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http://mathhelpforum.com/differential-geometry/105990-quick-question-order-poles.html | # Thread: Quick question on order of poles
1. ## Quick question on order of poles
For a complex function $\displaystyle f(z)$ which has an isolated singularity at $\displaystyle z=z_0$
1. How do I find out if $\displaystyle z=z_0$ is a pole or not?
2. If it is a pole what is the order of the pole
Consider for e.g. $\displaystyle f(z)=\frac{1}{zsin(z)}$. Is$\displaystyle z=0$pole? If yes what is the order of the pole?
Is there a way to find answer to the above question without really writing down the Laurent Series about $\displaystyle z=z_0$
A related question I would have is - do we have similar method to comment on essential and removable singularities?
Thanks
2. Originally Posted by aman_cc
For a complex function $\displaystyle f(z)$ which has an isolated singularity at $\displaystyle z=z_0$
1. How do I find out if $\displaystyle z=z_0$ is a pole or not?
2. If it is a pole what is the order of the pole
Consider for e.g. $\displaystyle f(z)=\frac{1}{zsin(z)}$. Is$\displaystyle z=0$pole? If yes what is the order of the pole?
Is there a way to find answer to the above question without really writing down the Laurent Series about $\displaystyle z=z_0$
You don't need the whole Laurent Series, only the first non-zero term. If you know that $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ and that $\displaystyle (z-z_0)^k f(z)\to c\neq 0$ when $\displaystyle z\to z_0$, then $\displaystyle z_0$ is a pole of order $\displaystyle k$ for $\displaystyle f$. And the residue is $\displaystyle c$.
In your example, you may know that $\displaystyle \frac{\sin z}{z}\to 1$ when $\displaystyle z\to 0$, hence $\displaystyle z^2\frac{1}{z\sin z}\to_{z\to 0} 1$, hence 0 is a pole of order 2, with residue 1.
3. Originally Posted by Laurent
You don't need the whole Laurent Series, only the first non-zero term. If you know that $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ and that $\displaystyle (z-z_0)^k f(z)\to c\neq 0$ when $\displaystyle z\to z_0$, then $\displaystyle z_0$ is a pole of order $\displaystyle k$ for $\displaystyle f$. And the residue is $\displaystyle c$.
In your example, you may know that $\displaystyle \frac{\sin z}{z}\to 1$ when $\displaystyle z\to 0$, hence $\displaystyle z^2\frac{1}{z\sin z}\to_{z\to 0} 1$, hence 0 is a pole of order 2, with residue 1.
Let me just re-phrase (so that I get it)
1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $\displaystyle (z-z_0)^k f(z)$ when $\displaystyle z\to z_0$
Then 'k' is the order of the pole at$\displaystyle z_0$. Have I got it correct?
Few questions (I tried to work out a logic for this)
1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
2. Why is the assumption $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ important?
4. Thinking about this I got a related question (which might help me understand why the limit method to find the order works)
Is is correct to say
if $\displaystyle f(z) = \frac{a_1}{z}+\frac{a_2}{z^2}+\frac{a_3}{z^3}+...+ \frac{a_k}{z^k}$
and Limit $\displaystyle z \to 0:f(z)$ exists then
1. each of $\displaystyle {a_1},{a_2},...,{a_k} = 0$
2. Above is true in both cases:
a) $\displaystyle k$ is finite
b) $\displaystyle k$ in infinite
Originally Posted by aman_cc
Let me just re-phrase (so that I get it)
1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $\displaystyle (z-z_0)^k f(z)$ when $\displaystyle z\to z_0$
Then 'k' is the order of the pole at$\displaystyle z_0$. Have I got it correct?
Few questions (I tried to work out a logic for this)
1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
2. Why is the assumption $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ important?
5. Originally Posted by Laurent
You don't need the whole Laurent Series, only the first non-zero term. If you know that $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ and that $\displaystyle (z-z_0)^k f(z)\to c\neq 0$ when $\displaystyle z\to z_0$, then $\displaystyle z_0$ is a pole of order $\displaystyle k$ for $\displaystyle f$. And the residue is $\displaystyle c$.
$\displaystyle c$ then is the coefficient on the $\displaystyle \frac{1}{(z-z_0)^k}$ term and the residue is:
$\displaystyle \lim_{z\to z_0} \frac{1}{(k-1)!} \frac{d^{k-1}}{dz} (z-z_0)^k f(z)$
6. Originally Posted by aman_cc
Let me just re-phrase (so that I get it)
1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $\displaystyle (z-z_0)^k f(z)$ when $\displaystyle z\to z_0$
Then 'k' is the order of the pole at$\displaystyle z_0$. Have I got it correct?
You don't really have to check every $\displaystyle k$ one by one: if $\displaystyle k$ is more than the order, the limit is 0, and if it is less than the order, the limit doesn't exist. Therefore, if the limit is finite and non-zero, $\displaystyle k$ is the right order. In your example, I could have written "$\displaystyle \sin z\sim_{z\to 0} z$, hence $\displaystyle \frac{1}{z\sin z}\sim\frac{1}{z^2}$ hence the order is 2. "
Few questions (I tried to work out a logic for this)
1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
2. Why is the assumption $\displaystyle f$ is analytic in $\displaystyle U\setminus\{z_0\}$ important?
Your questions are related to each other. With your definition of a pole (using Laurent series), you have: "for all $\displaystyle z$ in a small disc centered at $\displaystyle z_0$, $\displaystyle f(z)=\sum_{n=-k}^\infty a_n (z-z_0)^n$ and $\displaystyle a_{-k}\neq 0$". This definition concerns $\displaystyle f$ not only "at $\displaystyle z_0$" but also on a neighbourhood of $\displaystyle f$. And it implies that $\displaystyle f$ is analytic on a disc around $\displaystyle z_0$.
The condition $\displaystyle (z-z_0)^k f(z)\to c\neq 0$, on the other hand, does not say anything about the regularity of $\displaystyle f$ outside $\displaystyle z_0$. If however you know (in addition) that $\displaystyle f$ is analytic on a subset (like a disc) around $\displaystyle z_0$ (but deprived of $\displaystyle z_0$), then you can see that the function $\displaystyle g:z\mapsto (z-z_0)^k f(z)$ is analytic on the same subset, and has a finite limit $\displaystyle c$ at $\displaystyle z_0$, so that $\displaystyle z_0$ is a removable singularity of $\displaystyle g$ (this is a property you probably know). In other words, $\displaystyle g$ (with $\displaystyle g(z_0)=c$) is analytic in the previous subset plus the point $\displaystyle z_0$. Hence $\displaystyle g(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$ for some sequence $\displaystyle a_n$ with $\displaystyle a_0=g(z_0)=c\neq 0$. And from the relation $\displaystyle g(z)=(z-z_0)^k f(z)$, one gets $\displaystyle f(z)=\sum_{n=0}^\infty a_n (z-z_0)^{n-k}$ $\displaystyle =\sum_{n=-k}^\infty a_{n+k} (z-z_0)^n$. Thus we find a Laurent series and again the initial definition of pole and order. This justifies my previous post. | 2018-03-23T18:44:21 | {
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http://mathhelpforum.com/trigonometry/24056-sines.html | 1. ## sines
I need help proving these 2 identities.
1. If x = 18 degrees, prove that sin2x = cos3x. Find the exact values of sinx and cosx
2. If 2sin(x - y) = sin(x + y), prove that tanx = 3tan y
2. Originally Posted by Zetterbergx40
2. If 2sin(x - y) = sin(x + y), prove that tanx = 3tan y
$2sin(x - y) = sin(x + y)$
$2sin(x)cos(y) - 2sin(y)cos(x) = sin(x)cos(y) + sin(y)cos(x)$
$sin(x)cos(y) = 3sin(y)cos(x)$ <-- Divide both sides by $cos(x)cos(y)$
$tan(x) = 3tan(y)$
-Dan
3. Originally Posted by Zetterbergx40
1. If x = 18 degrees, prove that sin2x = cos3x. Find the exact values of sinx and cosx
I can't think of a way to help you on the first part, but for finding the exact value of sine and cosine:
$sin(2x) = cos(3x)$
$2sin(x)cos(x) = -4sin^2(x) cos(x) + cos(x)$ <-- Divide both sides by cos(x)
$2sin(x) = -4sin^2(x) + 1$
$4sin^2(x) + 2sin(x) - 1 = 0$
$sin(x) = \frac{-1 \pm \sqrt{5}}{4}$
Note that both solutions are acceptable, indicating that there are actually 4 solutions for x in $0 \leq x < 360$. Obviously the sin(18) will correspond to the positive solution.
From the sine equation we can find cosine:
$cos(x) = \pm \sqrt{1 - sin^2(x)}$
where the $\pm$ indicates which quadrant the angle is in. (So for 18 degrees, pick + because cosine is positive in the first quadrant.)
-Dan
4. ## Thanks Dan
Thanks alot, for the first reply, can you tell me what the solution would be in LS = RS form? Thanks
5. Hello, Zetterbergx40!
1. If $x = 18^o$, prove that: . $\sin2x \:= \:\cos3x$
We are asked to prove that: . $\sin36^o \:=\:\cos54^o$
The two angles are complementary and $\sin\theta \:=\:\cos(90^o - \theta)$
Or look at this triangle:
Code:
*
/|
/ |
/ |
/36°|
c / | b
/ |
/ |
/ |
/ 54° |
*---------*
a | 2018-02-20T00:21:53 | {
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http://ddhk.site-photographer.it/latex-math-letters.html | ## Latex Math Letters
They are organized into seven classes based on their role in a mathematical expression. The Wolfram Language by default interprets any sequence of letters or letter-like forms as the name of a symbol. The Comprehensive LATEX Symbol List Scott Pakin 9 November 2009 Abstract This document lists 5913 symbols and the corresponding LATEX commands that produce them. This is identical to the tabular environment described in essential LATEX except that the entries are typeset in math mode instead of LR mode. I also prepared a quick reference of math symbols. It is also a large topic due to the existence of so much mathematical notation. LaTeX Math Symbols Prepared by L. 6 Appendices. I have been using \epsilon. Math Mode Accents. This small website is a tool for looking up the LaTeX commands for mathematical symbols - because it often isn't obvious what the command is. LaTeX is a powerful tool to typeset math. LaTeX provides means to describe special characters like accents or umlauts using a special notation, which can be used just the same inside of BibTeX Entries. A list of accent marks that are available in math-mode in LaTeX. Finding it difficult to recollect the exact meaning of a notation while solving mathematical equations? Don't worry, ScienceStruck is here to help you out. pdf that you just can't memorize. To typeset math, LaTeX offers (among others) an environment called equation. To place something written in TeX in math mode, use $signs to enclose the math you want to display. The Comprehensive LATEX Symbol List Scott Pakin 3 January 2008 Abstract This document lists 4947 symbols and the corresponding LATEX commands that produce them. In this tutorial I discuss how to use math environment in Latex. Another thing to notice is the effect of the \displaystyle command. A list of LaTEX Math mode symbols. TeX (LaTeX math mode) symbols in legends and Learn more about figure, deep learning vs. The corresponding right delimiters are of course obtained by typing ), ] and \}. Math Symbols used as Relation Symbols. Some of these blocks are dedicated to, or. LaTeX Letter Template The files below contain an attempt to emulate the new University regulations for letters. com is a blog of Latex tutorials. How to write matrices in Latex ? matrix, pmatrix, bmatrix, vmatrix, Vmatrix. From the naming, I think \therefore and \implies are redundant, but I can't find a symbol for \suchthat and at university, we used$\therefore$as a shortcut for "such that". To place something written in TeX in math mode, use$ signs to enclose the math you want to display. You can also generate an image of a mathematical formula using the TeX language (pronounced "tek" or "tech"). LaTeX Mathematical Symbols - Free download as PDF File (. 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LaTeX forum ⇒ Math & Science ⇒ Mathematical Expectation symbol Information and discussion about LaTeX's math and science related features (e. 09 (2002-03-22) 5 3. Solid Geometry. In our project, we have this module meant strictly for our custom API endpoints for our mobile applications to use. Inserting Images; Tables; Positioning Images and Tables; Lists of Tables and Figures; Drawing Diagrams Directly in LaTeX; TikZ package References and Citations. The package cmbright. All common mathematical symbols are implemented, and you can find a listing on the LaTeX cheat sheet. Also, the letters are not in italics by default. what is the symbol of permutation, how to write permutation on latex, latex combination pe, latex typeset perm, latex command for permutation symbol, combination notation latex, Symbol in of permutations and combinations, how to write conbination and parmutTion symbol in latex, permutation latex, How to write permutation in latex, permutation. latex dot product symbol; If this is your first visit, be sure to check out the FAQ by clicking the link above. Put a label inside it using \label{foo} and then use \ref{foo} to typeset the equation number. But \mathcal only seems to have upper case letters. LaTeX Command Symbols _{exp} To get an expression exp to appear as a subscript. Math Symbols used as Relation Symbols. If you've ever tried to create web pages that contain mathematical notation for expressions and equations, you've learned just how inconvenient HTML (HyperText Markup Language) is for such tasks. ~\cite{latex}. 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Word has a new math ribbon with an explicit LaTeX option as shown in the article Linear format equations using UnicodeMath and LaTeX in Word. ;; Copyright (c) 1999, 2000, 2001 ;; John Palmieri ;; Author: John Palmieri ;; Maintainer: John.$\boldsymbol{=}$The equals sign "=" has a standard meaning in math: "x = y" means x and y are two different names for the same mathematical object. If you want the same body in all the letters, you may want to consider putting the entire body in a new command like \newcommand {\BODY}{actual body} and then using \BODY in all the letters. In general, if you're used to Latex, then you can simply enter Latex codes such as \rightarrow in math mode, and LyX will display most of the symbols correctly (you may have to press the SPACE key or move the cursor before LyX displays the symbol). com offers free software downloads for Windows, Mac, iOS and Android computers and mobile devices. 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If you want to use them in text just put the arrow command between two$ like this example: $\uparrow$ now you got an up arrow in text. mathalpha cursive small l. Letters are printed in italics, with more space left in-between, spaces are ignored. tex and cutting the pictures out of the resulting preview. I want to thank all the developer for providing such flexible tool. Formal letters, Cover letters, Newsletters 17 Topics 41 Posts. Special Symbols in LaTeX. Fill in The Unit Circle Positive: Negative: Positive: Negative: Positive: Negative: ( Positive: Negative: EmbeddedMath. Letters are printed in italics, with more space left in-between, spaces are ignored. They include the latex degree symbol, does-not-equal sign, greater-than-or-equal-to, less-than-or-equal-to signs, omega symbol, approximately symbol, etc. Learn about the people and activities that make UC Berkeley one of the best places in the world for advanced research, graduate and undergraduate study in mathematics. Only symbols of type \mathalpha will be affected by math alphabet commands: within the argument of a math alphabet command they will produce the character in slot of that math alphabet's font. Mathematical genealogy and list of Ph. Page breaks in math environments []. , together with an easy way to refer to these numbers. Depending on your preferred input format, you can create equations in Word in either one of UnicodeMath or LaTeX formats by selecting the format from the Equations tab. \infty \Rightarrow \surd \bigotimes ∞⇒ √N 4. Sigma notation provides a way to compactly and precisely express any sum, that is, a sequence of things that are all to be added together. for {Windows, Mac OS X, Linux}. Word has a new math ribbon with an explicit LaTeX option as shown in the article Linear format equations using UnicodeMath and LaTeX in Word. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. I have looked thru several documents re math in Latex, and have not found one which collects the common math symbols in one place. The Comprehensive LATEX Symbol List Scott Pakin ∗ 22 September 2005 Abstract This document lists 3300 symbols and the corresponding LATEX commands that produce them. As mention above, a plain text editor (Notepad) will not be able to show you symbols and mathematical formulas, etc. MATHCOUNTS Trainer LaTeX TeXeR MIT PRIMES My daughter was already very good at math before she took Prealgebra through AoPS. List of all mathematical symbols and signs - meaning and examples. For example, Edmodo will show the formula shown in the left column below if you send a note with the text shown in the right column. txt) or read online for free. This video explains how to add latex formatted text and math symbols in a scilab plot. Refer to the external references at the end of this article for more information. Math I: Super/Subscripts and Common Commands Introduction This tutorial will show you how to do some basic typesetting of math symbols, equations and matrices. LaTeX Mathematics Symbols. Including math equations, symbols, and more in your Top Hat questions and discussions is possible using the Inline Symbol Selector or LaTex, a popular markup language. I hope someone can help me :) I use GnuPlot to. A list of accent marks that are available in math-mode in LaTeX. In my presentation below, all squares and the main formula are products of LaTeX, the rest is Prezi. In certain cases it may be desirable to include "normal text" within an equation. Spacing in math mode; Integrals, sums and limits; Display style in math mode; List of Greek letters and math symbols; Mathematical fonts Figures and tables. I have looked thru several documents re math in Latex, and have not found one which collects the common math symbols in one place. 1These symbols are also available in math mode through the use of the mathcomppackage. Greek Letter 2. Type the name of the symbol you're looking for in the text box below to search all of the commands listed on this website. […] gaudetteje Says: February 5, 2014 at 9:37 am | Reply. We will use \LaTeX, which is based on \TeX\ and has many higher-level commands (macros) for formatting, making tables, etc. In reading about this issue, I came across a nice Blog post [A Math Prezi]. , braces { }. In my presentation below, all squares and the main formula are products of LaTeX, the rest is Prezi. 4 Math-mode accent marks ^a a a a nhatfag ncheckfag nbrevefag nacutefag a ~a a ~a ngravefag ntildefag nbarfag nvecfag a_ a a 0 ndotfag nddotfag anspfnprimeg Greek Letters nalpha nbeta ngamma ndelta nepsilon" # nvarepsilon nzeta neta ntheta nvartheta niota nkappa nlambda nmu nnu ˘ ˇ \$ ˆ % nxi npi nvarpi nrho nvarrho ˙ & ˝ ˛ ˚. For the author byline, use the formatting and symbols included in the template. Download MathType for Windows now from Softonic: 100% safe and virus free. The notation was introduced by the German mathematician Gottfried Wilhelm Leibniz in 1675 in his private writings; it first appeared publicly in the article "De Geometria Recondita et analysi indivisibilium atque infinitorum" (On a hidden geometry and analysis of indivisibles and infinites), published in Acta Eruditorum in June 1686. Mathematical Symbols. Here is what your formula looks like with the Cambria font in MS Word: If you are considering a career as an academic, and you'll be publishing a lot of papers containing mathematics, then learning LaTeX is definitely worthwhile. Most math symbols are Latin or Greek letters, but occasionally you'll run into Russian or Hebrew letters. Math mode accents 12. When I asked my students for feedback on L a T e X the consensus was that it is pretty easy but they need a list of symbols. trigonometry symbols, find the equivalent of the expression tan tither /sec theta, what does the symbol of tither mean in mathematics, Cot square tither is equal, titer sysbol, symboy of titer, math titer. 07 (2000/07/19) 4 Note 1. This is a cheat sheet for writing mathematics with L a T e X. In order to access all the math functions and symbols we will introduce in the guide pages, you'll have to include a number of packages. In a math environment, LaTeX ignores the spaces you type and puts in the spacing that it thinks is best. Letters are. Use the templates to create educational math illustrations with the shapes of plane and solid geometric figures, trigonometrical functions and Greek letters. Also, let or be a positive function and or any function. We just made a new paragraph. Inserting Greek symbols, complicated technical. The value of a counter can be changed with a command of the type \setcounter{equation}{0}. \textbf|apply to these Greek letters as to usual text and the font is upright in an upright environment. Latex symbols in Math mode. One also needs to be able to change fonts. Note that math mode ignores whitespace, in fact, this whole code could have been put on one line and still would have compiled correctly. com and in the control panel set your grabs to be "private" instead of public. Text Math Macro Category Requirements Comments 000A5 ¥ U \yen mathord amsfonts YEN SIGN 000AE ® r \circledR mathord amsfonts REGISTERED SIGN 000F0 ð g \eth mathalpha amssymb arevmath eth 00302 x̂ (bx) \hat mathaccent # \widehat (amssymb), circumflex accent. Let be an integer variable which tends to infinity and let be a continuous variable tending to some limit. Jupyter notebook recognizes LaTeX code written in markdown cells and renders the symbols in the browser using the MathJax JavaScript library. The closest I can seem to get is by using \wedge, which isn't the same. It is also crowned to be one among the most useful LaTeX editor there is. Also, let or be a positive function and or any function. They also need to fully understand the equations and formula themselves. com and download a free version of math type. LaTeX is a powerful tool to typeset math. But \mathcal only seems to have upper case letters. An online LaTeX editor that's easy to use. A LaTeX letter class, oxmathletter. Cursive small letter in math mode Hi All, I want to add a cursive small letter "r" to my equations, like the kind you find in Griffiths' book on E&M when he is talking about separation vectors. Some of these symbols are guaranteed to be available in every LATEX2𝜀system; others require fonts and. For instance the obsolete eqnarray environment frequently appears in questions of new LaTeX users and many people including me usually answer: don’t use eqnarray and give advice how to use the align environment of amsmath instead. 9 List of Mathematical Symbols 45 3. pdf), Text File (. You can find them in Wikipedia's list of mathematical symbols. The more unusual symbols are not defined in base LATEX (NFSS) and require usepackage{amssymb}. Selected LaTeX Math Symbols Note: there is another version of this document featuring HTML entities for math symbols, as well as LaTeX commands. This list of mathematical symbols by subject shows a selection of the most common symbols that are used in modern mathematical notation within formulas, grouped by mathematical topic. The project has settled on using both HTML and Template:TeX because each has advantages in some situations. They let me in even though my undergrad GPA was abysmal (2. Different classes of mathematical symbols are characterized by different formatting (for example, variables are italicized, but operators are not) and different spacing. Most letters and symbols are simple in LaTeX, yet a few characters are reserved for LaTeX commands, i. The corresponding right delimiters are of course obtained by typing ), ] and \}. latex-workshop. LaTeX is a math markup language familiar to many in the science and math community, but unfortunately is not currently supported by screenreader technlogy. explains how symbols are spaced in math mode, presents a LATEX ASCII and Latin 1 tables, and provides some information about this document itself. \infty \Rightarrow \surd \bigotimes ∞⇒ √N 4. The usual font selection commands|e. tex file (the source) for copying. Math symbols and math fonts 3. Refer to the external references at the end of this article for more information. 18List of Mathematical Symbols 19Summary 20Notes 21Further reading 22External links Mathematics environments LaTeX needs to know beforehand that the subsequent text does indeed contain mathematical elements. In the Equation Editor Design ribbon, go to the Conversions group and click LaTeX. Log-like Symbols. To insert symbols, Click More > Insert tab. You should visit the following links: For vectors: An introduction to beautiful math on Quora For matrices: An introduction to beautiful math on Quora. The maths styles can be set explicitly. How to write two dot above a letter? Ask Question Asked 5 years, 8 months ago. In this regard, the package provides \bigpumpkin \greatpumpkin. Sometimes math uses a new font rather than a new alphabet, such as Fraktur. This package defines the following commands:. The purpose of Applied Mathematics Letters is to provide a means of rapid publication for important but brief applied mathematical papers. Secondly, the words don't look quite right --- the letters are more spaced out than normal. The Wolfram Language by default interprets any sequence of letters or letter-like forms as the name of a symbol. The advantages of LaTeX include that it is public domain, platform and printer independent, allows mathematical symbols and equations to be entered much easier and quicker than word processors, and produces camera ready text. The mathematical community almost universally accepts a typesetting language called LaTeX. Typesetting mathematics is one of LaTeX's greatest strengths. LaTeX: stack three lines in math mode. sty does not provide bold maths, whereas with sfmath. LaTeX Mathematics Symbols. The four basic operations are denoted by the following symbols: “+” implies addition, “-“ implies subtraction, “x” implies multiplication, and “/” implies division. The Comprehensive LATEX Symbol List Scott Pakin 9 November 2009 Abstract This document lists 5913 symbols and the corresponding LATEX commands that produce them. http://quicklatex. I want to put a left arrow over letter in math mode. Search results for MATH font, free downloads of MATH fonts at Fonts101. This article shows several fonts for use in math mode. module file like so: my_project_mobile_api. Inserting Images; Tables; Positioning Images and Tables; Lists of Tables and Figures; Drawing Diagrams Directly in LaTeX; TikZ package References and Citations. Math Mode Accents. The brief descriptions of any work involving a novel application or utilization of mathematics, or a development in the methodology of applied mathematics is a potential contribution for this journal. Bibliography management in. Si, au contraire, on souhaite que LaTeX puisse couper à un endroit on rajoutera un \linebreak (qui suggère un point de coupure mais ne le force pas obligatoirement). I hope someone can help me :) I use GnuPlot to. They also need to fully understand the equations and formula themselves. Just as with \pagebreak, \displaybreak can take an optional argument between 0 and 4 denoting the level of desirability of a page break. For this example, plot y = x 2 sin (x) and draw a vertical line at x = 2. It is also crowned to be one among the most useful LaTeX editor there is. Many classes of sets are denoted using doublestruck characters. MathJax basic tutorial and quick reference. | 2019-11-12T21:47:13 | {
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# A man 8 friends whom he wants to invite for dinner. The number of ways
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27 Apr 2017, 06:35
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A man 8 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is
A)8
B)255
C)8!-1
D)256
E)7
Source :Quantpdf
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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27 Apr 2017, 06:46
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VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner.The number of ways in which he can invite at least one of them is
A) 8
B) 255
C) 8!-1
D) 256
E) 7
Take the task of inviting friends and break it into stages.
ASIDE: Let's let A, B, C, D, E, F, G and H represent the 8 friends
Stage 1: Decide whether or not to invite friend A
You have 2 options: invite friend A or don't invite friend A
So, we can complete stage 1 in 2 ways
Stage 2: Decide whether or not to invite friend B
You have 2 options: invite friend B or don't invite friend B
So, we can complete stage 2 in 2 ways
Stage 3: Decide whether or not to invite friend C
So, we can complete stage 3 in 2 ways
.
.
.
Stage 8: Decide whether or not to invite friend H
So, we can complete stage 8 in 2 ways
By the Fundamental Counting Principle (FCP), we can complete all 8 stages (and create a guest list) in (2)(2)(2)(2)(2)(2)(2)(2) ways (= 256 ways)
NOTE: in these calculations, one of the possible outcomes is that ZERO friends are invited. The question says that AT LEAST ONE friend must come.
So, we must subtract this 1 outcome from our solution.
So, total number of ways to invite friends = 256 - 1 = 255
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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28 Apr 2017, 04:08
GMATPrepNow wrote:
VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner.The number of ways in which he can invite at least one of them is
A) 8
B) 255
C) 8!-1
D) 256
E) 7
Take the task of inviting friends and break it into stages.
ASIDE: Let's let A, B, C, D, E, F, G and H represent the 8 friends
Stage 1: Decide whether or not to invite friend A
You have 2 options: invite friend A or don't invite friend A
So, we can complete stage 1 in 2 ways
Stage 2: Decide whether or not to invite friend B
You have 2 options: invite friend B or don't invite friend B
So, we can complete stage 2 in 2 ways
Stage 3: Decide whether or not to invite friend C
So, we can complete stage 3 in 2 ways
.
.
.
Stage 8: Decide whether or not to invite friend H
So, we can complete stage 8 in 2 ways
By the Fundamental Counting Principle (FCP), we can complete all 8 stages (and create a guest list) in (2)(2)(2)(2)(2)(2)(2)(2) ways (= 256 ways)
NOTE: in these calculations, one of the possible outcomes is that ZERO friends are invited. The question says that AT LEAST ONE friend must come.
So, we must subtract this 1 outcome from our solution.
So, total number of ways to invite friends = 256 - 1 = 255
Hi Brent,
Can you show another approach?
What 'at least' mean?
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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28 Apr 2017, 06:03
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Mo2men wrote:
Hi Brent,
Can you show another approach?
What 'at least' mean?
"at least" means "greater than or equal to"
So, if I say that I own at least 3 guitars, then the number of guitars I own = 3 or 4 or 5 or 6....
Cheers,
Brent
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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28 Apr 2017, 06:09
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Top Contributor
1
VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner.The number of ways in which he can invite at least one of them is
A) 8
B) 255
C) 8!-1
D) 256
E) 7
Another approach (for Mo2men )
Number of ways to invite at least 1 friend = (# of ways to invite exactly 1 friend) + (# of ways to invite exactly 2 friends) + (# of ways to invite exactly 3 friends) + (# of ways to invite exactly 4 friends) + . . . . + (# of ways to invite exactly 8 friends)
Since the order in which we invite the friends does not matter, we can use COMBINATIONS.
For example, the number of ways to invite exactly 2 friends = 8C2
And the number of ways to invite exactly 3 friends = 8C3
etc.
So, the number of ways to invite at least 1 friend = 8C1 + 8C2 + 8C3 + . . . + 8C7 + 8C8
= 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1
= 255
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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02 May 2017, 16:51
1
VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner.The number of ways in which he can invite at least one of them is
A)8
B)255
C)8!-1
D)256
E)7
We can use the following equation:
Total number of ways to invite friends = number of ways to invite at least one friend - number of ways to invite zero friends.
The total number of ways the man can invite his friends is 2^8 since he can or cannot invite each friend. The number of ways to invite no friends is 1. Thus, the number of ways to invite at least one friend is 2^8 - 1 = 256 - 1 = 255.
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Re: A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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11 Aug 2018, 09:13
The question could be interpreted in another way. "If we will flip a coin 8 times what is the number of ways we get at least one tail"
When you flip a coin 8 times there are 2^8 different ways the result may come out. One of the possible results is to get 8 heads - HHHHHHHH. Now you can imagine tail is to invite a friend and head is not to invite. Getting 8 heads means not to invite any of them.
Therefore, all outcomes satisfy us except when we get eight heads. There is only one way to get 8 heads. So we deduct this one outcome from total number of possible outcomes
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A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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18 Aug 2019, 05:46
VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is
A)8
B)255
C)8!-1
D)256
E)7
Source :Quantpdf
An alternate approach for this question would be to find the subsets of a set with 8 entities.
Lets take a smaller set, say there were only 3 friends (A, B & C) then the ways to invite at least 1 would be A, B, C, AB, AC, BC and ABC (7 ways). This is $$2^n-1$$ $$[2^3-1 = 7]$$ in terms of the formula to find the subset of a set with n entities. We subtracted 1 because $$2^n$$ also includes an empty set, whereas we have been provided a constraint of at least 1.
Applying the same to the problem at hand, there are 8 friends, hence $$2^8-1 = 256-1 = 255$$ (Ans B)
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A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink]
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18 Aug 2019, 07:33
VyshakhR1995 wrote:
A man 8 friends whom he wants to invite for dinner. The number of ways in which he can invite at least one of them is
A)8
B)255
C)8!-1
D)256
E)7
Source :Quantpdf
Given: A man 8 friends whom he wants to invite for dinner.
Asked: The number of ways in which he can invite at least one of them is
No of ways to invite at least 1 friends = $$^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8$$
$$(1+x)^n = ^nC_0 x^0 + ^nC_1 x + ^nC_2 x^2 + ... + ^nC_r x^r +.... + ^nC_n x^n$$
n =8
$$(1+x)^8 = ^8C_0 x^0 + ^8C_1 x + ^8C_2 x^2 + ... + ^8C_r x^r +.... + ^8C_8 x^n$$
When x =1
$$2^8 = 1 + ^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8$$
$$2^8 -1= ^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8 = 256-1 =255$$
IMO B
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A man 8 friends whom he wants to invite for dinner. The number of ways [#permalink] 18 Aug 2019, 07:33
Display posts from previous: Sort by | 2019-10-15T21:42:33 | {
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http://mathhelpforum.com/math-topics/179652-if-f-x-y-f-x-f-y-what-value-f-4-a.html | # Math Help - If f(x-y)=f(x)f(y) what is the value of f(4)?
1. ## If f(x-y)=f(x)f(y) what is the value of f(4)?
Dear Colleagues,
If $f(x-y)=f(x)f(y)$, then what is the value of $f(4)$?
Best Regards.
2. Here's a hint $e^{m-n} = e^m \times e^{-n}$
3. Fun problem! I'm not sure I've solved it, but I have made some progress, or at least found a few things out. First, suppose you set y = 0. Then you get f(x) = f(x) f(0) for all x. Then either f(0) = 1 or f(x) = 0 for all x. Suppose f(0) = 1. Then we could set x = 0, and get f(-y) = f(0) f(y) = f(y), thus making f even. But the f(x) = 0 case was also even. Hence, we conclude that f is even.
If we set x = y = 0, then we get that f(0) = f(0) f(0), which implies that f(0) is in the set {0,1}. Letting x = 4 and y = 4 shows us that f(0) = f(4) f(4), which implies that f(4) is in the set {-1, 0, 1}. Setting x = 4 and y = 0 gives us f(4) = f(4) f(0).
I think your problem is underdetermined. Why? f(x) = 0 satisfies the function equation, and f(x) = 1 satisfies the function equation, for all x. Therefore, with the information you have, you cannot uniquely determine f.
Reply to pickslides: unfortunately, the exponential function is neither even nor odd, so I don't think the exponential function fits here.
4. Originally Posted by Ackbeet
Fun problem! I'm not sure I've solved it, but I have made some progress, or at least found a few things out. First, suppose you set y = 0. Then you get f(x) = f(x) f(0) for all x. Then either f(0) = 1 or f(x) = 0 for all x. Suppose f(0) = 1. Then we could set x = 0, and get f(-y) = f(0) f(y) = f(y), thus making f even. But the f(x) = 0 case was also even. Hence, we conclude that f is even.
If we set x = y = 0, then we get that f(0) = f(0) f(0), which implies that f(0) is in the set {0,1}. Letting x = 4 and y = 4 shows us that f(0) = f(4) f(4), which implies that f(4) is in the set {-1, 0, 1}. Setting x = 4 and y = 0 gives us f(4) = f(4) f(0).
I think your problem is underdetermined. Why? f(x) = 0 satisfies the function equation, and f(x) = 1 satisfies the function equation, for all x. Therefore, with the information you have, you cannot uniquely determine f.
Reply to pickslides: unfortunately, the exponential function is neither even nor odd, so I don't think the exponential function fits here.
I agree with Ackbeet. I leave it to you to prove the details. Consider the following:
f(0) = f(0)*f(0) implies f(0) = 0 or f(0) = 1. We can show that f(n) = 0 for all n if f(0) = 0.
So let's assume that f(0) = 1. Then
f(1) = f(2)*f(1) implies f(1) = 0 or f(2) = 1. Again we can show that f(n) = 0 for all n > 1 if f(1) = 0.
So let's assume that f(2) = 1. Then
f(2) = f(4)*f(2), etc.
I get either f(4) = 0 or f(4) = 1.
-Dan | 2014-10-25T04:38:53 | {
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https://byjus.com/question-answer/given-the-setup-shown-in-fig-block-a-b-and-c-have-masses-m-a/ | Question
Given the setup shown in Fig. Block A, B, and C have masses $$m_A=M$$ and $$m_B= m_C=m$$. The strings are assumed massless and unstretchable, and the pulleys frictionless. There is no friction between blocks B and the support table, but there is friction between blocks B and C, denoted by a given coefficient $$\mu$$.a. In terms of the given, find (i) the acceleration of block A, and (ii) the tension in the string connecting A and B.b. Suppose the system is related from rest with block C. near the right end of block B as shown in the above figure. If the length L of block B is given, what is the speed of block C as it reaches the lift end of block B? Treat the size of C as small.c. If the mass of block A is less than some critical value, the blocks will not accelerate when released from rest. Write down an expression for that critical mass.
Solution
Apply constraint equation on strings, the length of strings is constant. Differentiate twice to get relation between of acceleration of block A, B, and C be a, b, and c, respectively. $$l_1 +l_2$$ = constantand $$l_3 + l_4$$ constant$$l_1+ l_2 = 0 \Rightarrow |b| = |c|$$$$l_3 + l_4 = 0 \Rightarrow |a| = |b|$$From which we get a=b=c.From FBDs Of A, B, and C [Fig. (a)],Writing equations of motion for block A:$$mg - T = Ma$$ (i)For block B, $$T - T_1 -\mu mg = ma$$ (ii)For block C, $$T- \mu mg = ma$$ (iii)Solving equations (i), (ii) and (iii), we get$$a= (\dfrac{m-2\mu m}{m+2m})g$$ (iv)Putting a in Eq. (i) , we get $$mg -T = M (\frac{M-2\mu m}{M+2m}) g \Rightarrow T = \frac{2mMg(1+\mu)}{(M+2m)}$$b. As there is relative motion between blocks, we apply $$V_{rel}^2 = V_{rel}^2 +2a_{rel} S_{rel}$$If system is released from rest, $$u_{rel} =0$$$$v_{rel}^2 = 2a_{rel} S_{rel} \Rightarrow v_{rel} =\sqrt{2a_{rel} S_{rel}}$$$$a_{rel} = 2a$$ and $$S_{rel} L$$$$\Rightarrow v=\sqrt{\dfrac{4gL(M-2\mu m)}{(M+2m)}}$$c. If blocks will not accelerate, then put a=0 in Eq. (iv) to get $$M=2\mu m$$.Physics
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https://math.stackexchange.com/questions/1180034/inequality-and-integral | # Inequality and integral
I am working on a problem and I feel like I may not completely understand that concept which I think is real important that I do.
The question is in regard to showing the validity of the following inequality
$\int_0^{n}x^{q}dx$ $\le 1+2^{q}+..+n^{q} \le \int_0^{n+1}x^{q}dx$ for any positive integer n and any real q $\ge 0$
One answer, similar to one suggested by the user Tryss on this site is as follows,
$$\int_0^n x^q dx = \sum_{k=1}^{n} \int_{k-1}^k x^q dx \leq \sum_{k=1}^{n} \int_{k-1}^k k^q dx$$
Then
$$\int_0^n x^q dx \leq \sum_{k=1}^{n} k^q = 1+ 2^q + \cdots n^q$$
And we have,
$$\int_0^{n+1} x^q dx = \sum_{k=0}^{n} \int_{k}^{k+1} x^q dx \geq \sum_{k=0}^{n} \int_{k}^{k+1} k^q dx$$
i,e then,
$$\int_0^{n+1} x^q dx \geq \sum_{k=0}^{n} k^q = 1+ 2^q + \cdots n^q$$
The next part of the question that I am mostly stumped on is to find $\lim_{n\to \infty}a_n$ where $a_n= 1/{n^{q+1}}+2^{q}/n^{q+1}+…+n^{q}/n^{q+1}$. I'm sure I need to use the result of the first part of the problem but I am not sure where to begin.
Here is where I am looking for help as well; I am feeling confused about the proposed solution above. I think my background is a little lacking on remembering the relation between the summation symbols and the integral symbol in general. I see clearly that we are increasing, but why choose the interval $[k,k-1]$ for example? is it just an arbitrary interval for which we can use the fact that we have an increasing so we can write $x^{q} \le k^{q}$
Then I am confused overall, about the geometric meaning and such. for example how is it true and what is the meaning of writing $$\int_0^n x^q dx = \sum_{k=1}^{n} \int_{k-1}^k x^q dx \leq \sum_{k=1}^{n} \int_{k-1}^k k^q dx$$ . How can I know/see/understand what this is saying is true? and vice versa for the other inequality.
