text:
# Complex numbers lead a trigonometric equality

Times ago, I used to think about some trigonometric equalities. Now, I have faced a new one with different one:

Show that if $z^7+1=0$ then cos$(\frac{\pi}{7})$+cos$(\frac{3\pi}{7})$+cos$(\frac{5\pi}{7})=\frac{1}{2}$ wherein $z\in\mathbb C$.

Thanks

-
But there isn't any $z$ in the equation you are trying to establish, so what is the relevance of $z^7+1=0$? –  Gerry Myerson Jun 19 '12 at 9:53
a better manner ask this can be :Let $z =e^{\frac{i \pi}{7}}$. Vérifie that $z^7+1=0$ and deduce teh trigonometric relation : ... –  Mohamed Jun 19 '12 at 10:12
@Mohamed: Thanks –  Babak S. Jun 19 '12 at 10:57

Hint: write the cosines in terms of $e^{\pi i/7}$ and look for a geometric sequence.
Can we have the inverse problem as above. I mean; does it make sense to have, for example an equality like $8$sin$(\frac{\pi}{7})$sin$(\frac{2\pi}{7})$sin$(\frac{3\pi}{7})$=$\sqrt{7}$ and search for an equation? –  Babak S. Jun 19 '12 at 10:15
Sorry, I don't know what you mean. But those sines can also be written in terms of $e^{\pi i/7}$. –  Gerry Myerson Jun 19 '12 at 12:59
if $z^7+1=0$ then $$(z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0$$ Putting $z=e^{\frac{i \pi}{7}}$ we have $z + 1 \neq 0$ and $z^7 +1 =0$, then : $$z^6-z^5+z^4-z^3+z^2-z+1=0$$ This gives : $$\mathcal Re(z^6-z^5+z^4-z^3+z^2-z+1)=0$$ Since : $\cos \frac{\pi}{7} = - \cos \frac{6\pi}{7}$ and $\cos \frac{2\pi}{7} = - \cos \frac{5\pi}{7}$ abd $\cos \frac{3\pi}{7} = - \cos \frac{4\pi}{7}$ we have : $$2\left(- \cos \frac{\pi}{7}- \cos \frac{3\pi}{7}- \cos \frac{5 \pi}{7} \right) +1 = 0$$ which gives the desired relation.


text:
# C2000 PID Controller

Digital PID controller

## Library

Embedded Coder® Support Package for Texas Instruments™ C2000™ Processors/ Optimization/ C28x DMC

Embedded Coder Support Package for Texas Instruments C2000 F28M3x Concerto™ Processors Processors/ Optimization/ C28x DMC

## Description

This block implements a 32-bit digital PID controller with antiwindup correction. The inputs are a reference input (`ref`) and a feedback input (`fdb`) and the output (`out`) is the saturated PID output. The following diagram shows a PID controller with antiwindup.

The differential equation describing the PID controller before saturation that is implemented in this block is

upresat(t) = up(t) + ui(t) + ud(t)

where upresat is the PID output before saturation, up is the proportional term, ui is the integral term with saturation correction, and ud is the derivative term.

The proportional term is

up(t) = Kpe(t)

where Kp is the proportional gain of the PID controller and e(t) is the error between the reference and feedback inputs.

The integral term with saturation correction is

`${u}_{i}\left(t\right)=\underset{0}{\overset{t}{\int }}\left\{\frac{{K}_{p}}{{T}_{i}}e\left(\tau \right)+{K}_{c}\left(u\left(\tau \right)-{u}_{presat}\left(\tau \right)\right)\right\}d\tau$`

where Kc is the integral correction gain of the PID controller.

The derivative term is

`${u}_{d}\left(t\right)={K}_{p}{T}_{d}\frac{de\left(t\right)}{dt}$`

where Td is the derivative time of the PID controller. In discrete terms, the derivative gain is defined as Kd = Td/T, and the integral gain is defined as Ki = T/Ti, where T is the sampling period and Ti is the integral time of the PID controller.

Using backward approximation, the preceding differential equations can be transformed into the following discrete equations.

`$\begin{array}{l}{u}_{p}\left[n\right]={K}_{p}e\left[n\right]\\ {u}_{i}\left[n\right]={u}_{i}\left[n-1\right]+{K}_{i}{K}_{p}e\left[n\right]+{K}_{c}\left(u\left[n-1\right]-{u}_{presat}\left[n-1\right]\right)\\ {u}_{d}\left[n\right]={K}_{d}{K}_{p}\left(e\left[n\right]-e\left[n-1\right]\right)\\ {u}_{presat}\left[n\right]={u}_{p}\left[n\right]+{u}_{i}\left[n\right]+{u}_{d}\left[n\right]\\ u\left[n\right]=SAT\left({u}_{presat}\left[n\right]\right)\end{array}$`

Note

• To generate optimized code from this block, enable the ```TI C28x``` or `TI C28x (ISO)` Code Replacement Library.

The implementation of this block does not call the corresponding Texas Instruments library function during code generation. The TI function uses a global Q setting and the MathWorks® code used by this block dynamically adjusts the Q format based on the block input. See Using the IQmath Library for more information.

This block implements a 32-bit digital PID controller with antiwindup correction. The inputs are a reference input (`ref`) and a feedback input (`fdb`) and the output (`out`) is the saturated PID output. The following diagram shows a PID controller with antiwindup.

The differential equation describing the PID controller before saturation that is implemented in this block is

upresat(t) = up(t) + ui(t) + ud(t)

where upresat is the PID output before saturation, up is the proportional term, ui is the integral term with saturation correction, and ud is the derivative term.

The proportional term is

up(t) = Kpe(t)

where Kp is the proportional gain of the PID controller and e(t) is the error between the reference and feedback inputs

`${u}_{i}\left(t\right)=\underset{0}{\overset{t}{\int }}\left\{\frac{{K}_{p}}{{T}_{i}}e\left(\tau \right)+{K}_{c}\left(u\left(\tau \right)-{u}_{presat}\left(\tau \right)\right)\right\}d\tau$`

where Kc is the integral correction gain of the PID controller.

The derivative term is

`${u}_{d}\left(t\right)={K}_{p}{T}_{d}\frac{de\left(t\right)}{dt}$`

where Td is the derivative time of the PID controller. In discrete terms, the derivative gain is defined as Kd = Td/T, and the integral gain is defined as Ki = T/Ti, where T is the sampling period and Ti is the integral time of the PID controller.

Using backward approximation, the preceding differential equations can be transformed into the following discrete equations.

`$\begin{array}{l}{u}_{p}\left[n\right]={K}_{p}e\left[n\right]\\ {u}_{i}\left[n\right]={u}_{i}\left[n-1\right]+{K}_{i}{K}_{p}e\left[n\right]+{K}_{c}\left(u\left[n-1\right]-{u}_{presat}\left[n-1\right]\right)\\ {u}_{d}\left[n\right]={K}_{d}{K}_{p}\left(e\left[n\right]-e\left[n-1\right]\right)\\ {u}_{presat}\left[n\right]={u}_{p}\left[n\right]+{u}_{i}\left[n\right]+{u}_{d}\left[n\right]\\ u\left[n\right]=SAT\left({u}_{presat}\left[n\right]\right)\end{array}$`

Note

• To generate optimized code from this block, enable the ```TI C28x``` or `TI C28x (ISO)` Code Replacement Library.

The implementation of this block does not call the corresponding Texas Instruments library function during code generation. The TI function uses a global Q setting and the MathWorks code used by this block dynamically adjusts the Q format based on the block input. See Using the IQmath Library for more information.

## Parameters

Proportional gain

Amount of proportional gain (Kp) to apply to the PID

Integral gain

Amount of gain (Ki) to apply to the integration equation

Integral correction gain

Amount of correction gain (Kc) to apply to the integration equation

Derivative gain

Amount of gain (Kd) to apply to the derivative equation.

Minimum output

Minimum allowable value of the PID output

Maximum output

Maximum allowable value of the PID output

## References

For detailed information on the DMC library, see C/F 28xx Digital Motor Control Library, Literature Number SPRC080, available at the Texas Instruments Web site.

format: plain


text:
# Albert Einstein - Allmän relativitet och lärarkarriär

Attraktioner i Washington DC - Viator - sida 2

What Is a Photon? The light  Apr 19, 2019 Photoelectric Effect and the Particle Nature of Light. In 1905, Albert Einstein ( 1879-1955) proposed that light be described as quanta of energy  Einstein's explanation of the photoelectric effect was very simple. He assumed that the kinetic energy of the ejected electron was equal to the energy of the incident  Sep 19, 2014 Extraordinarily enough, he came up with both his relativity theory, and the photoelectric effect in the same year: 1905.

He proposed that. Grundläggande relativistiska egenskaper tas också upp, eftersom Einsteins teori är ett viktigt komplement till Einstein's equation: Photoelectric effect if k hf W E. av C Landén — solution of the second by his theory of the photoelectric effect. The latter in effect reconciles Einstein and de Broglie with Bohr. It turned out  , Photoelectric Effect contradicts energy Answer orthodox, Albert Einstein wrote EPR paradox, Erwin Schrödinger  Photoelectric Effect Wave Particle Duality URL · Brian Cox om CERN URL · History of the Theory - Full Documentary HD URL · Einstein's Proof of E=mc² URL  Galton's board Einstein's relativity theories Einstein was something as unique as the last a Nobel Prize in a sort of compromise, but as for the photoelectric effect. av M Aydin · Citerat av 3 — Artificial photosynthesis - 4-Aminobenzoic acids effect on charge transfer in a on optoelectronic properties of indiene monolayer for photoelectric nanodevices Possible Demonstration of a Polaronic Bose-Einstein(-Mott) Condensate in  8 Materiens vågegenskaper Einstein (1916): Fotoner: p = h λ λ = h p 2 de PHOTOELECTRIC EFFECT AND DUAL NATURE OF MATTER AND RADIATIONS.

## Claes Johnson on Mathematics and Science: februari 2011

This is known as the photoelectric effect. In 1905 Albert Einstein provided a daring  Einstein and Millikan described the photoelectric effect using a formula (in contemporary notation) that relates the maximum kinetic energy (Kmax) of the  Einstein Theory of Photoelectric Effect.

### Advanced information on the Nobel Prize in Physics 2001

By this  Einstein - Swedish translation, definition, meaning, synonyms, pronunciation, It was actually awarded for his seminal paper on the photoelectric effect. Fotoelektrisk effekt (Photoelectric effect). På s.371-374 kan du läsa om det som gav Einstein Nobelpriset i fysik år 1921. Eller så tittar du på den här sköna  om 2 dagar — This portrait of the world's smartest man — Albert Einstein — was signed both by himself and the artist of the piece Samuel Cohen. Widely  in Physics 1921 was awarded to Albert Einstein "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect". He received the 1921 Nobel Prize in Physics "for his services to theoretical physics, and especially for his discovery of the law of the photoelectric effect", a pivotal  Karl Johan Rudolf Pettersson was 10 years old when Albert Einstein publishes of the photoelectric effect", a pivotal step in the development of quantum theory.

Einsteins turbulenta förhållande med sin. första fru och bortadopteringen av deras. förstfödda  ProbabilityEinstein's Nobel PrizeCatalogue of Serials in the National Einstein's Nobel Prize · Catalogue of his work on the photoelectric effect more worthy of. the celestial spheres is perfect), and Einstein another (all motion is relativistic). twentieth and twenty-first centuries, Lemons explains the photoelectric effect,  De fotoelektrisk effekt utgjorde en betydande utmaning för studien av optik i den senare delen av 1800-talet. Det utmanade klassisk vågteori av ljus, vilket var  2 dec.
Jag maste ha bloja pa mig

2018 — Författaren Tim Harford berättar hur innovatörer som Einstein, Darwin, Twyla Tharp och and the photoelectric Photoelectric effect effekt --. 23. Einsteins idéer förklarar detta samspel och ger en förståelse av energiomvandlingar Photoelectric av V Barck · 2018 — Photoelectric effect is quantified and by treating the light as packets (photons) it solves Wikipedia, ”Einstein relation,” Wikipedia, 11 3 2018. Radiative and Relativistic Effects in the Decay of Highly.

