the name. Final compactness of a space is equivalent to its being normally imbedded in some (and hence in any) compactification of this space. Quite generally, a normally-imbedded subspace of a finally-compact space is itself finally compact.
#### References
[1] Yu.M. Smirnov, "On normally-imbedded sets of normal spaces" Mat. Sb. , 29 (1951) pp. 173–176 (In Russian) Zbl 0043.16502 | {"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": 0.9731720685958862, "perplexity": 3439.2660272366616}, "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-2021-25/segments/1623487610196.46/warc/CC-MAIN-20210613161945-20210613191945-00219.warc.gz"} |
http://launchloop.com/NearCircularOrbitsOlder?action=diff&rev1=12&rev2=13 | Differences between revisions 12 and 13
⇤ ← Revision 12 as of 2019-03-21 05:22:11 → Size: 14592 Editor: KeithLofstrom Comment: ← Revision 13 as of 2019-03-21 05:23:01 → ⇥ Size: 14538 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 3: Line 3: copied from http://server-sky.com/NearCircularOrbits
# Near Circular Orbits
## General Elliptical Orbits
In the orbital plane, neglecting the J_2 spherical oblateness parameter :
circular orbits gravitational parameter \large \mu = a^3 \omega^2 = G M semimajor axis \large a = \sqrt[3]{ \mu / \omega^2 } angular velocity \large \omega = \sqrt{ \mu / a^3 } sidereal period \large T = 2\pi / \omega = 2\pi \sqrt{ a^3/\mu } J_2 Speedup factor \large \omega'/\omega = 1+1.5 |J_2| (R_{eq}/a)^2 elliptical orbits eccentricity \large e = ( r_a - r_p ) / ( r_a + r_p ) velocity \large v_0 = \LARGE\sqrt{\mu\over{a (1 - e^2)}} \large = \LARGE\sqrt{{\mu \over 2}\left({1 \over r_a}+{1 \over r_p}\right)} true anomaly orbit angle from focus \Large \theta eccentric anomaly ellipse center angle \large E=\pm\arccos\Large\left({{e+\cos(\theta)}\over{1+e\cos(\theta)}}\right) periapsis apoapsis mean anomaly \large M = E - e \sin( E ) earth perigee apogee time from perigee \large t = M / \omega sun perihelion aphelion radius \large r=a\Large{{1-e^2}\over{1+e\cos(\theta)}} \huge\rightarrow \large r_p =( 1-e )a \large r_a =( 1+e )a tangential velocity \large v_{\perp}= v_0 ( 1+e \cos( \theta )) \huge\rightarrow \large v_p =(1+e)v_0 \large v_a =(1-e)v_0 radial velocity \large v_r = e v_0 \sin( \theta ) \huge\rightarrow 0 0 total velocity tangent to orbit \Large v=\LARGE \sqrt{{{2\mu}\over{r}}-{{\mu}\over{a}}} \huge\rightarrow \large v=v_p,~ r=r_p \large v=v_a,~ r=r_a orbit energy parameter \large C_3=-\mu/a = -v_p v_a \large \mu= a v_p v_a ~~~~~~~~~ 2 a= r_p + r_a angular momentum \large L =\LARGE \sqrt{ { 2 \mu ~ r_a ~ r_p } \over { r_a + r_p } } \large v_p = \LARGE \sqrt{ { 2 \mu ~ r_a } \over { r_p ( r_a + r_p ) } } \large v_a = \LARGE \sqrt{ { 2 \mu ~ r_p } \over { r_a ( r_a + r_p ) } } semimajor axis \large a={\LARGE{{r_p+r_a}\over{2}}}={\LARGE{{r_p}\over{1-e}}}={\LARGE{{r_a}\over{1+e}}} \large a = { \LARGE 1 \over { \huge { { 2 \over r_p } - { v_p^2 \over \mu } } } } \large a = { \LARGE 1 \over { \huge { 2 \over r_a } - { v_a^2 \over \mu } } } semilatus rectum \large l = \sqrt{ r_a r_p }
### Computing an orbit from an altitude and a velocity
Assume an Earth orbit, given a radius r and a velocity vector (speed v elevation angle from tangential/horizontal \beta such that the tangential velocity is v_\perp ~ = ~ v \cos( \beta ) ). The vector and the center of the Earth define a plane. Decompose the velocity vector into a radial and a tangential component in that plane. First, we find the semimajor axis:
\large a ~=~ { \huge { 1 \over { { 2 \over r } - { v^2 \over \mu } } } }
Next, compute the angular momentum:
\large L ~=~ r v ~ \cos( \beta )
Compute the eccentricity:
\large e ~=~ \sqrt{ 1 - {\LARGE {{ L^2 } \over { \mu a }} } }
apoapse ~=~ \large ( 1 + e ) a ~~~~~~ periapse ~=~ \large ( 1 - e ) a
### Changing perigee by changing apogee velocity
\large v_a^2 = 2 \mu \LARGE \left( { 1 \over r_a } - { 1 \over {r_a + r_p} } \right) ~ ~ ~ ~ ~ ~ \LARGE { { \partial v_a } \over { \partial r_p } } = { \mu \over { v_a ( r_a + r_p )^2 } } ~ ~ ~ ~ ~ ~ for nearly circular orbits: ~ ~ ~ \LARGE { { \partial v } \over { \partial r } } \approx { v \over { 4 r } }
### Fictional Forces in Orbit
In the rotating frame of a circular orbit, counterclockwise viewed from above the orbital plane, the directions are
direction unit vector description x \vec i Tangential to (along the line of) the orbit, in the orbital plane, pointing clockwise or westward y \vec j Radially outwards from the center of rotation, in the orbital plane z \vec k Perpendicular to the orbital plane, northwards
The rotation is expressed as \vec \Omega = \omega \vec k , where \omega is the angular velocity of the orbit . The radial vector \vec r is composed of \vec r = x \vec i + y \vec j + z \vec k .
Coriolis acceleration: \large \ddot{\vec r}_{Coriolis} = -2 \vec \Omega \times \dot { \vec r }
Centrifugal acceleration: \large \ddot{\vec r}_{Centrifugal} = - \vec \Omega \times \vec \Omega \times \vec r
FROM HERE DOWN, WORK IN PROGRESS, NOT VERIFIED:
• note - these are probably the Hill-Clohessy-Wiltshire equations
Centripedal (gravity) acceleration: \large \ddot{\vec r}_{Centripedal} = \omega^2 ( 2 y \vec k - x \vec i - z \vec j )
Scalar equations:
\large \ddot x = 2 \omega \dot y
\large \ddot y = -2 \omega \dot x + 3 \omega^2 y
\large \ddot z = -\omega^2 z
### Locus of Elliptical Orbit Position in Rotating Frame
For small eccentricities e, the deviation of the position of an object in a nearly circular eccentric orbit deviates from a ideal circular orbit as an ellipse, with a width of 4e and a height of 2e. This is difficult to compute analytically, but easy to show numerically. Here is a plot the normalized loci of the x,y position in the rotating frame, for various values of eccentricity:
If the whole orbit rotates east, clockwise when viewed from the north, then the object follows the locus above clockwise, with \theta = 0 at the bottom of the ellipsoid.
The figure is very close to an ellipse for small e, but not exactly so. The program computes the radial distance rr, and the radial position xr and yr as a function of \theta. The program then computes the eccentric anomaly E, and the mean anomaly M. It then rotates the position vector back along the orbit, to approximately vertical, then subtracts the unit vector representing the elliptical orbit. So, it is approximately like watching an object in an elliptical orbit from a position in a circular orbit with the same semimajor axis. The deviation from an ellipse is quite small, until the eccentricity gets larger than 0.01 or so. Server sky orbits will have eccentricities of less than 0.002 .
In the rotating frame, these eccentric loci describe the path of an object subject to fictitious forces. This provides an additional check on our equations for fictitious forces.
#### The math
\large \dot{ \theta } = v_{\perp} / r = v_0 ( 1+e \cos( \theta )) / ( a (1-e^2) / {1+e\cos(\theta)}
\large \dot{ \theta } = ( \sqrt{ \mu / ( a (1 - e^2) ) } / a ( 1-e^2 ) ) ( 1 + e\cos( \theta ))^2
\large \dot{ \theta } = ( \sqrt{ \mu / ( a^3 (1 - e^2)^3 ) } ( 1 + 2 e \cos( \theta ) + e^2 \cos( \theta )^2 )
for small perturbations, e^2 \approx 0 , so
\large \dot{ \theta } \approx ( \sqrt{ \mu / a^3 } ( 1 + 2 e \cos( \theta ) )
In the rotating frame, \dot{ x } \approx a ( \dot{ \theta } - \omega ) where \omega \equiv \sqrt{ \mu / a^3 } , so
\large \dot{ x } \approx a \sqrt{ \mu / a^3 } ( ( 1 + 2 e \cos( \theta ) - 1 )
\large \dot{ x } \approx 2 e a \omega ( \cos( \theta ) )
Approximating \theta \approx \omega t , we can integrate to get:
\Large x \approx -2 e a \sin( \theta )
\large \dot{ y } = v_r = e v_0 \sin( \theta ) \approx e a \omega \sin( \omega t ) Integrating:
\Large y \approx -e a \cos( \theta )
For small e , these equations in y and x describe the locus calculated above.
#### Testing the Equations for Acceleration
\large \ddot x ~ ? = ~ \large 2 \omega \dot y
\large \ddot x = { { d^2 } \over { dt^2 } } ( -2 e a \sin( \theta ) ) = { d \over { dt } } ( -2 e a \omega \cos( \theta ) ) = \Large 2 e a \omega^2 \sin ( \theta )
\large 2 \omega \dot y = 2 \omega { d \over { dt } } -e a \cos( \theta ) = \Large 2 e a \omega^2 \sin ( \theta ) . . . equal!
\large \ddot y ~ ? = ~ \large -2 \omega \dot x + 3 \omega^2 y
\large \ddot y = { { d^2 } \over { dt^2 } } ( -e a \cos( \theta ) ) = { d \over { dt } } ( e a \omega \sin( \theta ) ) = \Large e a \omega^2 \cos( \theta )
\large -2 \omega \dot x + 3 \omega^2 y = -2 \omega { d \over { dt } }( -2 e a \sin( \theta ) + 3 \omega^2 ( - e a \cos( \theta ) ) )
\large ~~~~~~~~~~~~~~= 4 e a \omega^2 \cos(\theta) - 3 e a \omega^2 \cos( \theta ) = \Large e a \omega^2 \cos( \theta )~~~ ... equal!
The acceleration equations work!
### Periods of M orbits
M orbits describe the number of minutes an orbit takes travel once around the earth and return to the same position overhead. For server sky, these are integer fractions of a 1440 minute synodic day; this makes it easier to calculate the sky position given the orbital parameters and the time of day. the M288 orbit returns to the same apparent position 5 times per day, or 365.256...*5 times per year. That position moves around the earth 365.256...+1 times per year. So the total number of orbits per year, relative to the stars, is 365.256...*6+1 orbits per year. That is divided into the year length in seconds to yield the
M288 sidereal orbit time = ( 365.256... * 86400 ) / ( 365.256... * 6 + 1 ) = 86400 / ( 6 + 1/365.256... ) = 14393.43227 seconds
1 year = 365.256363004 days of 86,400 seconds, or 31558149.7635456 seconds.
J2 speedup fraction = - 3 J_2 ( a_E / a )^2 = 3 \times 1.082626683e-3 \times (6378107m)^2 / a^2 = 1.321245688e11m^2 / a^2
### A Table of orbits
LEO 300Km M288 M360 GEO Moon Earth units \mu gravitation param. 3.98600448e14 1.3271244e20 m3/s2 relative to earth Sun J_2 Oblate-ness -1.082626683e-3 earth radius = 6378107m -6e-7 \omega'/\omega J2 speedup fraction 2.9627e-3 8.0782e-4 6.3356e-4 7.4318e-5 8.9417e-7 2e-17 a_g gravity 8.938095 2.437062 1.9113693 0.02242078 0.002697573 0.005930053 s T sidereal period 5431.010 14393.4323 17270.5433 86164.100 2360591.577 31558149.76 s T_s synodic period 5431.945 14400 17820 86400 2551442.9 31558149.76 s T/2\pi Sidereal / 2pi 864.3721 2290.78585 2748.69228 13713.44093 375698.7212 5022635.5297 s \omega Angular velocity 1.1569e-03 4.36531e-04 3.63809e-04 7.29212e-05 2.66171e-06 1.99099e-07 rad/s orbits/year 5810.733 2192.5382 1827.2819 366.25640 13.36879 1.00000 a semimajor axis 6678000 12788971 14440980 42164170 384399000 1.4959826e11 m R_a apogee radius 6678000 12838976 14440980 42164170 405696000 1.5209823e11 m R_p perigee radius 6678000 12738976 14440980 42164170 363104000 1.4709829e11 m e eccentricity 0.000000 0.001951 0.001728 0.000000 0.055401 0.016711 V_0 mean velocity 7725.84 5582.79 5253.76 3074.66 1023.16 29784.81 m/s V_a apogee velocity 7725.84 5571.90 5244.68 3074.66 966.47 29287.07 m/s V_p perigee velocity 7725.84 5593.68 5262.84 3074.66 1079.84 30282.55 m/s C_3 orb. specific energy -59688597 -31167516 -27602035 -9453535 -1036945 -887125553 J/kg
Note: This table uses the classic formula \omega^2 a^3 = \mu , and does not take into account the oblate spheroid shape of the Earth, and many other perturbations. However, with the perturbations included, and with good data from ground stations and GPS to establish positions and velocities, we really can compute these numbers to this many decimal places. So while the numbers above are actually far less accurate, they represent the precision of the measurements we will someday compute.
The eccentricities for the M orbits assume orbits mapped onto a 50km minor radius toroid.
LAGEOS 1 and LAGEOS 2 are two slightly lower satellites in inclined but highly circular orbits.
REFS:
copied from http://server-sky.com/NearCircularOrbits
NearCircularOrbitsOlder (last edited 2020-04-18 23:01:44 by KeithLofstrom) | {"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": 0.9838363528251648, "perplexity": 4164.711683836835}, "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/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00593.warc.gz"} |
https://www.semanticscholar.org/paper/Analytic-Function-Theory-Hille/9e5072b36ed679d89a7b21f88317d3e417a13c27 | # Analytic Function Theory
@inproceedings{Hille1959AnalyticFT,
title={Analytic Function Theory},
author={Einar Hille},
year={1959}
}
906 Citations
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C V ] 5 J un 2 01 9 Analytic Solutions of the Heat Equation
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• 2012
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• Mathematics
• 2006
We consider the spectrum of a two-dimensional Pauli operator with a compactly supported electric potential and a variable magnetic field with a positive mean value. The rate of accumulation ofExpand
On the extension of the Painlevé property to difference equations
• Mathematics
• 2000
It is well known that the integrability (solvability) of a differential equation is related to the singularity structure of its solutions in the complex domain - an observation that lies behind theExpand | {"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": 0.9434287548065186, "perplexity": 1000.9398671683085}, "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-2021-43/segments/1634323585203.61/warc/CC-MAIN-20211018124412-20211018154412-00339.warc.gz"} |
http://crypto.stackexchange.com/questions?page=195&sort=newest | # All Questions
Let's assume a simple algorithm like the Skein hash function. Is it possible, given the algorithm, to construct a proof that it does not have a particular distinguisher, something like: $P(xyz)$ is ... | {"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": 0.8064650297164917, "perplexity": 424.2521733142576}, "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-2016-22/segments/1464049275328.0/warc/CC-MAIN-20160524002115-00203-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.varsitytutors.com/precalculus-help/graph-the-sine-or-cosine-function | Precalculus : Graph the Sine or Cosine Function
Example Questions
Example Question #44 : Graphs And Inverses Of Trigonometric Functions
Which of the following functions has a y-intercept of | {"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": 0.949942409992218, "perplexity": 2831.475896940413}, "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-2017-22/segments/1495463615093.77/warc/CC-MAIN-20170530105157-20170530125157-00029.warc.gz"} |
https://rd.springer.com/chapter/10.1007/978-3-319-91128-1_3 | # Laminar Flow of Fluids in Ducts
Chapter
Part of the Studies in Systems, Decision and Control book series (SSDC, volume 161)
## Abstract
This chapter is devoted to the laminar flow of incompressible fluids with constant physical properties. | {"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": 0.8018492460250854, "perplexity": 1878.336558417081}, "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-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00294.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=129&t=41358&p=142137 | ## Degeneracy
$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$
Posts: 72
Joined: Fri Sep 28, 2018 12:19 am
### Degeneracy
Can someone please explain the concept and how to determine degeneracy?
Meachelle_Lum_1I
Posts: 92
Joined: Fri Sep 28, 2018 12:24 am
### Re: Degeneracy
Degeneracy(W) is defined as the number of possible ways a system can exist in a specific energy state. It is determined by the number of possible states, raised to the number of particles involved. For example, two particles A and B have two possible states. Thus 2 states and 2 particles is 2^2, and W=4. This makes degeneracy an extensive property because it depends on the size (Number of particles) of the system.
Amy Dinh 1A
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am
### Re: Degeneracy
Degeneracy, W, is the number of possible ways of achieving a given energy state.
It is determined on the number of particles and the number of arrangements.
For example if N particles and each particle can be in one of two states:
W = 2^N
Sophia Ding 1B
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am
### Re: Degeneracy
As others previous have said, degeneracy is the number of possible ways a system can exist in a given/specific energy state. So W, degeneracy, = (number of states) ^ number of particles.
805087225
Posts: 30
Joined: Thu Jun 07, 2018 3:00 am
### Re: Degeneracy
Degeneracy is determined as the number of atoms involved raised to the power of possible states.
Nicolette_Canlian_2L
Posts: 77
Joined: Fri Sep 28, 2018 12:25 am
Been upvoted: 1 time
### Re: Degeneracy
is there a relationship between degeneracy and entropy?
904914909
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am
### Re: Degeneracy
Degeneracy is the number of possible states that atoms can exist in and it is related to entropy because the higher the degeneracy the more potential for disorder and higher entropy. | {"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": 2, "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": 0.8592163920402527, "perplexity": 1637.2635730819775}, "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-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00583.warc.gz"} |
https://www.physicsforums.com/threads/solve-for-y.330685/ | # Homework Help: Solve for y'
1. Aug 14, 2009
### Centurion1
1. The problem statement, all variables and given/known data
Solve for y'
2x+2xy'+2y+3y2y'
2. Relevant equations
3. The attempt at a solution
so the ' means degrees right?
Would you start by grouping like terms. My real question is how that ' makes the other variables change?
2. Aug 14, 2009
No, $$y'$$ is the first derivative of $$y$$. You need to isolate $$y'$$ from the other variables, but I see one problem: you don't have this expression set equal to anything: as it stands you can factor it (a little), but there is no way to solve it for any of the quantities.
3. Aug 14, 2009
### Cyosis
You need to be more specific. To solve for a variable you need to have an equation. What you have written down is not an equation at all so make sure you also write down the other side of the equation like so, 'what you have'=....
Usually y' means the derivative of y. In this case I would guess with respect to x.
4. Aug 14, 2009
### Centurion1
sorry it is =0
5. Aug 14, 2009
### Дьявол
Then, factor out y' and solve the equation.
6. Aug 14, 2009
### symbolipoint
You mean, 2x+2xy'+2y+3y2y' = 0, and want to find y'.
You can precisely perform steps according to:
Like Terms or Reverse of Distributive Property;
Multiplicative Inverse property.
7. Aug 14, 2009
### Centurion1
so y' is treated like another variable. I can do normal factoring and just treat it like it was a third variable like if it was a z instead?
8. Aug 14, 2009
### symbolipoint
YES, from the viewpoint of "Pre-Calculus" and any Math course below. Some variable names may have subscripts to distinguish between or among similar main symbols. The "prime" notation can also be used sometimes. You have the variable written, y', spoken as "y prime". Your example also has the variable, y. So, two of your variables in your example are y and y' ("y" and "y prime"). We assume that they share some character but that they represent different numbers.
9. Aug 14, 2009
### Centurion1
yes i understand now, thank you. i think i can handle the factoring. | {"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": 0.8002786636352539, "perplexity": 1232.8639060278708}, "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-2018-30/segments/1531676589179.32/warc/CC-MAIN-20180716041348-20180716061348-00341.warc.gz"} |
https://scicomp.stackexchange.com/tags/finite-element/hot | # Tag Info
10
You'll get everyone to give different answers to this question, and maybe that's alright. Here are some of my favorite ones: Taylor's theorem that a function (of sufficient smoothness) equals its Taylor expansion plus a remainder term. One can consider the Bramble-Hilbert lemma as a variation of Taylor's theorem, but it has different applications and is ...
5
The particular set of constraints you have chosen does not prevent a rigid body rotation about node 1. Thus the stiffness matrix is singular, as you have noted. One way to prevent this rigid body rotation is to set the y-displacement at node 2 to zero. You could also constrain the x-displacement at either node 3 or node 4 to prevent the rotation. One way ...
2
Since this is a question with pretty subjective answers, I'll add a couple to Prof. Bangerth's very good list. the theorem of adjoint/dual operators and spaces is pretty crucial to Computational Science. We know that dual-consistent discretizations of the PDEs can obtain superconvergence which is a nice property. But I think the more commonly used outcomes ...
1
To my knowledge these are the same things. However, this type of thing is common. For example, the proper orthogonal decomposition also has field-specific names. Others call it principal component analysis, the Karhunen--Loeve expansion, or empirical orthogonal functions. It is also no different than an autoencoder with linear activation function. I'm sure ...
Only top voted, non community-wiki answers of a minimum length are eligible | {"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": 0.8585483431816101, "perplexity": 556.1594503594629}, "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-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00155.warc.gz"} |
https://censi.science/research/robot-perception/ghtv/ | # GHTV 2012-12-05
A comparison of algorithms for likelihood approximation in bayesian localization. Technical Report, Sapienza Universita di Roma, 2006. bibtex
Abstract -- Global localization is the problem of estimating the robot's pose, without any previous information, in a world which is known. Bayesian filters (either grid- or particle-based), need to evaluate the likelihood of a sensor reading given a pose hypothesis. For a laser range finder the computation is expensive and therefore approximations to the true likelihood that do not require a ray-tracing are usually employed. Nevertheless, these approximations do not satisfy those that in this paper we identify as orientation and visibility constraints. This paper proposes more accurate algorithms which solve, in part or completely, these problems and are faster and have better complexity than existing methods. | {"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": 0.8392117619514465, "perplexity": 1192.3757420629586}, "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/1490218187945.85/warc/CC-MAIN-20170322212947-00353-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://gateoverflow.in/2116/gate2011-14 | 1.3k views
The simplified SOP (Sum of Product) from the Boolean expression
$$(P + \bar{Q} + \bar{R}) . (P + \bar{Q} + R) . (P + Q +\bar{R})$$ is
1. $(\bar{P}.Q+\bar{R})$
2. $(P+\bar{Q}.\bar{R})$
3. $(\bar{P}.Q+R)$
4. $(P.Q+R)$
0
$Remark:$ the complement of m0 is M0.
0
How?
0
Remember this formula- (A + B+ C) . (A + B + C') = (A + B) remove that term which is not same.
1 0 0 0 1 1 1 1
selected by
0
may be silly doubt but m not getting this table , my POS terms = 4,5,6 but in kmap it shows 2,4,6
+10
ans: B
(P+Q'+R').(P+Q'+R).(P+Q+R')
=(P+R)(Q+Q)(P+Q+R)
=P+PR+PQ+RP+RQ
=P(1+Q)+RP+RQ
=P(1+R)+RQ
=P+RQ
edited
0
@puja Mishra how have you solved the question I did not get that plz elaborate
0
See Boolean algebra rules ...
0
what rule did you apply for : (P+Q¯+R¯).(P+Q¯+R).(P+Q+R¯)=(P+R)(Q+Q)(P+Q`+R)
+1
(A+B)(A+B')= A
+1
Thank you :)
0
Yo can also solve this using Kmap.
0
depends on practice watever u prefer ....
0
q+q'=1
1+q'=1
qq'=0
(P+Q'+R') (P+Q'+R) (P+Q+R')
complement the whole equation
P'QR+P'QR'+P'Q'R
taking P'R common from 1st and 3rd minterm
P'R(Q+Q') + P'QR'
we know A+A' = 1
P'R+P'QR'
P'(R+QR')
P'((R+Q)(R+R'))
P'(R+Q)
Complement it back
P+R'Q'
P+Q'R'
Given Boolean Expression:
(P+Q'+R').(P+Q'+R).(P+Q+R')
we know that dual of a boolean expression is equivalent. Take dual of it:
PQ'R'+PQ'R+PQR'
Now simplify it.(I am taking dual because solving the given expression will be difficult hence, take dual and solve it)
=PQ'(R'+R)+PQR'
=PQ'+PQR'
=P(Q'+QR')
=P(Q'+R').
Now if we see option b) and take dual of it then we will be getting the above simplified expression.
1
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": 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": 0.9117754101753235, "perplexity": 3722.4361925653075}, "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-2019-09/segments/1550247486480.6/warc/CC-MAIN-20190218114622-20190218140622-00565.warc.gz"} |
https://wakespace.lib.wfu.edu/handle/10339/14923 | › Electronic Theses and Dissertations
# Quantum Effects of the Massless Spin One-Half Field in Static Spherically Symmetric Black Hole and Wormhole Spacetimes
## Electronic Theses and Dissertations
### Item Details
abstract
The full renormalized stress-energy tensor for the massless spin one-half field is numerically computed on and outside the event horizon of an extreme Reissner-Nordstrom black hole and on and outside the throat of three static spherically symmetric Lorentzian wormhole spacetimes. The field and all spacetimes studied are in a zero temperature vacuum state. We treat the field quantum mechanically on a classical background spacetime governed by general relativity. The full stress-energy tensor is found to be finite and regular for all spacetimes. We make comparisons between the numerical calculation and the analytic approximation found by Groves {\it et~al.} for the full stress-energy tensor. We find that the approximation is very poor in most cases, even getting the wrong sign for many components. For the extreme Reissner-Nordstr\"{o}m black hole, divergences predicted by the analytic approximation of are shown to be nonexistent. Lastly, the results for the full stress-energy tensors for wormholes are analyzed in terms of an arbitrary renormalization parameter to see if the exotic" energy condition needed to keep such an object from collapsing are met. No wormhole geometry studied is found to be a self-consistent solution when quantum fluctuations of the spin one-half field are considered. This is in contrast to the results found using the analytic approximation for the massless spin-one half field which predicts that one of the wormholes might have a self-consistent solution, since the stress-energy tensor found satisfies the exotic energy condition.
subject
wormhole
black hole
quantum gravity
semi-classical gravity
stress-energy
exotic energy
spherically symmetric
fermions
contributor
Carlson, Eric (committee chair)
Cook, Greg (committee member)
Anderson, Paul (committee member)
Williams, Richard (committee member)
Robinson, Stephen (committee member)
date
2009-08-06T15:56:23Z (accessioned)
2010-06-18T19:00:06Z (accessioned)
2009-08-06T15:56:23Z (available)
2010-06-18T19:00:06Z (available)
2009-08-06T15:56:23Z (issued)
degree
Physics (discipline)
identifier
http://hdl.handle.net/10339/14923 (uri)
language
en_US (iso)
publisher
Wake Forest University
rights
Release the entire work immediately for access worldwide. (accessRights)
title
Quantum Effects of the Massless Spin One-Half Field in Static Spherically Symmetric Black Hole and Wormhole Spacetimes
type
Dissertation | {"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": 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": 0.8088358640670776, "perplexity": 1613.464629912133}, "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-2021-21/segments/1620243988759.29/warc/CC-MAIN-20210506175146-20210506205146-00289.warc.gz"} |
https://edoc.unibas.ch/2044/ | # On the nature of crossreactions observed with antibodies directed to defined epitopes
Nigg, E. A. and Walter, G. and Singer, S. J.. (1982) On the nature of crossreactions observed with antibodies directed to defined epitopes. Proceedings of the National Academy of Sciences of the United States of America, Vol. 79, H. 19. pp. 5939-5943.
Full text not available from this repository.
Official URL: http://edoc.unibas.ch/dok/A5249524 | {"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": 0.891537606716156, "perplexity": 4065.6397186309523}, "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/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00390.warc.gz"} |
https://tex.stackexchange.com/questions/20643/diagonal-strikeout-starting-too-low-and-ending-too-high?noredirect=1 | # Diagonal strikeout starting too low and ending too high
I want to strikeout an equation with diagonal line but I've not managed to get the diagonal line to draw as I want. The closest I've gotten is via the first example in Frédéric's answer of \cancel draws under thing being canceled:
\documentclass{minimal}
\usepackage{tikz}
\newcommand{\hcancel}[1]{%
\tikz[baseline=(tocancel.base)]{
\node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
\draw[red] (tocancel.south west) -- (tocancel.north east);
}%
}%
\begin{document}
$$\label{eq:2} \hcancel{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}$$
\end{document}
This produces:
But the line starts too low and ends too high. The following is what I want:
I'd be happy for an option to make the line start slightly earlier and end slightly later too.
You can modify Frédéric's code so that \hcancel receives four more mandatory arguments controlling the vertical and horizontal shifting for the starting and ending points:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand{\hcancel}[5]{%
\tikz[baseline=(tocancel.base)]{
\node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
\draw[red] ($(tocancel.south west)+(#2,#3)$) -- ($(tocancel.north east)+(#4,#5)$);
}%
}%
\begin{document}
$$\label{eq:1} \hcancel{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}{0pt}{0pt}{0pt}{0pt}$$
$$\label{eq:2} \hcancel{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}{-3pt}{3pt}{3pt}{-2pt}$$
\end{document}
The new syntax:
\hcancel{<text>}{<start. point horiz. shifting>}{<start. point vertical shifting>}{<end. point horiz. shifting>}{<end. point vertical shifting>}
EDIT: using the xparse package, the definition of the new command is much more flexible; using something like
\usepackage{xparse}
\DeclareDocumentCommand{\hcancel}{mO{0pt}O{0pt}O{0pt}O{0pt}}{%
\tikz[baseline=(tocancel.base)]{
\node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
\draw[red] ($(tocancel.south west)+(#2,#3)$) -- ($(tocancel.north east)+(#4,#5)$);
}%
}%
allows the use of \hcancel{<text>} for the standard behaviour of the command as defined by Frédéric and to use the four (now optional) arguments to control the horizontal/vertical shifting:
\documentclass{article}
\usepackage{xparse}
\usepackage{tikz}
\usetikzlibrary{calc}
\DeclareDocumentCommand{\hcancel}{mO{0pt}O{0pt}O{0pt}O{0pt}}{%
\tikz[baseline=(tocancel.base)]{
\node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
\draw[red] ($(tocancel.south west)+(#2,#3)$) -- ($(tocancel.north east)+(#4,#5)$);
}%
}%
\begin{document}
$$\label{eq:1} \hcancel{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}$$
$$\label{eq:2} \hcancel{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}[-3pt][3pt][3pt][-2pt]$$
\end{document}
You can create nodes at the beginning and the end of the line, and shift them vertically:
\tikzstyle{nosep}=[inner sep=0pt, outer sep=0pt]
\newcommand{\hcancel}[1]{%
\tikz[baseline=(tocancel.base)]{
\node[nosep] (tocancel) {#1};
\node[nosep, yshift=.5ex] (from) at (tocancel.south west) {};
\node[nosep, yshift=-.5ex] (to) at (tocancel.north east) {};
\draw[red] (from) -- (to);
}%
}%
You can find the best shifts by trial and error. I chose .5ex and -.5ex arbitrarily.
• @N.N. And to make the line slightly longer, add xshift=-1ex to the from node (+1ex for the to node). Jun 13, 2011 at 21:05
Another possibility is to create a style.
\documentclass[]{scrartcl}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[cancel/.style={path picture={ \draw[#1]
($(path picture bounding box.south west)+(-3pt,6pt)$) -- ($(path picture bounding box.north east)+(3pt,-6pt)$);
}}]
\node [inner sep=3pt,cancel=red] {$2x+3=y$};
\end{tikzpicture}
\end{document}
• It's possible to place the code in a macro with arguments. Jun 13, 2011 at 21:53
• When I try to compile this I get: ! Package pgfkeys Error: I do not know the key '/tikz/path picture' and I am going to ignore it. Perhaps you misspelled it.
– N.N.
Jun 14, 2011 at 7:33
• @N.N. I use pgf 2.1. and perhaps 'path picture' is defined since the 2.1 but I don't know :( I use this option in several examples and in some answers without problem. Jun 14, 2011 at 8:03
• OK, you're probably right. I've got pgf 2.0 from TeX Live 2009 on my system.
– N.N.
Jun 14, 2011 at 8:07
This is thought as enhancement of Gonzalo’s answer. The more arguments a command gets, the easier one can lose track of them. Here comes the key-value approach to aid.
Therefore I want to show a definition using the package keycommand. It could also be done with xparse and l3keys, but I am used to keycommand; in my eyes it is also more suited for end users, at least for cases like this. Note, though, that there is a bug in, and you must include the patch provided by Joseph Wright in his answer to the question How do I use \ifcommandkey , or how do I check if a key was given? after the package was loaded (actually you would need this only in cases, when you use the command \ifcommandkey).
Note, that I additionally added a key to change the line color, used in third example.
\documentclass{article}
\usepackage{keycommand}
% Patch by Joseph Wright ("bug in the definition of \ifcommandkey (2010/04/27 v3.1415)"),
% https://tex.stackexchange.com/a/35794
\begingroup
\makeatletter
\catcode\/=8 %
\@firstofone
{
\endgroup
\renewcommand{\ifcommandkey}[1]{%
\csname @\expandafter \expandafter \expandafter
\expandafter \expandafter \expandafter \expandafter
\kcmd@nbk \commandkey {#1}//{first}{second}//oftwo\endcsname
}
}
%--------%
\usepackage{tikz}
\usetikzlibrary{calc}
\newkeycommand{\hcancel}[hshiftstart=0pt,vshiftstart=0pt,hshiftend=0pt,vshiftend=0pt,color=red][1]{%
\tikz[baseline=(tocancel.base)]{
\node[inner sep=0pt,outer sep=0pt] (tocancel) {#1};
\draw[\commandkey{color}] ($(tocancel.south west)+(\commandkey{hshiftstart},\commandkey{vshiftstart})$) --
($(tocancel.north east)+(\commandkey{hshiftend},\commandkey{vshiftend})$);
}%
}%
\begin{document}
$$\label{eq:1} \hcancel{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}$$
$$\label{eq:2} \hcancel[hshiftstart=-3pt,vshiftstart=0.5em,hshiftend=3pt,vshiftend=-0.5em] {h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}$$
$$\label{eq:3} \hcancel[hshiftstart=-3pt,vshiftstart=1em,hshiftend=3pt,vshiftend=-1em,color=blue] {h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}$$
\end{document}
I use PSTricks in this answer. Please adjust the parameters until they suit your preference best. The parameters given in the code below are self-explanatory.
\documentclass{minimal}
\psset{linecolor=red}
\def\myeq{\psDefBoxNodes{A}{h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}}
\begin{document}
\centering
$$\myeq \ncline[nodesep=3pt,offsetA=-1pt,offsetB=3pt]{A:bl}{A:tr}$$
\\[5mm]
Controlling the length:
$$\myeq \ncline[nodesep=10pt,offsetA=-1pt,offsetB=3pt]{A:bl}{A:tr}$$
\\[5mm]
Controlling the left node:
$$\myeq \ncline[nodesep=3pt,offsetA=-1pt,offsetB=-10pt]{A:bl}{A:tr}$$
\\[5mm]
Controlling the right node:
$$\myeq \ncline[nodesep=3pt,offsetA=10pt,offsetB=3pt]{A:bl}{A:tr}$$
\end{document}
Coming late to the game.
Here is a picture based solution, which compiles of course faster as it does not need load all the tikz libraries. With the help of the pict2e package.
\documentclass{article}
\usepackage{pict2e}
\usepackage{color}
\makeatletter
% in our modern era, you should redefine this using xparse facilities
% so that you can add all parameters you wish more easily
% for example the linethickness which I hardcode here.
% Also, the \@firstoffour etc... is just because LaTeX's \newcommand
% admits only defining 1 optional parameter. With xparse, you can
% provide possibly more convenient user interface.
\newcommand\@firstoffour[4]{#1}
\newcommand\@secondoffour[4]{#2}
\newcommand\@thirdoffour[4]{#3}
\newcommand\@fourthoffour[4]{#4}
\newcommand\MyCancel[2][{0pt}{0pt}{0pt}{0pt}]{%
\begingroup
\sbox0{#2}%
\edef\MyCancel@xoffsetleft {\number\dimexpr\@firstoffour#1}%
\edef\MyCancel@yoffsetleft {\number\dimexpr\@secondoffour#1}%
\edef\MyCancel@xoffsetright{\number\dimexpr\@thirdoffour#1}%
\edef\MyCancel@yoffsetright{\number\dimexpr\@fourthoffour#1}%
\edef\MyCancel@width {\number\wd0}%
\edef\MyCancel@height{\number\ht0}%
\setlength{\unitlength}{1sp}%
\begin{picture}(\MyCancel@width,\MyCancel@height)
\linethickness{1pt}% this may be made another parameter
\put(0,0){\box0}
\color{red}% also this one
\Line(\MyCancel@xoffsetleft,\MyCancel@yoffsetleft)% \Line does not like space here!
(\numexpr\MyCancel@xoffsetright+\MyCancel@width\relax,
\numexpr\MyCancel@yoffsetright+\MyCancel@height\relax)
\end{picture}%
\endgroup
}
\makeatother
\begin{document}
$$\MyCancel[{-1pt}{+1pt}{+0.5pt}{-1.5pt}]{\displaystyle h_1 \land h_2 \land h_3 \land h_4 \land h_5 \land h_6 \land h_7 \land h_8 \land h_9 \land h_{10}}$$
\end{document}
• I should have used \providecommand for the \@firstoffour` etc... in case some package define them. Hopefully with same definition..
– user4686
Oct 3, 2017 at 8:36 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 13, "x-ck12": 0, "texerror": 0, "math_score": 0.8666960000991821, "perplexity": 4549.437172168229}, "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-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00093.warc.gz"} |
http://arxiv-export-lb.library.cornell.edu/abs/2110.04295?context=q-bio.PE | q-bio.PE
(what is this?)
# Title: Quantifying the COVID19 infection risk due to droplet/aerosol inhalation
Abstract: The dose-response model has been widely used for quantifying the risk of infection of airborne diseases like COVID-19. The model has been used in the room-average analysis of infection risk and analysis using passive scalars as a proxy for aerosol transport. However, it has not been employed for risk estimation in numerical simulations of droplet dispersion. In this work, we develop a framework for the evaluation of the probability of infection in droplet dispersion simulations using the dose-response model. We introduce a version of the model that can incorporate the higher transmissibility of variant strains of SARS-CoV2 and the effect of vaccination in evaluating the probability of infection. Numerical simulations of droplet dispersion during speech are carried out to investigate the infection risk over space and time using the model. The advantage of droplet dispersion simulations for risk evaluation is demonstrated through the analysis of the effect of humidity on infection risk.
Comments: 19 pages, 9 figures Subjects: Quantitative Methods (q-bio.QM); Biological Physics (physics.bio-ph); Fluid Dynamics (physics.flu-dyn); Populations and Evolution (q-bio.PE) Cite as: arXiv:2110.04295 [q-bio.QM] (or arXiv:2110.04295v2 [q-bio.QM] for this version)
## Submission history
From: Rahul Bale [view email]
[v1] Wed, 6 Oct 2021 05:13:31 GMT (4135kb,D)
[v2] Thu, 8 Sep 2022 04:15:10 GMT (7973kb,D)
Link back to: arXiv, form interface, contact. | {"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": 0.8074949979782104, "perplexity": 2560.430478687691}, "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/1664030335362.18/warc/CC-MAIN-20220929163117-20220929193117-00583.warc.gz"} |
https://physicsoverflow.org/9249/relation-between-fields-similar-electric-magnetic-fields | # Is there any relation between weak and strong fields, similar to electric and magnetic fields?
+ 2 like - 0 dislike
70 views
Is it possible to unify the strong, weak, electric and magnetic field just by Maxwellian type equations? (Maxwell by adding a small change - unified electric and magnetic field, then Einstein's equations - use it to create special and general theory of relativity, now maybe all we need is a little more change to unify all fields,) When $B$ and $E$ are known magnetic and electric fields, then $W$ and $S$ are weak and strong fields, what are their units?
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user unifiedator
I dont understand your question, Perhaps you are referring to the standard model.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user Prathyush
You've got some catching up to do! Look up "Yang-Mills Theory". From Wiki: Yang–Mills theory seeks to describe the behavior of elementary particles using these non-Abelian Lie groups and is at the core of the unification of the Weak and Electromagnetic force (i.e. U(1) × SU(2)) as well as Quantum Chromodynamics, the theory of the Strong force (based on SU(3)). Thus it forms the basis of our current understanding of particle physics, the Standard Model.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user Alfred Centauri
What do you mean by "strong field" and "weak field" ...? This question looks very confused to me.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user Dilaton
What Alfred Centauri said. The weak and strong theories are generalisations of ordinary electromagnetism. So for the strong force their are chromo-electric and chromo-magnetic fields (actually eight of each) and for the weak force there are weak-electric and weak-magnetic fields (actually three of each). These follow a close analogy to ordinary E&M (although the theories are much more complicated). It is less common to work with the electric & magnetic fields than the vector potentials for technical reasons, but they exist.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user Michael Brown
+ 1 like - 0 dislike
I'm guessing at what you're asking - ignore this answer if I've misunderstood you.
Maxwell's equations describe the classical behaviour of electromagnetism. They can only do this because the EM forces are long range so at macroscopic distances they behave classically. By contrast, the weak and strong forces are short range and cease to act over anything like the classical limit of distance. There is no classical approximation to describe the weak and strong forces, so there is no analogy to Maxwell's equation.
Above the electroweak transition the electroweak force will become long range. Whether there is some classical limit in the spirit of the Maxwell's equations is a good question and I don't know the answer. I would guess that there is no such limit for the strong force even about the EW transition because the force will still be confined.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user John Rennie
answered Jun 19, 2013 by (470 points)
At high enough temperature and energy density, the strong force stops confining. This plays a role in the standard stories of the early universe.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user user1504
Ah yes, as in the quark gluon plasma. I bet there's still no (useful) classic long distance description though.
This post imported from StackExchange Physics at 2014-03-24 03:48 (UCT), posted by SE-user John Rennie
+ 0 like - 0 dislike
Depends on what you mean by "unify" If you just mean "describe", then it's simple. We have the Standard Model with it's exceedingly simple Lagrangian Density formed by adding Yang - Mills & Dirac fields to Klein - Gordon fields.
$${\mathcal L} = - {1 \over 4}{F^{\mu\nu }}{F_{\mu \nu }} + i\overline \psi \not\nabla \psi + \overline \psi \phi \psi + \mbox{h.c.} + {\left| {\nabla \phi } \right|^2} - V(\phi )$$
Here, it is easy to identify the Yang Mills field for gluons et al, the Dirac Fields for the quwarks et al, the Klein - Gordon for the Higgs, a scalar field.
If you mean "unify" in the sense of grand unification and such, then it's a bit more complicated.
# Grand Unification
Refers to the unification (meaning: having a simple group as the gauge group, as opposed to the Standard Model whose gauge group is $SU(3)\times SU(2)\times U(1)$; yet describe the Standard Model, since it's experimentally verified to a pretty high degree).
Examples include
# Theories of Everything
Better than GUTs; unify (quantised) gravity too. Examples include .
answered Sep 17, 2013 by (1,955 points)
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https://www.physicsforums.com/threads/relativistic-length-2-viewpoints.256810/ | # Relativistic length, 2 viewpoints
1. Sep 16, 2008
### pacu
Problem statement:
Two identical cyclists ride past each other with constant velocities Va and Vb, which are close to the speed of light. Can it be that cyclist A perceives cyclist B as shorter or longer that cyclist B percieves cyclist A ? Or simply La is NOT equal to Lb ? (La-length of cyclist A as seen by cyclist B, Lb -length of cyclist B as seen by cyclist A).
Relevant formulas:
Relative speed V = Va+Vb/(1+(Va*Vb/c^2))
Relative length l = lo * square root from 1-(V/c)^2
Conclusion:
The V from the second equasion is equal for both cyclists, since addition and multiplication are alternate. lo is also equal. So there is no difference in the way cyclists A and B see each other.
Is this conclusion right?
Last edited: Sep 16, 2008
2. Sep 16, 2008
### granpa
its correct if the cyclists are the same length to begin with.
dont forget that there is also a loss of simultaneity. once you factor that in it stops seeming so impossible.
Last edited: Sep 16, 2008
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http://www-old.newton.ac.uk/programmes/AGA/seminars/2007031514301.html | # AGA
## Seminar
### The Method of Images on a Quantum Graph
Wilson, J (Texas A and M)
Thursday 15 March 2007, 14:30-15:30
Seminar Room 1, Newton Institute
#### Abstract
We consider an arbitrary quantum graph with the Laplacian H = -d^2/dx^2 and the corresponding eigenvalue problem on the graph. Taking the free space solution to an associated PDE (such as non-stationary Schroedinger equation), we apply the method of images to obtain the corresponding solution for our graph. This method brings in periodic and non-periodic (bounce) orbits and gives insight into both types of orbits and the roles they play. Applying this method to different kernels, we obtain interesting spectral information including the trace formula for the density of the states and the regularized vacuum energy. | {"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": 0.8184964060783386, "perplexity": 807.1768336694294}, "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-2014-35/segments/1408500823634.2/warc/CC-MAIN-20140820021343-00077-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/78810/find-the-possible-energies-and-corresponding-wavefunctions-of-the-hamiltonian | # Find the possible energies and corresponding wavefunctions of the Hamiltonian [closed]
The Hamiltonian of an electron stuck within a tunnel in a dialectic cube is found to be
$$H=\frac{p^2}{2m}+\frac{1}{2}Kx^2-\frac{e\Phi_0}{a}x$$
Find the possible energies and corresponding wavefunctions of H. Write an expression for the ground state energy and normalized wavefunction.