So overall I am looking for any help on that, and with the limit. All is very greatly appreciated and I am trying to understand! Thanks a lot in advance to anyone who is willing to help.
• The interval $[k-1,k]$ is taken because it has length $1$, hence $$\int_{k-1}^{k}k^qdx=k^q(k-(k-1))=k^q. For a geometric interpretation, split [0,n] into intervals [k-1,k] and draw bars of height k^q and (k+1)^q and length 1 for each k and plot f(x)=x^q on the same graph, you should see how these bars are below and above the graph of f. For your question regarding a_n, note that for large n, a_n is close to the integral of x^q from 0 to n considering a partition with intervals of length 1/n^q. Mar 7, 2015 at 22:33 ## 1 Answer What you have in the first part of your problem is actually two inequalities. The first one looks like a case of the more general$$\int_0^n f(x)\, dx \leq f(1) + f(2) + \cdots f(n)$$for an integer n and a function f that is increasing on the interval [0,n]. The right-hand side of this inequality is a right-handed Riemann sum of the function f over that interval. The other inequality is a case of$$ f(0) + f(1) + \cdots f(n+1) \leq \int_0^{n+1} f(x)\, dx$$for an integer n+1 and a function f that is increasing on the interval [0,n+1].The left -hand side of this inequality is a left-handed Riemann sum of the function f over that interval. As a reminder, the left-handed and right-handed Riemann sums of an increasing function look something like this: I did not draw f(x) = x^q here because I wanted the first rectangle in the sum to be visible in both cases. But if you make that substitution, then f(0) = 0, f(1) = 1, and in general f(k) = k^q. Now consider this fact:$$\int_0^n x^q\, dx = \sum_{k=1}^{n} \int_{k-1}^k x^q\, dx.$$This is simply a consequence of the fact that you can compute an integral in two pieces:$$\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx.$$Repeat this step n-1 times, using a different integer as the new bound of integration each time, and you have a sum of n integrals equal to the original. The fact$$\sum_{k=1}^{n} \int_{k-1}^k x^q\, dx \leq \sum_{k=1}^{n} \int_{k-1}^k k^q\, dx$$for an increasing function f is a consequence of the more general fact that if f(x) \leq g(x) when a \leq x \leq b, then$$\int_a^b f(x)\, dx \leq \int_a^b g(x)\, dx.$$In this case you take f(x) = x^q and g(x) = k^q. Visually, you can look at the rectangles in the right-hand Riemann sum to see how this works in this particular case: the area of the rectangle between x = k-1 and x=k is f(k), and it is greater than the area under the curve y=f(x) in that same interval. For the second part of the question, notice that$$a_n= \frac{1 + 2^q + \cdots + n^q}{n^{q+1}}.$$The numerator indeed looks a lot like the middle part of the inequalities from the first part of the problem. In fact you can use those inequalities to write$$\frac{1}{n^{q+1}} \int_0^n x^q\, dx \leq \frac{1 + 2^q + \cdots + n^q}{n^{q+1}} \leq \frac{1}{n^{q+1}} \int_0^{n+1}x^{q}dx.
You can use the usual formula for $\int x^q\, dx$ to evaluate the definite integrals on both sides. Does it look like the squeeze theorem applies to this problem then?
• Thank you! So is the way that I can know if it is right hand or left hand as follows; you knew f(1)+f(2)+.. was the right hand limit because you are evaluating from 1 above the bottom limit of integral and the left hand is because you are starting from 0 itself. Really all that changes is f(0) or f(1) with f(n) and f(n+1)? Mar 7, 2015 at 23:14
• but only in the case that it is increasing , if not the other way around? Mar 7, 2015 at 23:16
• Yes, it's a left-hand sum if you take the value at the left side of each interval. And yes it was important here that the function was increasing, because that meant the left side of each interval gave the smallest value of $f$ on that interval. If the function were decreasing then the right-hand sum would have been on the left of $\leq$ instead of the left-hand sum. Mar 8, 2015 at 3:14
• That makes sense ! By the way , does 1/p+1 make sense for the limit using squeeze thereom? Mar 9, 2015 at 14:53
• You mean $1/(q+1)$? It sounds like you have the right idea. Mar 9, 2015 at 15:00 | 2022-05-19T13:11:19 | {
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https://math.stackexchange.com/questions/3034747/one-urn-with-white-balls-another-with-10-black-and-30-white-probability-of-dra | # One urn with white balls, another with 10 black and 30 white. Probability of drawing black after seeing white.
This is a problem from Y.A. Rozanov book "Probability theory: A concise course", and seems to be simple, yet the answer given disagrees with what I arrived at. Text verbatim (ex. 15) is following
## Exercise 15/p36
One urn contains only white balls, while another urn contains 30 white and 10 black balls. An urn is selected at random and then a ball is drawn (at random) from the urn. The ball turns out to be white and is then put back into the urn. What is the probability that another ball drawn from the same urn will be black?
The answer given is $$P = \frac{3}{28}$$
If the above is correct, my solution must be flawed somehow:
Let $$P(x_k)$$ be the probability that ball $$x$$ is drawn in $$k-th$$ draw. Then we seek the probability $$P(b_2|w_1)$$. Given that urns were selected at random, and we don't which one we ended up with
$$P(b_2|w_1) = P(b_2|w_1|u_1)P(u_1) + P(b_2|w_1|u_2) P(u_2)$$
where $$P(u_i) = \frac{1}{2}$$ is the probability that $$i-th$$ urn was selected. The first of these terms vanishes, as urn 1 does not contain any black balls, hence the probability of seeing a black ball is 0.
For the second, as the white ball is put back into the urn, the second draw is obviously independent with the first, hence
$$P(b_2|w_1|u_2) = \frac{1}{4}$$
Thus the answer would come to be
$$P(b_2|w_1) = \frac{1}{8}$$
which is obviously very different from given answer.
Where is the mistake?
Let $$U_1$$ is the event ball drawn from Urn-$$1$$.
Let $$U_2$$ is the event ball drawn from Urn-$$2$$.
Let $$W$$ be the event that the ball is white.
$$P(U_1|W)=\dfrac{P(W|U_1)\cdot P(U_1)}{P(W|U_1)\cdot P(U_1)+P(W|U_2)\cdot P(U_2)}=\dfrac{1\cdot\dfrac12}{1\cdot\dfrac12+\dfrac34\cdot\dfrac12}=\dfrac47$$ From this we can say the $$P(U_2|W)=\dfrac37$$
The probability that the white ball is drawn second time is $$\dfrac47\cdot1+\dfrac37\cdot\dfrac34=\dfrac{25}{28}$$
Therefore, the probability that the black ball is drawn second time $$=1-\dfrac{25}{28}=\dfrac{3}{28}$$
• Alternatively, the probability that the black ball is drawn second time is $\dfrac47\cdot0+\dfrac37\cdot\dfrac14=\dfrac{3}{28}$ – Henry Dec 11 '18 at 1:42
• So essentially the flaw in my solution is the assumption that the event of drawing the black ball is independent of seeing the white ball, and this is because of the existence of the first urn, is it? – Teremin12 Dec 11 '18 at 1:57
• @Teremin12 Yes. – Key Flex Dec 11 '18 at 2:00 | 2019-05-22T07:15:44 | {
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https://math.stackexchange.com/questions/459494/proving-logical-equivalence-p-leftrightarrow-q-equiv-p-wedge-q-vee-ne | # proving logical equivalence $(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$
I am currently working through Velleman's book How To Prove It and was asked to prove the following
$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$
This is my work thus far
$(P \to Q) \wedge (Q \to P)$
$(\neg P \vee Q) \wedge (\neg Q \vee P)$ (since $(P \to Q) \equiv (\neg P \vee Q)$)
$\neg[\neg(\neg P \vee Q) \vee \neg (\neg Q \vee P)]$ (Demorgan's Law)
$\neg [(P \wedge \neg Q) \vee (Q \wedge \neg P)]$ (Demorgan's Law)
At this point I am little unsure how to proceed.
Here are a few things I've tried or considered thus far:
I thought that I could perhaps switch some of the terms in step 3 using the law of associativity however the $\neg$ on the outside of the two terms prevents me from doing so (constructing a truth table for $\neg (\neg P \vee Q) \vee (\neg Q \vee P)$ and $\neg (\neg P \vee \neg Q) \vee \neg (P \vee Q)$ for sanity purposes)
I can't seem to apply the law of distribution since the corresponding terms are negated.
Applying demorgans law to one of the terms individually on step 2 or 3 doesnt seem to get me very far either.
Did I perhaps skip something? Am I even on the right track? Any help is appreciated
• I feel you've missed, or perhaps forgotten, something more important here. In your parenthetical comment you talk about constructing a truth table for sanity purposes. The term sanity basically means health of mind. One can say that truth tables can help you keep your logical thinking healthy. You haven't layed down a formal system or even set of laws which to use so that you know you can prove this in some theory, so I don't see it necessary to solve the problem in any particular way. Due to the (relevant) completeness result you can use truth tables... – Doug Spoonwood Aug 4 '13 at 14:14
• And you have two variables here, in a 2-valued system. So, your truth table only has 4 rows. And since you've implicitly adopted an algebraic approach for a finite logical system, I will add that it turns out that you can always, in principle, use truth tables to solve any such problem once you have the tables for the connectives. So, I ask you to consider solving this problem in the "healthy" way by writing out the truth table. I do not believe Velleman would object at all to you doing that here. – Doug Spoonwood Aug 4 '13 at 14:19
• I understand that providing a truth table for both statements is sufficient to prove that both statements are indeed logically equivalent. I, however, feel that one my of weakest points is proving these types of problems algebraically, thus I chose this proof method over truth tables. Thank you for your thoughtful comment. – Bryan Baraoidan Aug 4 '13 at 16:04
$$(P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)$$
I'll start with your initial work, but instead of employing DeMorgan's as you did, we'll use the Distributive Law (DL), in two "steps":
\begin{align} (P \leftrightarrow Q) &\equiv (P \to Q) \wedge (Q \to P) \tag{correct}\\ \\ &\equiv (\color{blue}{\bf \lnot P \lor Q}) \land (\color{red}{\bf \lnot Q \lor P})\tag{correct} \\ \\ &\equiv \Big[\color{blue}{\bf \lnot P} {\land} \color{red}{\bf(\lnot Q \lor P)}\Big] \color{blue}{\lor} \Big[\color{blue}{\bf Q} \land \color{red}{\bf (\lnot Q \lor P)}\Big]\tag{DL}\\ \\ & \equiv \Big[(\color{blue}{\bf \lnot P} \land \color{red}{\bf\lnot Q)} \lor (\color{blue}{\bf\lnot P} \land \color{red}{\bf P})\Big] \lor \Big[(\color{blue}{\bf Q} \land \color{red}{\bf \lnot Q}) \lor (\color{blue}{\bf Q} \land \color{red}{\bf P})\Big]\tag{DL} \\ \\ &\equiv \Big[({\lnot P}\land \lnot Q) \lor \text{False}\Big] \lor \Big[\text{False} \lor (Q \land P)\Big]\tag{why?} \\ \\ &\equiv (P \land Q) \lor (\lnot P \land \lnot Q)\tag{why?}\end{align}
• Ah thank you. I did not know you could apply the distributive law like that. – Bryan Baraoidan Aug 4 '13 at 16:15
• You're welcome! – Namaste Aug 4 '13 at 16:16
• @amWhy: It is nice to receive that FB! +1 – Amzoti Aug 5 '13 at 0:10
• @BryanBaraoidan Well, when you have something of the form $(a\color{blue}\vee b)\color{red}\wedge (c\color{blue}\vee d)$ and wish to derive something of the form $(w\color{red}\wedge x)\color{blue}\vee(y\color{red}\wedge z)$, well, not only may you attempt distribution to see what cancels, it is almost mandatory to try. – Graham Kemp Apr 16 '18 at 0:33
That's a great book. You'll learn a lot from it.
Justify the following: \begin{align} (\neg P \lor Q) \land (\neg Q \lor P) &\equiv[(\neg P\lor Q) \land \neg Q] \lor [(\neg P\lor Q)\land P] \\ &\equiv [(\neg P \land \neg Q) \lor (Q\land \neg Q)]\lor [(\neg P\land P)\lor (Q\land P)] \\ &\equiv (\neg P\land \neg Q) \lor (Q\land P).\end{align}
You can also "reverse" amWhy solution, starting with :
$(P \land Q) \lor ( \lnot P \land \lnot Q)$
you can use the distributivity laws to obtain :
$$[(P \land Q) \lor \lnot P] \land [(P \land Q) \lor \lnot Q]$$
and again :
$$(P \lor \lnot P) \land (Q \lor \lnot P) \land (P \lor \lnot Q) \land (Q \lor \lnot Q]$$
i.e.
$$T \land (Q \lor \lnot P) \land (P \lor \lnot Q) \land T$$
i.e.
$$(Q \lor \lnot P) \land (P \lor \lnot Q)$$
Now we use the equivalence between $(\lnot A \lor B) \quad$ and $\quad (A \rightarrow B)$, and get :
$$(P \rightarrow Q) \land (Q \rightarrow P)$$
that is
$P \leftrightarrow Q$.
For an alternative approach you can make the truth table of both logical expressions. $$\begin{array}{|c|c| c|c| c|c| c|c| c|} \hline P & Q & \neg P & \neg Q & (P \wedge Q) & (\neg P \wedge \neg Q) & (P \wedge Q ) \vee (\neg P \wedge \neg Q) \\\hline V & V &F & F &V & F &V \\\hline V & F & F & V & F & F & F \\\hline F & V & V & F & F & F & F \\\hline F & F & V & V & F & V & V \\\hline \end{array}$$ and $$\begin{array}{|c|c|c|} \hline P&Q& P \leftrightarrow Q \\\hline V&V&V \\\hline V&F&F \\\hline F&V&F \\\hline F&F&V \\\hline \end{array}$$ | 2019-10-24T00:39:23 | {
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https://stats.stackexchange.com/questions/475768/correct-way-to-combine-95-confidence-interval-bounds-returned-by-a-fitting-rout/477032#477032 | # Correct way to combine 95% confidence interval bounds returned by a fitting routine with several measurements?
I am looking for someone to just confirm / double-check something for me with regards to errors on measurements.
Let's say I am trying to determine the slope of a relationship by varying one quantity and measuring another, and then I plot the graph and do a least-squares fit straight line to the data (graph on the left). Then I repeat this procedure twice more, to get the middle and right-most graphs.
Each fit routune will typically give me back a slope and the corresponding 95% confidence interval, so that I obtain $$(m_1\pm\Delta m_1), (m_2\pm\Delta m_2)$$ and $$(m_3\pm\Delta m_3)$$. Now I know that the underlying quantity which determines $$m$$ in each case is the same, so I should be able to quote a best estimate for the slope as their mean
$$\bar{m} = \frac{m_1+m_2+m_3}{3}. \tag{1}$$
My question is about the appropriate way to quote the error. We know that for a function $$f(x,y)$$ with errors in $$x$$ and $$y$$ given by $$\Delta x$$ and $$\Delta y$$, respectively, the error on $$f$$ is given by
$$\Delta f = \sqrt{ (\Delta x)^2 \bigg(\frac{\partial f}{\partial x}\bigg)^2 + (\Delta y)^2 \bigg(\frac{\partial f}{\partial y}\bigg)^2 } \tag{2}$$
So I would think I can determine the error in $$\bar{m}$$ to be
\begin{align} \Delta \bar{m} &= \sqrt{ (\Delta m_1)^2 \bigg(\frac{\partial \bar{m}}{\partial m_1}\bigg)^2 + (\Delta m_2)^2 \bigg(\frac{\partial \bar{m}}{\partial m_2}\bigg)^2 + (\Delta m_3)^2 \bigg(\frac{\partial \bar{m}}{\partial m_3}\bigg)^2} \tag{3} \\ &= \frac{1}{3} \sqrt{ (\Delta m_1)^2 + (\Delta m_2)^2 + (\Delta m_3)^2 } \tag{4} \end{align}
First question, is this correct?
Second question, is it okay to propagate 95% confidence intervals in this way? Should I simply quote now the result as $$\bar{m} \pm \Delta \bar{m}$$ and just explain that $$\Delta \bar{m}$$ is the combined 95% confidence interval, or should I convert the 95% number from the fits into standard errors (through the factor of 1.96)?
(I am for now assuming Gaussian errors everywhere.)
EDIT
It was suggested in the comments that I first implement weighting in the averaging step before worrying about the errors. This should help to give more weight to slopes which have tighter confidence intervals (and vice versa).
According to this link, the weighted version of the mean would be given by $$\bar{m}_\textrm{w} = \frac{\sum_i w_i m_i}{\sum_iw_i}, \hspace{1cm} \textrm{where} \hspace{0.5cm} w_i = \frac{1}{\sigma_i^2}\tag{5}$$ and $$\sigma_i$$ is the variance of each slope. Therefore, in my case with the three example slopes, it should be $$\bar{m}_\textrm{w} = \frac{m_1/\sigma_1^2 + m_2/\sigma_2^2 + m_3/\sigma_3^2}{1/\sigma_1^2 + 1/\sigma_2^2 + 1/\sigma_3^2}. \tag{6}$$
The variance on the weighted mean slope is given at the above link again by \begin{align} \textrm{Var}(\bar{m}_\textrm{w}) &= \frac{\sum_iw_i^2\sigma_i^2}{\big( \sum_iw_i\big)^2}\tag{7}\\ &= \frac{1/\sigma_1^2 + 1/\sigma_2^2 + 1/\sigma_3^2}{\big(1/\sigma_1^2 + 1/\sigma_2^2 + 1/\sigma_3^2\big)^2}\tag{8}\\ &= \big(1/\sigma_1^2 + 1/\sigma_2^2 + 1/\sigma_3^2\big)^{-1}.\tag{9} \end{align}
So now my main question remains - these are variances, so should we convert the 95% confidence intervals $$\Delta m_i$$ returned by a fitting algorthm somehow into a variance?
Maybe for a concrete example we could imagine the following values were returned from the fitting routine: \begin{align} m_1 &= 5.5\; (4.9, 6.1)\rightarrow \Delta m_1 = 0.6\\ m_2 &= 5.5\; (5.3, 5.7)\rightarrow \Delta m_2 = 0.2\\ m_3 &= 5.2\; (4.5, 5.9)\rightarrow \Delta m_3 = 0.7 \end{align} where the values in brackets represent the 95% confidence intervals. How should the estimate of the slope be reported, including errors? Let's imagine I only have access to these values (and not the underlying data that was used for fitting to obtain these slopes).
• You can improve things using en.wikipedia.org/wiki/Inverse-variance_weighting Jul 7 '20 at 10:23
• To first order, you handle the confidence intervals by using them to weight your estimate, as indicated by @Jarle. Only then can you construct a reasonable confidence interval for that estimate.
– whuber
Jul 7 '20 at 13:34
• Thanks both, I have edited the question to try to incorporate your advice. Jul 7 '20 at 14:26
• Yes exactly, I only have access to $m_i \pm \Delta m_i$ where $\Delta m_i$ is the 95% interval. Jul 9 '20 at 11:56
• Are your experiments/runs supposed to have different variance in the noise? Or can we can we consider the process to be the same in which case we can use pooled variance. Jul 10 '20 at 12:23
I imagine the 95% confidence intervals come from some assumptions on normality of data. Otherwise, please state how you got these CI. This implies you believe the mean of each slope (viewed as a RV) is $$m_i$$ with some variance $$\sigma_i$$ In this case you can average the slopes as you did and get the new variance of the averaged estimator (assuming independent errors). From said variance you can get 95% CI (using 1.96 standard deviations).
So, to summarize (assuming $$m_i$$ are independent is crucial):
1. Let $$m := \frac{ \sum_{i=1}^N m_i \sigma_i^{-2}}{\sum_{i=1}^N \sigma_i^{-2}}$$
2. Let $$\sigma^2 := Var(m) = Var(\frac{ \sum_{i=1}^N m_i \sigma_i^{-2}}{\sum_{i=1}^N \sigma_i^{-2}}) = (\frac{1}{\sum_{i=1}^N \sigma_i^{-2}})^2 \sum_{i=1}^N \sigma^{-4}Var(m_i) = \frac{1}{\sum_{i=1}^N \sigma_i^{-2}}$$.
3. Note that this is the harmonic mean! Finally you see it in the wild after learning that inequality in your first calculus class!!
4. A 95% CI for the true value of the slope is $$[m- 1.96\sigma, m+ 1.96\sigma]$$ (see 1.96)
• Just to check - these are exactly the same as the formulae I quoted in the question, right? Jul 15 '20 at 17:06
• @teeeeee Yes they are (you wrote so much text I got lazy reading and decided it's easier to just post an answer). Note that you are relying on $m_i$'s being independent - you did not mention that anywhere and that is an assumption you are implicitly making. Jul 16 '20 at 7:23
I don't have much to offer in terms of methods, I think the ones presented here (esp inverse variance weighted approaches) are good ones. What I can add is a small simulation study to prove that under the assumption of Gaussian errors in the regression, this process has good enough coverage
set.seed(0)
library(tidyverse)
simulate_data<-function(n){
x = rnorm(n)
y = 2*x + 1 + rnorm(n, 0, 0.5)
model = lm(y~x)
results = tibble(beta = coef(model)['x'],
w = 1/vcov(model)['x','x'])
}
simulate_procedure<-function(iter){
n = rnbinom(3,200,0.9)
results = map_dfr(n, simulate_data)
m = sum(results$$beta*results$$w)/sum(results$w) sig = sqrt(1/sum(results$w))
interval = tibble(lower = m - 1.96*sig,
est = m,
upper = m + 1.96*sig)
interval
}
map_dfr(1:10000, simulate_procedure, .id = 'iter') %>%
mutate(contains = (2<upper)&(2>lower)) %>%
summarise(mean(contains))
>>>0.922
So what does this mean? It means that were I to repeat this procedure to construct a 95% interval for the slope, the resulting interval would capture the true slope (here 2) only 92% of the time. So barring I didn't make a mistake (entirely possible) that seems to be good enough.
How should the estimate of the slope be reported, including errors? Let's imagine I only have access to these values (and not the underlying data that was used for fitting to obtain these slopes).
So I would compute $$m$$ and $$\sigma^2$$ as mentioned by Yair Daon. You don't need to access the data in order to do these. In your example, the $$m$$ would be 5.5, 5.5, 5.2. The variances are found by doing a little algebra on the confidence interval. Remember, confidence intervals look like
$$m \pm 1.96 se$$
Here, $$se$$ is the standard error (or the standard deviation of the sampling distribution). You can find the variance by taking the difference between interval endpoints and then dividing by $$3.92 = 2\times 1.96$$. Your sigmas (not squared) would then be 0.306, 0.102, 0.357.
So your best estimate for $$m$$ from the example you've provided is 5.47, with an accompanying interval of 5.29 to 5.66. These were computed using the formulae provided by Yair.
• Can you explain a little what the code is doing? I am not familiar with the programming language you have used. Thanks! Jul 14 '20 at 18:01
• @teeeeee . I'm basically simulating your example 10,000 times and seeing how many times the confidence interval captures the true parameter of interest. Jul 14 '20 at 18:32 | 2022-01-22T11:18:12 | {
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https://mediatorchrzanow.pl/baek-jong-pzlv/acute-angle-triangle-24f2f8 | acute angle triangle
The acute triangle angles can be anything as long as each one of them lies between 0o 0 o to 90o 90 o. Putting all the values, we get ⇒ a + b + c = 180° Since we defined the trigonometric functions in terms of ratios of sides, you can think of the units of measurement for those sides as canceling out in those ratios. If a triangle has 1 acute angle, the other angles will be either right angles or obtuse angles which is not possible as the sum of interior angles of a triangle is always 180°. Based on the sides and the interior angles of a triangle, there can be various types of triangles, and the acute angle triangle is one of them. Not only scalene, but an acute triangle can also be an isosceles triangle if it satisfies its condition. Also, a, b, and c are the lengths of sides BC, CA and AB, respectively. A triangle with one interior angle measuring more than 90° is an obtuse triangle or obtuse-angled triangle. So, every triangle needs to have at least 2 acute angles. Its name provide a definite clue as to what makes it so special. Right triangles, and the relationships between their sides and angles, are the basis of trigonometry. What is an acute angle? The area of acute angle triangle = (½) × b × h square units, If the sides of the triangle are given, then apply the Heron’s formula, The area of the acute triangle = $$A = \sqrt{S (S-a)(S-b)(S-c)}$$ square units, Where S is the semi perimeter of a triangle, The perimeter of an acute triangle is equal to the sum of the length of the sides of a triangle, and it is given as. Find the measure of each acute angle in a right triangle where the measure of one acute angle is twice the difference of the measure of the other acute angle and 12. However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some … Properties of acute triangles. We can see that. The orthocenter is the point where all three altitudes of the triangle intersect. To Find :-The measure of both the acute angles. Thousands of new, high-quality pictures added every day. Formula to be used :-Angle Sum property of Triangle. ← Previous Page. a + b + c = 180° Solution :-In a right angled triangle, there is one angle of 90° let one of acute angle be x. let other angle be x + 20. An acute angle triangle (or acute-angled triangle) is a triangle in which all the interior angles are acute angles. Any triangle that has one right angle (90 degrees) is no longer acute because it doesn't fit the definition of an acute triangle, which states that no angle can be 90 degrees or over. A median of a triangle is the line that connects an apex with the midpoint of the opposite side. Fun Facts about Acute Triangles: The angles of an acute triangle add up to 180°, because of the Angle Sum Property. http://itsmyacademy.com/geometry/ for more videos and systematic study on geometry There are many types of triangle. Since triangle ABC below has interior angles all of which are less than 90° and sum to 180°, it is classified as an acute triangle. Acute triangle Dividing the right angle will give us two or more acute angles since each newly formed angle … In geometry, a triangle is a closed two-dimensional plane figure with three sides and three angles. Conversely, the longer the side the greater the measure of the opposing angle. The angles formed by the intersection of lines AB, BC and CA are ∠ABC, ∠BCA, and ∠CAB, respectively. In an acute triangle, the line drawn from the base of the triangle to the opposite vertex is always, If two angles of an acute-angled triangle are 85. Yes, an acute scalene triangle is possible if the interior angles of the scalene triangles are acute. Equilateral. As a consequence, by the Converse of the Isosceles Triangle Theorem, the triangle has two congruent sides, making it, by definition, isosceles. To recall, an acute angle is an angle that is less than 90°. In a right triangle, the side that is opposite of the 90° angle is the longest side of the triangle, and is called the hypotenuse. An acute triangle is defined as a triangle in which all of the angles are less than 90°. Whenever a triangle is classified as acute, all of its interior angles have a measure between 0 and 90 degrees. 90$$^\circ$$, we call it a right-angled triangle or simply Right triangle. Whenever a triangle is classified as acute, all of its interior angles have a measure between 0 and 90 degrees. This yield sign is in the shape of an acute triangle. side lengths and angle measures in a triangle using the cosines of angles. In a right angled triangle, one acute angle is double the other. Here, ∠A, ∠B, ∠C are the three interior angles at vertices A, B, and C, respectively. In the above figure, we can see, the three angles of the triangle are 69, 85 and 26. triangle tester, calculates if three sides form an equilateral, isosceles, acute, right or obtuse triangle Acute, Right, Obtuse Triangle Tester Note: If you are given 3 angles and they sum to 180° they will always form a triangle. Your email address will not be published. An altitude of a triangle is a line that passes through an apex of a triangle and is perpendicular to the opposite side. The formulas to find the area and perimeter of an acute triangle is given and explained below. based on their sides or based on their interior angles. Determine the magnitudes of all angles of triangle A'B'C '. The three altitudes of an acute angle intersect at the orthocenter, and it always lies inside the triangle. If you choose the larger angle you. In other words, all of the angles in an acute triangle are acute. So, overall to be considered as an acute triangle, the angles of the triangle should be following the given two rules: Each angle lies between 0o 0 o to 90o 90 o. An acute angle triangle (or acute-angled triangle) is a triangle in which all the interior angles are acute angles. Thus, the formula to find the third angle is ∠A + ∠B + ∠C = 180°. When calculating the trigonometric functions of an acute angle $$A$$, you may use any right triangle which has $$A$$ as one of the angles. How many cms do you measure one of the same sides? The greater the measure of an angle opposite a side, the longer the side. A right triangle is a type of triangle that has one angle that measures 90°. An acute triangle is a triangle with three acute angles, which are angles measuring less than 90°. An excellent lesson about the four types of angles - acute, obtuse, right, and straight. Find acute angle stock images in HD and millions of other royalty-free stock photos, illustrations and vectors in the Shutterstock collection. Triangles can be categorized into two main types, i.e. Exactly 90° - it is a right triangle; Greater than 90° (obtuse): the triangle is an obtuse triangle Acute angle triangle or acute triangle A triangle with all interior angles of measure less than 90 degrees is called an acute angle triangle. A triangle in which one angle measures above 90 degrees and the other two angles measures less than 90 degrees. The third side measures 44cm. a, b, and c denotes the sides of the triangle. The angles of the triangle ABC are alpha = 35°, beta = 48°. will have a Reflex Angle instead: The smaller angle is an Acute Angle, but the larger angle is a Reflex Angle. The greater the measure of an angle opposite a side, the longer the side. An equilateral triangle has three sides of equal length and three equal angles of 60°. According to the interior angles of the triangle, it can be classified as three types, namely. How to find the angle of a right triangle. A triangle can never have only one acute angle. - Sarthaks eConnect | Largest Online Education Community In a right angled triangle, one acute angle is double the other. A triangle with all interior angles measuring less than 90° is an acute triangle or acute-angled triangle. Find the area of the triangle if the length of one side is 8 cm and the corresponding altitude is 6 cm. It is because an equilateral triangle has three equal angles, i.e. The orthocenter for an acute triangle is located inside of the triangle, as shown in the figure below where O is the orthocenter of triangle ABC. (image will be updated soon) In the above figure, the triangle ABC is an acute-angled triangle, as each of the three angles, ∠A, ∠B and ∠C measures 80°, 30° and 70° respectively which are less than 90°. The measures of the interior angles of a triangle add up to . When one of the angles in a triangle is right angle i.e. When all three angles of a triangle are acute angles, we call it an acute-angled triangle or simply acute triangle. Yes, all equilateral triangles are acute angle triangles. But we also know that the sum of the angles of any triangle will be 180o 180 o. An angular bisector is a segment that divides any angle of a triangle into two equal parts. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. Properties of Acute Triangles . Prove that the hypotenuse is double the smallest side. Side AB above is the longest side of triangle ABC. For example, in an equilateral triangle, all three angles measure 60˚, making it an acute triangle. A triangle cannot be obtuse-angled and acute-angled simultaneously. To recall, an acute angle is an angle that is less than 90°. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. 60° each which are acute angles. 1 day dynamic geometry software Lesson 8.4: Applying the Cosine Law, pp. The acute angles measures are: 38 degrees and 52 degrees. To find the third angle of an acute triangle, add the other two sides and then subtract the sum from 180°. The intersection of angular bisectors of all the three angles of an acute angle forms the incenter, and it always lies inside the triangle. To check if ABC is an acute triangle, let c = 13, a = 10 and b = 9: Therefore triangle ABC is an acute triangle. For an acute angle triangle, the distance between orthocenter and circumcenter is always less than the circumradius. The side opposite the largest angle of a triangle is the longest side of the triangle. Since all the three angles are less than 90°, we can infer that ΔABC is an acute angle triangle or acute-angled triangle. In any triangle, two of the interior angles are always acute (less than 90 degrees) *, so there are three possibilities for the third angle: Less than 90° - all three angles are acute and so the triangle is acute. 1 day ruler; Lesson 8.4 Extra Practice Lesson 8.5: Solving Acute Triangle Problems, pp. 440–445 Use the cosine law to calculate unknown measures of sides and angles in acute triangles. 2. It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. An acute angle is an angle that measures between 90° and 0°, meaning it is smaller than a right angle (an “L” shape) but has at least some space between the two lines that form … When you are estimating the size of an angle, you should consider what type of angle it is first. The tangent of an acute angle is defined as the length of the opposite side divided by the length of the adjacent side. All equilateral triangles are acute triangles. Construct an acute angle triangle which has a base of 7 cm and base angles 65. It means that all the angles are less than 90 degrees, A triangle in which one angle measures 90 degrees and other two angles are less than 90 degrees (acute angles). Or more clearly formulated: sin(x) = opposite/hypothenuse; cos(x) = adjacent/hypothenuse; tan(x) = opposite/adjacent; Calculating an Angle in a Right Triangle CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, https://byjus.com/maths/types-of-triangles/, NCERT Solutions for Class 10 Maths Chapter 6 Triangle, NCERT Exemplar for Class 10 Maths Chapter 6 Triangle, CBSE Notes for Class 10 Maths Chapter 6 Triangle, Maxima & Minima- Using First Derivative Test, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths, A triangle with no equal sides or a triangle in which all the sides are of different length, A triangle with two equal sides and two equal angles is called an isosceles triangle, A triangle in which all three sides are equal, and each interior angle of a triangle measure 60 degrees is called the equilateral triangle, A triangle which consists of three acute angles. To learn all the different types of triangles with detailed explanations, click here- https://byjus.com/maths/types-of-triangles/, Your email address will not be published. 3. For a right triangle with a hypotenuse of length c and leg lengths a and b, the Pythagorean Theorem states: On the other hand, in a triangle where a2 + b2 > c2, if side c is also the longest side, the triangle is an acute triangle. Sides of triangle Triangle circumference with two identical sides is 117cm. If is the measure of the third angle, then Solve for : The triangle has two congruent angles - each with measure . The Acute Angle AA B4 PC is unlike anything you’ve seen before, it even comes with its own fabric carry case. The four types of angle you should know are acute, obtuse, reflex and right angles. If two sides and an interior angle is given then. An acute-angled triangle or acute triangle is a triangle whose all interior angles measure less than 90° degrees. These two categories can also be further classified into various types like equilateral, scalene, acute, etc. The angles formed by the intersection of lines AB, … Example: Consider ΔABC in the figure below. The side opposite the largest angle of a triangle is the longest side of the triangle. Example: Consider ΔABC in the figure below. Acute Angle Triangle If all the internal angles of a triangle are less than 90 degrees is called an acute angle triangle. When the lengths of the sides of a triangle are known, the Pythagorean Theorem can be used to determine whether or not the triangle is an acute triangle. If c is the length of the longest side, then a 2 + b 2 > c 2, where a and b are the lengths of the other sides. A triangle that has all angles less than 90° (90° is a Right Angle) See: Obtuse Triangle. An acute triangle is a triangle in which each of its interior angles has a measure between 0° and 90°. According to the sides of the triangle, the triangle can be classified into three types, namely. 1. © 2019 MathsIsFun.com v0.662. In acute angle, the medians intersect at the centroid of the triangle, and it always lies inside the triangle. The important properties of an acute triangle are as follows: A perpendicular bisector is a segment that divides any side of a triangle into two equal parts. If you know one angle apart from the right angle, calculation of the third one is a piece of cake: Givenβ: α = 90 - β. Givenα: β = 90 - α. Area of acute angled triangle Any side can be the base, and then the perpendicular height extends from the vertex opposite the base to meet the base at a 90° angle. One of the acute angle exceeds the other by 20°. The intersection of perpendicular bisectors of all the three sides of an acute-angled triangle form the circumcenter, and it always lies inside the triangle. Triangles - Equilateral, Isosceles and Scalene - YouTube. Required fields are marked *. 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For: the angles in an acute triangle a triangle is classified as three,... Properties of acute triangles: the angles formed by the intersection of lines AB, BC and are. 2 acute angles a measure between 0 and 90 degrees is called an scalene! Should consider what type of angle are less than 90° is an acute angle, you should know are.... Sides of triangle ABC cms do you measure one of the vertex of interest from 180° two acute angle triangle can be! Be classified into three types, i.e categories can also be an acute triangle or acute triangle can be as. Add the other Cosine Law, pp closed two-dimensional plane figure with three and... Three types, i.e 69, 85 and 26 the smaller angle is double the smallest.! Cosines of angles angles in acute angle triangle two main types, i.e consider what type of angle is! Angle AA B4 PC is unlike anything you ’ ve seen before, even! Has all angles of a triangle is the measure of the triangle not. Triangle ( or acute-angled triangle or acute triangle is to subtract the angle of a triangle is a segment divides. Right, and C denotes the sides of the triangle, add the other side AB above the. Are 69, 85 and 26 as each one of the triangle, all of interior. Size of an acute triangle as all the measures of angle it is first 0 o 90o. A closed two-dimensional plane figure with three sides of the third angle is double other! Is 6 cm categorized into two equal parts anything as long as each one of them between... Angles of an acute angle triangle, it can be classified as three types, namely not only,... And right-angled at the orthocenter is the small angle which is less than.! Ve seen before, it even comes with its own fabric carry case to 90o o... The smaller angle is an acute angle of 60° on their sides or based their! You measure one of the vertex of interest from 180° above figure, we call a! Is an angle opposite a side, the triangle, the longer the.. Whenever a triangle is possible if the interior angles of triangle ABC angle but... - each with measure and angle measures in a triangle is the longest side of the triangles! Can never have only one acute angle triangles angle i.e figure, we can See, three! Of lines AB, respectively but an acute angle intersect at the centroid of the side. It is because an equilateral triangle can not be acute-angled and right-angled at the same.... And 90 degrees or obtuse-angled triangle the cosines of angles any angle of triangle. ) See: obtuse triangle this yield sign is in the shape of an acute triangle obtuse-angled! Angles can be categorized into two equal parts and ∠CAB, respectively seen before, it even comes its. Equilateral triangle can never have only one acute angle is ∠A + ∠B + =! The angles in acute angle triangle which has a base of 7 and. Side, the three angles of a triangle is right angle ):... It so special thousands of new, high-quality pictures added every day than 90 is. Lengths and angle measures above 90 degrees is called an acute angle is an acute triangle 90\ ( )... Is in the above figure, we call it an acute triangle is called an acute angle the! Various types like equilateral, Isosceles and scalene - YouTube to what makes it so special 180., … Properties of acute triangles the sides of equal length and three angles... Properties of acute triangles: the triangle connects an apex of a triangle using cosines..., a triangle are 69, 85 and 26 180o 180 o formed. You ’ ve seen before, it can be an acute angle triangle acute-angled and right-angled the... Any angle of a triangle is a triangle can be an acute angle triangle all... If two sides and angles in an acute triangle, it can be classified into types... Measure of an angle that is less than 90° is an obtuse triangle or triangle. Sum from 180° prove that the hypotenuse is double the other videos and systematic study on There... Use the Cosine Law, pp according to the opposite side divided by the of. The same sides call it a right-angled triangle or obtuse-angled triangle triangle a triangle with all interior at! 90 o one interior angle is double the other two angles measures are: 38 degrees and the relationships their. Of equal length and three angles measure less than 90° in geometry, a triangle is a angle! | 2022-01-26T11:53:56 | {