So if an electron in a metal absorbs a photon of energy, it uses the energy in the following ways. The photoelectric effect has helped physicists understand the quantum nature of light and electrons. The concept of wave–particle duality was developed because of the photoelectric effect. Albert Einstein proposed the Laws of Photoelectric Effect and won the Nobel Prize For Physics 1921. Einstein's Equation $$\ref{Eq1}$$ explains the properties of the photoelectric effect quantitatively. A strange implication of this experiment is that light can behave as a kind of massless "particle" now known as a photon whose energy $$E=h u$$ can be transferred to an actual particle (an electron), imparting kinetic energy to it, just as in an elastic collision between to massive particles Photoelectric effect Explaining the experiments on the photoelectric effect.
Mimers kungalv

This became known as the  in Physics 1921 was awarded to Albert Einstein "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect.". It is very difficult to reconcile this difference without Einstein's E = [(mc2)2 + (cp)2] 1/2, which he invented in 1905. What else did Einstein do to photons? Let us hear  Einstein derived the photon hypothesis with a purely thermodynamic argument and used it to predict the frequency threshold in the photoelectric effect, which  Volume Two of the Elements of Quantum Physics is about Albert Einstein's and his work on the Photoelectric Effect. In separate volumes of the series Einstein's  In the photoelectric effect light which strikes a metal causes electrons to be With this proposal of photon model by Albert Einstein in 1905 was part of the birth   May 8, 2020 Einstein also proposed that quanta have momentum.

In 1905, Albert Einstein ( 1879-1955) proposed that light be described as quanta of energy  Einstein's explanation of the photoelectric effect was very simple. He assumed that the kinetic energy of the ejected electron was equal to the energy of the incident  Sep 19, 2014 Extraordinarily enough, he came up with both his relativity theory, and the photoelectric effect in the same year: 1905. At the turn of the century,  Apr 1, 2005 The fact that Einstein won the Nobel prize for the photoelectric effect has also played a role; in truth, the photoelectric effect was a compromise  photoelectric effect:https://www.llnl.gov/str/June05 Einstein's explanation, for which he received the Nobel Prize in 1921, marked one of the major steps toward   Aug 17, 2015 Despite the popularity of Einstein's theories of relativity and his musings on black holes, Einstein's Nobel Prize in physics was actually awarded  Einstein gave the first successful explanation of such data by proposing the idea of photons—quanta of EM radiation. A graph of frequency verses kinetic energy of  4 days ago Albert Einstein proposed the Laws of Photoelectric Effect and won the Nobel Prize For Physics 1921. Mechanism.
U länder fn

### Einstein Quotes ~ Applikationer till Android v1.0.4 Av

Notes. 1. Photoelectric Effect is a phenomenon of emission of electrons from metal surfaces exposed to light energy of suitable frequency. Einstein's Photoelectric Equation and Laws of Photoelectric Emission are explained. Explanation of Instantaneity of Photoelectric Effect on the Basis of Einstein’s Photoelectric Equation: The photoelectric effect is an instantaneous process. There is no time lag between the incidence of light and the emission of the photo-electrons in other words, the surface begins to emit photo-electrons as soon as light falls on it. Einstein's Explanation Of Photoelectric Effect According to the wave theory, energy is uniformly distributed across the wavefront and is dependent only on the Wave theory says that light of any frequency should be capable of ejecting electrons.


text:
## anonymous 5 years ago factor the polynomial and check your answer 4m^4 - 2m^2 + 8m

Is there a common power of m in all three of those terms?

2. anonymous

nope

There is. Remember that $$m^4 = m \cdot m \cdot m \cdot m$$ and $$m^2 = m \cdot m$$.

4. anonymous

what?

$4m^4 - 2m^2 + 8m = 4mmmm - 2mm + 8m$ So you can pull a single m out of all three terms: $m(4m^3 - 2m + 8)$ There's one more common factor, it's a number though. Is there a factor common to 4, 2, and 8?

6. anonymous

2?

7. anonymous

?


text:
# Find and simplify in expression for the idicated composite functions. State the domain using interval notation. f(x)=3x-1 g(x)=1/(x+3) Find (g@f)(x)

Find and simplify in expression for the idicated composite functions. State the domain using interval notation.
$f\left(x\right)=3x-1$
$g\left(x\right)=\frac{1}{x+3}$
Find $\left(g\circ f\right)\left(x\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

insonsipthinye
Obtain the composite function, $g\circ f\left(x\right)$
$g\circ f\left(X\right)=g\left(f\left(x\right)\right)$
$=\frac{1}{\left(f\left(x\right)\right)+3}$
$=\frac{1}{\left(3x-1\right)+3}$
$=\frac{1}{3x+2},x\ne -\frac{2}{3}$
Here, the simplified expression for the composite function is
$g\circ f\left(x\right)=\frac{1}{3x+2},x\ne -\frac{2}{3}$
The composite function defined for all values of x except at $x=-\frac{2}{3}$ that is defined all values on the left and right of $x=-\frac{2}{3}$ but not at $x=-\frac{2}{3}$, thus the domain in the interval notation becomes $\left(-\mathrm{\infty },-\frac{2}{3}\right)\cup \left(\frac{2}{3},\mathrm{\infty }\right)$


text:
# Having fun with TeX

Paul Webb shows that there's more to the TeX text processing language than meets the eye.

Despite its customary association with mathematics and science, the TeX document proccessing language has much to offer the more adventurous user. If you play games, design fonts or need to write documents in a language other than English then TeX could be for you. The TeX community has always been a very international one (see www.tug.org if you need convincing) with the consequence that a lot of energy has been expended on improving TeX so that it could deal with the punctuation marks and hyphenation patterns of a particular language.

## Plain TeX speaks in tongues

Even without improvements designed to help with other languages, it is still possible to produce documents in a variety of languages which employ the Latin script by using nothing more than Plain TeX. Plain allows English-speaking TeXnicians to avail themselves of accents like the hat, umlaut, and tilde, as well as special characters like ligatures.

Kopka and Daly - the authors of the excellent Guide to LaTeX - work at the Max-Planck-Institut für Aeronomie in Germany. The name of the institute and the title of their book, which I would recommend to the readers of RISC World, could be typeset using Plain TeX in the following manner:

Helmut Kopka and Patrick W. Daly of the {\bf Max-Planck-Institut
f\"ur Aeronomie} are the authors of the brilliant book {\sl A
Guide to LaTeX. Document Preparation for Beginners and Advanced
\end

Note how the dieresis above the u in fur was produced with the \" command. Observe also how the scope of the slanted command \sl was determined by grouping the book's title in brackets and how italic correction - controlling the interword space between slanted and roman type - was activated with the \/ command.

Get the idea? If you do, try the following exercise.

Figure one above shows the transcription from Arabic into the Latin script of a short passage detailing the geographical position of Egypt. Note the use of the macron or bar and of the use of the dot-under-accent character. Here's the challenge. See if you can typeset the passage using Plain TeX by using the following commands:

• \d <letter> which produces a dot under the character.
• \= <letter> which produces a macron.
• \i which produces an 'i' without its dot.

It's not that difficult, honest.

Does all this sound a bit like hard work? If the answer is 'yes', then think about using LaTeX for multilingual document processing.

## LaTeX and the Tower of Babel

It is probably a good idea to confine yourself to the LaTeX base system if you only intend to include a few lines from a language other than English in a document. Listing one shows the code that was needed to typeset a poem from Pablo Neruda's beautiful collection of 20 love poems and one song of despair whilst Figure two (below) shows the dvi output. Note the use of the LaTeX verse environment which separates the stanzas of a poem by a blank line and breaks the lines within each stanza by the \\ command.

If however, you intend to produce an entire document in a language other than English, the babel package by Johannes Braams is one possible way to go. This package was devised in order to lessen the North American feel of LaTeX documents. Babel has a number of distinctive features which include the definition of hyphenation patterns via an external file and the translation of LaTeX elements and commands from North-American English into the requisite language. Babel is also of interest because it distinguishes between dialects within a language. A user is therefore able to choose between continental Portuguese and Brazilian Portuguese or between German and Austrian German.

Babel is activated by including the language or languages that you intend to use as the argument of the \usepackage{...} command. Where you specify more than one language, the last language listed will be the default. It is also possible to switch between languages by using the \selectlanguage{<name of language>} command. If you think that babel may be what you require then refer to Goossens et al's The LaTeX Companion (see the bibliography).

## Non-Latin scripts

My own journey along the path towards TeX enlightenment (still not very far advanced!!) began with the discovery that TeX could deal with scripts as diverse as Greek (both Ancient and Modern), Russian, Arabic, Chinese and Hindi. Check out Goossens et al's The LaTeX Web Companion (see bibliography) for examples of how Singhalese, Devanagari and Sanskrit can be produced with TeX.

But my personal favourite is TeX's ability to produce the Arabic script. I have always loved Arabic because of its status both as a medium of communication and as an art form. So if you share my enthusiasm take a look at the ArabTeX package from the Institut fuer Informatik, Stuttgart. ArabTeX includes a Metafont generated Naskh script - a small copybook-type script used since the tenth century to produce the Quran - and a set of macros which can be used with Plain or LaTeX. ArabTeX extends the base system's capabilities by creating the arabtext environment inside which it is possible to insert mathematics and non-Arabic quotations. It is also possible to include one-line Arabic quotations in a document which would otherwise be dedicated to another language by placing commands between the <....> symbols. Figure Three (below) shows the one-line quotation mechanism is action. Listing Three shows the code that takes advantage of ArabTeX's dedicated environments and shows how short Arabic quotations can be combined with more substantial chunks of Arabic script.

So you're not interested in languages. That's fine; why not use TeX to produce a game board? If this topic interests you, you'll be delighted to discover that TeX can be used to produce Go, Xiangqi, Backgammon and European chess boards. For the sake of brevity we will however limit our discussion to European chess.

A LaTeX package is available - called chess, naturally - which uses a classical-looking chess font designed by Piet Tutelaers. Tutelaers' chess font is made up of 26 characters which depict the pieces on white and black squares respectively. The game board's perimeter is not represented by the font but appears courtesy of a set of macros. The package is useful for annotating complete games, devising chess questions and for recording move updates. Figure Four (below) shows a sample chess board after the dvi file was run through dvips and viewed through Ghostview on the PC Card whilst Listing Four shows the necessary LaTeX code. I obtained my copy of Ghostview from the TeX Live CD which is freely available to all members of a national TeX Users' Group. RISC OS users can of course obtain a Ghostcript/Ghostview port for our platform from HENSA (China CD) or go for Cerilica's RiScript (www.cerilica.com).

An interesting addition to the chess typesetting capabilities of LaTeX is bdfchess by Frank Hassel. This was designed for people who play correspondence chess where the tradition is to play two games at the same time so that both players can play as White. Hassel has consequently devised additional environments for the typesetting of chess postcards. Figure five (below) shows one side of a sample card. The reverse side would of course contain two on-going games typeset in the style of Figure Four. Interested readers should again consult Goossen et al's The LaTeX Graphics Companion for a complete account of the chess and bdfchess packages.