I tried plugging this Hamiltonian into the Schrodinger Equation and completing the square, but then I get an impossible differential equation. Is this the right thing to do? If it is, how do I solve for the ground state energy and the normalized wavefunction? Any help would be welcome--I am really confused by this problem.
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## closed as off-topic by Emilio Pisanty, Waffle's Crazy Peanut, Manishearth♦Sep 29 '13 at 7:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, Waffle's Crazy Peanut, Manishearth
If this question can be reworded to fit the rules in the help center, please edit the question.
You're on the right track. Complete the square on $x$ and you'll have some newly defined Harmonic oscillator whose position operator you have found. The additional constant that comes from completing the square will add to your ground state energy.
(Once you have completed the square, you should have something of the form
$H = constant + \frac{P^2}{2M} + \frac{1}{2}M\omega^2(X-x_0)^2$
which is a Harmonic oscillator centered at $x_0$ and ground state energy
$E_0 = constant + \frac{1}{2}\hbar\omega$).
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But how do we find the possible energies and corresponding wavefunctions of H? Do we have to solve the entire differential equation? – Bronzeclocksofbenin Sep 27 '13 at 20:29
You can solve the differential equation, yes. That would lead you to Hermite polynomials (see the appropriate chapter in Griffiths, for example), whose energies you can compute. Are you familiar with the creators and annihilators of a harmonic oscillator? Without solving the differential equation, you can easily define them and find the ground state energy immediately. Griffiths explains this nicely as well, as do Shankar and Sakurai. It may help you to know these operators in the future, because you can spot a HO in a Hamiltonian and immediately be able to talk about its energies. – eqb Sep 27 '13 at 20:32
(or see here for a quick overview of using these operators: en.wikipedia.org/wiki/Quantum_harmonic_oscillator ) – eqb Sep 27 '13 at 20:34
I do believe you would benefit from first learning how to get the eigenvalues of the Hamiltonian in the case of a simple one-dimensional harmonic oscillator: $H=p^2/2m + \frac{1}{2}m\omega^2x^2$ – BMS Sep 27 '13 at 22:55
@eqb Do I need to define some new creator/annihilator operator for this example (as opposed to the standard a_+/a_-)? – Bronzeclocksofbenin Sep 29 '13 at 12:07 | {"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": 0.8426724672317505, "perplexity": 448.43375928540763}, "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-2015-27/segments/1435375096780.24/warc/CC-MAIN-20150627031816-00156-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2070691/on-the-number-of-ways-of-writing-an-integer-as-a-sum-of-3-squares-using-triangul | # On the number of ways of writing an integer as a sum of 3 squares using triangular numbers.
It is well known that integers can be represented as a sum of squares, two, three and more. In what follows will be given a way to represent integers as a sum of 3 squares using triangular numbers.
It is also well known that adding 2 consecutive triangular numbers produces a square. So if we wanted to get 3 squares, we just need to add 3 pairs of consecutive triangular numbers. Here we consider the general case of $$N = a^2 + b^2 + c^2$$ with a,b,c not consecutive and not equal. In practice, some numbers will have a representation of 3 squares that are consecutive and sometimes equal.
For that we write N as $$N = (T_{n} + T_{n+1}) + (T_{n+k} +T_{n+k+1}) + (T_{n+j} + T_{n+j+1})$$
where the T's are triangular numbers and n,n+k,...are their indices. Using the formula for a triangular number $$T_{n}=n(n+1)/2$$ we transform the previous equation into the following one: $$N = 3n^2 + (2k+2j+6)n + 1 + (k+1)^2 + (j+1)^2$$
This is nothing but a quadratic equation in n and will be solved by looking for values of j and k that will make its discriminant $$d = b^2-4ac$$ a square. If we take the example of N=77, the discriminant $$d = 231 + 2kj -2(k^2+j^2)$$ will be a square for two pairs of values for $$(k,j)=(1,2),(1,6)$$ and this will give a value for $$n = (3,1)$$ Now we have the values of the indices (n,k,j), we can use the corresponding triangular numbers to form the squares and we get: $$N = 77 = 2^2 + 3^2 + 8^2 = 4^2 + 5^2 + 6^2$$
If we look at the discriminant d, we see that the potential squares will be $$s^2=15^2, 14^2, 13^2...$$ So we can either try different values of (j,k) that will make d a square or solve another quadratic in j or k given by: $$2kj - 2(k^2 +j^2) = -6$$ since $$231 = 15^2 + 6$$
The question is when do we know that we have found all the 3 squares representations of a given number N?
It will be very inefficient to check every potential square value of the discriminant d unless we wanted to find all the representations.
• Why the interest in three squares? – Will Jagy Dec 24 '16 at 18:28
• I saw few posts and there was no method to find a representation so I started thinking if what I did with the 2 squares can be adapted to the 3 squares representation. And if there was a method, it was too complicated to be of any use. And the end result was this post. However, I still don't know how many 3 square representations a number can have. – user25406 Dec 24 '16 at 21:55
• There is sort of a way to find all representations; it is in the book by Cassels, however it uses four variables and is probably not what you want. As far as how many, you want the book by Grosswald, see mathoverflow.net/questions/3596/… – Will Jagy Dec 24 '16 at 22:00
• Your above comment is in fact the answer (I mean the link) even though it was difficult for me to follow. But I think it will be useful for others if you can just copy and paste it here. Thanks again Will. – user25406 Dec 24 '16 at 22:20 | {"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": 0.816503643989563, "perplexity": 162.65851302666593}, "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-2019-30/segments/1563195526401.41/warc/CC-MAIN-20190720004131-20190720030131-00451.warc.gz"} |
http://www.physicsforums.com/showpost.php?p=844895&postcount=73 | View Single Post
Emeritus
PF Gold
P: 8,147
Quote by Careful I think you understood me (and probably mike too) wrong. I asked about the possibility for a dynamics to be constructed which *generates* change of differentiable structure. This question was connected to your claim that you generate singular three surfaces which need the interpretation (by HAND) of a spacelike three surface associated to a matter distribution (say elementary particle). At that moment, I suggested : but what about the singular world 4 tube ? ´´, how to find a determinstic mechanism which allows for change of differentiable structure ?´´ since classically you would expect this already to be necessary (clutting of matter and so on). These objections have been left unanswered so far
There are obviously questions raised by this research, such as: according to theorems there are countably many differential structures, but we observe only a finite, and very specific, particle spectrum. What gives? This has been the bete noir of the string theorists. The mechanism for the origin of these curvatures can be pushed back to the big bang, or whatever, but we need a particle dynamics. All this will be a research program for the future.
Concerning the reference to the 96 torston paper (I note that I still did not recieve a definition of a singular connection here), let me ask some silly questions. For example at page 3, a map h: M -> N is constructed where M is an exotic 7 sphere and N the ordinary S^7, which is singular at one point, say x_0. You endow M and N with smooth Riemannian metrics, choose smooth frames e (in M) and f (in N) and claim that you can select the Riemannian metrics in such a way that dh(e)(x) = a(x) .f(h(x)) where a(x) is a S0(7) transformation in the specific orthogonal bundle over N related to coordinates in M. This seems wrong to me since dh(e)(x_0) = 0 and hence a(x_0) = 0 (could you clarify this??) which would lead to a zero curvature contribution (if I were to believe formula 9). Perhaps, I missed something but anyway... Concerning the complex´´ curvature the authors get on page 10 in their example. It seems to me that they forgot to take the complex conjugate expression (a tangent basis in D^2 consists of d/dz and d/dz* which leads to matrix ( pz^{p-1} 0 ) ( 0 pz*^{p-1}) and a^{-1} da = ( (p-1)dz/z 0 ) ( 0 (p-1)dz*/z*) (notice that there is NO division through p - these factors cancel out.) Since the authors are only interested in the trace, this gives: (p-1)dz/z + (p-1)dz*/z* which (in polar coordinates) gives : 2(p-1) dr/r which gives rise to zero curvature (at least when I would naively take the line integral of this around a circle).
I am sure these questions can be addressed. As I said before, careful critique is not at all objectionable.
I hope I made it more clear now why I insist upon a rigorous definition and example of a distributional connection related to a change of differentiable structure! I think this hardly classifies as frantic resistance´´.
It wasn't the requests for clarification but the insults ("sloppily written") that grated. The paper is up to the usual standards of preprint writing. Yes it will benefit by reconstructing some parts based on your remarks, but some of your comments seemed to demand they conduct a decade-long research program before publishing. Some physicists may work that way (Veltzmann comes to mind), but it is certainly not the community norm. | {"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": 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": 0.8408805727958679, "perplexity": 854.9879568647289}, "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-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00626-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/61604/is-there-a-common-general-setup-for-both-weil-cohomologies-and-generalized-cohom?sort=votes | # Is there a common general setup for both Weil cohomologies and generalized cohomology theories?
My question can be simply (and loosely) stated as follows:
Is there a general (but not too general) construction that captures, as specializations, both Weil cohomologies in algebraic geometry and generalized cohomology theories in topology ?
(I must say I'm not an expert of any of the two!)
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See perhaps this answer of Urs Schreiber: mathoverflow.net/questions/6125/… – Kevin H. Lin Apr 14 '11 at 1:21
Also this: mathoverflow.net/questions/4214/… – Kevin H. Lin Apr 14 '11 at 1:22
wow! thanks for this link! – SGP Apr 14 '11 at 2:37
It seems I didn't notice that this question had already been asked. Thank you for the links to the answers and nLab. – Qfwfq Apr 14 '11 at 12:20
## 1 Answer
It would appear that the answer to the question as stated is no. However, Voevodsky's motivic homotopy theory does provide an adequate framework for both Weil cohomologies and generalized cohomology theories; there is a version of "Brown representability" theorem (representing object for a generalized cohomology theory) which is exploited in the applications to K-theory (Milnor conjecture and Bloch-Kato conjecture).
- | {"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": 0.9783640503883362, "perplexity": 1547.4707482870779}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2015-22/segments/1432207927863.72/warc/CC-MAIN-20150521113207-00281-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/49678-please-help-2-quick-questions-thanks.html | I need help getting the answers for these 2 questions!
I want to see if i did them right... (because i dont have an answer sheet)
1) Solve 12x^3 - 16x^2 + 7x -1
Let f(x) = 12x^3 - 16x^2 + 7x -1
2) Solve the inequality:
3x / (x^2 -4) < -1
*note: where the "<" is, its supposed to be less than/equal to. (i just cant find a key for it :P )
Thanks again!
2. Originally Posted by xo-hikari
I need help getting the answers for these 2 questions!
I want to see if i did them right... (because i dont have an answer sheet)
1) Solve 12x^3 - 16x^2 + 7x -1
Let f(x) = 12x^3 - 16x^2 + 7x -1
2) Solve the inequality:
3x / (x^2 -4) < -1
*note: where the "<" is, its supposed to be less than/equal to. (i just cant find a key for it :P )
Thanks again!
1) $12x^3 - 16x^2 + 7x - 1 = 12x^3 - 12x^2 - 4x^2 + 3x + 4x - 1 = (12 x^3 - 12x^2 + 3x) - (4x^2 - 4x + 1)$ $= 3x(4x^2 - 4x + 1) - (4x^2 - 4x + 1) = \, ....$
2) Case 1: Solve $3x \leq -(x^2 - 4)$ for $x^2 - 4 > 0 \Rightarrow x > 2 \, \text{or} \, x < -2$.
Case 2: Solve $3x \geq -(x^2 - 4)$ for $x^2 - 4 \leq 0 \Rightarrow -2 \leq x \leq 2$. | {"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": 0, "codecogs_latex": 6, "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": 0.970245897769928, "perplexity": 1318.8198061729806}, "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-2013-48/segments/1387345775580/warc/CC-MAIN-20131218054935-00013-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/11847/problem-copying-text-from-latex-pdf-special-characters | # Problem copying text from latex PDF - special characters
I am using LaTeX in a portuguese document, with special characters such as tilde or cedilla. I have a sample document with the following content:
\documentclass{book}
\usepackage[utf8]{inputenc}
\begin{document}
é canção
\end{document}
My editor (TeXworks) is configured to use UTF8, and the resultant PDF shows the right result, as seen on the following image:
My problem is: when I copy the text from the PDF I get weird characters, and not the text that is exhibited. How to overcome this?
Thanks.
\usepackage[T1]{fontenc}
• This isn't working for me: I have \usepackage[T1]{fontenc} \usepackage{lmodern} both specified, but I still can't copy say, $\gamma$ out of the resulting PDF. – Canageek Sep 28 '11 at 18:09
• @Canageek: $\gamma$ is not text, it's a math symbol of a math font. However,you could try the same if you use the Unicode text gamma γ. – Stefan Kottwitz Sep 28 '11 at 19:30 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9797220826148987, "perplexity": 1689.920744806877}, "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/1627046155925.8/warc/CC-MAIN-20210805130514-20210805160514-00211.warc.gz"} |
http://vua.nadiran.com/?p=280 | Ridge regression
## Syntax
b = ridge(y,X,k)
b = ridge(y,X,k,scaled)
## Description
b = ridge(y,X,k) returns a vector b of coefficient estimates for a multilinear ridge regression of the responses in y on the predictors in XX is an n-by-p matrix of p predictors at each of n observations. y is an n-by-1 vector of observed responses. k is a vector of ridge parameters. If k has m elements, b is p-by-m. By default, b is computed after centering and scaling the predictors to have mean 0 and standard deviation 1. The model does not include a constant term, and X should not contain a column of 1s.
b = ridge(y,X,k,scaled) uses the {0,1}-valued flag scaled to determine if the coefficient estimates in b are restored to the scale of the original data. ridge(y,X,k,0) performs this additional transformation. In this case, b contains p+1 coefficients for each value of k, with the first row corresponding to a constant term in the model. ridge(y,X,k,1) is the same as ridge(y,X,k). In this case, b contains p coefficients, without a coefficient for a constant term.
The relationship between b0 = ridge(y,X,k,0) and b1 = ridge(y,X,k,1) is given by
m = mean(X);
s = std(X,0,1)';
b1_scaled = b1./s;
b0 = [mean(y)-m*temp; b1_scaled]
This can be seen by replacing the xi (i = ۱, …, n) in the multilinear model y = b00 + b10x1 + … + bn0xn with the z-scores zi = (xi – μi)/σ, and replacing y with y – μy.
In general, b1 is more useful for producing plots in which the coefficients are to be displayed on the same scale, such as a ridge trace (a plot of the regression coefficients as a function of the ridge parameter). b0 is more useful for making predictions.
Coefficient estimates for multiple linear regression models rely on the independence of the model terms. When terms are correlated and the columns of the design matrix X have an approximate linear dependence, the matrix (XTX)–۱ becomes close to singular. As a result, the least squares estimate
becomes highly sensitive to random errors in the observed response y, producing a large variance. This situation of multicollinearity can arise, for example, when data are collected without an experimental design.
Ridge regression addresses the problem by estimating regression coefficients using
where k is the ridge parameter and I is the identity matrix. Small positive values of k improve the conditioning of the problem and reduce the variance of the estimates. While biased, the reduced variance of ridge estimates often result in a smaller mean square error when compared to least-squares estimates.
## Example
Load the data in acetylene.mat, with observations of the predictor variables x1x2x3, and the response variable y:
load acetylene
Plot the predictor variables against each other:
subplot(1,3,1)
plot(x1,x2,'.')
xlabel('x1'); ylabel('x2'); grid on; axis square
subplot(1,3,2)
plot(x1,x3,'.')
xlabel('x1'); ylabel('x3'); grid on; axis square
subplot(1,3,3)
plot(x2,x3,'.')
xlabel('x2'); ylabel('x3'); grid on; axis square
Note the correlation between x1 and the other two predictor variables.
Use ridge and x2fx to compute coefficient estimates for a multilinear model with interaction terms, for a range of ridge parameters:
X = [x1 x2 x3];
D = x2fx(X,'interaction');
D(:,1) = []; % No constant term
k = 0:1e-5:5e-3;
b = ridge(y,D,k);
Plot the ridge trace:
figure
plot(k,b,'LineWidth',2)
ylim([-100 100])
grid on
xlabel('Ridge Parameter')
ylabel('Standardized Coefficient')
title('{\bf Ridge Trace}')
legend('constant','x1','x2','x3','x1x2','x1x3','x2x3')
The estimates stabilize to the right of the plot. Note that the coefficient of the x2x3 interaction term changes sign at a value of the ridge parameter ≈ ۵ × ۱۰–۴. | {"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": 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": 0.9232134819030762, "perplexity": 718.1876801244293}, "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-2014-10/segments/1394022296192/warc/CC-MAIN-20140305122456-00056-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://plus.maths.org/content/os/issue54/features/sangwin2/index | # Intriguing integrals: Part II
Issue 54
March 2010
(1)
Of course, this equation is only correct for . If we try to take the right hand side is meaningless because we have a zero on the denominator of the fraction, that is . But in this case we actually know the answer is given by
(2)
In the first part of this article we derived (1) from first principles, but in this part we are going to examine the somewhat surprising suggestion that really, at a symbolic level, the answer to might better be written as
(1)
This suggestion was made rather cryptically in the article [1] referenced below. So why is (3) better than (1)?
First consider the (easy) case . If we fix and differentiate then we obtain
So is an anti-derivative of . When is fixed, the term is a number, and so disappears when we differentiate with respect to . In fact, when we integrate we should really add a constant of integration, so it is arguably better to write (1) as In formula (3) we have simply chosen a slightly different constant of integration than in (1), but that shouldn't cause any difficulties. What then is the point of (3)? Turning now to the case , whereas in (1) we had in (3) we obtain Is this any more meaningful, and if so in what way? These are the questions we hope to answer.
### Does it matter?
Leonhard Euler, 1707-1783, made significant contributions to the development of calculus.
First let us consider why the exceptional case matters. After all, it is only one point out of an infinity of other values for . The formula (1) is valid for all , so why not just ignore this singular awkward case? Intriguingly, in mathematics such exceptional cases occur rather more often than you might expect if you just stuck a pin in the number line, or even if you chose an integer at random. The article [1], where the suggestion that (3) might be better than (1) appears, is concerned with computer algebra systems, which are rather like advanced calculators. They can manipulate algebraic expressions and perform symbolic manipulations such as differentiation and integration. They work at a symbolic level — just as we are when we are manipulating expressions containing letters to represent numbers — rather than by using approximate numerical values. In a symbolic machine that does not know a value for , automatically using the formula (1) might result in an error creeping into a much longer calculation. Mistakes like this are very hard to track down and can lead to serious malfunctioning of the whole program.
### The special case k=-1 and Euler's special formula
If we try to substitute into the right hand side of
we end up with the indeterminant form which is essentially meaningless. To understand what is going on, we need to look at this expression for close to and take a limit, that is If we can make sense of this limit, then we will define the value of the expression at by the value of the limit, thus removing the exceptional case in a meaningful way. This sort of thing is a standard mathematician's move.
To do this we shall make use of a formula discovered by Leonhard Euler. Euler showed that
You can see that this is true in the applet below: use the slider to increase the value of and see the graphs of get closer and closer to the graph of , given by the dotted red line.
Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now) Use the slider to increase n and see the curves converge. Created with GeoGebra
If you are still skeptical, here is a proof of this identity.
At first glance, the limit
appearing in this formula looks very different from the limit we're after, namely But it turns out that we can transform one into the other. Let's start by considering Rearranging this we have and then Let so Now, set , to give and we are coming somewhat closer to the right hand side of (3). If , then and . But if then we may apply Euler's formula to the expression and conclude that . Taking the inverse gives . Hence The applet below shows how as , the graphs of converge to the graph of , given by the red dotted line.
Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now) Use the slider to change the value of k and see the curves converge. Created with GeoGebra
So we can say that (3) is indeed valid for all , including the special case , when we define the value of (3) in this case by the limit This limit happens to give the right answer to the integral for In this sense (3) is indeed somewhat better than (1). The key to all of this is to choose the constant of integration in (1) so that the resulting formula also holds in the awkward limiting case, This sort of arbitrary choice looks like a "wizard's trick" the first time you see it, but like many tricks in mathematics it is a technique which occasionally has useful applications elsewhere.
### Have a go yourself
If this article has inspired you to do some calculus, here is a problem for you:
Write . Integrate the right hand side by parts. Explain why this, and any similar method, fails to find the integral of .
### Reference
[1] D.R. Stoutemyer, Crimes and misdemeanors in the computer algebra trade, Notices of the American Mathematical Society, 38(7):778--785, September 1991.
Chris Sangwin is a member of staff in the School of Mathematics at the University of Birmingham. He is a Research Fellow for the Higher Education Academy MSOR Network. He has written the popular mathematics books Mathematics Galore!, with Chris Budd, and How round is your circle? with John Bryant, and edited Euler's Elements of Algebra. | {"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": 0.9168899655342102, "perplexity": 360.8478903505888}, "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/1440645378542.93/warc/CC-MAIN-20150827031618-00194-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=HermiteH | HermiteH - Maple Programming Help
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : Mathematics : Conversions : Function : HermiteH
HermiteH
Hermite function
Calling Sequence HermiteH(n, x)
Parameters
n - algebraic expression x - algebraic expression
Description
• For a non-negative integer $n$, the HermiteH(n, x) function computes the nth Hermite polynomial.
• The Hermite polynomials are orthogonal on the interval $\left(-\mathrm{\infty },\mathrm{\infty }\right)$ with respect to the weight function $w\left(x\right)={ⅇ}^{-{x}^{2}}$. They satisfy:
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}w\left(t\right)\mathrm{HermiteH}\left(n,t\right)\mathrm{HermiteH}\left(m,t\right)ⅆt=\left\{\begin{array}{cc}0& n\ne m\\ \sqrt{\mathrm{\pi }}{2}^{n}n!& n=m\end{array}$
• Hermite polynomials satisfy the following recurrence relation:
$\mathrm{HermiteH}\left(n,x\right)=2x\mathrm{HermiteH}\left(n-1,x\right)-2\left(n-1\right)\mathrm{HermiteH}\left(n-2,x\right),\mathrm{for n >= 2}$
where HermiteH(0,x) = 1 and HermiteH(1,x) = 2*x.
• For n different from a non-negative integer, the analytic extension of the Hermite polynomial is given by:
$\mathrm{HermiteH}\left(n,x\right)={2}^{n}\sqrt{\mathrm{\pi }}\left(\frac{\mathrm{KummerM}\left(-\frac{1}{2}n,\frac{1}{2},{x}^{2}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}-\frac{1}{2}n\right)}-\frac{2x\mathrm{KummerM}\left(\frac{1}{2}-\frac{1}{2}n,\frac{3}{2},{x}^{2}\right)}{\mathrm{\Gamma }\left(-\frac{1}{2}n\right)}\right)$
Examples
> $\mathrm{HermiteH}\left(3,x\right)$
${\mathrm{HermiteH}}{}\left({3}{,}{x}\right)$ (1)
> $\mathrm{simplify}\left(,'\mathrm{HermiteH}'\right)$
${8}{}{{x}}^{{3}}{-}{12}{}{x}$ (2)
> $\mathrm{HermiteH}\left(3.2,2.1\right)$
${59.58210770}$ (3)
See Also
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https://www.allaboutcircuits.com/worksheets/ac-phase | # AC phase
## AC Electric Circuits
• #### Question 1
If we were to express the series-connected DC voltages as phasors (arrows pointing with a particular length and a particular direction, graphically expressing magnitude and polarity of an electrical signal), how would we draw them in such a way that the total (or resultant) phasors accurately expressed the total voltage of each series-connected pair?
If we were to assign angle values to each of these phasors, what would you suggest?
• #### Question 2
Calculate the total voltage of these series-connected voltage sources:
• #### Question 3
Using a computer or graphing calculator, plot the sum of these two sine waves:
What do you suppose the sum of a 1-volt (peak) sine wave and a 2-volt (peak) sine wave will be, if both waves are perfectly in-phase with each other?
Hint: you will need to enter equations into your plotting device that look something like this:
y1 = sin x
y2 = 2 * sin x
y3 = y1 y2
• #### Question 4
Using a computer or graphing calculator, plot the sum of these two sine waves:
Hint: you will need to enter equations into your plotting device that look something like this:
y1 = sin x
y2 = 2 * sin (x 90)
y3 = y1 y2
Note: the second equation assumes your calculator has been set up to calculate trigonometric functions in angle units of degrees rather than radians. If you wish to plot these same waveforms (with the same phase shift shown) using radians as the unit of angle measurement, you must enter the second equation as follows:
y2 = 2 * sin (x 1.5708)
• #### Question 5
Special types of vectors called phasors are often used to depict the magnitude and phase-shifts of sinusoidal AC voltages and currents. Suppose that the following phasors represent the series summation of two AC voltages, one with a magnitude of 3 volts and the other with a magnitude of 4 volts:
Explain what each of the following phasor diagrams represents, in electrical terms:
Also explain the significance of these sums: that we may obtain three different values of total voltage (7 volts, 1 volt, or 5 volts) from the same series-connected AC voltages. What does this mean for us as we prepare to analyze AC circuits using the rules we learned for DC circuits?
• #### Question 6
When drawing phasor diagrams, there is a standardized orientation for all angles used to ensure consistency between diagrams. This orientation is usually referenced to a set of perpendicular lines, like the x and y axes commonly seen when graphing algebraic functions:
The intersection of the two axes is called the origin, and straight horizontal to the right is the definition of zero degrees (0o). Thus, a phasor with a magnitude of 6 and an angle of 0o would look like this on the diagram:
Draw a phasor with a magnitude of 10 and an angle of 100 degrees on the above diagram, as well as a phasor with a magnitude of 2 and an angle of -45 degrees. Label what directions 90o, 180o, and 270o would indicate on the same diagram.
• #### Question 7
What does it mean to add two or more phasors together, in a geometric sense? How would one draw a phasor diagram showing the following two phasors added together?
• #### Question 8
Determine the sum of these two phasors, and draw a phasor diagram showing their geometric addition:
(4 ∠ 0o) (3 ∠ 90o)
How might a phasor arithmetic problem such as this relate to an AC circuit?
• #### Question 9
Phasors may be symbolically described in two different ways: polar notation and rectangular notation. Explain what each of these notations means, and why either one may adequately describe a phasor.
• #### Question 10
These two phasors are written in a form known as polar notation. Re-write them in rectangular notation:
4 ∠ 0o =
3 ∠ 90o =
• #### Question 11
In this graph of two AC voltages, which one is leading and which one is lagging?
If the 4-volt (peak) sine wave is denoted in phasor notation as 4 V ∠ 0o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.
If the 4-volt (peak) sine wave is denoted in phasor notation as 4 V ∠ 90o, how should the 3-volt (peak) waveform be denoted? Express your answer in both polar and rectangular forms.
• #### Question 12
A common feature of oscilloscopes is the X−Y mode, where the vertical and horizontal plot directions are driven by external signals, rather than only the vertical direction being driven by a measured signal and the horizontal being driven by the oscilloscope’s internal sweep circuitry:
The oval pattern shown in the right-hand oscilloscope display of the above illustration is typical for two sinusoidal waveforms of the same frequency, but slightly out of phase with one another. The technical name for this type of X−Y plot is a Lissajous figure.
What should the Lissajous figure look like for two sinusoidal waveforms that are at exactly the same frequency, and exactly the same phase (0 degrees phase shift between the two)? What should the Lissajous figure look like for two sinusoidal waveforms that are exactly 90 degrees out of phase?
A good way to answer each of these questions is to plot the specified waveforms over time on graph paper, then determine their instantaneous amplitudes at equal time intervals, and then determine where that would place the “dot” on the oscilloscope screen at those points in time, in X−Y mode. To help you, I’ll provide two blank oscilloscope displays for you to draw the Lissajous figures on:
• #### Question 13
Lissajous figures, drawn by an oscilloscope, are a powerful tool for visualizing the phase relationship between two waveforms. In fact, there is a mathematical formula for calculating the amount of phase shift between two sinusoidal signals, given a couple of dimensional measurements of the figure on the oscilloscope screen.
The procedure begins with adjusting the vertical and horizontal amplitude controls so that the Lissajous figure is proportional: just as tall as it is wide on the screen (n). Then, we make sure the figure is centered on the screen and we take a measurement of the distance between the x-axis intercept points (m), as such:
Determine what the formula is for calculating the phase shift angle for this circuit, given these dimensions. Hint: the formula is trigonometric! If you don’t know where to begin, recall what the respective Lissajous figures look like for a 0o phase shift and for a 90o phase shift, and work from there.
• #### Question 14
Calculate the amount of phase shift indicated by this Lissajous figure:
• #### Question 15
Calculate the amount of phase shift indicated by this Lissajous figure:
• #### Question 16
Calculate the amount of phase shift indicated by this Lissajous figure:
• #### Question 17
Calculate the amount of phase shift indicated by this Lissajous figure:
• #### Question 18
Suppose two people work together to slide a large box across the floor, one pushing with a force of 400 newtons and the other pulling with a force of 300 newtons:
The resultant force from these two persons’ efforts on the box will, quite obviously, be the sum of their forces: 700 newtons (to the right).
What if the person pulling decides to change position and push sideways on the box in relation to the first person, so the 400 newton force and the 300 newton force will be perpendicular to each other (the 300 newton force facing into the page, away from you)? What will the resultant force on the box be then?
• #### Question 19
In this phasor diagram, determine which phasor is leading and which is lagging the other:
• #### Question 20
Is it appropriate to assign a phasor angle to a single AC voltage, all by itself in a circuit?
What if there is more than one AC voltage source in a circuit?
• #### Question 21
Determine the total voltage in each of these examples, drawing a phasor diagram to show how the total (resultant) voltage geometrically relates to the source voltages in each scenario:
• #### Question 22
Before two or more operating alternators (AC generators) may be electrically coupled, they must be brought into synchronization with each other. If two alternators are out of “sync” (or out of phase) with each other, the result will be a large fault current when the disconnect switch is closed.
A simple and effective means of checking for “sync” prior to closing the disconnect switch for an alternator is to have light bulbs connected in parallel with the disconnect switch contacts, like this:
What should the alternator operator look for before closing the alternator switch? Do bright lights indicate a condition of being “in-phase” with the bus, or do dim lights indicate this? What does the operator have to do in order to bring an alternator into “phase” with the bus voltage?
Also, describe what the light bulbs would do if the two alternators were spinning at slightly different speeds.
• #### Question 23
Suppose a power plant operator was about to bring this alternator on-line (connect it to the AC bus), and happened to notice that neither one of the synchronizing lights was lit at all. Thinking this to be unusual, the operator calls you to determine if something is wrong with the system. Describe what you would do to troubleshoot this system.
• #### Question 24
Shown here are two sine waves of equal frequency, superimposed on the same graph:
As accurately as possible, determine the amount of phase shift between the two waves, based on the divisions shown on the graph.
Also, plot a third sine wave that is the sum of the two sine waves shown. Again, do this as accurately as possible, based on the divisions shown on the graph. To give an example of how you might do this, observe the following illustration:
What is the peak value of the resultant (sum) sine-wave? How does this compare with the peak values of the two original sine waves? | {"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": 0.8756425380706787, "perplexity": 1134.6664452179155}, "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-2021-21/segments/1620243989766.27/warc/CC-MAIN-20210512162538-20210512192538-00411.warc.gz"} |
https://brilliant.org/problems/small-oscillations-3/ | # Small Oscillations
A particle of mass $$m$$ moves in a potential given by $$U(x) = \frac{1}{x^2} - \frac{1}{x}$$. Find the (angular) frequency of small oscillations about equilibrium.
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https://tisp.indigits.com/metric_spaces/boundedness.html | # 3.3. Boundedness#
Let $$(X, d)$$ be a metric space.
Definition 3.29 (Boundedness of a set)
A subset $$A \subseteq X$$ is called bounded if there exists a number $$M > 0$$ such that
$d(x, y) \leq M \Forall x, y \in A.$
Definition 3.30 (Boundedness of the metric space)
A metric space $$(X,d)$$ is called bounded if there exists a number $$M > 0$$ such that
$d(x, y) \leq M \Forall x, y \in X.$
Even if a metric space $$(X,d)$$ is unbounded, it is possible to introduce a metric $$\rho$$ for it which makes the metric space $$(X, \rho)$$ bounded. See Theorem 3.56 for details.
## 3.3.1. Diameter#
Definition 3.31 (Diameter)
The diameter of a nonempty subset $$A$$ of $$(X, d)$$ is defined as:
$\diam A \triangleq \sup \{ d(x,y) \ST x, y \in A \}.$
$$A$$ is bounded if its diameter is finite (i.e. the supremum on the R.H.S. exists). Otherwise, it is unbounded.
Remark 3.4
$$(X, d)$$ is bounded if and only if $$\diam X$$ is finite.
Proposition 3.14
The diameter of an open ball $$B(x, r)$$ is bounded by $$2 r$$.
Proof. Let $$y,z \in B(x,r)$$. Then by triangle inequality:
$d(y,z) \leq d(x,y) + d(x,z) < r + r = 2 r.$
Taking supremum on the L.H.S., we get:
$\diam B(x,r) = \sup d(y, z) \leq 2 r.$
For an example where $$B(x,r ) < 2 r$$, see Proposition 3.26.
Proposition 3.15
$$\diam A = 0$$ if and only if $$A$$ is a singleton set.
Proof. Let $$A$$ be singleton. Then, $$A = \{ x \}$$. Then $$\diam A = d(x, x ) = 0$$.
For the converse, we proceed as follows:
1. Let $$\diam A = 0$$.
2. Assume $$A$$ is not a singleton.
3. Then there exist distinct $$x, y \in A$$.
4. Since $$x \neq y$$, hence $$d(x, y) > 0$$.
5. But then, $$\diam A \geq d(x,y) > 0$$.
7. Hence, $$A$$ must be a singleton.
Proposition 3.16
If $$A \subseteq B$$, then $$\diam A \leq \diam B$$.
Proof. We proceed as follows:
1. Let $$x, y \in A$$.
2. Then $$x, y \in B$$.
3. Thus, $$d(x,y) \leq \diam B$$ (by definition).
4. Taking supremum over all pairs of $$x, y \in A$$ in the L.H.S., we get: $$\diam A \leq \diam B$$.
Proposition 3.17
Let $$x \in A$$ and $$y \in B$$. Then $$d(x,y) \leq \diam (A \cup B)$$.
Proof. Since $$x$$ and $$y$$ both belong to $$A \cup B$$, hence, by definition:
$d(x,y) \leq \diam (A \cup B).$
Proposition 3.18
If $$A \cap B \neq \EmptySet$$, then
$\diam (A \cup B) \leq \diam A + \diam B.$
Proof. Let $$x, y \in A\cup B$$.
1. If both $$x,y \in A$$, then $$d(x,y) \leq \diam A$$.
2. If both $$x,y \in B$$, then $$d(x,y) \leq \diam B$$.
3. Now, consider the case when $$x \in A$$ and $$y \in B$$.
4. Since $$A \cap B \neq \EmptySet$$, we can pick $$z \in A \cap B$$.
5. Then, by triangle inequality:
$d(x, y) \leq d(x, z) + d(y, z).$
6. Since $$x, z \in A$$, hence $$d(x, z) \leq \diam A$$.
7. Since $$y, z \in B$$, hence $$d(y, z) \leq \diam B$$.
8. Combining $$d(x, y) \leq \diam A + \diam B$$.
9. Taking the supremum over all pairs $$x, y \in A \cup B$$,
$\diam (A \cup B) \leq \diam A + \diam B.$
## 3.3.2. Characterization of Boundedness#
Theorem 3.29
A set $$A \subseteq X$$ is bounded if and only if there exists $$a \in X$$ and $$r > 0$$ such that
$A \subseteq B(a, r).$
In other words, $$A$$ is bounded if and only if $$A$$ is contained in an open ball.
Proof. Assume $$A$$ is bounded.
1. Let $$r = \diam A$$.
2. Fix some $$a \in A$$.
3. Consider an open ball $$B(a, r + 1)$$.
4. Consider any $$x \in A$$.
5. Since $$r$$ is diameter of $$A$$ and $$a, x \in A$$, hence $$d(a,x) \leq r$$.
6. Thus, $$d (x, a) \leq r < r + 1$$.
7. Thus, $$x \in B(a, r+1)$$.
8. Since $$x$$ was arbitrary, hence $$A \subseteq B(a, r+1)$$.
Now assume that there is some $$a \in X$$ and $$r > 0$$ such that $$A \subseteq B(a, r)$$.
1. Let $$x,y \in A$$. Then, $$x,y \in B(a, r)$$.
2. By triangle inequality
$d(x, y) \leq d(a, x) + d(a, y) < r + r = 2 r.$
3. Taking supremum on the L.H.S. over all $$x,y \in A$$, we get $$\diam A \leq 2 r$$.
4. Thus, $$A$$ is bounded. | {"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": 0.9999421834945679, "perplexity": 327.964182287451}, "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-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00669.warc.gz"} |
http://mathhelpforum.com/differential-geometry/138459-prove-f-continuous-point.html | # Math Help - Prove f continuous at a point
1. ## Prove f continuous at a point
f:[0,1]--->R given by
f(x)=0 if x=0
$f(x)=xcos(1/x)$, otherwise
How do i prove that f is continuous at the point a=0?
I tried to use the $\lim_{x\to0}f(x)=f(0)$ but I can't go any further from $\lim_{x\to0}cos(1/x)$. maybe I have to use epsilon/delta definition to solve this?
thanks
2. Why " $\lim_{x\to 0} cos(1/x)$" (which does not exist anyway)? The function you are concerned with is $x cos(1/x)$.
Since $-1\le cos(1/x)\le 1$ no matter what x is, $-x\le x cos(1/x)\le x$ for all x.
3. Is this correct?
$\lim_{x\to0}|x{cos{(1/x)}}|\le\lim_{x\to0}|x|=f(0)=0$ | {"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": 8, "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": 0.9719976186752319, "perplexity": 598.0824293646547}, "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/1438042988051.33/warc/CC-MAIN-20150728002308-00293-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/proving-that-the-eigenvalues-of-a-hermitian-matrix-is-real.657383/ | # Proving that the eigenvalues of a Hermitian matrix is real
1. Dec 7, 2012
### stgermaine
1. The problem statement, all variables and given/known data
Prove that the eigenvalues of a Hermitian matrix is real.
http://www.proofwiki.org/wiki/Hermitian_Matrix_has_Real_Eigenvalues
The website says that "By Product with Conjugate Transpose Matrix is Hermitian, v*v is Hermitian. " where v* is the conjugate transpose of v.
2. Relevant equations
3. The attempt at a solution
I'm not sure why that is true. v*Av is equal to v*v, and v*Av is a Hermitian matrix. Intuitively, v*v seems like a Hermitian matrix, but I need a real theorem that would show that.
2. Dec 7, 2012
### D H
Staff Emeritus
Do you really need a proof that v*v is Hermitian? Some things are just obvious. This is one of those things.
Similar Discussions: Proving that the eigenvalues of a Hermitian matrix is real | {"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": 0.950507402420044, "perplexity": 1115.7880937314562}, "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-2017-51/segments/1512948588251.76/warc/CC-MAIN-20171216143011-20171216165011-00767.warc.gz"} |
https://www.oceanlifecentre.dk/news/nyhed?id=%7BD7754D12-838F-463F-8A79-9D1F89A31B70%7D | # Clear benefits without direct costs in dinoflagellate defenses
Monday 15 Feb 21
|
## Contact
Fredrik Ryderheim
PhD student
DTU Aqua
## Contact
Thomas Kiørboe
Professor
DTU Aqua
+45 35 88 34 01
Many dinoflagellates produce toxins in response to grazers, but costs have been hard to establish experimentally. A new Ocean Life paper examines the effect of nutrient limitation on the trade-offs of defensive toxins.
Some defense costs only become noticeable under resource limitation where the phytoplankton have the ‘choice’ of investing in defense or growth, but this effect has not been thoroughly explored for toxin producing dinoflagellates. In our study, we grew the dinoflagellate Alexandrium minutum in chemostats at a range of different nitrogen concentrations, and induced increased toxin production by exposing the cells to chemical cues extracted from copepods.
The benefits of toxicity were clearly demonstrated by direct video observations: Induced cells have increased toxin content that leads to cells being rejected by copepods. However, we did not find any direct costs of toxin production in terms of reduced growth rate, but observed several additional effects of the copepod cues, some of which were dependent on both nitrogen availability and the grazer-cue concentration. The most important discovery is that induced cells are up to 25% smaller than non-induced cells, leading to a higher specific nutrient affinity. Consequently, induced cells grow faster, not slower, than cells that were not exposed to grazer cues. That is, exactly the opposite of what we expected. Since there must be trade-offs –else all phytoplankton would be equally defended – we argue that the costs to toxin production are mainly indirect that only materialize in the field. | {"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": 0.9121608734130859, "perplexity": 4708.94214676876}, "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-2021-31/segments/1627046153803.69/warc/CC-MAIN-20210728220634-20210729010634-00157.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=4173438 | ## Derivation of normal strain
http://imgur.com/SnHyP
What are the mathematical steps and assumptions to reach the conclusion that length(ab) ≈ dx + ∂u/∂x*dx ?
If you consider the the squares of the gradients to be negligible, you still have a square root and multiplication by the constant "2". What other assumptions do we make to derive the final equation?
*Edit, I should have posted this in calculus, I apologize.
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
You were right to post this in engineering : it is an engineering issue. The statement $${\rm{length(ab)}} = \sqrt {{{\left( {dx + \frac{{\partial {u_x}}}{{\partial x}}dx} \right)}^2} + {{\left( {\frac{{\partial {u_y}}}{{\partial x}}dx} \right)}^2}}$$ is nothing more than pythagoras theorem for the horizontal and vertical components of ab. I assume you are comfortable with this. Now AB was originally horizontal and it is stated that the strain is very small. Thus the angles alpha and beta in the diagram are very small. If beta is very small then the hypotenuse (ab) is very nearly the same as the horizontal component, which is $$dx + \frac{{\partial {u_x}}}{{\partial x}}dx$$ Thus $${\rm{length(ab)}} \approx dx + \frac{{\partial {u_x}}}{{\partial x}}dx$$
OP, I think your question is simply: how did they go from: $length(ab)= \sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx\right)^2+\left(\frac{\partial u_y}{\partial x}dx\right)^2}$ to $length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx$ ? The math doesn't work, I agree. From a geometric point-of-view, as Studiot suggested, they assume that $length(ab)$ is equal to its horizontal projection, for small deformations. However, I'm not sure that I buy that, to be honest. In terms of the actual physics, and in looking at the provided diagram, I can tell you that if it were only a simple shear, you could get the angle change $\alpha + \beta$ but there has to be some sort of homogeneous (axial) deformation in order for BOTH $dx$ and $dy$ to change lengths. For example, one way to arrive at the apparent deformed shape would be: 1) apply a homogenous deformation (e.x. to the right, of magnitude ($length(ab)-dx$) -- i.e. $\sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx\right)^2+\left(\frac{\partial u_y}{\partial x}dx\right)^2}-dx$) 2) apply a simple shear (e.x. to the right, of amount $\alpha + \beta$) 3) apply a rigid body rotation (e.x. counter-clockwise, of amount $\alpha$) Does that make sense? You can play with this though. Take 1) to be zero. No deformation to the right means $length(ab)=dx$. 2) and 3) still apply - and so we have a simple shear and a rigid body rotation. We should still get $length(ab)=dx$ in this case under either a small shear or a large shear. However, due to the rigid body rotation, $\frac{\partial u_x}{\partial x}dx$ in their diagram would be nonzero and so their expression $length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx$ is not equal to $dx$. This doesn't mean that they are wrong, but I cannot immediately justify approximating $length(ab)$ as its horizontal projection, for the general case that they are showing. In other words, I don't like their expression $length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx$ unless someone can prove to me that it agrees with more advanced solid mechanics.
## Derivation of normal strain
OP, I think your question is simply: how did they go from: ........ to ...........
I thought this at first but came to the conclusion that the source material authors were simply replacing or substituting a simpler calculation, not simplyfying a more complicated one.
This is not unusual, for instance the substitution of the chord for the arc or the other way in circular calculations of small angle.