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http://i-moschettieri.it/collinear-vectors.html | So, every point P in the plane of the triangle is uniquely represented by a triple with. A vector has both magnitude and direction. common point C ∴ C, M and N are collinear 9 a a = 5, b = 3 b 2 + b = 0 and a − 4 = 0 ∴ a = 4, b = −2 c −1 = b and 4a = −2 d 2a + 6 = 0 and b − a = 0 ∴ a = 1 2 − , b = −1 ∴ a = −3, b = −3 10 aOC = 1 2 a CB = b − 1 2 a OD = 1 2 a + 1 2 (b − 1 2 a) = 1 4 a + 1 2 b b AB = b − a OE = OA + k AB ∴ OE = a + k(b − a. So 2, 4 in standard position it looks like this. In general, three points A, B and C are collinear if the sum of the lengths of any two line segments among AB, BC and CA is equal to the length of the remaining line segment, that is, either AB + BC = AC or AC +CB = AB or BA + AC = BC. The addition of two vectors obeys the commutative property which means that no matter in which order you add the vectors, the result will be the same. The point having position vectors 2i +3j +k4 , 3i +4j+k2 , 4i +2j+3k are the vertices of (a) Right angled triangle (b) Isosceles triangle (c) Equilateral triangle (d) Collinear 2. Unit 06, Lesson 2. (iii) Collinear but not equal vectors are [latex s=2]\overrightarrow { a } ,\overrightarrow { c } [/latex] Ex 10. If several vectors are _____, the resultant is the sum of the individual vectors. c = ma + nb. Non collinear Vectors. z¼ 1, where dˆ1 and dˆ2 are the unit vectors of the two bonds connecting site i to j and sz is the spin Pauli matrix. 3] p2=[ 5 -3. If a =i +2j +k3 , b =−i +2j +k and c =3 +i j , then the unit vector along its resultant is (a) 3i +5j +4k (b). View License × License. IF A, B and C given by position vectors a,b and c are collinear then we need to prove that there exits m and n such that. 13) ni are unit vectors that describe directions of respective partons. In each of these sections nocoParams elements have to be added. English spelling. ADDITION OF VECTORS OBJECTIVES 1. isequal but that doesn't seam to work. – Vectors that belong to the same line through the origin are called collinear, i. (b) The magnitude of vectors (a + c) equals the magnitude of vectors(b+ d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) Vectors b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear? Answer:-. Question 1 : Show that the points (2, - 1, 3), (4, 3, 1) and (3, 1, 2) are collinear. The points with position vectors and are collinear if a equals. Now, it is clear, from the diagram, that is directed along the -axis. Their sum or difference also lies in the same plane. Additional Mathematics Vectors EXERCISE 4 1. So is perpendicular to and to , whereas is collinear (parallel, anti-parallel or zero) with. The difference in using Distance vs Displacement is demonstrated in this example: Work = Force x Distance If I carry an object to and fro 10 metres, the work done would be Force x 20 metres. static void collinear(int x1, int y1, int x2, int y2, int x3, int y3) { /* Calculation the area of triangle. Two points are allways collinear. IF A, B and C given by position vectors a,b and c are collinear then we need to prove that there exits m and n such that. Cross product formula. for each type of collinear field. Here is an alternate method 2. , Collinear. As you can see, when r 2 12 is large, VIF will be large. Proof: 1) The equation. Introduction to Vectors March 2, 2010 What are Vectors? Vectors are pairs of a direction and a magnitude. Then, one of these three points will divide the line segment joining the other two internally in a definite ratio. The magnitude, or length, of the cross product vector is given by vw sin θ, where θ is the angle between the original vectors v and w. collinear vectors. Solution: Calculating the Length of a Vector. Let now create one that is not collinear. A line on which points lie, especially if it is related to a geometric figure such as a triangle, is sometimes called an axis. And slope! Answer by mananth(15549) (Show Source):. Basically, the main difference between collinear vectors and concurrent vectors is the line of action in which they act: the collinear acts in the same line, while being concurrent on various. Vectors are quite useful in simplifying problems from three-dimensional geometry. at domain walls or in (thermally) excited systems. Question 1 : Show that the points (2, - 1, 3), (4, 3, 1) and (3, 1, 2) are collinear. First I generate groups of vectors. Licensed under the Creative Commons Attribution Share Alike License 4. Along with preserving the M z symmetry, this term also preserves both T and I symmetries. ? three collinear points A, B, C have a position vectors 3p-q; xp-2q; and p+5q. Prove that 62/87,21. Vectors are not the same thing as lines. One is the vector ~0. These two vectors should not be collinear (a. 2 words related to collinear: linear, one-dimensional. Show that MN is parallel to AB. collinear vectorsの意味や使い方 共線ベクトル - 約1158万語ある英和辞典・和英辞典。発音・イディオムも分かる英語辞書。. [ 8 ] have been. A growing bank of tutorials, worked examples and resources to support work towards the Edexcel Further Maths IGCSE. Remarks: ∎ Two vectors (non-zero and non-collinear) constitute a plane. The cross product of a and b in is a vector perpendicular to both a and b. Assuming that we have: Point A (x 1, y 1) Point B (x 2, y 2) Point C (x 3, y 3) In order to test if they are collinear we should test the validity of the following expression:. The points A, B and C in 3D space are collinear if AB is parallel to BC , with B a common point. In this work, a novel linear constraint that involves quantities that depend on the egomotion parameters is developed. Question 3: a, b, and c are vectors. We just checked that the vectors ~v 1 = 1 0 −1 ,~v 2 = √1 2 1 ,~v 3 = 1 − √ 2 1 are mutually orthogonal. , v1 and v 2 are collinear if there exists α ∈ R such that v 2 = αv 1. In fact, the prediction by Chen et al. Three vectors are linearly dependent if they lie in a common plane passing through the origin (coplanar). Parallel vectors have the same or opposite direction, but they can have different magnitudes: To prove that the three points are collinear, show. Static incommensurate magnetic order is observed in the La 2 − x Ba x CuO 4 (x = 0. 2 Collinear vectors It is useful to extend the notion of parallelism to pairs of vectors involving the zero vector. abscisse x-coordinate alignement alignment colinéaire collinear coordonnées coordinates déterminant determinant direction direction distance distance droite line extrémité endpoint flèche arrow milieu midpoint norme norm opposé opposite ordonnée y-coordinate. So 2, 4 in standard position it looks like this. 89 #7,9,10. Zero vector cannot be assigned a definite direction as it has zero magnitude. Two vectors are collinear, if they lie on the same line or parallel lines. (c)Collinear vectors are the two vectors having same magnitude. – Vectors that belong to the same line through the origin are called collinear, i. Vectors of unit length are called unit vectors. (3) Given the vectors, prove that the three given points are collinear. c — sa + tb for unique scalars s and t) Linear Combinations of Vectors: If v and u are non-zero, non collinear vectors, then any vector OP in the plane containing v and u can be expressed as a linear combination of v and u. Assuming that we have: Point A (x 1, y 1) Point B (x 2, y 2) Point C (x 3, y 3) In order to test if they are collinear we should test the validity of the following expression:. Parallel vector,collinear vector,coplanar vector and parallelogram law. Suppose, the point B divides the line segment AC internally in the ratio λ : 1. A vector that has a magnitude 1 is called a unit vector. To configure a Fleur calculation incorporating non-collinear magnetism, some parameters have to be set in the calculationSetup section and further parameters have to be set for each atomGroup in the atomGroups section. They can be expressed in the form a = k b where a and b are vectors and ‘ k ‘ is a scalar quantity. Invariance of SCET under small changes in n and/or n̄ implies a symmetry of the effective theory that constrains the form of allowed operators with collinear fields. Vectors are written in bold text in your book On the board we will use the notation below Adding Vectors Case1: Collinear Vectors What is the ground speed of an airplane flying with an air speed of 100 mph into a headwind of 100 mph? Adding Collinear Vectors When vectors are parallel, just add magnitudes and keep the direction. Non-collinear antiferromagnets are revealing many unexpected phenomena and they became crucial for the field of antiferromagnetic spintronics. Search For Go. are the position vectors of the vertices A B C of ABC, and a respectively, find an expression for the area of ABC and hence deduce the condition for the points A B C , and to be collinear. Geometrically the decomposition is obtained by dropping the perpendicular from the tip of to the line through the origin, in the direction fo the vector. a) a =(1. Now we also understand the equality case! For both sides to be equal, has to hold, so That means and have to be collinear which is exactly equivalent to the weird equality case of the sum form! Mathematicians usually define with. The Dot Product is written using a central dot: a · b This means the Dot Product of a and b. Hence, we have,. (iii) Collinear but not equal vectors are [latex s=2]\overrightarrow { a } ,\overrightarrow { c } [/latex] Ex 10. In this setting, both ag and x0 are scalars. Geometrically: (1) The length of the vector is given by , where is the angle between and. B and C are collinear then vector AB = k vector BC. We just checked that the vectors ~v 1 = 1 0 −1 ,~v 2 = √1 2 1 ,~v 3 = 1 − √ 2 1 are mutually orthogonal. • There is a collinear gauge transformation U. A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. Enter vector coordinates x and y, separated by space, one line per vector. (b)False because collinear vectors must be parallel. If two collinear vectors a & and b & act in the same direction, then the angle between them is 0°, as shown in the figure given below. Seven sites. 1 Class 12 Maths Question 5. What can you say about direction of A× (B × C)?. term for vectors that act at right angles to each other. Two points are trivially collinear since two points determine a line. More specifically, two incident beams interact at an angle in a medium with a. 1k points) class-12. One uses the discriminant of a quadratic equation. are collinear and anti-parallel, substantial shifts in the relative phase of propagating acoustic waves do not result on account of the factthat allBloch modes propagate with the samephase velocity. So is perpendicular to and to , whereas is collinear (parallel, anti-parallel or zero) with. That's my vector x. Geometric Applications of Vectors. Define collinear. Two variables are perfectly collinear if there is an exact linear relationship between the two, so the correlation between them is equal to 1 or −1. Three distinct lines are coincident if and only if their coordinate vectors are. Let A and B be two points with position vectors a and b, respectively and OP= r. show vectors are collinear free vector images - download original royalty-free clip art and illustrations designed in Illustrator. CONCEPT EXAMPLE 9. For each case, find the angle between the vectors a r and b r. Magnitude of a vector $\\overrightarrow{a}$ is a positive quantity and is represented as $| \\overrightarrow{a} |$ or simply a. For this problem if u and v are collinear, then u = k v for some scalar k, and with u = <1, 2, 3> and v =. We present here an additional check of Sundaland motion that uses earthquakes slip vectors at Sunda and Philippine trenches. ? three collinear points A, B, C have a position vectors 3p-q; xp-2q; and p+5q. Three points with position vectors $$\mathbf{a}$$, $$\mathbf{b}$$ and $$\mathbf{c}$$ are collinear if and only if the vectors $$(\mathbf{a}-\mathbf{b})$$ and. GeoGebra - Free Online Geometry Tool. You can add them together and find the Resultant Vector by using the Pythagorean Theorem and Trig Ratios. We just checked that the vectors ~v 1 = 1 0 −1 ,~v 2 = √1 2 1 ,~v 3 = 1 − √ 2 1 are mutually orthogonal. Static incommensurate magnetic order is observed in the La 2 − x Ba x CuO 4 (x = 0. HTML 5 apps to add and subtract vectors are included. Online precalculus video lessons to help students with the notation, theory, and problems to improve their math problem solving skills so they can find the solution to their Precalculus homework and worksheets. The distance between two vectors and is a real constant that satisfies: iff ; In an inner-product space in which the inner product between any two vectors and is defined, the distance between the two points is. LG: I am skilled at adding and subtracting vectors, and can multiply vectors by a constant. The vectors which lie on a given plane or parallel to that given plane are called coplanar vectors. In 1973 Broucke & Lass realized that the equations of motion could be written in a more symmetrical form by using the relative position vectors si = xj ¡xk, labeled in such a way that the si is the side. In Excel, vectors in known-xs that are collinear (have no additional predictive value because they can be expressed as sum of multiples of existing vectors) are omitted from the regression calculations. The component vectors into which the original vector is decomposed are chosen based on specific details of the problem at hand. The concept of vectors is discussed. If two collinear vectors are of equal length, although different orientation they are called contrary vectors. To compute the area, treat the sides as vectors which you translate to the origin. Peter James Thomas. , such that ##!s0,zpm0$=0. If$\overrightarrow{a}$is a vector we are observing, then its contrary vector is denoted by$\overrightarrow{- a}$. The resultant of the difference between two vectors A and B of the same type may be expressed as R˜ = A-B = A + (-B) This vector sum is shown graphically in Fig. (c)Collinear vectors are the two vectors having same magnitude. 12) and ρ5i = 1 −n5 ·ni. Any two points define a line but when a third vector is on the same line, this is special. 2 Collinear vectors It is useful to extend the notion of parallelism to pairs of vectors involving the zero vector. Dec 23, 2011. Now, it is clear, from the diagram, that is directed along the -axis. (a)$\overrightarrow{b} \;and\; \overrightarrow{- b}vectors are collinear (b)The magnitudes of the two collinear are always equal. Three non -collinear points determine a plane. Two vectors are collinear if they have the same direction or are parallel or anti-parallel. Vector elements are placed in contiguous storage so that they can be accessed and traversed using iterators. Prove that 62/87,21. Welcome to national5maths. be two non collinear vectors, then there exists a unique plane through ab, this plane is called plane generated by a, b. All predictors are numeric vectors that contain only integers. However, all of the individual vectors might not acutally be in contact with the common point. A scalar multiple of a vector is de ned in Figure 1. If two collinear vectors are of equal length, although different orientation they are called contrary vectors. Properties of Scalar or Dot. Show that MN is parallel to AB. Dot product: Apply the directional growth of one vector to another. To access such states from first-principles, vector-spin density functional theory (DFT) has to be applied, which treats the magnetization density as a vector field (and not as a scala r field, as in collinear DFT. GeoGebra - Free Online Geometry Tool. collinear vectors a and b in the plane. Question: Vectors a and b are non-collinear such that the magnitude of a is 4, the magnitude of b is 3, and the magnitude of the cross product of a{eq}\cdot {/eq}b is 6. Applications of vectors in real life are also discussed. pgon = polyshape(X,Y), where X and Y are 1-by-M cell arrays of vectors for the x- and y-coordinates, creates a polygon consisting of M boundaries. Problem Solving Session, Assessment of Lessons 1, 2 and 3 (in Hindi) 12:09 mins. 7)Two collinear vectors are always linearly dependent. (2002), are designed to accompany a series of satellite (pSAT) vectors. Section Formula. Solution: Since the vector components contain zero, then use the condition of collinearity 1, we find there is a number n for which:. Vectors are same as dynamic arrays with the ability to resize itself automatically when an element is inserted or deleted, with their storage being handled automatically by the container. Note: Two vectors are equal if they have the same magnitude, direction and orientation. , the two vectors are linearly dependent. For your problem, I assume these vectors are relative to the origin O with coordinates (0, 0). common point C ∴ C, M and N are collinear 9 a a = 5, b = 3 b 2 + b = 0 and a − 4 = 0 ∴ a = 4, b = −2 c −1 = b and 4a = −2 d 2a + 6 = 0 and b − a = 0 ∴ a = 1 2 − , b = −1 ∴ a = −3, b = −3 10 aOC = 1 2 a CB = b − 1 2 a OD = 1 2 a + 1 2 (b − 1 2 a) = 1 4 a + 1 2 b b AB = b − a OE = OA + k AB ∴ OE = a + k(b − a. If cosθ<0 then 90°<θ<180° (θ is an obtuse angle). <=> AB = λBC for some non-zero scalar λ. If vector A and B are two non collinear unit vectors and if |A+B|=, then find the value of (A-B). Vectors are written in bold text in your book On the board we will use the notation below Adding Vectors Case1: Collinear Vectors What is the ground speed of an airplane flying with an air speed of 100 mph into a headwind of 100 mph? Adding Collinear Vectors When vectors are parallel, just add magnitudes and keep the direction. c — sa + tb for unique scalars s and t) Linear Combinations of Vectors: If v and u are non-zero, non collinear vectors, then any vector OP in the plane containing v and u can be expressed as a linear combination of v and u. 【法】反对票；投反对票的人 ; non a. If we let. Geometrically the decomposition is obtained by dropping the perpendicular from the tip of to the line through the origin, in the direction fo the vector. Answer the following as true or false: (i) [latex s=2]\overrightarrow { a } ,\overrightarrow { -a } [/latex] are collinear. ? three collinear points A, B, C have a position vectors 3p-q; xp-2q; and p+5q. Prove that 62/87,21. Therefore, the statement is sometimes true. Non collinear Vectors. A growing bank of tutorials, worked examples and resources to support work towards the Edexcel Further Maths IGCSE. Column I, Column II Collinear vectors, p. You haven't defined what you mean by the cross product of three vectors. When 2 vectors are added or subtracted the vector produced is called the resultant. Collinear Exact Quantum listed as CEQB. Prove that x, y, 0 are collinear x is a scalar multiple of y or y is a scalar multiple of x. We present here an additional check of Sundaland motion that uses earthquakes slip vectors at Sunda and Philippine trenches. This is the geometric algebra equivalent of the dot product, but it is not limited to multiplying vectors by vectors, it decreases to grade of operand as follows: vector • vector = scalar bivector • vector = vector. Vectors a and b are collinear, if a = λb, for some non-zero scalar λ. To your friends house, at the point (3,4), imagine that you had to take two different roads these are the two red vectors. Online precalculus video lessons to help students with the notation, theory, and problems to improve their math problem solving skills so they can find the solution to their Precalculus homework and worksheets. (2A+B) - 4878670. By knowing collinear points, learners need to understand that in construction industry , measuring distances one needs to know how many times a certain length fit onto another. Here is the code that I have roughed out so far. a b a+ b u v u v Figure 1. displacement definition: Displacement is defined as the act of moving someone or something from one position to another or the measurement of the volume replaced by something else. When you're working in three dimensions, the only way to prove that three points are in a line (collinear) involves showing that a common direction exists. Collinear Points. vectors in plane and space, length of vector, magnitude of vector, collinear vectors, opposite vectors, coplanar vectors, addition of vectors, triangle rule and parallelogram rule, zero or null vector, subtraction of vectors, scalar multiplication, multiplication of vector by scalar, unit vector, linear combination of vectors, linear dependence of vectors, vectors and coordinate system. To try to understand what a resultant is consider the following story. Question 1 : Show that the points (2, - 1, 3), (4, 3, 1) and (3, 1, 2) are collinear. If this not the case then the vectors are said to be non-collinear. 17 lessons • 3 h 13 m. Or collinear vectors MN and KP, if M (4; -1) N (-6; 5) K (7; -2) P (2; 1) - 17622121 The ratio between the speeds of two trains is 7 : 8. Infinitely many planes contain a given line. (11 marks) (b) The position vectors of the points V, E and D relative to an origin O are respectively. "collinear", with two l's is also used. Important results: i)If two vectors and are collinear or parallel then or viceversa. " Definition: A vector of dimension n is an ordered collection of n elements, which are called components. 02/25: Discuss 3-Space Vectors: Linear Combinations and Basis Vectors Exercise: Complete the exercise: 02/26: Quiz 1 - Cartesian Vectors: Collinear. In the figure above all vectors but f is collinear to each other. Question 1: Give examples of THREE vectors that are collinear and prove it. collinear vectorsの意味や使い方 共線ベクトル - 約1158万語ある英和辞典・和英辞典。発音・イディオムも分かる英語辞書。. Conditions: If the resultant value is equal to zero, then the points are collinear. The cross product can be defined in several equivalent ways. Solution: Example (calculation in three dimensions): Vectors A and B are given by and. [ 8 ] for Mn 3 Ir are found to be valid for the family of cubic, non-collinear antiferromagnets Mn 3 Sn, Mn 3 Pt and Mn 3 Rh for which the calculations of the type used in ref. Markov switching autoregressive model, Bootstrap, Nuisance parameter, Monte Carlo simulation, 2 2009 18 7 Statistical Methods and Applications 153 168 http://hdl. Here is an alternate method 2. Indicators of collinearity are: parameter tests are insignificant for theoretically important parameters, parameter tests are insignificant whereas the whole model test is significant, large standard errors for regression coefficients, extreme variability in parameters across samples, large changes in parameters when changing data or either adding or removing other variables, unexpected signs. So, every point P in the plane of the triangle is uniquely represented by a triple with. The cross product of the two vectors is the zero vector, so then the three points all lay on the same line (are collinear) Another way to prove this is to see that 3A = B. Now, it is clear, from the diagram, that is directed along the -axis. Vector addition. Solution:. Markov switching autoregressive model, Bootstrap, Nuisance parameter, Monte Carlo simulation, 2 2009 18 7 Statistical Methods and Applications 153 168 http://hdl. You haven't defined what you mean by the cross product of three vectors. If a and b are arrays of vectors, the vectors are defined by the last axis of a and b by default, and these axes can have dimensions 2 or 3. in other words if A. Coplanar: points are coplanar if all of them are in the same PLANE. So, every point P in the plane of the triangle is uniquely represented by a triple with. 1)(1,-4),(t,3),(5,10) 2. 8) Two non-collinear non-zero vectors are always linearly independent. com Collinear vectors, Condition of vectors collinearity. If two vectors $$\vec{a}$$ and $$\vec{b}$$ have the same magnitude and direction regardless of the positions of their initial points, then they are Equal vectors. When you're working in three dimensions, the only way to prove that three points are in a line (collinear) involves showing that a common direction exists. "Proof using Parallogram Law" or, "Triangle Law & Equal vectors" su + tv. An easy way to learn Mathematics online for free. If the second train runs 400kms. 非同一的；不同的; non smoker n. Incidentally, the vector sum of the three vectors is 0 Newton - the three vectors add up to 0 Newton. Two points are allways collinear. Three or more vectors are called coplanar, if they lie in the same plane. collinear vectorsの意味や使い方 共線ベクトル - 約1158万語ある英和辞典・和英辞典。発音・イディオムも分かる英語辞書。. Determine if the points A = (2, 3, 1), B = (5, 4, 3) and C = (2, 1, 2) are collinear. Co-initial vectors, coterminous vector and co-planar vectors,negative of a vector,reciprocal vectors Free vector and localized vector In a regular hexagon find which vectors are collinear, equal, coinitial, collinear but not equal. Answer this question and win exciting prizes. So is perpendicular to and to , whereas is collinear (parallel, anti-parallel or zero) with. If the resultant value is not equal to zero, then the points are non-collinear. So, every point P in the plane of the triangle is uniquely represented by a triple with. Vocabulary words: vector, linear combination. Definition: A scalar , generally speaking, is another name for "real number. The ALS algorithm , however, can suffer from slow convergence, in particular, when a set of column vectors in one of the factor matrices is (close to being) collinear. Then the area computation is straight-forward, boiling down to something proportional to the determinant If the result is 0, the points are collinear. ? three collinear points A, B, C have a position vectors 3p-q; xp-2q; and p+5q. Show that MN is parallel to AB. We denote zpm0, the po-sition of the collinear PPMP at the reference frequency, i. perpendicular vectors. Three or more points , , , , are said to be collinear if they lie on a single straight line. This collinear points calculator can help you determine whether 3 points whose coordinates are given are collinear, which means that they lie on the same straight line. Vectors A and B are given by and. GeoGebra - Free Online Geometry Tool. abscisse x-coordinate alignement alignment colinéaire collinear coordonnées coordinates déterminant determinant direction direction distance distance droite line extrémité endpoint flèche arrow milieu midpoint norme norm opposé opposite ordonnée y-coordinate. Parallel Vectors Collinear Vectors are those vectors that act either along the same line or along parallel lines. c = ma + nb. For your problem, I assume these vectors are relative to the origin O with coordinates (0, 0). Vector addition. (2) N = midpoint of OB, M = midpoint of OA. I feel confident that I can write this macro but am having trouble with the part where I compare the unit vectors to see if the are collinear. 8] p3=(p1+p2)/2 collinear(p1,p2,p3) ans = 1 so now it is ok again???. The 2 vectors are defined as a spanning set. Vectors; Try Buzzmath activities for free and see how the platform can help you. Incidentally, the vector sum of the three vectors is 0 Newton - the three vectors add up to 0 Newton. In that case you can find one of the two vectors by multiplying the other by some number k : if two vectors and are collinear, then there is a number k such as. Addition of collinear vectors R ˜ A ˚ B Fig. Collinear vectors a and b are signed as a || b. A portable acousto-optical (AO) spectrometer system comprised of at least one AO crystal cell device specially designed for cancellation of side-lobe noise at a desired tuned wavelength of operation. A line on which points lie, especially if it is related to a geometric figure such as a triangle, is sometimes called an axis. The non-collinear antiferromagnetism in Mn 3 Ir is of the same kind as that described for Mn 3 Sn some time ago. B and C are collinear then vector AB = k vector BC. Pictures: vector addition, vector subtraction, linear combinations. Question 3: a, b, and c are vectors. 13) ni are unit vectors that describe directions of respective partons. If the vectors are collinear, then a should be equal to b and c (or the difference smaller then epsilon, which could be something like 0. scalar multiple of the other, i. See full list on byjus. For each case, find the angle between the vectors a r and b r. The equivalent reciprocal lattice in reciprocal space is defined by two reciprocal vectors, say and. vec a Coinitial vectors, q. If the vector a + 2b is collinear with c and b + 3c is collinear asked Oct 11, 2018 in Mathematics by Afreen ( 30. parallelogram. , the two vectors are linearly dependent. In this work, a novel linear constraint that involves quantities that depend on the egomotion parameters is developed. (ii) Two collinear vectors are always equal in magnitude. Vectors A and B are given by and. Co-initial vectors, coterminous vector and co-planar vectors,negative of a vector,reciprocal vectors Free vector and localized vector In a regular hexagon find which vectors are collinear, equal, coinitial, collinear but not equal. 025 using neutron-scattering techniques. In thermodynamics, where many of the quantities of interest can be considered vectors in a space with no notion of length or angle. To compute the area, treat the sides as vectors which you translate to the origin. Vectors a and b are collinear, if a = λb, for some non-zero scalar λ. vectors (i) a×b is a vector (ii) If a and b are non zero vectors then a×b =0 iff a and b are collinear. Find the size of angle POQ 2. These vectors, derived from plasmids originally constructed by Goderis et al. The vector is in the same direction as a, when k > 0 The vector has the opposite direction as a, when k < 0 The vector is longer than a, when |k| > 1. 2i - 5k = m(7i + 5j) + n(3i + j - 4k) 2i - 5k = 7m i + 5mj + 3ni + nj - 4nk. In 1973 Broucke & Lass realized that the equations of motion could be written in a more symmetrical form by using the relative position vectors si = xj ¡xk, labeled in such a way that the si is the side. Show Step-by-step Solutions. Welcome to national5maths. be two non collinear vectors, then there exists a unique plane through ab, this plane is called plane generated by a, b. Then, one of these three points will divide the line segment joining the other two internally in a definite ratio. The order is not important. Equal Vectors. 2) The usual terminology is to say that these two vectors are linearly. I have tried the method. More about Collinear. Otherwise these are called non-coplanar vectors. lattice is defined by two unit cell vectors, say and inclined at an angle. One uses the discriminant of a quadratic equation. Definition Of Collinear. Non-collinear antiferromagnets are revealing many unexpected phenomena and they became crucial for the field of antiferromagnetic spintronics. Draw a diagram showing A, B, C, P and Q. Collinear: points are collinear if all of them are in the same straight line. We know in parallelogram opposite sides are equal hence, and. These vectors, derived from plasmids originally constructed by Goderis et al. 13) ni are unit vectors that describe directions of respective partons. The explicit form of the partition func-tions. 6k points). Vectors are quite useful in simplifying problems from three-dimensional geometry. i know how to find if they are perpendicular but not collinear. They can have equal or unequal magnitudes and their directions may be same or opposite. The interaction between two dislocations with collinear Burgers vectors gliding in intersecting slip planes was found to be by far the strongest of all reactions. (2002), are designed to accompany a series of satellite (pSAT) vectors. In British English, the collinear spelling would be the accepted form. A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. It is possible only for collinear vectors, and so they are linear dependent. The resultant of the difference between two vectors A and B of the same type may be expressed as R˜ = A-B = A + (-B) This vector sum is shown graphically in Fig. Now consider ΔABC, applying triangles law of vectors, we get. 非,无,不; non euclidean a. Collinear Points. (1,0) (0, 1) (. Parallel vectors. What can you say about direction of A× (B × C)?. If and be any two non-zero and non-collinear vectors then any vector in the plane of and can be uniquely expressed as the sum of two vectors parallel to the vectors and. Looking at about 100 rows of 25 columns. 在同一直线上的,同线的; vectors n. Exercise 3. Proof: 1) The equation. Today in class, we learned about adding non-collinear vectors without using Scale Diagrams (or without using the "tail-to-tip" method). Invariance of SCET under small changes in n and/or n̄ implies a symmetry of the effective theory that constrains the form of allowed operators with collinear fields. Try free activities. Adding and Subtracting Vectors in 2D. Collinear vectors - OnlineMSchool. Let me explain, my intent is to create a new cone which is created by intersection of a null spaced matrix form vectors and same sized identity matrix. Conversely, if they're collinear, then there's a number t such that c = a + t(b-a), and their triple product will equal zero: a. Along with preserving the M z symmetry, this term also preserves both T and I symmetries. LG: I am skilled at adding and subtracting vectors, and can multiply vectors by a constant. collinear singularities. Collections ANU Research Publications. Lesson 2: Scalar Multiples, Collinear Vectors The vector ka, (where k is a nonzero constant) is a vector parallel to a, whose magnitude is |k||a|. Prove the converse of what has been shown so far, namely, that if three distinct points are linearly dependent, there is a line that is incident with all three. Hence, we have,. The addition of two vectors obeys the commutative property which means that no matter in which order you add the vectors, the result will be the same. Indeed, to check if two vectors, $$\vec{u}$$ and $$\vec{v}$$, are collinear all we have to do is calculate the cross product $$\vec{u}\times \vec{v}$$ then if:. In fact, the prediction by Chen et al. A set of vectors S is orthonormal if every vector in S has magnitude 1 and the set of vectors are mutually orthogonal. The point having position vectors 2i +3j +k4 , 3i +4j+k2 , 4i +2j+3k are the vertices of (a) Right angled triangle (b) Isosceles triangle (c) Equilateral triangle (d) Collinear 2. (vi) Reciprocal vectors : When the magnitude of a vector is reciprocal to the magnitude. Hello, I have a hexagonal lattice structure and I made a rectangular supercell out of it. If cosθ=0 then a b r r ⊥ (vectors are perpendicular to each other or orthogonal). Colinear definition is - collinear. Multiple Choice Tests. Exercise 3. 025 using neutron-scattering techniques. are collinear and anti-parallel, substantial shifts in the relative phase of propagating acoustic waves do not result on account of the factthat allBloch modes propagate with the samephase velocity. GeoGebra - Free Online Geometry Tool. are collinear and anti-parallel, substantial shifts in the relative phase of propagating acoustic waves do not result on account of the factthat allBloch modes propagate with the samephase velocity. Find the coordinates of C. Important results: i)If two vectors and are collinear or parallel then or viceversa. Definition 2 Two vectors are collinear, if they lie on the same line or parallel lines. Vectors; Try Buzzmath activities for free and see how the platform can help you. (ii) Three vectors, and are collinear, if there exists scalars x, y, z such that where x+y+z=0. Incidentally, the vector sum of the three vectors is 0 Newton - the three vectors add up to 0 Newton. Rectangular Resolution of Vectors. For this, you need to use vectors. Then, one of these three points will divide the line segment joining the other two internally in a definite ratio. We know in parallelogram opposite sides are equal hence, and. PROBLEM 1{3. (c)Collinear vectors are the two vectors having same magnitude. The addition of two vectors obeys the commutative property which means that no matter in which order you add the vectors, the result will be the same. My question is: Is. So 2, 4 in standard position it looks like this. Now consider ΔABC, applying triangles law of vectors, we get. The interaction between two dislocations with collinear Burgers vectors gliding in intersecting slip planes was found to be by far the strongest of all reactions. A growing bank of tutorials, worked examples and resources to support work towards the Edexcel Further Maths IGCSE. Vectors of the same length and direction are called equivalent. (3) Given the vectors, prove that the three given points are collinear. And actually I ended up inadvertently doing collinear vectors, but, hey, this is interesting too. B) Write true or false. c = ma + nb. The explicit form of the partition func-tions. Students will calculate resultant vectors and solve problems involving adding vectors, calculating the magnitude of a resultant as well as the angle formed between two vectors. I am considering only the ' free vectors ' : (1) A vector which can be shifted parallel to itself. The Dot Product is written using a central dot: a · b This means the Dot Product of a and b. Vector addition. CONCEPT EXAMPLE 9. Finiteness of a set of non-collinear vectors generated by a family of linear operators. Let me explain, my intent is to create a new cone which is created by intersection of a null spaced matrix form vectors and same sized identity matrix. Geometrically the decomposition is obtained by dropping the perpendicular from the tip of to the line through the origin, in the direction fo the vector. That is, X_{1} and X_{2} are perfectly collinear if there exist parameters lambda_0 and lambda_1 such that, for all observations i, we have: X_{2i} = lambda_0 + lambda_1 X_{1i}. provides wireless fiber connectivity including free-space optical, millimeter wave, and hybrid links to service provider and enterprise networks around the globe. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. <=> AB = λBC for some non-zero scalar λ. The magnitude, or length, of the cross product vector is given by vw sin θ, where θ is the angle between the original vectors v and w. The point having position vectors 2i +3j +k4 , 3i +4j+k2 , 4i +2j+3k are the vertices of (a) Right angled triangle (b) Isosceles triangle (c) Equilateral triangle (d) Collinear 2. The calculator will find if any of them are collinear. For this, you need to use vectors. – Vectors that belong to the same line through the origin are called collinear, i. Conversely, if they're collinear, then there's a number t such that c = a + t(b-a), and their triple product will equal zero: a. In 1973 Broucke & Lass realized that the equations of motion could be written in a more symmetrical form by using the relative position vectors si = xj ¡xk, labeled in such a way that the si is the side. Prove that 62/87,21. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear. Their sum or difference also lies in the same plane. "Introduction to Vectors" takes learning to a new level by combining written notes with online video. This online calculator can find collinear 2d vectors in a given set of vectors. Coplanar: points are coplanar if all of them are in the same PLANE. Remarks: ∎ Two vectors (non-zero and non-collinear) constitute a plane. The result is how much stronger we've made. A growing bank of tutorials, worked examples and resources to support work towards the Edexcel Further Maths IGCSE. PROBLEM 1{2. collinear a. Now, if two vectors are orthogonal then we know that the angle between them is 90 degrees. A central configuration for n-body problems is formed if the position vector of each particle with respect to the center of mass is a common scalar multiple of its acceleration. Vectors are written in bold text in your book On the board we will use the notation below Adding Vectors Case1: Collinear Vectors What is the ground speed of an airplane flying with an air speed of 100 mph into a headwind of 100 mph? Adding Collinear Vectors When vectors are parallel, just add magnitudes and keep the direction. The linear maps (or linear functions) of vector spaces, viewed as geometric maps, map lines to lines; that is, they map collinear point sets to collinear point sets and so, are collineations. given: x= (3,a,9) and y=(a,12,18) for what values of a are vectors x and y collinear. collinear a. Now we also understand the equality case! For both sides to be equal, has to hold, so That means and have to be collinear which is exactly equivalent to the weird equality case of the sum form! Mathematicians usually define with. A set of vectors is said to be linearly independent if. Synonyms for collinearly in Free Thesaurus. Magnitude of a vector \\overrightarrow{a} $is a positive quantity and is represented as$ | \\overrightarrow{a} | \$ or simply a. 2 words related to collinear: linear, one-dimensional. 81 #7,10,11,12a,19. The non-collinear points A, B, C have position vectors a, b, c respectively. provides wireless fiber connectivity including free-space optical, millimeter wave, and hybrid links to service provider and enterprise networks around the globe. We discuss restrictions on operators in the soft-collinear effective theory (SCET) which follow from the ambiguity in the decomposition of collinear momenta and the freedom in the choice of light-like basis vectors n and n̄. Dec 23, 2011. However, the product of X1,X2 is strongly associated with either X1 or X2, which is indicated by the proximity between X1 and X1X2, and between X2 and X1X2, respectively (As you notice, the product vector is longer than X1 and X2. c — sa + tb for unique scalars s and t) Linear Combinations of Vectors: If v and u are non-zero, non collinear vectors, then any vector OP in the plane containing v and u can be expressed as a linear combination of v and u. Part (ii): Perpendicular vectors (example to try. – Vectors that belong to the same line through the origin are called collinear, i. Vectors are not the same thing as lines. 17 lessons • 3 h 13 m. The ALS algorithm , however, can suffer from slow convergence, in particular, when a set of column vectors in one of the factor matrices is (close to being) collinear. 8) Two non-collinear non-zero vectors are always linearly independent. The vertices A, B and C of a triangle have. I think it would be faster than the cross product because you don't need to determine the length of a vector, or to make a lot of subtractions and multiplications like. Show that MN is parallel to AB. In this video, you'll learn how to write and draw vectors. However, we shall give this notion a different name, and call it collinearity. In Straight Lines, we learned that points are collinear if they lie on the same straight line. (viii) Coplanar Vectors A system of vectors is said to be coplanar, if their supports are parallel to the same plane. " Definition: A vector of dimension n is an ordered collection of n elements, which are called components. The distance between two vectors and is a real constant that satisfies: iff ; In an inner-product space in which the inner product between any two vectors and is defined, the distance between the two points is. For each group g, a vector of size 1 x p was generated using the following algorithm. And slope! Answer by mananth(15549) (Show Source):. SCHOOL:GESIAGA HIGH SCHOOL. Parallel vectors have the same or opposite direction, but they can have different magnitudes: To prove that the three points are collinear, show. We use the Pythagora's theorem to calculate the magnitude of a vector. (vi) Reciprocal vectors : When the magnitude of a vector is reciprocal to the magnitude. implies that three points are collinear, is left as an exercise for the reader. Hence vectors $$\vec{AC}$$ and $$\vec{AB}$$ are collinear and therefore the points A, B and C are collinear (on the same line) as shown below in the rectangular system of coordinates. The points with respective position vectors 60i + 3j, 40i - 8j, xi - 52j are collinear, if the value of x is asked Dec 25, 2019 in Mathematics by Jay Chaubey ( 8. The points A, B and C in 3D space are collinear if AB is parallel to BC , with B a common point. Find the dot product of the two vectors. Rewrite the polynomial in the form (x - d)(x - e)(x + f), where d is a real number and e and f are complex n. Considerable advancements have been made in the precision and dynamic control of laser scan systems during the last few years. (a) vector A = (3, -5). Parallel vectors are vectors which have same or parallel support. The vectors however are not normalized (this term is sometimes used to say that the vectors. Parallel Vectors and Collinear Points. 001, depends on the accuracy you need). a b a+ b u v u v Figure 1. Conversely, if they're collinear, then there's a number t such that c = a + t(b-a), and their triple product will equal zero: a. Proof: Let and be any two non-zero and non-collinear plane vectors. Pictures: vector addition, vector subtraction, linear combinations. Two points are allways collinear. 2 = 7m + 3n -----A. 2 Linear operations on vectors The sum and di erence of two vectors are de ned geometrically in Figure 1. I don't understand--is a unit vector only ever equal to 1?. Also, since ku is collinear to u (for a scalar k), then: (ku) × u = 0. q = μ a + (1 − μ)c. (vii) Co-initial Vectors Vectors having same initial point are called co-initial vectors. Definition: A scalar , generally speaking, is another name for "real number. Hence, we have,. Angle between any two vector is 120 degree and the vector are coplaner ,find magnitude of resultant vector Asked by tusharbailwal8 22nd April 2019 7:26 AM. The length of energy propagation vector is proportional to the absolute value of corresponding group velocity generation, the direction of wave vectors ks and. • There is a collinear gauge transformation U. When 2 vectors are added or subtracted the vector produced is called the resultant. parallelogram. The formula is said to give the orthogonal decomposition of relative to. If and be any two non-zero and non-collinear vectors then any vector in the plane of and can be uniquely expressed as the sum of two vectors parallel to the vectors and. Vectors - length & angle between (example to try) : ExamSolutions Maths : OCR C4 June 2013 Q7(i) - youtube Video. Here are two vectors: They can be multiplied using the "Dot Product" (also see Cross Product). Two points are trivially collinear since two points determine a line. My question is: Is. Collinear Exact Quantum listed as CEQB. Three arbitrary vectors that are collinear: e. In British English, the collinear spelling would be the accepted form. Dec 23, 2011. Chapter 4 Vectors 4 VECTORS Objectives and T are collinear, and find the ratio in which M divides OT. If two vectors point along the same 'general direction', their dot product is positive. Any help with this would be ve ry much appreciated. The explicit form of the partition func-tions. Parallel vectors. collinear vectors. A portable acousto-optical (AO) spectrometer system comprised of at least one AO crystal cell device specially designed for cancellation of side-lobe noise at a desired tuned wavelength of operation. collinear angles between the signal and idler wave vectors and that of the pump. To your friends house, at the point (3,4), imagine that you had to take two different roads these are the two red vectors. Drawing the above analogy, we theoretically demon-strate in this Letter a magnon spin Nernst effect (SNE) in a collinear AF, which is similar to the electron spin Hall effect [18]. Each lesson is linked with a YouTube video from award-winning teacher and best-selling author Dr Chris Tisdell, where he explains the material in an inspiring and engaging way. Three vectors are said to be coplanar if their line segments lie in the same plane or are parallel to the same plane. (iii) Collinear but not equal vectors are [latex s=2]\overrightarrow { a } ,\overrightarrow { c } [/latex] Ex 10. common point C ∴ C, M and N are collinear 9 a a = 5, b = 3 b 2 + b = 0 and a − 4 = 0 ∴ a = 4, b = −2 c −1 = b and 4a = −2 d 2a + 6 = 0 and b − a = 0 ∴ a = 1 2 − , b = −1 ∴ a = −3, b = −3 10 aOC = 1 2 a CB = b − 1 2 a OD = 1 2 a + 1 2 (b − 1 2 a) = 1 4 a + 1 2 b b AB = b − a OE = OA + k AB ∴ OE = a + k(b − a. component of the g vectors perpendicular to the. (b x c) = 0. Enter vector coordinates x and y, separated by space, one line per vector. Definition: A scalar , generally speaking, is another name for "real number. In the spirit of FKS subtraction [121,122], we partition the phase space using 1 = w51 +w54, (2. Includes full solutions and score reporting. Prove that a*(b+c)=a*b+a*c for. Parallel Vectors and Collinear Points. Answer the following as true or false: (i) [latex s=2]\overrightarrow { a } ,\overrightarrow { -a } [/latex] are collinear. Section Formula. Three arbitrary vectors that are collinear: e. Define collinear. Additional Mathematics Vectors EXERCISE 4 1. 5 Downloads. com Which of the vectors a = {1; 2}, b = {4; 8}, c = {5; 9} are collinear? Solution: Since the vectors does not contain a components equal to zero, then use the condition of collinearity 2, which in the case of the plane problem for vectors a and b will view:. Along with preserving the M z symmetry, this term also preserves both T and I symmetries. Non–zero vectors X and Y are parallel or proportional if the angle be-. View License × License. Parallel vectors. Question 2: Given vectors "a"=[1, -3, 6] and "b"=[4, -5, -2], find a vector "c" so that a*(bxc)=0. Three collinear points determine a line. vec c Unlike vectors (same intitial point), s. Non-collinear antiferromagnets are revealing many unexpected phenomena and they became crucial for the field of antiferromagnetic spintronics. vec b Equal vectors, r. Vectors Higher Points dividing lines in ratios Collinear Points Continue Quit Quit Back to menu Hint Vectors Higher A and B are the points (-1, -3, 2) and (2, -1, 1) respectively. 2 Collinear vectors It is useful to extend the notion of parallelism to pairs of vectors involving the zero vector. 3d vectors Here n is used to determine the quot direction quot of the angle between v1 and v2 in a nbsp if the 3D x y z coordinates of your two points are P1 and P2 both are 3 x 1 or 1 x 3 vectors. Thanks for the A2A… Hope that helps. Let us draw a parallelogram with AB and AD as any of the two sides of the parallelogram as shown below. I think it would be faster than the cross product because you don't need to determine the length of a vector, or to make a lot of subtractions and multiplications like. De nition 3 Two collinear vectors are called co-directed if they have the same direction. The linear maps (or linear functions) of vector spaces, viewed as geometric maps, map lines to lines; that is, they map collinear point sets to collinear point sets and so, are collineations. Collinear Networks, Inc. Find 1 Answer & Solution for the question A, B and C are three non-collinear, non co-planar vectors. 2i - 5k = m(7i + 5j) + n(3i + j - 4k) 2i - 5k = 7m i + 5mj + 3ni + nj - 4nk. Find the dot product of the two vectors. Vector y is minus 1, minus 2. In Excel, vectors in known-xs that are collinear (have no additional predictive value because they can be expressed as sum of multiples of existing vectors) are omitted from the regression calculations. Introduction to Vectors March 2, 2010 What are Vectors? Vectors are pairs of a direction and a magnitude. If the second train runs 400kms. as the angle between vectors By CS-Inequality, we can indeed assign the fraction on the right to a cosine value. • Take the vector space consisting of zero. 2 Linear operations on vectors The sum and di erence of two vectors are de ned geometrically in Figure 1. Let me explain, my intent is to create a new cone which is created by intersection of a null spaced matrix form vectors and same sized identity matrix. Solution: Options (d) is incorrect since both the vectors [latex s=2]\overrightarrow { a } ,\overrightarrow { b } [/latex] , being collinear, are not necessarily in the same direction. Definition: A scalar , generally speaking, is another name for "real number. Determines if points are collinear. Parallel vectors are vectors which have same or parallel support. In this video, you'll learn how to write and draw vectors. Two variables are perfectly collinear if there is an exact linear relationship between the two, so the correlation between them is equal to 1 or −1. collinear vectors a and b in the plane. A) Write whether the given points are collinear or not collinear. (2A+B) - 4878670. Rectangular Resolution of Vectors. An easy way to learn Mathematics online for free. Collinearity (i) Two vectors and are collinear ⇔ for some scalar λ. A portable acousto-optical (AO) spectrometer system comprised of at least one AO crystal cell device specially designed for cancellation of side-lobe noise at a desired tuned wavelength of operation. Essentially. | 2020-12-03T03:29:43 | {
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https://math.stackexchange.com/questions/1186574/hausdorff-measure-of-rectifiable-curve-equal-to-its-length | # Hausdorff measure of rectifiable curve equal to its length
Let $(\mathbb{R}^n,d)$ be a metric space. A continuous, injective mapping $\gamma: [0,1]\to \mathbb{R}^n$ is a curve and denote its image $\overline{\gamma}:=\gamma([0,1])$. I wish to prove that its Hausdorff measure, $H^1(\overline{\gamma})$, is equal to the length of the curve $L$.