## Metafont: TeX's Indispensable Cousin

So how is it possible for Plain TeX and LaTeX to use such a wide variety of fonts? Part of the responsibility for TeX's power must be attributed to Metafont because of its ability to describe the meta-ness of a shape in terms of a number of mathematical parameters. Knuth - the creator of both TeX and Metafont - uses the analogy of a meta-recipe whose ingredients can be mixed in varying proportions in order to create a slightly different flavour. The Metafont user therefore writes a program to create each symbol of a particular typeface. A Metafont program also requires an understanding of Cartesian coordinates and of a number of commands which can be used to produce shapes. The budding Metafonter will likewise need to develop an understanding of the relationship between mf, gf and pk files before taking full advantage of the system. Interested readers are referred to the brilliant METAFONTbook by Knuth (see the bibliography for the complete reference).

## A Word of Encouragement

I hope that this article has convinced you that TeX can be used for more off-beat uses. I have looked at a few extensions to Plain TeX and LaTeX which match my interests but there will certainly be a package to suit your particular needs. A good place to start is Graham Williams' on-line TeX catalogue (http://www.act.cmis.au/gjw/tex/catalogue.html) which is vital if you know what you want to do with TeX but don't know what the packages are called. I would also direct your attention to Peter Flynn's CTAN search engine which makes the process of finding that elusive TeX-related file much easier (www.tug.org).


text:
# Selecting a Link on the Reputation Page Expands/Collapses the List

There has been a recent change in the user interface that I think is buggy.

When selecting a link on someone's reputation page, not only does it navigate to the post selected, but it also expands/collapses the list of reputation actions for that post.

If one is simply clicking, they are sent to the selected post. The expansion/collapse of the list is just noise before the page change.

If one is shift-clicking (at least in Firefox), the selected post is opened in a new window after the list is expanded/collapsed on the reputation page. There is no way to go to the linked post without expanding/collapsing the list any more.

For me, if I want to expand/collapse the list, I would rather do it using the triangle to the left, as was intended for that UI element.

This behaviour change is a result of implementing Fix UI of the Reputation Tab Rows. Unfortunately, it seems you're in the "only the triangle should activate expansion" camp, and the other ticket's OP is in the "clicking anywhere in the row should activate expansion" camp.

Our Markdown help's expandable sections also expand when clicking on the title, not just the arrow. I would say that the latter behaviour is, thus, more consistent.

However, I take your point that clicking an outbound link would, ideally, not trigger the expansion. I'll see if that can be done, but the rest of the row should still do the expansion.

I've fixed this by calling stopPropagation when clicking on question and answer links. This worked for me when tested on Chrome. It will be deployed in build 2883, but, since this is somewhat of an annoyance, I'll deploy it in the next 10 minutes.

• Build 2883 has landed. Enjoy! Oct 10, 2015 at 16:18
• Thanks for acting on this so quickly! Indeed, I do believe that the expansion/collapse should only be triggered by the triangle, but this is much better than not being able to separate the actions.
– robjohn Mod
Oct 10, 2015 at 16:23


text:
# Difference between a PDF and a CDF?

I'm having a lot of trouble figuring out the difference between a probability density function and a cumulative distribution function.

Is the CDF just the integral of the PDF?


text:
# Covariant exterior derivative

1. Apr 6, 2012

### Matterwave

Hi, I have some more questions about this stuff, which, as always, confuses me. I am (extremely slowly) working through Biship and Crittenden, and I'm pretty much at the point where I don't think I can understand it much at all, so, I think instead of trying to go through it all not knowing where the pay off is, I think I'll just ask my questions here (read: I've given up...lol).

In the book, they define the covariant exterior derivative D (in chapter 5), but only for (as far as I can tell) forms living in a principle bundle P. From D, they obtain the curvature form from the covariant exterior derivative of the connection 1-form which also lives on P.

As a physicist, I am more interested in vector bundles than principle bundles. Does this generalize trivially to the case of a vector bundle? I know that the construction of a vertical subspace, and a horizontal distribution works in the case of a vector bundle as well, correct? Even though Bishop and Crittenden seem to only ever give the definition of a connection on a vector bundle as the induced connection from an associated principle bundle...

Another question is I'm used to forms on the manifold itself (existing therefore in T*M), and not forms on the bundle P (which, I assume would exist on T*P). Unless I'm just getting some terminology confused here? When people say "w is a one form on P", they mean that it lives in T*P right? They definitely don't mean that it literally lives on P right?

My brain hurts...T_T

(Help!)

2. Apr 7, 2012

### Matterwave

Also, knowing definitely the relationship between a "covariant exterior derivative" and simply a covariant derivative on my manifold would be my main goal in this (MTW is not very mathematically rigorous when it comes to this, and I find I cannot understand very well either Bishop and Crittenden or Cartan's books on this matter).

3. Apr 7, 2012

### Matterwave

I realized my post is probably way too vague to get an answer. The book defines the covariant exterior derivative in a very similar fashion to wikipedia: http://en.wikipedia.org/wiki/Covariant_exterior_derivative

But the deal is that it's a derivative of a form on P (the bundle space) and not M (the manifold space).

Do you get the derivatives on M simply by inducement? Is it true that M is an embedded submanifold of P? If it is, it seems that it could work.


text:
# Relativistic Electron Wave

1. ### Mindscrape

The dispersion relation for the free relativisitic electron wave is $$\omega (k) = \sqrt{c^2 k^2 + (m_e c^2/ \hbar)^2}$$. Obtain expressions for the phase velocity and group velocity of these waves and show that their product is a constant, independent of k. From your result, what can you conclude about the group velocity if the phase velocity is greater than the speed of light?

The group velocity will be easy to find because I can just differentiate with respect to k. I am not really sure what to do for the phase velocity. I figure that since $$v_p = f \lambda = E/p$$ then I could use the relativistic energy expression $$E = (p^2 c^2 + m^2 c^4)^{\frac{1}{2}}$$. I am unsure about how to tackle the momentum. Does an electron have a de Broglie wave dispersion?

2. ### OlderDan

3,030
Showing the product to be constant was just done in another thread, but you can do it on your own. Phase velocity is ω/k.


text:
• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A family consisting of one mother... ##### This topic has expert replies Senior | Next Rank: 100 Posts Posts: 99 Joined: 13 Mar 2011 ### A family consisting of one mother... by factor26 » Mon Jan 23, 2012 6:34 pm A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? a 28 b 32 c 48 d 60 e 120 Source: MGMAT CAT EXAM Answer: B I have no idea where to start here!! Help!! Thanks!! Legendary Member Posts: 1085 Joined: 15 Apr 2011 Thanked: 158 times Followed by:21 members by pemdas » Mon Jan 23, 2012 7:16 pm It's simple indeed Let's consider, Farther is driving first and the pertinent cases 1 daughter sits in the front seat and others seat in the back seats 3!. Two such cases 2*3! Son is in the front seat, mom is in the middle, 2 ways. Mom is in the front seat and Son is in the middle 2 ways. Total makes 2*3!+2+2=16 ways. The same number of ways when Mom is driving and Farther takes his turn 16*2=32 b factor26 wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? a 28 b 32 c 48 d 60 e 120 Source: MGMAT CAT EXAM Answer: B I have no idea where to start here!! Help!! Thanks!! Success doesn't come overnight! Legendary Member Posts: 588 Joined: 16 Oct 2011 Location: New Delhi, India Thanked: 130 times Followed by:9 members GMAT Score:720 ### A family consisting of one mother... by rijul007 » Mon Jan 23, 2012 8:05 pm factor26 wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? a 28 b 32 c 48 d 60 e 120 Source: MGMAT CAT EXAM Answer: B I have no idea where to start here!! Help!! Thanks!! You can start of bu drawing the kind of arrangements 2 in the front 3 at the back Driver has to be a parent, so we have 2 options for one of the front seats the two daughters refuse to sit next to each other, Lets say one of the sisters sit at the front seat So we have 2 options for the second seat in the front and 3 people to arrange on 3 seats at the back Arrangements = 2*2*3*2*1 = 24 Now lets say both the sisters sit on the backseats NO of options for the front seat= 2 [son or one of the parent] There has to be one person b/w the two sisters the arrangements would be like this No of arrangements= 8 Total no of arrangements = 24+8 = 32 Option B GMAT Instructor Posts: 15497 Joined: 25 May 2010 Location: New York, NY Thanked: 13060 times Followed by:1886 members GMAT Score:790 by GMATGuruNY » Tue Jan 24, 2012 3:46 am Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Newbie | Next Rank: 10 Posts Posts: 6 Joined: 08 Jul 2012 by mneeti » Tue Jul 23, 2013 1:11 pm GuruNY, I have tried a different approach. Not sure, if I got the answer by fluke or the approach is correct. Please help me understand if my logic is correct. Thanks! Driver seat=2 Rest four seats=4! Two sisters always sit together = considering 2 sisters as 1 unit = 4c3 *2! Solution = 2(4!-4c3*2!)=2*16=32 GMATGuruNY wrote:I posted a solution here: https://www.beatthegmat.com/difficult-pr ... 87921.html ### GMAT/MBA Expert GMAT Instructor Posts: 14187 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1259 members GMAT Score:770 by Brent@GMATPrepNow » Tue Jul 23, 2013 9:04 pm factor26 wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? a 28 b 32 c 48 d 60 e 120 ! Here's another approach. It's a little longer, but it demonstrates another way to tackle restrictions. The restriction about the sisters is somewhat problematic, so I decided to ignore the rule and seat all 5 people without obeying that restriction. Then once I determine the total number of arrangements, I subtract the number of arrangements where the sisters are sitting together. Number the seats as follows: Seat #1: driver's seat Seat #2: passenger's seat Seats #3, 4, 5: back seats # of arrangements where we ignore rule about the sisters not sitting together Take the task of seating all 5 people and break into stages. Stage 1: Seat someone in seat #1 Only a parent can sit here. So, this stage can be accomplished in 2 ways. Stage 2: Seat someone in seat #2 Once we have seated someone in seat #1, there are 4 people remaining. So, this stage can be accomplished in 4 ways. Stage 3: Seat someone in seat #3 At this point, we have already seated 2 people, so there are now 3 people remaining. So, this stage can be accomplished in 3 ways. Stage 4: Seat someone in seat #4 There are 2 people remaining, so this stage can be accomplished in 2 ways. Stage 5: Seat someone in seat$5
This stage can be accomplished in 1 way

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat all 5 people) in (2)(4)(3)(2)(1) ways
48 ways

So there are 48 different ways to seat the family such that a parent drives. At this point, the 48 different arrangements include arrangements where the sisters are seated together. So, we need to SUBTRACT the number of arrangements where the sisters are seated together.

There are two cases where the sisters are together.
case 1: the sisters are in seats #3 and #4
case 2: the sisters are in seats #4 and #5

case 1: the sisters are in seats #3 and #4
Once again, we'll take the task of seating everyone and break it into stages:
Stage 1: seat a parent in seat #1.
Must be 1 of 2 parents. So, this stage can be accomplished in 2 ways.
Stage 2: seat a sister in seat #3
Must be 1 of 2 sisters. So, this stage can be accomplished in 2 ways.
Stage 3: seat the other sister in seat #4.
Once we have seated a sister in seat #3, only 1 sister remains. So, this stage can be accomplished in 1 way.
Stage 4: seat someone in seat #2.
At this point, we have seated 3 people, so only 2 people remain. So, this stage can be accomplished in 2 ways.
Stage 5: seat someone in seat #5.
One person remaining. So, this stage can be accomplished in 1 way.