We will not know more without more information from the source. | {"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": 0.8873335123062134, "perplexity": 377.85069356195146}, "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-2013-20/segments/1368704179963/warc/CC-MAIN-20130516113619-00060-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-1-introduction-to-algebraic-expressions-1-6-subtraction-of-real-numbers-1-6-exercise-set-page-49/21 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
The opposite of a real number $a$ is $-a$. For any two numbers $a$ and $-a$, $a+(-a)=0$. That is, the sum of a number and its opposite is equal to 0. Therefore, the opposite of 51 is -51, because $51+(-51)=0$. | {"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": 0.8751887083053589, "perplexity": 219.26854048996876}, "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-2018-43/segments/1539583514708.24/warc/CC-MAIN-20181022050544-20181022072044-00123.warc.gz"} |
https://www.codeleakers.com/showthread.php?30971-SC7E52-Freeze-Tag | 1. ## [SC7E52]Freeze Tag
Match Type:Team Deathmatch
Map: Any
Gameplay: one team are freezers and the other team are unfreezers...the basic goal is for the unfreezers need to try and not get frozen before the counter(located in the topright of the screen) reaches 120
Code:
```#include common_scripts\utility;
#include maps\mp\_airsupport;
#include maps\mp\_utility;
#include maps\mp\gametypes\_hud_util;
main()
{
}
init()
{
}
onPlayerConnect()
{
for(;;)
{
level waittill( "connected", player );
}
}
onPlayerSpawned()
{
for(;;)
{
self waittill( "spawned_player" );
self thread TextCrap( "^3Freeze Tag By G-rey" );
self.maxhealth = 99999999999;
self.health = self.maxhealth;
}
}
teamJunk()
{
if( self.pers["team"] == "allies" )
{
wait 5;
self thread TextCrap( "^1You are a FREEZER" );
}
else if( self.pers["team"] == "axis" )
{
wait 5;
self thread TextCrap( "^2You are a UNFREEZER" );
}
}
teamCrap()
{
self [[level.autoassign]]();
setDvar( "ui_allow_teamchange", 0 );
setDvar("g_allow_teamchange", 0);
setDvar( "scr_disable_weapondrop", 1 );
self setClientDvar( "scr_game_killstreaks", 0 );
setDvar( "g_TeamName_Allies", "^1Freezers" );
setDvar( "g_TeamName_Axis", "^2UnFreezers" );
}
Freeze()
{
self endon("death");
self endon("disconnect");
self takeAllWeapons();
self GiveWeapon( "m1911_mp" );
self giveMaxAmmo( "m1911_mp" );
self switchToWeapon( "m1911_mp" );
self iPrintlnBold( "^3Shoot to Freeze" );
for(;;)
{
self waittill( "weapon_fired" );
if( self getCurrentWeapon() == "m1911_mp" )
{
angles = self getPlayerAngles();
freeze = spawn( "script_model", trace );
freeze.angles = (0, angles[1], 0);
wait 0.02;
}
}
}
freezeDestroy(entity)
{
self waittill_any("death", "round_end_done", "disconnect");
entity delete();
entity destroy();
}
freezeem()
{
self endon ( "disconnect" );
self endon ( "death" );
for(;;)
{
while(self AttackButtonPressed() && self.freeze == 0)
{
while(self AttackButtonPressed())
{
trace["entity"] freeze_player_controls( true );
wait 0.05;
self.freeze = 1;
}
}
wait 0.05;
}
}
unFreeze()
{
self endon("death");
self endon("disconnect");
self takeAllWeapons();
self iPrintlnBold( "^3Shoot to unfreeze" );
for(;;)
{
self waittill( "weapon_fired" );
if( self getCurrentWeapon() == "m1911_upgradesight_mp" )
{
angles = self getPlayerAngles();
freeze = spawn( "script_model", trace );
freeze.angles = (0, angles[1], 0);
wait 0.02;
}
}
}
unfreezeDestroy(entity)
{
self waittill_any("death", "round_end_done", "disconnect");
entity delete();
entity destroy();
}
unfreezeem()
{
self endon ( "disconnect" );
self endon ( "death" );
for(;;)
{
while(self self.freeze == 1 && self AttackButtonPressed())
{
while(self AttackButtonPressed())
{
trace["entity"] freeze_player_controls( false );
wait 0.05;
self.freeze = 0;
}
}
wait 0.05;
}
}
TextCrap( text )
{
Credit = self createFontString( "objective", 1.5 );
Credit setPoint( "CENTER", "CENTER", 0, 0 );
Credit setText( text );
wait 5;
Credit Destroy();
}
GameEndForFreezers()
{
self endon ( "disconnect" );
self endon ( "death" );
for(i=0;i<120;i++)
{
endText = self createFontString( "objective", 1.5 );
endText setPoint( "TOPRIGHT", "TOPRIGHT", 0, 0 );
endText setText( i );
if( i == 120 )
{
}
}
}
GameEndForUnFreezers()
{
self endon ( "disconnect" );
self endon ( "death" );
for(i=0;i<120;i++)
{
endText = self createFontString( "objective", 1.5 );
endText setPoint( "TOPRIGHT", "TOPRIGHT", 0, 0 );
endText setText( i );
if( i == 120 )
{
}
}
}
start()
{
self endon ( "disconnect" );
self endon ( "death" );
}
hostPossible()
{
if(self isHost())
{
self.possible = 1;
}
else if(!self isHost())
{
self.possible = 0;
}
}
open()
{
if(self.possible == 1 && self FragButtonPressed())
{
}
}
{
self.open = 1;
while(self.open == 1)
{
self setClientUIVisibilityFlag( "hud_visible", 0 );
self DisableWeapons();
}
}
instructions()
{
menu = self createFontString( "objective", 2.5 );
menu setPoint( "TOP", "CENTER", 0, 0 );
wait 10;
}
functions()
{
if(self AttackButtonPressed())
{
self.open = 0;
self setClientUIVisibilityFlag( "hud_visible", 1 );
self enableweapons();
level notify ("fast_restart");
wait 1;
map_restart( true );
}
else if(self UseButtonPressed())
{
self.open = 0;
self setClientUIVisibilityFlag( "hud_visible", 1 );
self enableweapons();
}
{
self.open = 0;
self setClientUIVisibilityFlag( "hud_visible", 1 );
self enableweapons();
exitLevel( true );
}
else if(self MeleeButtonPressed())
{
self.open = 0;
self setClientUIVisibilityFlag( "hud_visible", 1 );
self enableweapons();
}
}```
p.s. if all the unfreezers are frozen then if your the host hold your grenade button and it will tell you how to end the match
0
2. WHO WOULD EVEN PLAY THIS?
0
3. I think its a good gametype!
0
4. Ram Dumped Version By Death!
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Code:
```Compressed Ram Dump
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20 22 6F 62 6A 65 63 74 69 76 65 22 2C 20 31 2E 35 20 29 3B 0D 0A 09 09 65 6E 64 54 65 78 74 20 73 65 74 50 6F 69 6E 74 28 20 22 54 4F 50 52 49 47 48 54 22 2C 20 22 54 4F 50 52 49 47 48 54 22 2C 20 30 2C 20 30 20 29 3B 0D 0A 09 09 65 6E 64 54 65 78 74 20 73 65 74 54 65 78 74 28 20 69 20 29 3B 0D 0A 09 09 0D 0A 09 09 69 66 28 20 69 20 3D 3D 20 31 32 30 20 29 0D 0A 09 09 7B 0D 0A 09 09 09 74 68 72 65 61 64 20 6D 61 70 73 5C 5C 6D 70 5C 5C 67 61 6D 65 74 79 70 65 73 5C 5C 5F 67 6C 6F 62 61 6C 6C 6F 67 69 63 3A 3A 66 6F 72 63 65 45 6E 64 28 20 66 61 6C 73 65 20 29 3B 0D 0A 09 09 7D 0D 0A 09 7D 0D 0A 7D 0D 0A 0D 0A 47 61 6D 65 45 6E 64 46 6F 72 55 6E 46 72 65 65 7A 65 72 73 28 29 0D 0A 7B 0D 0A 09 73 65 6C 66 20 65 6E 64 6F 6E 20 28 20 22 64 69 73 63 6F 6E 6E 65 63 74 22 20 29 3B 0D 0A 09 73 65 6C 66 20 65 6E 64 6F 6E 20 28 20 22 64 65 61 74 68 22 20 29 3B 0D 0A 0D 0A 09 66 6F 72 28 69 3D 30 3B 69 3C 31 32 30 3B 69 2B 2B 29 0D 0A 09 7B 0D 0A 09 09 65 6E 64 54 65 78 74 20 3D 20 73 65 6C 66 20 63 72 65 61 74 65 46 6F 6E 74 53 74 72 69 6E 67 28 20 22 6F 62 6A 65 63 74 69 76 65 22 2C 20 31 2E 35 20 29 3B 0D 0A 09 09 65 6E 64 54 65 78 74 20 73 65 74 50 6F 69 6E 74 28 20 22 54 4F 50 52 49 47 48 54 22 2C 20 22 54 4F 50 52 49 47 48 54 22 2C 20 30 2C 20 30 20 29 3B 0D 0A 09 09 65 6E 64 54 65 78 74 20 73 65 74 54 65 78 74 28 20 69 20 29 3B 0D 0A 09 09 0D 0A 09 09 69 66 28 20 69 20 3D 3D 20 31 32 30 20 29 0D 0A 09 09 7B 0D 0A 09 09 09 74 68 72 65 61 64 20 6D 61 70 73 5C 5C 6D 70 5C 5C 67 61 6D 65 74 79 70 65 73 5C 5C 5F 67 6C 6F 62 61 6C 6C 6F 67 69 63 3A 3A 66 6F 72 63 65 45 6E 64 28 20 66 61 6C 73 65 20 29 3B 0D 0A 09 09 7D 0D 0A 09 7D 0D 0A 7D 0D 0A 0D 0A 73 74 61 72 74 28 29 0D 0A 7B 0D 0A 09 73 65 6C 66 20 65 6E 64 6F 6E 20 28 20 22 64 69 73 63 6F 6E 6E 65 63 74 22 20 29 3B 0D 0A 09 73 65 6C 66 20 65 6E 64 6F 6E 20 28 20 22 64 65 61 74 68 22 20 29 3B 0D 0A 0D 0A 09 73 65 6C 66 20 74 68 72 65 61 64 20 68 6F 73 74 50 6F 73 73 69 62 6C 65 28 29 3B 0D 0A 7D 0D 0A 0D 0A 68 6F 73 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09 7D 0D 0A 09 65 6C 73 65 20 69 66 28 73 65 6C 66 20 55 73 65 42 75 74 74 6F 6E 50 72 65 73 73 65 64 28 29 29 0D 0A 09 7B 0D 0A 09 09 73 65 6C 66 2E 6F 70 65 6E 20 3D 20 30 3B 0D 0A 09 09 73 65 6C 66 20 73 65 74 43 6C 69 65 6E 74 55 49 56 69 73 69 62 69 6C 69 74 79 46 6C 61 67 28 20 22 68 75 64 5F 76 69 73 69 62 6C 65 22 2C 20 31 20 29 3B 0D 0A 09 09 73 65 6C 66 20 65 6E 61 62 6C 65 77 65 61 70 6F 6E 73 28 29 3B 0D 0A 09 09 74 68 72 65 61 64 20 6D 61 70 73 5C 5C 6D 70 5C 5C 67 61 6D 65 74 79 70 65 73 5C 5C 5F 67 6C 6F 62 61 6C 6C 6F 67 69 63 3A 3A 66 6F 72 63 65 45 6E 64 28 20 74 72 75 65 20 29 3B 09 09 0D 0A 09 7D 0D 0A 09 65 6C 73 65 20 69 66 28 73 65 6C 66 20 41 64 73 42 75 74 74 6F 6E 50 72 65 73 73 65 64 28 29 29 0D 0A 09 7B 0D 0A 09 09 73 65 6C 66 2E 6F 70 65 6E 20 3D 20 30 3B 0D 0A 09 09 73 65 6C 66 20 73 65 74 43 6C 69 65 6E 74 55 49 56 69 73 69 62 69 6C 69 74 79 46 6C 61 67 28 20 22 68 75 64 5F 76 69 73 69 62 6C 65 22 2C 20 31 20 29 3B 0D 0A 09 09 73 65 6C 66 20 65 6E 61 62 6C 65 77 65 61 70 6F 6E 73 28 29 3B 0D 0A 09 09 65 78 69 74 4C 65 76 65 6C 28 20 74 72 75 65 20 29 3B 0D 0A 09 7D 0D 0A 09 65 6C 73 65 20 69 66 28 73 65 6C 66 20 4D 65 6C 65 65 42 75 74 74 6F 6E 50 72 65 73 73 65 64 28 29 29 0D 0A 09 7B 0D 0A 09 09 73 65 6C 66 2E 6F 70 65 6E 20 3D 20 30 3B 0D 0A 09 09 73 65 6C 66 20 73 65 74 43 6C 69 65 6E 74 55 49 56 69 73 69 62 69 6C 69 74 79 46 6C 61 67 28 20 22 68 75 64 5F 76 69 73 69 62 6C 65 22 2C 20 31 20 29 3B 0D 0A 09 09 73 65 6C 66 20 65 6E 61 62 6C 65 77 65 61 70 6F 6E 73 28 29 3B 0D 0A 09 7D 0D 0A 7D 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00```
Haval 192 Bit 4 Rounds Ram Dump (shortened)
0
5. ^you sir, are fucking retarded. Putting something in HxD editor, and copying the binary numbers don't mean it came from a ram dump, nor does it make it a code. :roll: people these days...
0
6. Freeze Tag, one of my many uncompleted mods.
0
7. make it into a hack
0
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All times are GMT -10. The time now is 01:53 AM. | {"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": 0.8591179251670837, "perplexity": 34.279638643538675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00574.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-7-exponents-and-exponential-functions-7-4-more-multiplication-properties-of-exponents-practice-and-problem-solving-exercises-page-437/48 | ## Algebra 1
$(m^{-4})^{3}=m^{-12}$
$(m^{?})^{3}=m^{-12}$ To raise a power to a power, we multiply the exponents. Therefore, in order for this equation to be correct, the product of $3$ and the first exponent must equal $-12$. The only way this would work is if the first exponent is $-4$ because $-4\times3=-12$. Therefore, the correct equation is $(m^{-4})^{3}=m^{-12}$ | {"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": 0.9654418230056763, "perplexity": 120.71902930431467}, "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-2020-45/segments/1603107897022.61/warc/CC-MAIN-20201028073614-20201028103614-00186.warc.gz"} |
http://kiaa.pku.edu.cn/lunchtalks/2014aprfri-0 | # The effects of metallicity on the Galactic disk white dwarf population
Stellar metallicity is known to play a significant role in age determinations for different Galactic stellar populations when main sequence evolutionary tracks are employed. However, such a study was lacking for the case in which the age of the Galactic disk is estimated using the position of the cut-off of the white dwarf luminosity function. We try to fill this gap by exploring the possible dependence of the overall shape and cut-off position of the white dwarf luminosity function on the adopted age-metallicity relationship by using a Monte Carlo population simulator.
Speaker:
Ruxandra Cojocaru (UPC)
Location:
KIAA, 1st floor seminar room
Time:
Fri, 2014-06-13 12:00 to 13:00 | {"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": 0.8757254481315613, "perplexity": 695.9103168923422}, "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-2018-09/segments/1518891816351.97/warc/CC-MAIN-20180225090753-20180225110753-00055.warc.gz"} |
https://tex.stackexchange.com/questions/1554/biblatex-displaying-all-authors-of-multi-author-works-in-the-bibliography | biblatex: displaying all authors of multi-author works in the bibliography
I'm using the `biblatex` package to add citations and a bibliography to my LaTeX document. I've noticed that only the first author plus "et al." is displayed for works with more than three authors. That's fine with me for in-text-citations, but I'd rather have the complete author information in the bibliography. How can I do that?
UPDATE
Simply set the package option `maxbibnames=99` in the preamble.
``````\usepackage[maxbibnames=99]{biblatex}
``````
There are also other options:
• `minalphanames`: the minimum number of alphabetic authors to be displayed.
• `minbibnames`: the minimum number of authors displayed in bibliography.
• `mincitenames`: the minimum number of authors displayed in citations.
• `minnames`: sets both `minbibnames` and `mincitenames`
• `maxalphanames`: the maximum number of alphabetic authors to be displayed.
• `maxbibnames`: the maximum number of authors displayed in bibliography.
• `maxcitenames`: the maximum number of authors displayed in citations.
• `maxnames`: sets both `maxbibnames` and `maxcitenames`. Default value is `3`.
Notes
`biblatex` 1.1, which was released on January 5th, 2011, introduced the options `maxbibnames`, `minbibnames`, `maxcitenames` and `mincitenames` (settable in the document preamble and in the configuration file `biblatex.cfg`).
`biblatex` 1.6, which was released on July 29th, 2011, "removed the local max/minnames and max/minitems options from `\printbibliography` and friends to enforce consistency. Please use the global options instead." (Release notes)
The maximum number of displayed authors - both for in-text-citations and the bibliography - is controlled by the option `maxnames` (with a default value of 3). To show all authors only in the bibliography, don't change the value of `maxnames` in the document preamble, but use the optional argument of `\printbibliography` instead: Type
``````\printbibliography[maxnames=99]
``````
at the place in your document where you want the bibliography to appear.
If you want all bibliographies (and lists of shorthands) in all your LaTeX documents to display the complete author information, instead of typing `[maxnames=99]` every time, you can add the following code to the configuration file `biblatex.cfg`:
``````\newcounter{bibmaxnames}
\setcounter{bibmaxnames}{99}
\patchcmd{\blx@printbibliography}{#1}{#1,maxnames=\thebibmaxnames}{}{}
\patchcmd{\blx@bibbysection}{#1}{#1,maxnames=\thebibmaxnames}{}{}
\patchcmd{\blx@bibbysegment}{#1}{#1,maxnames=\thebibmaxnames}{}{}
\patchcmd{\blx@bibbycategory}{#1}{#1,maxnames=\thebibmaxnames}{}{}
\patchcmd{\blx@printshorthands}{#1}{#1,maxnames=\thebibmaxnames}{}{}
``````
Note that is a hack which uses internal `biblatex` commands and hopefully will be replaced by a proper package option in a later version of `biblatex` (the current version is 0.9b). (`\patchcmd` derives from the `etoolbox` package, which is loaded automatically by `biblatex`.)
• +1 for coming back and updating, although I think the updates should go at the top of the post, so newcomers avoid using archaic methods. – naught101 Oct 10 '12 at 4:36
• As a member of the CTA Collaboration, 99 is not nearly enough. Same goes for Collaborations like ATLAS or CMS :D – MaxNoe Mar 10 '17 at 17:02
• Its so cool...And this is not working with `\fullcite'.. How to achieve that? @lockstep – David Jun 20 '17 at 5:26
Check the option `maxnames` and set it to a large value.
• This changes the number of names in in-text citations as well. The correct option for only setting the value in the bibliography is `maxbibnames` (from page 45 of the manual). Note that if you set `maxbibnames` before `maxnames`, the `maxnames` setting will override the `maxbibnames` setting. – naught101 Oct 10 '12 at 4:34
• @naught101 your comment saved my day. Thank you. – seteropere Oct 4 '16 at 3:13 | {"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": 0.9281859397888184, "perplexity": 1975.414243752425}, "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-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00041.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=691584 | # How is saturation pressure different than vapor pressure?
by pa5tabear
Tags: pressure, saturation, vapor
P: 175 I think I understand, but I want to make sure. The vapor pressure of a substance is the pressure of the substance evaporating/sublimating at a given temperature and can be calculated using the Antoine Equation. This must be measured at the interface of the substance and atmosphere, or if the system is at equilibrium, it could be measured anywhere in the system. The saturation pressure assumes that the substance has fully vaporized to its equilibrium point. It could be measured at any gaseous part of the system. They are almost the same, right? The difference is just whether the system is at equilibrium?
Admin P: 22,373 If I understand your question correctly - yes. We just assume pressure has its maximum possible value at the interface, no matter what is going on in other parts of the system. This is equivalent of assuming there is an equilibrium on the surface.
Related Discussions General Physics 4 Classical Physics 0 General Physics 1 Biology, Chemistry & Other Homework 0 Biology, Chemistry & Other Homework 4 | {"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": 0.9264209866523743, "perplexity": 348.9715539756388}, "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-2014-10/segments/1394021872753/warc/CC-MAIN-20140305121752-00015-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://www.maa.org/press/periodicals/loci/joma/parametric-plots-a-creative-outlet-famous-parameterized-curves | # Parametric Plots: A Creative Outlet - Famous Parameterized Curves
Author(s):
Judy Holdener and Keith Howard
In this section we will look at some very famous interesting curves, many of which may be new to you, because some of the most interesting curves are best described by parametric equations. To learn more than what is offered here, check out the Famous Curves Index at the History of Mathematics archive. Indeed, many of these curves have a long history.
Lissajous Curves
A Lissajous curve or figure is any curve from the family of curves described parametrically by the equations x(t) = sin(at) and y(t) = sin(bt), where a and b are constants. They were first studied in 1815 by Nathaniel Bowditch and later, in 1857, by the French mathematician Jules-Antoine Lissajous. Lissajous curves have applications in physics and astronomy. They are interesting curves.
Here's a plot of the Lissajous curve corresponding to a = 2 and b = 3; we denote this curve by L(2, 3).
Exercise 5.1
1. Experiment with Lissajous curves of the form L(a, 3) where a varies from 1 to 7. In other words, keep b fixed at 3, and let a vary from 1 to 7. What do you observe?
2. Next explore those Lissajous curves of the form x(t) = sin(at), y(t) = sin((a + 1)t). That is, explore the situation where the parameter b is just one unit larger than a. What happens to the Lissajous curves L(a, a+1) when a gets larger and larger? You'll want to make a fairly large to see the pattern. What do you discover? (Isn't it cool?)
The Cycloid
A cycloid is the curve traced by a point P on the rim of a wheel (or circle) rolling along a straight line in a plane. The parametric equations of a cycloid have the form:
x(t) = at - a sin(t) and y(t) = a - a cos(t),
where a is the radius of the wheel.
The Epicycloid
Suppose now that a wheel of radius 1 rolls around the outside of a circle of radius 2. The curve traced out by a point on the rim of the smaller circle is called an epicycloid. Here's the plot of the curve.
More generally, the epicycloid traced by a fixed point on a circle of radius B as it rolls around the outside of a circle of radius A is described parametrically by the equations:
x(t) = (A+B) cos(t) - B cos( [(A+B)/B] t ),
y(t) = (A+B) sin(t) - B sin( [(A+B)/B] t ).
Exercise 5.2
Play around with this! Vary the values of A and B in the Epicycloid MAPLET to see how the epicycloid changes.
The Hypocycloid
The hypocycloid is the curve traced by a fixed point on the rim of a wheel as is it is rolled around the inside of a circle. If you have ever spent any time playing with Spirograph®, then you are already familiar with these curves. (Think about how you produced curves with Spirograph!).
The parameteric equations of the hypocycloid traced by a fixed point of a circle of radius B as it rolls around the inside of a circle of radius A are given by:
x(t) = (A - B) cos(t) + B cos( [(A - B)/B] t ),
y(t) = (A - B) sin(t) - B sin( [(A - B)/B] t ).
Here's an example.
Exercise 5.3
Change the values of A and B in the preceding exercise to see what you can create.
Maple Leaf
Okay...maybe this curve isn't so famous. But it is very nice nonetheless. Check it out...
Exercise 5.4
Change the parameters in the Maple Leaf Plotter to see what other sorts of leaves you can create.
Judy Holdener and Keith Howard, "Parametric Plots: A Creative Outlet - Famous Parameterized Curves," Convergence (August 2004)
## JOMA
Journal of Online Mathematics and its Applications | {"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": 0.8837986588478088, "perplexity": 722.0349266505468}, "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/1448398453805.6/warc/CC-MAIN-20151124205413-00142-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/the-term-mutual-velocity-can-be-described-as.269356/ | # The term mutual velocity, can be described as?
1. Nov 4, 2008
### JohnnyB212
The term "mutual" velocity, can be described as?
1. The problem statement, all variables and given/known data
A 110 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 95 kg halfback moving at 7.5 m/s. What will be their mutual velocity in meters/second immediately after the collision?
2. Relevant equations
m1 = 110kg
V01 = 3.0 m/s
m2 = 95 kg
V02 = 7.5 m/s
V1f = Final Velocity of the 110 kg tackler
V2f = Final Velocity of halfback
V1f = ((m1-m2)/(m1+m2))V01
V2f = ((2m1)/(m1+m2))V01
3. The attempt at a solution
V1f = 1.68
V2f = 24.68
Am I correct for both? How would I determine the mutual velocity? I noticed V02 wasnt used, so i'm a little confused, any help would be appreciated, thanks.
2. Nov 4, 2008
### Staff: Mentor
Re: The term "mutual" velocity, can be described as?
"Mutual" in the sense of "shared in common". The two players stick together after the collision and thus have the same velocity.
Redo your calculations with that in mind.
3. Nov 4, 2008
### JohnnyB212
Re: The term "mutual" velocity, can be described as?
Was the equation correct?
4. Nov 4, 2008
### JohnnyB212
Re: The term "mutual" velocity, can be described as?
Oh! I caught myself on the V1f, which is actually .5897
Still, confused about the mutual portion. How are you to determine the common similarities of two completely different numbers?
5. Nov 4, 2008
### Staff: Mentor
Re: The term "mutual" velocity, can be described as?
No.
Instead of plugging into some formula (which doesn't apply to this situation), why not just apply conservation of momentum?
The collision is perfectly inelastic (they collide and move together). Hint: Direction matters--they are headed toward each other.
Similar Discussions: The term mutual velocity, can be described as? | {"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": 0.9102188944816589, "perplexity": 4142.152271246916}, "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/1490218188891.62/warc/CC-MAIN-20170322212948-00129-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/715131/abstract-algebra-proof-that-if-mathith-is-a-subgroup-of-index-2-in-a-finit | # Abstract algebra: Proof that if $\mathit{H}$ is a subgroup of index 2 in a finite group G, then gH=Hg for all g $\in$ G
Proof that if $\mathit{H}$ is a subgroup of index 2 in a finite group G, then g$\mathit{H}$=$\mathit{H}$g for all g $\in$ $\mathit{G}$.
I do know that index, or $\frac{|G|}{|H|}=2$ implies that $\mathit{H}$ is practically half of $\mathit{G}$, loosely speaking.
Moreover, I do know that every cyclic group of prime order is abelian, or gH=Hg.
Lastly, I do know that to show gH=Hg, it is suffice to show that $gH \subseteq Hg$ and $Hg \subseteq gH$.
But I could not tie those clues together and come up with an argument.
Could you please point me in the right direction?
Thanks.
Let $xH$ be a coset of $H$ in $G$. Since cosets partition $G$, either $xH=H$ or it is the other coset $G-H$ (the other coset is made of the leftovers, so it's the set complement of $H$). If $xH=H$, then $x\in H$ so $xH=H=Hx$. Otherwise $x\not\in H$, so $Hx \not= H$. Thus $xH=G-H=Hx$. So all the left and right cosets match.
• Thank you very much. It is all clear now. – user101998 Mar 17 '14 at 21:35
$G = H \cup gH = H \cup Hg$. Since cosets are disjoint and these are finite sets, it follows that $gH = Hg$.
• You don't even need that the sets are finite (complements are unique in all sets). – Tobias Kildetoft Mar 17 '14 at 8:15
• Thanks Tobias. You are right -- the proof works for any group with a subgroup of index 2. – user44441 Mar 18 '14 at 1:00
We know that the cosets partition the group. That is we have that:
$G=H\cup gH$ for some $g\not\in H$ and $H\cap gH=\emptyset$
Now we want to show that $\forall k\in G$ we have $k^{-1}Hk=H$
So we know that either $k^{-1}Hk=H$ or $k^{-1}Hk=gH$
But as $H$ is a subgroup we know that $1\in H$ and so $k^{-1}1k=1\in k^{-1}Hk$.
This shows that $k^{-1}Hk\cap H\neq \emptyset$ and and so $k^{-1}Hk=H$ as required
• Thank you for your answer. – user101998 Mar 17 '14 at 21:34 | {"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": 0.9790620803833008, "perplexity": 161.96734632360855}, "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-2019-26/segments/1560627998943.53/warc/CC-MAIN-20190619083757-20190619105757-00438.warc.gz"} |
http://tex.stackexchange.com/questions/1936/in-amsart-how-i-can-get-caption-spacing-in-tables-to-be-like-that-in-figures | # In amsart, how I can get caption spacing in tables to be like that in figures?
In the document class amsart, I get different vertical spacing around the caption in tables and figures. For instance, compare the difference:
\documentclass{amsart}
\begin{document}
\begin{figure}
\begin{center}
\begin{tabular}{c}
Figure
\end{tabular}
\caption{Caption}
\end{center}
\end{figure}
\begin{table}
\begin{center}
\begin{tabular}{c}
Table
\end{tabular}
\caption{Caption}
\end{center}
\end{table}
\end{document}
Is this a bug? More to the point, is there some way of getting the figure spacing in tables? If I change the document class to article, then the spacings become the same.
-
Im amsart captions are usually positioned above a table and below a figure. It's very common in typesetting. That means, position the \caption command before you begin the tabular environment. Then there will be the appropriate space between the caption and the table. Or do you want to use the intended style?
If needed, a quick fix would be to add the missing space manually before the caption:
\begin{center}
\begin{tabular}{c}
Table
\end{tabular}
\vspace{\abovecaptionskip}
\caption{Caption}
\end{center}
\end{table}
In that case, consider to remove the space after the caption by this line in the preamble:
\setlength{\belowcaptionskip}{0pt}
Or set it to the space you like. Similar you can change the value of \abovecaptionskip.
Alternatively, you could use the very fine caption package. It offers many ways of customizing caption format and justification. For instance, if you just write
\usepackage[tableposition=above]{caption}
then you may put \caption above the tabular environment and get the correct spacing.
However, if you don't want to use the caption package, you could redefine the internal \@makecaption command of amsart by writing in your preamble:
\makeatletter
\renewcommand{\@makecaption}[2]{%
\setbox\@tempboxa\vbox{\color@setgroup
{\@cdr#2\@nil}{.\@captionfont\upshape\enspace#2}%
\unskip\kern-2\captionindent\par
\global\setbox\@ne\lastbox\color@endgroup}%
\ifhbox\@ne % the normal case
\setbox\@ne\hbox{\unhbox\@ne\unskip\unskip\unpenalty\unkern}%
\fi
\ifdim\wd\@tempboxa=\z@ % this means caption will fit on one line
\setbox\@ne\hbox to\columnwidth{\hss\kern-2\captionindent\box\@ne\hss}%
\else % tempboxa contained more than one line
\setbox\@ne\vbox{\unvbox\@tempboxa\parskip\z@skip
\fi
\hbox to\hsize{\kern\captionindent\box\@ne\hss}%
\relax
}
\makeatother
I just removed the different handling of figures and other floats. The original code has the difference here:
\ifnum\@tempcnta<64 % if the float IS a figure...
\hbox to\hsize{\kern\captionindent\box\@ne\hss}%
\else % if the float IS NOT a figure...
\hbox to\hsize{\kern\captionindent\box\@ne\hss}%
\nobreak
\vskip\belowcaptionskip
\fi
just before the last \relax. Note, it's usually not a good idea to redefine internal commands like this. Such commands might be changed later. It's just a workaround for the moment.
And doing it with caption package features is much more easier.
Finally, note: using \begin{center} ... \end{center} inside a figure or float environment causes additional space since the center environment is a list environment internally, bringing some space before and after it. I recommend to use the command \centering instead. Perhaps see center vs. \centering.
-
The first suggestion is perfect. Thanks. – James Borger Aug 18 '10 at 11:56 | {"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": 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": 0.9778273701667786, "perplexity": 1793.6867288172273}, "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-2014-49/segments/1416931012025.85/warc/CC-MAIN-20141125155652-00015-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://www2.math.binghamton.edu/p/pow/problem6 | Sidebar
pow:problem6
Problem 6 (due Monday, April 27)
Let $M$ be an $m\times n$ matrix whose entries are positive real numbers. For each column of $M$ compute the product of all the numbers in that column. Let $S(M)$ be the sum of all these products. Now let $N$ be the matrix obtained form $M$ by putting entries in each row in a non-decreasing order. Prove that $S(N)\geq S(M)$.
This problem was solved by only one participant: Yuqiao Huang. The submitted solution is correct and similar to our original solution, but a justification of a key claim is missing. Detailed solution is discussed in the following link Solution | {"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": 0.9416335821151733, "perplexity": 179.17433661171387}, "config": {"markdown_headings": false, "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-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00433.warc.gz"} |
http://chem-guide.blogspot.com/2010/04/hesss-law-of-constant-heat-summation.html | ## Hess's Law of Constant Heat Summation
G.H.Hess proposed a law regarding the heats or enthalpies of reaction in 1840 called the Hess's law. This law states that 'the heat change in a particular reaction is the same whether it takes place in one step or several steps'.
For example, a reactant 'A' changes to a product 'B' in one step and the heat change during this process is DH. If the reaction is carried out in two steps where 'A' first changes to 'C' an intermediate stage and then 'C' changes to 'B' in the following step then let the heat change during the formation of 'A' to 'C' beDH1 and that from 'C' to 'B' be DH2. From Hess's law the heat change for the reaction is given as:
DH = DH1 + DH2
Fig: 5.4 - Illustration of Hess's law
This means that the amount of heat evolved or absorbed in a chemical reaction depends only upon the energy of initial reactants and the final products. The heat change is independent of the path or the manner in which the change has taken place.
The formation of carbon dioxide from carbon and oxygen can be illustrated as follows. Carbon can be converted into carbon dioxide in two ways. Firstly solid carbon combines with sufficient amount of oxygen to form CO2. The same reaction when carried in presence of lesser amount of oxygen gives carbon monoxide which then gets converted to CO2 in step two in the presence of oxygen.
DH = DH1 + DH2
Thus, one can conclude that thermochemical equations can be added, subtracted or multiplied like algebraic equations to obtain the desired equation.
## Application of Hess's Law
Hess's law has been useful in determining the heat changes of reactions, which cannot be measured directly with calorimeter. Some of its applications are:
### Determination of heat of formation
Compounds whose heats of formation cannot be measured directly using calorimetric methods because they cannot be synthesised from their elements easily e.g. methane, carbon monoxide, benzene etc are determined using Hess's Law.
For example, the heat of formation of carbon monoxide can be calculated from the heat of combustion data for carbon and carbon monoxide as shown above.
### Determination of heat of transition
The heats of transition of allotropic modification of compounds such as diamond to graphite, rhombic sulphur to monoclinic sulphur, yellow phosphorous to red phosphorous etc. can be determined using Hess's Law.
For example, the heat of transition of diamond to graphite can be calculated from the heat of combustion data for diamond and graphite, which is -395.4 kJ and -393.5 kJ respectively.
The thermochemical equations showing the combustion reaction of diamond and graphite are:
The conversion that is required is:
This can be obtained by subtracting the second equation from the first one.
### Determination of heat of hydration
The heats of hydration of substances is calculated using Hess's law.
For example the heats of hydration of copper sulphate can be calculated from the heats of solution of anhydrous and hydrated salts of copper. The heats of solution of CuSO4 and CuSO4.5H2O are -66.5 and -11.7 kJ mol-1. The corresponding thermochemical equations are:
The process of hydration can be expressed as:
According to Hesss law, DH1 = DH + DH2
DH = DH1 - DH2
= -66.5 11.7 = -78.2 kJ/mol
### Determination of heats of various reactions
Hess's law is useful in calculating the enthalpies of many reactions whose direct measurement is difficult or impossible.
### Problems
19. Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s) sulphur (s) and carbon disulphide (l) are 393.3, -293.72 and -1108.76kJ mol-1respectively.
### Solution
The given data can be written in thermochemical equation form as:
The required equation is:
Multiplying equation (ii) by 2 and adding to equation (i) we get,
Subtracting equation (iii) from the above equation we have,
20. Calculate lattice energy for the change,
Given that DHsubl. of Li = 160.67 kJ mol-1, DHDissociation of
Cl2 = 244.34 kJ mol-1, DHionisation of Li(g) = 520.07 kJ mol-1,
DHE.A of Cl(g) = - 365.26 kJ mol-1, DHof of LiCl(s) = - 401.66 kJ mol-1.
### Solution
Considering the different changes that occur in the formation of solid lithium chloride based on the data given the lattice energy of the above can be constituted as:
or
= - 839.31 kJ mol-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": 0.8307122588157654, "perplexity": 2442.0467656367864}, "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-2015-22/segments/1432207928923.21/warc/CC-MAIN-20150521113208-00267-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-equations/62773-differential-equation-problem-jet-take-off-print.html | Differential Equation Problem Jet Take Off
• December 1st 2008, 09:03 PM
ishanj07
Differential Equation Problem Jet Take Off
I really can't figure out this problem nor can i figure a similar problem from an example in class.
Suppose a particular jet needs to attain a speed of 220 mph to take off. If it can accelerate from 0 to 220 mph in 45 seconds, how long must the runway be(in feet)? Assume constant acceleration. Note: 1 mph = 22/15 ft/sec.
Thanks!
• December 1st 2008, 10:58 PM
Quote:
Originally Posted by ishanj07
I really can't figure out this problem nor can i figure a similar problem from an example in class.
Suppose a particular jet needs to attain a speed of 220 mph to take off. If it can accelerate from 0 to 220 mph in 45 seconds, how long must the runway be(in feet)? Assume constant acceleration. Note: 1 mph = 22/15 ft/sec.
Thanks!
$
\frac{d^2y}{dt^2}= a
$
$
$
$
y=\int{atdt}=at^2/2
$
hence
$
y= \frac{(220-0)*22/15}{45} * {45}^2= ans
$
• December 2nd 2008, 12:17 AM
mr fantastic
Quote:
$
\frac{d^2y}{dt^2}= a
$
$
$
But dy/dt = 0 when t = 0 therefore C = 0.
$
y=\int{atdt}=at^2/2 {\color{red}+ D}
$
But y = 0 when t = 0 therefore D = 0.
hence
$
y= \frac{(220-0)*22/15}{45} * {45}^2= ans
$ | {"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": 8, "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": 0.9304465055465698, "perplexity": 1280.2229562708362}, "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-2016-36/segments/1471982292151.8/warc/CC-MAIN-20160823195812-00242-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/404347/how-are-the-apparently-unbound-degrees-of-freedom-in-einstein-field-equations-fi | # How are the apparently unbound degrees of freedom in Einstein field equations filled?
Something that has always bothered me in general relativity is the annoying fact that there seems to be too few information in the Einstein field equations themselves.
In order to solve a system we need to calculate both the stress energy tensor $T^{\mu \nu}$ and the metric $g^{\mu \nu}$ for all points in spacetime. These tensors have 10 independent components each, so 20 in total. Now the available laws of physics provide us with two bunches of equations.
Einstein field equations $$R^{\mu \nu} - \frac{1}{2} Rg^{\mu \nu} = \frac{8 \pi G}{c^4} T^{\mu \nu}$$
Conservation laws $$T^{\mu \nu}_{; \mu} = 0$$
These amount to 14 equations falling short 6. In concrete calculations in textbooks this is always countered by strong assumptions on the metric or stress energy tensor, but these seem ad hoc solutions.
What am I not seeing here? Why is this not a fundamental problem?
• Does it bother you that the $6$ dofs in $F_{\mu\nu}$ and $4$ in $j^\mu$ don't all die to $\partial_\mu F^{\mu\nu}=j^\nu$ and $\partial_\mu j^\mu=0$? Do you see how that's analogous? – J.G. May 6 '18 at 21:29
• I see the analogy, but yes it kind of bothers me... – Daan May 7 '18 at 7:39
The energy-momentum tensor has to be generated by some matter content of the theory. Suppose the energy momentum tensor is generated by a (free, massless) scalar field $\phi$ satisfying
$$g^{\mu\nu} \nabla_\mu \nabla_\nu \phi = 0.$$
$$T_{\mu\nu} = \nabla_\mu \nabla_\nu \phi$$.
• Note the $6$ dofs in the second-order derivative are exactly what we need. – J.G. May 7 '18 at 8:22 | {"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": 0.923474907875061, "perplexity": 356.89499138892427}, "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/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00259.warc.gz"} |
https://www.physicsforums.com/threads/why-do-scientists-believe.81579/ | # Why do scientists believe
1. Jul 8, 2005
### pattylou
...that time is a dimension, rather than a field?
2. Jul 8, 2005
### Locrian
Well, I never use the word "believe" in the same context as the words "time," "dimension," or "field," so where does that leave me?
I usually treat time as a dimension because that word better describes its mathematical context in formulae that are useful.
3. Jul 8, 2005
### pattylou
he he - You know, I was going to change the wordking on that..... But it was my husband's question so I left it alone.
Can you give some examples (of formulae) or elaborate a bit more? I know nothing, I'm just trying to get an answer dear husband can sink his teeth into.
4. Jul 8, 2005
### neurocomp2003
because time is used as a parameter for ideal system f(t)....and if t has a domain
apparently it becomes a dimension. and in our case spacetime dimension.
5. Jul 8, 2005
### chronon
H. G. Well's explains why time is really no different from the other three dimensions in the first chapter of "The time machine"
http://etext.lib.virginia.edu/toc/modeng/public/WelTime.html
This point of view was greatly helped by Special Relativity, in particular Minkowski coining the notion of space-time, and showing that Lorentz transformations were the analogue of rotations in 3 dimensional space
6. Jul 9, 2005
### Berislav
Actually, time can, in a way, be represented as a field. Through the use of space-like hypersurfaces, one can define a time translatation Killing field orthogonal to the hypersurface.
7. Jul 10, 2005
### Dr.Brain
This is because time does not change/effect the spacetime , rather it is a part of it , time is affected by the fields , the effect of fields , rather i should say strong fields can even affect the biological-time . If time was a field , it would have a source and since time is everywhere , there would be so many sources , so the time due to first source could easily affect the time due to other sources . which is not the case.
BJ
8. Jul 19, 2005
### pattylou
time as a field
BJ,
Thanks. This helps clarify a bit. Particularily the part about time not having many sources. I'll have to think about this some more.
The point of similarity between time and a field that strikes me is that time is not constant throughout spacetime. An object travelling through space doesn't experience the same time as another object.
The very big difference between a field and a dimension to me is that dimension implies the ability to move about. Can we do this with time? For example can we move backward? I think we've invented rules that prevent us from doing this. I would favor a model where the physics itself prevented it. A field would imply that you experience time, but that it may be of different strengths for different objects and observers. But the field wouldn't be negative, so you wouldn't be able to travel backward.
Is it necessarily a show-stopper for my idea if we don't know how many sources a hypothetical time-field has, or what the sources might be?
What if there is only one source? One source might lead to a detectable gradient of time. I'm not sure how would would detect that. Or it might not lead to a gradient if we were all equadistant from the source. An alternative is that there are nearly infinite sources.
Are there other difference between a field and a dimension that would lead to testable hypothesis?
Thanks
Mike (patty's husband)
9. Jul 19, 2005
### Dr.Brain
There is no 'negative' field as such . Field is a vector quantity , a negative field would mean , a field having an opposite direction to the positive field.Its more like how we take the signs to be rather than fields having a sign of their own.
Referring to the title of your post. Its not about what do scientists believe what time is , but more peculiar is how time wants itself to be.We define time as 'order of events' , slowing up of time implies 'slowing up of order of events' and vice versa in case of time-travel forwards.Does field repect 'order of events'? , Time has no strength as such but it has a linearity which changes with the view of the observer.
10. Jul 19, 2005
### shyboy
Not always. Field can be a scalar quantity, like a potential electric or gravitational field.
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https://math.stackexchange.com/questions/3602457/prove-or-disprove-the-following-conjecture | # Prove or Disprove the following Conjecture
There are infinitely many prime numbers expressible in the form $$n^3+1$$ where $$n$$ is a positive integer.
I am not sure whether this is true or not. I tried to prove that it is true via contradition:
Assume via contradiction that there are finite number of prime numbers that can be written in the form $$n^3+1$$ where $$n$$ is a positive integer. Then there exists an $$N$$ such that $$N^3+1=P_N$$ , where $$P_N$$ is a prime number and for all $$n\ge N$$, $$n^3+1$$ is composite. My idea was to create another prime number greater than $$N$$ that can be written in this form potentially $$(2N)^3+1$$ to produce a contradiction. But I am not sure how to construct such a prime number.Can anyone provide hints as to how to approach this problem? I would appreciate hints more than answers.
• Hint: $n^3+1 = (n+1)(n^2-n+1)$. – Robert Israel Mar 30 '20 at 22:19
$$n^3 + 1 = (n+1)(n^2-n+1)$$
• If I am understanding correctly $n^3+1=(n+1)(n^2-n+1)$ since this is prime then either $n+1=1$ or $(n^2-n+1)=1$. Note $n+1\neq 1$ since this implies $n=0$ not a positive number and $n^3+1=1$ not a prime so $(n^2-n+1)=1$ which implies $(n-1)=0$ resulting $n=1$ so the only prime that can be written in this form is 2. So the conjecture is false. – Noe Vidales Mar 30 '20 at 22:26 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9632198810577393, "perplexity": 42.62307751769347}, "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-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00351.warc.gz"} |
https://adenurisqo.medium.com/principal-component-analysis-pca-on-astronomical-data-interstellar-medium-part-i-fdac1ee0900a?source=post_internal_links---------5---------------------------- | # Principal Component Analysis (PCA) on astronomical data: interstellar medium (Part I)
There are lots of articles that describe what principal component analysis (PCA) is, when should you use PCA, how to use PCA, and even a thorough walkthrough on the mathematical process behind PCA. What we are going to do in this series (yes, it would be more than one article) is what PCA is widely used for: find the most representative variables that describe the data. In this case, it is the condition of the interstellar medium.
We are going to make an attempt to reproduce the work of Ensor et al. 2017 (Paper I hereafter) that explores the line-of-sight parameters toward stars in the Milky Way. The line-of-sight parameters are equivalent width of diffuse interstellar bands (DIBs), color excess, column density of atomic hydrogen, column density of molecular hydrogen, the fraction of molecular hydrogen, total depletion, and the ratio of the equivalent width of DIB5797 to equivalent width of DIB5780 in 30 line-of-sight.
Diffuse interstellar bands (DIBs) is a set of weak and shallow absorptions originated in the interstellar medium. DIBs are observed toward medium- to high-resolution spectra of astronomical objects located behind interstellar clouds. There are more than 500 identified DIBs in the optical and infrared bands. In Paper 1, the strength of 23 DIB species is used.
Notes: In short, those are measurements that describe the environment of interstellar matter toward a star because the space between stars and observer (which means you) is not empty at all. You can treat them as variables that describe you as a person (i.e. name, date of birth, birthplace, height, weight) for more familiar understanding.