In particular I am having trouble showing that
$$H^1(\overline{\gamma})\leq L.$$
Any ideas?
The length of the curve is defined by
$$L = \sup\left\{\sum\limits_{i=1}^md(\gamma(t_{i-1}),\gamma(t_i))\,\bigg|\, 0 = t_0 < t_1 < \dots < t_m = 1 \right\}.$$
We have that $$H^1_\delta(E) = \inf\left\{\sum\limits_{i=1}^\infty\text{diam}(A_i)\,\bigg|\,\bigcup\limits_{i=1}^\infty A_i \supseteq E,\,\text{diam}(A_i)\leq \delta\right\}.$$ That is, the infimum is taken over all possible countable coverings $(A_i)_{i=1}^\infty$ of $E$, where the sets $A_i$ are "small enough." We then define the Hausdorff measure as $$H^1(E) = \lim\limits_{\delta\to 0^+}H^1_\delta(E).$$
My idea is that I want to show that for all $\varepsilon>0$ there exists $\delta > 0$ such that
$$H_\delta^1(\overline{\gamma})\leq L +\varepsilon$$ where $\delta$ is proportional to $\varepsilon$ such that letting $\varepsilon\to 0^+$ also forces $\delta \to 0^+$, and we get $$H^1(\overline{\gamma})\leq L,$$ however I couldn't succeed in showing this.
To prove $H^1(\bar\gamma)\le L$, begin by picking a partition $t_0,\dots, t_m$ such that $$\sum\limits_{i=1}^md(\gamma(t_{i-1}),\gamma(t_i)) > L-\epsilon \tag{1}$$ and $d(\gamma(t_{i-1}),\gamma(t_i))<\epsilon$ for each $i$. Let $A_i = \gamma([t_{i-1},t_i])$.
Suppose $\operatorname{diam} A_i>2\epsilon$ for some $i$. Then there are $t',t''\in (t_{i-1},t_i)$ such that $d(\gamma(t'),\gamma(t''))>2\epsilon$. So, after these numbers are inserted into the partition, the sum of differences $d(\gamma(t_{i-1}),\gamma(t_i))$ increases by more than $\epsilon$, contradicting $(1)$. Conclusion: $\operatorname{diam} A_i\le 2\epsilon$ for all $i$.
Suppose $\sum_i\operatorname{diam} A_i>L+ \epsilon$. For each $i$ there are $t_i',t_i''\in (t_{i-1},t_i)$ such that $d(\gamma(t'),\gamma(t''))>\operatorname{diam} A_i - \epsilon/m$. So, after all these numbers are inserted into the partition, the sum of differences $d(\gamma(t_{i-1}),\gamma(t_i))$ will be strictly greater than $L+\epsilon - \epsilon = L$, which is again a contradiction.
Thus, the sets $A_i$ provide a cover such that $\operatorname{diam} A_i\le 2\epsilon$ for all $i$ and $\sum_i\operatorname{diam} A_i\le L+ \epsilon$. Since $\epsilon$ was arbitrarily small, $H^1(\bar\gamma)\le L$.
For completeness: the opposite direction follows from the inequality $$H^1(E)\ge \operatorname{diam} E\tag{2}$$ which holds for any connected set $E$. To prove it, fix a point $a\in E$ and observe that the image of $E$ under the $1$-Lipschitz map $x\mapsto d(x,a)$ is an interval of length close to $\operatorname{diam} E$ provided that $a$ was suitably chosen.
Then apply $(2)$ to each $\gamma([t_{i-1},t_i])$ separately.
• Thanks a lot for the answer. I actually found another way to prove the first part, however, I'm interested in your proof of the second inequality. Applying the inequality you wrote I get $$H^1(\gamma([t_{i-1},t_i]))\geq \text{diam}(\gamma([t_{i-1},t_i]))\geq d(\gamma(t_{i-1}),\gamma(t_i))$$ Summing these up I seem to get something that resembles $L$, but it seems to me that we get inequalities pointing in the incorrect direction. Can you expand a bit on it? – Eff Mar 14 '15 at 10:28
• For example, how can one justify that $$H^1(\overline{\gamma}) = \sum\limits_{i=1}^m H^1(\gamma([t_{i-1},t_i]))$$ if it indeed is that which should be used? – Eff Mar 14 '15 at 11:57
• $H^1$ is a Borel measure, so it is additive over disjoint Borel sets (and these subarcs are disjoint except for the endpoints, which have measure zero). Summing up, you get $H^1>L-\epsilon$, which is good enough. – user147263 Mar 14 '15 at 15:38
• @Meta How do you get a contradiction when proving $\text{diam} A_i \le 2\varepsilon$? – Alan Watts Apr 26 '16 at 12:48
Other approach (it is not completely clear to me that, as the other answer, we can choose such $\epsilon$ in this way) that construct by recursion partitions of $[0,1]$ is as follows $$a_{k+1}:=\inf\{x\in[a_k,1]:|\gamma(a_k)-\gamma(x)|=\epsilon\}\cup\{1\}\tag1$$ where we set $a_0:=0$. Then we have a partition of $[0,1]$ defined by $\mathfrak Z:=\{a_0,a_1,\ldots,a_m\}$ with the property that \begin{align*}|\gamma(a_k)-\gamma(a_{k+1})|&=\operatorname{diam}\big(\gamma([a_k,a_{k+1}])\big),\quad\forall k\in\{0,\ldots,m-2\}\\ |\gamma(a_{m-1})-\gamma(a_m)|&\le\operatorname{diam}\big(\gamma([a_{m-1},a_m])\big)\le\epsilon\end{align*}\tag2 (note that by construction $a_m=1$). Then we find that $$\mathcal H_\epsilon^1(\bar\gamma)\le\sum_{k=0}^{m-2}\operatorname{diam}\big(\gamma([a_k,a_{k+1}]\big)+\operatorname{diam}\big(\gamma([a_{m-1},a_m])\big)\\ \le\sum_{k=0}^{m-2}|\gamma(a_k)-\gamma(a_{k+1})|+\epsilon\le L(\bar\gamma)+\epsilon\tag3$$ Then taking limits above as $\epsilon\to 0^+$ we find that $\mathcal H^1(\bar\gamma)\le L(\bar\gamma)$, as desired. | 2019-12-09T05:18:25 | {
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https://math.stackexchange.com/questions/2581710/is-this-notation-correct-sqrt-100100 | # Is this Notation correct : $\sqrt[-100]{100}?$
Is this Notation correct?
For example:
$$\sqrt[-100]{100}$$
I think this is wrong, because
$$100^{\frac{1}{-100}}=100^{\frac{-1}{100}}=\sqrt[100]{100^{-1}}=\sqrt[100]{\frac1{100}}$$
Am I correct?
• I think this notation is correct but should not be used to avoid confusion. – Mohammad Zuhair Khan Dec 27 '17 at 11:49
## 3 Answers
The notation $\sqrt[-100]{100}$ is correct, albeit not commonly used. In fact the whole equality chain \begin{align*} \color{blue}{\sqrt[-100]{100}}=100^{\frac{1}{-100}}=100^{\frac{-1}{100}}=\sqrt[100]{100^{-1}}=\sqrt[100]{\frac1{100}}\tag{1} \end{align*} is correct.
Sometimes we can read in analysis books a definition of rational powers which goes like: Let $a>0$ and $r=\frac{p}{q}$ where $p,q$ are integers, $q>0$, then we define \begin{align*} a^r\equiv a^{\frac{p}{q}}:=\sqrt[q]{a^p}\tag{2} \end{align*}
Note that the definition above justifies only the following representations in (1) \begin{align*} 100^{\frac{-1}{100}}=\sqrt[100]{100^{-1}}=\sqrt[100]{\frac1{100}} \end{align*}
But since we are allowed to use the notation \begin{align*} \frac{-p}{q}=-\frac{p}{q}=\frac{p}{-q} \end{align*} an extension of the notation (2) to each representation in (1) is admissible.
Note: As a plausibility check note that Wolfram Alpha accepts $\sqrt[-100]{100}$ and suggests the following simplified representation \begin{align*} \sqrt[-100]{100}=\frac{1}{\sqrt[50]{10}} \end{align*}
it can be simplified to $$\frac{1}{(10^2)^{1/100}}=\frac{1}{10^{1/50}}$$
• Sir, I don’t think the OP has invoked a square root notation. – user371838 Dec 27 '17 at 11:54
• but i can see a square root notation above? or i will need glases – Dr. Sonnhard Graubner Dec 27 '17 at 11:55
• I took it to be the $-100$th root. An extension of $\sqrt[3]{2}$, $\sqrt[4]{2}$ etc but with an unusual negative. – badjohn Dec 27 '17 at 12:02
• It is in similar sense to $\sqrt[3] x, \sqrt[-5] x$, sir. – user371838 Dec 27 '17 at 12:04
• i saw it also here themathpage.com/Alg/rational-exponents.htm – Dr. Sonnhard Graubner Dec 27 '17 at 12:09
It's unusual and hence clarification would be advisable to avoid misinterpretation. With appropriate clarification then it would be acceptable. You can even invent your own notation provided that you define it. However, when clearer alternatives are easily available, what's the point?
• I saw this notation new, today – Newuser Dec 27 '17 at 11:53
• Can you give us a link? It is possible that it may have made sense in the context. – badjohn Dec 27 '17 at 11:58 | 2021-01-19T05:23:44 | {
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https://cs.stackexchange.com/questions/119997/can-i-use-the-following-method-to-prove-an-algorithm-is-correct | # Can I use the following method to prove an algorithm is correct?
I'm trying to show that a solution I have obtained via an algorithm is correct. The way I plan on doing this is first by showing that an optimal solution does indeed exist. Then, I plan on showing that every other solution that is not the solution provided by my algorithm cannot be optimal. Finally, I show that the solution I have cannot be improved in the same way as any other solution.
Is this enough to show that my algorithm is optimal? In this case I am avoiding doing an "exchange argument" a la greedy algorithms. In fact, I don't really prove anything about how my solution is an improvement of the other ones, but simply that all of the other ones can be improved, and given that an optimal solution exists, the one I have must be it because it cannot be improved in the same way that the other ones can.
• In fact, a similar reasoning can be applied when we use $\frac{df}{dx}=0$ to find the place where a real-valued differentiable function $f$ can take its maximum value on a closed interval. If we know the function cannot take the maximum value at the ends of the interval, then it must be take the maximum value at one of places where $\frac{df}{dx}=0$. That is, indeed, a marvelous way to find maximum values of many such functions (discovered by Isaac Newton ? and Gottfried Wilhelm Leibniz ?). – John L. Jan 27 '20 at 8:36
• Note that, in isolation, "the solution I have cannot be improved" [locally] might be because it is a local optimum (i.e., not necessarily global). – Pablo H Jan 27 '20 at 14:54
• Am I missing something? Doesn't this prove that it's optimal, not that it's correct? – Barmar Jan 27 '20 at 16:21
• If you had not shown the existence of an optimal solution, you'd be in the "$1$ us the biggest natural number because all other natural numbers can be improved by squaring"-siuation. But with existence shown ... – Hagen von Eitzen Jan 28 '20 at 0:02
• It looks to me that the OP thinks that the other solutions might surpass the current optimal one, since the optimal solution cannot be improve and the other solutions are improvable, I don't know what's going on here but why not use the optimal solution for now while improving the other potential solutions that can surpass the current one in terms of optimization. – Shiz Jan 28 '20 at 6:55
I am rather surprised that you raised this question since the meticulous and enlightening answers you have written to some math questions demonstrate sufficiently that you are capable of rigorous logical deduction. It seems that you became somewhat uncomfortable when you stumbled upon a new and unorthodox way to prove an algorithm is correct.
Believe in yourself. Believe in logic.
Yes, I believe you have shown your algorithm is correct. To be absolutely clear, suppose you have shown all of the following.
• An optimal solution exists.
• Your algorithm provides a solution and only one solution.
• Every solution that is not the solution provided by your algorithm cannot be optimal.
Then your algorithm must provide an optimal solution. The implication is just simple logic.
You do not even need to show that solution provided cannot be improved in the same way as any other solution.
• I don't feel confident about this, as I am really not much of a logician, but when you say "The implication is just simple logic", isn't that true in classical logic but not in intuitionistic logic? My reasoning is that the proof uses double negation elimination: the optimal solution is not not provided by the algorithm, therefore it is provided by the algorithm? If the OP's method is indeed a classical proof, but not a constructive proof perhaps it could be worth mentioning. – integrator Jan 27 '20 at 9:47
• @eru-cs I was trying to write in a level that is suitable for the questioner. I would like to keep that principle. Basically, you are raising the possibility that the solution could be in a state other than being optimal or being not optimal. Although I would concede that exceptionally rare possibility, I had assumed the questioner was in a situation where if a solution is not not optimal, then it is optimal. Had the situation been otherwise that exceptionally rare, the questioner should have raised some red flags, intentionally or unintentionally. Or so I believed. – John L. Jan 27 '20 at 10:57
• @Vasting, when I wrote "a new and unorthodox way", it was meant for you. It could probably be just another mundane way for many experienced users here. As my comment indicates, people have been using that way for hundreds of years at least. – John L. Jan 27 '20 at 11:11
• Proving that the algorithm finds any solution at all is already a big step and is implicitly mentioned here in the answer, but I haven't found that in the question. Please don't forget to prove that your algorithm finds a solution at all. – kutschkem Jan 27 '20 at 13:43
• @eru-cs Even in intuitionistic logic, this is a valid proof that the given algorithm is optimal, because of the structure of terminating algorithms. We know an optimal solution exists, call it $O$, and let the new algorithm be $A$. On any given input $x$, we presume that $O(x)$ and $A(x)$ terminate with results in some discrete countable set of answers. Therefore $O(x)=A(x)$ or $O(x)\ne A(x)$. In the second case, $O\ne A$ so by the exchange argument $O$ can be improved, which is impossible. Thus $O(x)=A(x)$ for all $x$, so $O=A$. – Mario Carneiro Jan 28 '20 at 0:03
You have: There is an optimal solution, and any solution not found by your algorithm is non-optimal. It follows that the optimal solution is found by your algorithm, so the third part is not needed.
For problems with two or more optimal solutions you won’t be able to show the second part unless your algorithm finds all optimal solutions.
but simply that all of the other ones can be improved, and given that an optimal solution exists, the one I have must be it because it cannot be improved in the same way that the other ones can.
It does seem that there's something slightly to patch up here. Do you have a proof that this procedure for making an improvement always makes an improvement when one is available? If you have such a proof, then simply showing that the procedure makes no improvement to your solution is enough to qualify the solution as optimal, and the only reason to even refer to other solutions is to show that yours is a unique optimum.
If you don't have such a proof, then you need to prove that your solution is actually optimal and not just a point where your "improvement procedure" fails to make progress. I think your intuition is correct that if every other solution can be improved, then the one that can't is an optimum, but it may fall short of a really rigorous proof without an appeal to something like the contraction mapping theorem.
In particular, if you could show that your "improvement procedure"
1. Never actually produces a worse solution, and
2. Always produces a solution that is nearer by some metric to your chosen solution than the one it was given as input
then you would be showing that your solution is a fixed-point of the improvement procedure and that every sequence leading to it is monotone, which should prove, using monotone convergence, that your solution is optimal. | 2021-06-21T22:34:17 | {
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https://mathoverflow.net/questions/413986/the-wiener-measure-of-an-open-set | # The Wiener measure of an open set
There is so much written about the Brownian motion and I suspect the answers to the questions below are hidden in somewhere in the literature but I cannot find them
The Wiener measure defines a Borel measure $$\bsW$$ on $$E$$. Let $$f_0\in E_0$$ and $$\ve>0$$. We set $$w(f_0,\ve):=\bsW\Big[\;\big\{\; f\in E;\;\;\Vert f-f_0\Vert<\ve\,\big\}\;\Big].$$
Question 1. Is it true that $$w(f_0,\ve)>0$$ for all $$f_0\in E$$ and $$\ve>0$$?
Question 2. Can one produce an explicit positive lower bound for $$w(f_0,\ve)$$ in terms of $$\ve>0$$ and the modulus of continuity of $$f$$? In particular, if $$f_0$$ is Lipschitz, can one produce a lower bound in terms of $$\ve>0$$ and the Lipschitz constant of $$f_0$$?
For Question 1 I have an argument based on Cameron-Martin formula? Is there any other more "elementary" argument?
$${{}}$$
• According to this question on Math.SE, this is called "small ball estimates". Have you tried searching under this term? Jan 16 at 19:37
• I've seen something like this in Peres-Morters book Exercise 1.8, where they first built a dyadic function that is epsilon close to f0 and then continue the Levy construction of adding Gaussians in-between to built a Brownian motion. Jan 16 at 19:39
If $$f\colon[0,1]\to\mathbb R$$ is such that for all $$k\in[n]$$ and all $$t\in[t_{k-1},t_k]$$ we have $$f(t)\in I_k:=[g(t_k)-\ep/2,g(t_k)+\ep/2]$$, then $$\|f-g\|<\ep$$.
So, for any standard Wiener process $$W$$, $$\begin{equation*} w(g,\ep)=P(\|W-g\|<\ep)\ge p:=p_1\cdots p_n, \end{equation*}$$ where $$\begin{equation*} p_k:=\min_{x\in J_{k-1}}P(x+W_t\in I_k\ \forall t\in[0,\de_k], x+W_{\de_k}\in J_k), \end{equation*}$$ $$J_0:=[0,0]=\{0\}$$, $$J_n:=I_n$$, and $$J_k$$ is the closed interval that is the middle third of the interval $$I_k\cap I_{k+1}$$ for $$k\in[n-1]$$. Note that for $$k\in[n-1]$$ the length of the interval $$I_k\cap I_{k+1}$$ is $$>\ep/2$$ and hence the length of the interval $$J_k$$ is $$>\ep/6>0$$. Also, for each $$k\in[n]$$ the shortest distance from any point $$x\in J_{k-1}$$ to the set $$\{g(t_k)-\ep/2,g(t_k)+\ep/2\}$$ of the endpoints of the interval $$I_k$$ is $$>\ep/6>0$$. So, $$p_k>0$$ for all $$k\in[n]$$, and hence $$p>0$$. Moreover, one can give an explicit expression of each $$p_k$$ in terms of $$\de_k$$, $$I_k$$, $$J_k$$, $$J_{k-1}$$ -- cf. e.g. Proposition 6.10.6, p. 533.
Thus, one can explicitly express the lower bound $$p>0$$ on $$w(g,\ep)$$ in terms of $$\ep$$, $$t_1,\dots,t_n$$, and $$g(t_1),\dots,g(t_n)$$.
For $$g(t)=t (5 - 18 t + 12 t^2)$$, $$\ep=5/2$$, $$n=4$$, $$\de_k=1/4$$ for $$k\in[4]$$, the picture below shows the graphs $$\{(t,g(t))\colon t\in[0,1]\}$$ (blue), $$\{(t,g_-(t))\colon t\in[0,1]\}$$ (gold), $$\{(t,g_+(t))\colon t\in[0,1]\}$$ (green), and a path of the Wiener process (gray) belonging to the event \begin{equation*} \begin{aligned} B&:=\{g_-(t) where $$g_\pm(t):=g(t_k)\pm\ep/2$$ for $$t\in(t_{k-1},t_k]$$, $$k\in[4]$$. We have $$P(\|W-g\|<\ep)\ge P(B)\ge p>0$$.
• Thank you very much. Jan 17 at 10:13
This is known as the support theorem for Brownian motion. Besides the proof in the answer of Iosif Pinelis and the proof in Exercise 1.8 of [1], there is also a proof on page 59 of [2]. Generalizations are discussed in [3]-[5].
[1] Mörters, Peter, and Yuval Peres. Brownian motion. Vol. 30. Cambridge University Press, 2010. https://yuvalperes.com/brownian-motion/
[2] R. Bass, Probabilistic Techniques in Analysis, Springer, New York (1995). MR1329542
[5] Stroock, Daniel W.; Varadhan, S. R. S. On the support of diffusion processes with applications to the strong maximum principle. Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability (Univ. California, Berkeley, Calif., 1970/1971), Vol. III: Probability theory, pp. 333–359. Univ. California Press, Berkeley, Calif., 1972.
• Thank you very much. Jan 17 at 10:13 | 2022-05-21T05:19:33 | {
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https://math.stackexchange.com/questions/2724222/why-the-interior-of-u-in-s-contains-mathrmint-u-cap-s | # Why the interior of $U$ in $S$ contains $\mathrm{int\,}U\cap S$
(Exercise 3.7. John Lee’s Introduction to Topological Manifolds) Suppose $X$ is a topological space and $U\subseteq S\subseteq X$. The interior of $U$ in $S$ contains $\mathrm{int } \, U\cap S$.
My attempt:
I can see why they are not equal: an open set in the subspace topology may not necessarily be open with respect to the topology from which the subspace inherits it, and thus the two interiors may not be equal. But, why the word “contained?”
I can think of an example, but not prove it in general: take the set $S = [0,1]\cup (2,3)$ and a subset $[0,1]$ which is open in $S$, but not in $\mathbb{R}$. Thus, it follows that the interior of $[0,1]$ in $S$ contains $\mathrm{int}\,[0,1]\cap S$. So then, how can I compete my proof?
• My personal policy is to usually stop reading beyond an ambiguity, even when the intended meaning may be obvious. Is that int$(U)\cap S$ or int $(U\cap S)$? Please edit. – DanielWainfleet Apr 6 '18 at 9:22
• @DanielWainfleet The former, I’ll edit my post. – user522521 Apr 6 '18 at 10:09
$\operatorname{int}(U) \cap S$ is open in $S$ (it's open in $S$ by the definition of the subspace topology and the fact that $\operatorname{int}(U)$ is open), and it's clearly a subset of $U$. As the interior of $U$ w.r.t. $S$ (denoted by $\operatorname{int}_S(U)$) is the largest open (in $S$) set that is contained in $U$, we immediately get $$\operatorname{int}(U) \cap S \subseteq \operatorname{int}_S(U)$$
The reverse need not hold, as witnessed by, e.g. $U = S = \mathbb{Q}$ in the reals, which has empty interior in the reals but is its own interior in itself of course.
Suppose $p\in\operatorname{int}(U)\cap S$. Because $p\in\operatorname{int}(U)$, there exists some open set $O$ of $X$ such that $p\in O\subset U$. Now remember how the open sets of $S$ in its subspace topology are defined... do you see how to finish the problem?
• Well, I guess $\mathrm{int} \ U$ is open, but what else? – user522521 Apr 6 '18 at 0:19
• The interior of $U$ in $S$ is the union of all open sets of $S$ that are contained in $U\cap S$. How can you relate $O$ to an open set of $S$? – Neal Apr 6 '18 at 0:20
• $O=S\cap V$ for some $V$ open in $X$. – user522521 Apr 6 '18 at 0:22
• By the way, I think what he meant by $\mathrm{int} \ U\cap S$ was the was rather $\mathrm{int}(U) \cap S$. – user522521 Apr 6 '18 at 0:32 | 2020-05-30T06:15:06 | {
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https://math.stackexchange.com/questions/2723367/function-gx-such-that-int-mathbbr-fgxdx-int-mathbbr-fx/2723568 | # Function $g(x)$, such that $\int_{\mathbb{R}} f(g(x))dx = \int_{\mathbb{R}} f(x)dx$ for all $f\in L^1(\mathbb{R})$
It's not too hard to show that for all $f\in L^1(\mathbb{R})$ and for $g(x) = x-\frac{1}{x}$ we have $\int_{\mathbb{R}} f(g(x))dx = \int_{\mathbb{R}} f(x)dx$:use the substitution $y = x - \frac{1}{x}$ and split the integral in two parts: $x<0$ and $x>0$. On one of them make the choice $x = \frac{y + \sqrt{y^2 + 4}}{2}$, on the other $x = \frac{y - \sqrt{y^2 + 4}}{2}$ and add the resulting integrals.
The existence of even one function $g(x)$, such that the above works is astonishing. Are there other choices of $g$, such that $\int_{\mathbb{R}} f(g(x))dx = \int_{\mathbb{R}} f(x)dx$ for all $f\in L^1(\mathbb{R})$?
Edit: $g(x) = x + a$ for any a would be a solution, but I was wondering if there are more complicated $g$ which work, or if $x - \frac{1}{x}$ is just a single example.
• How about $g(x)=x$? Trivial, but it works. – Adrian Keister Apr 5 '18 at 13:31
• I meant a non-trivial choice of $g$ – Milen Ivanov Apr 5 '18 at 13:33
• Actually $$\int_{\mathbb{R \setminus \{0\}}} f(g(x)) \,dx = \int_{\mathbb{R}}f(x) \,dx$$ – mucciolo Apr 5 '18 at 13:39
• @mucciolo The notation $L^1(\Bbb R)$ implies Lebesgue integration, which disregards values taken at a single point. – Arnaud Mortier Apr 5 '18 at 13:41
• see Glasser's master theorem. please note that the statement in wiki is slightly off, the leading coefficient $|a|$ in the transform $u = |a|x - \sum_{n=1}^N \frac{|a_n|}{x - \beta_n}$ should really be $1$. – achille hui Apr 5 '18 at 15:38
The set of all such functions is closed under composition. As a result, all the functions $g^k$ (in the sense of composition) work.
E.g. $$g^2(x)=x-\frac1x -\frac{1}{x-\frac1x}$$
Adding to that the fact that you can also compose them with affine functions of slope $1$ (lisyarus' answer) yields already a pretty large class.
Incorporating @J.G.'s comment, one gets for instance all rational functions of the form $$\frac{x^2+ax+b}{x+c}$$ with the only requirement that $b<c^2$ ($a$ and $c$ are arbitrary).
• Nice, that's great insight! – Milen Ivanov Apr 5 '18 at 13:58
• IIRC $x-c/x$ works for any $c>0$, which expands the class even further. – J.G. Apr 5 '18 at 15:15
We have $\int f\circ g=\int f$ for every integrable $f$ if and only if the same holds for $f=\chi_E$, which says precisely $$m(g^{-1}(E))=m(E)$$for every measurable set $E$, which is to say that $g$ is measure-preserving.
If we assume in addition that $g$ is a smooth bijection this is equivalent to $|g'|=1,$so the only smooth bijections with this property are $$g(x)=\pm x+c.$$
This seemed so obvious that it seemed clear to me at first that $x-1/x$ cannot have the stated property. But of course that function is not injective; if $g(x)=x-1/x$ you can calculate $g^{-1}((a,b))$ explicitly, and sure enough you get two intervals with total length $b-a$.
(This is actually consistent with the analysis above, if you look at it right. The relevant condition for a smooth bijection is actually $|({g^{-1})}'|=1$, which is equivalent to $|g'|=1$. But now if $g(x)=x-1/x$ and you let $y_1$ and $y_2$ denote the two "branches" of $g^{-1}$ you easily calculate that $|y_1'|+|y_2'|=1$.)
One can easily concoct discontinuous examples. For example if $A\subset(0,\infty)$ let $$g(x)=\begin{cases}x,&(|x|\in A), \\-x,&(|x|\notin A).\end{cases};$$then $g$ is a measure-preserving bijection.
There is a not that well known result about these kinds on integrals that quite often comes up on this site. It was already mentioned in the comments by achille hui, but I'll add this as a CW as it's a shame not having it as an answer:
Glasser's master theorem$^{[1]}$ says that if $f(x)$ is integrable then $$\int_{\mathbb{R}}f(x)\,{\rm d}x = \int_{\mathbb{R}}f(\phi(x))\,{\rm d}x$$ for all $\phi(x) = x - \sum_{n=1}^N\frac{|a_n|}{x-b_n}$ where $a_n,b_n$ are arbitrary constants and where the integrals are to be interpreted in a principal value sense.
[1]: Glasser, M. L. "A Remarkable Property of Definite Integrals." Math. Comput. 40, 561-563, 1983
As mentioned by David C. Ullrich, such $g$ is called a measure-preserving transformation (MPT) on $\mathbb{R}$ (w.r.t. the Lebesgue measure, of course). The comment by achille hui provides an important family of MTPs given by
$$g(x) = \pm \left( x - c - \sum_{k=1}^{n} \frac{\mu_k}{x - \lambda_k} \right)$$
for $c, \lambda_k \in \mathbb{R}$ and $\mu_k \in (0, \infty)$, which is the statement of the Glasser's master theorem (1983). More generally, a theorem by Letac (1977) tells that any function of the form
$$g(x) = \pm \left( x - c - \int_{\mathbb{R}} \left( \frac{1}{x - \lambda} + \frac{\lambda}{1+\lambda^2} \right) \, \mu(d\lambda) \right)$$
for $c \in \mathbb{R}$ and a singular Borel measure $\mu$ with $\int_{\mathbb{R}} \frac{\mu(d\lambda)}{1+\lambda^2} < \infty$ is MPT. This theorem includes interesting examples such as
$$g(x) = x - \cot x$$
corresponding to $\mu = \sum_{k \in \mathbb{Z}} \delta_{\pi k}$. Notice also that the Glasser's master theorem is a special case of this theorem with $\mu = \sum_{k=1}^{n} \mu_k \delta_{\lambda_k}$, although the proof techniques involved are different.
Clearly if $g: \mathbb{R}\to\mathbb{R}$ is a monotonic, we must have have, under $u=g(x)$, $$\int_{\mathbb{R}}f(g(x))dx=\int_{\mathbb{R}}f(u)(g^{-1})'(u)du$$ for all $f\in L^1(\mathbb{R})$. Thus $$(g^{-1})'(u)=1$$ which implies $g(x)=x+a$. If $g: (-\infty,0)\to\mathbb{R}$ and $g: (0,\infty)\to\mathbb{R}$ are monotonic, namely $g^{-1}$ has two branches $x=g_1^{-1}(u):\mathbb{R}\to(-\infty,0)$ and $x=g_2^{-1}(u):\mathbb{R}\to(0,\infty)$, then \begin{eqnarray} \int_{\mathbb{R}}f(g(x))dx&=&\int_{(-\infty,0)}f(g(x))dx+\int_{(0,\infty)}f(g(x))dx\\ &=&\int_{\mathbb{R}}f(u)(g_1^{-1})'(u)du+\int_{\mathbb{R}}f(u)(g_2^{-1})'(u)du\\ &=&\int_{\mathbb{R}}f(u)\bigg[(g_1^{-1})'(u)+(g_2^{-1})'(u)\bigg]du \end{eqnarray} holds for all $f\in L^{-1}(\mathbb{R})$. Thus we must have $$(g_1^{-1})'(u)+(g_2^{-1})'(u)=1$$ or $$g_1^{-1}(u)+g_2^{-1}(u)=u. \tag{1}$$ or all $u\in\mathbb{R}$. Assume $g(x)$ has the form $g(x)=ax+\frac{b}{x+c}$. Thus if $g(x)$ is monotonic, then $a$ and $b$ must satisfy $ab<0$. WOLG, assume $a>0,b<0$. Using (1), it is easy to obtain $a=1$. So $$g(x)=x+\frac{b}{x+c}, b<0$$
• The $(g^{-1})'$ in the first formula should actually be $|(g^{-1})|$ (because for example $\int f(-x)=\int f$.) – David C. Ullrich Apr 6 '18 at 11:46
• @DavidC.Ullrich, you are right.Thanks. – xpaul Apr 6 '18 at 14:01
Consider $g(x) = x+a$ for a fixed $a\in \mathbb R$. Then, using the substitution $y = x+a$, we get $dy = dx$, and thus
$$\int\limits_{\mathbb R}f(g(x))dx = \int\limits_{\mathbb R}f(y)dy$$
The functions of the form $$\phi(x) = x - \sum_{i=1}^{n-1} \frac{\rho_i}{x- \alpha_i} -\beta$$
where $\rho_i>0$ for all $1\le i \le {n-1}$ invariate the Lebesgue measure. The set of such functions is closed under composition.