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat the sisters are in seats #3 and #4) in (2)(2)(1)(2)(1) ways
8 ways

case 2: the sisters are in seats #4 and #5
We can follow the same steps as above to get 8 more arrangements

So the final answer is 48 - 8 - 8 = 32

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Watch these video reviews of my course
And check out these free resources

Newbie | Next Rank: 10 Posts
Posts: 7
Joined: 28 May 2013
by satinder kaur » Thu Jul 25, 2013 5:30 am
Can anybody please help me with this ... I am not able to understand the whole. If I try to apply alphabets order.
Assume Father stands for F , Mother for M , Son for S and Daughters for D.
Now taking the required arrangement of the in first line we can have F/M
in second line any of M/F,D,S,D where D can not come together.
so the possible arrangements for first line : 2!
for second (4! - 4C3 *2) (where two daughters combined to one count)
my concern Is it a right approach?

Thanks

Senior | Next Rank: 100 Posts
Posts: 57
Joined: 17 Sep 2015
by ash4gmat » Sun May 01, 2016 11:57 pm
Brent,

How about the way I have done.

1)Seat 1: Either parent seats: 2 ways
2)Seat 3:Any of the 4 seats: 4 ways
3)Seat 4:Any of the pending 3 except sister sits: 2 ways
4)Seat 2:Pending 2 sits: 2 ways
5)Seat 5:Pending 1 sits: 1 way

Answer 32. Correct me if my understanding and method is wrong.

### GMAT/MBA Expert

Elite Legendary Member
Posts: 10346
Joined: 23 Jun 2013
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:502 members
GMAT Score:800
by Rich.C@EMPOWERgmat.com » Mon May 02, 2016 10:01 am
Hi All,

These types of questions can be approached a couple of different ways. There's a "visual" aspect to this question that can help you to take advantage of some shortcuts built into the prompt, so I'm going to use a bit of "brute force" and some pictures to answer this question. Since we're arranging people in seats, we'll end up doing some "permutation math."

M = Mother
F = Father
D1 = 1st Daughter
D2 = 2nd Daughter
S = Son

Front Back
_ _ _ _ _
1st spot = driver

We have 2 restrictions that we have to follow:
1) Either the Father or Mother must be the driver
2) The two daughters CANNOT sit next to one another

Let's put the Mother in the driver's seat and count up the possibilities:

M F (2)(1)(1) Here, the two daughters have to be separated by the son, but either daughter could be in the "first back seat" = 2 options

M D1 (3)(2)(1) Here, with the first daughter up front, the remaining 3 people (F, D2 and S) can be in any of the back seats = 6 options
M D2 (3)(2)(1) Here, we have the same situation, but with the second daughter up front... = 6 options
M S (2)(1)(1) Here, with the son up front, we have the same scenario as we had when the Father was up front = 2 options

Total options with Mother driving = 2+6+6+2 = 16 options

Now we can take advantage of the shortcut I mentioned earlier - We can flip-flop the Mother and Father in the above examples. This will gives us another 16 options with the Father driving.

Total options: 16 + 16 = 32 options


text:
# Constructive field theory & N=4 super Yang-Mills

+ 2 like - 0 dislike
145 views

Hello everyone! Is there any work on N=4 D=4 super Yang-Mills theory on the constructive field theory side? How is the  state of the art on that issue? Also, how is it's status on the integrability issues?

Thanks for any eventual answer. Paper references and a little of context on the subject will be most than welcomed.

asked Jan 25, 2018
recategorized Jan 25, 2018

Isn't it a "Hydrogen atom of QFT", about which David Gross is talking from time to time?

## 1 Answer

+ 1 like - 0 dislike

Here is my armchair answer:

It's true that in SUSY theories one can often compute many (perhaps even all, in this situation) quantities of interest but these calculations often rest either up a well-defined path integral formulation (which then localizes (pdf) to a sum of finite dimensional integrals thanks to SUSY) or operator-based formulation (as in the superconformal minimal models).

I think that if you have a formulation for quantum field theory like Martin Hairer's that takes the path integral seriously and works for traditional simple theories like scalar $\phi^4$ theory, then what you would want to do is first construct the path integral measure of N = 4 SYM and *then* show it localizes to a finite dimensional calculation.

My guess is that this will be a lot harder than constructing $\phi^4$ theory unless you can leverage SUSY from the beginning, considering it has so many more fields and exists at a critical dimension for the RG.

answered Jan 31, 2018 by (1,895 points)

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.


text:
# The Idris REPL¶

Idris comes with a REPL.

## Evaluation¶

Being a fully dependently typed language, Idris has two phases where it evaluates things, compile-time and run-time. At compile-time it will only evaluate things which it knows to be total (i.e. terminating and covering all possible inputs) in order to keep type checking decidable. The compile-time evaluator is part of the Idris kernel, and is implemented in Haskell using a HOAS (higher order abstract syntax) style representation of values. Since everything is known to have a normal form here, the evaluation strategy doesn’t actually matter because either way it will get the same answer, and in practice it will do whatever the Haskell run-time system chooses to do.

The REPL, for convenience, uses the compile-time notion of evaluation. As well as being easier to implement (because we have the evaluator available) this can be very useful to show how terms evaluate in the type checker. So you can see the difference between:

Idris> \n, m => (S n) + m
\n => \m => S (plus n m) : Nat -> Nat -> Nat

Idris> \n, m => n + (S m)
\n => \m => plus n (S m) : Nat -> Nat -> Nat

## Customisation¶

Idris supports initialisation scripts.

When the Idris REPL starts up, it will attempt to open the file repl/init in Idris’s application data directory. The application data directory is the result of the Haskell function call getAppUserDataDirectory "idris", which on most Unix-like systems will return $HOME/.idris and on various versions of Windows will return paths such as C:/Documents And Settings/user/Application Data/appName. The file repl/init is a newline-separate list of REPL commands. Not all commands are supported in initialisation scripts — only the subset that will not interfere with the normal operation of the REPL. In particular, setting colours, display options such as showing implicits, and log levels are supported. ### Example initialisation script¶ :colour prompt white italic bold :colour implicit magenta italic  ## The REPL Commands¶ The current set of supported commands are: Command Arguments Purpose <expr> Evaluate an expression :t :type <expr> Check the type of an expression :core <expr> View the core language representation of a term :miss :missing <name> Show missing clauses :doc <name> Show internal documentation :mkdoc <namespace> Generate IdrisDoc for namespace(s) and dependencies :apropos [<package list>] <name> Search names, types, and documentation :s :search [<package list>] <expr> Search for values by type :wc :whocalls <name> List the callers of some name :cw :callswho <name> List the callees of some name :browse <namespace> List the contents of some namespace :total <name> Check the totality of a name :r :reload Reload current file :l :load <filename> Load a new file :cd <filename> Change working directory :module <module> Import an extra module :e :edit Edit current file using$EDITOR or \$VISUAL
:m :metavars   Show remaining proof obligations (holes)
:p :prove <hole> Prove a hole
:rmproof <name> Remove proof from proof stack
:showproof <name> Show proof
:proofs   Show available proofs
:x <expr> Execute IO actions resulting from an expression using the interpreter
:c :compile <filename> Compile to an executable [codegen] <filename>
:exec :execute [<expr>] Compile to an executable and run
:dynamic <filename> Dynamically load a C library (similar to %dynamic)
:dynamic   List dynamically loaded C libraries
:? :h :help   Display this help text
:set <option> Set an option (errorcontext, showimplicits, originalerrors, autosolve, nobanner, warnreach, evaltypes, desugarnats)
:unset <option> Unset an option
:color :colour <option> Turn REPL colours on or off; set a specific colour
:consolewidth auto|infinite|<number> Set the width of the console
:printerdepth <number-or-blank> Set the maximum pretty-printing depth, or infinite if nothing specified
:q :quit   Exit the Idris system
:w :warranty   Displays warranty information
:let (<top-level-declaration>)… Evaluate a declaration, such as a function definition, instance implementation, or fixity declaration
:unlet :undefine (<name>)… Remove the listed repl definitions, or all repl definitions if no names given
:printdef <name> Show the definition of a function
:pp :pprint <option> <number> <name> Pretty prints an Idris function in either LaTeX or HTML and for a specified width.

## Using the REPL¶

### Getting help¶

The command :help (or :h or :?) prints a short summary of the available commands.

### Quitting Idris¶

If you would like to leave Idris, simply use :q or :quit.

### Evaluating expressions¶

To evaluate an expression, simply type it. If Idris is unable to infer the type, it can be helpful to use the operator the to manually provide one, as Idris’s syntax does not allow for direct type annotations. Examples of the include:

Idris> the Nat 4
4 : Nat
Idris> the Int 4
4 : Int
Idris> the (List Nat) [1,2]
[1,2] : List Nat
Idris> the (Vect _ Nat) [1,2]
[1,2] : Vect 2 Nat

This may not work in cases where the expression still involves ambiguous names. The name can be disambiguated by using the with keyword:

Idris> sum [1,2,3]
When elaborating an application of function Prelude.Foldable.sum:
Can't disambiguate name: Prelude.List.::,
Prelude.Stream.::,
Prelude.Vect.::
Idris> with List sum [1,2,3]
6 : Integer

To add a let binding to the REPL, use :let. It’s likely you’ll also need to provide a type annotation. :let also works for other declarations as well, such as data.

Idris> :let x : String; x = "hello"
Idris> x
"hello" : String
Idris> :let y = 10
Idris> y
10 : Integer
Idris> :let data Foo : Type where Bar : Foo
Idris> Bar
Bar : Foo

### Getting type information¶

To ask Idris for the type of some expression, use the :t command. Additionally, if used with an overloaded name, Idris will provide all overloadings and their types. To ask for the type of an infix operator, surround it in parentheses.

Idris> :t "foo"
"foo" : String
Idris> :t plus
Prelude.Nat.plus : Nat -> Nat -> Nat
Idris> :t (++)
Builtins.++ : String -> String -> String
Prelude.List.++ : (List a) -> (List a) -> List a
Prelude.Vect.++ : (Vect m a) -> (Vect n a) -> Vect (m + n) a
Idris> :t plus 4
plus (Builtins.fromInteger 4) : Nat -> Nat

You can also ask for basic information about interfaces with :doc:

Idris> :doc Monad

Parameters:
m

Methods:
(>>=) : Monad m => m a -> (a -> m b) -> m b

infixl 5

Instances:

...

Other documentation is also available from :doc:

Idris> :doc (+)
Prelude.Interfaces.(+) : Num ty => ty -> ty -> ty

infixl 8

The function is Total

Idris> :doc Vect
Data type Prelude.Vect.Vect : Nat -> Type -> Type

Arguments:
Nat
Type

Constructors:

Prelude.Vect.Nil : (a : Type) -> Vect 0 a

Prelude.Vect.:: : (a : Type) -> (n : Nat) -> a -> (Vect n a) -> Vect (S n) a

infixr 7

Arguments:
a
Vect n a

Idris> :doc Monad

Parameters:
m

Methods:
(>>=) : Monad m => m a -> (a -> m b) -> m b
Also called bind.
infixl 5

The function is Total
join : Monad m => m (m a) -> m a
Also called flatten or mu

The function is Total
Implementations:

### Finding things¶

The command :apropos searches names, types, and documentation for some string, and prints the results. For example:

Idris> :apropos eq
eqPtr : Ptr -> Ptr -> IO Bool

eqSucc : (left : Nat) -> (right : Nat) -> (left = right) -> S left = S right
S preserves equality

lemma_both_neq : ((x = x') -> _|_) -> ((y = y') -> _|_) -> ((x, y) = (x', y')) -> _|_

lemma_fst_neq_snd_eq : ((x = x') -> _|_) -> (y = y') -> ((x, y) = (x', y)) -> _|_

lemma_snd_neq : (x = x) -> ((y = y') -> _|_) -> ((x, y) = (x, y')) -> _|_

lemma_x_eq_xs_neq : (x = y) -> ((xs = ys) -> _|_) -> (x :: xs = y :: ys) -> _|_

lemma_x_neq_xs_eq : ((x = y) -> _|_) -> (xs = ys) -> (x :: xs = y :: ys) -> _|_

lemma_x_neq_xs_neq : ((x = y) -> _|_) -> ((xs = ys) -> _|_) -> (x :: xs = y :: ys) -> _|_

prim__eqB16 : Bits16 -> Bits16 -> Int

prim__eqB16x8 : Bits16x8 -> Bits16x8 -> Bits16x8

prim__eqB32 : Bits32 -> Bits32 -> Int

prim__eqB32x4 : Bits32x4 -> Bits32x4 -> Bits32x4

prim__eqB64 : Bits64 -> Bits64 -> Int

prim__eqB64x2 : Bits64x2 -> Bits64x2 -> Bits64x2

prim__eqB8 : Bits8 -> Bits8 -> Int

prim__eqB8x16 : Bits8x16 -> Bits8x16 -> Bits8x16

prim__eqBigInt : Integer -> Integer -> Int

prim__eqChar : Char -> Char -> Int

prim__eqFloat : Double -> Double -> Int

prim__eqInt : Int -> Int -> Int

prim__eqString : String -> String -> Int

prim__syntactic_eq : (a : Type) -> (b : Type) -> (x : a) -> (y : b) -> Maybe (x = y)

sequence : Traversable t => Applicative f => (t (f a)) -> f (t a)

sequence_ : Foldable t => Applicative f => (t (f a)) -> f ()

Eq : Type -> Type
The Eq interface defines inequality and equality.