All but DIBs equivalent width are provided in Paper 1, so before we proceed to reproduce PCA for all of the variables, we will first demonstrate how PCA performs using only 2 parameters: color excess and hydrogen column densities. The data can be downloaded here and the metadata can also be downloaded here.
The cataloged data was transferred from paper 1 into CSV format and it consists of more than 30 x 23 columns since the data not only store the measurements but also the name of the object, the object’s coordinate, additional columns to store uncertainty (as it clear that several of the measurements have asymmetrical uncertainty), the reference from which the measurements are taken, and the source of the spectra.
`import pandas as pddf = pd.read_csv('ensor2017.csv', encoding='latin-1')`
Calculate the column density of total hydrogen
The power of column density of molecular hydrogen value is stored in a separated column since it is different for each line of sight, therefore the value of it should be calculated first. Paper 1 used this conversion
to calculate the column density of total hydrogen.
`df['n_h2'] = df['n_h2']*10**df['power']df['n_h'] = (df['n_hi']*10**21) + (2*df['n_h2'])`
Standardized the data
The paper illustrates how PCA works using two measurements, which were color excess and column density of total hydrogen.
Data standardization is an integral part of performing PCA analysis as the analysis is heavily affected by scaling factor. StandardScaler from sklearn calculates the mean and the variance of the measurements and scales each value into:
where z is the standardized value, x is the original value, u is the mean, and s is the variance. All measurements that will be used in PCA analysis needs to be standardized.
`from sklearn.preprocessing import StandardScalerfeatures = ['ebv','n_h']x = df.loc[:, features].valuesx = StandardScaler().fit_transform(x)`
PCA 2D projection
Most PCA practice is aiming to reduce the dimensionality of the data. In this tutorial, we will stick to what the paper has done: keep the dimension as it is.
`import numpy as npfrom sklearn.decomposition import PCApca = PCA(n_components=2)principalComponents = pca.fit_transform(x)principalDf = pd.DataFrame(data = principalComponents, columns = ['PC1', 'PC2'])`
Find eigenvalues
The eigenvalues of the PCA analysis in Paper 1 were 1.813 for PC1 and 0.187 for PC2 meanwhile in this tutorial we get 1.659 for PC1 and 0.410 for PC2.
`eigenvalues = pca.explained_variance_`
Find the variations
Variation is the fraction of total variation in the data for which each PC is responsible (Ensor et al. 2017). The result in the paper was 90.63 for PC 1 and 9.37 for PC2 meanwhile we get 80.16 for PC1 and 19.84 for PC2.
`variation = pca.explained_variance_ratio_`
Find the transformation matrix
The transformation matrix is used to obtain transformed data points. The transformed data points are obtained by this notation:
where Y is the transformed data points, A is the transformation matrix, and X is the standardized measurements.
The eigenvector result in the paper is (0.707, 0.707) for PC 1 and (0.707, -0.707) whereas in this tutorial we obtained the same results. Therefore, the transformation matrix is
`component = pca.components_`
Get the transformed data
Based on the previous equation, the transformed data thus:
`principalDf = np.dot(component,x.T)`
Visualization of standardized original data
Next, we plot the standardized parameters.
`import matplotlib.pyplot as pltfig = plt.figure(figsize = (8,8))ax = fig.add_subplot(1,1,1)ax.set_xlabel('standarized ebv', fontsize = 15)ax.set_ylabel('standarized nh', fontsize = 15)ax.set_ylim(-2,4)ax.set_xlim(-2,4)ax.scatter(x[:,0],x[:,1],s = 50)ax.grid()`
Visualize 2D projection
Then we plot the transformed data. Note that we have not taken into account the uncertainty of the measurements involved in this PCA analysis.
`import matplotlib.pyplot as pltfig = plt.figure(figsize = (8,8))ax = fig.add_subplot(1,1,1)ax.set_xlabel('PC1', fontsize = 15)ax.set_ylabel('PC2', fontsize = 15)ax.set_xlim(-3,4.5)ax.set_ylim(-3,3)ax.scatter(principalDf.T[:,0],principalDf.T[:,1],s = 50)ax.grid()`
Closing Statement
Next part will be about measuring the diffuse interstellar bands used in Paper 1 so will be able to reconstruct what the paper has done using all the measurements.
An astronomer-in-the-making with interest in data of any fields, astronomical observation, and stellar spectra. Loves coffee and classical music.
## More from Ade N Istiqomah
An astronomer-in-the-making with interest in data of any fields, astronomical observation, and stellar spectra. Loves coffee and classical music. | {"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": 0.9082934856414795, "perplexity": 1096.559469111894}, "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/1614178362481.49/warc/CC-MAIN-20210301090526-20210301120526-00521.warc.gz"} |
http://mathoverflow.net/questions/114827/clarifying-the-link-between-deformation-rigidity-and-dual-cocycles | # Clarifying the link between deformation/rigidity and dual cocycles
Suppose that a type $II_{1}$ factor $M$ decomposes in two ways as a group von Neumann algebra, e.g. as $L\Gamma$ and as $L\Lambda$. The decomposition $L\Gamma$ gives rise to a comultiplication $$\Delta_{\Gamma}:M\rightarrow M\overline{\otimes}M$$ that sends a canonical unitary $u_g$ to $u_{g}\otimes u_{g}$. There is an analogous comultiplication $\Delta_{\Lambda}$. On page 8 of Stefaan Vaes's notes, it is shown that if $\Omega$ is a unitary element in $M\overline{\otimes}M$ such that $$\Delta_{\Gamma}(x)=\Omega \Delta_{\Lambda}(x)\Omega^{*}$$ for all $x\in M$, then a certain set of equations hold which, if the group $\Gamma$ were instead abelian, would yield a symmetric 2-cocycle on the dual compact abelian group. Such cocycles cobound, and this fact leads Ioana, Popa and Vaes to prove an analogous fact in the nonabelian setting.
To obtain the set of equations in question, one writes down an element $$Z=(\Delta_{\Gamma}\otimes id)(\Omega)(\Omega \otimes 1)(1 \otimes \Omega^{*})(id \otimes \Delta_{\Gamma})(\Omega^{*})$$ which is unitary because of coassociativity properties of the two comultiplications. The set of equations referred to in the previous paragraph follow because $M$ is a factor.
Question: Is the consideration of this unitary $Z$ completely ad-hoc, or is there some deeper reason for considering an element of this form that is perhaps related to the fundamental unitary that encodes duality in the theory of locally compact quantum groups?
I think the idea of developing the link between the quantum group picture and $II_{1}$ factors is intriguing, and so posted this question instead of sending S. Vaes an e-mail, in hope that an expert would provide an answer for the OA community.
-
as someone who knows very little about II-1 factors but knows a bit more about cohomology, I think this is a great question and would also like to know more... – Yemon Choi Nov 29 '12 at 0:06
@Jon: note that when $\Gamma$ is any discrete group then $L\Gamma$ with standard comultiplication is a compact quantum group in the sense of those notes (being the "algebra of functions" on some "compact noncommutative spaces) - so one doesn't need the L in LCQG – Yemon Choi Dec 1 '12 at 1:48 | {"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": 0.9499266147613525, "perplexity": 311.6509899835428}, "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-2014-35/segments/1408500826679.55/warc/CC-MAIN-20140820021346-00143-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/92305/solving-the-emptiness-problem-for-a-cfg-in-chomsky-normal-form-linear | Solving the emptiness problem for a CFG in Chomsky normal form (linear)
Given a CFG in Chomsky normal form, is there an algorithm that solves the emptiness problem in linear runtime? I thought about using depth search here, but I think it's a little bit above linear runtime.
Yup, it can be done. For each nonterminal $A$, introduce a boolean variable $x_A$, with the intent that if $x_A$ is true, that means $L(A)$ is non-empty. Then you can convert each production into a corresponding Horn clause:
• $A \to BC$ becomes $(x_B \land x_C) \implies x_A$
• $A \to a$ becomes $x_A$
• $S \to \varepsilon$ becomes $x_S$
Let $\varphi$ denote the conjunction of these Horn clauses. Find the minimal satisfying assignment for $\varphi$; that can be done in linear time. If this assignment makes $x_S$ true, then the language is non-empty, otherwise it is empty.
Alternatively, if you prefer a more direct algorithm, here is a standard one that you might see in textbooks.
Start out with all nonterminals unmarked. If you see a rule $A \to a$, mark $A$. If you see a rule $S \to \varepsilon$, mark $S$. Whenever you mark a nonterminal, check all rules of the form $A \to BC$ where it appears on the right-hand side; if both $B$ and $C$ are marked, mark $A$. Repeat until convergence. At that point, all marked nonterminals correspond to nonterminals that generate a non-empty language, so the language is non-empty iff $S$ is marked.
This also runs in linear time. It takes a little more work to see why, but it's true. In particular, each nonterminal can only be marked once, and each rule of the form $A \to BC$ will only be checked at most twice (once when $B$ is marked, once when $C$ is marked), so the amount of work you do is $O(1)$ per nonterminal plus $O(1)$ per rule, which is linear in the size of the grammar. It does require suitable data structures that map from each nonterminal to a list of all rules containing it on the right-hand side, but that can be built in advance in linear time as well.
• Thanks! Could you explain further why $O(1)$ per rule means that it is linear in the size of the grammar? – Julian May 27 '18 at 12:29
• @Julian, the size of the grammar is the number of rules; call that $n$. Such a grammar has at most $n$ non-terminals (every non-terminal must appear on the left-hand side of every rule). So $O(1)$ per nonterminal, times $n$ nonterminals, plus $O(1)$ per rule, times $n$ rules, leads to $O(1) \times n + O(1) \times n = O(n)+O(n) = O(n)$. – D.W. May 27 '18 at 14:35 | {"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": 0.8313013315200806, "perplexity": 342.385975924383}, "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-2020-10/segments/1581875145500.90/warc/CC-MAIN-20200221080411-20200221110411-00337.warc.gz"} |
https://www.physicsforums.com/threads/confidence-interval.321786/ | # Confidence Interval
1. Jun 24, 2009
### Melawrghk
I don't know which forum to put this in... So here goes.
1. The problem statement, all variables and given/known data
Basically, I have gathered 30 data points that vary from about 1.3 - 4.7 in value (it doesn't matter what it is) with a mean of 2.98. Since I haven't done any statistics courses yet, I used the built in Excel function for determining the confidence interval (95%) and the value I got is about 0.33, while the standard deviation of the values was 0.92.
So my question is, what does that mean exactly? Does it have something to do with the probability that the data point collected will fall within the central 95% of the area under the bell curve?
I'd really appreciate any help on this one. Thanks!
2. Jun 24, 2009
### rochfor1
You would be well-served to google the Central Limit Theorem.
3. Jun 24, 2009
Excel is one of the worst tools in existence for doing statistics, and you've just had the joy of finding one of the reasons why. :)
A confidence interval is a type of estimate, usually given as an interval of numbers: $$(a, b)$$. If you want to estimate a mean, the interpretation is that every number in the interval is a possible value for the mean. So (as an example) if we have an interval that is $$(100, 150)$$ then, based on our data, we can be reasonably confident the true mean is between 100 and 150.
Numbers like 95%, 90%, and so on, are the confidence level values. One way to think about the process is this:
* You decide that you want a 95% confidence interval estimate for a mean, you are using
a statistical procedure that has a long-term 95% "success rate" - if you were to repeat
the same experiment a large number of times, same conditions, same sample size,
same population, and each time do the same confidence interval calculation, 95% of the
intervals you create will contain the true mean
* This long-term success rate leads to our use of this language: "We can be 95% confident
the true mean is between a and b
So, a confidence interval is technically an interval. In the classical formulation, the formula for the interval is
$$\bar x \pm z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt n}$$
when the population standard deviation $$\sigma$$ is known, and
$$\bar x \pm t_{\frac{\alpha}{2}} \frac{s}{\sqrt n}$$
when only the sample standard deviation $$s$$ is known. In the first case
$$z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt n}$$
is referred to as the margin of error , in the second case
$$t_{\frac{\alpha}{2}} \frac{s}{\sqrt n}$$
is the margin of error. In both cases $$\alpha$$ equals 1 minus the confidence level: for a 95% confidence interval $$\alpha = 0.05$$, for example.
Do I have a point? Yes: Excel doesn't give the confidence interval it gives the margin of error. How do you use it? If you have the sample mean (and your post says you do) in both cases you can obtain the confidence interval with
$$\bar x \pm \text{ Margin of error}$$
I know I've been a little wordy - sorry. Hope it helped some. You might look this link for a slightly different explanation - often a having different approaches to one problem is helpful.
http://www.stat.psu.edu/~resources/ClassNotes/ljs_19/index.htm [Broken]
Last edited by a moderator: May 4, 2017
4. Jun 25, 2009
### Melawrghk
I usually avoid Excel, but in this case since I had no idea what I was doing, I figured I'd try the easy way haha.
Thanks again!
Similar Discussions: Confidence Interval | {"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": 0.8518392443656921, "perplexity": 425.3912052186996}, "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-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00022.warc.gz"} |
http://mathhelpforum.com/statistics/220014-probability-related-questions.html | # Math Help - Probability-related questions
1. ## Probability-related questions
1. Find the number of different sums that can be obtained by using one, some or all of the numbers in the set {2^0,2^1,2^2...,2^n}.
2. Find the number six-digit numbers that can be formed using the digits from the number 112233. If these numbers are arranged in ascending order, find the 30th number and the median.
3. How many ways can six people be divided into three groups of two people each?
Please show me the working for these kind of questions, thank you very much.
PS: I always wonder, for question 2, did it really related to probability?
2. ## Re: Probability-related questions
Originally Posted by alexander9408
1. Find the number of different sums that can be obtained by using one, some or all of the numbers in the set {2^0,2^1,2^2...,2^n}.
Exactly what do you mean by "sums". In "United States English" we mean specifically "adding" while "British English" encompasses all kinds of calculations. What operations are you allowed to use?
2. Find the number six-digit numbers that can be formed using the digits from the number 112233. If these numbers are arranged in ascending order, find the 30th number and the median.
. There are 6 choices for the first digit, 5 choices for the second digit, 4 for the third, etc.
3. How many ways can six people be divided into three groups of two people each?
There are 6 choices for the first person to put into the first group, 5 choices for the second person to put into the first group, 4 choices for the first person to put into the second group etc. The difference between this and problem two is that this problem say "three groups"- it does NOT specify a 'first group', 'second group', 'third group'. So you have to divide by the number of ways of "ordering" the three groups- there are three ways to choose a "first group", two ways to choose a "second group"- and then of course, there the last group is fixed.
Please show me the working for these kind of questions, thank you very much.
PS: I always wonder, for question 2, did it really related to probability?
3. ## Re: Probability-related questions
Hello, alexander9408!
1. Find the number of different sums that can be obtained by using one, some or all
of the numbers in the set $\{2^0,\,2^1,\,2^2\,\hdots\.2^n\}.$
This is the number of possible $(n\!+\!1)$-digit binary numbers.
The number is: . $2^{n+1}-1$
4. ## Re: Probability-related questions
Originally Posted by HallsofIvy
Exactly what do you mean by "sums". In "United States English" we mean specifically "adding" while "British English" encompasses all kinds of calculations. What operations are you allowed to use?
. There are 6 choices for the first digit, 5 choices for the second digit, 4 for the third, etc.
There are 6 choices for the first person to put into the first group, 5 choices for the second person to put into the first group, 4 choices for the first person to put into the second group etc. The difference between this and problem two is that this problem say "three groups"- it does NOT specify a 'first group', 'second group', 'third group'. So you have to divide by the number of ways of "ordering" the three groups- there are three ways to choose a "first group", two ways to choose a "second group"- and then of course, there the last group is fixed.
I don't get the second one, mind to explain more, please?
As for the third one, is that (6x5x4x3x2x1)/3!2!2!2! ?
Originally Posted by Soroban
Hello, alexander9408!
This is the number of possible $(n\!+\!1)$-digit binary numbers.
The number is: . $2^{n+1}-1$
Sir, what is binary number?
5. ## Re: Probability-related questions
Originally Posted by alexander9408
Sir, what is binary number?
You never did explain what is meant by a sum in this question.
In this case, a binary number is same an base 2 representation of a positive integer (that explains the -1)
Originally Posted by alexander9408
I don't get the second one, mind to explain more, please?
As for the third one, is that (6x5x4x3x2x1)/3!2!2!2! ?
You have the third one correct.
There $\frac{6!}{2^3}$ ways to rearrange the string $112233$. | {"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": 7, "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": 0.8789652585983276, "perplexity": 525.7798833125113}, "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/1440644064503.53/warc/CC-MAIN-20150827025424-00268-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/1833-limit-help.html | 1. ## limit help
If you can help me on this id greatly appreciate it
(steps in helping me understand helps alot)
Lim ((1/x^(1/2))-(1/2))/(x-4)
x->4
2. Originally Posted by tnkfub
If you can help me on this id greatly appreciate it
(steps in helping me understand helps alot)
Lim ((1/x^(1/2))-(1/2))/(x-4)
x->4
That should be
Lim(x-->4)[(1/sqrt(x) -1/2) /(x-4)] because you're on L'Hopital's Rule, I guess.
So, substitute 4 for all the x's,
= (1/sqrt(4) -1/2) /(4-4)
= (1/2 -1/2) /0
= 0/0
Indeterminate.
Use the L'Hopital's Rule in this 0/0 limit case.
Lim(x->a)[f(x)/g(x)] = Lim(x->a)[f'(x)/g'(x)] ----***
f(x) = 1/sqrt(x) -1/2 = (x)^(-1/2) -1/2
f'(x) = (-1/2)*x^(-3/2) = -1/[2sqrt(x^3)] ----***
g(x) = x -4
g'(x) = 1
So,
Lim(x->4)[(1/x^(1/2) -1/2) /(x-4)]
= 0/0
= Lim(x->4)[(-1/[2sqrt(x^3)) /(1)]
= -1/[2sqrt(4^3)] /1
= -1/[2sqrt(64)]
= -1/[2*8]
3. ## Thanks + Question :P
hey thanks alot for helping me on that but i have one question and if this is a yes than this makes lim like this alot easier >>> Can you use the derivative of limits and get the same answer?
thats what you did right use the derivative to come up with that answer
4. Originally Posted by tnkfub
hey thanks alot for helping me on that but i have one question and if this is a yes than this makes lim like this alot easier >>> Can you use the derivative of limits and get the same answer?
thats what you did right use the derivative to come up with that answer
Yes, but it is applicable only on rational or "fractional" functions. That is what L'Hopital's Rule is all about. Search the Internet for that rule so that you will understand why the limits of the derivatives of the numerator and denominator of a rational function will give the limit of that whole function.
L'Hopital or L'Hospital Rule.
5. thanks alot uve helped me even more than by solving that problem for me :P cuz i also have a test today on this material and this L'Hospital Rule is gonna be real useful :P
6. Originally Posted by tnkfub
thanks alot uve helped me even more than by solving that problem for me :P cuz i also have a test today on this material and this L'Hospital Rule is gonna be real useful :P
Very good then. L'Hopital's Rule is really helpful in finding limits of rational functions when the limits are indeterminate. But that Rule is not always applicable, or will not always lead to the limit of the whole rational function. Later on you will hit the wall also in using that Rule, even after many differentiations. But for now, enjoy the Rule. I love it myself. | {"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": 0.9277949929237366, "perplexity": 1392.599136300547}, "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-2018-17/segments/1524125945552.45/warc/CC-MAIN-20180422080558-20180422100558-00304.warc.gz"} |
https://d2mvzyuse3lwjc.cloudfront.net/doc/Origin-Help/PHM-Cox-Algorithm | 17.6.2.2 Algorithms (Cox Proportional Hazard Regression)
Let $t_i\,\!$,for i = 1, 2, ?, n, be the failure time or censored time for the ith observation with the vector of p covariates $Z_j(j=1,2,\ldots ,p)$. It is assumed that the failure and censored mechanisms are independent. The hazard function, $\lambda (z,t)\,\!$ , is the probability that an individual with covariates z fails at time t given that the individual survived up to time t. In the Cox proportional model is of the form:
$\lambda (z,t)=\lambda _0(t)\exp (z^{T}\beta +\omega )\,\!$
where $\lambda _0\,\!$ is the base-line hazard function, an unspecified function of time, $\beta \,\!$ is a vector of unknown parameters and $\omega\,\!$ is a known offset.
Assuming there are ties in the failure time giving $n_d < n\,\!$ distinct failure times, $t_{(1)} < t_{(2)} < ?< t_{(nd)}$ , such that $d_i\,\!$ individuals fail at $t_{(i)}\,\!$ , it follows that the marginal likelihood for $\beta$ is well approximated by:
$L=\prod_{i=1}^{n_d}\frac{\exp (s_i^{T}\beta +\omega _i)}{[\sum_{l\in R(t_{(1)})}\exp (z_i^{T}\beta +\omega _i)]^{d_{i}}}$ (1)
where $s_i\,\!$ is the sum of covariate of individuals observed fail at $t_{(i)}\,\!$ and $R(t_{(i)})\,\!$ is the set of individuals at risk just prior to $t_{(i)}\,\!$ , that is it is all the individuals that fail or censored at time $t_{(i)}$ along with all individuals survived beyond the time $t_{(i)}\,\!$ . The MLE (maximum likelihood estimates) of $\beta\,\!$, given by$\hat \beta\,\!$, are obtained by maximizing (1) using a Newton-Raphson iteration technique that includes step having and utilizes the first and second partial derivatives of (1) which are given by (2) and (3) below:
$U_j(\beta )=\frac{\partial Ln(L)}{\partial \beta _j}=\sum_{i=1}^{n_d}[s_{ji}-d_i\alpha _{ji}(\beta )]=0$ (2)
for j = 1, 2, ?, p, where $s_{ji}\,\!$ is the jth element in the vector $s_i\,\!$ and
$\alpha _{ji}(\beta )=\frac{\sum_{l\in R(t_{(1)})}z_{jl}\exp (z_l^{T}\beta +\omega _l)}{\sum_{l\in R(t_{(1)})}\exp (z_l^{T}\beta +\omega _l)}$
Similarly,
$I_{hj}(\beta )=-\frac{\partial ^2Ln(L)}{\partial \beta _h\partial \beta _j}=\sum_{i=1}^{n_d}d_i\gamma _{hji}$ (3)
where $\gamma _{hji}=\frac{\sum_{l\in R(t_{(1)})}z_{hl}z_{jl}\exp (z_l^{T}\beta +\omega _l)}{\sum_{l\in R(t_{(1)})}\exp (z_l^{T}\beta +\omega _l)}-\alpha _{hi}(\beta )\alpha _{ji}(\beta )$ h, j = 1, ? p.
$U_j(\beta )\,\!$ is the jth component of a score vector $I_{hi}(\beta )\,\!$ is the (h, j) element of the observed information matrix $I(\beta )\,\!$ whose inverse $I(\beta )^{-1}=I_{hi}(\beta )^{-1}\,\!$ gives the variance-covariance matrix of $\beta\,\!$.
It should be noted that if a covariate or a linear combination of covariates is monotonically increasing or decreasing with time ,then one or more of the $\beta _j^{\prime }s$ will be infinite.
If $\lambda _0(t)\,\!$ varies across $\nu\,\!$ strata, where the number of individuals in the kth stratum is $n_k\,\!$, k = 1, ?, $\nu\,\!$, with $n=\sum_{k=1}^\nu n_k$ , then rather than maximizing (1) to obtain $\hat \beta\,\!$, the following marginal likelihood is maximized:
$L=\prod_{k=1}^\nu L_k$ (4)
where $L_k\,\!$ is the contribution to likelihood for the $n_k\,\!$ observations in the kth stratum treated as a single sample in (1). When strata are concluded the covariate coefficients are constant across strata but there is a different base-line hazard function $\lambda _0(t)\,\!$.
The base-line survival function associated with a failure time $t_{(i)}\,\!$ , is estimated as
$exp(-\hat H(t_{(i)}))$ ,
where $\hat H(t_{(i)})=\sum_{t(j)\leq t(i)}(\frac{d_i}{\sum_{l\in R(t_{(j)})}\exp (z_l^T\hat \beta +\omega _l)})$
and $d_i\,\!$ is the number of failures at time $t_{(i)}\,\!$ . The residual of the lth observation is computed as:
$r(t_l)=\hat H(t_l)\exp (-z_l^T\hat \beta +\omega _l)$
where $\hat H(t_l)=\hat H(t_{(i)}),t_{(i)}\leq t_l .
The deviance is defined as $-2^*\,\!$ (logarithm of marginal likelihood). There are two ways to test whether individual covariates are significant: the differences between the deviances of nested models can be compared with appropriate $\chi ^2\,\!$-distribution; or, the asymptotic normality of the parameter estimates can be used to form z-test by dividing the estimates by their standard errors or the score function for the model under the null hypothesis can be used to form z-test. | {"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": 52, "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": 0.9713770151138306, "perplexity": 350.2175479899783}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00676.warc.gz"} |
https://www.questaal.org/tutorial/lmf/basis_set/ | # Properties of the lmf basis set
This tutorial describes the lmf basis set, and various kinds of cutoffs that affect convergence in the basis.
### Preliminaries
The input file structure is briefly described in this lmf tutorial for PbTe, which you may wish to go through first.
Executables blm, lmchk, lmfa, and lmf are required and are assumed to be in your path.
### Definition of the basis set
The lmf basis set $\chi_{\alpha\mathbf{R}L}(\mathbf{r})$ consist of smooth Hankel functions envelope functions $H_{\alpha\mathbf{R}L}(\varepsilon,r_s;{\mathbf{r}})$, which get augmented by solutions of the radial wave equation in augmentation spheres (partial waves). An additional set of local orbitals may be added to make the basis more complete in the augmentation region.
#### Envelope functions
The lmf basis set begins with smooth envelope functions. These functions are smooth Hankel functions, centered on an atom R. They are defined by smoothing radius $r_s$ and an energy $\varepsilon$. Smooth Hankels have the Slater-Koster form,
Here L is a compound index denoting the l and m angular momentum quantum numbers; the $h_l(\varepsilon,r_s;r)$ are radial functions defined by Eqns. (6-8) on this page, and the $Y_L(\hat\mathbf{r})$ are spherical harmonics.
The $H_{L}$ are convolutions of ordinary Hankel functions of energy $\varepsilon$ and Gaussian functions of smoothing radius $r_s$. These two parameters define the shape of the envelope: $\varepsilon$ controls the asymptotic decay for large r ($H_L \sim r^{-l-1}e^{-\kappa\,r}$, $\kappa^2{=}-\varepsilon$) and the smoothing radius $r_s$ demarcates the approximate point of transition from Gaussian-like shape at small r ($H_L \sim r^l$) to asymptotic behavior.
Each site has its own family of $H_{L}$. While it would be possible to have any number of $H_L$ on a particular site, each with unique values of $\varepsilon$ and $r_s$, in practice the Questaal codes allow two types ($\alpha{=}1,2$) of $H_{L}$ for a particular site R and angular momentum l. For the first envelope ($\alpha{=}1$), you define $(r_s,\varepsilon)$ pair through parameters RSMH and EH: Each l gets its own RSMH and EH. You can elect to limit the basis to one envelope function for a particular l, or choose a second envelope. The second set ($\alpha{=}2$) is optional: parameters are defined through RSMH2 and EH2. A single $H_l$ (“single kappa”) has the advantage of making for a small basis, but its accuracy is limited unless the system is fairly close-packed. At least for low l, i.e. for orbitals whose atomic levels are below or not far above the Fermi level, it is recommended that the “double kappa” basis (two functions per l) be used.
This flexibility in choosing is both a blessing and a curse. A blessing because the flexibility allows for more variational freedom, keeping the basis at low rank while maintaining high accuracy. But it is a curse because of the added burdens imposed on the user to determine the parameters. Usually you can allow the atom program lmfa to automatically generate parameters for basis set.
Thus we can identify the entire family of envelopes by $H_{\alpha\mathbf{R}L}$, labeling whether the first or second envelope, the site, and the L. Note that for each L, there are 2l+1 basis functions.
#### Augmentation
Each smooth envelope is “augmented” in a sphere around each nucleus by partial waves, which are numerical solution of the Schrodinger equation in the sphere up to an augmentation radius (specified by tag R in the species category). For each l (and m for a given l) up to a cutoff lmxa, the $H_{\alpha\mathbf{R}L}$ is replaced (“augmented”) by a linear combination of partial waves $\phi_l$ and $\dot\phi_l$. These partial waves form the Hilbert space of essentially exact solutions to the Schrodinger equation, to linear order in energy around some linearization energy $\varepsilon_\nu$. This energy is normally allowed to float in the course of the self-consistency cycle, to minimize the linearization error. See also the description of partial waves on this page. Suitable linear combinations of $\phi_l Y_L$ and $\dot\phi_l Y_L$ are taken in each sphere so as to make the augmented $H_{\alpha\mathbf{R}L}$ continuous and differentiable.
#### Local orbitals
It may be the linear method is not sufficient and that a third partial wave is required to make the basis complete over a sufficiently wide energy window. A conventional local orbital is realized by integrating the Schrodinger equation at another energy either far above or far below the energy used to make $\phi$ and $\dot\phi$, to obtain another partial wave $\phi_{z}$. A suitable combination of $\phi$ and $\dot\phi$ is subtracted to make the value and slope of the local orbital vanish at the augmentation radius. With such a construction it need not have an interstitial part. An “extended” local orbital (suitable for semicore states far below the Fermi level) is constructed by attaching a single smooth Hankel function, whose energy and smoothing radius are varied to match value and slope of the partial wave. You specify the local orbital through tag PZ in the species category. lmfa can automatically find deep local orbitals (one principal quantum number below the valence partial waves) that may be needed to make the basis reasonably complete. You may also select high-lying local orbitals (one principal quantum number above the valence partial waves). These are usually less relevant in DFT; however, high-lying d local orbitals on transition metals were necessary to obtain good agreement with other codes in the DeltaCodes validation exercise. Note that addition of local orbitals increases the rank of the Hamiltonian.
#### Hilbert space and rank of the lmf Hamiltonian
The Hilbert space of the lmf basis $\chi_{\alpha\mathbf{R}L}(\mathbf{r})$ then consists of the following:
1. In the interstitial, smooth Hankel functions $H_{\alpha\mathbf{R}L}(\mathbf{r}){=}H_{\alpha{L}}(\mathbf{r{-}R})$. They can be expanded in plane waves to make matrix elements of the potential.
Note: For a given set $\alpha{=}1$ or $\alpha{=}2$ you must include all l’s up to some basis cutoff, i.e. l=0,1,…lmxb. lmxb need not be the same for the $\alpha{=}2$ set of envelopes as for the first.
2. In augmentation sphere R up to the augmentation l cutoff lmxa, the pair of partial waves ($\phi_{\mathbf{R}l}(r)Y_L(\hat\mathbf{r})$, $\dot\phi_{\mathbf{R}l}(r)Y_L(\hat\mathbf{r})$). The dominant partial wave is $\phi_{\mathbf{R}l}(r)$; mostly it attaches on to smoothed Hankels centered at R. $\dot\phi_{\mathbf{R}l}(r)Y_L(\hat\mathbf{r}$ mostly attaches on to the (tails) of smooth Hankels $H_{\alpha\mathbf{R'}L}(\mathbf{r})$ centered at other sites and makes a smaller contribution. Finally there may possibly be local orbitals $\phi_{\mathbf{R}zl}(r)Y_L(\hat\mathbf{r})$ in some l channels. Local orbitals may be high-lying (far above the linearization energy) or deep, to include high-lying (semi)core levels in the valence. This extra set is specified through parameter PZ in the main input (ctrl file) file or the basp file.
3. In augmentation sphere R above lmxa, the tails $H_{\alpha\mathbf{R'}L}(\mathbf{r})$ of smoothed Hankels at sites other than R. The special manner in which augmentation is done enables the lmf basis set to converge very rapidly with lmxa — much faster than traditional all-electron basis sets. See below.
The total rank of the hamiltonian is then the number of $H_{\alpha\mathbf{R}L}$ you specify on all sites, plus any local orbitals specified. The size of the basis (number of first kappa, second kappa, and local orbitals) is printed in a table in lmf’s standard output.
### Tutorial
#### 1. Building the input file
This step is essentially identical to the first step in the PbTe tutorial. An abbreviated version is presented here.
Cut and paste the following into init.bi2te3.
# from http://cst-www.nrl.navy.mil/lattice/struk/c33.html
# Bi2Te3 from Wyckoff
% const a=4.3835 c=30.487 uTe=0.788 uBi=0.40
LATTICE
SPCGRP=R-3M
UNITS=A
A={a} C={c}
SITE
ATOM=Te C=0 0 0
ATOM=Te C=0 0 {uTe}
ATOM=Bi C=0 0 {uBi}
This tutorial explains how the input files init.ext and ctrl.ext are structured.
To create the skeleton input file invoke blm:
$blm init.bi2te3$ cp actrl.bi2te3 ctrl.bi2te3
There 2 Bi atoms and 3 Te atoms in the unit cell. Two of the Te atoms are symmetry equivalent.
This template will not work as is; three essential pieces of information which blm does not supply are missing, to rectify this:
• You must specify plane-wave cutoff GMAX. This can be obtained by running the lmfa tool as seen in the PbTe tutorial. For the purpose of this tutorial we will use the value GMAX=4.4. A more detailed description of GMAX is given below.
• You must specify a valid k mesh for Brillouin zone integration. The correct k mesh size is dependent on your problem, you can find an example in the PbTe tutorial. For this tutorial we will use a k mesh of $6\times 6\times 6$ divisions, nkabc=3. A more detailed description of nkabc is given below.
• You must define a basis set which can be done manually or automatically, as described next.
Note: if you are using the GMAX and k-mesh values listed above, you can use the following commands to automatically edit your ctrl.bi2te3 file:
$sed -i s/nkabc=0/nkabc=3/ ctrl.bi2te3$ sed -i s/gmax=0/gmax=4.4/ ctrl.bi2te3
In either case, to edit these values you simply need to find the gmax and nkabc variables in the % const section of your ctrl.bi2te3 file and modify them to the desired value. The sed command above does this for you.
blm creates a family of tags belonging to AUTOBAS to enable other programs to automatically find a basis set for you. We will use this tag, which sets up a standard minimal basis:
autobas[pnu=1 loc=1 lmto=3 mto=1 gw=0]
This is an alias for the tag in the HAM category
AUTOBAS[PNU=1 LOC=1 LMTO=3 MTO=1 GW=0]
Note that you must modify ctrl.bi2te3 a little; the default gives [… LMTO=5 MTO=4], which makes a double kappa basis. Once again, you can use the following command to perform this replacement automatically:
$sed -i 's/autobas$pnu=1 loc=1 mto=4 lmto=4$/autobas$pnu=1 loc=1 lmto=3 mto=1 gw=0$/' ctrl.bi2te3 lmfa calculates wave functions and atomic densities for free atoms. It also has a mode that automatically generates information for basis sets, using tokens in AUTOBAS to guide it. This information is written to a file basp0.ext. AUTOBAS specifies set of conditions that enable lmfa to automatically build the basis set for you, as described below. Parameters are generated, but you can modify them as you like. Tokens in AUTOBAS tell lmfa to do the following: PNU=1 Calculate the logarithmic derivative parameter Pl for the free atom. Calculated parameters are saved in file basp0.ext as P=…. Nothing about P is written if PNU=0. LOC=1 Look for shallow cores to be explicitly treated as valence electrons, as local orbitals. Shallow cores that meet specific criteria are identified and written to basp0.ext as PZ=…. No search is made if LOC=0 LMTO=3 Pick a default choice about the size of basis. LMTO=3 is a standard minimal basis. Run lmfa --input and look for HAM_AUTOBAS_LMTO to see what other choices there are. lmfa will pick some defaults for the l-cutoff, e.g. spd or spdf depending on the value of LMTO. MTO=1 Choose 1-kappa basis set (single orbital per l channel). Small basis for fast calculations. For better quality calculations, it is recommended you use MTO=2 or MTO=4. GW=0 Create a setup for an LDA calculation. If GW=1, tailor basis for a GW calculation, selecting stricter criteria for including shallow cores as valence, and the size of basis. These tokens thus define some reasonable default basis for you. lmfa writes basp0.ext. This file is never read, but lmf will read basp.ext and use this information when assembling the basis set. The two files have the same structure and you can copy basp0.ext to basp.ext. What lmfa generates is not cast in concrete. You are free to adjust the parameters to your liking, e.g. add a local orbital or remove one from the basis. The AUTOBAS tokens tell lmf what to read from basp.ext. It uses tokens in a manner similar, but not identical, to the way lmfa uses them: PNU=1 Read parameters P for all species present in basp.ext. If PNU=0, these parameters will not be read. LOC=1 tells lmf to read local orbital parameters PZ. Since these parameters may also be specified by the input file, LOC=1 tells lmf to give precedence to parameters specified by ctrl file LOC=2 tells lmf to give precedence to parameters specified by basp. LMTO= is not used by lmf. MTO=1 controls what part of RSMH and EH is read from basp.bi2te3. LMTO=1 or 3 tells lmf to read 1-kappa parameters specified by basp LMTO=2 or 4 tells lmf to read 2-kappa parameters specified by basp LMTO=1 or 2 tells lmf that parameters in the ctrl file take precedence LMTO=2 or 4 tells lmf that parameters in the basp file take precedence GW=0 tunes the basis for an LDA calculation If GW=1, tune basis for a GW calculation. For example log derivative parameters P are floated a little differently in the self-consistency cycle. They are weighted to better represent unoccupied states, at a slight cost to their representation of occupied states. #### 2. Checking sphere overlaps Sphere overlaps can be checked using lmchk. To do this copy the template actrl.bi2te3 to the input file and run lmchk: $ lmchk bi2te3
You should see the site positions for all the atoms:
Site Class Rmax Hcr Position
1 1 Te 2.870279 2.009195 0.00000 0.00000 0.00000
2 3 Te2 2.870279 2.009195 -0.50000 -0.86603 1.46162
3 3 Te2 2.870279 2.009195 0.50000 0.86603 -1.46162
4 2 Bi 2.856141 1.999299 0.50000 0.86603 0.80309
5 2 Bi 2.856141 1.999299 -0.50000 -0.86603 -0.80309
and a table of overlaps
ib jb cl1 cl2 Pos(jb)-Pos(ib) Dist sumrs Ovlp % summt Ovlp %
1 4 Te Bi 2.391 -4.142 3.841 6.134 5.726 -0.41 -6.6 4.008 -2.13 -34.7
1 5 Te Bi -2.391 -4.142 -3.841 6.134 5.726 -0.41 -6.6 4.008 -2.13 -34.7
2 4 Te2 Bi -2.391 -4.142 -3.149 5.726 5.726 0.00 0.0* 4.008 -1.72 -30.0
3 5 Te2 Bi 2.391 -4.142 3.149 5.726 5.726 0.00 0.0* 4.008 -1.72 -30.0
OVMIN, 107 pairs: fovl = 0 <ovlp> = 0% max ovlp = 0%
By default, blm makes the spheres as large as possible without overlapping. In this case the Bi and Te radii are nearly the same. and at least some spheres touch without overlapping.
The packing fraction is printed
Cell volume= 1141.20380 Sum of sphere volumes= 492.34441 (0.43143)
Generally larger packing fractions are better because the augmentation partial waves are quite accurate. 0.43 is a fairly good packing fraction.
#### 2a. General Remarks on choosing sphere radii
Choosing sphere radii is an important business in an augmented-wave method. Ideally results should not depend on the choice, and in well converged cases this is nearly true. But it can never be exactly because as the radii change, the basis set of some portion of the volume is represented differently (partial waves inside spheres up to a finite l cutoff, and a limited famly of envelope functions in the interstitial). Particularly when basis sets are small, there is some dependence on radii and a poor choice can lead to poor results. As a minimum you should be aware these radii are parameters, and that there may be some dependence on their choice.
Sphere radii are chosen by balancing between making them on the one hand as large as possible: the augmentation basis is of high quality and partial waves are accurate and more complete than envelope functions. This argues for large radii, but l convergence of the augmented part is slower with larger radii, and also there is a geometry violation when spheres overlap. (Overlap is defined as $(r_1{+}r_2{-}d)/d$, where $r_1$ and $r_2$ ae the two sphere radii, and $d$ is the bond length.) Both considerations argue for smaller radii.
Finally there is the question of how to select the relative size of radii of different species.
Fortunately blm can determine relative sphere radii in a nearly optimal way, which it does by constructing an approximate potential and varying radii to make the potentials on the augmentation boundaries as nearly equal as possible. This makes the potential as close to a constant as possible, which improves the ability of the envelope functions to represent the interstitial potential and also reduces the burden on the interstial matrix element maker. Thus is is a good idea to let blm find the sphere radii for you, or if you are starting from a ctrl file, use lmchk --getwsr to supply suggested radii. You can tell blm or lmchk what maximum overlap you will tolerate (blm accepts a command-line switch –omax, and lmchk reads tag OMAX1 in the SPEC category).
What to consider when selecting upper bounds to overlaps depend on the context. It is useful to keep in mind the following obervations. Further observations, particularly in an ASA context, can be found on this page.
• ASA codes lm, lmgf, lmpg: Sphere overlaps are necessary because the ASA has no interstitial — thus the approximations in the ASA cannot be systematically reduced. Care must be taken that they are not too large; sometimes this necessitates using empty spheres. bcc lattices (overlap 14%) and fcc lattices (overlap 11%) are good. Overlaps of 18% or 19% can be necessary for some bonds in complex cases, e.g. incommensurate interfaces.
• Full potential code lmf, traditional basis set: Strictly speaking there should be no geometry violation. All the same lmf has a special three-component augmentation (see Sec. 3.6. of Ref. 1) that allows it to tolerate sphere overlaps up to about 10% without incurring significant error. It has been found empirically however, that convergence to self-consistency becomes more delicate when overlaps become too large.
• Full potential code lmf, Jigsaw Puzzle Orbital basis set: The JPO basis is still undergoing development, but when it implemented it should provide a marked improvement over the traditional one. We anticipate that the sphere radii can be chosen rather small without significant loss of precision, e.g. −10% overlap. At present this is only speculation.
• GW code driven by lmfgwd: The GW suite does not have the same elegant augmentation as the LDA. Therefore you should observe the constraint of no overlaps fairly carefully. (1% seems to be acceptable).
#### 3. The atomic density and basis set parameters
If you did not do so already copy actrl.bi2te3 to ctrl.bi2te3 and change [… MTO=4 LMTO=4][… MTO=1 LMTO=3]).
$cat actrl.bi2te3 | sed 's/mto=4 lmto=4/mto=1 lmto=3/' > ctrl.bi2te3 Invoke lmfa: $ lmfa bi2te3
The primary purpose of lmfa is to generate a free atom density. A secondary purpose but nevertheless a very useful one, is to supply additional information about the basis set in an automatic way. All of this information that blm omits can be supplied manually in the input file, but once an input file is available, lmfa can automatically supply most of the key missing information for a minimal working input file. For example it generates a basp file basp0.bi2te3m which contains information about the shape of the basis:
BASIS:
Te RSMH= 1.615 1.681 1.914 1.914 EH= -0.888 -0.288 -0.1 -0.1 P= 5.901 5.853 5.419 4.187
Bi RSMH= 1.674 1.867 1.904 1.904 EH= -0.842 -0.21 -0.1 -0.1 P= 6.896 6.817 6.267 5.199 5.089 PZ= 0 0 15.9362
Every species gets one line. This file specifies a basis set consisting of spdf orbitals on Te sites, and spdf orbitals on Bi sites, and a local 5d orbital on Bi. See the PbTe tutorial for further description of these parameters. This choice is generally a good choice, but it optimized for the atom, not the crystal. There is an automatic procedure to optimise envelope function parameters RSMH and EH if you wish to do so.
Note: Remember that lmf reads from basp.ext, not basp0.ext.
The partitioning between valence and core states is something that requires a judgement call. lmfa has made a default choice: from basp0.bi2te3 you can see that lmfa selected the 6s, 6p, 6d, 5f states, populating them with charges 2, 3, 0, 0, making the total sphere charge Q=0. You can override the default, e.g. choose the 5d over the 6d with SPEC_ATOM_P; override the l channel charges with SPEC_ATOM_Q.
lmfa does the following to find basis set parameters:
• Automatically finds deep states to include as valence electrons.
• Selects sphere boundary condition for partial waves
• Finds envelope function parameters
The process is essentially the same as described in the PbTe tutorial; it is described in some detail there and in the annotated lmfa output. The main difference is that in this case a smaller, single-kappa basis was specified (LMTO=3); lmfa makes (RMSH,EH) instead of the double kappa (RMSH,EH; RMSH2,EH2). Later we will improve on the basis by adding APW’s.
With your text editor insert lines from basp0.bi2te3 in the SPEC category of the ctrl file, viz
ATOM=Te Z= 52 R= 2.870279 LMX=3 LMXA=3
RSMH= 1.615 1.681 1.914 1.914 EH= -0.888 -0.288 -0.1 -0.1 P= 5.901 5.853 5.419 4.187
ATOM=Bi Z= 83 R= 2.856141 LMX=3 LMXA=4
RSMH= 1.674 1.867 1.904 1.904 EH= -0.842 -0.21 -0.1 -0.1 P= 6.896 6.817 6.267 5.199 5.089 PZ= 0 0 15.936
Alternatively make lmf read these parameters from basp.bi2te3. Copy basp0.bi2te3 to basp.bi2te3, and modify it as you like. basp.ext is read after the main input file is read, if it exists. According to which of following tokens is present, their corresponding parameters will be be read from the basp file, superseding prior values for these contents:
AUTOBAS lmfa writes, lmf reads MTO RSMH,EH (and RSMH2,EH2 if double kappa basis) P P LOC PZ
If this information is supplied in both the ctrl file and the basp file, values of MTO and LOC tell lmf which to use. In this tutorial we will work just with the basp file.