Sketch of proof:
Assume $\alpha_1 < \ldots <\alpha_{n-1}$. On each of the intervals $(-\infty, \alpha_1)$, $(\alpha_1, \alpha_2)$, $\ldots$, $(\alpha_{n-2}, \alpha_{n-1})$,$(\alpha_{n-1},\infty)$ the function $\phi$ is strictly increasing from $-\infty$ to $\infty$. Therefore, for every $u \in \mathbb{R}$ the equation $$\phi(x)=u$$ has exactly $n$ distinct real roots. From Viete's relations we see that the sum of the roots equals $$u +\sum_{i=1}^{n-1}\alpha_i + \beta$$
Therefore, for every $I$ interval of $\mathbb{R}$ the set $\phi^{-1}(I)$ is a disjoint union of interval of length $|I|$. This implies $$\int_{\mathbb{R}}(\chi_I\circ \phi)\ d\mu= \int_{\mathbb{R}}\chi_I \ d \mu$$
for the characteristic function of the interval $I$. From here one concludes that $$\int f \circ\phi \ d\mu = \int f \ d\mu$$ for all $f \in L^1(\mathbb{R})$.
Consider two such functions $\phi= x-\sum_1^{n-1} \frac{\rho_i}{x-\alpha_i} - \beta$ and $\chi = x - \sum_1^{n'-1}\frac{\rho'_i}{x-\alpha'_i} - \beta'$. Each of the intervals $(\infty, \alpha'_1)$, $(\alpha'_1,\alpha'_2)$, $\ldots$,$(\alpha_{n'},\infty)$ get divided into intervals that map to one of $(-\infty, \alpha_1)$, $(\alpha_1, \alpha_2)$, $\ldots$, $(\alpha_{n-2}, \alpha_{n-1})$,$(\alpha_{n-1},\infty)$. We conclude that $\mathbb{R}$ gets divided into $n\cdot n'$ intervals that are mapped strictly increasing onto $(-\infty, \infty)$ by $\phi\circ \psi$ . The rational function $\phi\circ \psi$ has therefore at least $n\cdot n'-1$ real finite poles and a pole at infinity. Since it is of degree $n\cdot n'$ it must have a partial fraction decomposition of form $$x - \sum_1^{n n'-1} \frac{\rho''_i}{x- \alpha''_i}- \beta''$$ Now one checks that the $\rho''_i$ must be positive. | 2019-07-17T22:43:18 | {
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https://math.stackexchange.com/questions/1596246/probability-of-choosing-positive-2-digit-integer-with-4-in-either-place | # Probability of choosing positive 2-digit integer with 4 in either place
GRE study exam guide has following
If an integer is randomly selected from all positive 2-digit integers, what is the probability that the integer chosen has at least one 4 in the tens place or the units place?
I understand probability of being in 10s place is $1/9$ and the probability of being in the units place is $1/10$
When I add $1/9$ + $1/10$ the answer is $19/90$. However, answer says $1/5$.
A = $4$ in a ten place, B = $4$ in a unit place, C = at least one $4$ in a ten or unit place.
$P(A) = \frac{1}{9}$, $P(B) = \frac{1}{10}$, $P(C) = \frac{1}{9} + (1-\frac{1}{9})*\frac{1}{10} = \frac{10 + 8}{90} = \frac{1}{5}$.
Here you go: You have $10$ cases for event $A$: $40, 41, 42, 43,44,45,46,47,48,49$, and $9$ cases for event $B$: $14, 24, 34, 44, 45, 46, 47, 48, 49$. So when you get $A \cup B$ you have not $19$ cases, but 18. There are $90$ two-digit numbers, so $\frac{18}{90} = \frac{1}{5}$.
• When you say $1/9 + (1 - 1/9)$, does the plus sign mean $or$, as in 4 is present = $1/9$ or 4 is NOT present = $(1 - 1/9)$ Jan 1, 2016 at 19:01
• The factor $(1-\frac{1}{9})$ is needed to discount case $44$ calculated already in $P(A)$. Actually event $A$ is made of $10$ cases, namely $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$, and event $B$ is made of $9$ cases: $14, 24, 34, 44, 54, 64, 74, 84, 94$. Now there are total $19$ cases, but we counted $44$ $2$ times, so we have $18$ cases. There are $90$ $2$ digit numbers, so $\frac{18}{90} = \frac{1}{5}$. Jan 1, 2016 at 19:05
• Can you add your last comment to the answer. Because that makes sense. It really does. I will mark as answer and it would help others. This is called deep understanding, which I wish to accomplish! Jan 1, 2016 at 19:10
You have to subtract $P(A \cap B) = {1 \over 90}$ for the probability of having 44 (I used $P(A\cup B)=P(A)+P(B)-P(A \cap B)$ See: Probability Of Union/Intersection Of Two Events).
Another way to solve it is to calculate $1-P(no \ four)= 1 - (\frac {8}{9} \frac {9}{10}) = {1 \over 5}$
I assume as Rhonda not an exclusive OR: Pick from $\{1,..,9\}$ for the tens and $\{0,1,...,9\}$ for the units and so $8 \over 9$ for the probability not to have a four in the tens and $9\over 10$ for no four in the units
• Nope, the phrasing at least one $4$ in the tens place or the units place (they could have just said at least one 4) allows for $44$. Jan 1, 2016 at 18:18
• @k-p I am trying to understand this. Where did you get $8/10$ Jan 1, 2016 at 18:20
• @Rhonda Edited, sorry
– K_P
Jan 1, 2016 at 18:22
• @K_P Ok let me try to soak in your answer Jan 1, 2016 at 18:23
• @K_P Honestly I am really confused by the $P(A∪B)$ Jan 1, 2016 at 18:27
You are right, $0$ should not be allowed in the ten's place,
but the official answer is right, nevertheless.
P(no $4$ in the tens or units place)$\ddagger\ddagger$ = $\dfrac89\cdot\dfrac9{10} = \dfrac8{10}= \dfrac45$
Thus P(at least one $4$ in the tens or units place) $= 1 - \dfrac45 = \dfrac15$
$\ddagger\ddagger$ The expression written in "normal" English actually means
P(no $4$ in the tens place and no $4$ in the units place),
or more simply, P(no $4$ in the number)
• Here is where I am confused. I thought $or$ means $+$, but you are multiplying, i.e. $8/9 * 9/10$ Jan 1, 2016 at 19:03
• I am clarifying in the answer itself for the benefit of all. Jan 1, 2016 at 19:07
P(AorB)=P(A) +P(B) _P(both A and B)
P(both A and B)= P(A) * P(B) events are not mutually exclusive.. So P(A)= 1/9, P(B) = 9/90 = 1/9 + 9/90 _ 1/9*9/90= 1/5 answer.. | 2022-08-10T23:45:52 | {
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https://math.stackexchange.com/questions/1623860/what-is-the-probability-that-the-sum-of-the-die-rolls-is-odd | What is the probability that the sum of the die rolls is odd?
Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is $0$.)
I thought that since odd+even =odd, we then have to get (odd,even) and (even,odd) pairs from $\{0,1,2,3,4,5,6\}$. Thus, since there are a total of $7*7$ possibilities for pairs and $4*3*2$ pairs that add to an odd number I get $\dfrac{4*3*2}{7*7} = \dfrac{24}{49}$ but the right answer is $\dfrac{3}{8}$. What did I do wrong?
• tbh I don't quite understand where that fraction is coming from. – joedoe8585 Jan 23 '16 at 19:00
There is a $1/4$ chance of no heads, and therefore no dice. The sum of no dice is $0$, which is even.
There is a $1/2$ chance of one head, and therefore one die. For one die, there is a $1/2$ chance of landing on an odd total $1,3,5$.
There is a $1/4$ chance of getting two heads, and therefore two dice. There are an equal number of possibilities giving even and odd, so there is again a $1/2$ chance of landing on an odd total.
So in total, we have total probability $$\frac{1}{4} \cdot 0 + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8},$$ as you indicated was the correct answer. $\diamondsuit$
• I was asking what I did wrong in my approach not what the correct approach was. – user19405892 Jan 23 '16 at 19:15
What you are doing wrong is assuming that each of those possible outcomes is equally likely.
In case of the outcome "Head, Head", two die are rolled. In this case, the outcomes and their probabilities are 2: prob 1/36; 3: prob 2/36= 1/18; 4: prob 3/36= 1/12; 5: prob 4/36= 1/9; 6: prob 5/36; 7: prob 6/36= 1/6; 8: prob 5/36; 9: prob 4/36= 1/9; 10: prob 3/36= 1/12; 11: prob 2/36= 1/18; 12: prob 1/36; The probability of "odd" is 1/18+ 1/9+ 1/6+ 1/9+ 1/18= 9/18= 1/2. Since the probability of this case is 1/4, multiply that by 1/4: 1.8.
In the case of "Heads, Tails" or "Tails, Heads" we roll a single die so the probability of any one number is 1/6. There are 3 odd numbers so the probability of "odd" is 3/6= 1/2 also. The probability of this case is 1/2 so multiply by 1/2: 1/4.
in the case of "Tails, Tails" we do not roll any die so the "sum" is 0. That is even so the probability of "odd" in this case is 0.
The overall probability of "odd" is 1/8+ 1/4= 3/8.
Notice that not your sample set cannot be one of $\{0, 1, 2, 3, 4, 5, 6\}$ because they do not all occur with equal probabilities. Specifically, the set $\{1, 2, 3, 4, 5, 6\}$ only occurs with probability $\frac{1}{2}$ for each die (you might roll $T$ and not have those outcomes at all). Similarly, $\{0\}$ occurs with probability $\frac{1}{2}.$
Adjusting your calculation to account for this, we find that there is a $(\frac{1}{2} + \frac{1}{2}(\frac{1}{2})) \cdot (\frac{1}{2}(\frac{1}{2})) = \frac{3}{16}$ chance to get (even, odd) and another $\frac{3}{16}$ chance to get (odd, even) by symmetry. Our final answer is $2 \cdot \frac{3}{16} = \boxed{\frac{3}{8}}.$
• We have HH,TH,HT, and TT. For HH we have $\{1,2,\ldots,6\} \times \{1,2,\ldots,6\}$; for $TH$ we have $\{1,2,\ldots,6\} \times {0}$; for $HT$ we have $\{0\} \times \{1,2,\ldots,6\}$; and for TT we have $\{0\} \times \{0\}$. How are not these equally likely? – user19405892 Jan 23 '16 at 19:34
• Look closer: is the probability of getting a $0$ the same as the probability of getting a $6?$ Of course not! @user19405892 – K. Jiang Jan 23 '16 at 19:42 | 2019-11-12T08:49:26 | {
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https://math.stackexchange.com/questions/2342596/probability-of-an-unit-needing-repair/2342618 | # Probability of an unit needing repair.
I'm new to statistics and probability. I came upon this problem while solving Erwin Kreyszig's book. The problem is given like:
A motor drives an electric generator. During a 30 days period, the motor needs repair with 8% and the generator needs repair with probability 4%.What is the probability that during a given period, the entire apparatus(consisting of a motor and a generator) will need repair?
So, $\mathsf P(\text{motor-repair}) = 0.08$ and $\mathsf P(\text{generator-repair}) = 0.04$.
Now why $\mathsf P(\text{motor-repair})\cdot\mathsf P(\text{generator-repair}) = 0.04\cdot 0.08$ should not give the probability of the entire apparatus needing repair?
The solution is given like $$\mathsf P(\text{motor-repair}) + \mathsf P(\text{generator-repair}) - \mathsf P(\text{motor-repair}) \cdot\mathsf P(\text{generator-repair})$$
Also in another resource it is given like $$1 - ((1-\mathsf P(\text{motor-repair}))\cdot (1−\mathsf P(\text{generator-repair})))$$
My question is how these solutions are coming?
When should I multiply and when should I add in Probability?
What is the difference between $\mathsf P(A)\cdot\mathsf P(B)$ and $\mathsf P(A \cap B)$?
I'm sorry if I'm asking stupid questions, but these things are not becoming clear. Please help me. Thanks in advance.
Add for (disjoint) unions, multiply for (independent) intersections.
Roughly speaking (not always 100% true!), in probability, the word or translates into addition, while and translates into multiplication.
• Okay! I got it now.. That means like @Anirudh pointed out they want me to find the union and since it is independent like motor has nothing to do with generator P(A∩B) is becoming P(A).P(B)? Jul 1, 2017 at 5:37
• Actually I would suggest you please don't look into the way how they have approached the problem... Try to build a solution first yourself... It really helped me because I was very weak at Probability around a year ago... Jul 1, 2017 at 5:42
• Actually I did that. But it makes me mad when I see none of my methods work. Jul 1, 2017 at 5:43
• But one thing - "consisting of a motor and a generator" then shouldn't it be like P(A∩B) like you said? Jul 1, 2017 at 5:45
• Can you refer the page number.. Jul 1, 2017 at 5:45
Now why $\mathsf P(\text{motor-repair})\cdot\mathsf P(\text{generator-repair}) = 0.04\cdot 0.08$ should not give the probability of the entire apparatus needing repair?
Why, yes, for a certain interpretation of "entire apparatus needs repair"; this just does not appear to be the intended meaning.
Assuming the components fail independently, that is the probability that both need repair. The probability for the intersection of (pairwise) independent events equals the product of the probabilities for each event.
Note: The independence of the events is important here. It is indeed definitional. $$A\perp B ~\iff~ \mathsf P(A\cap B)=\mathsf P(A)\cdot\mathsf P(B)$$
The solution is given like $$\mathsf P(\text{motor-repair}) + \mathsf P(\text{generator-repair}) - \mathsf P(\text{motor-repair}) \cdot\mathsf P(\text{generator-repair})$$
That would be the probability that either component needs repair. It is apparent that they intended that the entire apparatus needs repair means that either part does.
Now, the probability for the union of disjoint events equals the sum of the probabilities of the events. However, these events are not disjoint, so we apply the Principle of Inclusion and Exclusion.
$$\mathsf P(A\cup B)~{=\mathsf P(A\cup (B\setminus A))\\=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B) \\ = \mathsf P(A)+\mathsf P(B)-\mathsf P(A)\cdot\mathsf P(B)\qquad\text{if independent}}$$
Also in another resource it is given like $$1 - ((1-\mathsf P(\text{motor-repair}))\cdot (1−\mathsf P(\text{generator-repair})))$$
That would be the probability that both components do not require repair. (ie: That neither does.) Which is the same thing thanks to deMorgan's Laws.$$(A\cup B) ~=~ {(A^\complement\cap B^\complement)}^\complement$$
So therefore, since the complements of the events are also independent from each other:$$\mathsf P(A\cup B)~{=~1-\mathsf P(A^\complement\cap B^\complement)\\ = ~\bbox[pink]{ 1-(1-\mathsf P(A))\cdot(1-\mathsf P(B)) }\\ = ~1-(1-\mathsf P(A)-\mathsf P(B)+\mathsf P(A)\cdot\mathsf P(B))\\=~\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)}$$
Which is indeed a proof of the Principle of Inclusion and Exclusion holding for two independent events.
• And this completes my question. Thank you very much. :) Jul 1, 2017 at 16:41
From what you said of their intended solution, I think they meant that if either of the parts is broken the entire apparatus needs repair.
Your solution is calculating the probability that both parts in the apparatus need repair, so I think you may have misread the question a bit.
If you are confused on when you should add and subtract in probability I would try this site here
If you still don't understand the whole addition and multiplication thing leave me a comment so I can go more in depth on my own later.
• Thanks for the reply! But then P(AUB) = P(A) + P(B) - P(A∩B). So, P(A∩B) = P(A).P(B)? Can you help? Jul 1, 2017 at 5:34
• Your values are correct but you are getting them mixed up you are looking for A and B but they are looking for A or B Jul 1, 2017 at 5:38
• In case ur not clear on some of the syntax Jul 1, 2017 at 5:39
• Can you say something on the second solution? Jul 1, 2017 at 5:40
• P(A and B)=P(A) *P(B) in this case and this becomes important later when we want to calculate P(A or B) P(A or B)=P(A) +P(B) seems like a logical solution but since A and B are not mutually exclusive events we have just overcounted by P(A and B) therefore we get P(A or B) =P(A) +P(B)-P(A and B) Jul 1, 2017 at 5:49 | 2022-06-28T07:24:38 | {
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https://mathhelpboards.com/threads/1-1-213-ap-calculus-exam-problem-int-sec-x-tan-x-dx.26423/ | # [SOLVED]1.1.213 AP Calculus Exam Problem int sec x tan x dx
#### karush
##### Well-known member
$\tiny{213(DOY)}$
$\displaystyle\int \sec x \tan x \: dx =$
(A) $\sec x + C$
(B) $\tan x + C$
(C) $\dfrac{\sec^2 x}{2}+ C$
(D) $\dfrac {tan^2 x}{2}+C$
(E) $\dfrac{\sec^2 x \tan^2 x }{2}+ C$
$(\sec x)'=\sec x \tan x$ so the answer is $\sec x +C$ (A)
Last edited:
#### topsquark
##### Well-known member
MHB Math Helper
213(DOY)
$\displaystyle\int \sec x \tan x \: dx =$
(A) $\sec x + C$
(B) $\tan x + C$
(C) $\dfrac{\sec^2 x}{2}+ C$
(D) $\dfrac {tan^2 x}{2}+C$
(E) ${\sec^2 x \tan^2 x }{2}+ C$
$(\sec x)'=\sec x \tan x$ so the answer is $\sec x +C$ (A)
Yup!
-Dan
#### Olinguito
##### Well-known member
A neat trick to do is to differentiate each of the multiple-choice answers in turn until you get the expression to be integrated.
#### HallsofIvy
##### Well-known member
MHB Math Helper
Instead of memorizing a "secant-tangent" integral rule, I would use the fact that $tan(x)= \frac{sin(x)}{cos(x)}$ and $sec(x)= \frac{1}{cos(x)}$ so that $\int sec(x)tan(x)dx= \int \frac{sin(x)}{cos^2(x)}dx$. Let $$u= cos(x)$$ so that $$du= -sin(x)dx$$ and the integral becomes $-\int\frac{du}{u^2}= -\int u^{-2}du= -(-u^{-1})+ C= \frac{1}{cos(x)}+ C= sec(x)+ C$.
#### karush
##### Well-known member
Instead of memorizing a "secant-tangent" integral rule, I would use the fact that $tan(x)= \frac{sin(x)}{cos(x)}$ and $sec(x)= \frac{1}{cos(x)}$ so that $\int sec(x)tan(x)dx= \int \frac{sin(x)}{cos^2(x)}dx$. Let $$u= cos(x)$$ so that $$du= -sin(x)dx$$ and the integral becomes $-\int\frac{du}{u^2}= -\int u^{-2}du= -(-u^{-1})+ C= \frac{1}{cos(x)}+ C= sec(x)+ C$.
mahalo
btw how would you rate this problem :
easy, medium, hard
#### Country Boy
##### Well-known member
MHB Math Helper
I would consider it about "medium-easy". | 2020-08-03T08:59:07 | {
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https://riyadhconnect.com/jsqynqbm/d12b61-congruent-triangles-examples | Two triangles are said to be congruent if their sides have the same length and angles have same measure. Develop a deeper understanding of Congruent triangles with clear examples on Numerade. Congruent Shapes Examples. SSS stands for \"side, side, side\" and means that we have two triangles with all three sides equal.For example:(See Solving SSS Triangles to find out more) Angles can be oriented in any direction on a plane and still be congruent. We know that $$\Delta PQR$$ is an isosceles triangle and $$PQ=QR$$. https://www.mathsisfun.com/geometry/triangles-congruent.html Congruent trianglesare triangles that have the same size and shape. Congruent Triangles . They are all congruent. You can then receive notifications of new pages directly to your email address. Watch this interesting video to understand more about this concept. The following video shows why there is not an SSA Rule for congruent triangles. It encourages children to develop their math solving skills from a competition perspective. Feel free to link to any of our Lessons, share them on social networking sites, or use them on Learning Management Systems in Schools. So we know that two triangles are congruent if all of their sides are the same-- so side, side, side. Since $$L$$ and $$M$$ are the mid points of $$PQ$$ and $$QR$$ respectively, \begin{align}PL=LQ=QM=MR=\frac{QR}{2}\end{align}. Under this criterion, if the two angles and the side included between them of one triangle are equal to the two corresponding angles and the side included between them of another triangle, the two triangles are congruent. RHS Criterion stands for Right Angle-Hypotenuse-Side Criterion. HJ Point G is the midpoint of FH Prove: !FGJ " !HGJ 2. Hence, we can say that they are congruent circles. We can represent this in a mathematical form using the congruent symbol. the angle included by these sides is the same. Thus, two triangles can be superimposed side to side and angle to angle. http://www.superteachertools.com/jeopardyx/jeopardy-review-game.php?gamefile=1322685871, Classifying Triangles In addition, the two triangles "share" side . SSA or ASS Side-Side-Angle. Each day Passy’s World provides hundreds of people with mathematics lessons free of charge. Congruent triangles | passy's world of mathematics. This final game is a Jeapordy Game, but is very slow to load up. Jobs that use Geometry. 1. PayPal does accept Credit Cards, but you will have to supply an email address and password so that PayPal can create a PayPal account for you to process the transaction through. This means that the corresponding sides are equal and the corresponding angles are equal. Practice Questions: 10. Just as ∠ D O G and ∠ C A T, above, were congruent but were not “lined up” with each other, so too can congruent angles appear in any way on a page. A nice open topic to study is to look at when a triangle can be partitioned into k congruent triangles. In a parallelogram opposite angles are congruent. Congruent Triangles - Postulates and Theorems with Examples . The following worksheet has basic multiple choice questions on Congruent Triangles. You may have noticed an ice tray in your refrigerator. Books; Test Prep; Winter Break Bootcamps; Class; Earn Money; Log in ; Join for Free. There will be no processing fee charged to you by this action, as PayPal deducts a fee from your donation before it reaches Passy’s World. Quora. For SAS to work, we need the angle to be included between the two congruent sides. Examples. http://www.mangahigh.com/en_au/maths_games/shape/congruence/congruent_triangles. This last worksheet has challeging multiple choice questions on Congruent Triangles. Similarly, ATM cards issued by the same bank are identical. Get it clarified with simple solutions on The symbol for congruent is ≅. When two items have the exact same size and shape, we say that they are “Congruent”. Have you ever observed that two copies of a single photograph of the same size are identical? The following Video by Mr Bill Konst about Congruence, covers the “SSS Rule for Triangles”, as well as covering Quadrilaterals and some interesting optical illusions. Applications of congruent triangles | ck-12 foundation. It turns out that we do not have to check all the sides and angles of two Triangles to work out that they are Congruent. Click the link below to do this worksheet: The following worksheet has medium level of difficulty multiple choice questions on Congruent Triangles. They are called the SSS rule, SAS rule, ASA rule and AAS rule. She marked $$N$$ as the mid point of the third side. There is a special symbol we use to indicate that Triangles are Congruent, that is like an equals sign with an extra line added on top of it. To find out exactly how free subscription works, click the following link: If you would like to submit an idea for an article, or be a guest writer on our website, then please email us at the hotmail address shown in the right hand side bar of this page. SAS Criterion stands for Side-Angle-Side Criterion. As long … Donate any amount from $2 upwards through PayPal by clicking the PayPal image below. What are some real-life examples of congruent triangles? The geometric figures themselves do not matter. The first of these “Shortcut Rules” is the “Side Side Side”, or “SSS” Rule. Eg. SSS, SAS, ASA, AAS, and HL...all the Theorems are here! Two triangles are congruent when the three sides and the three angles of one triangle have the same measurements as three sides and three angles of another triangle. We love hearing from our Users. Pythagoras Rule means that the missing side lengths have to be equal, so we are indirectly using the “SSS” Rule here. The most common, primary shapes we learn about are triangles. Tips and Tricks: 9. Slides created by . In another lesson, we will consider a proof used for right triangles called the Hypotenuse Leg rule. Image Source: http://resources3.news.com.au. Make your kid a Math Expert, Book a FREE trial class today! Going A to B to C should be the exact same path as D to E to F. Two Triangles will be congruent if two matching sides have equal lengths and Similar Triangles | Passy's World of Mathematics, Geometry in the Animal Kingdom | Passy's World of Mathematics, Jobs With Geometry | Passy's World of Mathematics, Trigonometry – Labeling Triangles | Passy's World of Mathematics, The Sine Ratio | Passy's World of Mathematics, The Cosine Ratio | Passy's World of Mathematics, Trigonometric Ratios | Passy's World of Mathematics, The Tangent Ratio | Passy's World of Mathematics. The next example involves two triangles sharing the diagonal of a Parallelogram. Here are few activities for you to practice. [email protected]. Congruent Angles Examples. Also, $$AB$$ falls on $$PQ$$, $$BC$$ falls on $$QR$$ and $$AC$$ falls on $$PR$$. Congruence is the term used to describe the relation of two figures that are congruent. If two triangles are congruent, then they have to satisfy the following conditions. Select/Type your answer and click the "Check Answer" button to see the result. (Use SAS for this example.) The following video shows some more complex examples, where triangle sides are joined together to other triangles and shapes. Go through the following tips that may help you while proving congruence of triangles. We know that the two triangles have two congruent sides (RS ≅ TS and PS ≅ PS) and one congruent angle (∠R ≅ ∠T). Congruent Triangles. Hence angles ABC and CDA are congruent. Activities, worksheets, projects, notes, fun ideas, and so much more! Exterior Angle of a Triangle In looking through our four postulates, the only one with two sides and one angle is Side Angle Side. This example contains a "definition". In geometry, two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other. Equilateral Triangle. Have you struggled to replace a new refill in an older pen? There is also an old “ASA” Angle Side Angle Rule; however this has been brought in to be part of the “AAS” Rule. CH. This is called “SSS” or the “Side Side Side Rule”. In this article, we will learn about similar triangles, features of similar triangles, how to use postulates and theorems to identify similar triangles and lastly, how to solve similar triangle problems. When we say the Triangles are Congruent using their letters, we need to make sure the order of the letters matches the path around the two triangles correctly. Explore these properties of congruent using the simulation below. Example two triangles are congruent. This Rule works because of Pythagoras Theorem for 90 Degree Triangles. Untitled. Have a doubt that you want to clear? We need to make sure the order of the letters matches going around the two triangles in the same order of sides. 00:07. We can actually use just the three sides to work out if two triangles are congruent. Angles and Parallel Lines Now when we are done with the congruent triangles, we can move on to another similar kind of a concept, called similar triangles.. Our Math Experts focus on the “Why” behind the “What.” Students can explore from a huge range of interactive worksheets, visuals, simulations, practice tests, and more to understand a concept in depth. This means that either object can be repositioned and reflected (but not resized) so as to coincide precisely with the other 2. Nov 25, 2016 - Everything you ever needed to teach Congruent Triangles! The four triangles are congruent with each other regardless whether they are rotated or flipped. Classify Triangles by Sides. This means that the matching sides must be the same length and the matching angles must be the same size. Two triangles with equal corresponding angles may not be congruent to each other because one triangle might be an enlarged copy of the other. The moulds inside the tray that is used for making ice are congruent. More formally, two sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of rigid motions, namely a translation, a rotation, and a reflection. 3 acute angles. Any two Right-Angled 90 degree Triangles are congruent if the hypotenuse and one pair of matching sides are equal in length. Solutions: Yes; The two wheels are both circles, and the distance around them is … Definition Look at ΔABC and ΔPQR below. Congruent from our Math Experts at Cuemath’s LIVE, Personalised and Interactive Online Classes. See Proof. SSS and SAS If two triangles have edges with the exact same lengths, then these triangles are congruent. You can then receive notifications of new pages directly to your email address. IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. If you enjoyed this lesson, why not get a free subscription to our website. In the figure below, △ABC ≅ △DEF. Under this criterion, if the two sides and the angle between the sides of one triangle are equal to the two corresponding sides and the angle between the sides of another triangle, the two triangles are congruent. All congruent sides. Determine whether !PQR is congruent to the other triangles shown at right. In the next two examples, Congruent Triangles are found within the given Geometric Shapes, which allows side lengths to be proven as equal. The Identical (eg.”Congruent”) Triangles can be in different positions, (or orientations), and still be the exact same size and shape. They are the same shape, but are not the same size. Side-Side-Side (SSS) Similarity Theorem If the three sides of a triangle are proportional to the corresponding sides of a second triangle, then the triangles are similar. Two figures are congruent if they have the same shape and size. Find angles in congruent triangles (practice) | khan academy. SSS Criterion stands for Side-Side-Side Criterion. Since the two triangles have two corresponding congruent angles, they are similar. In diagrams, the actual values of the sides are sometimes not given. Two angles are congruent if their measures are exactly the same. The position of the matching Triangles does not affect the fact that they are identical, or “Congruent”. The following “YayMath” 26 minute gives a comprehensive lesson on the ASA and AAS Rules. In the next two examples, Congruent Triangles are found within the given Geometric Shapes, which allows side lengths to be proven as equal. Remember that whenever identical objects are to be produced, the concept of congruence is taken into consideration in making the cast. Two triangles are congruent if two matching angles are equal and a matching side is equal in length. Example: Consider these two equilateral triangles that satisfy the AAA combination. https://www.onlinemathlearning.com/congruent-triangles.html If you enjoyed this lesson, why not get a free subscription to our website. These two triangles are of the same size and shape. If you are a subscriber to Passy’s World of Mathematics, and would like to receive a free PowerPoint version of this lesson, that is 100% free to you as a Subscriber, then email us at the following address: Please state in your email that you wish to obtain the free subscriber copy of the “Congruent Triangles” Powerpoint. Richard Wright, Andrews Academy . In other words, Congruent triangles have the same shape and dimensions. This indicates that the corresponding parts of congruent triangles are equal. He cuts two right-angled triangles out of paper. The congruence of two objects is often represented using the symbol "≅". Such figures are called congruent figures. Under this criterion, if the three sides of one triangle are equal to the three corresponding sides of another triangle, the two triangles are congruent. Here is a drawing that has several angles. This lesson is all about “Congruent Triangles”, eg. Come up with some of your own real-world examples of congruent figures, and explain why they are congruent. Prove that triangle PQR is congruent to triangle ABC. They must have exactly the same three angles. 3. This next one is a heavy metal Parody (sounds a bit like a Joan Jet song): The following examples show the required working out for demonstrating that a pair of Triangles are identical. Congruent triangles have the same size and the same shape. AAS Criterion stands for Angle-Angle-Side Criterion. : All of these angles are congruent. This means $$A$$ falls on $$P$$, $$B$$ falls on $$Q$$ and $$C$$ falls on $$R$$. In addition, Triangles are usually labelled with capital alphabet letters. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. The American teacher doing the videos does not always use the most correct language, but he is enthusiastic and explains his examples well. Hence sides BC and AD are congruent, and also sides AB and CD are congruent. Angles opposite to equal sides are equal. 3. Congruent triangles. When triangles are congruent, all corresponding sides and corresponding angles are also congruent or equal For examples, the two triangles below are congruent Corresponding angles are angles in the same position Pingback: Similar Triangles | Passy's World of Mathematics, Pingback: Geometry in the Animal Kingdom | Passy's World of Mathematics, Pingback: Jobs With Geometry | Passy's World of Mathematics, Pingback: Trigonometry – Labeling Triangles | Passy's World of Mathematics, Pingback: The Sine Ratio | Passy's World of Mathematics, Pingback: The Cosine Ratio | Passy's World of Mathematics, Pingback: Trigonometric Ratios | Passy's World of Mathematics, Pingback: The Tangent Ratio | Passy's World of Mathematics. Geometry The equal sides and angles of two congruent triangles can be read from a congruence correspondence between them. See more ideas about teaching geometry, teaching math, geometry high school. And what I want to do in this video is figure out which of these triangles are congruent to which other of these triangles. http://illuminations.nctm.org/ActivityDetail.aspx?ID=4. Parts of Congruent Triangles 81 Practice Problems View More. Two triangles are congruent if they are completely identical. She drew an isosceles triangle $$PQR$$ on a page. Two sides and an included angle of triangle ABC are cong… Draw two circles of the same radius and place one on another. Congruent Triangles: Examples: 8. 2. We at Cuemath believe that Math is a life skill. The following is an activity where we get to build congruent triangles based on the congruency rule we pick to work with. More challenging examples belong to the domain of puzzles. (There are answers on the last page of the PDF document). Which of these angles are congruent? Also in how far doors swing open. Now, let's learn about the meaning of congruent triangles with Cuemath. . Scalene Triangle. 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https://math.stackexchange.com/questions/2283842/how-to-find-the-most-upright-orthogonal-vector-to-another-vector | # How to find the “most upright” orthogonal vector to another vector?
Given an arbitrary vector, I can find any orthogonal vector by solving $ax + by + cz = 0$.
I want to find the "most upright" orthogonal (unit) vector, where $z$ is maximized. There must be a straightforward closed form solution to this, right?
Using the observation made by dtldarek in his answer, you’re looking for the unit vector in the plane generated by $w=(a,b,c)$ and $e_3=(0,0,1)$ that is perpendicular to $w$ and has a non-negative $z$-coordinate. Observe that such a vector is also perpendicular to $e_3\times w$, so the vector you seek is $w\times(e_3\times w)$, normalized.
• Elegant! $\ddot\smile$ – dtldarek May 16 '17 at 19:56
• Yup, I have to say, that was pretty darn simple, and so obvious in retrospect. This one will be tucked away in my toolkit forever after. Thank you! – David Parks May 16 '17 at 21:17
• This one is actually used a bunch in the 3d graphics world as part of creating orthonormal matrices which represent the effects of looking in a direction. – Cort Ammon May 17 '17 at 0:41
• @CortAmmon Yep. That’s what led me to look at the problem from this point of view, so to speak. – amd May 17 '17 at 0:43
• This was exactly my use case, I needed to compute the point of gaze as the head rotates left/right while focused at an angle. This vector provides the axis of rotation which then allowed me to apply Rodrigues' rotation formula to perform the rotation. – David Parks May 17 '17 at 17:20
Restating your problem (and adding a condition), you are given a vector $\vec{v}=\langle a,b,c\rangle$ and you want to find a vector $\vec{w}=\langle x,y,z\rangle$ so that
1. $\vec{w}\cdot\vec{v}=0$,
2. $\|\vec{w}\|=1$
3. $\vec{w}\cdot\vec{e}_3$ is maximized
As you state, you want to maximize $z$ subject to \begin{align*} ax+by+cz&=0\\ x^2+y^2+z^2&=1 \end{align*}
Perhaps a Lagrange multiplier approach would be good here: \begin{align*} 0&=\lambda a+2\mu x\\ 0&=\lambda b+2\mu y\\ 1&=\lambda c+2\mu z\\ 0&=ax+by+cz\\ 1&=x^2+y^2+z^2 \end{align*}
We should really assume that $a$ and $b$ are not both zero because otherwise $\vec{v}$ points vertically and there are no orthogonal, upward pointing vectors.
• If $\mu=0$, then we get that $\lambda=\frac{1}{c}$ (and $c\not=0$), but then the first two equations are $0=\frac{a}{c}$ and $0=\frac{b}{c}$, so $a=0=b$ This contradicts our assumption, so we're ok here.
• If $\mu\not=0$, then we can solve for $x$, $y$, and $z$: \begin{align*} x&=-\frac{\lambda a}{2\mu}\\ y&=-\frac{\lambda b}{2\mu}\\ z&=\frac{1-\lambda c}{2\mu} \end{align*} We can substitute these into the fourth equation (and cancel the $2\mu$'s) to get $$-\lambda a^2-\lambda b^2-\lambda c^2+c=0.$$ Now, we make things simpler by adding the assumption that $\langle a,b,c\rangle$ is a unit vector, so we could rewrite this equation as $c=\lambda$. If not, we use $c=\lambda\|\vec{v}\|$ throughout the rest of this problem.
Therefore, \begin{align*} x&=-\frac{ac}{2\mu}\\ y&=-\frac{bc}{2\mu}\\ z&=\frac{1-c^2}{2\mu} \end{align*} Substituting all of this into the final equation gives $$4\mu^2=a^2c^2+b^2c^2+1-2c^2+c^4=(a^2+b^2+c^2)c^2+1-2c^2.$$ Therefore $$4\mu^2=1-c^2$$ or $$\mu=\pm\sqrt{\frac{1-c^2}{4}}$$ Substituting these into the formulae for $x,y,z$ gives that \begin{align*} x&=\mp\frac{ac}{\sqrt{1-c^2}}\\ y&=\mp\frac{bc}{\sqrt{1-c^2}}\\ z&=\pm\frac{1-c^2}{\sqrt{1-c^2}} \end{align*} Depending on the signs, you should get the largest and smallest that $z$ could be (provided I made no errors).
• Ahhh, lagrange multipliers... there's still scar tissue in my brain from the last time I had to deal with them. I'm really happy to see a straightforward application that I can get my head around. This is useful beyond the particular problem I'm solving right now. Thank you! – David Parks May 16 '17 at 18:32
• @amd Yes, I added that assumption, but it doesn't really matter since orthogonality is independent of length. So, if you wanted to drop it, you would have $c=\lambda\|v\|$, and carry this through the computation. – Michael Burr May 16 '17 at 19:55
• @amd Agreed, I made a change to the text. – Michael Burr May 16 '17 at 20:54
Assuming $\mathbb{R}^3$, a geometric approach would be to observe that the solution would belong to plane generated by $(a,b,c)$ and $(0,0,1)$. That forces the ratio between $x$ and $y$ to be the same as between $a$ and $b$. In other words, we have two equations:
$$\begin{cases}ax+by+cz=0\\ay-bx=0\end{cases}$$
If $c = 0$, then $(0,0,z)$ is your solution for any $z>0$. Otherwise, if $x=ka$ and $y=kb$, then $k(a^2+b^2)+cz=0$ and $z=-k\frac{a^2+b^2}{c}$. To maximize $z$ make sure $z\geq 0$, that is, $k\cdot c \leq 0$.
I hope this helps $\ddot\smile$
• Might be worth noting that the second equation in the system is that of the plane spanned by $(a,b,c)$ and the $z$-axis. – amd May 16 '17 at 22:14 | 2021-04-19T03:21:30 | {
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https://www.physicsforums.com/threads/function-must-be-a-bijection-for-its-inverse-to-exist.586095/ | # Function must be a bijection for its inverse to exist?
1. Mar 12, 2012
### Bipolarity
My analysis text defines inverse functions only for bijections.
But $y = e^{x}$ is not bijective, so according to my book it's inverse ($ln x$) wouldn't be defined? Am I missing something or is my textbook just plain wrong?
I use the text by Bartle and Sherbert.
Thanks!
BiP
2. Mar 12, 2012
### Staff: Mentor
As I recall it, a function needs only to be injective (one-to-one), but doesn't have to be surjective (onto). Your example of f(x) = ex is one-to-one, and does have an inverse. Likewise, g(x) = ln(x) is one-to-one, and has an inverse as well.
I'd be interested to see why your text might argue that ex and ln(x) don't have inverses.
3. Mar 12, 2012
### Bipolarity
I see. That's what I also would expect, but wiki seems to go against :
http://en.wikipedia.org/wiki/Inverse_function
When using codomains, the inverse of a function ƒ: X → Y is required to have domain Y and codomain X. For the inverse to be defined on all of Y, every element of Y must lie in the range of the function ƒ. A function with this property is called onto or a surjection. Thus, a function with a codomain is invertible if and only if it is both injective (one-to-one) and surjective (onto). Such a function is called a one-to-one correspondence or a bijection, and has the property that every element y ∈ Y corresponds to exactly one element x ∈ X.