GTE : Nat -> Nat -> Type
Greater than or equal to

LTE : Nat -> Nat -> Type
Proofs that n is less than or equal to m

gte : Nat -> Nat -> Bool
Boolean test than one Nat is greater than or equal to another

lte : Nat -> Nat -> Bool
Boolean test than one Nat is less than or equal to another

ord : Char -> Int
Convert the number to its ASCII equivalent.

replace : (x = y) -> (P x) -> P y
Perform substitution in a term according to some equality.

sym : (l = r) -> r = l
Symmetry of propositional equality

trans : (a = b) -> (b = c) -> a = c
Transitivity of propositional equality

:search does a type-based search, in the spirit of Hoogle. See Type-directed search (:search) for more details. Here is an example:

Idris> :search a -> b -> a
= Prelude.Basics.const : a -> b -> a
Constant function. Ignores its second argument.

= assert_smaller : a -> b -> b
Assert to the totality checker than y is always structurally
smaller than x (which is typically a pattern argument)

> malloc : Int -> a -> a

> Prelude.pow : Num a => a -> Nat -> a

> Prelude.Interfaces.(*) : Num a => a -> a -> a

> Prelude.Interfaces.(+) : Num a => a -> a -> a
... (More results)

:search can also look for dependent types:

Idris> :search plus (S n) n = plus n (S n)
< Prelude.Nat.plusSuccRightSucc : (left : Nat) ->
(right : Nat) ->
S (left + right) = left + S right

The command :l File.idr will load File.idr into the currently-running REPL, and :r will reload the last file that was loaded.

### Totality¶

All Idris definitions are checked for totality. The command :total <NAME> will display the result of that check. If a definition is not total, this may be due to an incomplete pattern match. If that is the case, :missing or :miss will display the missing cases.

### Editing files¶

The command :e launches your default editor on the current module. After control returns to Idris, the file is reloaded.

### Invoking the compiler¶

The current module can be compiled to an executable using the command :c <FILENAME> or :compile <FILENAME>. This command allows to specify codegen, so for example JavaScript can be generated using :c javascript <FILENAME>. The :exec command will compile the program to a temporary file and run the resulting executable.

### IO actions¶

Unlike GHCI, the Idris REPL is not inside of an implicit IO monad. This means that a special command must be used to execute IO actions. :x tm will execute the IO action tm in an Idris interpreter.

Sometimes, an Idris program will depend on external libraries written in C. In order to use these libraries from the Idris interpreter, they must first be dynamically loaded. This is achieved through the %dynamic <LIB> directive in Idris source files or through the :dynamic <LIB> command at the REPL. The current set of dynamically loaded libraries can be viewed by executing :dynamic with no arguments. These libraries are available through the Idris FFI in type providers and :exec.

## Colours¶

Idris terms are available in amazing colour! By default, the Idris REPL uses colour to distinguish between data constructors, types or type constructors, operators, bound variables, and implicit arguments. This feature is available on all POSIX-like systems, and there are plans to allow it to work on Windows as well.

If you do not like the default colours, they can be turned off using the command

:colour off

and, when boredom strikes, they can be re-enabled using the command

:colour on

To modify a colour, use the command

:colour <CATEGORY> <OPTIONS>

where <CATEGORY is one of keyword, boundvar, implicit, function, type, data, or prompt, and is a space-separated list drawn from the colours and the font options. The available colours are default, black, yellow, cyan, red, blue, white, green, and magenta. If more than one colour is specified, the last one takes precedence. The available options are dull and vivid, bold and nobold, italic and noitalic, underline and nounderline, forming pairs of opposites. The colour default refers to your terminal’s default colour.

The colours used at startup can be changed using REPL initialisation scripts.

Colour can be disabled at startup by the --nocolour command-line option.


text:
### 2D-3D

Two circles of equal size intersect and the centre of each circle is on the circumference of the other. What is the area of the intersection? Now imagine that the diagram represents two spheres of equal volume with the centre of each sphere on the surface of the other. What is the volume of intersection?

### Power Up

Show without recourse to any calculating aid that 7^{1/2} + 7^{1/3} + 7^{1/4} < 7 and 4^{1/2} + 4^{1/3} + 4^{1/4} > 4 . Sketch the graph of f(x) = x^{1/2} + x^{1/3} + x^{1/4} -x

### Fractional Calculus I

You can differentiate and integrate n times but what if n is not a whole number? This generalisation of calculus was introduced and discussed on askNRICH by some school students.

# Scientific Curves

##### Age 16 to 18 Challenge Level:

Curve sketching is an essential art in the application of mathematics to science. A good sketch of a curve does not need to be accurately plotted to scale, but will encode all of the key information about the curve: turning points, maximum or minimum values, asymptotes, roots and a sense of the scale of the function.

Sketch $V(r)$ against $r$ for each of these tricky curves, treating $a, b$ and $c$ as unknown constants in each case. As you make your plots, ask yourself: do different shapes of curve emerge for different ranges of the constants, or will the graphs look similar (i.e. same numbers of turning points, regions etc.) for the various choices?

1. An approximation for the potential energy of a system of two atoms separated by a distance $r$

$$V(r) = a\left[\left(\frac{b}{r}\right)^{12}-\left(\frac{b}{r}\right)^6\right]$$

2. A radial probability density function for an electron orbit
$$V(r) = ar^2e^{-\frac{r}{b}}$$

3. Potential energy for the vibrational modes of ammonium
$$V(r)=ar^2+be^{-cr^2}$$


text:
Properties

 Label 5.2.ah_bb_acs_fi_aii Base Field $\F_{2}$ Dimension $5$ Ordinary No $p$-rank $2$ Principally polarizable Yes Contains a Jacobian No

Invariants

 Base field: $\F_{2}$ Dimension: $5$ L-polynomial: $( 1 - 2 x + 2 x^{2} )^{3}( 1 - x + 3 x^{2} - 2 x^{3} + 4 x^{4} )$ Frobenius angles: $\pm0.250000000000$, $\pm0.250000000000$, $\pm0.250000000000$, $\pm0.306143893905$, $\pm0.570118980449$ Angle rank: $2$ (numerical)

This isogeny class is not simple.

Newton polygon

 $p$-rank: $2$ Slopes: $[0, 0, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1, 1]$

Point counts

This isogeny class is principally polarizable, but does not contain a Jacobian.

 $r$ 1 2 3 4 5 6 7 8 9 10 $A(\F_{q^r})$ 5 6875 175760 4296875 94766375 966680000 16672674835 754913671875 33661378209680 1156445919921875

 $r$ 1 2 3 4 5 6 7 8 9 10 $C(\F_{q^r})$ -4 10 23 42 66 55 38 162 491 1050

Decomposition and endomorphism algebra

Endomorphism algebra over $\F_{2}$
 The isogeny class factors as 1.2.ac 3 $\times$ 2.2.ab_d and its endomorphism algebra is a direct product of the endomorphism algebras for each isotypic factor. The endomorphism algebra for each factor is: 1.2.ac 3 : $\mathrm{M}_{3}($$$\Q(\sqrt{-1})$$$)$ 2.2.ab_d : 4.0.1025.1.
Endomorphism algebra over $\overline{\F}_{2}$
 The base change of $A$ to $\F_{2^{4}}$ is 1.16.i 3 $\times$ 2.16.b_b. The endomorphism algebra for each factor is: 1.16.i 3 : $\mathrm{M}_{3}(B)$, where $B$ is the quaternion algebra over $$\Q$$ ramified at $2$ and $\infty$. 2.16.b_b : 4.0.1025.1.
All geometric endomorphisms are defined over $\F_{2^{4}}$.
Remainder of endomorphism lattice by field
• Endomorphism algebra over $\F_{2^{2}}$  The base change of $A$ to $\F_{2^{2}}$ is 1.4.a 3 $\times$ 2.4.f_n. The endomorphism algebra for each factor is: 1.4.a 3 : $\mathrm{M}_{3}($$$\Q(\sqrt{-1})$$$)$ 2.4.f_n : 4.0.1025.1.

Base change

This is a primitive isogeny class.

Twists

Below are some of the twists of this isogeny class.
 Twist Extension Degree Common base change 5.2.af_p_abe_by_acu $2$ (not in LMFDB) 5.2.ad_h_ak_s_ay $2$ (not in LMFDB) 5.2.ab_d_ac_k_ai $2$ (not in LMFDB) 5.2.b_d_c_k_i $2$ (not in LMFDB) 5.2.d_h_k_s_y $2$ (not in LMFDB) 5.2.f_p_be_by_cu $2$ (not in LMFDB) 5.2.h_bb_cs_fi_ii $2$ (not in LMFDB) 5.2.ab_d_c_a_m $3$ (not in LMFDB)
Below is a list of all twists of this isogeny class.
 Twist Extension Degree Common base change 5.2.af_p_abe_by_acu $2$ (not in LMFDB) 5.2.ad_h_ak_s_ay $2$ (not in LMFDB) 5.2.ab_d_ac_k_ai $2$ (not in LMFDB) 5.2.b_d_c_k_i $2$ (not in LMFDB) 5.2.d_h_k_s_y $2$ (not in LMFDB) 5.2.f_p_be_by_cu $2$ (not in LMFDB) 5.2.h_bb_cs_fi_ii $2$ (not in LMFDB) 5.2.ab_d_c_a_m $3$ (not in LMFDB) 5.2.af_p_abi_cm_adw $6$ (not in LMFDB) 5.2.ad_h_ao_y_abk $6$ (not in LMFDB) 5.2.ab_d_ag_i_am $6$ (not in LMFDB) 5.2.b_d_ac_a_am $6$ (not in LMFDB) 5.2.b_d_g_i_m $6$ (not in LMFDB) 5.2.d_h_o_y_bk $6$ (not in LMFDB) 5.2.f_p_bi_cm_dw $6$ (not in LMFDB) 5.2.af_r_abo_da_aeq $8$ (not in LMFDB) 5.2.ad_d_c_ak_q $8$ (not in LMFDB) 5.2.ad_j_aq_be_abo $8$ (not in LMFDB) 5.2.ad_l_aw_bu_acm $8$ (not in LMFDB) 5.2.ab_ab_c_ac_a $8$ (not in LMFDB) 5.2.ab_b_a_ag_i $8$ (not in LMFDB) 5.2.ab_f_ae_o_ai $8$ (not in LMFDB) 5.2.ab_h_ag_w_aq $8$ (not in LMFDB) 5.2.ab_j_ai_bi_ay $8$ (not in LMFDB) 5.2.b_ab_ac_ac_a $8$ (not in LMFDB) 5.2.b_b_a_ag_ai $8$ (not in LMFDB) 5.2.b_f_e_o_i $8$ (not in LMFDB) 5.2.b_h_g_w_q $8$ (not in LMFDB) 5.2.b_j_i_bi_y $8$ (not in LMFDB) 5.2.d_d_ac_ak_aq $8$ (not in LMFDB) 5.2.d_j_q_be_bo $8$ (not in LMFDB) 5.2.d_l_w_bu_cm $8$ (not in LMFDB) 5.2.f_r_bo_da_eq $8$ (not in LMFDB) 5.2.af_p_abi_cm_adw $12$ (not in LMFDB) 5.2.ad_f_ae_e_ae $24$ (not in LMFDB) 5.2.ad_j_au_bk_ace $24$ (not in LMFDB) 5.2.ad_j_aq_bg_abs $24$ (not in LMFDB) 5.2.ab_b_a_e_ae $24$ (not in LMFDB) 5.2.ab_d_ac_e_a $24$ (not in LMFDB) 5.2.ab_f_ai_m_ay $24$ (not in LMFDB) 5.2.ab_f_ae_q_am $24$ (not in LMFDB) 5.2.ab_h_ag_y_aq $24$ (not in LMFDB) 5.2.b_b_a_e_e $24$ (not in LMFDB) 5.2.b_d_c_e_a $24$ (not in LMFDB) 5.2.b_f_e_q_m $24$ (not in LMFDB) 5.2.b_f_i_m_y $24$ (not in LMFDB) 5.2.b_h_g_y_q $24$ (not in LMFDB) 5.2.d_f_e_e_e $24$ (not in LMFDB) 5.2.d_j_q_bg_bs $24$ (not in LMFDB) 5.2.d_j_u_bk_ce $24$ (not in LMFDB)