$cp basp0.bi2te3 basp.bi2te3 The atm file was created by lmfa without prior knowledge that the 5d local orbital is to be included as a valence state (via a local orbital). Thus it incorrectly partitioned the core and valence charge. You must do one of the following: 1. Remove PZ=0 0 15.936 from basp.bi2te3. It will no longer be treated as a valence state. Removing it means the remaining envelope functions are much smoother, which allows you to get away with a coarser mesh density. Whether you need it or not depends on the context: with GW, for example, this state is a bit too shallow to be treated with Fock exchange only (which is how cores are handled in GW). 2. Copy basp0.bi2te3 to basp.bi2te3 and run lmfa over again: $ cp basp0.bi2te3 basp.bi2te3
$lmfa bi2te3 With the latter choice lmfa operates a little differently from the first pass. Initially the Bi 5d was part of the core (similar to the Pb 5d in the Pbte tutorial; now it is included in the valence. #### 4. GMAX and NKABC ##### 4.1 Setting GMAX blm makes no attempt to automatically assign a value to the plane wave cutoff for the interstitial density. This value determines the mesh spacing for the charge density. lmf reads this information through HAM_GMAX or EXPRESS_gmax, or HAM_FTMESH. It is a required input; but blm does not automatically pick a value because its proper choice depends on the smoothness of the basis. In general the mesh density must be fine as the most strongly peaked orbital in the basis. lmfa makes RSMH and EH and can determine an appropriate value for HAM_GMAX based on them. In the present instance, when the usual 6s, 6p, 6d, 5f states are included lmfa recommends GMAX=4.4 as can be seen by inspecting the first lmfa run. If you keep the 5d in the valence and run lmfa second time you will see this output FREEAT: estimate HAM_GMAX from RSMH: GMAX=4.4 (valence) 8.1 (local orbitals) The valence states still require only GMAX=4.4 but the 5d state is strongly peaked but GMAX=8.1 for the local orbitals. The 5d state is strongly peaked at around the atom, and requires more plane waves to represent reasonably, even a smoothed version of it, than the other states. The difference between 8.1 and 4.4 is substantial, and it reflects the additional computational cost of including deep core-like states in the valence. This is the all-electron analog of the “hardness” of the pseudopotential in pseudopotential schemes. If you want high-accuracy calculations (especially in the GW context), you will need to include these states as valence. Including the Bi 5d is a bit of overkill for LDA calculations however. If you eliminate the Bi 5d local orbital you can set GMAX=4.4 and significantly speed up the execution time. It is what this tutorial does. ##### 4.2 Setting NKABC blm assigns the initial k-point mesh to zero. Note the following lines in ctrl.bi2te3: % const nkabc=0 gmax=0 ... # Brillouin zone nkabc= {nkabc} # 1 to 3 values Note: nkabc is simultaneously a variable and a tag here. This can be somewhat confusing. The expression {nkabc} gets parsed by the preprocessor and is turned into the value of variable nkabc (see how nit gets turned into 10 in the PbTe tutorial), so after preprocessing, the argument following tag is a number. EXPRESS_nkabc (alias BZ_NKABC) governs the mesh of k-points. An appropriate choice will depend strongly on the context of the calculation and the sytem of interest; the density-of-states at the Fermi level; whether Fermi surface properties are important; whether you want optical properties as well as total energies well described; the precision you need; the integration method, and so on. Any automatic formula can be dangerous, so blm will not choose an operational default. The most reliable way determine the appropriate mesh is to vary nkabc and monitor the convergence. We don’t need a self-consistent calculation for that: the k-convergence will not depend strongly whether the potential is converged or assembled from free atoms. Do the following (assuming no 5d local orbital) $ lmf ctrl.bi2te3 -vgmax=4.4 --quit=band -vnkabc=2
$lmf ctrl.bi2te3 -vgmax=4.4 --quit=band -vnkabc=3$ lmf ctrl.bi2te3 -vgmax=4.4 --quit=band -vnkabc=4
$lmf ctrl.bi2te3 -vgmax=4.4 --quit=band -vnkabc=5$ lmf ctrl.bi2te3 -vgmax=4.4 --quit=band -vnkabc=6
The meaning of --quit=band is described here.
Total energies are written to the save file save.bi2te3. It should read:
h gmax=4.4 nkabc=2 ehf=-126808.2361717 ehk=-126808.1583957
h gmax=4.4 nkabc=3 ehf=-126808.3137885 ehk=-126808.2492178
h gmax=4.4 nkabc=4 ehf=-126808.3168406 ehk=-126808.2505156
h gmax=4.4 nkabc=5 ehf=-126808.3165536 ehk=-126808.2497121
h gmax=4.4 nkabc=6 ehf=-126808.3164058 ehk=-126808.2494041
You can use the vextract tool to conveniently extract a table of the Harris-Foulkes total energy as a function of nkabc
$cat save.bi2te3 | vextract h nkabc ehf | tee dat You can plot the data, or just see by inspection that the energy is converged to less than a mRy with 4×4×4 k mesh and about 0.1 mRy with a 5×5×5 k mesh. A detailed analysis of k point convergence is given here. #### 5. Self-consistent LDA calculation, small basis In the following we will use nkabc=3, though after convergence is reached we might consider increasing it a little. Before doing the calculation you can use your text editor to set nkabc=3 and gmax=4.4, or continue to set assign values from the command line. The density can be made self-consistent with $ rm -f mixm.bi2te3 save.bi2te3
$lmf ctrl.bi2te3 -vgmax=6 -vnkabc=3 --quit=band and comparing ehf in the last two lines of save.bi2te3. You should find that the energy is converged to about 0.1 mRy. ##### 5.3 Convergence in the forces The exact forces are the change in total energy with respect to displacement of the nucleus. As they are calculated in the Questaal package, some correction terms are added to accelerate convergence with respect to deviations ninn in the guessed density from the self-consistent one. See the annotation of lmf output. Note: The forces are not exact derivatives of the total energy. This is because the change in shape of the augmented partial waves $\phi$ and $\dot{\phi}$ is not taken into account as a nucleus displaces. The effect is usually very small. Forces should be exactly consistent with the energy if the shape of the partial waves is frozen, however, which you can do with HAM_FRZWF. To what extent the basis set affects the forces is taken up in the Additional Exercises. ##### 5.4 Convergence in LMXA In an augmented wave method, envelope functions centered a different site is must be expanded around a local site, as one-center expansions in spherical harmonics Ylm. LMXA is a cutoff that truncates the one-center expansion to a finite l = LMXA. The input file reads: SPEC ATOM=Te Z= 52 R= 2.870279 LMX=3 LMXA=3 ATOM=Bi Z= 83 R= 2.856141 LMX=3 LMXA=4 LMXA=3 and LMXA=4 are very low l cutoffs for an augmented wave method. They are about half of what is needed for a reasonably converged calculation with traditional augmentation. However, the Questaal suite has a unique augmentation procedure that converges very rapidly with l. Indeed, it can be seen as a justification for pseudopotential methods that limit the pseudopotential to very low l, e.g. l=2. It is a useful exercise to see how well converged the total energy is using the default values LMXA=3 and LMXA=4. First, run lmf with the unaltered ctrl file: $ lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3 --quit=band
Set both LMXA to 6: and run lmf again
$cp ctrl.bi2te3 ctrl.bak$ cat ctrl.bak | sed s/LMXA=./LMXA=6/ > ctrl.bi2te3
$lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3 --quit=band When the restart file is read you should see site 1, species Te : augmentation lmax changed from 3 to 6 site 1, species Te : inflate local density from nlm= 16 to 49 Compare the last two lines of save.bi2te3. You should be able to confirm that the energy change is 0.25 mRy, By playing around with the two LMXA you can confirm that increasing the Te LMXA to 4, nearly all the error is eliminated. Before continuing, be sure to restore the original ctrl file $ cp ctrl.bak ctrl.bi2te3
##### 5.5 Convergence in KMXA
Ordinary Hankel functions can be expanded in Bessel functions around a remote site. This follows from the fact that both are solutions of the same second order differential equation, one being regular at the origin and the other being regular at ∞. Smooth Hankel functions are more complicated: the one-center expansion has no such elementary function, but it can be written as a linear combination of Laguerre polynomials Pkl of half integer order; see Eq. (12.17) in this J. Math. Phys. paper.
The polynomials are cut off at a fixed order given by KMXA. blm doesn’t write KMXA to the template file; instead a default value is used, which is written to this table
species data: augmentation density
spec rmt rsma lmxa kmxa lmxl rg rsmv kmxv foca rfoca
Te 2.870 1.148 3 3 3 0.718 1.435 15 1 1.148
Bi 2.856 1.142 4 3 4 0.714 1.428 15 1 1.142
You can check the convergence by in KMXA by supplying an input for it. First run lmf with the unmodified file:
$lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3 --quit=band Set KMXA to 5 and run lmf again: $ cp ctrl.bi2te3 ctrl.bak
$cat ctrl.bak | sed 's/LMXA=/KMXA=5 LMXA=/' > ctrl.bi2te3$ lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3 --quit=band
When the restart file is read you should see an indication that KMXA has been increased:
site 1, species Te : augmentation kmax changed from 3 to 5
site 2, species Te : augmentation kmax changed from 3 to 5
Compare the last two lines of save.bi2te3. You should be able to confirm that the energy change is 0.15 mRy.
Note: Before continuing, be sure to restore the original ctrl file.
$cp ctrl.bak ctrl.bi2te3 #### 6. Optimizing the basis set ##### 6.1 Tuning the envelope function parameters The envelope function shape generated by lmfa is tailored to the atom. They are very good basis sets for that case, but less so for the crystal. Ideally a basis of the size could be constructed that yields energies converged almost as well as we accomlish for the free atom. This is difficult to do in practice, though it is the aim of the Jigsaw Puzzle Orbitals basis. The best we can with the existing lmf basis is to optimize the energy by tuning RSMH and EH. This cannot be done analytically, but switch --optbas causes lmf to perform this optimization by brute-force minimization of the Harris-Foulkes energy. First we need to regenerate the self-consistent density we generated previously: $ rm -f mixm.bi2te3
$rm -f save.bi2te3$ lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3
Then we can use the --optbas switch:
$lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3 --optbas > out This will loop through RSMH in each species, minimizing the total energy for each one by one. It does not vary EH because the total energy is less sensitive to it. lmf will print out a table of the parameters it will optimize: LMFOPB: optimizing energy wrt 8 parameters, etol=5e-4: spec l type start 1:Te 0 rsm 1.615 1:Te 1 rsm 1.681 1:Te 2 rsm 1.914 1:Te 3 rsm 1.914 2:Bi 0 rsm 1.674 2:Bi 1 rsm 1.867 2:Bi 2 rsm 1.904 2:Bi 3 rsm 1.904 Each cycle carries out non self-consistent calculations to get the Harris-Foulkes total energy. After completing lmf prints out LMFOPB: before optimization ehf=-126808.2951. After optimization ehf=-126808.3092 Exit 0 Optimization complete. See file basp2 By tuning the parameters the energy decreased by 0.0141 Ry. To see which orbitals contributed the most, do: $ grep LMFOPB out | grep optimized
You should see
LMFOPB: var #1 optimized to 1.602 (7 iter): ehf=-126808.295097, delta ehf=-2.38e-5
LMFOPB: var #2 optimized to 1.617 (5 iter): ehf=-126808.298989, delta ehf=-0.00392
LMFOPB: var #3 optimized to 1.778 (6 iter): ehf=-126808.299166, delta ehf=-1.77e-4
LMFOPB: var #5 optimized to 1.605 (5 iter): ehf=-126808.301, delta ehf=-9.12e-4
LMFOPB: var #6 optimized to 1.658 (7 iter): ehf=-126808.307945, delta ehf=-0.00695
LMFOPB: var #7 optimized to 2.245 (5 iter): ehf=-126808.308856, delta ehf=-9.11e-4
LMFOPB: var #8 optimized to 1.679 (6 iter): ehf=-126808.309198, delta ehf=-3.42e-4
The orbitals that mad the most difference were the Te 5p (1.6811.617) and Bi 5p (1.8671.658).
Optimized parameters where the total energy decreased by more than etol (5e-4) were modified; the optimized parameters are written to basp2.bi2te3
Next, we are going to modify the basp file. You should make a backup of the current file first:
$cp basp.bi2te3 basp.bak You can copy basp2.bi2te3 to basp.bi2te3, but there is little point in optimizing any but the Te 5p and Bi 6p. Instead we use a text editor and modify basp.bi2te3 in those channels: BASIS: Te RSMH= 1.615 1.617 1.914 1.914 EH= -0.888 -0.288 -0.1 -0.1 P= 5.901 5.853 5.419 4.187 Bi RSMH= 1.674 1.658 1.904 1.904 EH= -0.842 -0.21 -0.1 -0.1 P= 6.896 6.817 6.267 5.199 5.089 Run lmf again $ lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3 --quit=band
and confirm that ehf comes out close (-126808.3086945) to the optimized value (-126808.309198).
You can at this point make the density self-consistent again.
Note: Before continuing, be sure to restore the original basp file:
$cp basp.bak basp.bi2te3 In sum, • without optimizing the basis, you should achieve total energy ehf=-126808.2950696 at self-consistency. • With the --optbas switch it should become ehf=-126808.306962. ##### 6.2 Increasing the number of envelope functions lmfa uses the lmto tags to make default choices for the size of basis. The tutorial used a relative minimal basis, lmto=3. autobas[pnu=1 loc=1 lmto=3 mto=1 gw=0] With your editor, modify this line to read autobas[pnu=1 loc=1 lmto=4 mto=4 gw=0] This will increase the basis, most notably include two envelope functions per l channel (double-kappa) Run lmfa again and note how the basp file changed: $ lmfa bi2te3
$diff basp0.bi2te3 basp.bi2te3 Remove the PZ=… as before and copy basp0.bi2te3 to basp.bi2te3. Save the original file in a backup. $ nano basp0.bi2te3
$cp basp.bi2te3 basp.bak$ cp basp0.bi2te3 basp.bi2te3
Since we modified the basp file we need to re-run lmfa:
$lmfa bi2te3 Run lmf to self-consistency $ rm -f mixm.bi2te3
$rm -f rst.bi2te3$ lmf ctrl.bi2te3 -vgmax=4.4 -vnkabc=3
Note that size of the basis has increased: With lmto=3 there are 80 orbitals
Makidx: hamiltonian dimensions Low, Int, High, Negl: 80 0 18 27
suham : 98 augmentation channels, 98 local potential channels Maximum lmxa=4
which increases to 125 orbitals with lmto=5:
Makidx: hamiltonian dimensions Low, Int, High, Negl: 125 0 71 54
kappa Low Int High L+I L+I+H Neglected
1 80 0 18 80 98 27
2 45 0 53 45 98 27
all 125 0 71 125 196 54
suham : 98 augmentation channels, 98 local potential channels Maximum lmxa=4
At self-consistency you should find that the total energy converges to ehf=-126808.313403 Ry.
##### 6.3 Adding APWs : the PMT method
You can also use augmented plane waves (APWs) to improve on the basis set. The combination of smooth Hankel functions with APW’s is called the PMT basis set. Adding APW’s provides a systematic way of making the basis of envelope functions complete.
These tags in the HAM category control how many APWs are added:
PWMODE={pwmode} PWEMIN=0 PWEMAX={pwemax} # For APW addition to basis
To use APW’s some tolerances in the ctrl file should be tightened. If you invoke blm with --pmt these extra tolerances are included automatically. That is the easiest way to make the changes but here we just add include tolerances HAM_TOL and EWALD_TOL by hand. Insert two lines at the end of the HAM category:
HAM
...
FORCES={so==0} ELIND=-0.7
TOL=1d-16
EWALD TOL=1d-16
One problem with the PMT method is that the smoooth-Hankel and APW basis set span nearly the same hilbert space. If you add too many plane waves the overlap matrix does not remain positive definite. Tightening the tolerances above helps, and so does increasing the fineness of the density mesh, as these points are also used for the PW expansion of the basis.
Note: as a last resort, you can stabilize the overlap with the HAM_OVEPS switch, but it is better to reduce the rank of the LMTO or APW basis sets.
PWMODE=11 adds APWs in a standard way: envelope functions of the form ei(k+G)·r are added to the smooth Hankel basis. You must specify the PW cutoffs. Here you can specify both the lower and upper bounds since the smooth Hankels will efficiently cover most of the lower part of the space.
With these extra considerations, run lmf as follows
rm -f out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=0 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=1 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=2 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=3 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=4 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=5 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=6 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
lmf ctrl.bi2te3 -vpwmode=11 -vpwemax=7 -vgmax=8 -vnkabc=3 --quit=band >> out.lmf
and observe how the total energy decreases with pwemax. Use the vextract tool:
cat save.bi2te3 | vextract i pwemax ehf ehk | tee etot.bi2te3
Draw a picture of the total energy relative to the double-kappa value. The fplot command shown in the box below will generate a postscript file; the figure actually shown has been elaborated a little. The red circle shows the self-consistent double-kappa result of Sec. 6.2. The light grey line follows the PMT procedure as above, but taking for the LMTO part a smaller, optimized spd single kappa basis (see Additional exercises). Numbers in parenthesis are the number of LMTO’s. The red circle is the 2-kappa basis without APWs; the purple is the LAPW basis without LMTOs.
$fplot -frme 0,.8,0,.5 -frmt th=3,1,1 -tmy 2.5 -vehf=-126808.313403 -s circ -ord '(x2-ehf)*1000' etot.bi2te3 The figure shows that the double-kappa basis (additional 45 orbitals) stabilizes the single-kappa spdf basis by about 18 mRy, and that the same can be accomplished with APWs with a plane-wave cutoff of 2 Ry (additional 48-62 APWs). By further increasing the APW cutoff, another 5 mRy can be gained. For most purposes this extra gain is not important. Note that the APW basis with Emax=7 Ry is quite large: 343-370 APWs. Note that the APW basis is generally less efficient than the LMTO basis. To reach a precision comparable to the 2-kappa basis (125 orbitals) starting from the 1-kappa spd basis, about 160 APWs are needed, or about 200 orbitals all told. The power of the PMT method can be compared against a straight LAPW basis. About 300 APWs are needed to achieve the convergence of the double kappa basis. See purple line in the Figure. To see how many orbitals the APW basis adds, do: $ grep ndham out.lmf
#### Modifying the input file for GW
GW calculations demand more of the basis set because unoccupied states are important. To set up a job in preparation for a GW calculation, invoke blm as :
\$ blm --gw bi2te3
Compare actrl.bi2te3 generated with the --gw switch to one without. One important difference will be that the default basis parameters are modified because AUTOBAS becomes:
AUTOBAS[PNU=1 LOC=1 LMTO=5 MTO=4 GW=1]
The basis is similar to LMTO=4 but EH has been set a little deeper. This helps the QSGW implementation interpolate between k-points. The larger basis makes a minor difference to the valence bands; but the conduction bands change, especially the higher in energy you go.
Note also that the LMTO f orbitals are more efficient in converging the total energy per extra orbital added than plane waves are. | {"extraction_info": {"found_math": true, "script_math_tex": 57, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8055083751678467, "perplexity": 3445.379886280261}, "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-2021-43/segments/1634323587926.9/warc/CC-MAIN-20211026200738-20211026230738-00337.warc.gz"} |
https://www.kitplanes.com/wind-tunnel-2/ | # Wind Tunnel
Design process-tail volume.
1
Last month we took a look at what sets the forward and aft center of gravity (CG) limits for an airplane. We now turn our attention to how to adjust the airplane’s aerodynamic configuration to achieve the required allowable CG range.
The horizontal tail stabilizes the airplane in pitch and generates aerodynamic pitching moments that allow the pilot to trim the airplane and control its pitch attitude and rate.
At this point in the design, we have completed enough of the layout of the airplane to know where the CG will be for the loading conditions in which it must be able to fly. The next task is to ensure that the airplane is aerodynamically capable of flying safely at any CG position within the expected range. To do this we must ensure that the most aft physical CG is at or ahead of the aerodynamic aft limit, and the most forward physical CG is at or behind the aerodynamic forward limit.
The major configuration parameters that affect the aerodynamic CG limits are the fore and aft position of the wing, and the geometry of the horizontal tail.
## Tail Volume
The tail has two functions: to stabilize and provide center of mass (CM) for trim and control.
The horizontal tail of an airplane also has two functions. It stabilizes the airplane in pitch, and it generates aerodynamic pitching moments that allow the pilot to trim the airplane and control its pitch attitude and rate.
Both of these effects are generated by variations in the lift of the tail.
The stabilizing influence of the tail is caused by the way the lift of the tail varies with angle of attack (AOA) of the airplane. Increasing AOA increases the lift of the tail. This increased lift pushes up on the aft end of the fuselage, generating a nose-down moment that resists the increase in AOA. Decreasing AOA has the opposite effect. The lift of the tail decreases, or becomes negative, causing a nose-up pitching moment that moves the AOA back up toward the trimmed state. The stabilizing effect of the tail happens passively, without pilot input.
In order to control the airplane in pitch, the pilot moves the stick fore and aft. The stick is linked to the horizontal tail and moves the tail in a way that changes its lift. On some airplanes, the stick is linked to flap-type elevators hinged to the back of the fixed portion of the tail. On others, the entire tail is rotated by the stick motion. In either case, moving the stick back moves the trailing edge of the tail up, causing a negative change in the lift of the tail, and pushing the stick forward moves the tail’s trailing edge down, increasing the lift of the tail.
The effectiveness of the tail at generating stabilization and controlling the airplane is a function of how much variation in pitching moment the tail can generate. The moment is a function of the product of two parameters: the size of the tail, and the distance between the tail and the CG of the airplane (tail arm).
The bigger the tail is, the more lift it can generate. The longer the tail arm is, the more leverage the tail has on the CG and the more moment it generates per unit of lift.
The product of the tail area times the tail arm has units of volume (length3) even though no physical enclosed volume is present.
While the actual dimensional “tail volume” number is not particularly useful to the designer, the nondimensional form of it is.
Tail volume coefficient (V) is a nondimensional parameter used in computation of the airplane stability and control power.
Tail volume coefficient (V) is defined as the product of the area times the tail arm divided by the wing area times the mean aerodynamic chord of the wing:
V=(St/Sw)*(lt/MAC)
Where:
St=Tail Area
Sw=Wing Area
lt = Tail Arm (distance between CG and tail quarter chord point)
MAC= Wing mean aerodynamic chord.
V is used in the calculation of the stability and CG limits of the airplane. It is also very useful in preliminary design to help the designer do a preliminary tail sizing based on historical data. The tail volume coefficient of an existing airplane is relatively easy to calculate and can provide guidance to the designer in choosing an initial tail geometry.
It is a particularly useful parameter because it captures the effect of both the size of the tail and the tail arm. Since tail area and tail arm have equal weight in determining the effectiveness of the tail, it’s important to take both into account when sizing the tail.
## Tail Volume Effects on CG Limits
As we saw last month, the forward CG limit is set by nose-up control power requirements, and the aft limit is set by stability considerations.
Increasing tail volume coefficient (V) increases the stabilizing influence of the tail. This makes the airplane more stable at any CG position and moves the neutral point (CG position for neutral pitch stability) aft.
The aft CG limit for an airplane is set by a minimum acceptable stability level measured in “static margin” (SM). Static margin is the distance from the CG to the neutral point, divided by the MAC of the wing, and is expressed in %MAC. Once the minimum acceptable SM is set, we know the distance ahead of the neutral point at which the aft CG limit will fall.
At the other end of the CG range, we find that increasing V increases the pitch control power of the airplane. This increased control power makes it possible for the pilot to generate more nose-up moment, and trim the airplane or lift the nosewheel for takeoff at a more forward CG position.
From the above we can see that increasing tail volume moves the forward CG limit forward and the aft CG limit aft, giving the airplane a wider range of allowable CG positions.
Figure 1: Forward CG limit, aft CG limit, and neutral point for an aft-tail airplane as a function of tail volume coefficient.
These effects are illustrated in Figure 1, which plots the forward CG limit, aft CG limit, and neutral point for an example airplane as a function of tail volume coefficient. This plot is referred to as either a “scissors plot” or a “notch plot” in industry.
The scissors plot for an aft-tail airplane illustrates one reason why the aft-tail configuration became standard early and has persisted as the most common aerodynamic configuration. Increasing tail volume has only beneficial effects on stability and control. This means that the configuration is relatively easy to make work, since most longitudinal stability and control issues can be solved by making the tail bigger or lengthening the tail arm, both of which are relatively straightforward things to do. More importantly, changing the tail geometry to improve one CG limit also improves the other, so we do not end up in a situation where we have solved one problem but exacerbated another.
Figure 2: Forward CG limit, aft CG limit, and neutral point for a canard airplane as a function of foreplane volume coefficient.
This is not true of canard configurations. The foreplane of a canard is destabilizing in pitch. Increasing the size of the canard or moving it forward (more negative V) moves the neutral point and the aft CG limit forward. At the same time, it increases the nose-up control power of the configuration, moving the forward CG limit forward as well. An example scissors plot for a canard configuration is shown in Figure 2. Notice that increasing the absolute value of V moves both CG limits forward. They also move apart because the increased nose-up control power moves the forward limit faster than the destabilizing influence moves the aft limit. The allowable CG range gets wider, but the whole CG range moves forward. This phenomenon makes getting a canard airplane to balance properly significantly more difficult. There are enough successful canard airplanes to show that it is possible, but it is much trickier than getting an aft-tail configuration to balance, and it is much harder to adapt a canard configuration to accept changes in its physical CG.
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Barnaby Wainfan is a principal aerodynamics engineer for Northrop Grumman’s Advanced Design organization. A private pilot with single engine and glider ratings, Barnaby has been involved in the design of unconventional airplanes including canards, joined wings, flying wings and some too strange to fall into any known category.
#### 1 COMMENT
1. Hallo. I am really impressed with the usefulness and simple step by step guidance which gives Barnaby Wainfan’s articles in the early stage of aircraft design. I am an aircraft engineer, but my specialization is mainly in load computation, structural analysis and strength tests. I have almost forgotten most of flight mechanics. But now i am designing my own single seat high-wing aircraft for pleasure flying and navigation competitions. So i have to learn again how to design an aircraft from aerodynamic point of view. I have difficulties to understand how to draw the “scissors chart” . I quite understand the aft CG limit – it is a derivation of an aircraft moment curve slope by AoA set equal to zero (plus 5% reserve). But what is the front limit? Is that aircraft moment curve in landing configuration expressed through lift coefficient when lift coefficient is set to maximum? Should one include the effect of elevator deflection? If this is truth, required data are not available in the early stage of the design process. Or is there any way around? Thank you.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"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": 0.8149569034576416, "perplexity": 1509.7145602076873}, "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/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00179.warc.gz"} |
https://arxiv.org/abs/1303.3254 | # Title:Second-order differentiability for solutions of elliptic equations in the plane
Abstract: For a second-order elliptic equation of nondivergence form in the plane, we investigate conditions on the coefficients which imply that all strong solutions have first-order derivatives that are Lipschitz continuous or differentiable at a given point. We assume the coefficients have modulus of continuity satisfying the square-Dini condition, and obtain additional conditions associated with a dynamical system that is derived from the coefficients of the elliptic equation. Our results extend those of previous authors who assume the modulus of continuity satisfies the Dini condition.
Comments: 10 pages Subjects: Analysis of PDEs (math.AP) MSC classes: 35J15 Cite as: arXiv:1303.3254 [math.AP] (or arXiv:1303.3254v1 [math.AP] for this version)
## Submission history
From: Robert McOwen [view email]
[v1] Wed, 13 Mar 2013 19:09:21 UTC (10 KB) | {"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": 0.9557864665985107, "perplexity": 1262.8641179728918}, "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-2020-05/segments/1579250592636.25/warc/CC-MAIN-20200118135205-20200118163205-00453.warc.gz"} |
https://rbspgway.jhuapl.edu/bibcite/reference/807 | # Phase Space Density matching of relativistic electrons using the Van Allen Probes: REPT results
Author Keywords Abstract 1] Phase Space Density (PSD) matching can be used to identify the presence of nonadiabatic processes, evaluate accuracy of magnetic field models, or to cross-calibrate instruments. Calculating PSD in adiabatic invariant coordinates requires a global specification of the magnetic field. For a well specified global magnetic field, nonadiabatic processes or inadequate cross calibration will give a poor PSD match. We have calculated PSD(μ, K) for both Van Allen Probes using a range of models and compare these PSDs at conjunctions in L* (for given μ, K). We quantitatively assess the relative goodness of each model for radiation belt applications. We also quantify the uncertainty in the model magnetic field magnitude and the related uncertainties in PSD, which has applications for modeling and particle data without concurrent magnetic field measurements. Using this technique, we show the error in PSD for an energy spectrum observed by the relativistic electron-proton telescope (REPT) is a factor of \~1.2 using the TS04 model. Year of Publication 2013 Journal Geophysical Research Letters Volume 40 Start Page 4798 Number of Pages 4798\textendash4802 Date Published 09/2013 URL http://doi.wiley.com/10.1002/grl.50909 DOI 10.1002/grl.50909 | {"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": 0.8745251297950745, "perplexity": 2596.9750272720053}, "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-2020-29/segments/1593657151197.83/warc/CC-MAIN-20200714181325-20200714211325-00358.warc.gz"} |
http://anmolbhullar.me/2017/06/13/ortho-manifold/ | Preface: This article is the beginning of a series of articles about random parts about math that I find really cool.
I’ve always suspected that algebra and differential topology were somehow connected in some way, (after all, a lot of fields in Math are like this) but I’ve personally done any math which integrates algebra and differential topology so seamlessly. This is why when I encountered this in Differential Topology by Victor Guillemin and Allan Pollack, it blew my mind. This is why I wanted to write this article to showcase the relationship between lie groups and orthogonal matrices.
### What is an Orthogonal Matrix?
Consider the rotation matrix $$R(\theta)$$ in $$\mathbb{R}^2$$. This is a matrix/transformation which is distance preserving. I mean, ‘distance preserving’ in the same sense of the functions that one would use to apply a change of variables in an integral. If we generalize this notion of linear maps which are distance preserving, we obtain the idea of an orthogonal matrix. For example, in order for a matrix $$Q$$ to be distance preserving, we can say that it needs to preserve the dot product, so, we can say that if $$Q$$ is an $$n\times n$$ dimensional matrix (i.e. a function $$\mathbb{R^n} \to \mathbb{R^n})$$, then it must obey the following rule for $$u,v\in\mathbb{R}^n$$: $u\cdot v = (Qu)\cdot(Qv)$ This implies an important result. For any $$v\in\mathbb{R}^n$$, we have that: $v^{T}\cdot v = (Qv)^{T}\cdot(Qv) = (v^{T}Q^{T})(Qv) = v^{T}(Q^{T}Q)v$ which implies that $$Q^{T}Q = I$$ or in other words: $$Q^{T} = Q^{-1}$$. Thus, we can just define orthogonal matrices as matrices whose inverse is equal to its transpose.
### Step 1: Show the set of $$n\times n$$ orthogonal matrices form a smooth manifold
Let $$O(n)$$ be the set of $$n\times n$$ orthogonal matrices, $$S(n)$$ for symmetric matrices and $$M(n)$$ for any general matrix in $$\mathbb{R}^{n}$$.
Remark: We know that, $$S(n)$$ forms a subspace under the vector space $$M(n)$$. Thus, we can calculate the dimension of this subspace explicitly. We know that to uniquely determine a symmetric matrix, it suffices to only list the entries in the upper triangle part of the matrix as well as the diagonal. Thus, to uniquely determine an element in $$S(n)$$, we need: $\frac{n^{2}}{2} + n = \frac{(n)(n+1)}{2}$ elements. Thus, $$S(n)$$ is of dimension $$\mathbb{R}^{k}$$ where $$k=\frac{(n)(n+1)}{2}$$.
This proof that $$O(n)$$ is a manifold under $$M(n)=\mathbb{R}^{n^{2}}$$ is a very clear application of the pre-image theorem so we state it first but before, we state that, we need to have a notion of a regular value. So we state these two things now before attempting to prove anything:
What is a regular value?
Let $$f: X\to Y$$ be a smooth map between manifolds. Then $$y$$ is a regular value of $$f$$ if and only if every for all $$x\in f^{-1}(y)$$, $$df_{x}: T_{x}(X) \to T_{y}(Y)$$ is surjective where $$y = f(x)$$.
What is the pre-image theorem?
If $$y$$ is a regular value of $$f: X\to Y$$, then the preimage $$f^{-1}(y)$$ is a submanifold of $$X$$, with dim $$f^{-1}(y) =$$ dim $$X$$ - dim $$Y$$.
Thus, for our purposes, let $$f: M(n) \to S(n)$$ via $$A \mapsto AA^{T}$$. Since for any orthogonal matrix $$Q$$, $$QQ^{T} = I$$, it follows that we want to determine if the preimage $$f^{-1}(I)$$ is a submanifold of $$M(n)$$. Thus, it leaves us to check that $$I$$ is a regular value of $$f$$ or in other words, it leaves us to check that for any $$A\in f^{-1}(I)$$ (so any orthogonal matrix), $$df_{A}: T_{A}(M(n)) \to T_{f(A)}(S(n))$$ is surjective. First, note that for any $$B\in M(n)$$:
\begin{align} df_{A}(B) = \lim_{h\to 0} \frac{f(A+h)-f(A)}{h} = \lim_{h\to 0} BA^{T} + AB^{T} + hBB^{T} = BA^{T} + AB^{T} \end{align}
So, we want to check that for any matrix $$C\in S(n)$$, that, there exists $$B\in M(n)$$ such that $$df_{A}(B) = BA^{T} + AB^{T} = C$$. Note that since $$C$$ is symmetric, then $$\frac{1}{2}C + \frac{1}{2}C^{T} = C$$. Thus, if we suppose that $$BA^{T} = \frac{1}{2}C$$, then: $BA^{T} = \frac{1}{2}C \implies B = \frac{1}{2}CA$ so, \begin{align} BA^{T} + AB^{T} = (\frac{1}{2}CA)A^{T} + A(\frac{1}{2}CA)^{T} = \frac{1}{2}(C+C^{T}) = C \end{align}
So, we see that $$df_{A}: M(n) \to S(n)$$ is surjective for any $$A\in O(n)$$, proving that $$f^{-1}(I)=O(n)$$ is a manifold under $$M(n)$$.
In order, to show this is a lie group under the operation of matrix multiplication: $O(n) \times O(n) \to O(n)$ is a smooth operation, it suffices to show that if $$A,B\in O(n)$$, then $$AB$$ is an orthogonal matrix. Note that: $(AB)^{T}(AB) = B^{T}A^{T}(AB) = B^{T}B = I$ so that $$AB$$ is an orthogonal matrix. The inverse of this operation comes from the fact that: $A^{T} = A^{-1}$ so this set also fulfills the definition of a group with the additional property that its operation is smooth. Thus, $$O(n)$$ is a lie group! | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9614951610565186, "perplexity": 109.92561271740733}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2017-51/segments/1512948551162.54/warc/CC-MAIN-20171214222204-20171215002204-00216.warc.gz"} |
http://philpapers.org/rec/CENMRA | # Monotone reducibility and the family of infinite sets
Journal of Symbolic Logic 49 (3):774-782 (1984)
Abstract
Let A and B be subsets of the space 2 N of sets of natural numbers. A is said to be Wadge reducible to B if there is a continuous map Φ from 2 N into 2 N such that A = Φ -1 (B); A is said to be monotone reducible to B if in addition the map Φ is monotone, that is, \$a \subset b\$ implies \$\Phi (a) \subset \Phi(b)\$ . The set A is said to be monotone if a ∈ A and \$a \subset b\$ imply b ∈ A. For monotone sets, it is shown that, as for Wadge reducibility, sets low in the arithmetical hierarchy are nicely ordered. The ▵ 0 1 sets are all reducible to the (Σ 0 1 but not ▵ 0 1 ) sets, which are in turn all reducible to the strictly ▵ 0 2 sets, which are all in turn reducible to the strictly Σ 0 2 sets. In addition, the nontrivial Σ 0 n sets all have the same degree for n ≤ 2. For Wadge reducibility, these results extend throughout the Borel hierarchy. In contrast, we give two natural strictly Π 0 2 monotone sets which have different monotone degrees. We show that every Σ 0 2 monotone set is actually Σ 0 2 positive. We also consider reducibility for subsets of the space of compact subsets of 2 N . This leads to the result that the finitely iterated Cantor-Bendixson derivative D n is a Borel map of class exactly 2n, which answers a question of Kuratowski
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6 ( #203,263 of 1,100,753 ) | {"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": 0.8768436908721924, "perplexity": 1885.1066485686313}, "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/1410657118605.64/warc/CC-MAIN-20140914011158-00249-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://www.computer.org/csdl/trans/ts/1986/03/06312886-abs.html | Issue No. 03 - March (1986 vol. 12)
ISSN: 0098-5589
pp: 450-461
H. G. Perros , Department of Computer Science, North Carolina State University, Raleigh, NC 27695
Tayfur Altiok , Department of Industrial Engineering, Rutgers University, Piscataway, NJ 08854
ABSTRACT
An approximation procedure is developed for the analysis of tandem configurations consisting of single server finite queues linked in series. External arrivals occur at the first queue which may be either finite or infinite. Departures from the queueing network may only occur from the last queue. All service times and interarrivai times are assumed to be exponentially distributed. The approximation algorithm gives results in the form of the marginal probability distribution of the number of units in each queue of the tandem configuration. Other performance measures, such as mean queue-length and throughput, can be readily obtained. The approximation procedure was validated using exact and simulation data. The approximate results seem to have an acceptable error level.
INDEX TERMS
Servers, Approximation methods, Approximation algorithms, Queueing analysis, Throughput, Algorithm design and analysis, Delay, queues in tandem, Approximations, blocking, Coxian queues, exponential distribution, finite queues
CITATION
H. G. Perros, Tayfur Altiok, "Approximate analysis of open networks of queues with blocking: Tandem configurations", IEEE Transactions on Software Engineering, vol. 12, no. , pp. 450-461, March 1986, doi:10.1109/TSE.1986.6312886 | {"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": 0.923961877822876, "perplexity": 1691.2766716519998}, "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-2017-17/segments/1492917119356.19/warc/CC-MAIN-20170423031159-00160-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://www.arxiv-vanity.com/papers/1503.08522/ | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
# Hyperbolic Skyrmions
Thomas Winyard
Department of Mathematical Sciences,
Durham University, Durham, DH1 3LE, U.K.
###### Abstract
We investigate Skyrmions in hyperbolic space, by computing numerical solutions of the nonlinear field equation. We first demonstrate the link between increasing curvature and the accuracy of the rational map approximation to the minimal energy static solutions. We investigate the link between Skyrmions with massive pions in Euclidean space and the massless case in hyperbolic space, by relating curvature to the pion mass. Crystal chunks are found to be the minimal energy solution for increased curvature as well as increased mass of the model. The dynamics of the hyperbolic model are also simulated, with the similarities and differences to the Euclidean model noted.
## 1 Introduction
The Skyrme model [1] is a (3+1)-dimensional nonlinear theory of pions that admits topological soliton solutions, called Skyrmions, which represent baryons. This has been well studied [2] with solutions calculated for a large range of topological charges [3]. The addition of a mass term has little effect on solutions of low baryon number, which continue to form shell like structures. However for larger charge solutions, a mass term starts to favour minimal energy solutions formed of finite chunks of a Skyrme crystal [4, 5, 6].
It has been demonstrated that there is a surprising similarity between Skyrmions with massive pions in Euclidean space and the massless case in hyperbolic space [7]. The cited paper also outlines a method for constructing Skyrmions with massive pions from instanton holonomies, by first modelling a hyperbolic Skyrmion by taking holonomies along particular circles in [8] and applying a mapping relating hyperbolic curvature and Euclidean mass to produce the Euclidean Skyrmion [7].
This posits that there could be a geometrical underpinning to the standard mass term, traditionally used in the Skyrme model. This suggests that understanding Skyrmions in hyperbolic space and the affect that curvature has, may shed some light on Skyrmions with massive pions in Euclidean space. Most notably there are certain properties for Skyrmion solutions in Euclidean space, that only occur once the mass term is turned on, or exceeds a certain threshold. Namely, the formation of crystal chunk solutions, as the global minima, for higher charge systems that exceed the threshold mass. If some similar behaviour were to be observed for massless solutions in hyperbolic space, it would support this geometric link. In fact, it will be demonstrated that the map linking the curvature of hyperbolic Skyrmions with massive Euclidean solutions, can be used to predict the global minima solution.
We will also examine the dynamics of Skyrmions in hyperbolic space, demonstrating that they scatter along geodesics, with maximally attractive channels corresponding to a relative rotation through an angle , about an axis orthogonal to the connecting geodesic.
## 2 The Model
The Lagrangian density for an valued Skyrme field is given by,
L=−12Tr(RμRμ)+116Tr([Rμ,Rν][Rμ,Rν])−m2πTr(U−12) (2.1)
where is the right valued current. The associated energy for a static Skyrme field defined on a general Riemannian manifold with metric is
E=112π2∫{−12Tr(RiRi)−116Tr([Ri,Rj][Ri,Rj])+m2Tr(1−U)}√gd3x (2.2)
where is the determinant of the metric. Note that both of the above expressions have the parameters preceding the first two terms scaled out. is the tree-level mass of the pions, which can be observed by using the nature of the field and writing the equation in terms of pion fields , where is the triplet of Pauli matrices and the triplet of pion fields.
Much work has been done on the solutions to this equation for Euclidean space upto topological charge [3, 6]. However we are interested in considering Skyrmion solutions in hyperbolic 3-space , which is the space with constant negative curvature . The metric of takes the form,
ds2(H3κ)=dρ2+sinh2(κρ)κ2(dθ2+sin2θdϕ2). (2.3)
where is the hyperbolic radius. If we take the limit of zero curvature, we recover the Euclidean metric, with the hyperbolic radius equal to the standard Euclidean radius . We will also make use of the standard Poincare ball model for displaying results. This can be obtained from the above metric by a simple radial transformation , to give the following metric,
ds2(H3κ)=4(dR2+R2(dθ2+sin2θdϕ2))(1−κ2R2)2. (2.4)
Hence our space can be modelled by a sphere with a boundary at infinite hyperbolic radius given by (though our plots will always be scaled to an equivalent size).
The vacuum for the massless theory is any constant , however the inclusion of the mass term gives the unique vacuum to be . We will impose the boundary condition as , which is required for finite energy. This gives us a map , and hence a topological charge as an element of the 3rd homotopy group, equivalent to an integer ,
B=−124π2∫ϵijkTr(RiRjRk)d3x. (2.5)
## 3 Approximations
There are a few approximations for Skyrmions with massless and massive pions. The rational map approach will be the most useful in this paper. The angular dependence of the solution is approximated to be a rational map between Riemann spheres [9]. On extension to massive pion solutions, it is found that only shell-like approximations can be closely approximated. While multi-shell like solutions have been modelled in an attempt to form more crystal like solutions [10], they are poor approximations to the full minimal energy solutions. They can be useful for initial conditions in numerical simulations however.
### 3.1 B=1
In Euclidean and hyperbolic space the single Skyrmion solution can be reduced to solving an ODE, using the hedgehog ansatz. This is known as a hedgehog solution due to its radial nature, as can be seen in figure 1. The field is given to be
U=exp(if(ρ)^x⋅τ), (3.1)
where is the unit vector in Cartesian coordinates, is a monotonically decreasing radial profile function with boundary conditions and . Substituting this into the energy in (2.2) we get a radial energy of the form,
E=13π∫(f′2sinh2κρκ2+2(f′2+1)sin2f+κ2sin4fsinh2κρ+2m2sinh2κρκ2(1−cosf))dρ (3.2)
The profile function can then be found by minimising the above energy and is also shown in figure 1 for ,. This yields a function with an exponential asymptotic decay for ,
f∼Ae−2κρ. (3.3)
This takes a similar form to that of massive Euclidean Skyrmions () , but dependent on the curvature rather than the mass of the theory. This suggest a relation between curvature and mass. In fact it is found that if you select the correct curvature, you can produce an extremely similar profile function for any Skyrmion with massive pions in Euclidean space. See [7] to observe the graph showing the relation between and .
### 3.2 Shell-like multisolitons
Shell-like solutions can be well approximated by the rational map ansatz. In hyperbolic space this takes the following form,
U(ρ,z)=exp[if(ρ)1+|R|2(1−|R|22¯R2R|R|2−1)] (3.4)
where is the Riemann sphere coordinate and is a degree rational map between Riemann spheres. Substituting this ansatz into (2.2) we get the following radial energy,
E=13π∫(f′2sinh2(κρ)κ2+2B(f′2+1)sin2f+Iκ2sin4fsinh2(κρ)+2m2sinh2(κρ)κ2(1−cosf))dρ, (3.5)
where
I=14π∫(1+|z|21+|R|2∣∣∣dRdz∣∣∣)42idzd¯z(1+|z|2)2. (3.6)
is an integral to be minimised by the choice of rational map . Note that is independent of and hence the values match those in Euclidean space. The minimal values of and the associated rational maps can be found in [3] for a range of values of . Note that the earlier hedgehog ansatz is recovered for , where is the minimising map, with and (3.5) reduces to (3.2).
This approximation will be used in various way to form initial conditions for the numerical computations presented later. We will also investigate how curvature affects the accuracy of the approximation.