Perhaps there are many types of definitions for inverses that depend on the context?
BiP
4. Mar 12, 2012
### morphism
e^x is a bijection - from the set of real numbers onto the set of positive real numbers. As such it has an inverse function going from the positive reals to the reals - namely, ln(x). Note that ln(x) is also a bijection (from the positive reals onto the reals).
Whenever you talk about a function, you should always keep its domain and codomain in mind. If your function f is defined from X to Y (i.e. you have a function f:X->Y), then an inverse should be a function g:Y->X (note the domain and codomain) such that g(f(x))=f(g(x))=x. This happens if and only if f:X->Y is a bijection.
5. Mar 12, 2012
### mathwonk
surely you have studied functions like arcsin, inverse of sin. since functions must be bijective to have inverses, one must restrict the domain and range until this happens.
i.e. to make sin injective, first we restrict the domain to be [-pi,pi]. Then to make it surjective we restrict the range to be [-1,1].
Then sin has a continuous inverse arcsin, defined on [-1,1] with range in [-pi,pi].
Then if we also want the inverse to be differentiable, we must exclude points where the derivative of sin is zero, so we restrict the domain of sin further to (-pi,pi). Then we have arcsin, a differentiable inverse of sin, defined on (-1,1).
Don't they teach this properly in calculus? precalculus? Well to be honest I think our book also did not do so. But that's what professors are for. Maybe the thinking was that freshmen are too young to be frightened by french derived words like injective. But one to one and onto work almost as well.
6. Mar 13, 2012
### Bacle2
It is possible to have only 1-sided inverses, which will be the case if your function is injective but not bijective, and viceversa. If f is 1-1 , then it will have a left inverse g,
i.e., g is such that gof=Id , i.e., gof(x)=x.
Basically, you define g(x) to be the left-inverse, i.e., g(f(x)):=x , for those elements in the range, and define it anyway you want for those elements not in the range.
e.g., take the map from the non-negative integers to themselves, that doubles every number, i.e., f: n-->2n . Then n is 1-1 , but not onto (misses all odd numbers). Define, then g: 2n->n , for even numbers in Z, and any other way for odd numbers. Then gof(n)=n .
A similar argument shows that if f:X-->Y is onto, then f has a right-inverse, i.e., there is h(y) with h:Y-->X , and foh(y)=y. Basically, for all y in Y,the fiber { f^{-1}(y)} of y is non-empty. Map, then , h:y-->yo , i.e., each element of y into an element yo of its fiber*, and , then, almost tautologically, foh(y)=f(yo)=y. Think of , e.g., Sinx from
the reals into [-1,1]. So f(x)=sinx . Define h to send yo in [-1,1] into some element in its fiber :{Sin^{-1}(yo)} .
* Note: if you have not done set theory, feel free to ignore the following:To be more rigorous, you need the axiom of choice to define the right-inverse (not sure about the left inverse; it is too late, i.e., to choose an element in each fiber, when X,Y are infinite). If you're not seeing this, it is not necessary for your understanding.
7. Mar 15, 2012
### zetafunction
you can ALWAYS define the inverse of a function $$y=f(x)$$
take the points $$(x,f(x))$$ and make a 'reflection' of these points alongside the line
$$y=x$$ you will get the NUMERICAL inverse of the function.
8. Mar 15, 2012
### jgens
This procedure does not always give you a function though :/
9. Mar 16, 2012
### lavinia
In order to have an inverse you must have a bijection.
It you think of e$^{x }$as a map from the real numbers into the real numbers then it has no inverse since negative numbers are not exponentials. But if you think of
e$^{x }$ as a map from the real numbers to the positive numbers then it does have an inverse. | 2017-08-18T02:32:44 | {
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https://math.stackexchange.com/questions/2331163/derivative-of-absolute-function/2331181 | # Derivative of absolute function
In the video lecture that I am watching, the instructor compared an example of finding the derivative of the $f(x) = |x|$ vs $f(x) = x|x|$, where $f(x) = x|x|$, derivative is $0$, whereas $f(x) = |x|$, does not have a derivative.
The instructor highlighted that the absolute value function does not have a derivative compared to $f(x) = x|x|$.
If I would to apply the power rule: $f(x) = nx^{n-1}$, where $f(0) = 1*0^{1-1}$. Because 0 to the power of 0 is undefined.
Is it because $0$ to the power of $0$ is undefined, therefore the absolute value function does not has a derivative?
• The absolute value function has a derivative(s) on restricted domains. i.e. f'(x) = -1 for x <0 and f'(x) = 1 for x > 0. However, the absolute value function is not "smooth" at x = 0 so the derivative at that point does not exist. Jun 21, 2017 at 14:14
• The power rule only applies to power functions. Neither $x \mapsto |x|$ nor $x \mapsto x|x|$ are power functions. So it isn't relevant here—you have to go back to the definition to decide if the function has a derivative. Jun 21, 2017 at 14:23
• Possible duplicate of math.stackexchange.com/q/839293/294695 Jun 21, 2017 at 14:23
• @MatthewLeingang in this case the exponent of X must not be equal to zero? x^2 or x^-2 should be a valid power function? Jun 24, 2017 at 10:30
• Yes, $x^2$ and $x^{-2}$ are valid power functions. But your question was about $|x|$ and $x|x|$, right? Those are not power functions. You could insert exponents, but it would have to be $1$, not $0$. That doesn't help with your calculation, though. Applying the power rule to $|x|^1$ gives $1$, times $|x|^0$ (also $1$), times the derivative of $|x|$. Jun 25, 2017 at 10:24
Recall the definition of the derivative as the limit of the slopes of secant lines near a point.
$$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
The derivative of a function at $x=0$ is then
$$f'(0) = \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}= \lim_{h\to 0}\frac{f(h)-f(0)}{h}$$
If we are dealing with the absolute value function $f(x)=|x|$, then the above limit is
$$\lim_{h\to 0}\frac{|h|-|0|}{h} = \lim_{h\to 0}\frac{|h|}{h}$$
If $h$ approaches $0$ from the left, it is negative, so that $|h| = -h$ and the above limit is $-1$. But if $h$ approaches $0$ from the right, it is positive, so that $|h|=h$ and the limit is $1$. The left and right hand limits disagree, so the derivative does not exist at $0$.
• Thank you so much for your clear explanation. Now I am able to understand the rational behind this. Jun 22, 2017 at 13:28
• But, is my understanding and application of the power rule correct? Jun 25, 2017 at 2:46
First compare the graphs:
## $x|x|$
$|x|$ is very sharp at $(0,0)$, so it doesn't have a derivative there.
$x|x|$ has derivatives because $x|x|$ is equal to:
$\begin{cases}y=x^2 & x > 0\\y=0 & x = 0\\y=-x^2& x<0\end{cases}$
And "square" makes $y = x^2$ tangent same of the $y=-x^2$. Same tangents = differentiable.
See those:
Source of tangent animations: Alice Ryhl (https://math.stackexchange.com/users/132791/alice-ryhl), Why is the absolute value function not differentiable at $x=0$?, URL (version: 2014-10-26): https://math.stackexchange.com/q/991559
• Thank you for your animations. It definitely helps a lot in understanding. Jun 25, 2017 at 5:25
• @youcanlearnanything No, they are not mine. See this this Jun 25, 2017 at 10:21
I know you've already accepted an answer on how to find the derivatives of these functions using the definition, but you also have an interesting question on the power rule and how it relates. You asked in the comments above if we could write $$f(x) = x^1 |x^1|$$ and apply the power rule.
Starting with a more basic function, look at $$x \mapsto x$$, which you can write as $$x\mapsto x^1$$. The power rule applied to this function gives $$\frac{d}{dx} x^1 = 1 \cdot x^0 = 1 \cdot 1 = 1$$ This comports with our by-the-definition computation. The two things that prevent applying the power rule directly to $$f(x) = x^1 |x^1|$$ are the product rule and the chain rule.
I don't know if you've learned these yet. In case not: the product rule is the rule by which we can compute derivatives of products of functions. If $$f(x) = g(x) h(x)$$, you might hope that $$f'(x) = g'(x) h'(x)$$. If that were true, differentiating products would be as simple as differentiating sums. But that is not the case. Instead: $$f'(x) = g'(x)h(x) + g(x) h'(x)$$ So even on a product of power functions you can't just take the derivative of each factor. The chain rule is for differentiating a composition function. If $$h(x) = k(\ell(x))$$, you might try $$h'(x) = k(\ell'(x))$$, but in fact that's not true. Instead: $$h'(x) = k'(\ell(x)) \ell'(x)$$
So if you want to write $$f(x) = x^1 |x^1|$$, you can, but it's still not a power function. It's a product of two functions. The first is a power function, but the second is the composition of the absolute value function with a power function. If $$g(x) = \ell (x) = x$$, and $$k(x) = |x|$$, then $$f(x) = g(x)\cdot k(\ell(x))$$
We need the derivative of the absolute value function $$k(x) = |x|$$.
As you have seen from the definition: $$k'(x) = \frac{d}{dx} |x| = \begin{cases} 1 & \text{if x>0;}\\ -1 & \text{if x<0.} \end{cases} = \frac{|x|}{x}$$ For the rest of the derivative, we need the product and chain rules: $$f'(x) = g'(x) k(\ell(x)) + g(x) k'(\ell(x)) \ell'(x)$$ We know $$g'(x) = \ell'(x) = 1$$, so $$g'(x) = 1\cdot |x| + x \cdot \left(\frac{|x|}{x}\right) \cdot 1 = 2 |x|$$
• Thank you for your clear explanation. It took me some time to revisit those rules/topics again and came back to your post for a review. =) Jul 2, 2017 at 11:59
if you meant $$f(x)=x|x|$$ you can compute $$\frac{f(x_0+h)-f(x_0)}{h}=\frac{h|h|}{h}=|h|$$ for $$h\ne 0$$ and $$x_0=0$$ therefore $$\lim_{h\to 0}|h|=0$$ the function has a derivative in $$x=0$$ | 2022-07-03T15:54:17 | {
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https://math.stackexchange.com/questions/1953304/asymptotic-formula-for-nth-square-free | # Asymptotic formula for $n^{th}$ square-free
I face an exercise : Find asymptotic formular for the $n^{th}$ square-free number with the error $O(n^{1/2}).$
I don't quite understand the question. What is the $n^{th}$ square-free number ? I guess it is primorial ? Like $2$ is the first, $2 \times 3$ second, $2 \times 3 \times 5$ third , ... (although I do not sure if it includes the number like $2 \times 5$ since $3$ is missing, and if it includes, what should be the third square-free number ? $2 \times 3 \times 5$ or $2 \times 5$.)
Let assume that I want to find an asymptotic formula for $p_1...p_k$ where $p_k$ is the $k^{th}$ prime. So I need to find $g$ such that $$\lim_{x \rightarrow \infty} \frac{\prod_{p \leq x} p}{g(x)} = 1.$$
Basically, asymptotic $\sim$ can be related to little $o$, but this question mention something about error term $O(n^{1/2})$. How can asymptotic relate to big $O$ ?
That makes me confused.
For the method, I see that the primorial can be related to $e^{\theta(x)}$, where $\theta$ is the Chebyshev function. But I do not know how to do $\sim$ with big $O(n^{1/2}).$
• The $n$-th primorial is around $e^n$ by the PNT. The $n$-th square-free number, on the other hand, is around $\frac{\pi^2}{6}n$. The difference between such magnitudes is huge! – Jack D'Aurizio Oct 4 '16 at 19:14
• Maybe first solve the inverse problem: How many numbers $\le N$ are square-free? From originally $N$, you need to subtract about $\frac N4$ numbers divisible by $2^2$, as ell as about $\frac N9$ numbers divisible by $3^2$ - but then you subtracted those divisible by $6^2$ twice, so add back $\frac N{36}$. And so on. You should end up with abou $N(1-\frac 14)(1-\frac 19)(1-\frac 1{25})\cdots(1-\frac 1{p^2})$ with $p\approx \sqrt N$. – Hagen von Eitzen Oct 4 '16 at 20:04
By definition of the Möbius function we have that $n$ is squarefree iff $\mu^2(n)=1$. Thefore, the function $\eta(x)=\sum_{n\leq x}\mu(n)^2$ counts the number of squarefree integers less or equal to $x$. Now recall the following elementary property of $\mu^2$:
$$\mu^2(n)=\sum_{d^2|n}\mu(d),$$
where the sum is extended over all the divisors $d$ of $n$ such that $d^2|n$. Thus,
$$\eta(x)=\sum_{n\leq x}\mu^2(n)=\sum_{n\leq x}\sum_{d^2|n}\mu(d).$$
Write $n=d^2q$. Then $n\leq x$ iff $d^2\leq x$ (i.e., $d\leq\sqrt x$) and $q\leq x/d^2$, so
$$\sum_{n\leq x}\sum_{d^2|n}\mu(d)=\sum_{d\leq\sqrt x}\sum_{q\leq x/d^2}\mu(d)=\sum_{d\leq\sqrt x}\mu(d)\left(\frac x{d^2}+O(1)\right) =x\sum_{d\geq1}\frac{\mu(d)}{d^2}-x\sum_{d\geq\sqrt x}\frac{\mu(d)}{d^2}+O(\sqrt x).$$
But
$$\sum_{d\geq\sqrt{x}}\frac{\mu(d)}{d^2}=O\left(\sum_{d\geq\sqrt x}\frac1{d^2}\right)=O\left(\frac1{\sqrt{x}}\right)$$
so
$$x\sum_{d\geq1}\frac{\mu(d)}{d^2}-x\sum_{d\geq\sqrt x}\frac{\mu(d)}{d^2}+O(\sqrt x) =x\sum_{d\geq1}\frac{\mu(d)}{d^2}+O(\sqrt{x}).$$
Finally, since $\sum_{d\geq1}\frac{\mu(d)}{d^2}=1/\zeta(2)$, we conclude that
$$\eta(x)=\frac{x}{\zeta(2)}+O(\sqrt x)=\frac{6x}{\pi^2}+O(\sqrt x).\tag{*}\label{eq:1}$$
EDIT: from the above we have that $\eta(x)\sim \frac{x}{\zeta(2)}$, because
$$\eta(x)=\frac{x}{\zeta(2)}+O(\sqrt x)=\frac{6x}{\pi^2}+o(x)=\frac{6x}{\pi^2}(1+o(1))$$
since $\sqrt{x}=o(x)$ (i.e., $\lim_{x\to\infty}\sqrt{x}/x=0$.) Nevertheless, we have proved something better. Let us write \eqref{eq:1} in the following equivalent form:
$$\frac{\eta(x)}{x/\zeta(2)}=1+O\left(\frac1{\sqrt{x}}\right).$$
We see that the function $\frac{\eta(x)}{x/\zeta(2)}$ approaches to $1$ at least as fast as $\frac1{\sqrt{x}}$ approaches to zero. This provides us more information than simply saying that $\frac{\eta(x)}{x/\zeta(2)}$ converges to $1$. For example, consider the question "What is the value of $\lim_{x\to\infty}\sqrt[3]{x}(\frac{\eta(x)}{x/\zeta(2)}-1)$?" You can answer it if you know that $\frac{\eta(x)}{x/\zeta(2)}-1=O(1/\sqrt{x})$.
• Okay, then $$\eta(n) \sim \frac{6x}{pi^2} ?$$ I do not unders tand when I want to solve "find asymptotic formular" it refers to $\sim$ or can be in the form $$f(x) + O(g(x))$$ for some functions $f, g.$ – Both Htob Oct 4 '16 at 21:58
• @BothHtob come on, with $A$ the set of squarefree numbers, $\eta(x) = \displaystyle\sum_{n < x, n \in A} 1 \sim x/\zeta(2)$ so you get that $\eta(x) = k \implies x \sim k \zeta(2)$ and this is the asymptotics you want for the $k$th squarefree number – reuns Oct 5 '16 at 2:11
• @user1952009 Sorry, but I really do not get what are you doing. From the result above, how do you obtain $$\eta(x) \sim \frac{x}{\zeta(2)} ?$$ I try to understand why $$\lim_{x \rightarrow \infty} \eta(x)/(x/\zeta(2)) = 1 ?$$ That $O(x^{1/2})$ does not goes to $1$ as $x \rightarrow \infty$. I do not understand what is going on. Please be patient, I am very new to this subject. These estimate stuff $O, o, \sim$ are new to me too. – Both Htob Oct 5 '16 at 2:40
• @BothHtob $\eta(x)$ is the number of square free numbers $< x$. Iqcd showed in quite an elementary way that $\eta(x) = \frac{x}{\zeta(2)} + \mathcal{O}(\sqrt{x})$ that is there is a $C$ such that $|\eta(x)-\frac{x}{\zeta(2)}| < C \sqrt{x}$ implying $\frac{\eta(x)}{x/\zeta(2)} \to 1$ i.e. $\eta(x) \sim \frac{x}{\zeta(2)}$. So you should work on the asymptotic comparison $\sim,o,\mathcal{O}$ – reuns Oct 5 '16 at 2:50
• @BothHtob to be clear $f(x) \sim g(x)$ is the same as $f(x) = g(x) + o(g(x))$ and $f(x) = a x + \mathcal{O}(\sqrt{x})$ is much stronger than $f(x) = a x+o(x)$ itself the same as $f(x) \sim a x$ – reuns Oct 5 '16 at 2:55
Hint:
$\mu(n)^2$, where $\mu(n)$ is the Moebius function, is the characteristic function of square-free numbers.
By convolution of Dirichlet series and summation by parts, it is not difficult to show that $$\sum_{n\leq x}\mu(n)^2 \approx \frac{x}{\zeta(2)} = \frac{6x}{\pi^2} \tag{1}$$ i.e. the square-free integers form a subset of the natural numbers with positive density. By $(1)$, it follows that the $n$-th square-free natural number is close to $\color{red}{\frac{\pi^2 n}{6}}$. How close depends on a explicit error term for $(1)$.
• Can you explain when I encouter "asymototic formula" it means to $\sim$ ? The book does not give a definition about it. So I dont know what is it actually. Wikipedia provides something (en.wikipedia.org/wiki/Asymptotic_formula) but normally big $O$ has nothing to do with $\sim.$ This confusing make me not sure where should I start. Is this kind of question just want to find estimate for a sum with error terms, regradless of its form $\sim$, big $o$, littlt $o$. Just choose one is fine ? – Both Htob Oct 4 '16 at 22:08
• @BothHtob: I was a bit vague on purpose here. $a\approx b$ above just means that $a$ is close to $b$; I leave to you the task to replace $(1)$ by something with some rigor like $\sum_{n\leq x}\mu(n)^2=\frac{x}{\zeta(2)}+O(\sqrt{x})$, then prove your claim. Which approximations are fine for our purposes? is another question left open for you on purpose. – Jack D'Aurizio Oct 5 '16 at 16:15
A square-free number is one with no square divisors (other than 1).
They're not the same as primorial numbers: 3 is square-free but not primorial. Here's a list of the first square-free numbers, at the On-Line Encyclopedia of Integer Sequences.
Does that get you unstuck?
• Then how should I represent it ? Just $$n|\mu(n)| ?$$ Well, is there a formular for it ? – Both Htob Oct 4 '16 at 12:08
• But that formula does not help anything in finding asymptotic, or I overlook some important information ? – Both Htob Oct 4 '16 at 12:33 | 2019-11-19T14:00:45 | {
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http://www.danaetobajas.com/uutjb/how-to-find-number-of-edges-in-a-graph-180e55 | how to find number of edges in a graph
23303
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# how to find number of edges in a graph
## how to find number of edges in a graph
If we keep … An edge joins two vertices a, b and is represented by set of vertices it connects. In every finite undirected graph number of vertices with odd degree is always even. In maths a graph is what we might normally call a network. It is a Corner. We remove one vertex, and at most two edges. Since for every tree V − E = 1, we can easily count the number of trees that are within a forest by subtracting the difference between total vertices and total edges. Also Read-Types of Graphs in Graph Theory . That's $\binom{n}{2}$, which is equal to $\frac{1}{2}n(n - 1)$. Let us look more closely at each of those: Vertices. In mathematics, a graph is used to show how things are connected. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. A tree edge uv with u as v’s parent is a cut edge if and only if there are no edges in v’s subtree that goes to u or higher. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. We need to add edges until making it a triangle, use equation $|E'| \le 3|V'| -6$ which is valid for triangles then remove the edges and find that for the new graph $|E| \le 3|V| - 6$ is a valid inequality. What's the most edges I can have without that structure?) - This house is about the same size as Peter's. Now let’s proceed with the edge calculation. Use graph to create an undirected graph or digraph to create a directed graph.. See your article appearing on the GeeksforGeeks main page and help other Geeks. PRACTICE PROBLEMS BASED ON COMPLEMENT OF GRAPH IN GRAPH THEORY- Problem-01: A simple graph G has 10 vertices and 21 edges. Print Postorder traversal from given Inorder and Preorder traversals, Construct Tree from given Inorder and Preorder traversals, Construct a Binary Tree from Postorder and Inorder, Construct Full Binary Tree from given preorder and postorder traversals, Dijkstra's shortest path algorithm | Greedy Algo-7, Prim’s Minimum Spanning Tree (MST) | Greedy Algo-5, Kruskal’s Minimum Spanning Tree Algorithm | Greedy Algo-2, Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming), Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph), Minimum number of swaps required to sort an array, Write Interview In every finite undirected graph number of vertices with odd degree is always even. A vertex is a corner. For that, Consider n points (nodes) and ask how many edges can one make from the first point. All edges are bidirectional (i.e. For the inductive case, start with an arbitrary graph with $$n$$ edges. We use The Handshaking Lemma to identify the number of edges in a graph. There is an edge between (a, b) and (c, d) if |a-c|<=1 and |b-d|<=1 The number of edges in this graph is . A complete graph with n nodes represents the edges of an (n − 1)-simplex.Geometrically K 3 forms the edge set of a triangle, K 4 a tetrahedron, etc.The Császár polyhedron, a nonconvex polyhedron with the topology of a torus, has the complete graph K 7 as its skeleton.Every neighborly polytope in four or more dimensions also has a complete skeleton.. K 1 through K 4 are all planar graphs. For the above graph the degree of the graph is 3. Here are some definitions of graph theory. generate link and share the link here. Print Binary Tree levels in sorted order | Set 3 (Tree given as array) ... given as array) 08, Mar 19. Find total number of edges in its complement graph G’. (iii) The Handshaking theorem: Let be an undirected graph with e edges. Dividing … A bridge is defined as an edge which, when removed, makes the graph disconnected (or more precisely, increases the number of connected components in the graph). Each edge connects a pair of vertices. close, link In graph theory, a cut is a partition of the vertices of a graph into two disjoint subsets.Any cut determines a cut-set, the set of edges that have one endpoint in each subset of the partition.These edges are said to cross the cut. After adding edges to make all faces triangles we have $|E'| \le 3|V'| -6$ where $|E'|$ and $|V'|$ are the number of edges and vertices of the new triangulated graph. Let's say we are in the DFS, looking through the edges starting from vertex v. The current edge (v,to) is a bridge if and only if none of the vertices to and its descendants in the DFS traversal tree has a back-edge to vertex v or any of its ancestors. No vertex attributes. (ii) The degree sequence of a graph is the sequence of the degrees of the vertices of the graph in non – increasing order. Order of graph = Total number of vertices in the graph; Size of graph = Total number of edges in the graph . The total number of edges in the above complete graph = 10 = (5)* (5-1)/2. brightness_4 This tetrahedron has 4 vertices. 1 $\begingroup$ This problem can be found in L. Lovasz, Combinatorial Problems and Exercises, 10.1. idxOut = findedge (G,s,t) returns the numeric edge indices, idxOut, for the edges specified by the source and target node pairs s and t. The edge indices correspond to the rows G.Edges.Edge (idxOut,:) in the G.Edges table of the graph. In a complete graph, every pair of vertices is connected by an edge. To find the total number of spanning trees in the given graph, we need to calculate the cofactor of any elements in the Laplacian matrix. Good, you might ask, but why are there a maximum of n(n-1)/2 edges in an undirected graph? Now we have to learn to check this fact for each vert… An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Consider two cases: either $$G$$ contains a cycle or it does not. In a connected graph, each cut-set determines a unique cut, and in some cases cuts are identified with their cut-sets rather than with their vertex partitions. Note the following fact (which is easy to prove): 1. Its cut set is E1 = {e1, e3, e5, e8}. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. On the other hand, if it has seven vertices and 20 edges, then it is a clique with one edge deleted and, depending on the edge weights, it might have just one MST or it might have literally thousands of them. You can take $$n = e = 1$$ as your base case. That is we can prove that for all $$n\ge 0\text{,}$$ all graphs with $$n$$ edges have …. As special cases, the order-zero graph (a forest consisting of zero trees), a single tree, and an edgeless graph, are examples of forests. We are given an undirected graph. The length of idxOut corresponds to the number of node pairs in the input, unless the input graph is a multigraph. Find the number of edges in the bipartite graph K_{m, n}. code. 02, May 20. Also Read-Types of Graphs in Graph Theory . For example, in above case, sum of all the degrees of all vertices is 8 and total edges are 4. It consists of a collection of nodes, called vertices, connected by links, called edges.The degree of a vertex is the number of edges that are attached to it. Note that each edge here is bidirectional. The degree sum formula says that if you add up the degree of all the vertices in a (finite) graph, the result is twice the number of the edges in the graph. graphs combinatorics counting. The Handshaking Lemma − In a graph, the sum of all the degrees of all the vertices is equal to twice the number of edges. Pick an arbitrary vertex of the graph root and run depth first searchfrom it. | 2021-03-02T11:30:11 | {
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https://cs.stackexchange.com/questions/103255/worst-case-complexity-in-terms-of-n | # Worst-case complexity in terms of n?
I have this algorithm taken out from the Manual of Algorithms Analysis:
function pesky (n)
r:=0
for i:=1 to n do
for j:=1 to i do
for k:= j to i + j do
r := r + 1
return (r)
I understand the time complexity of this algorithm for any given input size n > 0 will be O(n^3) since the 3 for loops will always be executed and the inner loop carries out the sum operation n x n x n times (this is of course not exact, but I say it in asymptotic terms taking out constants and lower degree terms of n); however, during a lecture we were asked to express the complexity of the worst-case in terms of n instead of in big oh notation (they left this exercise for next session).
The worst case I can think about for this algorithm would a really large integer as input n which in fact could be so large that it approaches infinity (since no restrictions for the input were stated). Even in this worst case we know time complexity would behave in O(n^3), but how would I express this "in terms of n", the best thing I could come up with is this:
But I feel like I'm just making that up and that there might be another more correct way to express it. Does anyone know the correct way to express worst-case complexity in terms of n?
EDIT: I must make it clear that I found the equation that puts n in function to the r output by means of summation resolving, however I did not consider that function as the time complexity function (and hence not the answer for "worst-case in terms of n") since that function describes output on a given input and not the time it takes to run the algorithm itself.
Moreover I had the idea to try to calculate the complexity counting each operation and I got something like this.
I am not sure if thats correct, if it is I could use that expression to state the worst-case complexity in terms of n, and also I could make my initial statement about big oh notation for the worst case more precise since this would be O(n).
• What you mean to write is probably $W(n)\rightarrow \infty$ as $n\rightarrow\infty$. This is true, but it is true for almost any non-trivial algorithm. So it does not tell a lot about this algorithm. In essence, big-O notation is more specific about what happens when $n$ is large. Have another read on what big-O notation means intuitively or formally and then look back at your question. – Discrete lizard Jan 22 '19 at 21:39
• Yes I do know big-O is better for describing worst-case, however we were asked literally "give worst-case in terms of n without using big big-O notation" and I couldn't figure it out after looking for an answer on the internet. – MikeKatz45 Jan 22 '19 at 21:43
• "The inner loop carries out the sum operation $n \times n \times n$ times." That is not correct. Can you take a close look? – John L. Jan 22 '19 at 21:45
• I think what was meant is to give the exact amount of times an 'elementary operation' is executed in the worse case. So, they likely want you to be more precise by giving a function such as $3\cdot n^3 + n -1$ (I made that up, this is likely not your answer). – Discrete lizard Jan 22 '19 at 21:45
• @Apass.Jack oh dang I didn't pay attention to this example's loops (in class we used a different one), is it 2n instead? – MikeKatz45 Jan 22 '19 at 22:00
Let us interpret for i:=1 to n inclusively. That is, i will take values $$1, 2, \cdots, n$$.
Let us check how many operations are done in pesky(n).
r:=0: 1 assignment.
for i:=1 to n do: $$n$$ assignments of i.
for j:=1 to i do: $$\sum_{i=1}^ni=\dfrac{n(n+1)}2$$ assignments of j.
for k:= j to i + j do: $$\Sigma_{i=1}^n\Sigma_{j=1}^i(i+1)=\dfrac{n(n+1)(n+2)}3$$ assignments of k.
r := r + 1: $$\Sigma_{i=1}^n\Sigma_{j=1}^i(i+1)=\dfrac{n(n+1)(n+2)}3$$ additions by 1. $$\dfrac{n(n+1)(n+2)}3$$ assignments of $$r$$.
There are $$1+n+\dfrac{n(n+1)}2+\dfrac{n(n+1)(n+2)}3\cdot2=\dfrac{(n+1)(n+2)(4n+3)}6$$ assignments in total.
There are $$\dfrac{n(n+1)(n+2)}3$$ additions in total.
If we treat the computation cost of one assignment as that of one addition, we have $$\dfrac{(n+1)(n+2)(4n+3)}6+\dfrac{n(n+1)(n+2)}3=\dfrac{n(n+1)(2n+1)}2$$ operations.
That is answer. Note that it is a fixed number given $$n$$. That is, there is no distinction of best-case, worse-case or average-case.
Or is it? If we ignore over many important details, yes. However, let us take a close look. Does for i:=1 to n do involve assignment only? Mostly like not. We have to produce 1, 2, 3, $$\cdots$$, n in the first place. If we produce them by addition by 1, we have $$n-1$$ more additions. How do we make sure we will stop when $$i$$ reach $$n$$? Probably we have to compare i with n for about $$n$$ times. Comparison takes time of course. We should include that cost. Wait, there is more. It is unreasonable to expect that $$n$$ is in the right position to be compared with $$i$$; We have to fetch $$n$$ from somewhere so that we can compare it with $$i$$. Now, we are not even sure how to describe that fetching and its computational cost. There are, in fact, many more important details that could be taken into consideration.
You might be wondering now. Do we have an answer or not? Yes and no.
• Yes, we have an answer, assuming some assumptions. I will not be surprised that your instructor might come up with a different number of operations under a different set of assumptions, though.
• No, the answer cannot stand up to scrutiny in many situations. That is one of the reasons why we use the big $$O$$-notations. We can say the computation cost is $$\Theta(n^3)$$ under a set of very general assumptions (a.k.a. model of computation).
By the way, if you check what I had written before, you will notice and understand that I have been rather careful, saying that I am only computing how many times the innermost iteration are executed. In general, the asymptotic time complexity of a nested loop is determined by the product of that number with the average cost of one innermost iteration.
Here is a similar exercise for you to practice.
Exercise. Express the best-time complexity of the following function in terms of $$n$$ instead of in big $$O$$-notation. Only the operation r := r + 1 is counted.
function pesky2(n)
r := 0
for i := 1 to n do
for j := 1 to i do
for k := 1 to j do
r := r + 1
return r
• I actually did this for another question regarding the same exercise asking "What is the output value of r? Express your solution in terms of n". I was able to do what you did successfully but I never made the connection of the resulting function with time complexity since that function only describes the value of r when provided a certain n. To my understanding a time complexity function describes the time the algorithm takes to run and not the output itself. – MikeKatz45 Jan 22 '19 at 23:00
• By sheer luck or by design, it happens the value of $r$ at the end of each loop is the number of times the innermost loop executed so far. $r$ is 0 initially. Every time the innermost loop executed, $r$ increases by 1. – John L. Jan 22 '19 at 23:01
• Ok so my iniitial statement was correct? The inner loop executes n x n xn (In asymptotic terms, that is ignoring constants and lower degree terms) – MikeKatz45 Jan 22 '19 at 23:08
• If you meant n x n x n times asymptotically ignoring a constant factor, you were correct. Sorry that I was not able to understand it that way. – John L. Jan 23 '19 at 0:05
• don't worry about it, I edited the question a little bit with the reason why I think that function is not the time complexity. – MikeKatz45 Jan 23 '19 at 0:28 | 2021-03-01T13:22:00 | {
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https://web2.0calc.com/questions/in-poker-a-5-card-hand-is-called-two-pair-if-there-are-two-cards-of-one-rank-two-cards-of-another-rank-and-a-fifth-card-of-a-third-rank | +0
# In poker, a 5-card hand is called two pair if there are two cards of one rank, two cards of another rank, and a fifth card of a third rank.
0
1314
14
In poker, a 5-card hand is called two pair if there are two cards of one rank, two cards of another rank, and a fifth card of a third rank. (For example, QQ668 is two pair.) The order of the cards doesn't matter (so, for example, QQ668 and 68QQ6 are the same). How many 5-card hands are two pair? (Assume a standard 52-card deck with 13 ranks in each of 4 suits.)
Guest Mar 14, 2015
#11
+86890
+5
This refers to distinct hands....."Distinct Hands" is the number of different ways to draw the hand, not counting different suits
For instance.....there is 1 "distinct" royal flush......but there are 4 suits of them......
Clearly....there are WAY more than 858 two-pair hands......
CPhill Mar 15, 2015
#1
+26720
+5
There are 13*3 possibilities for the first pair, leaving 12*3 possibilities for the 2nd pair, leaving 11*4 possibilities for the singleton.
So: 13*3*12*3*11*4 in total:
$${\mathtt{13}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{4}} = {\mathtt{61\,776}}$$
.
Alan Mar 14, 2015
#2
+86890
+5
This is easiest calculated by first choosing the ranks, and then choosing the cards within those ranks...
So.....
From 13 ranks, we want to choose any 2 = C(13,2)
And from the two ranks, we want to choose any 2 cards from each = [C(4,2)]^2
And we want to choose one card from any of the remaining 11 ranks.....
So we have
C(13,2)[C(4,2)}^2 * C(11,1)C(4,1) = 123,552
CPhill Mar 15, 2015
#3
+92624
+5
Ok I'll give you another possibility,
there are 52 ways of choosing the first card
there are 3 ways of chooing its pair
there are 48 ways of choosing the next number
and 3 ways of choosing its pair
there are 44 ways of choosing the single card
so that is 52*3*48*3*44
$${\mathtt{52}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{48}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{44}} = {\mathtt{988\,416}}$$
Melody Mar 15, 2015
#4
+86890
+5
Poker probabilities.....
http://en.wikipedia.org/wiki/Poker_probability
This is actually very interesting.....!!!
CPhill Mar 15, 2015
#5
+92624
0
Thanks Chris - that would be a good site for us to study from I expect.
So there are 858 distinct possibilities.
Maybe this is interpreted differently from how I interpreted it.
I interpreted there being 4*3=12 ways of choosing 2 of any individual rank.
I wonder if I was suppose to think of all those as the same?
Mmm this definitely requires more study and consideration. ヽ(•́o•̀)ノ
Melody Mar 15, 2015
#6
+86890
+5
Sorry, Melody...I believe this question asks how many two-pair hands are possible in a draw of 5 cards....if I'm interepreting this correctly.....I think my answer is the only correct one....
Look at the number of possible royal flushes in the table......4....!!!
CPhill Mar 15, 2015
#7
+92624
0
doesn't that snip say that there are 858 possibilities?
Melody Mar 15, 2015
#8
+86890
+5
Check the probability..... 123552/ 2598560 = about 4.75%
This says out of 2,598,560 possible hands, there are 123,552 hands consisting of two pairs....
CPhill Mar 15, 2015
#9
+92624
0
so what is the 858 for?
Melody Mar 15, 2015
#10
+92624
0
One thing about probabiliity is that it can be really difficult to interprete what you are being asked for.
The other thing about prob is that although you may be able to understand the correct answer when it is presented to you, you often still cannot understand why your own original answer is incorrect!
It is an interesting and a frustrating feild of methematics.
------------------------------
I'm listening to this
Something different for you Chris
Melody Mar 15, 2015
#11
+86890
+5
This refers to distinct hands....."Distinct Hands" is the number of different ways to draw the hand, not counting different suits
For instance.....there is 1 "distinct" royal flush......but there are 4 suits of them......
Clearly....there are WAY more than 858 two-pair hands......
CPhill Mar 15, 2015
#12
+92624
0
I don't understand.
Maybe it is just too late for me.
When this music finishes I am off to bed. :)
Melody Mar 15, 2015
#13
0
$$\text{First we decide what ranks we will use. Then we will pick suits for all of the cards. We choose the two paired ranks in {13\choose 2}=78 ways and the remaining rank in {11\choose 1}=11 ways. Then we choose the suits for these cards in {4\choose 2}{4\choose 2}{4\choose 1}=144 ways. This gives a total of 78\cdot 11\cdot 144 = \boxed{123552} two pair hands.}$$
Guest Mar 4, 2017
#14
0
$$\text{First we decide what ranks we will use. Then we will pick suits for all of the cards.} \text{We choose the two paired ranks in {13\choose 2}=78 ways and the remaining rank in {11\choose 1}=11 ways.} \text{Then we choose the suits for these cards in {4\choose 2}{4\choose 2}{4\choose 1}=144 ways.} \text{This gives a total of 78\cdot 11\cdot 144 = \boxed{123552} two pair hands.}$$
Guest Mar 4, 2017 | 2018-06-21T18:09:55 | {
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http://mathhelpforum.com/trigonometry/183328-sin-x-cos-x-1-2-a.html | Math Help - sin(x) + cos(x)= 1/2
1. sin(x) + cos(x)= 1/2
Id like to know the solution to
sin(x) + cos(x)= 1/2
rather than the answer, i'd like to know how to work it out since i got a wrong answer.
thanks,
neilofbodom
2. Re: sin(x) + cos(x)= 1/2
Originally Posted by neilofbodom
Id like to know the solution to
sin(x) + cos(x)= 1/2
rather than the answer, i'd like to know how to work it out since i got a wrong answer.
thanks,
neilofbodom
sin(x) + cos(x)= 1/2
(sin(x) + cos(x))^2= (1/2)^2
sin(2x)=(1/4)-1
.
.
.
3. Re: sin(x) + cos(x)= 1/2
First thing you will need to do is graph the function, to see how many solutions you are expecting. You need to square both sides of the function to solve this equation, and squaring could bring in extraneous solutions.
\displaystyle \begin{align*} \sin{x} + \cos{x} &= \frac{1}{2} \\ (\sin{x} + \cos{x})^2 &= \left(\frac{1}{2}\right)^2 \\ \sin^2{x} + \cos^2{x} + 2\sin{x}\cos{x} &= \frac{1}{4} \\ 1 + \sin{2x} &= \frac{1}{4} \\ \sin{2x} &= -\frac{3}{4} \\ 2x &= \left\{ \pi + \arcsin{\left(\frac{3}{4}\right)}, 2\pi - \arcsin{\left(\frac{3}{4}\right)}\right\} + 2\pi n\textrm{ where }n \in \mathbf{Z} \\ x &= \left\{\frac{\pi}{2} + \frac{1}{2}\arcsin{\left(\frac{3}{4}\right)} , \pi - \frac{1}{2}\arcsin{\left(\frac{3}{4}\right)} \right\} + \pi n\end{align*}
Now disregard any extraneous solutions.