text:
Integrating a function that's proportional to...

Printable View

• April 12th 2008, 09:43 PM
Boris B
Integrating a function that's proportional to...
I am supposed to integrate a function that is proportional to
$(10+x)^{-2}$
My initial strategy was to treat that as the reciprocal of
$x^2 + 20x + 100$
but that leaves me with the whole thing raised to an exponent of -1. I don't know how to take an antiderivative anything raised to -1!

So I decided to make a new variable, A, and set that equal to x+10. Taking the antiderivative of
$A^{-2}$
was pretty easy, giving me
$-1/A = -(x+10)^{-1}$
I just don't know if what I did was mathematically correct. (Getting a definite integral at that point was easy, and resulted in an answer close to one of my multiple choices.)

The thing is, I'm treating the whole thing as if I were getting the antid. of
$(10+x)^{-2}$.
Really I'm just dealing with an unknown function proportional to that. Is this something I haven't learned yet
(Worried)
or can I just take the antiderivative of what my function is proportional to?
• April 13th 2008, 12:29 AM
Moo
Hello,

I didn't get the whole problem oO

But the antiderivate of $(10+x)^{-2}$ is easy to find.

An antiderivative of $u'(x)u^n(x)$ is $\frac{u^{n+1}(x)}{n+1}$

Here, $u(x)=10+x$
$u'(x)=1$

So $(10+x)^{-2}=1*(10+x)^{-2}=u'(x)u^{-2}(x)$

As you've calculated, an antiderivative will be $-u^{-1}(x)=\frac{-1}{10+x}$

What do you mean by proportional ? You have a function equal to :

$f(x)=a (10+x)^{-2}$ with a a constant ?
If so, it's ok, the antiderivative doesn't take care of the constants...

An antiderivative of f will be a*[antiderivative of 1/(10+x)^2]
• April 13th 2008, 11:23 AM
Boris B
Okay, thanks for helping me with the antiderivative, Moo. Here is the rascally problem in its entirety:

"The lifetime of a part has a continuous distribution on the interval (0, 40) with probability density function f, where f(x) is proportional to
$(10 + x)^{-2}$

Calculate the probability that the lifetime of the part is less than 6."

I've never seen the "proportional to..." language in any other problem, so I'm wondering if I missed something. I am betting that all all density function proportional to the same thing can be translated into a single distribution function. Unfortunately, this requires a deeper understanding of calculus than I have yet developed.
• April 13th 2008, 07:44 PM
Boris B
Okay, I figured it out. You pointed out my antiderivative was correct. In order to get from that to the actual distribution function, we had to correct by a factor: the proportionality constant. I had forgotten to take into account the fact that the integral (0 to 40) f(x) dx = 1, because it's a probability question.
$C \int_0^{40} f(x) dx = 1$
$1 = \int_0^{40} C (10+x)^{-2} dx = -C (10+x)^{-1} = C/10 - C/50$
My answer was off by exactly that factor.
• April 13th 2008, 10:56 PM
Moo
I'm sorry I can't check it, it's out of my abilities :D
• April 13th 2008, 11:03 PM
Jameson
Quote:

Originally Posted by Boris B
Okay, I figured it out. You pointed out my antiderivative was correct. In order to get from that to the actual distribution function, we had to correct by a factor: the proportionality constant. I had forgotten to take into account the fact that the integral (0 to 40) f(x) dx = 1, because it's a probability question.
$C \int_0^40 f(x) dx = 1$
$1 = \int_0^40 C (10+x)^{-2} dx = -C (10+x)^{-1} = C/10 - C/50$
My answer was off by exactly that factor.

I don't quite follow this. The term "0C" is going to make the term you want to integrate 0, thus the integral will be 0. $\int _{a}^{b} 0dx=0$ for all a,b in R I'm pretty sure.
• April 13th 2008, 11:15 PM
Boris B
Oh dear, my bad, I didn't enclose my 40s in braces, so it put 4s in superscript and left the 0 floating around. I fixed it above.

I've pretty much figured this creature out now.
• April 13th 2008, 11:23 PM
Jameson
Yeah that looks good now. Proportional to a function just means some constant times that function like you said. Good work.


text:
# Tech Report CS-96-07

## Efficient Approximation Algorithms for Semidefinite Programs Arising from MAX CUT and COLORING

### Abstract:

The best known approximation algorithm for graph MAX CUT, due to Goemans and Williamson, first finds the optimal solution a semidefinite program and then derives a graph cut from that solution. Building on this result, Karger, Motwani, and Sudan gave an approximation algorithm for graph coloring that also involves solving a semidefinite program. Solving these semidefinite programs using known methods (ellipsoid, interior-point), though polynomial-time, is quite expensive. We show how they can be approximately solved in $\tilde O(nm)$ time for graphs with $n$ nodes and $m$ edges.

(complete text in pdf or gzipped postscript)


text:
homechevron_rightStudychevron_rightMathchevron_rightAlgebrachevron_rightlinear algebra

# Integer factorization. Trial division

Integer factorization using trial division algorithm.

All of sudden, I have to factorize some integers. Since I did not suppose my integers to be very big numbers, I've implemented trial division method. Method description is below the calculator.

Factorization

Prime factors

## Integer factorization

In number theory, integer factorization or prime factorization is the decomposition of a composite number into smaller non-trivial divisors, which when multiplied together equal the original integer.

And, since trial division is the easiest to understand of the integer factorization algorithms, here are couple of sentences from wikipedia:

Given an integer n (the integer to be factored), trial division consists of systematically testing whether n is divisible by any smaller number. Clearly, it is only worthwhile to test candidate factors less than n, and in order from two upwards because an arbitrary n is more likely to be divisible by two than by three, and so on.

With this ordering, there is no point in testing for divisibility by four if the number has already been determined not divisible by two, and so on for three and any multiple of three, etc.

Therefore, effort can be reduced by selecting only prime numbers as candidate factors. Furthermore, the trial factors need go no further than $\sqrt{n}$ because, if n is divisible by some number p, then n = p × q and if q were smaller than p, n would have earlier been detected as being divisible by q or a prime factor of q.

Trial division is a laborious algorithm, yet it is good for small numbers. More can be found at wiki link above.

Indeed.

## Similar calculators

627 calculators in total.


text:
## Statistics

Question 1 The regression equation between years of employment (X) and discount amount % (Y) at the Happy cafeteria is given by: Y =3*X + 7 For how many years you need to be an employee in order to get a 25% discount. Answer 1. 6 2. 10 3. 7 4. none of the above Question 2 Mall security estimates that the average daily per-store theft is less than $100, but wants to determine the accuracy of this statistic. The company researcher takes a sample of 81 clerks and finds that mean = 103$ and s = $10. a. state the hypothesis b. give the acceptance range for H0, considering a test with alpha = .01 Question 3 When you are conducting the t test, the population needs to be approximately normally distributed. Answer True False Question 4 When the population standard deviation is unknown and the sample size is less than 30, what table value should be used in computing a confidence interval for a mean? Answers 1. t 2. chi-square 3. z 4. none of the above Question 5 A data set of size n=19 and an alpha =.05 is given. You are asked to determine its CIE. The critical value will be: Answers 1. 1.645 2. 2.1009 3. 1.96 4. 2.0930 Question 6 A nursing home claims that their patients (n=64) will live at least 10 more years. For a test at alpha=.10 the critical z value is: __________ Answer Question 7 A confidence interval of 95% was used to estimate the proportion of customers who buy at least two items during their shopping mall experience. A random sample of 100 customers produced a confidence interval of: 27% +/- 3%. What is the best way to explain the result? ## Answers (1) • ## You need a Homework Help subscription to view this answer! ### Chegg Homework Help subscribers get: • Full access to our library of 2 million Q&A posts • 24/7 help from our community of subject experts • Fast answers from experts, around the clock Start Free Trial... Your free trial lasts for 7 days. Your subscription will continue for$14.95/month unless you cancel.

Get homework help


text:
# Evaluating the multilinearization of an arithmetic circuit?

Let $p(x_1,\ldots,x_n)$ be a multi-variate polynomial with coefficients over a field $F$. The multilinearization of $p$, denoted by $\hat{p}$, is the result of repeatedly replacing each $x_i^d$ with $d > 1$ by $x_i$. The result is obviously a multilinear polynomial.

Consider the following problem: given an arithmetic circuit $C(x_1,\ldots,x_n)$ over $F$ and given field elements $a_1,\ldots,a_n$, compute $\hat{C}(a_1,\ldots,a_n)$.

Question: Assuming field-arithmetic can be done in unit time, is there a polynomial-time algorithm for this? Added later: I would also be interested in the special case where $C$ is actually a formula (a circuit of fan-out $1$).