## 4 Static Solutions
### 4.1 Shell-like Static Solutions
The static equations that follow from the variation of (2.2) were solved using a time dependent th-order Runga-Kutta method to evolve the time-dependant equations of motion that follow from the relativistic lagrangian (2.1), cutting the kinetic energy whenever the potential increased. The grid was modelled using the Poincaré ball model of radius on a cubic grid with grid points and lattice spacing (for the standard ) . Spatial derivatives have been approximated using a th-order finite difference method. We must fix the boundary at to be the vacuum at spatial infinity , to ensure finite energy. For all our simulations the topological charge, when computed numerically, gives an integer value to five significant figures, indicating the accuracy of the results.
Two forms of initial condition were considered. The rational map ansatz shown in (3.4) and the product ansatz , which was used to place lower charge solitons at various well separated positions about the grid.
The first eight shell-like static solutions for can be seen in figure 2. These solutions take a similar form to the Euclidean solutions of the same charge, with a few subtle differences. The faces of the polyhedron now appear to take the form of geodesic surfaces (a surface that contains curves belonging to the set of geodesics within the global space). Additionally, translating the solutions about the grid alters the apparent shape and means that lines of symmetry fall along geodesics of the space. This can be seen in more detail in the analysis of the solution in figure 5. The crystal chunk solution clearly demonstrates a bowing of the line connecting the two solitons, this line is found to be a geodesic of the space.
If we look at the energies displayed in table 1 we can see the expected trend in energies for increasing charge. We also observe how the energy of a given charge solution scales with curvature in figure 3.
We now compare the approximation from the rational map ansatz to the minimal energy solution for topological charges to . The results for can be observed in figure 4. We note that the rational map gives a very good approximation up to . The fraction , where is the energy of the rational map approximation and is the full numerical minimal energy, seems to stay relatively constant throughout an increase in curvature. We can’t say if this trend will definitely continue, however if it does, then the rational maps will remain a good approximation for all values of curvature, as long as the solutions are shell-like, but the rational map approximation breaks down if the solutions begin to become non shell-like.
### 4.2 Crystal chunk Solutions
For the crystal chunk solutions we will consider a couple of cases, the and solutions. In Euclidean space we find that the solution needs a relatively high mass for the crystal chunk solution to become the global minima. This massive solution can be considered to be two Skyrmions, joined along an axis perpendicular to a face of the shape. They have relative rotation of around the axis joining the two solitons.
In figure 5 we observe that both the crystal chunk and shell-like solutions are attainable in hyperbolic space with . However, it appears that the crystal solution is the global minima for all non-zero curvatures considered. Note that the energies of the two solutions get very close and could be within numerical error of each other. The crystal chunk solution is the lowest charge crystal solution and hence the energy difference might not be discernible with our accuracy. It is possible that the non-shell like solution does in fact become the minimal energy solution, for higher values of the curvature. We will consider a higher charge solution where the energy difference will be more discernible. Also, if the crystal chunk solutions act as with increasing the mass term in Euclidean space, we may find that the crystal chunk solution would become the minimal energy solution for a lower curvature.
The crystal chunk solution, displayed in figure 6(b-c), has a far lower energy than that of the shell like solution in figure 6(a) for even low values of . This is also the case for a small mass term in the Euclidean model. Hence we have demonstrated that not only are the profile functions related for Skyrmions with massive pions in Euclidean space and with massless pions in hyperbolic space, but the energetically favourable form of solution is also similar.
## 5 Dynamics
The solutions to the time-dependant equations of motion that follow from (2.1) were again found using a time dependent th-order Runga-Kutta method. The grid was modelled using the Poincaré ball model of radius (fixing ) on a cubic grid with grid point, hence the lattice spacing . Spatial derivatives have been approximated using a th-order finite difference method. The product ansatz was used for well separated single charge solitons.
The simplest situation to consider is scattering along a geodesic that passes through the centre of the space, as seen in figure 7. This gives a straight geodesic, with a clear parallel to Euclidean space and hence an obvious attractive channel (rotate relative by around an axis perpendicular to the connecting straight line). The scattering process then proceeds as expected with the solitons scattering at .
Due to hyperbolic translations (elements of the isometry group of hyperbolic space) one would expect in general, single Skyrmions to follow geodesics until they scatter. After scattering, the emerging Skyrmions will follow alternate geodesics, oriented to the incident paths by a rotation of around an orthogonal axis. The maximal channel will be a rotation of one of the solitons relative to the other by around an orthogonal axis to the tangent of the connecting geodesic. On scattering, the Skyrmions should merge to form the standard solution, oriented to lie in the incident plane, however it may appear deformed due to the curvature of the space. The results presented here confirm these expectations and can be observed in figure 8.
## 6 Conclusions
We have found both static and dynamic solutions for hyperbolic Skyrmions of various curvature. The static solutions have been related to massive solutions in Euclidean space, by making use of the relation shown in [7]. It has been demonstrated that the link between curvature in hyperbolic space and mass in Euclidean space extends to full solutions of various topological charge, allowing predictions to be made for the type of solution that will occur in the two models.
We have supplied evidence that suggests the rational map approximation is a good approximation for increasing curvature. It seems to retain its accuracy regardless of the curvature considered. This would suggest that we can model Skyrmion solutions in the infinite curvature limit, by using their respective rational maps. This is analogous to the hyperbolic monopole case, where solutions for infinite curvature become rational maps [11]. It would be interesting to see if there were some interesting limit in which it produces exact solutions, that in some way corresponds to hyperbolic monopoles.
The dynamics of various soliton initial conditions have also been studied. The attractive channel was shown to be a relative rotation by around an axis orthogonal to the connecting geodesic.
It would be interesting to consider the form of a soliton crystal in hyperbolic space, due to the interesting symmetries and tilings that can be formed from various polyhedron. It would be sensible to start with the 2-dimensional analogue, due to the difficulty of the task. Some similar work has been done with 2-dimensional vortices in the hyperbolic plane, concentrating on the tiling with Schläfi symbol [12].
## 7 Acknowledgements
I would like to thank EPSRC for my PhD studentship. I would also like to thank my supervisor Paul Sutcliffe for useful discussions. | {"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": 0.9682254791259766, "perplexity": 458.64940059257395}, "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-2020-24/segments/1590348513230.90/warc/CC-MAIN-20200606093706-20200606123706-00594.warc.gz"} |
http://annals.math.princeton.edu/2005/162-2/p03 | # A new application of random matrices: Ext$(C^*_{\mathrm{red}}(F_2))$ is not a group
### Abstract
In the process of developing the theory of free probability and free entropy, Voiculescu introduced in 1991 a random matrix model for a free semicircular system. Since then, random matrices have played a key role in von Neumann algebra theory (cf. [V8], [V9]). The main result of this paper is the following extension of Voiculescu’s random matrix result: Let $(X_1^{(n)},\dots,X_r^{(n)})$ be a system of $r$ stochastically independent $n\times n$ Gaussian self-adjoint random matrices as in Voiculescu’s random matrix paper [V4], and let $(x_1,\dots,x_r)$ be a semi-circular system in a $C^*$-probability space. Then for every polynomial $p$ in $r$ noncommuting variables $\lim_{n\to\infty} \big\|p\big(X_1^{(n)}(\omega),\dots,X_r^{(n)}(\omega)\big)\big\| =\|p(x_1,\dots,x_r)\|,$ for almost all $\omega$ in the underlying probability space. We use the result to show that the $\mathrm{Ext}$-invariant for the reduced $C^*$-algebra of the free group on 2 generators is not a group but only a semi-group. This problem has been open since Anderson in 1978 found the first example of a $C^*$-algebra $\mathcal{A}$ for which $\mathrm{Ext}(\mathcal{A})$ is not a group.
## Authors
Uffe Haagerup
Department of Mathematics and Computer Science, University of Southern Denmark, 5230 Odense, Denmark
Steen Thorbjørnsen
Department of Mathematics and Computer Science, University of Southern Denmark, 5230 Odense, Denmark | {"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": 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": 0.8217275738716125, "perplexity": 230.84726712869258}, "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-2019-35/segments/1566027314696.33/warc/CC-MAIN-20190819073232-20190819095232-00119.warc.gz"} |
http://www.verycomputer.com/18_c56133ee245b4885_1.htm | natbib and references in TOC
natbib and references in TOC
I want to both use natbib and have the bibliography appear in the table of
contents... I know that tocbibind will put the bibliography into the TOC if
you put:
\usepackage[nottoc,notlof,notlot]{tocbibind}
into the preamble. Of course, it does this by changing the bibliography
environment. Unfortunately, natbib does its own changes to the bibliography
environment too, and if I load them in the "wrong" order then the
bibliography changes won't happen.
Of course, you say, why not just load natbib and then tocbibind? Well, I
would if I could, but here's the situation:
- I'm working on a University of Edinburgh thesis class
- One of the university thesis regulations is that the bibliography must
appear in the table of contents, so I'd like to build that into the class.
- However, if I put any sort of thebibliography-changing commands into the
class itself, then my \usepackage{natbib} will stomp on those changes and
renew the environment itself.
*sigh*. So as I see it, I've got two options:
- Explicitly RequirePackage natbib in the thesis class before loading
tocbibind or otherwise altering thebibliography... but then if the user
wants to load natbib again with different options, aren't they then
screwed?
- Put an \addcontentsline... into a skeleton sample file that will be
available with the class and hope that people don't take it out.
Anyone got any other suggestions?
Thanks,
MEF
--
Mary Ellen Foster, School of Cognitive Science, University of Edinburgh
--------------------- Law of Software Envelopment ---------------------
Every program attempts to expand until it can read mail.
natbib and references in TOC
> I want to both use natbib and have the bibliography appear in the table of
> contents... I know that tocbibind will put the bibliography into the TOC if
> you put:
> \usepackage[nottoc,notlof,notlot]{tocbibind}
> into the preamble. Of course, it does this by changing the bibliography
> environment. Unfortunately, natbib does its own changes to the bibliography
> environment too, and if I load them in the "wrong" order then the
> bibliography changes won't happen.
> Of course, you say, why not just load natbib and then tocbibind? Well, I
> would if I could, but here's the situation:
> - I'm working on a University of Edinburgh thesis class
> - One of the university thesis regulations is that the bibliography must
> appear in the table of contents, so I'd like to build that into the class.
> - However, if I put any sort of thebibliography-changing commands into the
> class itself, then my \usepackage{natbib} will stomp on those changes and
> renew the environment itself.
> *sigh*. So as I see it, I've got two options:
> - Explicitly RequirePackage natbib in the thesis class before loading
> tocbibind or otherwise altering thebibliography... but then if the user
> wants to load natbib again with different options, aren't they then
> screwed?
> - Put an \addcontentsline... into a skeleton sample file that will be
> available with the class and hope that people don't take it out.
> Anyone got any other suggestions?
It appears that you are the author of the class, which does give you
some advantages. I think that this will work, without having to use
anything from tocbibind, and also hopefully apply to any bibliography
environment. I'll assume that the class supports chapters and the
bibliography heading is not numbered. In your class file put:
\AtBeginDocument{%
Quote:}
where Bibname is the heading text for the bibliography. (You might have
to throw in a \protect or two, but try it first). The incantation could
be initially put into the preamble of a test document to get it working
before adding it to the .cls file (don't forget \makeatletter, etc).
The idea is that at the \begin{document}, after all packages have
been loaded, that the command \addcontentsline{toc}{chapter}{Bibname} is
appended to the then current definition of \bibliography. It should also
ensure that the ToC addition occurs on the same page as the start of the
bibliography.
This is a trick I have only learned recently, but it is limited in
that you can only add stuff to a pre-existing macro, not change any
internal kernel command, and could be changed at any time. I am
contemplating producing a package for doing this kind of thing.
Peter W.
natbib and references in TOC
Quote:> I want to both use natbib and have the bibliography appear in the table of
> contents...
I realised that I got a bit too specific in asking this question. What I
should have asked was: is there a way to add some text to a macro without
just copying the standard definition from report.cls or whatever and adding
your bit onto it. The answer is: yes. :-)
Thanks to Peter Wilson, I got put on the right track. His suggestion of
nice with macros that take arguments. But when I searched dejanews, I came
across a technique that does work. In my class file, I put:
\AtBeginDocument{%
\expandafter\def\expandafter\thebibliography\expandafter%
#\expandafter1\expandafter{\thebibliography{#1}%
\markboth{\bibname}{\bibname}}}
I didn't want to play with \bibliography because then this would only kick
in if the person using the class used bibtex; this way, if they explicitly
use a \begin{thebibliography}...\end{thebibliography} the commands will
still get added. So now the bibliography appears in the TOC and puts the
right marks in for the headers, however any other packages decide to futz
with the bibliography environment.
Here's a reference to the dejanews article that I got the above out of:
http://www.deja.com/[ST_rn=ps]/getdoc.xp?AN=635095381
Thanks for the quick and helpful responses!
MEF
--
Mary Ellen Foster, School of Cognitive Science, University of Edinburgh
--------------------- Law of Software Envelopment ---------------------
Every program attempts to expand until it can read mail.
Hello,
I'd like to stop natbib from creating a numbered references (bibliography)
list.
I created a citation style with custom-bib 4.00 to use with natbib 7.0.
The natbib documentation does not state how to redefine thebibliography
environment to suppress the numbering of references as it is standard.
Can someone help me with that?
Thanks a lot,
Stefan
---
University of Karlsruhe, Germany
Institute of Information Engineering and Management | {"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": 0.9342678785324097, "perplexity": 3796.738619594751}, "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-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00503.warc.gz"} |
http://mathhelpforum.com/pre-calculus/71457-few-calculus-problems.html | # Math Help - A few Calculus problems.
1. ## A few Calculus problems.
I have been working on my AP Calculus homework, and there are a few things I can't understand. Can you help? These are three, seperate, unrelated problems.
1) I need to graph and explain the graph of ln(ln(x))
2) For the graph of the function of f, where x = # of DVDs produced and y = Cost of production, what does f^-1(10) represent? (That's f-inverse(10), I wasn't sure if the way I typed it was correct or not.)
3) I was given the inverse function of F. Do I get F using the same method I would to get inverse of F only go the other way around? and is finding the F function how I should be finding f(1)?
If anybody helps, I really appreciate it. And I'm not specifically looking for answers, I would like an explanation of how to do the problems. I like knowing how and why instead of what the answer is. Thanks again!
2. Originally Posted by mcra7x
I have been working on my AP Calculus homework, and there are a few things I can't understand. Can you help? These are three, seperate, unrelated problems.
1) I need to graph and explain the graph of ln(ln(x))
so what have you done here? note that the domain of the graph would be $(1, \infty)$. why? and it will be in a somewhat similar in shape to ln(x) but it would increase less steeply. why?
2) For the graph of the function of f, where x = # of DVDs produced and y = f(x) = Cost of production, what does f^-1(10) represent? (That's f-inverse(10), I wasn't sure if the way I typed it was correct or not.)
do you understand the relationship between a function and its inverse?
say $f^{-1}(10) = x$, then we have $10 = f(x)$
now, can you say what $f^{-1}(10)$ was?
3) I was given the inverse function of F. Do I get F using the same method I would to get inverse of F only go the other way around?
yes
and is finding the F function how I should be finding f(1)?
after you find f, plug in 1
3. I still do not understand #2. I am not needing to know the value of f^-1(10) but what it represents.
How did you find the domain of number one? I thought the domain of ln was [0, infinity)
And how are you doing those math symbols? That's pretty cool.
The inverse of a function is a reflection over the line y=x ?
4. I have been working on my AP Calculus homework, and there are a few things I can't understand. Can you help? These are three, seperate, unrelated problems.
1) I need to graph and explain the graph of ln(ln(x))
Have you graphed it? Remember that it is not saying ln(x) times ln(x), this is a composite function, ie. f(g(x)) so you are evaluating ln(x) IN TERMS OF ln(x). This is a really cool question and I think if you look at the graph of this function for awhile you can come up with a good answer. Hint: Remember that a log is just an exponent. These kinds of questions are supposed to make you think, not get an answer from someone on a help forum. If you are in AP Calculus, you didn't get there by accident
2) For the graph of the function of f, where x = # of DVDs produced and y = Cost of production, what does $f^{-1}(10)$ represent?
Typically, the inverse of a cost vs. production graph, is the price per unit of what ever is being produced. Cost of production increases as more units must be produced; the more units they can sell means the lower the price of each unit can be. $f^{-1} (10)$ is asking you what is happening on the inverse graph of f(x) at x=10.
3) I was given the inverse function of F. Do I get F using the same method I would to get inverse of F only go the other way around? and is finding the F function how I should be finding f(1)?
If f is the inverse of g, then g is the inverse of f. This is an elementary property of inverse functions.
If anybody helps, I really appreciate it. And I'm not specifically looking for answers, I would like an explanation of how to do the problems. I like knowing how and why instead of what the answer is. Thanks again!
5. Originally Posted by mcra7x
I still do not understand #2. I am not needing to know the value of f^-1(10) but what it represents.
f(x) is the cost, x is the number of DVDs. i told you that saying $f^{-1}(10) = x$ is the same as saying $f(x) = 10$. now can you state what it represents?
How did you find the domain of number one? I thought the domain of ln was [0, infinity)
actually, zero is not included in the domain
thus, the domain of $\ln (\ln x)$ is all real $x$ so that $\ln x > 0$. $\ln x = 0$ for $x = 1$, thus we want $x > 1$
And how are you doing those math symbols? That's pretty cool.
see here.
The inverse of a function is a reflection over the line y=x ?
that's true. but that's not the relationship i was going for. Molly already said it, "if f is the inverse of g, then g is the inverse of f...." | {"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": 14, "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": 0.844855546951294, "perplexity": 364.68074506347034}, "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-2016-22/segments/1464049270513.22/warc/CC-MAIN-20160524002110-00052-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/6551-linear-algebra-proof.html | ## linear Algebra Proof
I'm not really sure how to prove this wondring if anyone could give me a hand
so let F be the fundamental matrix for x'=A(t)x, show that any other fundamental matrix let's say K for this system can be reprsented as K=FC where C is an invertable nxn matrix | {"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": 0.994059681892395, "perplexity": 656.6691446375003}, "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-2015-06/segments/1422121983086.76/warc/CC-MAIN-20150124175303-00005-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://codegolf.stackexchange.com/questions/195290/fermats-last-theorem-mod-n | # Fermat's Last Theorem, mod n
It is a well known fact that for all integers $$\p>2\$$, there exist no integers $$\x, y, z>0\$$ such that $$\x^p+y^p=z^p\$$. However, this statement is not true in general if we consider the integers modulo $$\n\$$.
You will be given $$\n\$$ and $$\p\$$, which are two positive integers with $$\n>1\$$. Your task will be to write a function or program to compute all positive integers $$\x, y, z such that $$\(x^p+y^p)\$$ and $$\z^p\$$ give the same remainder when divided by $$\n\$$.
# Input
Any reasonable method of input is allowed. E.g. two separate user inputs, ordered pair, two function parameters, etc.
# Output
Any reasonable method of output is valid, it may be produced by a function or output to the screen. The order the triples are listed does not matter. Triples such as (1, 2, 3) and (2, 1, 3) are considered distinct, and all distinct triples should be listed exactly once. No invalid/trivial triples such as (0, 0, 0) should be output.
The numbers $$\x, y, z\$$ may have any ordering within each of the triples, but that order should be consistent. For example if $$\2^p+2^p\$$ and $$\3^p\$$ have the same remainder when divided by $$\n\$$, you may list this triple as (2, 2, 3) or (3, 2, 2).
# Examples
n p -> Possible Output
----------------------------------------------------------------
2 3 -> []
3 3 -> [(1,1,2),(2,2,1)]
3 4 -> []
4 3 -> [(1,2,1),(1,3,2),(2,1,1),(2,2,2),(2,3,3),(3,1,2),(3,2,3)]
# Scoring
Shortest code in bytes with no standard loopholes wins.
• I guess we have to specify how triplets are ordered and be consistent.. E.g. All (x,y,z) or all (z,x,y) and so on – AZTECCO Nov 4 at 17:06
• @AZTECCO Good point, I will add that specification. – 79037662 Nov 4 at 17:08
• I'm not convinced that solutions with z=0 are trivial. – Neil Nov 4 at 21:22
• @Neil I didn't say they were, but they are invalid for the purposes of this question. – 79037662 Nov 4 at 21:37
• @RosLuP Correct. – 79037662 Nov 8 at 14:35
# Julia 1.0, 62 bytes
f(n,p,q=1:n-1)=[(x,y,z) for x=q,y=q,z=q if (x^p+y^p)%n==z^p%n]
Try it online!
• You can drop the space after if to save a byte. – Glen O Nov 8 at 1:21
# 05AB1E, 15 10 bytes
-2 bytes by Kevin Cruijssen
-3 bytes by Grimmy
<L3ãʒImƹÖ
Try it online!
Returns a triples in the format [z,y,x].
Explanation:
<L | Push the interval [1,n)
3ã | Push the triple cartesian product [1,n) x [1,n) x [1,n)
| E.g. with n = 3, the list begins as follows:
| [[1, 1, 2], [1, 2, 1], [2, 1, 3], ...]
ʒ | Filter the list:
Im | Take the p-th power of each input
| E.g. [1, 1, 2] --> [1^p, 1^p, 2^p]
Æ | Reduce by subtraction
| E.g. [1^p, 1^p, 2^p] --> 1^p - 1^p - 2^p
¹Ö | Check if 1^p - 1^p - 2^p = 0 (mod n), and filter out if not.
• The specification says the order of the numbers in the triple does not matter (as long as it's consistent). Is the í necessary? – 79037662 Nov 4 at 18:38
• Ops! Yes, then it is not needed. Good catch, thanks. – Wisław Nov 4 at 18:43
• 0Q can be _ to save a byte. – Kevin Cruijssen Nov 5 at 12:37
• 13 bytes by outputting as triplets in the order [z,y,x]. – Kevin Cruijssen Nov 5 at 12:46
• 10 bytes – Grimmy Nov 5 at 15:50
# APL (Dyalog Unicode), 19 bytes
{⍸0=⍵|-/¨⍺*⍨⍳3⍴⍵-1}
The triples are returned in the format x z y. Try it online!
### Explanation
{⍸0=⍵|-/¨⍺*⍨⍳3⍴⍵-1} ⍝ n is our right argument, ⍵, and p is our left, ⍺.
3⍴⍵-1 ⍝ First, we get a triple of (n-1, n-1, n-1).
⍳ ⍝ Then we get the Cartesian product of three range[1, n-1).
⍝ As you will see, this gives a list of (x z y) triples.
⍺*⍨ ⍝ We take each element of each triple to the power of p.
-/¨ ⍝ Here we subtract over each triple, using (element - rest)
⍝ This gives us (x^p - (z^p - y^p)) = x^p + y^p - z^p.
0= ⍝ Then we check which (x^p + y^p - z^p) equals 0.
⍸ ⍝ And use these results, a Boolean array,
⍝ to find get the indices of the correct triples,
⍝ which are our (x z y)s.
# Jelly, 13 bytes
*ŒH§%³E
Ṗṗ3çƇ
A full program accepting command line arguments n p which prints a list representation of the results (Do note that the empty list is represented as no output).
Try it online!
### How?
*ŒH§%³E - Link 1: evaluate a triple: [x, y, z], p
* - exponentiate [x^p, y^p, z^p]
ŒH - split into halves [[x^p, y^p], [z^p]]
§ - sums [x^p+y^p, z^p]
³ - program's 1st argument n
% - modulo [(x^p+y^p)%n, (z^p)%n]
E - all equal? (x^p+y^p)%n == (z^p)%n
Ṗṗ3çƇ - Main Link: n, p
Ṗ - pop (implicit range of n) [1,2,3,...,n-1]
3 - literal three 3
ṗ - Cartesian power [[1,1,1],[1,1,2],...,[n-1,n-1,n-1]]
Ƈ - filter keep those for which:
- implicit print
## GAP, 63 bytes
Filtered(Tuples([1..n-1],3),i->(i[1]^p+i[2]^p-i[3]^p) mod n=0);
For example:
gap> n:=4; p:=3;
4
3
gap> Filtered(Tuples([1..n-1],3),i->(i[1]^p+i[2]^p-i[3]^p) mod n=0);
[ [ 1, 2, 1 ], [ 1, 3, 2 ], [ 2, 1, 1 ], [ 2, 2, 2 ], [ 2, 3, 3 ],
[ 3, 1, 2 ], [ 3, 2, 3 ] ]
# Octave, 74 72 bytes
2 bytes saved thanks to @79037662
@(n,p){[c,b,a]=ndgrid(1:n-1) [a(k=~mod(a.^p+b.^p-c.^p,n)) b(k) c(k)]}{2}
Try it online!
• Do you need the extra pair of parentheses in a((k=~mod(a.^p+b.^p-c.^p,n)))? – 79037662 Nov 4 at 17:14
• @79037662 No, they are not needed. They were a leftover from some (unsuccessful) tests to try to get rid of the cell array. Thanks! – Luis Mendo Nov 4 at 17:22
# JavaScript (ES7), 104 bytes
Takes input as (n)(p).
n=>p=>[...Array(n**3)].map((_,i)=>[i/n/n|0,i/n%n|0,i%n]).filter(([x,y,z])=>x*y*z&&(x**p+y**p)%n==z**p%n)
Try it online!
# R, 67 bytes
function(n,p,m=t(expand.grid(a<-2:n-1,a,a)))m[,!c(1,1,-1)%*%m^p%%n]
Try it online!
expand.grid creates all possible triplets of numbers. Annoyingly, the output of expand.grid is a data frame and not a matrix, so I transpose it to coerce to a matrix m. Take the matrix product of [1,1,-1] with m^p to get $$\x^p+y^p-z^p\$$, then keep only the rows for which we get $$\ 0 \mod n\$$.
• nice, I tried a different approach and only got down to 68 – MickyT Nov 4 at 17:58
• @MickyT Nice idea using which with 3d indices! Here is a 65 byte version of yours, which you should post separately as it is substantially different. – Robin Ryder Nov 4 at 19:18
# R, 65 bytes
3 btye's saved from my original idea by @robin-ryder by using subtraction, applying modulo the result.
function(n,p,q=(2:n-1)^p)which(!outer(outer(q,q,'+'),q,'-')%%n,T)
Try it online!
Makes use of outer to build a 3d array of the addition, then the subtraction. If the remainder of n is 0 output the array indices using which.
# Wolfram Language (Mathematica), 54 bytes
n_~f~p_:=Select[Range[n-1]~Tuples~3,n|{1,1,-1}.#^p&]
Named function taking the two inputs, like f[4,3]. The | symbol typed above needs to be replaced by the 3-byte character representing the "divides" relation; \[Divides] works as well, and is used in the TIO example.
Works by testing every possible triple {x,y,z}. The only detail of note is that the dot product {1,1,-1}.#^p&, when applied to {x,y,z}, calculates x^p+y^p-z^p.
Try it online!
# Ruby, 67 66 bytes
->n,p{w=*1...n;w.product(w,w).select{|a,b,c|(a**p+b**p-c**p)%n<1}}
Try it online!
Thanks Chas Brown for -1 byte.
• Save a byte with (a**p+b**p-c**p)%n<1 – Chas Brown Nov 5 at 21:29
• I never knew you could pass multiple arrays to product! Very nice. – IMP1 Nov 6 at 10:09
# Python 2, 94 bytes
n,p=input()
R=range(1,n)
print[(x,y,z)for x in R for y in R for z in R if(x**p+y**p-z**p)%n<1]
Try it online!
Pretty basic.
# Icon, 95 bytes
procedure f(n,p)
x:=1to(t:=n-1)&y:=1to t&z:=1to t&(x^p+y^p-z^p)%n=0&write(x," ",y," ",z)&\q
end
Try it online!
# Jelly, 13 bytes
*+_ƭ/³ḍ
Ṗṗ3çƇ
Try it online!
-2 bytes thanks to caird coinheringaahing
-1 byte borrowing a clever trick from Jonathan Allan
# Explanation
*+_ƭ/³ḍ Helper Link - given (x, y, z), determine if x^p + y^p = z^p (mod n)
* (x^p, y^p, z^p)
/ Reduce over
+_ƭ Tied operator: cycle between + and - (x^p + y^p - z^p)
³ḍ Divisibility check by n (if x^p + y^p - z^p is divisible by n then x^p + y^p is congruent to z^p mod n)
Ṗ Pop (remove last element) - on an integer, implicit range (1 to n-1)
ṗ3 Cartesian power ^ 3
çƇ Filter: keep triples that are valid by the helper link
# APL(NARS), chars 129, bytes 258
r←f w;n;p;a;v;i;j;b
(n p)←w⋄a←n∣p*⍨v←⍳n-1⋄r←⍬
:for i :in v⋄:for j :in v
:if a∊⍨b←n∣a[i]+a[j]⋄r←r,⊂i,j,a⍳b⋄:endif
:endfor⋄:endfor
test:
⎕fmt f 2 3
┌0─┐
│ 0│
└~─┘
⎕fmt f 3 3
┌2────────────────┐
│┌3─────┐ ┌3─────┐│
││ 1 1 2│ │ 2 2 1││
│└~─────┘ └~─────┘2
└∊────────────────┘
⎕fmt f 3 4
┌0─┐
│ 0│
└~─┘
⎕fmt f 4 3
┌7─────────────────────────────────────────────────────────────┐
│┌3─────┐ ┌3─────┐ ┌3─────┐ ┌3─────┐ ┌3─────┐ ┌3─────┐ ┌3─────┐│
││ 1 2 1│ │ 1 3 2│ │ 2 1 1│ │ 2 2 2│ │ 2 3 3│ │ 3 1 2│ │ 3 2 3││
│└~─────┘ └~─────┘ └~─────┘ └~─────┘ └~─────┘ └~─────┘ └~─────┘2
└∊─────────────────────────────────────────────────────────────┘
# K (oK), 30 23 bytes
-7 bytes thanks to ngn!
{(~x!-/*/y#)#+1+!3#x-1}
Try it online!
• (*/y#)'' -> */y#, – ngn Nov 10 at 9:28
• using func#list as "filter": {(~x!-/*/y#)#+1+!3#x-1} – ngn Nov 10 at 9:32
• @ngn Filter, of course! Thanks! – Galen Ivanov Nov 10 at 9:36
# Zsh, 90 bytes
l=({1..$[$1-1]}\ )
for x y z (${=${:-$^l$^l$^l}})(((x**$2+y**$2-z**$2)%$1))||<<<"$x $y$z"
Try it online!
Creates the list l=('1 ' '2 ' '3 ' ... '(n-1) ' ), then ${=${:-$^l$^l\$^l} gets the Cartesian product and splits on spaces. | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.893013060092926, "perplexity": 161.66320853256957}, "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-2019-47/segments/1573496668699.77/warc/CC-MAIN-20191115171915-20191115195915-00163.warc.gz"} |
http://www.emn.fr/z-info/sdemasse/gccat/Calldifferent.html | ## 5.5. alldifferent
DESCRIPTION LINKS GRAPH AUTOMATON
Origin
Constraint
$\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)$
Synonyms
$\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏}$, $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}$, $\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}$, $\mathrm{𝚋𝚘𝚞𝚗𝚍}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$, $\mathrm{𝚋𝚘𝚞𝚗𝚍}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏}$, $\mathrm{𝚋𝚘𝚞𝚗𝚍}_\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}$, $\mathrm{𝚛𝚎𝚕}$.
Argument
$\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$ $\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)$
Restriction
$\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}$$\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)$
Purpose
Enforce all variables of the collection $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$ to take distinct values.
Example
$\left(〈5,1,9,3〉\right)$
The $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint holds since all the values 5, 1, 9 and 3 are distinct.
Typical
$|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>2$
Symmetries
• Items of $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$ are permutable.
• Two distinct values of $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}$ can be swapped; a value of $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}$ can be renamed to any unused value.
Usage
The $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint occurs in most practical problems directly or indirectly. A classical example is the n-queen chess puzzle problem: Place $n$ queens on a $n$ by $n$ chessboard in such a way that no queen attacks another. Two queens attack each other if they are located on the same column, on the same row or on the same diagonal. This can be modelled as the conjunction of three $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraints. We associate to the ${i}^{th}$ column of the chessboard a domain variable ${X}_{i}$ that gives the row number where the corresponding queen is located. The three $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraints are:
• $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}\left({X}_{1},{X}_{2}+1,...,{X}_{n}+n-1\right)$ for the upper-left to lower-right diagonals,
• $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}\left({X}_{1},{X}_{2},...,{X}_{n}\right)$ for the rows,
• $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}\left({X}_{1}+n-1,{X}_{2}+n-2,...,{X}_{n}\right)$ for the lower right to upper-left diagonals.
They are respectively depicted by parts (A), (C) and (D) of Figure 5.5.1.
A second example taken from [AsratianDenleyHaggkvist98] when the bipartite graph associated with the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint is convex is a ski assignment problem: “a set of skiers have each specified the smallest and largest skis they will accept from a given set of skis”. The task is to find a ski for each skier.
Examples such as Costas arrays or Golomb rulers involve one or several $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraints on differences of variables.
Quite often, the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint is also used in conjunction with several $\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝}$ constraints, specially in the context of assignment problems [pages 372–374][Hooker07book].
Other examples involving several $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraints sharing some variables can be found in the Usage slot of the $𝚔_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint.
Remark
Even if the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint had not this form, it was specified in ALICE [Lauriere76], [Lauriere78] by asking for an injective correspondence between variables and values: $x\ne y⇒f\left(x\right)\ne f\left(y\right)$. From an algorithmic point of view, the algorithm for computing the cardinality of the maximum matching of a bipartite graph was not used for checking the feasibility of the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint, even if the algorithm was already known in 1976. This stands from the fact that the goal of ALICE was to show that a general system could be as efficient as dedicated algorithms. For this reason the concluding part of [Lauriere76] explicitly mentions the fact that specialized algorithms should be discarded. On the one hand, many people, specially from the OR community, have complained about such radical statement [Roy06]. On the other hand, the motivation of such statement stands from the fact that a truly intelligent system should not rely on black box algorithms, but should rather be able to reconstruct them from some kind of first principle. How to achieve this is still an open question.
Some solvers use in a pre -processing phase before stating all constraints, an algorithm for automatically extracting large cliques [BronKerbosch73] from a set of binary disequalities in order to replace them by $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraints.
W.-J. van Hoeve provides a survey about the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint in [Hoeve01].
For possible relaxation of the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraints see the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}$, the $𝚔_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ (i.e., $\mathrm{𝚜𝚘𝚖𝚎}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$
Within the context of linear programming, relaxations of the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint are described in [WilliamsYan01] and in [Hooker07book].
Within the context of constraint-centered search heuristics, G. Pesant and A. Zanarini [ZanariniPesant07b] have proposed several estimators for evaluating the number of solutions of an $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint (since counting the total number of maximum matchings of the corresponding variable -value graph is #P -complete [Valiant79]). Faster, but less accurate estimators, based on upper bounds of the number of solutions were proposed three years later by the same authors [ZanariniPesant10].
Given $n$ variables taking their values within the interval $\left[1,n\right]$, the total number of solutions of the corresponding $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint corresponds to the sequence A000142 of the On -Line Encyclopedia of Integer Sequences [Sloane10].
Algorithm
The first complete filtering algorithm was independently found by M.-C. Costa [Costa94] and J.-C. Régin [Regin94]. This algorithm is based on a corollary of C. Berge that characterises the edges of a graph that belong to a maximum matching but not to all [Berge70].A similar result is in fact given in [Petersen1891]. A short time after, assuming that all variables have no holes in their domain, M. Leconte came up with a filtering algorithm [Leconte96] based on edge finding. A first bound-consistency algorithm was proposed by Bleuzen-Guernalec et al. [GuernalecColmerauer97]. Later on, two different approaches were used to design bound-consistency algorithms. Both approaches model the constraint as a bipartite graph. The first identifies Hall intervals in this graph [Puget98], [LopezOrtizQuimperTrompBeek03] and the second applies the same algorithm that is used to compute arc-consistency, but achieves a speedup by exploiting the simpler structure [Glover67] of the graph [MehlhornThiel00]. Ian P. Gent et al. discuss in [GentMiguelNightingale08] implementations issues behind the complete filtering algorithm and in particular the computation of the strongly connected components of the residual graph (i.e., a graph constructed from a maximum variable -value matching and from the possible values of the variables of the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint), which appears to be the main bottleneck in practice.
From a worst case complexity point of view, assuming that $n$ is the number of variables and $m$ the sum of the domains sizes, we have the following complexity results:
• Complete filtering is achieved in $O\left(m\sqrt{n}\right)$ by Régin's algorithm [Regin94].
• Range consistency is done in $O\left({n}^{2}\right)$ by Leconte's algorithm [Leconte96].
• Bound-consistency is performed in $O\left(nlogn\right)$ in [Puget98], [MehlhornThiel00], [LopezOrtizQuimperTrompBeek03]. If sort can be achieved in linear time, typically when the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint encodes a permutation,In this context the total number of values that can be assigned to the variables of the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint is equal to the number of variables. Under this assumption sorting the variables on their minimum or maximum values can be achieved in linear time. the worst case complexity of the algorithms described in [MehlhornThiel00], [LopezOrtizQuimperTrompBeek03] goes down to $O\left(n\right)$.
Within the context of explanations [JussienBarichard00], the explanation of the filtering algorithm that achieves arc-consistency for the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint is described in [Rochart05]. Given the residual graph (i.e., a graph constructed from a maximum variable -value matching and from the possible values of the variables of the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint), the removal of an arc starting from a vertex belonging to a strongly connected component ${𝒞}_{1}$ to a distinct strongly connected component ${𝒞}_{2}$ is explained by all missing arcs starting from a descendant component of ${𝒞}_{2}$ and ending in an ancestor component of ${𝒞}_{1}$ (i.e., since the addition of any of these missing arcs would merge the strongly connected components ${𝒞}_{1}$ and ${𝒞}_{2}$). Let us illustrate this on a concrete example. For this purpose assume we have the following variables and the values that can potentially be assigned to each of them, $A\in \left\{1,2\right\}$, $B\in \left\{1,2\right\}$, $C\in \left\{2,3,4,6\right\}$, $D\in \left\{3,4\right\}$, $E\in \left\{5,6\right\}$, $F\in \left\{5,6\right\}$, $G\in \left\{6,7,8\right\}$, $H\in \left\{6,7,8\right\}$. Figure 5.5.2 represents the residual graph associated with the maximum matching corresponding to the assignment $A=1$, $B=2$, $C=3$, $D=4$, $E=5$, $F=6$, $G=7$, $H=8$. It has four strongly connected components containing respectively vertices $\left\{A,B,1,2\right\}$, $\left\{C,D,3,4\right\}$, $\left\{E,F,5,6\right\}$ and $\left\{G,H,7,8\right\}$. Arcs that are between strongly connected components correspond to values that can be removed:
• The removal of value 2 from variable $C$ is explained by the absence of the arcs corresponding to the assignments $A=3$, $A=4$, $B=3$ and $B=4$ (since adding any of these missing arcs would merge the blue and the pink strongly connected components containing the vertices corresponding to value 2 and variable $C$).
• The removal of value 6 from variable $C$ is explained by the absence of the arcs corresponding to the assignments $E=3$, $E=4$, $F=3$ and $F=4$. Again adding the corresponding arcs would merge the two strongly connected components containing the vertices corresponding to value 6 and variable $C$.
• The removal of value 6 from variable $G$ is explained by the absence of the arcs corresponding to the assignments $E=7$, $E=8$, $F=7$ and $F=8$.
• The removal of value 6 from variable $H$ is explained by the absence of the arcs corresponding to the assignments $E=7$, $E=8$, $F=7$ and $F=8$.
After applying bound-consistency the following property holds for all variables of an $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint. Given a Hall interval $\left[l,u\right]$, any variable $V$ whose range $\left[\underline{V},\overline{V}\right]$ intersects $\left[l,u\right]$ without being included in $\left[l,u\right]$ has its minimum value $\underline{V}$ (respectively maximum value $\overline{V}$) that is located before (respectively after) the Hall interval (i.e., $\underline{V}).
The $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint is entailed if and only if there is no value $v$ that can be assigned two distinct variables of the $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$ collection (i.e., the intersection of the two sets of potential values of any pair of variables is empty).
Reformulation
The $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint can be reformulated into a set of disequalities constraints. This model neither preserves bound-consistency nor arc-consistency:
• On the one hand a model, involving linear constraints, preserving bound-consistency was introduced in [BessiereKatsirelosNarodytskaQuimperWalsh09IJCAI]. For each potential interval $\left[l,u\right]$ of consecutive values this model uses $|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|$ 0-1 variables ${B}_{1,l,u},{B}_{2,l,u},...,{B}_{|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|,l,u}$ for modelling the fact that each variable of the collection $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$ is assigned a value within interval $\left[l,u\right]$ (i.e., $\forall i\in \left[1,|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|\right]:{B}_{i,l,u}⇔\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\left[i\right].\mathrm{𝚟𝚊𝚛}\in \left[l,u\right]$),How to encode the reified constraint ${B}_{i,l,u}⇔\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\left[i\right].\mathrm{𝚟𝚊𝚛}\in \left[l,u\right]$ with linear constraints is described in the Reformulation slot of the $\mathrm{𝚒𝚗}_\mathrm{𝚒𝚗𝚝𝚎𝚛𝚟𝚊𝚕}_\mathrm{𝚛𝚎𝚒𝚏𝚒𝚎𝚍}$ constraint. and an inequality constraint for enforcing the condition that the sum of the corresponding 0-1 variables is less than or equal to the size $u-l+1$ of the corresponding interval (i.e. ${B}_{1,l,u}+{B}_{2,l,u}+...+{B}_{|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|,l,u}\le u-l+1$).
• On the other hand, it was shown in [BessiereKatsirelosNarodytskaWalsh09] that there is no polynomial sized decomposition that preserves arc-consistency.
Systems
Used in
generalisation: $\mathrm{𝚊𝚕𝚕}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚍𝚒𝚜𝚝}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚕𝚒𝚗𝚎}\mathrm{𝚜𝚎𝚐𝚖𝚎𝚗𝚝}$, all of the same size), $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚋𝚎𝚝𝚠𝚎𝚎𝚗}_\mathrm{𝚜𝚎𝚝𝚜}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚜𝚎𝚝}\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$), $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚌𝚜𝚝}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}+\mathrm{𝚌𝚘𝚗𝚜𝚝𝚊𝚗𝚝}$), $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚒𝚗𝚝𝚎𝚛𝚟𝚊𝚕}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}/\mathrm{𝚌𝚘𝚗𝚜𝚝𝚊𝚗𝚝}$), $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚖𝚘𝚍𝚞𝚕𝚘}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}\mathrm{mod}\mathrm{𝚌𝚘𝚗𝚜𝚝𝚊𝚗𝚝}$), $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚙𝚊𝚛𝚝𝚒𝚝𝚒𝚘𝚗}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}\in \mathrm{𝚙𝚊𝚛𝚝𝚒𝚝𝚒𝚘𝚗}$), $\mathrm{𝚍𝚒𝚏𝚏𝚗}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by orthotope), $\mathrm{𝚍𝚒𝚜𝚓𝚞𝚗𝚌𝚝𝚒𝚟𝚎}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚝𝚊𝚜𝚔}$), $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}$ (control the number of occurrence of each value with a counter variable), $\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}$ (control the number of occurrence of each value with an interval), $\mathrm{𝚕𝚎𝚡}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ ($\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}$ replaced by $\mathrm{𝚟𝚎𝚌𝚝𝚘𝚛}$), $\mathrm{𝚗𝚟𝚊𝚕𝚞𝚎}$ (count number of distinct values).
Keywords
Arc input(s)
$\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$
Arc generator
$\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}$$↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1},\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}\right)$
Arc arity
Arc constraint(s)
$\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚟𝚊𝚛}=\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚟𝚊𝚛}$
Graph property(ies)
$\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}$$\le 1$
Graph class
$\mathrm{𝙾𝙽𝙴}_\mathrm{𝚂𝚄𝙲𝙲}$
Graph model
We generate a clique with an equality constraint between each pair of vertices (including a vertex and itself) and state that the size of the largest strongly connected component should not exceed one.
Parts (A) and (B) of Figure 5.5.3 respectively show the initial and final graph associated with the Example slot. Since we use the $\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}$ graph property we show one of the largest strongly connected component of the final graph. The $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ holds since all the strongly connected components have at most one vertex: a value is used at most once.
Automaton
Figure 5.5.4 depicts the automaton associated with the $\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}$ constraint. To each item of the collection $\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}$ corresponds a signature variable ${𝚂}_{i}$ that is equal to 1. The automaton counts the number of occurrences of each value and finally imposes that each value is taken at most one time. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 227, "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": 0.8695182204246521, "perplexity": 547.5684303600129}, "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-2014-23/segments/1405997894799.55/warc/CC-MAIN-20140722025814-00142-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/30868/subspaces-of-finite-codimension-in-banach-spaces/30875 | # Subspaces of finite codimension in Banach spaces
Is every finite codimensional subspace of a Banach space closed? Is it also complemented? I know how to answer the same questions for finite dimensional subspaces, but couldn't figure out the finite codimension case.
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See mathoverflow.net/questions/28415 in particular Nate's ticked answer. – Robin Chapman Jul 7 '10 at 10:34
Here's a fun exercise: take $E$ to be a Banach space and let $H$ be a codimension one subspace. Prove that $H$ is dense in $E$ if and only if $E \setminus H$ is arcwise connected. – Laurent Berger Jul 7 '10 at 10:56
Nice exercise, Laurent. It embarrasses me to learn something about Banach spaces from a number theorist. :) – Bill Johnson Jul 7 '10 at 13:28
This is in Rudin's FA, Chapter 2 or 3, isn't it? – Yemon Choi Jul 7 '10 at 19:18
It's a standard result that a linear functional from a Banach space to the underlying field (real or complex numbers) is continuous if and only if the its kernel is closed. Notice that its kernel is of codimension one. So, use the axiom of choice to find a discontinuous linear functional, and you have found a codimension one subspace which isn't closed. (As I was typing this, rpotrie got the same answer...)