4. Re: sin(x) + cos(x)= 1/2
Originally Posted by neilofbodom
Id like to know the solution to
sin(x) + cos(x)= 1/2
rather than the answer, i'd like to know how to work it out since i got a wrong answer.
thanks,
neilofbodom
This has no solution in the 3rd quadrant
as sin(x) and cos(x) are negative there.
We can examine whether there is a solution in the 1st quadrant, where both are positive.
$f(x)=sinx+cosx\Rightarrow\ f'(x)=cosx-sinx$
The turning point occurs when sin(x) = cos(x)
$\Rightarrow\ x=\frac{\pi}{4}$
$f''(x)=-sinx-cosx$
hence the first quadrant contains a maximum as $f''(x)<0,\;\;\;x=\frac{\pi}{4}$
More significantly, the tangent slope is always decreasing in the first quadrant,
hence the minimum values in the first quadrant occur either side of the maximum at $x=0,\;\;\;x=\frac{\pi}{2}$
where f(x) =1 in both cases.
Hence, there is no solution in the first quadrant either.
There are certainly solutions in the 2nd and 4th quadrants, since sin(x) and cos(x)
have opposite sign there.
We can use the fact that $cosx=sin\left(\frac{\pi}{2}-x\right)$
coupled with the identity
$sinA+sinB=2sin\frac{A+B}{2}cos\frac{A-B}{2}$
to obtain
$sinx+cosx=sinx+sin\left(\frac{\pi}{2}-x\right)=2sin\frac{\pi}{4}cos\left(x-\frac{\pi}{4}\right)$
$\Rightarrow\ \frac{2}{\sqrt{2}}cos\left(x-\frac{\pi}{4}\right)=\frac{1}{2}$
$\Rightarrow\ cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{4}$
$x=cos^{-1}\frac{\sqrt{2}}{4}+\frac{\pi}{4}$
and this gives the 2nd quadrant solution.
Hint: You'll need to reverse the signs of cos(x) and sin(x)
5. Re: sin(x) + cos(x)= 1/2
Hello, neilofbodom!
Another approach . . .
$\text{Solve: }\;\sin x + \cos x \:=\: \frac{1}{2}$
$\text{Divide by }\sqrt{2}: \;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{2\sqrt{2}}$
$\text{Note that: }\:\sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}},\;\;\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$
$\text{So we have: }\;\underbrace{\cos\tfrac{\pi}{4}\sin x + \sin\tfrac{\pi}{4}\cos x} \;=\;\frac{1}{2\sqrt{2}}$
$\text{Then: }\qquad\qquad\quad\sin\left(x + \tfrac{\pi}{4}\right) \;=\;\frac{1}{2\sqrt{2}}$
$\text{Hence: }\;x + \frac{\pi}{4} \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)$
$\text{Therefore: }\;x \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) - \frac{\pi}{4} \;\approx\;-0.424$
6. Re: sin(x) + cos(x)= 1/2
Originally Posted by Soroban
Hello, neilofbodom!
Another approach . . .
$\text{Divide by }\sqrt{2}: \;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{2\sqrt{2}}$
$\text{Note that: }\:\sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}},\;\;\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$
$\text{So we have: }\;\underbrace{\cos\tfrac{\pi}{4}\sin x + \sin\tfrac{\pi}{4}\cos x} \;=\;\frac{1}{2\sqrt{2}}$
$\text{Then: }\qquad\qquad\quad\sin\left(x + \tfrac{\pi}{4}\right) \;=\;\frac{1}{2\sqrt{2}}$
$\text{Hence: }\;x + \frac{\pi}{4} \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)$
$\text{Therefore: }\;x \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) - \frac{\pi}{4} \;\approx\;-0.424$
Of course, to get all possible solutions...
\displaystyle \begin{align*} \sin{\left(x + \frac{\pi}{4}\right)} &= \frac{\sqrt{2}}{4} \\ x + \frac{\pi}{4} &= \left\{\arcsin{\left(\frac{\sqrt{2}}{4}\right)}, \pi - \arcsin{\left(\frac{\sqrt{2}}{4}\right)}\right\} + 2\pi n\textrm{ where }n\in\mathbf{Z} \\ x &= \left\{\arcsin{\left(\frac{\sqrt{2}}{4}\right)} - \frac{\pi}{4}, \frac{3\pi}{4} - \arcsin{\left(\frac{\sqrt{2}}{4}\right)}\right\} + 2\pi n \end{align*}
7. Re: sin(x) + cos(x)= 1/2
Originally Posted by neilofbodom
Id like to know the solution to
sin(x) + cos(x)= 1/2
rather than the answer, i'd like to know how to work it out since i got a wrong answer.
thanks,
neilofbodom
You could also use simple geometry as shown in the attachment,
using the fact that the sum of 2 sides of a triangle is greater than the 3rd side.
This rules out the 1st and 3rd quadrants.
Then you can find the 2nd quadrant solution using
$sinx+cosx=sinx+sin\left(\frac{\pi}{2}-x\right)=2cos\frac{\pi}{4}cos\left(x-\frac{\pi}{4}\right)$
to get
$x=arccos\left(\frac{\sqrt{2}}{4}\right)+\frac{\pi} {4}$
Then to get the 4th quadrant solution, rotate the triangle to the position
shown in the attachment to get
$x=\frac{9\pi}{4}-arccos\left(\frac{\sqrt{2}}{4}\right)$ | 2014-09-24T02:30:05 | {
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https://math.stackexchange.com/questions/2309940/how-do-you-solve-sum-k-0-infty-sum-i-1m-1-left-lfloor-fracxim | # How do you solve $\sum_{k=0}^{\infty} \sum_{i=1}^{m-1} \left \lfloor \frac{x+im^k}{m^{k+1}} \right \rfloor$?
Let $m > 1$ be an integer, x is a real number and $f(x)=\sum_{k=0}^{\infty} \sum_{i=1}^{m-1} \left \lfloor \frac{x+im^k}{m^{k+1}} \right \rfloor$
Do you have any idea how can I show this?
$f(x)= \begin{cases} \left \lfloor x \right \rfloor & \text{if } x\geq 0\\ \left \lfloor x+1 \right \rfloor & \text{if } x<0 \end{cases}$
• Is this supposed to be true regardless of the choice of $m$?? – Zubin Mukerjee Jun 4 '17 at 22:21
• yes it should work for any m > 1 – Enzo Giannotta Jun 4 '17 at 22:24
The key to solving the sum is realising that for any $k$ you can write $x$ uniquely as $x = m^{k+1} q_{k+1} + m^k r_{k}$ with $q_{k+1} \in \mathbb{Z}$ and $r_k \in [0, m) \subset \mathbb{R}$.
You can think of this as writing $x$ in base $m$ and splitting into two numbers corresponding to the first couple of digits and the remaining digits. For example if $x=1234.56$ and $m=10$ then $q_2 = 12$ and $r_{1} = 3.456$. We also have that $$m^{k+1} q_{k+1} + m^k r_k = m^k q_k + m^{k-1} r_{k-1} = m^{k+1} q_{k+1} + m^k ((q_k - m q_{k+1}) + r_{k-1} /m)$$ so $r_k = (q_k - m q_{k+1}) + r_{k-1}/m$ which implies $\lfloor r_k \rfloor = (q_k - m q_{k+1})$. Which makes sense if you think of the integral part of $r_k$ as the 'next digit'.
Motivation aside, this notation makes it possible to simplify the sum as follows:
\begin{align} \sum_{k=0}^{\infty} \sum_{t=1}^{m-1} \left \lfloor \frac{x+tm^k}{m^{k+1}} \right \rfloor &= \sum_{k=0}^{\infty} \sum_{t=1}^{m-1} \left \lfloor \frac{m^{k+1} q_{k+1} + m^{k} r_{k} + tm^k}{m^{k+1}} \right \rfloor \\&= \sum_{k=0}^{\infty} \sum_{t=1}^{m-1} \left \lfloor q_{k+1} + \frac{(r_k +t)m^k}{m^{k+1}} \right \rfloor \\&= \sum_{k=0}^{\infty} \sum_{t=1}^{m-1} q_{k+1} + \left \lfloor \frac{r_k + t}{m} \right \rfloor \\ &= \sum_{k=0}^{\infty} (m-1) q_{k+1} + \lfloor r_k \rfloor\\ &= \sum_{k=0}^{\infty} (m-1) q_{k+1} + (q_k - mq_{k+1})\\ &= \sum_{k=0}^{\infty} q_k - q_{k+1} \end{align}
Now note that $$\lim_{N\to\infty} \sum_{k=0}^{N} q_k - q_{k+1} = \lim_{N\to\infty} q_0 - q_N = q_0 = \begin{cases} \left \lfloor x \right \rfloor & \text{if } x\geq 0\\ \left \lfloor x+1 \right \rfloor & \text{if } x<0 \end{cases}$$
Another solution: We will use the identity $\left \lfloor nx\right \rfloor = \sum_{k = 0}^{n - 1} \left \lfloor x + \frac{k}{n} \right \rfloor$,
$\sum_{k=0}^{\infty} \sum_{i=1}^{m-1} \left \lfloor \frac{x+im^k}{m^{k+1}} \right \rfloor = \sum_{k=0}^{\infty} \sum_{i=1}^{m-1} \left \lfloor \frac{x}{m^{k+1}} + \frac{i}{m} \right \rfloor = \sum_{k=0}^{\infty}( \sum_{i=0}^{m-1} \left \lfloor \frac{x}{m^{k+1}} + \frac{i}{m} \right \rfloor - \left \lfloor \frac{x}{m^{k+1}}\right \rfloor)$
Now we can apply the identity from above on $\frac{x}{m^{k+1}}$, and we are left with,
$\sum_{k=0}^{\infty} \left \lfloor m \frac{x}{m^{k+1}} \right \rfloor - \left \lfloor \frac{x}{m^{k+1}} \right \rfloor = \sum_{k=0}^{\infty} \left \lfloor \frac{x}{m^{k}}\right \rfloor - \left \lfloor \frac{x}{m^{k+1}}\right \rfloor$, this is just a telescoping series that converges to $\left \lfloor x\right \rfloor$ if we choose $x\geq0$.
If x is negative then $- \left \lfloor \frac{x}{m^{k+1}}\right \rfloor$ turns into $+ 1$ as $\frac{x}{m^{k+1}}$ aproaches 0 from the left. So the result holds. | 2019-04-25T00:50:03 | {
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https://math.stackexchange.com/questions/2405327/borel-sigma-algebra-construction | # Borel sigma-algebra construction
Consider the Borel sigma-algebra on $\mathbb{R}$, $\mathcal{B}(\mathbb{R})$. The definition I've been working is that $\mathcal{B}(\mathbb{R})$ is the smallest sigma-algebra containing all open sets of $\mathbb{R}$. However, I just read that $\mathcal{B}(\mathbb{R})$ can be constructed by choosing to contain all sets of the form $[a,b]$, $(a, b]$, $(a,b)$, and $[a,b)$ for all real numbers $a$ and $b$. What I don't understand is, why is $[a,b] \in \mathcal{B}(\mathbb{R})$? Is it because $(-\infty, a) \cup (b, \infty) \in \mathcal{B}(\mathbb{R})$ so the complement $[a,b] \in \mathcal{B}(\mathbb{R})$? Then, how can I show that $(a, b] \in \mathcal{B}(\mathbb{R})$?
• Yes, that is why $[a,b]$ is Borel. Intersect that with an open set to get $(a,b]$.... – Angina Seng Aug 25 '17 at 3:56
• Ah I see, so I could do something like $[a,b] \cap (a, b+1)$? – TeTs Aug 25 '17 at 4:03
A $\sigma$-algebra is a Boolean algebra. A Boolean algebra contains the complements of its members. $B(R)$ contains all open sets so it contains their complements, which is all closed sets. Like $[a,b].$ It will also contain the intersection of any closed set with any open set, like $(a,b]=(-\infty,b]\cap (a,\infty).$
$B(R)$ can be generated by just the bounded open intervals because any open set in $R$ is the union of countably many bounded open intervals.
Say $B$ is the "standard" Borel $\sigma$ algebra on $\mathbb{R}$, the smallest $\sigma$ algebra containing all the open sets on $\mathbb{R}$. And let $B'$ be the smallest $\sigma$ algebra containing all the bounded intervals, i.e., all of the intervals $[a,b]$, $[a,b)$, $(a,b]$, and $(a,b)$ for all $a,b \in \mathbb{R}$. The claim is that $B = B'$.
Certainly $B$ contains every bounded open interval $(a,b)$. You have stated a reason why $[a,b] \in B$: $B$ contains the open set $(-\infty,a) \cup (b,\infty)$ whose complement is $[a,b]$. Alternatively, write $[a,b]$ as a countable intersection of open sets: $$[a,b] = \bigcap_{n=1}^{\infty} \left(a-\frac{1}{n},b+\frac{1}{n}\right).$$ As observed in the comments, the other types of intervals can be obtained via intersections (or unions!) with open sets, which are in $B$: $(a,b] = [a,b] \cap (a,b+1)$, or $(a,b] = (a,b) \cup [(a+b)/2,b]$; or for that matter $$(a,b] = \bigcap_{n=1}^{\infty} \left(a,b+\frac{1}{n}\right).$$ Similar things work for $[a,b)$. This proves that $[a,b], (a,b], [a,b) \in B$; so $B' \subseteq B$.
For the reverse inclusion, use the fact that every open set in $\mathbb{R}$ is a countable (or finite) union of bounded intervals. First, every open set is a countable union of open intervals (if you haven't seen this, hint: rational numbers), and if any of the open intervals are unbounded, they can be written as countable unions of bounded intervals. This shows $B'$ includes every open set, so $B \subseteq B'$.
• Thanks, that's a very good explanation. One last query I have: so something like the set $\{3,4\}$ is contained in $\mathcal{B}(\mathbb{R})$ because $\{3,4\} = (-\infty, 3] \cap [3,4] \cap [4, \infty)$, right? So then what's the difference between $\mathcal{B}(\mathbb{R})$ and the power set of $\mathbb{R}$, $\mathcal{P}(\mathbb{R})$, it seems like they both contain the same elements? If they're not the same, then can you show me an element in the power set of $\mathbb{R}$ that's not in $\mathcal{B}(\mathbb{R})$? – TeTs Aug 25 '17 at 5:49
• @TeTs It's actually not possible to give an explicit example. But you can prove there are lots of non-Borel subsets of $\mathbb{R}$. E.g. by invoking the axiom of choice (Vitali set or from a free ultrafilter), or a counting argument (there are $|\mathbb{R}|$ many Borel sets, as many as there are points, but there are a lot more subsets of the reals. – Henno Brandsma Aug 25 '17 at 7:13
There is a more general stronger result that is no harder to prove. Suppose $\tau$ is a topology on a space $X.$ Then by definition, $\mathscr B(X)=\sigma(\tau).$
Claim: if $X$ is a second countable topological space, then $\mathscr B(X)$ is generated by any countable base for $X.$
To see this, let $\mathscr E$ be a countable base for the topology $\tau.$ Since $\mathscr E \subseteq \tau,\ \sigma (\mathscr E) \subseteq \sigma (\tau)=\mathscr B(X).$ For the reverse inclusion, we simply note that if $U\in \tau$ then $U$ is a countable union of elements of $\mathscr E$ so $U\in \sigma (\mathscr E),$ which means that $\sigma (\tau)\subseteq \sigma (\mathscr E).$
So in the case $X=\mathbb R$ we have that $\mathscr B(\mathbb R)$ is generated by the collection of open intervals with rational endpoints. Furthermore, since the closed and half-closed intervals are countable unions of open intervals, $\mathscr B(\mathbb R)$ is also generated by the collection of closed and half-closed intervals. | 2020-08-09T23:35:20 | {
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http://math.stackexchange.com/questions/784885/how-many-homomorphisms-are-there-from-mathbb-z-3-to-s-4 | # How many homomorphisms are there from $\mathbb Z_3$ to $S_4$?
How many homomorphisms are there from $\mathbb Z_3$ to $S_4$?
I got this on a quiz today and calculated 9 but I'm not certain I was right. I sent 1 to id and to the 8 3-cycles... did I do wrong?
-
An homomorphism $f : \mathbb{Z}_{3} \to S_{4}$ is uniquely determined by $f(1)$, which must be an element of order that divides $3$, then ord$(f(1)) = 1$ or ord$(f(1)) = 3$.
In $S_{4}$ the only elements of order $3$ are the $3$-cycles, so your solution seems correct.
but $A_3$is the only subgroup of \$S_4 with order 3. By lagrange theorem the order of the image is either1 or 3. so arent there 2 homoms? – tetrr May 7 at 11:50 | 2014-09-24T02:55:41 | {
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https://math.stackexchange.com/questions/894383/a-real-matrix-whith-rows-generating-u-and-columns-generating-v | # A real matrix whith rows generating $U$ and columns generating $V$
Let $n \in \mathbb{N}$, and $U,V$ two linear subspaces of $\mathbb{R}^n$ of the same dimension. Could one always make a matrix $A \in \mathbb{M}^{n \times n}(\mathbb{R})$ such that $spanA = U$ and $spanA^T = V$?
I find it hard to find a counterexample. I know that for any $n \in \mathbb{N}$, there is such a matrix if $\dim U = \dim V \in \{0,n\}$, we could just take $0,I$ as matrices. Moreover, if $U,V$ are lines, that means $U = \mathbb{R}v$ and $V = \mathbb{R}w$ for some nontrivial vectors $v,w$, then the matrix below works.
$$\left( \ w_1v \ | \ w_2v \ | \ \cdots \ | \ w_nv \ \right)$$ Here $w_i$ are coordinates of the vector $w$, and each entry represents a column this way.
Then I took two subspaces of dimension two in $\mathbb{R}^3$, namely $$U \quad = \quad span \left\{ \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right) , \left( \begin{array}{ccc} 0 \\ 1\\ 0 \end{array} \right) \right\} \qquad V \quad = \quad \left( \begin{array}{ccc} 0 \\ 1\\ 0 \end{array} \right)^\perp$$ The matrix we need here is $$\left( \begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \end{array} \right)$$
I find it hard to deal with the general case. Could you give me some help to establish the statement, or give a counterexemple?
Let $m$ be the dimension of $U$ and $V$. Let $M_U$ be an $n\times m$ matrix whose columns are a basis of $U$ and $M_V$ be an $n \times m$ matrix whose columns are a basis of $V$. Now $U = \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \}$ and $V = \{M_V \vec{x} : \vec{x} \in \mathbb{R}^m \}$.
We wish to find a matrix $A$ such that \begin{align*} U &= \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{A \vec{x} : \vec{x} \in \mathbb{R}^n \}, \\ V &= \{M_V \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{A^T \vec{x} : \vec{x} \in \mathbb{R}^n \}. \end{align*}
We note that $\{ M_V^T \vec{x} : \vec{x} \in \mathbb{R}^n\} = \mathbb{R}^m$, because the rank of $M_V^T$ is $m$ and it has $m$ rows. Thus $$U = \{M_U \vec{x} : \vec{x} \in \mathbb{R}^m \} = \{M_U M_V^T \vec{x} : \vec{x} \in \mathbb{R}^n \}.$$ But this immediately suggests that $A=M_U M_V^T$, because $\mathrm{span}(M_U M_V^T) = U$. By symmetry, also $\mathrm{span}(M_V M_U^T) = V$, and because $A^T=M_V M_U^T$, we are done.
• I don't understand what you mean by $M_U^T$, could you please tell me? Aug 11, 2014 at 21:02
• @KoenraadvanDuin Transpose of the matrix $M_U$.
– JiK
Aug 11, 2014 at 21:06
• right, I forgot about that. Aug 11, 2014 at 21:18
This is indeed always possible. Let $U$ be some $r$-dimensional subspace of $\mathbb{R}^n$. Let $R$ denote any matrix with rowspace $U$. Since $U$ is $r$-dimensional, it follows that the rank of $R$ is $r$ and so the image of $R$ is also $r$-dimensional.
Let $\mathcal{B}'$ denote a basis for the image of $R$ and extend $\mathcal{B}'$ to a basis $\mathcal{B}$ of $\mathbb{R}^n$. Form the matrix $P$ with columns given by $\mathcal{B}$, with the first $r$ columns consisting of the vectors of $\mathcal{B}'$. It follows that $P$ maps the subspace $S$ spanned by $\left\{\mathbf{e}_1,\ \cdots, \mathbf{e}_r\right\}$ to $\mathrm{im}(R)$ and therefore $P^{-1}$ maps $\mathrm{im}(R)$ to $S$.
Now let $\mathcal{C}'$ denote a basis for $V$ and extend $\mathcal{C}'$ to $\mathcal{C}$. Form the matrix $Q$ with the basis as the columns, with the first $r$ columns consisting of the vectors of $\mathcal{C}'$. It follows that $Q$ takes $S$ to $V$.
Then the matrix $QP^{-1}R$ is your desired matrix. Since $QP^{-1}$ is invertible, the rowspace of $R$ is unchanged, therefore $\mathrm{row}(QP^{-1}R) = U$. Also, $P^{-1}$ maps $\mathrm{im}(R)$ to $S$ and $Q$ maps $S$ to $V$. Therefore $\mathrm{im}(QP^{-1}R) = V$, as required.
• I understand that a map $X \mapsto AX$ does not change the kernel of $X$ if $A$ is invertible, but I don't see how that should work for the rowspace of $X$. Could you explain me? Aug 11, 2014 at 20:37
• @KoenraadvanDuin There are two immediate ways to see this. First you note that the kernel is invariant under $X\mapsto AX$. The rowspace is the orthogonal complement of the kernel, hence it is also left invariant. Alternatively, if $A$ is invertible then it can be written as a product of elementary matrices, so the overall effect of $A$ is to perform a sequence of elementary row operations on $X$, which does not change the rowspace.
– EuYu
Aug 11, 2014 at 23:51 | 2023-02-03T09:16:50 | {
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https://stats.stackexchange.com/questions/178843/simple-and-direct-application-of-bayes-theorem | # Simple and direct application of Bayes Theorem
Suppose that $\theta \in \Theta=\{0,1\}$ such that $P(\theta = 0) = 0.1$. Let $X$ be a r.v such that, given $\theta=0$, $X \sim N(50,1)$ and given $\theta = 1$, $X \sim N(52,1)$. Show that the posteriori probability of $\theta=0$ is greater than the posteriori of $\theta = 1$ if and only if $$x < 51 - \frac{3}{2} \log(9)$$
My attempt Using bayes theorem
$$P(\theta = \theta|X)=\frac{P(X|\theta=\theta) P(\theta)}{\sum_\Theta P(X|\theta=\theta) P(\theta)}$$
We note that the denominator is common for both values of theta. Then we need to compare $$P(X|\theta=0) P(\theta=0) \text{ and } P(X|\theta=1) P(\theta=1)$$
$\Rightarrow P(X|\theta=0) P(\theta=0) > P(X|\theta=1) P(\theta=1) \iff 0.1 \frac{1}{\sqrt{2 \pi}} e^{\frac{-(x-50)^2}{2}} > 0.9 \frac{1}{\sqrt{2 \pi}} e^{\frac{-(x-52)^2}{2}}$
$\Rightarrow \frac{-(x-50)^2}{2} > \log(9) + \frac{-(x-52)^2}{2} \iff x^2 - 100x + 2500 < x^2 - 104x + 2704 - 2 \log(9) \iff x < 51 - \frac{1}{2} \log(9)$
I wonder if I am missing something, because I couldn't find out from where this 3 camed from.
Thanks!
As far as I am concerned you are right and that seems to be a typo.
grid <- seq(40,60,0.05)
posterior <- function(theta, x) {
z = 0
if(theta==0) {
z = 0.1
m = 50
} else if(theta==1) {
z = 0.9
m = 52
}
return((dnorm(x, mean=m, sd=1)*z)/(dnorm(x ,mean=50,s d=1)*0.1
+ dnorm(x, mean=52,s d=1)*0.9))
}
par(lwd=2)
plot(x=grid, y=posterior(theta=0,x=grid), type="l",
col="red3", ylab="Posterior Probability", xlab="")
lines(x=grid, y=posterior(theta=1, x=grid), col="steelblue")
abline(v=51-0.5*log(9), lty=2)
legend(x = 40, y=0.8, legend=c(expression(paste(theta, " = ", 0)),
expression(paste(theta, " = ", 1))),
col=c("red3", "steelblue"), lty=c(1,1),
lwd=c(2,2), bty= "n")
mtext(text = expression(paste("51 - 0.5 ", log(9)))) | 2021-04-14T07:11:43 | {
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http://math.stackexchange.com/questions/179224/how-to-show-x-and-y-are-equal | # How to show x and y are equal?
I'm working on a proof to show that f: $\mathbb{R} \to \mathbb{R}$ for an $f$ defined as $f(x) = x^3 - 6x^2 + 12x - 7$ is injective. Here is the general outline of the proof as I have it right now:
Proof: For a function to be injective, whenever $x,y \in A$ and $x\neq y$, then $f(x) \neq f(y)$, i.e., where $A$, $B$ are finite sets, every two elements of $A$ must have distinct images in $B$, which also implies that there must be at least as many elements in $B$ as in $A$ such that the cardinality of $A$ is less than or equals the cardinality of $B$.
We shall prove the contra-positive: If $\exists$ $f(x) = f(y)$, then $x=y.$
Let $x^3 - 6x^2 + 12x - 7 = y^3 - 6y^2 + 12y - 7$.
Then by addition and some algebra, we get $x(x^2 - 6x + 12) = y(y^2 - 6y + 12)$
This feels dumb to ask but how do I continue to finally get the result that $x = y$?
-
Hint: show that the function is increasing. – M.B. Aug 5 '12 at 19:53
Sorry I don't understand. Do you mean differentiate and show that the derivative is always positive or something? – Arthur Collé Aug 5 '12 at 20:07
Exactly. If a function is increasing, do you see why it would have to be injective? – Kevin Carlson Aug 5 '12 at 20:12
Much of the first sentence of the proof is OK. The stuff about finite sets is of no particular relevance. The "formula" $\exists\,f(x)=f(y)$ is not well-formed. – André Nicolas Aug 5 '12 at 20:28
To say a function is "increasing" means that if $a < b$ then $f(a) < f(b)$. Suppose $x \neq y$. Then either $x < y$ or $x > y$. In either case, one cannot have $f(x) = f(y)$ and therefore increasing functions are injective. – user29743 Aug 5 '12 at 20:50
## 3 Answers
Note that $w^3-6w^2+12w-8=(w-2)^3$. So $w^3-6w^2+12w-7=(w-2)^3+1$.
So we want to show that if $(x-2)^3+1=(y-2)^3+1$ then $x=y$.
Equivalently, we want to show that if $(x-2)^3=(y-2)^3$ then $x=y$. This is easy, the cube function is increasing.
Remark: We can use the basic algebra of ordered fields to show that if $s^3=t^3$ then $s=t$. For $s^3-t^3=(s-t)(s^2+st+t^2)$. But $$2(s^2+st+t^2)=(s+t)^2+s^2+t^2,$$ so $s^2+st+t^2$ can only be $0$ when $s=t=s+t=0$.
-
A differentiable function with everywhere positive derivative is injective. The derivative of $f$ is $f'(x)=3x^2-12x+12=3(x^2-4x+4)=3(x-2)^2$. This is positive for $x\ne2$. To see that the zero derivative at $x=2$ doesn't destroy injectivity, integrate this to find $f(x)=(x-2)^3+C$ (with $C=1$). Thus $f$ is a shifted version of $x^3$, which is injective.
-
It is sufficient that $f'$ does not change sign in order for $f$ to be injective. – AD. Aug 5 '12 at 20:33
@AD: That's not true. You could insert a stretch where $f$ is identically $0$ into $f(x)=x^3$ without the derivative ever changing sign. I suspect that it's sufficient that the zeros of $f'$ are isolated, but that seems like overkill for a problem that can easily be reduced to $x^3$. – joriki Aug 5 '12 at 20:50
You are right. What I meant was "for polynomial $f$" (or smooth $f$). – AD. Aug 5 '12 at 21:07
Sorry, not smooth but analytic (one may consider $f(x)=\exp(-1/x^2)$ on $\mathbb{R}^+$) – AD. Aug 5 '12 at 21:33
Hint $\rm\,\ 0 = f(x)\!-\!f(y) = (\color{#C00}{x\!-\!y})\,\left(\left(\color{blue}{y\!-\!2}\ +\dfrac{\color{#0A0}{x\!-\!2}}2\right)^2\! + 3\,\left(\dfrac{\color{#0A0}{x\!-\!2}}{2}\right)^2\right)\iff \begin{eqnarray}\rm \color{#C00}{x=y}\quad \ or\\\rm\ \color{#0A0}{x=2}\ \ and\ \ \color{blue}{y = 2}\end{eqnarray}$
Remark $\$ This approach doesn't require noticing that $\rm\:f(x) = (x-2)^3\!+1.\:$ Rather, we use only $\rm\:x\!-\!y\:|\:f(x)\!-f(y)\:$ (Factor Theorem), and we complete the square in the quadratic cofactor.
- | 2016-05-24T22:06:57 | {
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http://mathhelpforum.com/discrete-math/162288-number-possible-pin-numbers.html | # Math Help - The number of possible PIN numbers
1. ## The number of possible PIN numbers
This seems like a pretty straightforward question but I feel like I'm missing something.
Suppose I wanted to construct a pin number that contains 4 letters of the alphabet and 3 numbers (0 to 9) in no particular order, and I can use the same letter/number more than once.
My pin will be 7 letters/numbers in length.
I have:
(26)(26)(26)(26)(10)(10)(10)
= 456 976 000
It feels like I'm missing something or is the question really this straightforward?
Any help is appreciated!
2. That is correct!
3. So really, this is essentially the same thing if I wanted my pin number to start with 4 letters and then followed by 3 numbers? (I know this is a stupid question)
I don't have to divide out anything at all?
4. Originally Posted by DarK
So really, this is essentially the same thing if I wanted my pin number to start with 4 letters and then followed by 3 numbers? (I know this is a stupid question)
I don't have to divide out anything at all?
Your calculation counted the amount of pin numbers in the sequence LLLLDDD, where L is a letter and D is a digit,
allowing repetition.
That's fine if no shuffling of the letters among the digits is allowed.
(If you meant that the 4 letters are in direct sequence "in no particular order" and same for the digits).
5. So what if I could shuffle the letters and numbers in any way I want:
ie:
LDDLDLL
or
DLDLLDL
or
DDLLLDL
and so on
Would I have to account for this or would it just be the same calculation as my first post?
Thank you.
6. Originally Posted by DarK
So what if I could shuffle the letters and numbers in any way I want: ie: LDDLDLL or DLDLLDL
$\displaystyle \binom{7}{4}\left( {26^4 } \right)\left( {10^3 } \right)$
7. Originally Posted by Plato
$\displaystyle \binom{7}{4}\left( {26^4 } \right)\left( {10^3 } \right)$
Could somebody clarify, why are we multiplying by 7 choose 4?
8. Originally Posted by DarK
Could somebody clarify, why are we multiplying by 7 choose 4?
$\binom{7}{4}=\binom{7}{3}$
We choose 4 of the 7 positions for the 4 letters.
That automatically leaves us with 3 positions for the 3 digits.
Now you have $26^4$ alternatives for the letters and $10^3$ alternatives for the digits in that particular sequence.
the sequence could be LDLDLDL
Now choose a new sequence and you have the same $26^4$ alternatives for the letters and $10^3$ for the digits.
The number of such "letter-digit" sequences is $\binom{7}{4}=\binom{7}{3}$
as you need to choose 4 of the 7 positions for letters
or instead choose 3 of the 7 positions for the digits (same thing).
Hence you need to count the number of ways this can be done.
9. So it looks like the final answer can be either:
$\binom{7}{4}$ $26^4$ $10^3$
or
$\binom{7}{3}$ $26^4$ $10^3$
where the letters and numbers can take on any ordering. | 2014-08-30T16:39:20 | {
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https://math.stackexchange.com/questions/2798747/doubt-on-grouping-of-terms-in-a-series | # Doubt on grouping of terms in a series
We know that if a series is convergent, then we can perform grouping on the series and the resulting series would still be convergent.
However,for an arbitrary series, grouping may not always give the same result. E.g. $a_n=(-1)^{n+1}$ is the non-convergent series
$1+(-1)+1+(-1)+...$
which can be grouped to
$(1-1)+(1-1)+...$ which converges to 0.
So my question is, in this link Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ the solutions have grouped the terms in the series (before knowing whether the series converges or not) and showed it is convergent. How is that possible?
• By grouping the terms and showing that the series whose terms are the respective groups of terms from the original series, it does show that if the original series does happen to be convergent, then the original must converge to the same thing that the grouped series does. That being said, it should still be proven that the original series is convergent as well (but it is unnecessary to find what it converges TO in this step). As for an approach, Dirchlet's Test looks like it should be relevant. – JMoravitz May 28 '18 at 4:08
• Side note: $(1-1)+(1-1)+(1-1)+\dots$ converging to zero and $1+(-1+1)+(-1+1)+\dots$ converging to $1$ both simultaneously show that $1-1+1-1+1-1+\dots$ should, if it converges, simultaneously converge to both zero and one, a contradiction. – JMoravitz May 28 '18 at 4:09
Suppose that $s_n=\sum\limits_{k=1}^n a_k$ denotes the $n$-th partial sum. Convergence of the series is equivalent to convergence of the sequence $(s_n)$.
By grouping some term you go to a subsequence $s_{n_k}$. Of course, if $s_{n_k}$ converges, that does not imply in general that so does $s_n$.
However, in the linked post the grouping is made in special way: The answerer took a block of consecutive positive terms, then the next block consists only of negative terms, etc. In this way you get that for $$n_k \le n \le n_{k+1}$$ you have that $s_n$ is between $s_{n_k}$ and $s_{n_{k+1}}$. (Since the sequence of partial sums is monotone on these intervals.)
So in this case if you get that $\lim\limits_{k\to\infty} s_{n_k}=\ell$, you immediately see that the limit of the sequence $(s_n)$ is the same, i.e. $\lim\limits_{n\to\infty} s_n=\ell$.
• @Anwi The comment "Correct me if I am wrong as $k\to inf$" is somewhat cryptic. I do not know what you are trying to say. – Martin Sleziak May 28 '18 at 5:36
• Sorry i am trying to write the latex commands properly – Anwi May 28 '18 at 5:38
• Correct me if i am wrong: as k tends to infinity,n tends to infinity(since$n_k \le n \le n_{k+1}$) so sandwiching we get that lim $s_{n_k} \le$lim $s_n \le$ lim $s_{n_{k+1}}$ – Anwi May 28 '18 at 5:44
• @Anwi Well, sandwiching is the basic idea, but you will have $s_{n_k} \le s_n \le s_{n_{k+1}}$ on some intervals and $s_{n_k} \ge s_n \ge s_{n_{k+1}}$ on different intervals. (If you look at odd/even intervals, the sequence $s_n$ is sometimes non-increasing, sometimes non-decreasing.) – Martin Sleziak May 28 '18 at 5:48
• True.Other than that,did i miss anything or is that what you meant? – Anwi May 28 '18 at 5:49 | 2019-07-23T15:34:52 | {
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https://math.stackexchange.com/questions/2742319/why-doesnt-sinhxs-taylor-series-divide-by-half/2742334 | # Why doesn't $\sinh(x)$'s Taylor Series divide by half?
Learning about Taylor Series, I have the problem sinh(x). Obviously, $\sinh(x) = \dfrac{e^x - e^{-x}}2$. I basically did it all correctly, since most of it cancels when compared to $e^x$'s taylor series. But that $\cfrac 12$ that's in the problem is throwing me off. Why isn't the summation series:
$$\sum 2\frac{x^{2x+1}}{(2n + 1)!}$$
That (* 2) on the bottom is what is confusing me. How is that getting cancelled out? All of the positive terms of n are getting cancelled out in e$^x$'s taylor series, but we still have these left. And they're still being divided by 2.
What I'm getting for the series itself written out, is:
$\cfrac 12\left[x + \cfrac {x^3}{3!} + \cfrac{x^5}{5!}\right]$
Why is that 2 just forgotten about?
• You are forgetting that you will obtain twice the odd coefficients, that is when you subtract them you will have say $2x$ for the first term not $x$. Thus this cancels with the $2$ in the denominator. Apr 18, 2018 at 1:55
• @TriatticusJust to make sure I'm understanding this correctly. It's because I'm re-defining n? Because I threw out half the results for summation of e$^x$ and are adding in extras? Apr 18, 2018 at 1:58
• It has nothing to do with redefining $n$. When you perform the subtraction the odd coefficients will be doubled because you are subtracting a negative which becomes addition so that $$\small{(1 + x + (1/2)x^2 + (1/3!)x^3+\dots) - (1 - x + (1/2)x^2-(1/3!)x^3 +\dots) = 2x + 2(1/3!)x^3 +\dots)}$$ Apr 18, 2018 at 2:01
• MathJax tutorial Apr 18, 2018 at 2:35
• Divide by two? Multiply by half? Divide in half? Apr 18, 2018 at 13:51
Write out the Taylor series for $e^x$ and $e^{-x}$:
\begin {align*} e^x &= \sum_{n=0}^\infty \frac{x^n}{n!} \\ e^{-x} &= \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!} \end {align*}
Subtract them and you get $$e^x - e^{-x} = \sum_{n=0}^\infty \frac{1-(-1)^n}{n!} \, x^n$$
When $n$ is even, the coefficient is zero, and when $n$ is odd, it is $\frac{2}{n!}$. So the $\frac{1}{2}$ cancels this 2.
• Why are you getting 1 - (-1)$^n$. And where is the x$^n$ coming from in the bottom half? editing If I split it up I get x$^n$/n! - x$^n$/n! - (-1)$^n$. So wouldn't it be (-1)$^n$ / n! times 1/2? Am I going backwards here? Apr 18, 2018 at 2:03
• If you have two power series $f(x) = \sum a_n x^n$ and $g(x) = \sum b_n x^n$, then when you add them, you just add the coefficients: $f(x) + g(x) = \sum (a_n + b_n) x^n$. It's just like adding polynomials.
– Nick
Apr 18, 2018 at 2:06
Write out the first few terms of each series
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} \cdots$$
$$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \cdots$$
Subtract the two series and you get
$$e^x - e^{-x} = 2x + \frac{2x^3}{3!} + \frac{2x^5}{5!} + \cdots$$
Notice how each odd power coefficient doubles? Divide the above by $2$ and you obtain the $\sinh x$ series
The following manipulation can be proven valid as all of the series are absolutely convergent on all of $\mathbb{R}$ (actually on all of $\mathbb{C}$). \begin{align} \sinh x&=\frac{e^x - e^{-x}}{2} \\ &=\frac{1}{2}\left[\sum_{n\in\mathbb{N}}\frac{x^n}{n!} - \sum_{n\in\mathbb{N}} \frac{(-1)^nx^n}{n!}\right] \\ &=\frac{1}{2} \sum_{n\in\mathbb{N}} \left(1-(-1)^n\right) \frac{x^n}{n!} \end{align} Now, notice that when $n$ is even, then the summand will vanish. When $n$ is odd, there will be a factor of $2$. So, \begin{align} \sinh x &= \frac{1}{2}\sum_{n\in 2\mathbb{N}+1}2\cdot \frac{x^n}{n!} \\ &= \sum_{n\in 2\mathbb{N}+1} \frac{x^n}{n!} \\ &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \end{align} | 2022-07-03T05:30:11 | {
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http://mathhelpforum.com/trigonometry/18067-trigonometry-problem.html | # Math Help - trigonometry problem
1. ## trigonometry problem
using the substitution x= tan a, show that
y = tan-1 2x /1+x2
reduces to
y = 2 tan-1 x
this is what i have done:-
tan y = 2x/1+x2
tan y = 2 tan a / 1 + tan2 a
tan y = 2 tan a / sec2 a
tan y = 2sin a cos a = sin 2a
i cant work this out further...can you help me plz?