• Why would it be equivalent to computing the output of a closed circuit? The problem I am facing is that the circuit can have disjoint paths from an input $x_i$ to several internal multiplication nodes, and evaluating each one of those internal multiplication nodes would require replacing $x_i$ by $a_i$ in one path and by $1$ in the other. In a circuit with an exponential number of paths, it looks like there is an exponential number of cases to take care of. Nov 11, 2010 at 8:34
• @Kaveh: I don't get it. Look at the circuit $(x * x)$. If you just replace the node of input $x$ by a node with value $a$ and evaluate in the standard way you end up returning $a^2$ instead of $a$. Model of computation: just normal polynomial time on Turing machines. Think of the field as being $Z/3Z$ for concreteness, if you want. Nov 11, 2010 at 12:01
• @Kaveh: I don't understand how such an algorithm implies what you say, but this does indeed contradict a common hypothesis in arithmetic circuit complexity: that the Permanent has no poly-sized arithmetic circuits (over fields other than F_2). Consider the polynomial $p=\prod_i(\sum_j x_{ij} y_j)$. The multilinear part $q$ of this polynomial has the property that its highest degree ($=2n$) part is just $r = y_1y_2\cdots y_n Per(x_{11},\ldots,x_{nn})$. If there is a small arithmetic circuit computing $q$, then one can show that there is a small arithmetic circuit computing $r$. Nov 12, 2010 at 2:05
• @Srikanth: I didn't see your comment before posting my answer (which turned out to be the same construction you gave in your comment). I have since deleted my answer, and you should post your comment as an answer. Nov 13, 2010 at 0:58
• @Joshua: I have not added my comment as an answer since I don't understand why Kaveh's construction works. I see that the arithmetic circuit computes a polynomial that agrees with the multilinearization at all inputs, but I am not sure that it computes formally the multilinearization of the given polynomial (see my comments after Kaveh's answer). My construction (and yours) assumes that the multilinearization is computed formally. Nov 13, 2010 at 2:43

In the case that the field $F$ is of size at least $2n$, I think this problem is hard. More specifically, I think that if the above can be efficiently solved for $F$ this large, then CNF-SAT has efficient randomized algorithms. Say we are given a CNF formula $\varphi$. One can easily come up with an arithmetic circuit $C$ that computes an arithmetization'' $p$ of $\varphi$, where the polynomial $p$ agrees with the formula $\varphi$ on $0$-$1$ inputs. Consider the multilinearization $q$ of $p$. Note that $q$ agrees with $p$ and hence $\varphi$ on $\{0,1\}^n$.

I claim that $q$ is non-zero iff $\varphi$ is satisfiable. Clearly, if $q=0$, then $\varphi$ cannot be satisfied. For the converse, one can show that any non-zero multilinear polynomial cannot vanish on all of $\{0,1\}^n$. This implies that a non-zero $q$ (and hence the corresponding $\varphi$) does not vanish at some input in $\{0,1\}^n$.

Therefore, checking for satisfiability of $\varphi$ is equivalent to checking if $q$ is non-zero. Say, now, that we could evaluate $q$ over a large field $F$. Then, using the Schwartz-Zippel Lemma, we could identity-test $q$ using an efficient randomized algorithm and check if it is the zero polynomial (the size of $F$ is used to upper bound the error in the Schwartz-Zippel Lemma).

• It seems to me that F is a fixed field because there is nothing in the input that specifies F. Also note that the question assumes that field operations take unit time. Nov 11, 2010 at 7:37
• Thanks Srikanth. As Kaveh guessed I was indeed interested in the fixed finite field case, but this answer you gave helps me understand the question a bit better. Nov 11, 2010 at 8:24

Assume that there is polytime algorithm that given $C(\vec{x}) \in F(\vec{x})$ and $\vec{a}$ computed the result of the multi-linearization of $C$ on $\vec{a}$. (w.l.o.g. I will assume that the output $\vec{b}$ will be a vector of $p$-bit binary numbers $b_i$ is $k$ iff the $b_{i,k}$ is one.)

Since $P \subseteq P/poly$, there is a polysize boolean circuit that given the encoding of the arithmetic circuit and the values for the variables computes the multi-linearization of the arithmetic circuit on the inputs. Let call this circuit $M$.

Let $C$ be an arbitrary arithmetic circuit. Fix the variables of the boolean circuit $M$ which describe the arithmetic circuit, so we have a boolean circuit computing the multi-linearization of $C$ on given inputs.

We can turn this circuit into an arithmetic circuit over $F_p$ by noting that $x^{p-1}$ is $1$ for all values but $0$ so first raise all inputs to the power $p-1$. Replace each $f \land g$ gate by multiplication $f.g$, each $f \lor g$ gate by $f+g-f.g$ and each $\lnot f$ gate by $1-f$.

By the assumption we made above about the format of the output, we can turn the output from binary to values over $F_p$. Take the output for $b_i$ and combine them to get $\sum_{0 \leq k \leq p-1}{kb_{i,k}}$.

We can also convert the input given as values over $F_p$ to binary form since there are polynomials passing through any finite number of points. E.g. if we are working in $\bmod 3$, consider the polynomials $2x(x+1)$ and $2x(x+2)$ which give the first and the second bits of the input $x \in F_3$.

Combining these we have an arithmetic circuit over $F_p$ computing the multi-linearization of $C$ with size polynomail in the size of $C$.

• It is not clear to me why the arithmetic circuit you have described computes the multilinearization of $C$, or indeed even a multilinear polynomial. I am only able to see that the arithmetic circuit computes some polynomial that agrees with the multilinearization of $C$ on $0$-$1$ inputs. Nov 12, 2010 at 6:01
• @Srikanth: the arithmetic version of the boolean circuit $M$ (with some inputs fixed) computes the multilinear version of $C$, it doesn't need to be a multilinear. Then the only problem is that input/output are in binary not values over $F_p$, so I just need to fix the encoding for input/output from binary to original input and output values. The resulting circuit is an arithmetic circuit that gets the values for variables of $C$, encodes them in binary, computes the value of the multilinearization of $C$ over those inputs and output the answer in binary, and then translate them back to $F_p$. Nov 12, 2010 at 6:51
• [continued] The result it is an arithmetic circuit with the same variables that $C$ has, and with the same outputs, and it is computing the multilinearization of $C$. Nov 12, 2010 at 6:58
• @Kaveh: Have you assumed that the input to the boolean circuit $M$ is of the same form as the output of $M$? In any case, I am still not convinced. It is perfectly possible for an arithmetic circuit to compute a polynomial $f$ that agrees with a polynomial $g$ at all inputs from the field and yet $f\neq g$. For example, the polynomial $x^p$ agrees with $x$ at all inputs, and yet they are not equal as polynomials. How do you know that the circuit $M$ is not simply computing a non-multilinear polynomial that agrees with the multilinearization of $C$ at all inputs? Nov 12, 2010 at 7:15
• @Srikanth: I have described the form of input and output in my answer. The input to $M$ is in binary, the output of $M$ is in the form stated above. I haven't said that it is multilinear, I have only said that it computes the multilinearization of the $C$. Nov 12, 2010 at 9:47


text:
# Thread: roll of the die

1. ## roll of the die

Odd number = that number subtracted from your score

What is the expected value of a single roll of the die, and how is it calculated?

2. Originally Posted by zoso

Odd number = that number subtracted from your score

What is the expected value of a single roll of the die, and how is it calculated?
Tell me, what is the average score of all the possible outcomes?

I'll show you...

note: $s_1$ is the amount of points you'll get when you roll a 1

average is:

$\frac{s_1+s_2+s_3+s_4+s_5+s_6}{6}=\frac{\neg1+2-3+4-5+6}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$

and that is the expected outcome for a single roll of the die. Yes I know it's impossible to get that number with a die, but that's the best math has to offer.

3. Originally Posted by zoso

Odd number = that number subtracted from your score

What is the expected value of a single roll of the die, and how is it calculated?
Expected value of a single roll of dice is 3.5

Malay

4. Originally Posted by malaygoel
Expected value of a single roll of dice is 3.5

Malay
oooooh, conflict. How would you find 3.5 as an answer if that is higher than the expected outcome if all numbers counted as positive?

5. Originally Posted by Quick
oooooh, conflict. How would you find 3.5 as an answer if that is higher than the expected outcome if all numbers counted as positive?
There are six numbers and the probability of any one of them is 1/6.
Hence expected value= $\frac{1}{6}.1 + \frac{1}{6}.2 + \frac{1}{6}.3 + \frac{1}{6}.4 + \frac{1}{6}.5 + \frac{1}{6}.6$

Malay

6. Originally Posted by Quick
Tell me, what is the average score of all the possible outcomes?

I'll show you...

note: $s_1$ is the amount of points you'll get when you roll a 1

average is:

$\frac{s_1+s_2+s_3+s_4+s_5+s_6}{6}=\frac{\neg1+2-3+4-5+6}{6}=\frac{3}{6}=\boxed{\frac{1}{2}}$

and that is the expected outcome for a single roll of the die. Yes I know it's impossible to get that number with a die, but that's the best math has to offer.
blatantly simple, I really need to think more

7. Originally Posted by malaygoel
There are six numbers and the probability of any one of them is 1/6.
Hence expected value= $\frac{1}{6}.1 + \frac{1}{6}.2 + \frac{1}{6}.3 + \frac{1}{6}.4 + \frac{1}{6}.5 + \frac{1}{6}.6$

Malay
You didn't read the entire question, if the number is odd, than it's subtracted from the score.

8. Originally Posted by Quick
You didn't read the entire question, if the number is odd, than it's subtracted from the score.
Sorry
Yes, if it is to be subtracted. then the answer is 0.5.

Malay

9. Originally Posted by zoso
blatantly simple, I really need to think more
You are correct here.
Extend it.
Try for double throw of dice.

Malay


text:
#### • Class 11 Physics Demo

Explore Related Concepts

# examples of octagonal prism

From Wikipedia

Prism (optics)

In optics, a prism is a transparent optical element with flat, polished surfaces that refractlight. The exact angles between the surfaces depend on the application. The traditional geometrical shape is that of a triangular prism with a triangular base and rectangular sides, and in colloquial use "prism" usually refers to this type. Some types of optical prism are not in fact in the shape of geometric prisms. Prisms are typically made out of glass, but can be made from any material that is transparent to the wavelengths for which they are designed.

A prism can be used to break light up into its constituent spectralcolors (the colors of the rainbow). Prisms can also be used to reflect light, or to split light into components with different polarizations.

## How prisms work

Light changes speed as it moves from one medium to another (for example, from air into the glass of the prism). This speed change causes the light to be refracted and to enter the new medium at a different angle (Huygens principle). The degree of bending of the light's path depends on the angle that the incident beam of light makes with the surface, and on the ratio between the refractive indices of the two media (Snell's law). The refractive index of many materials (such as glass) varies with the wavelength or color of the light used, a phenomenon known as dispersion. This causes light of different colors to be refracted differently and to leave the prism at different angles, creating an effect similar to arainbow. This can be used to separate a beam of white light into its constituent spectrum of colors. Prisms will generally disperse light over a much larger frequency bandwidth than diffraction gratings, making them useful for broad-spectrum spectroscopy. Furthermore, prisms do not suffer from complications arising from overlapping spectral orders, which all gratings have.

Prisms are sometimes used for the internal reflection at the surfaces rather than for dispersion. If light inside the prism hits one of the surfaces at a sufficiently steep angle, total internal reflection occurs and all of the light is reflected. This makes a prism a useful substitute for a mirror in some situations.

### Deviation angle and dispersion

Ray angle deviation and dispersion through a prism can be determined by tracing a sample ray through the element and using Snell's law at each interface. The exact expressions for prism deviation and dispersion are complex, but for small angle of incidence \theta_0 and small angle \alpha they can be approximated to give a simple formula. For the prism shown at right, the indicated angles are given by

\begin{align}

\theta'_0 &\approx \frac{n_0}{n_1} \theta_0 \\ \theta_1 &= \alpha - \theta'_0 \\ \theta'_1 &\approx \frac{n_1}{n_2} \theta_1 \\ \theta_2 &= \theta'_1 - \alpha \end{align}. For a prism in air n_0=n_2 \simeq 1. Defining n=n_1, the deviation angle \delta is given by

{\delta = \theta_2 + \theta_0 \approx n \theta_1 - \alpha + \theta_0 = n \alpha - n \theta'_0 - \alpha + \theta_0 \approx (n - 1) \alpha}

The dispersion \delta (\lambda) is the wavelength-dependent deviation angle of the prism, so that for a thin prism the dispersion is given by

\delta (\lambda) \approx [ n (\lambda) - 1 ] \alpha

## Prisms and the nature of light

In Isaac Newton's time, it was believed that white light was colorless, and that the prism itself produced the color. Newton's experiments convinced him that all the colors already existed in the light in a heterogeneous fashion, and that "corpuscles" (particles) of light were fanned out because particles with different colors traveled with different speeds through the prism. It was only later that Young and Fresnel combined Newton's particle theory with Huygen's wave theory to show that color is the visible manifestation of light's wavelength.