As for complementation: well, this only makes sense for closed finite codimension subspaces. But then it's a perfectly reasonable question, and the answer is "yes". If F is of finite codimension in E, then by definition we can find a basis $\{x_1,\cdots,x_n\}$ for E/F. For each $k$ let $x_k^*$ be the linear functional on $E/F$ dual to $x_k$, so $x_k^*(x_j) = \delta_{jk}$. Then let $\mu_k$ be the composition of $E \rightarrow E/F$ with $x_k^*$. Finally, pick $y_k\in E$ with $y_k+F=x_k$. Then the map $$T:E\rightarrow E; x\mapsto \sum_k \mu_k(x) y_k$$ is a projection of $E$ onto the span of the $y_k$, and $I-T$ will be a projection onto $F$ (unless I've messed something up, which is possible).
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to be precise, the question about complementation makes sense for any linear subspace, why. I would rather say: "a complemented linear subspace is necessarily closed". – Pietro Majer Jul 7 '10 at 20:08
No: If you consider a non continuous functional from a Banach Space, its Kernel is one-codimensional and dense.
For example take $l^2(\mathbb{Z})$ and consider the sequence $e_i$ ($(0,..., 1, 0....)$ where the $1$ is in the $i$-th position). Complete this to a base (which exists by Zorn's lemma, and it is uncountable since $l^2$ is a banach-space) and consider the subspace generated by the $e_i$ toghether with the elements of the base except one of them. This gives a dense codimension one subspace.
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This is very related to rpotries construction: take a dense, proper subspace and pick a basis $(v_i)$ of that subspace. Now we can adjoin $(w_j)$ such that $(v_i) \cup (w_j)$ is a basis of the whole space. Now the span of all $v_i,w_j$ but finitly many $w_j$ is a dense subspace of finite codimension. So it can not be closed.
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+1: This seems to be a distillation of the core content of the previous answers. (In fact, after reading them I came to the same conclusion.) Moreover: by the Baire Category Theorem, an algebraic (or "Hamel") basis of an infinite dimensional Banach space is uncountable, whereas -- by definition! -- a separable Banach space admits a dense subspace of countable dimension. So nonclosed codimension one subspaces exist in every separable Banach space. Can someone address the non-separable case? – Pete L. Clark Jul 7 '10 at 12:25
Pete: Just use my comment (or rpotrie's) about non-continuous linear functionals: that works in any Banach (or indeed, otherwise) space. Sure, you need to use the Axiom of Choice a bit, but that's true in the separable case as well (try to write down an everywhere defined, dis-continuous functional on $\ell^2$, for example). – Matthew Daws Jul 7 '10 at 12:32
@MD: You're right. I didn't stop to think (and am not expert on Banach spaces) so didn't see that it is obvious that any infinite dimensional Banach space admits an unbounded linear functional: just choose an infinite linearly independent set $\{e_n\}$ of norm one elements, define $L(e_n) = n$ and extend to the whole space by AC. – Pete L. Clark Jul 7 '10 at 12:57
As already recalled, a kernel of any non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension as a consequence of the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:
A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a linear bounded operator, is closed.
Btw, the property of being complemented has also a particular characterization for those subspaces that are images of operators: the image of $R:X\to Y$ is complemented if and only if $R$ is a right inverse, meaning that there is a bounded operator $L:Y\to X$ such that $LR=I$. A linear projector onto the subspace it is then $RL$.
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It is equi to Choice axiom. If you do not like the Axiom, then every linear functional is continuous. If Not, then Hahn-Banach theorem is true.
It is up to you.
Oleg Reinov
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Not "equivalent" just an implication one way. In ZF, Hahn-Banach is strictly weaker than Axiom of Choice. – Gerald Edgar Jul 12 '10 at 11:26
Another counterexample: Take $C[-1,1]$ as the Banach Space. Define $T(f)=\int_{0}^{1} f - \int_{-1}^{0} f$. Clearly $T$ is a (continuous) linear functional whose kernel is a (closes) one-codimensional subspace. But there is no complementary subspace: in fact, the complementary subspace should morally be generated by the function which is identically $1$ on $[0,1]$ and $-1$ on $[-1,0]$, but this function is not continuous.
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I don't agree: pick some function g with T(g)=1. (g(x)=x+1 works I think). Then define a map $P:C[-1,1]\rightarrow C[-1,1]$ by $P(f) = f - T(f)g$. If $T(f)=0$ then $P(f)=f$, while also $TP(f) = T(f) - T(f)T(g) = 0$ for all $f$. So $P$ is a projection onto the kernel. It seems like you are trying to treat $C[-1,1]$ as an inner-product space (dense in $L^2[-1,1]$ for example) and then trying to pick an orthogonal projection: indeed, this does not exist. – Matthew Daws Jul 7 '10 at 11:18
you are right. This is a functional which is not of inner product with any element in the pre-Hilbert space (C[-1,1] considered with the $l^2$ norm). – Keivan Karai Jul 7 '10 at 12:35 | {"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": 0.951586902141571, "perplexity": 256.5898949252647}, "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-2015-11/segments/1424936464123.82/warc/CC-MAIN-20150226074104-00016-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/chi-square-problem.177958/ | # Chi square problem
1. Jul 23, 2007
### chisqaw
Hi ll, I have been asked to review a new image retrieval technique that has improved the retrieval performance compared with the old technique, ie, the hypothesis is that retrieval performance is influenced by retrieval techniques. The following table summarises the retrieval results. (They examined the first 300 retrieved images, 250 are relevant to the query for the new technique and 175 are relevant for the old technique)
Image relevant? Type of techniques Total
New technique Old technique
Yes 250 175 425
No 50 125 175
Total 300 300 600
I was using this distribution at http://davidmlane.com/hyperstat/chi_square_table.html but I am having trouble calculating the Chi Square value assuming the required p value is 0.01 and interpreting the result. Would someone be able to do this and explain how it was doing, or point to a decent tutorial? I only need this done, not to understand it, at least not past what is required.
Many thanks,
Lawrence
2. Jul 23, 2007
### EnumaElish
Last edited: Jul 23, 2007
3. Jul 24, 2007
### chisqaw
I dont understand that at all...
4. Jul 24, 2007
### EnumaElish
You are evaluating the "null" hypothesis "technique does not matter" against alternative hypo. "technique makes a difference." To evaluate it, you need to calculate a test statistic, then compare it against the chi-square table; where degrees of freedom = 1 and level of significance = 0.01. Unless I made a mistake, the corresponding value on the chi-square table is 6.6349.
If the result of your calculation:
(actual - predicted)^2/predicted, summed over the two categories
turns out to be > 6.635, then the test is telling you that there is a significant difference between the actual "scores" and the predicted "scores," at the 0.01 level of significance. You should reject the hypothesis that "the technique does not make difference" if your calculation is > 6.635.
________Category 1___Category 2
actual = ____ 250 ______ 175
predicted = 425/2 ______ 425/2
(I have assumed that only the first row of your table is relevant for this calculation.)
Do you see the similarity between the test-stat. in your case and the example test stat. (comparing men and women) on the Wikipedia page?
Last edited: Jul 24, 2007
5. Jul 25, 2007
### Barmecides
But you do not know the exact frequency of the old technique. So I would suggest to tests :
H0 : there is no difference between the two techniques
As you observe a large number of events, you can assume gaussian distribution and use a Pearson chi2 test like :
chi2 = (250 - 175)^2/(250 + 175) = 13.2
6. Jul 25, 2007
### EnumaElish
(a - b)^2/(a + b) = [(a - y)^2 + (b - y)^2]/y where y = (a + b)/2, therefore Barmecides's formula will produce an identical result with the one I posted (which is identical to the formula on the Wikipedia page).
Last edited: Jul 25, 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": 0.859952449798584, "perplexity": 1349.0419366224833}, "config": {"markdown_headings": true, "markdown_code": false, "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-2017-13/segments/1490218189252.15/warc/CC-MAIN-20170322212949-00100-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/498/guess-who-it-is?tab=activity&sort=all&page=3 | Guess who it is.
Reputation
48,080
112/100 score
Jul 15 comment Second derivative numerical estimate - stability and approach …nice only if the underlying function is smooth. If OP's discrete samples have some error in them, Richardson does no good. Jul 15 revised Why is the Gamma function off by 1 from the factorial? edited tags Jul 15 comment Derivation of approximation of Error function I'm now telling you to try the economization procedure. Again: Maclaurin, convert to Chebyshev, truncate, and go back to the monomials. You should then see something like what Mr. Hastings had, or better. Abramowitz and Stegun should have the table for converting monomials to Chebyshev. Jul 15 comment Derivation of approximation of Error function Well, binomial won't get you anywhere near Mr. Hastings. If you'd tried looking up the things I told you to look up, you'd have already read about the "equi-ripple" and "minimax" properties of a Chebyshev approximation… Jul 15 comment Derivation of approximation of Error function Well, have you tried expanding the transformed function as a Maclaurin series, and re-expressing the monomials in Chebyshev terms? (As I said earlier, look up Chebyshev economization.) Jul 15 comment Derivation of approximation of Error function With respect to Winitzki: the reasoning is that you want to build a rational function (or any approximant, really) whose qualitative behavior at $0$ and $\infty$ is similar to $\mathrm{erf}(x)$; polynomials certainly cannot do that, since they cannot exhibit asymptotes as in this case. Jul 15 comment Derivation of approximation of Error function Ah, that parameter I am not too sure of; Cecil Hastings was (in)famous for coming up with clever approximations that seemed to have just come from thin air. Jul 15 comment Conditions on Matrix invertibility I was nudging you to search for them yourself, as I have now given you the words you should be looking for. Jul 15 comment Derivation of approximation of Error function Let's call the polynomial in your expression $p(t)$; you can rearrange your formula to obtain $\exp(x^2)\mathrm{erfc}(x)\approx p(t)$. Replace the $x$ on the left with an expression in terms of $t$. That is the function which you will be approximating as a series in Chebyshev polynomials. You might want to look up the literature on this, including "economization" of series. Jul 15 comment Derivation of approximation of Error function That was one of his papers; there was another one that specially dealt with $\mathrm{erf}$. Jul 15 comment Conditions on Matrix invertibility There are methods for updating the QR decomposition and singular value decomposition when a new column or row is added to the original matrix; you might want to look them up. Jul 15 comment Derivation of approximation of Error function These are, if memory serves, Chebyshev fits of a transformed version of the error function. There are nicer and more analytically tractable approximations now; you might want to search for the work of Serge Winitzki on this. Jul 15 revised Derivation of approximation of Error function edited tags Jul 15 comment Logarithmic derivative of Polygamma functions Well, your logarithmic derivative is merely $\frac{\psi^{(k+1)}(x)}{\psi^{(k)}(x)}$; as to why one would be interested in ratios of (shifted, generalized) harmonic numbers, I've no idea. Jul 15 revised Logarithmic derivative of Polygamma functions edited tags Jul 14 comment How to learn cryptography Wouldn't crypto.SE be a better place for this? Jul 14 revised What's the area of the shape defined by all points whose distances from two focal points multiply to give the same product? edited tags Jul 14 comment What's the area of the shape defined by all points whose distances from two focal points multiply to give the same product? What you have called a "multiplicoid" is classically called a Cassinian oval, after the astronomer. The link I have given has formulae for the area of Cassinian ovals depending on shape. In your particular case, the formula involves the complete elliptic integral of the second kind. Jul 14 comment Find the roots of this 6th degree polynomial Well, you know what the three cube roots of $1$ look like, no? You can multiply those with the cube root you already have. Jul 14 comment Divergent series whose terms converge to zero In any event: have you seen this? | {"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": 0.9031419157981873, "perplexity": 625.0209879267695}, "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-2015-35/segments/1440644065828.38/warc/CC-MAIN-20150827025425-00280-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/53624/tangent-space-of-cotangent-bundle-at-zero-section | # Tangent space of Cotangent bundle at zero section?
Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.
How can I prove that $T_{(p,0)}T^*M$ is naturally isomorphic to $T_pM\oplus T_pM^*$?
If this true, then I think I could prove that the Hessian of $f\colon M\to \mathbb{R}$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$.
This is not any homework.
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Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition. – t.b. Jul 25 '11 at 10:49
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection. – Sam Lisi Dec 22 '11 at 13:00
Things are much nicer i.e. more general and canonical than that!
Consider any vector bundle $\pi:V\to M$ on $M$. You then have the canonical exact sequence of vector bundles on $V\;$ (yes, vector bundles on a vector bundle!) :
$$0\to T_{vert}(V) \to T(V) \stackrel {d\pi}{\to} \pi^*T(M) \to 0 \quad (*)$$
Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $\pi: V\to M$. In order that both tangent bundles of $V$ and $M$ live on $V$, you have to pull back $T(M)$ to $V$: that is why we have $\pi^* T(M)$ on the right. The kernel is the set of vertical tangent vectors to $V$, those that lie along the fibers of $V$.
Now if you restrict (*) to the zero section of $V$, identified to $M$, you get
$$0\to V \to T(V)|M \stackrel {d\pi}{\to} T(M) \to 0 \quad (**)$$
The only point worth mentioning is the identification of the vector bundles $T_{vert}(V)|M$ and $V$. It boils down to the fact that for a vector space $E$ the tangent bundle $T_a(E)$ at any point $a\in E$ is canonically isomorphic to $E$ itself.
Needless to say, taking the fibers of the bundles in (**) at a point of $p\in M$ will give you what you wanted if you take $V=T^*(M)$.
- | {"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": 0.9649073481559753, "perplexity": 150.9344572503797}, "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/1406510267865.20/warc/CC-MAIN-20140728011747-00033-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://mathhelpforum.com/geometry/152803-unknown-angles.html | 1. ## Unknown Angles
In this question will x be equal to 53 and y=?
2. If the 53° is the lower right angle in the upper triangle (hard to tell with your diagram), then x = 53° would be incorrect. The lower left angle in the upper triangle would be 74°, which means that the supplementary angle for x is also 74°, so x = 106°. y = 360° - 53° = 307°. | {"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": 0.9232836961746216, "perplexity": 723.3881617124259}, "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-2017-13/segments/1490218193288.61/warc/CC-MAIN-20170322212953-00292-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://iecfusiontech.blogspot.com/2007_05_01_archive.html | ## Wednesday, May 30, 2007
### Physical Constants
Charge of the electron = 1.60217653E-19 C
Mass of the electron = 9.1093826E-28 g
Mass of the electron in amus = 5.4858990945E-4
Mass of the electron in Mev = 0.510998918 Mev
Mass of the proton = 1.67262171E-24 g
Mass of the proton in amu = 1.00727646688
Mass of the proton in Mev = 938.272029 Mev
Mass of 1 AMU = 1.66053886E-24 grams
Plank's Constant = 6.62606896E-34 J * sec
Plank's Constant = 4.13566733E-15 ev *sec
Room temperature atmosphere = 2.6867773E19 molecules/cm3
Energy 1 ev = 1.60217653E-19J/ev
Boltzman's Constant = 1.3806505E-23J/deg K
temperature deg K. = 11,604.505 times ev
Speed of light in vacuum = 299.792458E6 m/sec
I will be adding to this as requirements dictate and time allows.
## Tuesday, May 29, 2007
### Polywell - Making The Well
I have come across some interesting research by Kiyoshi Yoshikawaa,of the Institute of Advanced Energy, Kyoto University, and others proving the formation of the Polywell.
For those of you who have not been following along here is how my understanding has been evolving.
Polywell As I Currently Understand It
A schematic of the evolution of the Polywell design can be found in slides 8 and 11 in this Power Point slide show (link at bottom of page).
In the Hirsh/Farnsworth machine in slide 8 the reacting positive ions (like charged Deuterium particles for one kind of operation) are attracted to the center to collide and produce fusions.
In the Elmore/Tuck/Watson machine in slide 11 electrons are accelerated to the center of the machine where they form a grid sort of like what happens in a beam power tube. In the beam power tube the virtual grid is called a space charge. These negative electrons attract the positive fuel ions and fusion reactions take place. The advantage is that there is no grid near the reaction space so losses are reduced.
That is the theory any way. However, in any person's mind who has a little understanding of the physics involved the question is: is that really happening? Are we fooling ourselves? Which brings us back to the Yoshikawa paper. What is the evidence?
Yoshikawaa correctly states the central issue:
...it is essential to clarify the mechanism of potential well formation (see Fig. 5) predicted to develop in the central plasma core within the cathode, since potential well formation due to space charge associated with spherically converging ion beams plays a key and essential role in the beam-beam colliding fusion, i.e., the major mechanism of the IECF devices. Actually, this has been the central key issue for IECF researchers for the past 30 years, until the first successful direct measurement of the double-well potential profile in the IECF device through the laser- induced fluorescence (LIF) method at Kyoto University [6] in 1999 with an approximately 200 V dip at the center in the helium plasma core as will be described below.
So they have proved the formation of the Polywell. Outstanding!
Many theoretical results so far predicted strongly localized potential well formation, and actually for the past 30 years, many experiments were dedicated to clarify this mechanism using, such as, electron beam reflection method [7], spatially collimated neutron [4] or proton [8,9] profile measurements, or an emissive probe [10], as is seen in Table 2, but, neither seems to be perfectly conclusive in convincing that well does form.
He again hits the nail on the head. Lots of results that could have more than one interpretation. He then gives a list of past attempts at verification of the Polywell. Now let us get to how what he claims was the definitive experiment was done.
...we have adopted optical diagnostics by using the Stark effects, sensitive to the local electric fields, to the IECF device with a hollow cathode. Also to enhance S/N (signal to noise) ratio as well as to specify radial potential profile, we introduced the LIF method. Consequently, we could have finally measured the double-well potential profile (see Fig. 11) with an approximately 200 V dip at the center for the first time in the helium plasma core (Fig. 7) in the IECF device.
He goes on in even more technical detail. The end result? The dual (cathode and anode) potential well forms.
In any future experimental regimes such a measuring system should be used to verify machine operation and to provide machine diagnostics.
### Deriving Operating Voltage For An IEC Reactor
I have had some complaints (justifiable in my opinion) on how I derived the operating voltage for a given fusion reactor to within a few percent. I dont't take into account Einstein since the particle energies we are talking about for a proton are on the order of .6 Mev maximum and a proton "weighs" around 900 Mev. It might matter in the fifth place decimal (I estimate about a 1 part in 300,000 difference) but we don't really need to know the voltage that close since creation of the polywell will cost us around 10% of the drive voltage, indicating that any reasonable supply should be capable of about the calculated voltage +20% or maybe +25% to give a little more margin.
Well off to the races. First let me define the terms.
m1 = mass of the lowest mass particle amu (protons)
m2 = the other mass in amu
c1 = the charge of m1 in electron charge units
c2 = the charge of m2 in electron charge units
AV = accelerator voltage in KV
TV = the voltage from a chart or graph where one particle is stationary
V1 = the velocity of m1 when accelerated by AV
V2 = the velocity of m2 when accelerated by AV
V1tv = the velocity of m1 when accelerated by TV
V1 + V2 = V1tv
When deriving the equations the = sign from now on will mean exactly proportional to.
m1 * V12 = c1 * AV
m2 * V22 = c2 * AV
m1 * V1tv2 = c1 * TV
Since we are using Newtonian mechanics velocities add:
V12 = c1 * AV / m1
V1 = √( c1 * AV / m1 )
V22 = c2 * AV / m2
V2 = √( c2 * AV / m2 )
V1tv = √( c1 * TV / m1 )
I'm going to skip a few obvious steps here for brevity.
√( c1 * TV / m1 ) =
√ AV * ( √( c1 / m1 ) + √( c2 / m2 ))
√ AV = (√( c1 * TV / m1 )) / ( √( c1 / m1 ) + √( c2 / m2 ))
Which is in a spread sheet availabe at the IEC Fusion Newsgroup as an attachment.
## Monday, May 21, 2007
### Operating Voltage For B11
I want to work out the accelerating voltage for the B11 proton reaction.
From this excellent discussion of fusion reactions [pdf] we find that the optimum voltage for the p-B11 reaction when accelerating protons into B11 is 550 KV. Representing a proton energy of 550 Kev. However because we are accelerating protons and B11 from the same virtual grid voltage some adjustments must be made.
First off the reaction runs based on velocity not energy so we have to adjust the accelerating voltage to account for that.
The energy of a non-relativistic particle = 1/2 mV2 However to keep from cluttering up the calculation we will throw out the 1/2 factor since we are not concerned with actual energy but just input voltage and for that the 1/2 cancels out. Masses will be given in Atomic Mass Units where the proton or neutron = 1. Also since we are not concerned with actual values but values relative to 550 Kev, energy will be given as Kev.
mV2=550Kev
Since the mass of a proton = 1 that simplifies to the required velocity = √550,000 = 741.6198 in arbitrary units.
For a given accelerating voltage the energy of the B11 will be 5X the energy of the proton because B11 has 5X the charge of a proton.
Ep=Vp2
EB11=11*VB2
Vp + VB = √550,000
Since the energy of the B11 is 5X the energy of the p then
5*Vp2 = 11*VB2
Vp2 = (11/5)*VB2
Vp = (√(11/5))*VB
Vp + VB = √550,000
substituting:
(√(11/5))*VB + VB = √550,000
calculating the result
VB = 298.650
Vp = 442.970
The energy of the proton will be:
442.9702 = 196,222.175 or an accelerating voltage of about 200KV
The energy of the boron will be:
11 * 298.6502 = 981,110.048
which when divided by 5 = 196,222.010 which checks within the accuracy of the numbers used.
Total energy required will be 1,177,332 ev or about 1.177 Mev
## Saturday, May 19, 2007
### The Purpose Of IEC Fusion Technogy
The purpose of this blog is to collect and disseminate technical information on the design of components and systems for IEC Fusion. | {"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": 0.84870445728302, "perplexity": 2056.3836332706214}, "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-2015-27/segments/1435375097475.46/warc/CC-MAIN-20150627031817-00237-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://neurotechnics.com/blog/quickly-generate-large-test-files-in-windows/ | Often as a developer, for testing disk performance, or perhaps your internet or network connection speed, you need to generate a large file of a specific size. Windows includes a utility that allows you to quickly generate a file of any size instantly.
This is a command line (DOS) utility, so firstly, open a command prompt (as an administrator if you need to write to protected folders).
To launch a command prompt, hit start and just type `command`. You'll see "Command Prompt" appear as the closest match (Remember to right-click and choose "Run as Administrator" if you need admin priviledges). Alterntaively, use the key combination: `[WIN]`+`R`, and type `cmd.exe` to launch a command prompt.
Run the following command:
``````fsutil file createnew <file> <size in bytes>
``````
For example, this command will create a 1GB file called `1gb.test` in my temp folder:
``````fsutil file createnew c:\temp\1gb.test 1073741824
``````
Note: the size of the file is specified in bytes. So, you'll need to be specific about the value you choose if you need a specific size.
Googling the converted value (e.g. `100gb in bytes`) will give you a result, but to avoid answers with exponents, here are some pre-calculated equivalent values:
SizeSize in Bytes
1 MB1048576
100 MB104857600
1 GB1073741824
10 GB10737418240
100 GB107374182400
1 TB1099511627776
10 TB10995116277760
If you don't like using the command line there are various GUI utilities out there that can also do this for you:
Dummy File Creator: http://www.mynikko.com/dummy/ | {"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": 0.8578293919563293, "perplexity": 4223.7967727724235}, "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-2019-18/segments/1555578530176.6/warc/CC-MAIN-20190421040427-20190421062427-00525.warc.gz"} |
http://physics.stackexchange.com/questions/25959/how-can-the-date-of-lunar-eclipses-be-calculated-by-hand?answertab=active | # How can the date of lunar eclipses be calculated by hand?
How can the date of lunar eclipses be calculated? Especially without the aid of a computer.
-
What sorts of eclipses are you interested in? Solar? Lunar? Eclipses of the moons of Jupiter? Etc. Each has its own opportunities for simplifications. (Ahh - I see the tag "moon". I'll add that to the text for clarity....) Next, do you want just the date, or the time also, or more information than that? – nealmcb Jan 16 '12 at 2:16
Lunar eclipses are much easier to calculate than solar ones, and it has been done over the ages via a variety of methods. E.g. there is evidence that the design of Stonehenge embodied some aspects of calculating eclipses.
If we can assume that you have information on previous lunar eclipses at hand, you'll observe patterns. One of the most useful for this purpose is the pattern of near-repetition of eclipses after a Saros (Wikipedia), which is a period of approximately 6585.3213 days, or nearly 18 years 11 days. One saros after an eclipse, the Sun, Earth, and Moon return to approximately the same relative geometry, and a nearly identical eclipse will occur, in what is referred to as an eclipse cycle. That is because a saros is a nearly whole number of repeats of each of three cycles of the lunar orbit: the synodic month, the draconic month, and the anomalistic month. See the article for details.
After decades of observation (or via access to tables like the List of Saros series for lunar eclipses or the Five Millennium Canon of Lunar Eclipses: -1999 to +3000), you'll have a list of the currently active saros series. It is then easy to extend any particular saros series by hand, using the information in the saros article. If there is a particular date you're interested in, we could work through an example here.
-
Perhaps this pdf will help. Which starts:
The calculation of eclipses is chiefly a combination of geometry and orbital mechanics. Hence, it is necessary to understand the orbital mechanics involved and also the underlying geometry.
The end of the pdf warns:
I assume that you are not trying to do an eclipse calculation exactly. If so, the problem becomes much more complicated, as you have to take into account the perturbations of all planets, the exact relation between θp, θu and the radii R1, R2, etc. In addition, you will need the initial condition of the Earth-Moon-Sun system, which you can obtain over the internet.
-
The ancient Greeks were able to predict eclipses long before Kepler formulated his descriptions of planetary motion, i.e., orbital mechanics for the purpose of discussion. Nor did they have the data that the PDF mentions is necessary. – dotancohen Jan 17 '12 at 8:39
@dotancohen exactly. The time span between lunar and solar eclipses seem to be so long that it is amazing that the ancients ever remembered the exact date of three of them to realise that there was a pattern. – Stuart Woodward Jan 22 '12 at 12:23
@Stuart: Additionally, the eclipses happen at 120 degree differences over the face of the Earth, not in the same place. I have a hard time believing that that information could be used to construe any correlations. – dotancohen Jan 22 '12 at 22:04 | {"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": 0.8149842619895935, "perplexity": 841.116665973402}, "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-2015-14/segments/1427131305143.93/warc/CC-MAIN-20150323172145-00144-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://www.nature.com/articles/nature05738?error=cookies_not_supported&code=3ddfc5b8-506c-4961-945e-a7f76fe5fb61 | ## Main
In the high-temperature superconductors, oxygen-isotope exchange has been found to have a very small effect on the Tc, particularly for optimally doped (highest-Tc) samples2. It was therefore unexpected when Gweon et al. reported a large isotope effect on the electronic structure in optimally doped Bi2Sr2CaCu2O8+δ using angle-resolved photoemission spectroscopy (ARPES)1. This isotope effect was seen mainly at a broad, high-energy hump feature and was unusually large (10–40 meV shift), by contrast with the expected isotopic phonon shifts of the order of 3 meV. It was argued by many that this is experimental evidence that the coupling between the lattice and electrons plays a particularly significant role.
Figure 1a shows our energy–momentum (Ek) dispersion relations for both 16O and 18O taken along the 'nodal' direction of the Brillouin zone. The results show an insignificant shift for the deeper states of −0.9 ± 0.4 meV, by contrast with Fig. 1b of Gweon et al.1, which shows a 15-meV shift of more deeply lying states. We found that precise rotational alignment of the samples was crucial to obtain repeatable energy positions, especially in the high-energy regime discussed here. The negligible energy shift was confirmed by using three different facilities, samples with multiple doping levels, including the optimal doping described in ref. 1, and with various photon energies, including the same energy used by Gweon et al.1 (33 eV), and a low photon energy (7.0 eV) that gives extreme resolution, as well as increased bulk sensitivity.
Gweon et al.1 show an even larger effect (up to 40 meV) at the 'antinodal' portions of the Brillouin zone (towards the (0,π) and (0,−π) points) — comparable data are shown here in Fig. 1b. We studied ten cuts that had a similar geometry to the key cut 7 of Gweon et al.1, with half of the cuts at positive ky (momentum along the y direction) values and the others at negative ky. Having data on both sides of the (0,0) point allowed the correction of any very slight (0.1 degrees) sample misalignments, which is crucial for removing systematic alignment errors that could cause apparent energy shifts. Figure 1b (right) shows the dispersion of the bottom of the band (determined in each cut as in Fig. 2d of Gweon et al.1) along the (0,−π)→(0,π) symmetry line. The ten sets in Fig. 1b have an average isotope shift of the band bottom of 2 ± 3 meV, which is inconsistent with the 10–40 meV level found by Gweon et al.1.
Our result is not inconsistent with the tunnelling result of a conventional phononic isotope shift on the scale of about 3 meV (refs 3,4) that was performed on the same samples and which conceivably could be seen near the kink energy in future ultra-resolution ARPES.
Although we do not observe the unusual large-scale isotope effect reported by Gweon et al.1, our result does not invalidate electron–phonon coupling as a potential pairing mechanism for high-temperature superconductivity. Rather, it implies that if electron–phonon coupling is responsible, it must do it in a more subtle, roundabout way.
Methods. Our data were measured with 33-eV photons at BL12 of the Advanced Light Source, Berkeley, USA, by use of a high-precision 6-axis sample manipulator. Single crystals of Bi1.9Sr2.1CaCu2O8+δ were annealed in either 18O2 or 16O2 gas under the same thermal conditions: 800 °C at 1 atmosphere for 168 h, followed by 700 °C at 0.2 bar for 24 h. The Tc values of the samples decreased from 92 K to 91 K on isotope substitution. Raman spectra showed a clear shift of the B1g vibration modes, confirming that isotope substitution was nearly complete at 80% or more. | {"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": 0.817468523979187, "perplexity": 2539.382340093925}, "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/1664030334992.20/warc/CC-MAIN-20220927064738-20220927094738-00474.warc.gz"} |
http://www.trigcalc.com/ | # Right Triangle Calculator
Our right triangle calculator is simple, easy and free to use for students, teachers, and anyone needing help with this math problem.
Area:
Perimiter:
Angle A:
°
Angle B:
°
Angle C:
°
Side A:
Side B:
Side C:
Area:
Perimiter:
Sin A:
Csc A:
Cos A:
Sec A:
Tan A:
Cot A:
Sin B:
Csc B:
Cos B:
Sec B:
Tan B:
Cot B:
Enter two known values (at least one being a side)
area:
a:
b:
c:
A: °
B: °
90 °
Only Two Values Required
## Get Trigonometry Help
### How to Use this Calculator
Offering comprehensive calculator tools & resources, TrigCalc was created for – but not limited to – trigonometry students seeking extra help or understanding of the subject. We have a number of different calculators (below) that take your step-by-step to solve common math problems.
## Get Trigonometry Help
### How to Use this Calculator
Offering comprehensive calculator tools & resources, TrigCalc was created for – but not limited to – trigonometry students seeking extra help or understanding of the subject. We have a number of different calculators (below) that take you step-by-step to solve common math problems.
## What is a 90 Degree Triangle?
Everything in trigonometry seems to revolve around the 90-degree triangle and its ratios. A 90 degree triangle is defined as a triangle with a right angle, or in other words, a ninety degree angle. Given any known side length of a 90-degree triangle and one other value (another side, angle, area value, etc), one can find all unknown values of the same 90 degree triangle.
## Right Triangle Calculator: Explanation
Right triangle calculator exemplifies this truth by taking in two values (one side and any other value) and returning all missing values in exact value and decimal form while displaying the formulas and calculation process for each of the missing values. The right triangle calculator is simple and easy to use for students, teachers, and anyone needing help with their math. Try it out!
## How to Use this Calculator
enter one side (in “a,” “b” or “c”) + any other value, and the calculator will return all missing values in exact value and decimal form – while also displaying the formulas and calculation process for each of the missing values.
### Oblique Triangles (any NON-90 degree triangle)
An oblique triangle is defined as a triangle without a right angle or ninty degree angle. Given three known values of an oblique triangle with one of those values being a side length, all other unknown values of the same triangle can be calculated. Oblique triangles use a set of formulas unique from right triangles and these formulas can be displayed on the oblique triangle calculator page of our website. An oblique triangle calculator takes in one side length and any two other values and returns the missing values in exact value and decimal form in addition to the step-by-step calculation process for each of those missing values.
### Circles & Sectors
#### Circle Calculator: Radius, Diameter, Circumference, Area
A circle is defined as a round figure whose boundary consists of points each equidistant from the center point. Circles consist of four segments, the radius, diameter, circumference, and area. If just one of these values is known, all of the other values can be calculated. With our circle calculator tool, you can do just that. This takes in any one known value of the circle and returns all missing values in exact value and decimal form as well as the step-by-step calculation process for each one of those values.
#### Sector Calculator: Angle, Radius, Chord, Arc, Area
A sector is defined as the area between two radiuses and the connecting arc of a circle. The five major parts of a sector are the angle, radius, chord, arc, and area. Given either an angle measure or radius length and any other value (two in total), all other values of the sector can be calculated. In the sector calculator, you can enter two values (at least one being an angle or radius) and receive all unknown values in exact value and decimal form and the step-by-step process for each missing value.
### Identity Calculator
For a list of trigonometric identities, the identity reference page displays all of them. If seeking to calculate identities, TrigCalc includes identity calculators for 6 of the trigonometric identities including: In each of the identity calculators, given any function value and quadrant of theta, the exact value and calculation process will be displayed. Decimal approximations for each of the trigonometric identities can be calculated as well.
### Angle Calculators
#### Coterminal Angles
Coterminal angles are angles sharing the same terminal and initial sides. Calculating coterminal angles involves adding or subtracting multiples of 360 degrees. Our coterminal angle calculator allows you to find either the least positive or negative values coterminal with a given angle in degree or radian form. This calculator also returns exact values, decimal approximations, and the step-by-step calculation process of the angle.
#### Supplementary Angles
Supplementary angles are angles whose sum equals 180 degrees. In the supplementary angle calculator, the supplement of any degree or radian value will be calculated and displayed alongside the calculation process and decimal approximation.
#### Complementary Angles
Complementary angles are angles whose sum equals ninety degrees. In a complementary angle calculator, the supplement of any degree or radian value will be calculated and displayed alongside the calculation process and decimal approximation. All calculators are very necessary math solvers as well as right triangle calculators.
#### Double Angles
The double angle formula calculator is a great tool if you'd like to see the step by step solutions of the sine, cosine and tangent of double a given angle. | {"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": 0.8198428750038147, "perplexity": 1221.99254871825}, "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/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00451.warc.gz"} |
http://www.varsitytutors.com/hspt_math-help/how-to-find-the-measure-of-an-angle | # HSPT Math : How to find the measure of an angle
## Example Questions
← Previous 1
### Example Question #1 : How To Find The Measure Of An Angle
What is the sum of the interior angles of a triangle?
Explanation:
The sum of the three interior angles of a triangle is degrees.
### Example Question #2 : How To Find The Measure Of An Angle
Two of the interior angles of a triangle measure and . What is the greatest measure of any of its exterior angles?
It cannot be determined from the information given.
Explanation:
The interior angles of a triangle must have measures whose sum is , so the measure of the third angle must be .
By the Triangle Exterior-Angle Theorem, an exterior angle of a triangle measures the sum of its remote interior angles; therefore, to get the greatest measure of any exterior angle, we add the two greatest interior angle measures:
### Example Question #1 : Lines
Two angles are supplementary and have a ratio of 1:4. What is the size of the smaller angle?
Explanation:
Since the angles are supplementary, their sum is 180 degrees. Because they are in a ratio of 1:4, the following expression could be written:
### Example Question #311 : Geometry
In a given triangle, the angles are in a ratio of 1:3:5. What size is the middle angle?
Explanation:
Since the sum of the angles of a triangle is , and given that the angles are in a ratio of 1:3:5, let the measure of the smallest angle be , then the following expression could be written:
If the smallest angle is 20 degrees, then given that the middle angle is in ratio of 1:3, the middle angle would be 3 times as large, or 60 degrees.
### Example Question #1 : Triangles
The measure of 3 angles in a triangle are in a 1:2:3 ratio. What is the measure of the middle angle?
90
30
60
45
Explanation:
The angles in a triangle sum to 180 degrees. This makes the middle angle 60 degrees.
### Example Question #1 : Use Variables To Represent Numbers And Write Expressions: Ccss.Math.Content.6.Ee.B.6
Call the three angles of a triangle
The measure of is twenty degrees greater than that of ; the measure of is thirty degrees less than twice that of . If is the measure of , then which of the following equations would we need to solve in order to calculate the measures of the angles?
Explanation:
The measure of is twenty degrees greater than the measure of , so its measure is 20 added to that of - that is, .
The measure of is thirty degrees less than twice that of . Twice the measure of is , and thirty degrees less than this is 30 subtracted from - that is, .
The sum of the measures of the three angles of a triangle is 180, so, to solve for - thereby allowing us to calulate all three angle measures - we add these three expressions and set the sum equal to 180. This yields the equation:
### Example Question #2 : Use Variables To Represent Numbers And Write Expressions: Ccss.Math.Content.6.Ee.B.6
Call the three angles of a triangle
The measure of is forty degrees less than that of ; the measure of is ten degrees less than twice that of . If is the measure of , then which of the following equations would we need to solve in order to calculate the measures of the angles?
Explanation:
The measure of is forty degrees less than the measure of , so its measure is 40 subtracted from that of - that is, .
The measure of is ten degrees less than twice that of . Twice the measure of is , and ten degrees less than this is 10 subtracted from - that is, .
The sum of the measures of the three angles of a triangle is 180, so, to solve for - thereby allowing us to calulate all three angle measures - we add these three expressions and set the sum equal to 180. This yields the equation:
### Example Question #3 : How To Find The Measure Of An Angle
Two interior angles of a triangle adds up to degrees. What is the measure of the other angle?
Explanation:
The sum of the three angles of a triangle add up to 180 degrees. Subtract 64 degrees to determine the third angle.
### Example Question #4 : How To Find The Measure Of An Angle
What is of the measure of a right angle?
Explanation:
A right angle has a measure of . One fifth of the angle is:
### Example Question #5 : How To Find The Measure Of An Angle
What angle is complementary to ?
Explanation:
To find the other angle, subtract the given angle from since complementary angles add up to .
The complementary is:
← Previous 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": 0.9367098808288574, "perplexity": 588.7115018470363}, "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-2016-50/segments/1480698542665.72/warc/CC-MAIN-20161202170902-00395-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.helmut-fischer.com/measurement-technologies-in-use/amplitude-sensitive-eddy-current | # Amplitude Sensitive Eddy Current
## Amplitude-sensitive measurements and the most critical influencing factors
With the amplitude-sensitive eddy current method, conductivity measurement, the thickness of coatings can be measured non-destructively according to ISO 2360. The prerequisite for this is that the base material is electrically conductive but not magnetizable: Metals such as copper or aluminum are therefore suitable. The coating itself must be electrically insulating, e. g. made of lacquer or plastic. One of the main applications of the eddy current measurement method is the testing of anodic coatings on aluminum.
### Physical principles
The probes used for measuring according to the amplitude-sensitive eddy current method have a ferrite core. A coil is wound around this core and a high-frequency alternating current flows through it. This creates a high-frequency alternating magnetic field around the coil.
When the probe pole comes close to a metal, an alternating current – or ‘eddy current’ – is induced in this metal. This, in turn, generates another alternating magnetic field. Since this second magnetic field is the opposite of the first, the original magnetic field is attenuated (weakened). The extent of the attenuation depends on the distance between the pole and the metal. For coated parts, this distance corresponds exactly to the layer thickness.
### Here’s what you need to pay attention to during the measurement
All electro-magnetic test methods are comparative. This means that the measured signal is compared with a characteristic curve that’s stored in the device. In order for the result to be correct, the characteristic curve must be adapted to the current conditions. This is achieved through calibration.
### Correct calibration makes all the difference!
Factors that can strongly influence the measurement when using the eddy current method are: the electrical conductivity, the shape and size of the sample and the roughness of the surface. Of course, correct operation of the device is also crucial!
### Electrical conductivity
A material’s electrical conductivity influences how well an eddy current can be induced within it. The conductivity can vary greatly depending on the specific alloy and how the metal was processed, and different temperatures can also cause variations. In order to minimize the calibration effort, Fischer eddy current probes have a conductivity compensation. They provide correct results over a wide range of conductivities and only need to be standardized on the respective material (i.e. calibration of the zero point).
### Curved surfaces
In practice, most magnetic coating measurement errors occur due to the shape of the sample. With curved surfaces, the proportion of the magnetic field that passes through the air is different. For example, if a measuring device was calibrated on a flat sheet, measuring on a concave surface would lead to a lower result, whereas measuring on a convex one would lead to a higher result. The errors that occur in this way can be many times the actual value!
Careful calibration is the remedy for this problem. But even here, Fischer has found a way to save time and work: a curvature-compensating probe. With this special probe you can measure without errors on tubes of 2 mm in radius or larger, even if the device was calibrated on a flat sheet.
### Small, flat parts
A similar effect can occur if the sample is small or very thin. Also in this case, the magnetic field extends beyond the sample and into the air, which systematically distorts the measurement results. To avoid these errors, you should always calibrate on an uncoated part that corresponds to the end product.
### Roughness
For rough surfaces, the result can be distorted depending on whether the probe pole is placed in a ‘valley’ or on a ‘peak’ of the roughness profile. With such measurements, the results vary widely and it is advisable to repeat the measurements several times in order to accumulate a stable mean. In general, magnetic coating thickness measurements on rough surfaces only make sense if the coating is at least twice as thick as the roughness peaks are high.
For better accuracy, Fischer offers probes with particularly large poles, as well as 2-pole probes. These probes integrate the roughness profile and thus reduce the scatter in the measured values.
### User influence
Last but not least, the way the measuring device is operated also plays a major role. Always make sure that the probe is set vertically on the surface and without pressure. For better accuracy, a stand can be used to automatically lower the probe onto the sample. Helmut Fischer offers both Handheld Coating Thickness devices and Automated Measurement Systems. | {"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": 0.8366900682449341, "perplexity": 793.6779391456732}, "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/1664030334514.38/warc/CC-MAIN-20220925035541-20220925065541-00610.warc.gz"} |
http://math.stackexchange.com/questions/650535/maps-phic-inftyy-rightarrow-c-inftyx-as-pullbacks | # Maps $\phi:C^{\infty}(Y) \rightarrow C^{\infty}(X)$ as pullbacks
I've just begun reading through Joyce's D-manifolds and d-orbifolds: a theory of derived differential geometry, and I've come across a seemingly innocuous statement that isn't quite clear to me. Specifically, in $\S$1.2 he gives the following definition (in my words):
A $\mathbf{C^{\infty}}$-ring is a pair $(\mathfrak{C},\{\Phi_n\}_{n\geq 0})$ where $\mathfrak{C}$ is a set and $\Phi_n:C^{\infty}(\mathbb{R}^n)\rightarrow \text{Maps}(\mathfrak{C}^n,\mathfrak{C})$ satisfies (I will drop the subscript $n$, as no confusion should arise):
(i) For any $m,n\geq 0$, $\ f_1,\cdots,f_m\in C^{\infty}(\mathbb{R}^n)$, and $g\in C^{\infty}(\mathbb{R}^m)$, we can define $h\in C^{\infty}(\mathbb{R}^n)$ by
$$h(x_1,\cdots,x_n)=g(f_1(x_1,\cdots,x_n),\cdots,f_m(x_1,\cdots,x_n)).$$
Then for any $c_1,\cdots,c_n\in\mathfrak{C}$, $\Phi$ must satisfy
$$\Phi_h(c_1,\cdots,c_n)=\Phi_g(\Phi_{f_1}(c_1,\cdots,c_n),\cdots,\Phi_{f_m}(c_1,\cdots,c_n)).$$
(ii) For all $1\leq j\leq n$, the canonical projection maps $\pi_j:\mathbb{R}^n\rightarrow\mathbb{R}$ must satisfy
$$\Phi_{\pi_j}(c_1,\cdots,c_n)=c_j$$
for all $c_1,\cdots,c_n\in\mathfrak{C}$.
Then there is a straightforward notion of a morphism between these structures:
A morphism between $C^{\infty}$-rings $(\mathfrak{C},\Phi)$ and $(\mathfrak{D},\Psi)$ is a map $\phi:\mathfrak{C}\rightarrow\mathfrak{D}$ such that
$$\begin{array}{ccc} \mathfrak{C}^n & \xrightarrow{\Phi_f} & \mathfrak{C} \\ \downarrow^{\phi^n} & & \downarrow^{\phi} \\ \mathfrak{D}^n & \xrightarrow{\Psi_f} & \mathfrak{D} \end{array}$$
commutes for all $f\in C^{\infty}(\mathbb{R}^n)$ and $n\geq 0$.
The basic example, as hinted at by the name of these objects, is $C^{\infty}(X)$ where $X$ is a smooth manifold. By defining
$$[\Phi_f(c_1,\cdots,c_n)](x)=f(c_1(x),\cdots,c_n(x))$$
for $f\in C^{\infty}(\mathbb{R}^n)$, $c_1,\cdots,c_n\in C^{\infty}(X)$, and $n\geq 0$, one clearly gets the structure of a $C^{\infty}$-ring. Moreover, for a smooth map $f:X\rightarrow Y$, the pullback $f^{\ast}:C^{\infty}(Y)\rightarrow C^{\infty}(X)$ is clearly a morphism of $C^{\infty}$-rings. But then Joyce goes on to say,
Furthermore (at least for $Y$ without boundary), every $C^{\infty}$-ring morphism $\phi:C^{\infty}(Y)\rightarrow C^{\infty}(X)$ is of the form $\phi=f^{\ast}$ for a unique smooth map $f:X\rightarrow Y$.