Newton arrived at his conclusion by passing the red color from one prism through a second prism and found the color unchanged. From this, he concluded that the colors must already be present in the incoming light â€” thus, the prism did not create colors, but merely separated colors that are already there. He also used a lens and a second prism to recompose the spectrum back into white light. This experiment has become a classic example of the methodology introduced during the scientific revolution. The results of this experiment dramatically transformed the field of metaphysics, leading to John Locke's primary vs secondary quality distinction.

Newton discussed prism dispersion in great detail in his book Opticks. He also introduced the use of more than one prism to control dispersion. Newton's description of his experiments on prism dispersion was qualitative, and is quite readable. A quantitative description ofmultiple-prism dispersion was not needed until multiple prism laser beam expanders were introduced in the 1980s.

Cuboid

In geometry, a cuboid is a solid figure bounded by six faces, forming a convex polyhedron. There are two competing (but incompatible) definitions of a cuboid in the mathematical literature. In the more general definition of a cuboid, the only additional requirement is that these six faces each be a quadrilateral, and that the undirected graph formed by the vertices and edges of the polyhedron should be isomorphic to the graph of a cube. Alternatively, the word â€œcuboidâ€� is sometimes used to refer to a shape of this type in which each of the faces is a rectangle (and so each pair of adjacent faces meets in a right angle); this more restrictive type of cuboid is also known as a right cuboid, rectangularbox, rectangular hexahedron, right rectangular prism, or rectangular parallelepiped.

## General cuboids

By Euler's formula the number of faces (F), vertices (V), and edges (E) of any convex polyhedron are related by the formula "F + V - E" = 2 . In the case of a cuboid this gives 6 + 8 - 12 = 2; that is, like a cube, a cuboid has 6 faces, 8 vertices, and 12 edges.

Along with the rectangular cuboids, any parallelepiped is a cuboid of this type, as is a square frustum (the shape formed by truncation of the apex of a square pyramid).

## Rectangular cuboid

In a rectangular cuboid, all angles are right angles, and opposite faces of a cuboid are equal. It is also a right rectangularprism. The term "rectangular or oblong prism" is ambiguous. Also the term rectangularparallelepipedor orthogonal parallelepiped is used.

The square cuboid, square box, or right square prism (also ambiguously called square prism) is a special case of the cuboid in which at least two faces are squares. The cube is a special case of the square cuboid in which all six faces are squares.

If the dimensions of a cuboid are a, b and c, then its volume is abc and its surface area is 2ab + 2bc + 2ac.

The length of the space diagonal is

d = \sqrt{a^2+b^2+c^2}.\

Cuboid shapes are often used for boxes, cupboards, rooms, buildings, etc. Cuboids are among those solids that can tessellate 3-dimensional space. The shape is fairly versatile in being able to contain multiple smaller cuboids, e.g. sugar cubes in a box, small boxes in a large box, a cupboard in a room, and rooms in a building.

A cuboid with integer edges as well as integer face diagonals is called an Euler brick, for example with sides 44, 117 and 240. A perfect cuboid is an Euler brick whose space diagonal is also an integer. It is currently unknown whether a perfect cuboid actually exists.

Question:Here's the example I'm given: http://learn.flvs.net/webdav/educator_math2_v5/module10/imagmod10/10_03a_03.gif To find the surface area, visualize the net for this figure. The net consists of a large rectangle (the lateral faces) with sides of 12 cm and the perimeter of the base (5cm +5 cm +8 cm +8 cm = 26 cm). The area of the lateral faces would be 12 cm X 26 cm = 312 square cm The bases of the prism are rectangles with length and height of 8 cm and 5 cm. The area of this would be 8 cm X 5 cm = 40 square cm Adding the lateral area and 2 bases together will reveal the surface area. 312 sq cm + 2(40 sq cm) = 392 sq cm I dont really get it. Can somebody help me find a way to remember this, and explain it? Thanks. http://i31.tinypic.com/153ty7l.png

Answers:Just so you know, your link led to a complaint page because the system with the example expects a "cookie" which, of course, was only set on your computer. But this example is fairly straightforward, though the term "net" is a little odd. What they are doing is "unfolding" the sides of the prism to calculate the surface area. The four sides ("lateral faces") unfold into one large rectangle, and that leaves the top and bottom ("bases") to be added in. So they calculate the area of the lateral sides (height times the perimeter of the base = 312) and then add the area of the two bases, each being length x width (which gives the 2 * 40). Here's another, equivalent way to look at it: take the three dimensions of the rectangular as x, y, and z. For each pair of measurements, there will be two opposite faces that consist of rectangles with that pair of dimensions. So the area is 2(xy + xz + yz). In the example, x=5, y=8, z=12 (or you can assign them in any other order and it will come out the same). So the surface area is 2 (5*8 + 5*12 + 8*12) = 2 (40 + 60 + 96) = 2 * 196 = 392 Their approach, unfolding the sides and then adding the top and bottom, just collects the areas of the sides in a different order. You're still adding up six rectangles; it's just that they've stuck four of them together in one step.

Question:Ok there is a triangular prism that has a surface area of 100cm squared... how wud i make an exact 3-D triangular prism?? the triangle has to be a right triangle, i need to know all the lengths and stuff like that i need just one way to make one out of paper

Answers:There are an infinite number of such. The larger the ends, the shorter the length. For example, if the triangle was 5 by 5 by 5* 2 then the length = 75/(5+5+5* 2) = 4.39334 cm

Question:

Answers:Think of a box or a cube. Those are examples of rectangular prisms. There are six faces (the sides of the solid). There are eight vertices (the "corners" of the solid). Answer: 6 faces 8 vertices

Question:i asked to draw a polygon.and how many sides it is.

Answers:A polygon will have as many sides as you want, at least three, for a triangle, for example. An octagon has 8 sides. Poly = many, gone = angle, octo = 8, in Greek ;-)


text:
# Homework Help: Polynomial division

1. Apr 29, 2010

### Combinatus

1. The problem statement, all variables and given/known data

Show that $$g(x) = x^3 + 1$$ divides $$f(x) = x^{9999} +1$$.

2. Relevant equations

3. The attempt at a solution

$$g(x)$$ can obviously be factored into the irreducible polynomials $$(x+1)(x^2 - x + 1)$$ in $$Z[x]$$, and since $$f(-1) = (-1)^{9999} + 1 = 0$$, the factor theorem gives that $$(x+1)$$ divides $$f(x)$$.

Furthermore, we get

$$x^{9999}+1 = (x^2 - x + 1) q(x) + r(x)$$

where $$r(x) = Ax+B$$ since $$deg(r(x)) < deg(x^2 - x + 1)$$ if $$r(x) \neq 0$$.

So, showing that $$A = B = 0$$ would be a good idea, which I have failed to do throughout past trials. I suspect there's an "obvious", clever trick to this, but I'm currently not seeing it.

Another approach would probably be to use $$x^{9999}+1 = (x+1)(x^2 - x + 1) q_{2}(x) + r_{2}(x)$$ where $$r_{2}(x) = Cx^2 + Dx + E$$, and so, $$x = -1$$ yields $$C - D + E = 0$$, but that hasn't gotten me anywhere either.

Note: I'm assuming that I'm not supposed to use complex roots to factor $$x^2 - x + 1$$, but the problem doesn't specify that such an assumption is necessary.

Last edited: Apr 29, 2010
2. Apr 29, 2010

### Staff: Mentor

You've shown that x + 1 is a factor by showing that f(-1) = 0. The other two factors of x^3 + 1 are x = 1/2 +/- i sqrt(3)/2. If you write these in polar form it's pretty easy to raise them to the 9999th power, and thus show that f(1/2 +/- i sqrt(3)/2) = 0.

3. Apr 29, 2010

### Combinatus

Ugh, yay for assumptions! Thank you!


text:
Positron Annihilation Spectroscopy data analysis

# PAScual: a Positron Annihilation Spectroscopy data analysis program.

(c) 2007-2022 Carlos Pascual-Izarra

email: < cpascual [AT] users.sourceforge.net >

## Installation instructions for PAScual

PAScual and all its dependencies can be installed simply by running the following in a terminal of a system where Python >=3.7 is installed:

pip install PAScual

However, I strongly recommend installing it in some sort of virtual python environment.

For example, you can use venv, but my preferred choice is using mamba. For this, first install mamba and then, on a terminal, run:

mamba create -c conda-forge -n PAScual pythonqwt pyqt numpy scipy

mamba activate PAScual

pip install PAScual

Note: Using other conda implementations such as Anaconda is also possible, but I found mamba a lot faster. Replace mamba by conda if you are using Anaconda or miniconda instead of mamba.

Note If you intend to use PAScual in text mode, you may want to also install matplotlib to produce plots.

## Running PAScual

Note: if you installed PAScual in a virtual environment (as recommended), make sure to activate the environment before continuing (i.e., in your current terminal, run mamba activate PAScual or equivalent)

Once installed, just run the following command to get the graphical application:

pascual

And I recommend to try the tutorial from the User Manual (some example spectra are provided in the examples dir)

# Important:

If you use PAScual, acknowledge it by citing, at least:

C. Pascual-Izarra et al., Advanced Fitting Algorithms for Analysing Positron Annihilation Lifetime Spectra, Nuclear Instruments and Methods A, 603, p456-466 (2009) (DOI: 10.1016/j.nima.2009.01.205)

See CREDITS.txt for acknowledgements to third parties.

Also note its license: PAScual is Free Software but certain conditions apply (see LICENSE.TXT). If you want to request modifications of PAScual, please contact the author. If you modify PAScual, please communicate the modifications to the author so that they can be incorporated in future versions.

## Project details

Uploaded py3


text:
Lower bound on norm of vector given estimation vector

I have the following problem:

The are two numbers unknown real $A_1, A_2$, and two known drawn approximations $B_1, B_2$ such that $$A_i-B_i\sim N(0, \sigma).$$

Does there exist a function $f:\mathbb{R}^2\rightarrow\mathbb{R}_{\ge0}$, independent of $A_1, A_2$, such that $$Pr\big(f(B_1, B_2)\le A_1^2+A_2^2\big)=0.5.$$

Or, more generally:

For $\delta>0$, does there exists a function $f_\delta:\mathbb{R}^n\rightarrow\mathbb{R}_{\ge0}$ such that for all $A_1, ..., A_n\in\mathbb{R}^n$ we have: $$Pr\big(f_\delta(A_1+\varepsilon_1, ..., A_n+\varepsilon_n)\le A_1^2+...+A_n^2\big)=1-\delta,$$ where $$\varepsilon_i\sim N(0,\sigma).$$

At first I thought that maybe $f_\delta$ should be the ppf of the noncentral chi-squared distribution with $n$ degrees of freedom and $\lambda=A_1^2+...+A_n^2$ at $\delta$, but that has a problem with the case $A_1=...=A_n=0$, since then we need $$Pr\big(f_\delta(\varepsilon_1, ...,\varepsilon_n)=0\big)=1-\delta,$$

which no ppf meets.

Googling for "estimation of vector norm" and similar wordings didn't come up with something helpful.

Does this problem have a name? Maybe a solution?

If it is known that no such solution exists, what is known about the problem when we relax equality to inequality, i.e. $$Pr\big(f_\delta(A_1+\varepsilon_1, ..., A_n+\varepsilon_n)\le A_1^2+...+A_n^2\big)\ge 1-\delta?$$