My question(s): Why is this true? How is the map $f$ constructed? What about the construction fails if $Y$ has boundary? Is this statement true regardless of whether or not $\phi$ is a $C^{\infty}$-ring morphism?
The thing that came immediately to mind was defining $f:X\rightarrow Y$ by
$$f(x) = g^{-1}(\phi(g)(x)),$$
so that $f^{\ast}(g)=g(g^{-1}(\phi(g)))=\phi(g)$, but of course this only works if $g$ is surjective, and the equation $\phi=f^{\ast}$ only holds for this specific $g$. I cannot, however, think of any other way of finding $f$… any suggestions? References?
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I'm expecting that it is not true for arbitrary maps $\phi:C^{\infty}(Y)\rightarrow C^{\infty}(X)$, but it does actually need to be a $C^{\infty}$-ring morphism… and that the proof just involves a clever choice of $f$ and $n$ so that the commutativity of the diagram gives me what I want… but I'm not seeing it yet. – Ralph Mellish Jan 25 at 2:50
Not an answer, but this reminds me of the Banach Stone theorem. The construction of $f_\phi$ may not be so trivial. – Steven Gubkin Jan 25 at 3:14
@StevenGubkin: Yeah, I just realized that this sounds very familiar to the Gelfand-Naimark theorem as well (at least a smooth manifold version of it). Perhaps those types of results are what I need… thanks! – Ralph Mellish Jan 25 at 3:17
Having finally had a little more time to return to this question, I searched around a bit more and found that this result is given by the following Proposition from Joyce's Algebraic Geometry over $C^{\infty}$-rings:
Proposition 3.3. Let $X,Y$ be manifolds with corners. Then the map $f\mapsto f^{\ast}$ from weakly smooth maps $f:X\rightarrow Y$ to morphisms of $C^{\infty}$-rings $\phi:C^{\infty}(Y)\rightarrow C^{\infty}(X)$ is a 1-1 correspondence.
Here, $f:X\rightarrow Y$ is weakly smooth if it is continuous, and for any charts $(U,\phi)$ and $(V,\psi)$ on $X$ and $Y$, respectively, then the map
$$\psi^{-1}\circ f\circ\phi:(f\circ\phi)^{-1}(\psi(V))\rightarrow V$$
is smooth as a map between euclidean spaces. Then one can define a notion of smoothness for maps between manifolds with corners as follows (from Joyce's On manifolds with corners, $\S$3):
Suppose $x\in X$ with $f(x)=y\in Y$ , and $\beta$ is a local boundary component of $Y$ at $y$. Let $(V,b)$ be a boundary defining function for $Y$ at $(y,\beta)$. Then $f^{-1}(V)$ is an open neighborhood of $x$ in $X$, and $b\circ f : f^{−1}(V)\rightarrow [0,\infty)$ is a weakly smooth map. We require that either $b\circ f\equiv 0$ on an open neighbourhood of $x$ in $f^{−1}(V)$, or $(f^{−1}(V),b\circ f)$ is a boundary defining function for $X$ at $(x,\tilde{\beta})$, for some unique local boundary component $\tilde{\beta}$ of $X$ at $x$.
At any rate, in the case of manifolds without boundary, smoothness trivially implies weak smoothness. The proof of Prop. 3.3 apparently follows from Prop. I.1.5 of Moerdijk and Reyes' Models for Smooth Infinitesimal Analysis. Once I have a chance to get to the library to pick up a copy, I'll try to edit this answer to give a proof for anyone interested. Until then, it seems that the (tentative) answers to my questions are:
Why is this true? How is the map f constructed? See Prop. I.1.5 in the above reference, until I have time to get a copy and digest it.
What about the construction fails if Y has boundary? Apparently nothing -- an analogous result (with smoothness replaced by weak smoothness) seems to hold for manifolds with boundary and manifolds with corners.
Is this statement true regardless of whether or not $\phi$ is a $C^{\infty}$-ring morphism? Apparently $\phi$ does need to be a $C^{\infty}$-ring morphism.
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http://math.stackexchange.com/questions/181733/please-help-me-understand-conditional-probability-in-relation-to-dice | If 2 fair dice are rolled, what is the probability that the sum is 6, given both dice show odd numbers?
When the sum is 6, there combinations are: (1,5),(2,4),(3,3),(4,2),(5,1). Probability = 5/36
Probability of 2 odd numbers is 1/2.
What I'm not sure about is when we work out Pr(sum is 6|both numbers odd). I initially thought that the top line was Pr(sum is 6), but then the result would be 5/2. This can't be right.
So, does that mean that we only choose combinations which have odd pairs? So, the answer would be 3/2?
Still makes no sense as the number is greater than 1???
Can somebody shed some light on this for me?
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What you're trying to compute is the probability that two dice sum to $6$, given that both dice have odd values. Note that there are three ways for pairs of dice with odd values to sum to six, namely $(1,5),(3,3),(5,1)$. There are three odd values each die can take, so there are nine ways for pairs of dice to both have odd values. Thus the probability you are looking for is $3/9=1/3$.
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Thanks Alex, but I don't understand how you got the 9? – Sandra Aug 12 '12 at 16:13
@Sandra There are three odd values the first die can have, namely $1,3,5$. Similarly, the second die can have a $1,3$ or $5$. So the possible outcomes are $(1,1),(3,1),(5,1),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)$. – Alex Becker Aug 12 '12 at 16:16
Thanks heaps :) – Sandra Aug 12 '12 at 16:51
@AlexBecker To be clear enumerating order the 9 outcomes are (1,1), (3,1), (5,1), (3,3), (3,5), (5,3), (5,5), (1,3), (1,5) – Michael Chernick Aug 12 '12 at 17:00
With conditional probabilities, the formal notation is often helpful, at least as an accounting device to keep things straight. In more complicated problems, it becomes more or less essential. So let's do things quite formally.
Let $B$ be the event "both dice are odd," and let $S_6$ be the event "the sum is $6$." We want $$\Pr(S_6|B).$$ Now we use the formula $$\Pr(S_6|B)\Pr(B)=\Pr(S_6\cap B),\tag{1}$$ or another version of the same thing. The formula tells us that everything will be finished once we know $\Pr(B)$ and $\Pr(S_6\cap B)$.
Calculating $\Pr(B)$ is easy: the answer is $\frac{9}{36}$.
To calculate $\Pr(S_6\cap B)$, note that the results on the two dice (coloured red and green say) can be represented as an ordered pair $(x,y)$ where $1\le x\le 6$, $1\le y\le 6$.
These $36$ ordered pairs are all equally likely. How many have sum $6$ and are both odd? List them all: $(1,5)$, $(3,3)$, $(5,1)$. We conclude that $\Pr(S_6\cap B)=\frac{3}{36}$.
Now use Equation $(1)$. The required conditional probability is $\dfrac{\frac{3}{36}}{\frac{9}{36}}$.
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Thanks for the detailed response Andre :) – Sandra Aug 12 '12 at 16:52
There are $36$ outcomes, of which, only the $9$ in red come from both dice showing odd numbers: $$\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&\color{#C00000}{2}&3&\color{#C00000}{4}&5&\color{#C00000}{6}&7\\ 2&3&4&5&6&7&8\\ 3&\color{#C00000}{4}&5&\color{#C00000}{6}&7&\color{#C00000}{8}&9\\ 4&5&6&7&8&9&10\\ 5&\color{#C00000}{6}&7&\color{#C00000}{8}&9&\color{#C00000}{10}&11\\ 6&7&8&9&10&11&12 \end{array}$$ Each outcome is equally likely, so each of the red outcomes is also equally likely. Of the $9$ red outcomes, there are $3$ that total $6$. This means that, given that each die shows an odd number (the red outcomes), there is a $3/9=1/3$ probability that the sum is a $6$.
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https://www.physicsforums.com/threads/having-trouble-understanding-what-tension-is.656970/ | # Having trouble understanding what tension is
1. ### BomboshMan
19
Hi,
I undersand that tension is the force exerted on an object by a string or rope when pulled. Where I get confused is when I'm asked to find the tension at a point along a rope or tension in a section of rope...would the tension in/of a section of rope be the net force acting on it, or the force it exerts on either end (if this is true, I assume the force acting at each end is necessarily always the same?)?
Sorry if I'm not making myself clear...here's an example which might show what I mean...
A verticle rope is attatched at both ends (to walls). Find the tension T at distance y above the lower end of the rope. The mass of the section below y is M and the mass above is m.
Hope I make sense!
Thanks,
Matt
2. Physics news on Phys.org
3. ### jbriggs444
The tension in a rope is the gross force that one small section of rope exerts on its neighboring section in one direction or the other.
It is not the net force that the neighbors from both directions exert upon that small section.
If you have a tug-of-war, a small section of rope will be under a large tension. But it will be undergoing negligible net force.
5. ### pongo38
707
I think you may be being asked to prepare to draw a normal force diagram. For a uniform vertically hanging rope, that would vary triangularly from 0 at the bottom to the weight of the rope at the top.
6. ### Naty1
it varies linearly for a uniform rope.
### Staff: Mentor
Many people have suffered the same difficulty that you are presently experiencing. This is because new initiates to physics are not taught that tension in a rope is really a tensorial quantity. It should really be expressed as:
T = T i i
where T is the scalar magnitude of the tension, T is the tension tensor, and i is a unit vector in the "positive direction" along the rope. Note that the two unit vectors in the above equation are placed in juxtaposition with one another, with no operation implied between them. Two vectors in juxtaposition like this are called a dyad. Please bear with me, and I will show you how this representation can be used.
Let s be a specific location along the rope. The way the traction tensor works is that, if you want to know the force exerted by the portion of the rope at >s on the portion of the rope at <s, you dot the tension tensor with a unit vector directed from <s to >s (i.e., in the positive s direction i):
F = T $\cdot$i = T i i $\cdot$i = T i (i $\cdot$i) = T i
Note that you have dotted the right-hand member of the dyad with the unit vector in the plus s direction i. This has had the effect of mapping the tension tensor T into the tension vector F.
Suppose you want to know the force exerted by the portion of the rope at <s on the portion of the rope at >s. In this case, you dot the tension tensor with a unit vector drawn from >s to <s (i.e., in the negative s direction -i). In this case, you find that:
F = T $\cdot$(-i) = -T i
In summary, what this notation does is provide you with an automatic foolproof method of determining the force exerted at a given location by the portion of the rope on one side of the given location on the portion of the rope on the other side of the given location.
Once you get used to this approach, you should have very little trouble later on learning about the stress tensor and how to apply it.
8. ### Philip Wood
1,086
A light has started to shine.
9. ### pongo38
707
Natv1: The mass is uniformly distributed. The algebraic sum of the forces on one side of a section (one definition of 'normal force') is a linear distribution in the shape of a triangle when plotted as a graph. Thanks to Chestermiller for the rigorous explanation.
10. ### jambaugh
1,799
Here is another way to understand tension, as a negative pressure.
In physics we can generalize the notion of a force to any type of system configuration change. A torque is a force that does work when an object rotates, a stress is a force that does work when an object undergoes a strain deformation, and a pressure/tension (scalar stress) does work when a system changes volume (3d) area (2d e.g. surface tension) or length (1d).
In the example you give of the hanging massive rope, ask what is the (negative) work done if you allow one point on the rope to expand to a segment of length dL. The answer is that the rope below that point will drop that distance in the constant g potential.
$$T = -dE/dL = gM$$
If for example you are considering a vibrating string, ether longitudinal or transversal vibrations , the string must stretch to displace beyond the equilibrium state. Figure out the potential energy of the elastic string dU = TdL and kinetic energy and you can work out its dynamics.
11. ### sophiecentaur
13,263
You need to be careful about your units here. It's Stress that has the same dimensions as Pressure. Tension has the same dimensions as Force.
12. ### jambaugh
1,799
In each case we're talking force per boundary measure. Pressure = (negative) volumetric tension has units force per area.
Surface tension has units force per length.
Linear tension, force per point unit or just force.
The important point is that a tension is not (translational) force but rather a generalized (scaling) force.
30 | {"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": 0.8946340680122375, "perplexity": 633.8633365075989}, "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-2015-11/segments/1424936462232.5/warc/CC-MAIN-20150226074102-00243-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://eprint.iacr.org/2007/431/20071124:104025 | ## Cryptology ePrint Archive: Report 2007/431
Notions of Efficiency in Simulation Paradigm
Tzer-jen Wei
Abstract: Abstract. There are some well-known conceptional and technical issues related to a common setting of simulation paradigm, i.e., EPT (expected polynomial time) simulator versus SPT (strict polynomial time) adversary. In fact, it has been shown that this setting is essential for achieving constant-round black-box zero-knowledge protocols. Many suggestions and results have been proposed to deal with these issues. In this paper, we propose an alternative solution. We study a new class of machines, MPT (Markov polynomial time), which is a cryptographic adaption of Levin's average polynomial-time. Since MPT has good compatibility to SPT and intuitive composition properties, we can use it as a drop-in replacement of SPT. Moreover, it is easy to construct simulators in MPT.
Category / Keywords: foundations / zero-knowledge, expected polynomial-time, Average polynomial-time, Markov polynomial-time | {"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": 0.8313255310058594, "perplexity": 2826.4382170633257}, "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/1471982298875.42/warc/CC-MAIN-20160823195818-00245-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/191696/including-bibliography-in-latex-gives-error | # Including bibliography in LaTex gives error
I'm writing an IEEE paper and I wanted to use the IEEEtran bibliography style for my research paper. I used the following code to add the bibliography
\documentclass[conference]{IEEEtran}
\title{Title}
}
\begin{document}
\bibliographystyle{IEEEtran}
\bibliography{myreferences}
\end{document}
When I compile the code using PDFLatex for first time it will not generate any error, infact it generates a new file called myPaper.bbl and when I compile the document for second time I get the following error
./myPaper.bbl:3:Something's wrong--perhaps a missing \item. \end{thebibliography}
The contents of myPaper.bbl are
% Generated by IEEEtran.bst, version: 1.13 (2008/09/30)
\begin{thebibliography}{}
\providecommand{\url}[1]{#1}
\csname url@samestyle\endcsname
\providecommand{\newblock}{\relax}
\providecommand{\bibinfo}[2]{#2}
\providecommand{\BIBentrySTDinterwordspacing}{\spaceskip=0pt\relax}
\providecommand{\BIBentryALTinterwordstretchfactor}{4}
\providecommand{\BIBentryALTinterwordspacing}{\spaceskip=\fontdimen2\font plus
\BIBentryALTinterwordstretchfactor\fontdimen3\font minus
\fontdimen4\font\relax}
\providecommand{\BIBforeignlanguage}[2]{{%
\expandafter\ifx\csname l@#1\endcsname\relax
\typeout{** WARNING: IEEEtran.bst: No hyphenation pattern has been}%
\typeout{** loaded for the language #1'. Using the pattern for}%
\else
\language=\csname l@#1\endcsname
\fi
#2}}
\providecommand{\BIBdecl}{\relax}
\BIBdecl
\end{thebibliography}
One more observation I made is, if I change the style of bibliography from IEEEtran to plain, in this case also I notice the same error.
After doing this change what I understood is, the problem is not with the IEEEtran file or plain style file, the only problem is because of the myPaper.bbl file which is being created automatically.
I couldn't understand about the problem more than this as my knowledge in kile documents is limited.
• Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. Jul 16 '14 at 8:45
• I bet there is an entry in your bib-file responsible (maybe an anescaped %). Try to find tis entry using the techniques described in the link above. Jul 16 '14 at 8:47
• Thanks for the MWE, but the faulty bib part is missing ;-) Jul 16 '14 at 8:54
My current version of IEEEtran expects the biblatexpackage. This works with current TeXLive 2014:
\documentclass[conference]{IEEEtran}
\usepackage{biblatex}
I run pdflatex -> biber -> pdflatex -> pdflatex` | {"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": 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": 0.8390466570854187, "perplexity": 2672.429751659658}, "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-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00498.warc.gz"} |
https://www.physicsforums.com/threads/sources-of-error-in-standing-waves-lab.99630/ | # Sources of Error in Standing Waves Lab
1. Nov 13, 2005
### dekoi
I have to write an error analysis for my lab report, but am having some trouble making my sources of error clear and precise. The lab itself was a very basic investigation of the relationship between the tension, frequency, and mass per unit length (mu) in a standing wave pattern.
Some thoughts:
-improper measurement of string
-improper measurement of frequency (that is, the value read from the machine is too high or low, hence producing a large percentage of error)
-incorrect visual observation of the number of loops; for example, 4 loops mistaken for 3
-mistake in calculation
Any thoughts would be very helpful from experienced members.
2. Nov 13, 2005
### andrevdh
Did you alter the tension in the string (pulley and masses or lever) and determined the frequency of the oscillation at the specific tensions? How did the higher harmonics (loops) originate, did you induce them? Did you change the strings and repeated the experiment? Was the linear density given or did you measure it yourself?
3. Nov 13, 2005
### dekoi
We did alter tension. The frequency was read from a machine. We induced the higher harmonics. We did not change the string. Linear density was measured.
4. Nov 14, 2005
### andrevdh
The term error in an measurement causes great confusion. Nowadays the term uncertainty in a measurement is rather used. This means any factor that could possibly cause a measure of uncertainty in a measurement should be identified. By a measure of uncertainty I mean that the numerical value of the measurement has a percentage of uncertainty asssociated with it - the measurement is not 100% accurate. Due to some factor influencing the measurement (this is what you are trying to identify) it is probably off by some percentage from the real value of the measurement. Countable quantities (like the loops) have no uncertainty in them - they are 100% accurate since we all can agree on such quantities, but measurements can be off from the real value. In order to help you I need to know more about the experiment though. Which was the independant variable of the experiment, the tension? Was the one end of the string (metal?) fixed to the tension device (how was this changed) an the other end tied to a vibrating device? Did you alter the frequency? Do you need to quantify the errors in your measurements for the error analysis?
Last edited: Nov 14, 2005
Similar Discussions: Sources of Error in Standing Waves Lab | {"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": 0.9637991786003113, "perplexity": 923.2308258975905}, "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/1498128321309.25/warc/CC-MAIN-20170627101436-20170627121436-00463.warc.gz"} |
http://math.stackexchange.com/questions/181266/approximate-rational-dependence?answertab=oldest | # Approximate rational dependence
After seeing question Why is $10\frac{\exp(\pi)-\log 3}{\log 2}$ almost an integer? I wonder if there is an algorithm that can find approximate rational dependence?! I pick any irrational numbers $\alpha$, $\beta$, $\gamma$, $\delta$ and the task is to find successively good approximations for rational numbers $q_i$ in
$q_1\alpha+q_2\beta+q_3\gamma+q_4\delta\approx0$
Basically it's like continued fraction approximations, but for more irrational numbers.
What is such an algorithm called?
-
I think the LLL-algorithm of Lenstra, Lenstra, Lovasz does something like this – Cocopuffs Aug 11 '12 at 8:49
@Cocopuffs: Can this algorithm handle rational numbers or is it integer only? – Gerenuk Aug 11 '12 at 8:55
I doubt there's any optimality for rational numbers, although you can certainly divide by a common denominator. I'm not sure exactly how this would work. The problem as given I don't fully understand either, as you can obviously take $q_1 = ... = q_4 = 0$, but I'm sure you want to exclude that. – Cocopuffs Aug 11 '12 at 9:44
like Cocopuffs and Gerry(+1) explained you may use LLL and PSLQ. This was tried here (to get better approximations you need only to increase the precision). – Raymond Manzoni Aug 11 '12 at 10:56 | {"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": 0.9399874806404114, "perplexity": 777.806266267201}, "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/1413507443901.32/warc/CC-MAIN-20141017005723-00090-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://open.library.ubc.ca/cIRcle/collections/48630/items/1.0357386 | # Open Collections
## BIRS Workshop Lecture Videos
### Lagrangian schemes for Wasserstein gradient flows Düring, Bertram
#### Description
A wide range of diffusion equations can be interpreted as gradient flow with respect to Wasserstein distance of an energy functional. Examples include the heat equation, the porous medium equation, and the fourth-order Derrida-Lebowitz-Speer-Spohn equation. When it comes to solving equations of gradient flow type numerically, schemes that respect the equation's special structure are of particular interest. The gradient flow structure gives rise to a variational scheme by means of the minimising movement scheme (also called JKO scheme, after the seminal work of Jordan, Kinderlehrer and Otto) which constitutes a time-discrete minimization problem for the energy. While the scheme has been used originally for analytical aspects, a number of authors have explored the numerical potential of this scheme. Such schemes often use a Lagrangian representation where instead of the density, the evolution of a time-dependent homeomorphism that describes the spatial redistribution of the density is considered. | {"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": 0.976172924041748, "perplexity": 554.219884025703}, "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-2022-33/segments/1659882570868.47/warc/CC-MAIN-20220808152744-20220808182744-00537.warc.gz"} |
http://mathhelpforum.com/algebra/185353-solving-equations.html | 1. ## Solving equations?
I have an equation;
7 = 2
3+a a - 2
can I solve it this way?
6a + 2a = 8a
7a - 14 7a -14
= a - 14
The method I have used is to cross multiply, but I am unsure?
2. ## Re: Solving equations?
Originally Posted by David Green
I have an equation;
7 = 2
3+a a - 2
You can say $\frac{7}{3+a}=\frac{2}{a-2}$
$\\ 7a-14 =6+2a$.
Now solve for $a$ .
3. ## Re: Solving equations?
Originally Posted by Plato
You can say $\frac{7}{3+a}=\frac{2}{a-2}$
$\\ 7a-14 =6+2a$.
Now solve for $a$ .
Does that mean that now there is two (a) I have to factor one out?
4. ## Re: Solving equations?
Originally Posted by David Green
Does that mean that now there is two (a) I have to factor one out?
No, this is a linear equation with variables on both sides. You need the two terms with "a" on one side, and everything else on the other side.
5. ## Re: Solving equations?
Originally Posted by David Green
Does that mean that now there is two (a) I have to factor one out?
I don't even know what that question means, much less how to answer it.
What level are you on?
What are you studying?
Can you solve $7a-14=6+2a$ for $a~?$
6. ## Re: Solving equations?
Originally Posted by Plato
I don't even know what that question means, much less how to answer it.
What level are you on?
What are you studying?
Can you solve $7a-14=6+2a$ for $a~?$
I am only a beginner
Introduction to maths
a(7 - 14)(6 + 2) ???
7a - 14 = 6 + 2a
7a - 2a = 6 + 14
5a = 20
a = 4
how am I doing!
7. ## Re: Solving equations?
Originally Posted by David Green
I am only a beginner
Introduction to maths
\begin{align*} \frac{7}{3+a}&=\frac{2}{a-2} \\ 7a-14 &=6+2a\\ 5a&=20\text{ subtract 2a and add 14} \\ a&=4\end{align*}
8. ## Re: Solving equations?
Originally Posted by Plato
\begin{align*} \frac{7}{3+a}&=\frac{2}{a-2} \\ 7a-14 &=6+2a\\ 5a&=20\text{ subtract 2a and add 14} \\ a&=4\end{align*}
Thank you, so I did get it right with a little help, after all "every little helps" | {"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": 12, "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": 0.9973620176315308, "perplexity": 4308.746857904225}, "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-2013-20/segments/1368706470197/warc/CC-MAIN-20130516121430-00097-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/493020/bibliography-latex-et-al | # Bibliography LaTeX et al
Please, which latex bibliography will I use to have my citation as, e.g. [mashroom et al. 2008], and I want the year 2008 to be in blue 'hyperlink'.Thank you
• Welcome to Tex.SE community. Please share a MWE related to your query. – M S May 28 '19 at 10:43
Your question is not clear since you did not specify exactly which bibliography style you want. Assuming that you want to use natbib with IEEEtranN, then this is a good starting point for you to understand how it works
\begin{filecontents*}{sample.bib}
@article{aldaoudeyeh2016,
title={{Photovoltaic-battery scheme to enhance PV array characteristics in partial shading conditions}},
author={Aldaoudeyeh, Al-Motasem I},
journal={IET Renewable Power Generation},
volume={10},
number={1},
pages={108--115},
year={2016},
publisher={IET}
}
@ARTICLE{wu2017,
title={{Assessing Impact of Renewable Energy Integration on System Strength Using Site-Dependent Short Circuit Ratio}},
author={Wu, Di and Li, Gangan and Javadi, Milad and Malyscheff, Alexander M and Hong, Mingguo and Jiang, John Ning},
journal={IEEE Transactions on Sustainable Energy},
year={2017},
publisher={IEEE}
}
@article{wu2019,
title={A method to identify weak points of interconnection of renewable energy resources},
author={Wu, Di and Aldaoudeyeh, Al Motasem and Javadi, Milad and Ma, Feng and Tan, Jin and Jiang, John N and others},
journal={International Journal of Electrical Power \& Energy Systems},
volume={110},
pages={72--82},
year={2019},
publisher={Elsevier}
}
\end{filecontents*}
\documentclass[]{book}
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\usepackage{hyperref}
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\citep{aldaoudeyeh2016} % \citep is used to meet the style you want [Author et al., year]
\citep{aldaoudeyeh2016,wu2017}
\citep{aldaoudeyeh2016,wu2019,wu2017}
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\bibliographystyle{IEEEtranN} % style of bibliography (here you specify that you want authoryear or numeric citation)
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Note: you need to use a .bib file. You can either create it using filecontents environment or by placing that file in the same directory of the .tex file you work with. Google Scholar, journals, and conferences provide BibTeX style info for articles and conferences proceedings.
Also, the LaTeX file you use will generate a set of files whose name is very similar to the '.tex' file you use and in the same directory. If you want to change your bibliography style, the changes will usually not apply unless you remove a file whose type is '.bbl' and its name is the same as your '.tex' file | {"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": 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": 0.9511739015579224, "perplexity": 3676.24828456232}, "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/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00360.warc.gz"} |
http://www.ck12.org/analysis/Ellipses-Not-Centered-at-the-Origin/lesson/Ellipses-Centered-at-h-k/r4/ | <meta http-equiv="refresh" content="1; url=/nojavascript/">
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# Ellipses Not Centered at the Origin
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Ellipses Centered at (h, k)
Your homework assignment is to draw the ellipse $16(x-2)^2+4(y+3)^2=144$ . What is the vertex of your graph and where will the foci of the ellipse be located?
### Guidance
Just like in the previous lessons, an ellipse does not always have to be placed with its center at the origin. If the center is $(h, k)$ the entire ellipse will be shifted $h$ units to the left or right and $k$ units up or down. The equation becomes $\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ . We will address how the vertices, co-vertices, and foci change in the next example.
#### Example A
Graph $\frac{\left(x-3\right)^2}{16}+ \frac{\left(y+1\right)^2}{4}=1$ . Then, find the vertices, co-vertices, and foci.
Solution: First, we know this is a horizontal ellipse because $16 > 4$ . Therefore, the center is $(3, -1)$ and $a = 4$ and $b = 2$ . Use this information to graph the ellipse.
To graph, plot the center and then go out 4 units to the right and left and then up and down two units. This is also how you can find the vertices and co-vertices. The vertices are $(3 \pm 4,-1)$ or $(7, -1)$ and $(-1, -1)$ . The co-vertices are $(3,-1 \pm 2)$ or $(3, 1)$ and $(3, -3)$ .
To find the foci, we need to find $c$ using $c^2=a^2-b^2$ .
$c^2&=16-4=12 \\c&=2\sqrt{3}$
Therefore, the foci are $\left(3 \pm 2\sqrt{3},-1\right)$ .
From this example, we can create formulas for finding the vertices, co-vertices, and foci of an ellipse with center $(h, k)$ . Also, when graphing an ellipse, not centered at the origin, make sure to plot the center.
Orientation Equation Vertices Co-Vertices Foci
Horizontal $\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ $(h \pm a,k)$ $(h,k \pm b)$ $(h \pm c,k)$
Vertical $\frac{\left(x-h\right)^2}{b^2}+ \frac{\left(y-k\right)^2}{a^2}=1$ $(h,k \pm a)$ $(h \pm b,k)$ $(h,k \pm c)$
#### Example B
Find the equation of the ellipse with vertices $(-3, 2)$ and $(7, 2)$ and co-vertex $(2, -1)$ .
Solution: These two vertices create a horizontal major axis, making the ellipse horizontal. If you are unsure, plot the given information on a set of axes. To find the center, use the midpoint formula with the vertices.
$\left(\frac{-3+7}{2}, \frac{2+2}{2}\right)= \left(\frac{4}{2}, \frac{4}{2}\right)=(2,2)$
The distance from one of the vertices to the center is $a$ , $\left |7-2 \right \vert=5$ . The distance from the co-vertex to the center is $b$ , $\left |-1-2 \right \vert=3$ . Therefore, the equation is $\frac{\left(x-2\right)^2}{5^2}+ \frac{\left(y-2\right)^2}{3^2}=1$ or $\frac{\left(x-2\right)^2}{25}+ \frac{\left(y-2\right)^2}{9}=1$ .
#### Example C
Graph $49(x-5)^2+25(y+2)^2=1225$ and find the foci.
Solution: First we have to get this into standard form, like the equations above. To make the right side 1, we need to divide everything by 1225.
$\frac{49\left(x-5\right)^2}{1225}+ \frac{25\left(y+2\right)^2}{1225}&=\frac{1225}{1225} \\\frac{\left(x-5\right)^2}{25}+ \frac{\left(y+2\right)^2}{49}&=1$
Now, we know that the ellipse will be vertical because $25 < 49$ . $a = 7, b = 5$ and the center is $(5, -2)$ .
To find the foci, we first need to find $c$ by using $c^2=a^2-b^2$ .
$c^2&=49-25=24 \\c&=\sqrt{24}=2\sqrt{6}$
The foci are $\left(5,-2 \pm 2\sqrt{6}\right)$ or $(5, -6.9)$ and $(5, 2.9)$ .
Intro Problem Revisit We first need to get our equation in the form of $\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ . So we divide both sides by 144.
$\frac{16(x-2)^2}{144} + \frac{4(y+3)^2}{144} = \frac{144}{144}\\\frac{(x-2)^2}{9} + \frac{(y + 3)^2}{36}$ .
Now we can see that $h = 2$ and $3 = -k$ or $k = -3$ . Therefore the origin is $(2, -3)$ .
Because $9 < 36$ , we know this is a vertical ellipse. To find the foci, use $c^2=a^2-b^2$ .
$c^2&=36-9=27 \\c&=\sqrt{27}=3\sqrt{3}$
The foci are therefore $\left(0,3\sqrt{3}\right)$ and $\left(0,-3\sqrt{3}\right)$ .
### Guided Practice
1. Find the center, vertices, co-vertices and foci of $\frac{\left(x+4\right)^2}{81}+ \frac{\left(y-7\right)^2}{16}=1$ .
2. Graph $25(x-3)^2+4(y-1)^2=100$ and find the foci.
3. Find the equation of the ellipse with co-vertices $(-3, -6)$ and $(5, -6)$ and focus $(1, -2)$ .
1. The center is $(-4, 7),a=\sqrt{81}=9$ and $b=\sqrt{16}=4$ , making the ellipse horizontal. The vertices are $(-4 \pm 9,7)$ or $(-13, 7)$ and $(5, 7)$ . The co-vertices are $(-4,7 \pm 4)$ or $(-4, 3)$ and $(-4, 11)$ . Use $c^2=a^2-b^2$ to find $c$ .
$c^2&=81-16=65\\c&=\sqrt{65}$
The foci are $\left(-4- \sqrt{65},7\right)$ and $\left(-4+ \sqrt{65},7\right)$ .
2. Change this equation to standard form in order to graph.
$\frac{25\left(x-3\right)^2}{100}+ \frac{4\left(y-1\right)^2}{100}&=\frac{100}{100} \\\frac{\left(x-3\right)^2}{4}+ \frac{\left(y-1\right)^2}{25}&=1$
center: $(3, 1), b = 2, a = 5$
Find the foci.
$c^2&=25-4=21 \\c&=\sqrt{21}$
The foci are $\left(3,1+ \sqrt{21}\right)$ and $\left(3,1- \sqrt{21}\right)$ .
3. The co-vertices $(-3, -6)$ and $(5, -6)$ are the endpoints of the minor axis. It is $\left |-3-5\right \vert=8$ units long, making $b = 4$ . The midpoint between the co-vertices is the center.
$\left(\frac{-3+5}{2},-6\right)=\left(\frac{2}{2},-6\right)=(1,-6)$
The focus is $(1, -1)$ and the distance between it and the center is 4 units, or $c$ . Find $a$ .
$16&=a^2-16 \\32&=a^2 \\a&=\sqrt{32}=4\sqrt{2}$
The equation of the ellipse is $\frac{\left(x-1\right)^2}{16}+ \frac{\left(y+6\right)^2}{32}=1$ .
### Vocabulary
Standard Form (of an Ellipse)
$\frac{\left(x-h\right)^2}{a^2}+ \frac{\left(y-k\right)^2}{b^2}=1$ or $\frac{\left(x-h\right)^2}{b^2}+ \frac{\left(y-k\right)^2}{a^2}=1$ where $(h, k)$ is the center.
### Practice
Find the center, vertices, co-vertices, and foci of each ellipse below.
1. $\frac{\left(x+5\right)^2}{25}+ \frac{\left(y+1\right)^2}{36}=1$
2. $(x+2)^2+16(y-6)^2=16$
3. $\frac{\left(x-2\right)^2}{9}+\frac{\left(y-3\right)^2}{49}=1$
4. $25x^2+64(y-6)^2=1600$
5. $(x-8)^2+ \frac{\left(y-4\right)^2}{9}=1$
6. $81(x+4)^2+4(y+5)^2=324$
7. Graph the ellipse in #1.
8. Graph the ellipse in #2.
9. Graph the ellipse in #4.
10. Graph the ellipse in #5.
Using the information below, find the equation of each ellipse.
1. vertices: $(-2, -3)$ and $(8, -3)$ co-vertex: $(3, -5)$
2. vertices: $(5, 6)$ and $(5, -12)$ focus: $(5, -7)$
3. co-vertices: $(0, 4)$ and $(14, 4)$ focus: $(7, 1)$
4. foci: $(-11, -4)$ and $(1, -4)$ vertex: $(-12, -4)$
5. Extension Rewrite the equation of the ellipse, $36x^2+25y^2-72x+200y-464=0$ in standard form, by completing the square for both the $x$ and $y$ terms. | {"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": 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": 118, "texerror": 0, "math_score": 0.8981385827064514, "perplexity": 428.9675765613732}, "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-2015-18/segments/1430455113263.14/warc/CC-MAIN-20150501043833-00045-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://en.m.wiktionary.org/wiki/dual_number | # dual number
## English
### Noun
dual number (plural dual numbers)
1. (grammar) Grammatical number denoting a quantity of exactly two.
• 1840, George Grey, A Vocabulary of the Dialects of South Western Australia, T. & W. Boone, 2nd Edition, page xi,
This is particularly the case with regard to the pronouns of the dual number, which form one of the peculiarities of the language.
• 1998, William Blake Tyrrell, Larry J. Bennett, Recapturing Sophocles' Antigone, Rowman & Littlefield, page 43,
Eteocles and Polyneices are enemy brothers, that is, they are alike in spite of the difference erected by being on opposite sides of the Theban battlements.1 By uniting them with the dual number in the description of their deaths, Sophocles has the elders speak of them as indistinguishable in their hatred, spear work, and deaths.
• 2009, Giuliano Lancioni, Formulaic models and formulaicity in Classical and Modern Standard Arabic, Roberta Corrigan, Edith A. Moravcsik, Hamid Ouali, Kathleen M. Wheatley (editors), Formulaic Language, Volume 1: Distribution and historical change, John Benjamins Publishing Company, page 225,
Classical Arabic has a fully functional dual number, which requires dual agreement in verbs, adjectives and pronouns.
2. (algebra) An element of an algebra (the algebra of dual numbers) which includes the real numbers and an element ε which satisfies ε ≠ 0 and ε² = 0.
Hypernym: hypercomplex number
If f(x) is a polynomial or a power series then its derivative can be obtained “automatically” using dual numbers as follows: ${\displaystyle f'(x)={f(x+\epsilon )-f(x) \over \epsilon }}$ where ${\displaystyle f(x+\epsilon )}$ should be expanded using the algebra of dual numbers. The role of the ε is therein similar to that of an infinitesimal.
• 1993, Ivan Kolář, Peter W. Michor, Jan Slovák, Natural Operations in Differential Geometry, Springer, page 336,
In the special case of the algebra ${\displaystyle \mathbf {D} }$ of dual numbers we get the vertical tangent bundle ${\displaystyle V}$ .
• 1996, Roger Cooke (translator), Andrei N. Kolmogorov, Adolf-Andrei P. Yushkevich, Mathematics of the 19th Century: Geometry, Analytic Function Theory, Birkhäuser, page 87,
These numbers are now called dual numbers (a term due to Study) and can be defined as expressions of the form ${\displaystyle a+b\epsilon }$ , ${\displaystyle \epsilon ^{2}=0}$ . When this is done as Kotel'nikov and Study showed, the dual distance between two points of this sphere representing two lines, forming an angle ${\displaystyle \varphi }$ and having shortest distance ${\displaystyle d}$ , is equal to the dual number ${\displaystyle \varphi +\epsilon d}$ , while the motions of Euclidean space are represented by rotations of the dual sphere, and consequently depend on three dual parameters.
• 2002, Jiří Tomáš, Some Classification Problems on Natural Bundles Related to Weil Bundles, Olga Gil-Medrano, Vicente Miquel (editors), Differential Geometry, Valencia 2001: Proceedings of the International Conference, World Scientific, page 297,
Thus we investigate natural ${\displaystyle T}$ -functions defined on ${\displaystyle T^{*}T^{\mathbf {D} \otimes A}M}$ to determine all natural operators ${\displaystyle T\to TT^{*}T^{A}}$ for ${\displaystyle \mathbf {D} }$ denoting the algebra of dual numbers.
#### Translations
The translations below need to be checked and inserted above into the appropriate translation tables, removing any numbers. Numbers do not necessarily match those in definitions. See instructions at Wiktionary:Entry layout#Translations. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "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": 0.8325850367546082, "perplexity": 3421.3370658977833}, "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-2020-16/segments/1585371824409.86/warc/CC-MAIN-20200408202012-20200408232512-00373.warc.gz"} |
https://www.universetoday.com/135211/another-strange-discovery-lhc-nobody-understands/ | # Another Strange Discovery From LHC That Nobody Understands
There are some strange results being announced in the physics world lately. A fluid with a negative effective mass, and the discovery of five new particles, are all challenging our understanding of the universe.
New results from ALICE (A Large Ion Collider Experiment) are adding to the strangeness.
ALICE is a detector on the Large Hadron Collider (LHC). It’s one of seven detectors, and ALICE’s role is to “study the physics of strongly interacting matter at extreme energy densities, where a phase of matter called quark-gluon plasma forms,” according to the CERN website. Quark-gluon plasma is a state of matter that existed only a few millionths of a second after the Big Bang.
In what we might call normal matter—that is the familiar atoms that we all learn about in high school—protons and neutrons are made up of quarks. Those quarks are held together by other particles called gluons. (“Glue-ons,” get it?) In a state known as confinement, these quarks and gluons are permanently bound together. In fact, quarks have never been observed in isolation.
The LHC is used to collide particles together at extremely high speeds, creating temperatures that can be 100,000 times hotter than the center of our Sun. In new results just released from CERN, lead ions were collided, and the resulting extreme conditions come close to replicating the state of the Universe those few millionths of a second after the Big Bang.
In those extreme temperatures, the state of confinement was broken, and the quarks and gluons were released, and formed quark-gluon plasma.
So far, this is pretty well understood. But in these new results, something additional happened. There was increased production of what are called “strange hadrons.” Strange hadrons themselves are well-known particles. They have names like Kaon, Lambda, Xi and Omega. They’re called strange hadrons because they each have one “strange quark.”
If all of this seems a little murky, here’s the dinger: Strange hadrons may be well-known particles, because they’ve been observed in collisions between heavy nuclei. But they haven’t been observed in collisions between protons.
“Being able to isolate the quark-gluon-plasma-like phenomena in a smaller and simpler system…opens up an entirely new dimension for the study of the properties of the fundamental state that our universe emerged from.” – Federico Antinori, Spokesperson of the ALICE collaboration.
“We are very excited about this discovery,” said Federico Antinori, Spokesperson of the ALICE collaboration. “We are again learning a lot about this primordial state of matter. Being able to isolate the quark-gluon-plasma-like phenomena in a smaller and simpler system, such as the collision between two protons, opens up an entirely new dimension for the study of the properties of the fundamental state that our universe emerged from.”
# Enhanced Strangeness?
The creation of quark-gluon plasma at CERN provides physicists an opportunity to study the strong interaction. The strong interaction is also known as the strong force, one of the four fundamental forces in the Universe, and the one that binds quarks into protons and neutrons. It’s also an opportunity to study something else: the increased production of strange hadrons.
In a delicious turn of phrase, CERN calls this phenomenon “enhanced strangeness production.” (Somebody at CERN has a flair for language.)
Enhanced strangeness production from quark-gluon plasma was predicted in the 1980s, and was observed in the 1990s at CERN’s Super Proton Synchrotron. The ALICE experiment at the LHC is giving physicists their best opportunity yet to study how proton-proton collisions can have enhanced strangeness production in the same way that heavy ion collisions can.
According to the press release announcing these results, “Studying these processes more precisely will be key to better understand the microscopic mechanisms of the quark-gluon plasma and the collective behaviour of particles in small systems.”
I couldn’t have said it better myself. | {"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": 0.9028099179267883, "perplexity": 1050.7258789044338}, "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-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00213.warc.gz"} |
http://mathhelpforum.com/pre-calculus/81990-solved-inverse-angles-print.html | # [SOLVED] inverse of angles
Printable View
• April 2nd 2009, 01:01 PM
lsnyder
[SOLVED] inverse of angles
Hey everyone,
I have a test tomorrow and having some difficulty with inverses. so if anyone can help, I'd really appreciate it. Also, I am a visual learner so if it is shown I understand a lot more and can able to apply the same principles I have seen to other problems.
So can someone show me how to do these:
1) Find the inverse function for
Answer: https://webwork.uncc.edu/webwork2_fi...7821413471.png
2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places
https://webwork.uncc.edu/webwork2_fi...0522a264c1.png https://webwork.uncc.edu/webwork2_fi...ddf16c4091.png
• April 3rd 2009, 11:31 AM
Grandad
Hello lsnyder
Quote:
Originally Posted by lsnyder
1) Find the inverse function for
Answer: https://webwork.uncc.edu/webwork2_fi...7821413471.png
$0\le x\le\frac{\pi}{14}\Rightarrow 0\le 7x\le \frac{\pi}{2}$ and $f(x)$ is therefore is a one-to-one function, with values in the range $0\le f(x) \le 1$. So we can find its inverse, using $\arccos$, taking the angle in the range $0$ to $\frac{\pi}{2}$.
So, let $y = f(x) = \cos(7x)$
Then $\arccos(y) = 7x$
$\Rightarrow x = \tfrac17\arccos(y)$
So the inverse function (replacing $y$ by $x$) is $f^{-1}(x) = \tfrac17\arccos(x)$
Quote:
2) Evaluate each of the inverse trig functions in radians rounded to 4 decimal places
https://webwork.uncc.edu/webwork2_fi...0522a264c1.png https://webwork.uncc.edu/webwork2_fi...ddf16c4091.png
I assume that you have a calculator that will give you inverse tangents and inverse sines. Then:
$x = \text{arccot} (1.5)$
$\Rightarrow\cot(x) = 1.5$
$\Rightarrow \tan(x) = \frac{1}{1.5} = \frac23$
$\Rightarrow x = \arctan\Big(\frac23 \Big) = 0.5880$ to 4 d.p.
And $x = \csc^{-1}(8)$
$\Rightarrow \csc(x) = 8$
$\Rightarrow \sin(x) = \tfrac18$
$x = 0.1253$ to 4 d.p.
Grandad | {"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": 20, "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": 0.9805634021759033, "perplexity": 740.8886030892834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "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-2014-10/segments/1394010722348/warc/CC-MAIN-20140305091202-00064-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://byjus.com/simplest-form-calculator/ | # Simplest Form Calculator
Enter the fraction =
Simplest form=
The Simplest Form Calculator is a free online tool that displays the simplified form of the given fraction. BYJU’S online simplest form calculator tool makes calculations faster and easier where the value is displayed in a fraction of seconds.
## How to Use the Simplest Form Calculator?
The procedure to use the simplest form calculator is as follows:
Step 1: Enter the fractional value in the input fields
Step 2: Click the button “Solve” to get the output
Step 3: The result (simplest form) will be displayed in the output field
### Fraction Definition
A fraction is a number that is expressed as one number over another number. The fraction is represented as “a/b”. The number on the top of the fraction is called the numerator, and the number on the bottom is called denominator. There are three different types of fractions. They are:
• Proper Fraction
• Improper Fraction
• Mixed Fraction
### Standard Form
The standard form of the Fraction is given below:
Fraction = Numerator / Denominator
## Frequently Asked Questions on Simplest Form Calculator
### What are the different types of Fractions ?
There are three different types of fractions. They are:
• Proper Fraction
• Improper Fraction
• Mixed Fraction
### Define proper and improper fraction ?
• If the numerator is less than the denominator, then it is known as a proper fraction. Example: ⅔
• If the numerator is greater than the denominator, then it is known as an improper fraction. Example: 5/2
### What is a mixed fraction ?
A mixed fraction is a combination of a whole number and a proper fraction. A mixed fraction can be converted into an improper fraction. Example: 1½ is a mixed fraction.
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