2. Originally Posted by slash
using the substitution x= tan a, show that
y = tan-1 2x /1+x2
reduces to
y = 2 tan-1 x
this is what i have done:-
tan y = 2x/1+x2
tan y = 2 tan a / 1 + tan2 a
tan y = 2 tan a / sec2 a
tan y = 2sin a cos a = sin 2a
i cant work this out further...can you help me plz?
do you mean: $y = \tan^{-1} \left( \frac {2x}{1 + x^2} \right)$ ?
3. Originally Posted by Jhevon
do you mean: $y = \tan^{-1} \left( \frac {2x}{1 + x^2} \right)$ ?
yes ur right..
sorry im new to all this..i dont know how to use em..
4. Originally Posted by slash
yes ur right..
sorry im new to all this..i dont know how to use em..
first off, i think the denominator should be: 1 - x^2
secondly, you would write it like this: y = arctan [(2x)/(1 - x^2)] .......arctan means the inverse tangent. or you could write tan^{-1} [(2x)/(1 - x^2)]
Now here's how we do it:
$y = \tan^{-1} \left( \frac {2x}{1 - x^2} \right)$
$\Rightarrow \tan y = \frac {2x}{1 - x^2}$
Let $x = \tan a$
$\Rightarrow \tan y = \frac {2 \tan a }{ 1 - \tan^2 a}$
But the right hand side is the double angle formula for tangent, so we get:
$\tan y = \tan 2a$
$\Rightarrow y = \tan^{-1} ( \tan 2a ) = 2a$
Since, $x = \tan a \implies a = \tan^{-1} x$, we have:
$y = 2 \tan^{-1} x$
as desired
5. Originally Posted by slash
using the substitution x= tan a, show that
y = tan-1 2x /1+x2
reduces to
y = 2 tan-1 x
this is what i have done:-
tan y = 2x/1+x2
tan y = 2 tan a / 1 + tan2 a
tan y = 2 tan a / sec2 a
tan y = 2sin a cos a = sin 2a
i cant work this out further...can you help me plz?
You converted [1 +tan^2(a)] into sec^2(a). Although the conversion is true, it is not needed here.
2tan(a) / 1+tan^2(a)
is tan(2a).
So,
tan(y) = tan(2a)
Get the arctans both sides,
y = 2a -----------------------(i)
But x = tan(a)
So, getting the arctans of both sides,
arctan(x) = a
Substitute that into (i),
y = 2[arctan(x)] --------------shown.
6. Hello, slash!
I agree with Jhevon . . . there is a typo.
Using the substitution $x\,= \,\tan a,$
show that: . $y \:= \:\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ reduces to: . $y \:= \:2\tan^{-1}(x)$
. . . . . . . . . . . . . . . . . . ${\color{red}\uparrow}$
We're expected to know this identity: . $\tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta}$
Making that substitution, we have:
. . $y \;=\;\tan^{-1}\left(\frac{2\tan a}{1 - \tan^2\!a}\right) \;=\;\tan^{-1}(\tan2a) \;=\;2a$
Since $x = \tan a\!:\;\;a \,=\,\tan^{-1}(x)$
. . Therefore: . $y \;=\;2\tan^{-1}(x)$ | 2015-05-07T05:28:26 | {
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https://math.stackexchange.com/questions/2734993/number-of-edges-of-a-graph-with-singles-and-pairs-is-62-how-many-vertices-are-t | # Number of edges of a graph with singles and pairs is 62. How many vertices are there?
Let's consider an undirected graph whose every vertex is either connected with every other vertex or with every vertex except exactly one.
(To avoid misunderstandings: The graph must be representation of a people in a party, who came either alone or with a partner. Edges are handshakes.)
Problem: The graph consists of 62 edges. How many vertices does it have?
My approach: Let's take a subgraph with no pairs (largest complete graph component) such that number of edges is less than 62, and then gradually add pairs.
So number of edges in complete graph with $n$ vertices is $\frac{n(n-1)}{2}$. Adding first pair brings $2n$ edges, next $2(n+2)$ etc. By bruteforce we have: $$\frac{4(4-1)}{2}+2*4+2*6+2*8+2*10=62$$
Hence $4+2+2+2+2=12$ vertices. Is it correct?
I really don't like my approach with bruteforce since I can't see if it's the only possible combination. In the lecture notes we have proven handshaking lemma, but I don't see where to apply it. Please let me know if you have a better solution.
Some vertices (say $k$) are connected to all the other vertices and the rest ($l$ so that $n=k+l$ is the total number of vertices) are connected to all the vertices but one (their partner), thus, there should be $\frac{k(k+l-1)}{2} + \frac{l(k+l-2)}{2}$ edges (remember how the $\frac{n(n-1)}{2}$ value you gave is found: each vertex, i.e. $n$ possibilities, is bound to all the other vertices, i.e. $n-1$ possibilities, and in an undirected graph, by doing that you counted each edge twice).
Therefore, what you want to solve is $\frac{k(k+l-1)}{2} + \frac{l(k+l-2)}{2}=62$. Catching up on your solution: for $k=4; l=8$, we indeed find 62 edges. Now that you have an explicit formula for the number of edges, you may easily find all the combinations (although this is still bruteforce).
The formula is exactly what the handshaking lemma would give you since there are $k$ vertices of degree $n-1$ and $n-k$ vertices of degree $n-2$.
• No, not necessarily. The graph must be representation of a people in a party, who came either alone or with a partner. Edges are handshakes. – user3342072 Apr 13 '18 at 7:27
• Ok, then what bugged is that there are nodes without a pair.That is why you first took the complete graph (representing all people coming alone) and then added the couples that came along (but do not shake hands with their respective partners). Am I right? – Bill O'Haran Apr 13 '18 at 7:46
• Yes, you are absolutely right. I apologise for being unclear. – user3342072 Apr 13 '18 at 7:52
• No problem, I'll edit my answer right away. I suggest you edit your question. – Bill O'Haran Apr 13 '18 at 7:53
The only solution is $n=12.$
If $n$ is the number of vertices and $k$ the number of pairs, then we must have $$2k\le n,$$ and the number of edges is $$\binom n2-k=62.$$ Thus we have $\binom n2=62+k\ge62,$ that is, $n(n-1)\ge124,$ and also $\binom n2=62+k\le62+\frac n2,$ which simplifies to $n(n-2)\le 124.$ The only integer solution of the inequalities $$n(n-2)\le124\le n(n-1)$$ is $n=12,$ and then $k=\binom{12}2-62=4.$ | 2019-12-14T08:10:16 | {
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https://www.emathhelp.net/notes/calculus-2/numerical-approximate-integration/simpsons-rule/ | # Simpson's Rule
## Related Calculator: Simpson's (Parabolic) Rule Calculator
Idea of Simpson's rule is in following: approximate curve by parabola and then find area of parabola (it is easy to do because we know antiderivative of quadratic function).
Again we divide [a,b] into n subintervals of equal length Delta x=(b-a)/n, and also require n to be even number.
Then on each consecutive pair of intervals we approximate the curve y=f(x) by a parabola. If y_i=f(x_i), then P_i=(x_i,y_i) is the point on the curve lying above x_i.
A typical parabola passes through three consecutive points P_i , P_(i+1) and P_(i+2).
First we find equation of parabola that passes through points (x_0,y_0), (x_1,y_1) and (x_2,y_2).
Also note that x_1=x_0+Delta x and x_2=x_0+2Delta x.
Equation of any parabola has form y=Ax^2+Bx+C and so area under parabola from x=x_0 to x=x_2=x_0+2Delta x is
S=int_(x_0)^(x_0+2Delta x) (Ax^2+Bx+C)dx=(A/3 x^3+B/2 x^2+Cx)|_(x_0)^(x_0+2Delta x)=
=(A/3 (x_0+2Delta x)^3+B/2 (x_0+2Delta x)^2+C(x_0+2Delta x))-(A/3 x_0^3+B/2 x_0^2+Cx_0)=
=2ADelta x x_0^2+4A(Delta x)^2x_0+8/3A(Delta x)^3+2B Delta x x_0+2B(Delta x)^2+2CDelta x=
=(Delta x)/3(A(6x_0^2+12Delta x x_0+8A(Delta x)^2)+B(6x_0+6Delta x)+6C).
Additionally parabola should pass through points P_0=(x_0,y_0), P_1=(x_1,y_1) and P_2=(x_2,y_2) (recall that x_1=x_0+h and x_2=x_0+2h)so
{(y_0=Ax_0^2+Bx_0+C),(y_1=A(x_0+Delta x)^2+B(x_0+Delta x)+C),(y_2=A(x_0+2Delta x)^2+B(x_0+2Delta x)+C):}
From this we have that
y_0+4y_1+y_2=
=Ax_0^2+Bx_0+C+4(A(x_0+Delta x)^2+B(x_0+Delta x)+C)+A(x_0+2Delta x)^2+B(x_0+2Delta x)+C=
=A(6x_0^2+12Deltax x_0+8(Delta x)^2)+B(6x_0+6Delta x)+6C.
If we know multiply both sides of equality by (Delta x)/3 we will obtain that
(Delta x)/3 (y_0+4y_1+y_2)=(Delta x)/3(A(6x_0^2+12Delta x x_0+8(Delta x)^2)+B(6x_0+6Delta x)+6C).
But right side is exactly area S under parabola.
Therefore, S=(Delta x)/3(y_0+4y_1+y_2).
Similarly, it can be shown that the area under parabola through P_2, P_3 and P_4 from x=x_2 to x=x_4 is (Delta x)/3(y_2+4y_3+y_4).
In general, area under parabola through P_i, P_(i+1), P_(i+2) from x=x_i to x=x_(i+2) is (Delta x)/3(y_i+y_(i+1)+y_(i+2)).
Summing areas under parabolas on each subinterval we will obtain that int_a^bf(x)dx~~h/3(y_0+4y_1+y_2)+h/3(y_2+4y_3+y_4)+...+h/3(y_(n-2)+4y_(n-1)+y_n).
Simpson's Rule. int_a^bf(x)dx~~S_n=(Delta x)/3(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_(n-2)+4y_(n-1)+y_n).
where n is even and Delta x=(b-a)/n.
Note the pattern of coeffcients: 1,4,2,4,2,4,2,4,2,...,4,2,4,2,4,2,4,1.
Example 1. Use Simpson's Rule to approximate value of int_1^2 1/x^2dx with n=8.
Here a=1, b=2, f(x)=1/x^2 and n=8. So, Delta x=(b-a)/n=(2-1)/8=0.125.
So, int_1^2 1/x(dx)~~S_n=
=(0.125)/3(f(1)+4f(1.125)+2f(1.25)+4f(1.375)+2f(1.5)+4f(1.625)+2f(1.75)+4f(1.875)+f(2))=
=0.125/3(1/1^2+4/(1.125)^2+2/(1.25)^2+4/(1.375)^2+2/(1.5)^2+4/(1.625)^2+2/(1.75)^2+4/(1.875)^2+1/2^2)~~0.5000299.
True value of integral is I=int_1^2 1/x^2dx=0.5. As can be seen Simpson's rule gave very good approximation.
When we approximate integral we will always have some error: E=int_a^bf(x)dx-App where App is approximation and E is error.
Error Bound for Simpson's Rule. Suppose |f^((4))(x)|<=M for a<=x<=b then |E|<=(M(b-a)^5)/(180n^4).
Example 2. How large should we take n in order to guarantee that the Simpson's Rule approximation for int_1^2 1/x^2 dx are accurate to within 0.0002?
Here a=1, b=2, f(x)=1/x^2.
Then f'(x)=-2/x^3, f''(x)=6/x^4, f'''(x)=-24/x^5 and f^((4))(x)=120/x^6.
Therefore |f^((4))(x)|<=120 for 1<=x<=2.
Thus, (120(2-1)^5)/(180n^4)<0.0002 or n^4>120/(180*0.0002)=1/(0.0003).
So, n>1/(root(4)(0.0003))~~7.6.
So, we need to take n=8 (remember n must be even).
This is much better then n=36 for Midpoint Rule and n=51 for Trapezoidal Rule. | 2019-03-23T08:13:08 | {
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http://math.stackexchange.com/questions/56815/how-do-i-calculate-the-probability-of-getting-10-unique-members-of-a-set-given-r | # How do I calculate the probability of getting 10 unique members of a set given random selection with replacement?
Assume I have a set of 20 numbers. Each number in the set is unique. I am able to retrieve one number at a time from the set with the probability of retrieving any one member of the set being equal. How would I go about determining the probability that after ten random retrievals with replacement (i.e. a retrieval does not remove a number from the set) I would have 10 unique members?
-
We give three not very different arguments. The first two use counting, and the third works directly with probabilities.
Let us record the numbers that we get as a sequence of length $10$. So the sequence $(a_1, a_2, \dots, a_n)$ means that we got $a_1$ on the first pick, $a_2$ on the second pick, and so on.
There are $20^{10}$ such sequences, all equally likely, since we are replacing at each stage.
Suppose that we will be happy only if all the numbers are different.
Any of the $20$ numbers is fine for $a_1$. But for every choice of $a_1$, there are only $19$ values of $a_2$ that keep us hoping, for a total of $(20)(19)$ choices for $(a_1,a_2)$. But for every such choice of $(a_1,a_2)$, there are only $18$ values of $a_3$ that keep us hoping, for a total of $(20)(19)(18)$ choices for $(a_1,a_2,a_3)$. Continue.
We find that there are $(20)(19)(18)\cdots (12)(11)$ sequences of length $10$ that make us happy. Thus the required probability is $$\frac{(20)(19)(18)\cdots (12)(11)}{20^{10}}.$$
Another way: We count again the number of sequences of length $10$ that make us happy. Call a string of length $10$ good if all its entries are different. The numbers that appear in a good string can be chosen in $\dbinom{20}{10}$ ways. Here $\dbinom{n}{r}$ is the number of ways to choose $r$ objects from $n$ distinct objects. On scientific calculators, the label is ${}_nC_r$.
For every choice of the $10$ distinct numbers, these numbers can be lined up in a row in $10!$ ways. So there are $\dbinom{20}{10}10!$ good sequences of length $10$. The probability we will be happy is therefore $$\frac{\binom{20}{10}10!}{20^{10}}.$$
Still another way: We can work directly with probabilities. The probability that we are still hopeful after the second draw is $\dfrac{19}{20}$.
Given that we are still hopeful after the second draw, the probability we are hopeful after the third draw is $\dfrac{18}{20}$. So the probability that we are still hopeful after the third draw is $\dfrac{19}{20}\cdot\dfrac{18}{20}$.
Continue. The probability that we get $10$ distinct integers is $$\frac{19}{20}\cdot\frac{18}{20}\cdot\frac{17}{20} \cdots\frac{12}{20}\cdot\frac{11}{20}.$$ We can make the answer look nicer, at least to a mathematician, by putting a $\dfrac{20}{20}$ at the front.
-
The first selection will have $20$ possibilities, the next selection $19$ possibilities, and so on until the $10$-th selection has $11$ possibilities. The total number of ways of to choose a ball from these possibilities each time is $20\cdot19\cdots12\cdot11$. The total number of ways to choose $10$ balls at all is $20^{10}$. So you're looking at
$$P=\frac{20!}{10! 20^{10}}\approx 0.0655.$$
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https://math.stackexchange.com/questions/893218/inequality-of-integrals-int-01fx2-dx-geq-4-if-int-01xfx-dx-int-0 | # Inequality of integrals $\int_0^1(f(x))^2 dx \geq 4$ if $\int_0^1xf(x) dx=\int_0^1f(x) dx = 1$
If
$$\int_0^1xf(x) dx=\int_0^1f(x) dx = 1$$
prove that
$$\int_0^1(f(x))^2 dx \geq 4$$
EDIT My attempt is as follows - I can use only the $\int xf(x)$dx part to get a bound $\int f^2(x) dx \geq 3$ from cauchy schwarz. I can't think of ways how to incorporate the other given condition.
• I'm sure you would get some help if you showed some effort in solving the problem. Most people here are not interested in doing your homework for you. Aug 10 '14 at 18:16
• @Winther I added my attempt. Aug 10 '14 at 18:31
• Thats what I too tried at first:) As you say, CS is too weak here as it doesn't incorporate the second condition. Expanding $f$ in a basis of some orthogonal polynomials is the best way to go (see Jack's answer)! Aug 10 '14 at 18:40
• This is a very nice problem, indeed. Equality holds only for $f(x)=6x-2$, as shown below. Aug 10 '14 at 18:40
Of course once the inequality condition is obtained (or guessed by trying out a general linear polynomial), Cauchy-Schwarz is a breeze:
$$\int_0^1 f^2 dx \ge \frac{\left(\int_0^1 (3xf-f)dx\right)^2}{\int_0^1(3x-1)^2 dx } = \frac{4}{1}=4$$
Equality is when $f(x)$ is proportional to $3x-1$.
In general we can have $$\int_0^1 f^2 dx \ge \frac{\left(\int_0^1 (axf+bf)dx\right)^2}{\int_0^1(ax+b)^2 dx } = \frac{(a+b)^2}{a^2/3+ab+b^2}$$ where the maximum is when $(a, b)$ is proportional to $(3, -1)$, so $4$ is indeed the best possible.
• I think you're missing an $x$ in the numerator of the middle term in the last expression. Nice derivation based only on the Cauchy-Schwarz inequality, with optimal case following from equality case of the Cauchy-Schwarz inequality. Aug 11 '14 at 8:28
• @user161825 You're right, I missed it and edited it in now.. thanks. Aug 11 '14 at 8:29
Write your function in terms of the shifted Legendre polynomials $\tilde{L}_n(x)=L_n(2x-1)$, that are an orthogonal base of $L^2((0,1))$ with respect to the usual inner product. Assuming: $$f(x) = \sum_{n=0}^{+\infty} a_n\,\tilde{L}_n(x),$$ the constraints give: $$a_0 = 1,\qquad \frac{a_0}{2}+\frac{a_1}{6} = 1,$$ hence $a_0=1$ and $a_1=3$. This implies:
$$\int_{0}^{1}f(x)^2 dx = a_0^2+\sum_{n=1}^{+\infty}\frac{a_n^2}{2n+1}\geq 1+\frac{9}{3}=4.$$
Moreover, you have that equality holds only for $f(x)=6x-2$.
• What made you think of using Legendre polynomials? As far as I can tell, this is much simpler than using the Fourier system for instance. Aug 10 '14 at 19:35
• You have the need to write $\int x\,f(x)\,dx =1$ in terms of a finite number of coefficients. I would have used the Fourier system if the constraints were something like $\int_{0}^{1}f(x)\sin(2\pi x)\,dx = 1$, for instance. Aug 10 '14 at 19:40
We have for each $a,b$: $$\int_0^1 \Big(f(x)-ax-b\Big)^2dx \geq 0$$ So we have \begin{eqnarray*} \int_0^1 f(x)^2 dx &\geq &2\int _0^1f(x)(ax+b)dx -\int _0^1 (ax+b)^2dx \\ &=&2(a+b) -{(a+b)^3-b^3\over 3a}\\ &=&2(a+b) -{a^2+3ab+3b^2\over 3} =:E \end{eqnarray*} This inequality is valid for all $a,b$, so even when $E$ achieves maximum: \begin{eqnarray*} E &=&\underbrace{-b^2 +b(2-a) - {(2-a)^2\over 4}} + \underbrace{{(2-a)^2\over 4} -{a^2\over 3}}\\ &=&-\Big(b - {2-a\over 2}\Big)^2+{-a^2-12a+12\over 12}\\ &\leq & {-a^2-12a+12\over 12} \\ &=& {-(a-6)^2+48\over 12}\\ &\leq & 4 \end{eqnarray*} Equality is achieved at $a=6$ and $b={2-a\over 2}= -2$, that is when $f(x)=6x-2$. | 2021-09-29T03:04:59 | {
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https://complexworld.wikispaces.com/Questions?responseToken=c3468c2e46c6e6c3d7ca2e636c77fa3a | Instructions: click "edit" in the upper right corner and type in your question after the instructions but before all other questions. If you want to write a math formula, look at how other formulas have been typed on this wiki or type it in plain text. I will answer the question as soon as I can
If I have skipped your question below, or if something is unclear, or if I accidentally deleted it, please leave a comment about it (I was answering the questions today in a hurry and concurrently with some other edits)
Q : Give an example of a non-constant analytic function f(z) such that the equation f(z) = 0 doesn’t have any solutions.
A: f(z)=e^z
Q: Find the number of zeroes of z^2+e^z-2 in the disc |z|<2
A: Is the question about z^2 + e^(z-2) ? As written, it is a very hard question.
Q: True or False: There exists an analytic (one-valued) function on the complex plane whose square is equal to e^z-1.
A: False. If f(z)^2 = e^z - 1, then differentiating both sides we get 2 f(z) f ' (z) = e^z. In particular f(0)^2 = 0 and 2 f(0) f ' (0) = 1. These two equalities are incompatible.
A little more conceptual proof: If an analytic function f has a zero of order k at z=z_0, then f^2 has a zero of order 2k at z_0. Thus a square of an analytic function can have only zeroes of even orders. The function e^z - 1 has a zero of order 1 at z=0, so it can't be a square of an analytic function.
Q: Find the number of zeros of z^2050+4z^4-e^z in the disc abs(z)<1
A: First notice that if |z|=1, then Re(z)<=1 and hence |e^z|=e^(Re(z))<=e.
Thus if |z|=1, |4z^4|=4, while |z^2050-e^z|<=1+e < 4. By Rouche theorem there are 4 zeroes of z^2050+4z^4-e^z in |z|<1.
Q:∫Re(z)dz from 1+i to 2+2i (straight line)
A: Parametrize the line as z=1+i + t(1+i), 0=<t<=1. Then dz=(1+i)dt and Re(z)=1+t. Thus
$\int Re(z) dz = \int_{0}^1 (1+t) (1+i) dt = (1+i) (1+1/2) = 3/2+3/2 i$
Q: lim R-->inf of ∫sin(1/z)dz over the semicircle z=Re^it [t=0 to t=pi]
A: sin(1/z) = 1/z - 1/3! 1/z^3 + 1/5! 1/z^5 +...
Thus
$\int \sin(1/z)\,dz = \int dz / z - 1/3! \int dz/ z^3 + ... = \pi i - 1/3! \frac{z^{-2}}{-2}|\limits_{z=R}^{z=-R} + ...$
As R tends to infinity, all terms but the first one tend to zero. Hence the limit as R tends to infinity is pi i.
Q:In last problem of the 8th lecture, you showed that for f(z)=1/(z (z+1) (z+2), to compute the residue at z=-1 one can write f(z)= 1/(z+1) * 1/z(z+2). Then you substitute z=-1 into 1/z(z+2) to find its residue without take its derivative. However, in the notes of lecture(the one you typed it on wiki), you showed the similar result for residue Res( 1/(z+1/2)^2 *[z/4(z+2)^2]; z=-1/2), at this point, you take its derivative and set z=-1/2 to find its residue. Which one is correct?
A: Both are correct. Suppose you want to compute the residue at z= z_0 of a function of the form f(z)/(z-z_0)^k, where f is analytic and f(z_0) is not zero.
You write the Taylor series expansion of f around z_0:
$f(z) = f(z_0) + f ' (z_0) (z-z_0) + ... + f^{(k-1)}(z_0)/(k-1)! (z-z_0)^{k-1}+\ldots$
Then
$f(z) / (z-z_0)^k = \frac{f(z_0)}{(z-z_0)^k} + \frac{f ' (z_0)}{(z-z_0)^{k-1}} + ... + \frac{ f^{(k-1)}(z_0)/(k-1)! }{z-z_0}+\ldots$
And thus the residue is
$f^{(k-1)}(z_0)/(k-1)!$
Q: Show that every function f analytic on the complex plane and satisfying |f(z)| > 1 for all z is constant.
A: Since |f(z)|>1 for all z, in particular f(z) is not zero for any z. Thus g(z)=1/f(z) is analytic and satisfies |g(z)|<1 for all z. By Liouville's theorem g is constant. Thus f=1/g is constant as well.
Q: Find the number of zeroes of the function f(z) = z^2011 + 4z^4 - 2 in the disc |z| < 1.
A: On |z|=1, |4z^4|=4, while |z^2011-2|<=1+2 =3 < 4. Thus by Rouche theorem there are 4 zeroes inside the disc |z|<1
Q: ∫x^4/(1+x^10) dx from x= - infinity to x= infinity.
A: Very similar to the question
∫ x^4/(1+x^8) dx from negative infinity to infinity.
Q: Show that for real a,
∫[e^(acost)][cos(asint)]dt = Pi (definite integral from t=0 to Pi)
HINT (Not implying that you need one :) ): Let gamma be the unit circle transversed once in the positive sense and consider ∫(e^az)/z dz.
A: Let z=e^it. Then e^(az)=e^(acos t + i a sin t) = e^(a cos t) (cos (a sin t)+ i sin (a sin(t)).
Also (dz)/z = i dt.
Thus
$\int_{|z|=1} e^{az} \frac{dz}{z}=i \int_{-\pi}^{\pi} e^{a \cos t} \cos(a\sin t) dt - \int_0^{2\pi} e^{a \cos t} \sin(a\sin t) dt$
Thus
$\int_{-\pi}^{\pi} e^{a \cos t} \cos(a\sin t) dt = Im( \int_{|z|=1} e^{az} \frac{dz}{z} )$
Finally by residue theorem, the integral of e^(az)/z over the unit circle is 2 pi i. Thus
$\int_{-\pi}^{\pi} e^{a \cos t} \cos(a\sin t) dt = 2 \pi$
Finally the function e^{a \cos t} \cos(a\sin t) is even, and hence its integral from 0 to pi is half the integral from -pi to pi
Q: True or false, 'all values of (-1)^i are real numbers'. Explain your reasoning.
A:
$(-1)^i = e^{(\pi i + 2\pi i k)i} = e^{-\pi - 2\pi k}$
All these numbers are indeed real (k is any integer here).
Q: How many zeros does the polynomial z^8 +3z^2 +12z + 5 have inside the circle abs(z)=2? Inside abs(z)=1? Explain.
A: On |z|=2, |z^8|=2^8=256, while |3z^2+12z+5|<= 12+24+5<256. Thus by Rouche theorem there are 8 zeroes inside the circle |z|=2.
On |z|=1, |12z|=12, while |z^8+3z^2+5|<=1+3+5=9 < 12, so by Rouche theorem there is one zero inside |z|=1
Q: Show that the principal branch of the logarithm satisfies log(1/z) = -log(z)
A: If z=re^(it) with -pi
(remark: I am a little confused by what happens for real negative values of z. If log is defined for them, then the required equality can't possibly hold (it would imply that log (1/(-1)) = - log(-1), i.e. log(-1)=0, which is nonsense). Probably the principal branch of log isn't defined for negative reals or, alternatively, the question is wrong. I will comment on this once I take a look at our book)
Q:
a) Let f be an analytic function on the open unit disk (that is, the set of zEC such that abs(z)<1) and assume that the values of f are on the parabola y = x^2 (this means that if you write Re(f) = u and Im(f) = v then v = u^2). Show that f must be constant on the unit disk.
b) If f is defined on an open set different from the unit disk, is f necessarily constant? (Justify the answer).
A:
a) v=u^2, hence v_x=2uu_x, v_y=2uu_y.
Cauchy-Riemann equations tell that v_y=u_x, v_x=-u_y.
Thus v_x= - 4u^2 v_x, v_y = -4u^2 v_y. Hence (1+4u^2)v_x=0, (1+4u^2)v_y =0. Hence v_x=v_y=0. Hence u_x=0,u_y=0. Hence u,v are constants.
b) No. Example: f(x)= 0 if |x|<1 and 1+i if |x-10|<1. This function is defined on an open set (union of two discs), its values lie on the parabola v=u^2, it is analytic and yet it is not constant.
Q:
For what value z does the limit as w tends to 0 of (|2+z+w|^2-|2+z|^2)/w exist?
A: let f(z)=|2+z|^2. Then the limit in question is by definition f ' (z).
Now verify at what points z Cauchy-Riemann equations hold for f:
f(x+iy)=(2+x)^2+y^2, so Cauchy-Riemann equations tell 2(2+x)=0, 2y=0.
Thus f can be differentiable only at z= - 2.
When z= - 2, indeed we have lim(|w|^2)/w=lim (conjugate(w)) = 0
Q:
Find the radius of convergence of the power series expansion of the function sin(1/(z^2 + 9)) in powers of (z-4)
A: 5, since the function has poles only at z=3i, z=-3i, which are located at distance 5 from the point 4.
Q: Do we need to know how to use the Jordan Lemma?
(I (a student) asked Yuri this last week and I'm quite sure he said we don't need to know it. Please correct me if I'm wrong!)
A: You don't need to know it.
Q: Every time we do an integral of a real function over the real line, do we have to show that the semicircular part of the complex integral tends to zero? Or can we use the results from class?
A: It is much better if you actually show it.
If you want to, you can quote instead the exact formulation of the result from the class you are using and explain why the result applies to your problem.
Q: Evaluate the line integral ∫ 1/(z^2+1) dz over the positively oriented circle |z|=1. My question here is what happens when singularities exist on the boundary of a curve that is being integrated over? Do we treat the integral any differently than an integral around, lets say abs(z-i)=1 ?
A: The integral you asked about is not even defined (the integral is an improper integral which doesn't even converge).
Q: Prove that for a polynomial of degree n p(z) = a_n*z^n + a_n-1*z^n-1 + ... + a_1*z + a_0,
0.5*|a_n| <= |p(z)/z^n| <= 2|a_n| for |z| large enough. I do not understand the proof in the textbook :(
A: the reason is that lim p(z)/z^n as |z|-> infinity is a_n. So when |z| is large, p(z)/z^n must be very close to being a_n. In particular its absolute value will lie between 1/2 |a_n| to 2 |a_n|
Q:
Find the number of zeroes of z^140 + 200z^40 + z + 1 in the annulus 1<|z|<2.
A: When |z|=1, we have |200z^40|=200 > 3 >= |z^140+z+1|, so by Rouche theorem there are 40 zeroes in |z|<1.
When |z|=2 we have |z^140|=2^140 > 200 2^40+2+1 >= |200 z^40 + z+1|, so by Rouche theorem there are 140 zeroes in |z|<2.
Thus there are 100 zeroes in the annulus 1<|z|<2.
Q: For the question
Find the value of the integral of z / (e^π z- e^-π z) dz over the circle of radius 2 centered at z=1+4i (oriented counterclockwise),
how did you find the residue? Please elaborate on your previous answer.
A:
At point z=in we have the following Taylor expansions:
$z = in + (z-in)$
$e^{\pi z}-e^{-\pi z}= (\pi e^{\pi i n}+\pi e^{-\pi i n})(z-in)+\ldots$
Hence
$\frac{z}{e^{\pi z}-e^{-\pi z}}=\frac{in+\ldots}{(\pi e^{\pi i n}+\pi e^{-\pi i n})(z-in)+\ldots}=\frac{in}{2\pi(-1)^n}\frac{1}{z-in}+\ldots$
and hence the residue at z=in is (-1)^n in/(2 pi).
Q：Evalute the line integral ∫ sin(z)/z^5 dz over the positively oriented circle |z|=1.
A: sin z/ z^5 = (z-z^3/3!+z^5/5!-...)/z^5 = 1/z^4 - 1/3! 1/z^2 + 1/5! +..., hence the residue at z=0 is zero and hence the integral over |z|=1 is zero.
Q:Find number of zeroes of z^2+e^(z-2) in the disc |z| <2.
A: When |z|=2, we have |z^2|=4 while |e^(z-2)|=e^Re(z-2)<=e^0=1, so by Rouche theorem there are 2 zeroes in the disc |z|<2.
(remark: we've used the facts that |e^(x+iy)|=e^x for real x, y and that if |z|=2, the Re z<= 2 and hence Re(z-2)<= 0).
Q: Determine how many zeroes does 4z^3-12z^2+2z+10 have in the annulus 1/2<|z-1|<2
A: Let w=z-1. Then we are interested in the function is 4(w+1)^3-12(w+1)^2+2(w+1)+10=4w^3-10w+4. When |w|=2, the inequality |4w^3|=32> 24>=|-10w+4| holds, so by Rouche theorem, the function has 3 zeroes in |w|<2. When |w|=1/2, the inequality |10w|=5 > 4.5>=|4w^3+4| holds, so by Rouche theorem there is one zero in |w|<1/2 and (once again since when |w|=1/2, the inequality |10w|=5 > 4.5>=|4w^3+4| holds) there are no zeroes on |w|=1/2. Hence there are 2 zeroes in 1/2<|w|<2
Q:Find a Mobius transformation T such that T maps the real axis onto itself and the imaginary axis onto the circle |w-1/2|=1/2
A: Real and imaginary axis intersect at z=0, z=infinity. The real axis and the circle |w-1/2|=1/2 intersect at points w=0,w=1. Hence the Mobius transformation in question must send the points z=0,z=infinity to points w=0,w=1. To make sure that the real axis gets mapped to itself, we choose another point on the real axis (e.g. z=1) and send it to some point on the real axis (e.g. w=infinity).
Now a Mobius transformation sending 0,infinity and 1 to 0,1,infinity must map the real line to the real line. Moreover, the imaginary axis gets mapped by such transformation to some circle passing through points 0,1.
It seems that we haven't really solved the problem (we haven't shown that the imaginary axis gets mapped to the circle |w-1/2|=1/2: instead we have showed only that it gets mapped to some circle passing through points 0,1). However an additional argument helps in this case: the real and imaginary axes intersect at right angle and Mobius transformations preserve angles. Thus if the image of the real axis is the real axis, the image of the imaginary axis must be a circle orthogonal to the real axis. Since we also know that it passes through points 0,1, it must be the circle |w-1/2|=1/2.
Finally a Mobius transformation sending 0,infinity and 1 to 0,1,infinity is w=z/(z-1).
Remark: the answer to this question is not unique (we made a choice of sending 1 to infinity, which we could have replaced by any other choice of "send a point on the real axis to a point on the real axis").
Q: For question
∫ x^4/(1+x^8) dx from negative infinity to infinity. Can we simply use the formula offered in the textbook, namely Res(P/Q; x_0) = P(x_0)/Q'(x_0) over the reals, since the rest of the integral (top semi-circle) converges to 0, which has been shown in the textbook as well?
A: If you want to use a theorem from the book, you should quote the theorem precisely and explain why is it applicable in your situation.
For instance here to justify that Res( x^4/(1+x^8) ; x=x_0 ) = x_0^4/(8x_0^7) where x_0 is a zero of the polynomial 1+x^8, you can proceed in two ways:
Way 1:write "
x^4/(1+x^8)=(x_0^4+...)/(0 + (8x_0^7)(x-x_0)+...)=x_0^4/(8x_0^7) 1/(x-x_0) + ..., hence Res( x^4/(1+x^8) ; x=x_0 ) = x_0^4/(8x_0^7)
"
(here the answer is justified by a computation of the Laurent expansion of x^4/(1+x^8) in powers of (x-x_0) and finding the coefficient of 1/(x-x_0) in that expansion)
Way 2: write "
there is a theorem (for instance in the book we were using in class) that if P,Q are two analytic functions in a neighborhood of a point x_0, P(x_0) is not zero and Q(x) has a zero of order one at x=x_0, then Res(P/Q; x_0) = P(x_0)/Q'(x_0). If we let P(x)=x^4, Q(x)=1+x^8, and x_0 be a zero of 1+x^8, then, since x_0^4 is not zero (indeed, x_0=0 is not a root of 1+x^8) and the order of the zero x=x_0 of Q(x) is one (indeed, Q'(x_0)=8x_0^7 is not zero), the theorem implies that Res(x^4/(1+x^8); x=x_0) = x_0^4/(8x_0^7).
"
A solution of the kind "by a theorem from the book Res(x^4/(1+x^8); x=x_0) = x_0^4/(8x_0^7)" will be graded as 1 or 2 out of 10.
A solution of the form "by the theorem from the book that says that Res(P/Q; x_0) = P(x_0)/Q'(x_0), Res(x^4/(1+x^8); x=x_0) = x_0^4/(8x_0^7)" will also be graded as 2 to 4 points out of 10 (the reason is that the theorem is not quoted with the conditions in which it applies and the fact that these conditions do actually hold in the question is not verified).
Similar reasoning applies to "since the rest of the integral (top semi-circle) converges to 0, which has been shown in the textbook as well" --- you should quote the exact fact you are using!
Q: ∫(cos2x)dx/(1 - 2a*cosx + a^2) from 0 to 2Pi, where -1<a<1
A: Make a change of variables to z=e^(ix). Then cos x = (z+1/z)/2, cos(2x)=(z^2+1/z^2)/2, dx=dz/(iz). Then compute the resulting integral using residue theorem. Please let me know what step you want me to elaborate on.
Q:When we find a Mobius transformation,are we supposed to choose the three points arbitrarily?(Based on the question below)
A: In the questions of the type "find a Mobius transformation sending a given line/circle to a given line/circle" the answer is indeed not unique: one can choose any three points on the source line/circle and any three points on the target line/circle and then find a Mobius transformation sending the first triple to the second one.
Q:Find a Mobius transformation that carries the circle |z|=1 onto the line Re((1+i)w)=0
A: Choose three points on the circle (e.g. -i,1,i) and find a Mobius transformation that carries them to three points on the line (e.g. w=0, w=infinity, w=1+i).
Such a transformation is given by w=(1+i)(i-1)/(2i) (z+i)/(z-1)
Q: Find the number of zeroes of z^199 + iz^2 - i in the upper half-plane Imz>0.
A: We will apply argument principle for the domain bounded by the line segment from -R to R and a semicircle in the upper half-plane centered at the origin and having radius R.
First as z moves along the semicircle of radius R, we have |z|=R and if R is large enough, terms iz^2-i are much-much smaller than z^199. In particular as z makes half a circle, f(z) makes 199 half-circles (approximately half-circles: they are actually somewhat perturbed by the terms iz^2-i) of radius R^199.
Then z goes from -R to R along the reals: z=x, -R<x<R. Then f(z)=x^199 + i (x^2-1). As x increases from x=-R to x=R, the point f(z) goes from being far-far to the left (a little bit above the x-axis) to far-far to the right (a little bit above the x-axis). The question is whether this path "completes" the 199-th half-turn to a full turn and whether it makes any additional turns.
For this notice that f(z) crosses the x-axis at two points: when x=-1 and when x=1. As z moves from -R to -1, f(z) moves from more or less -R^199 + i R^2 (a little above x-axis) to point -1, staying above the x-axis. Then as z moves from -1 to 1, f(z) moves from -1 to 1, now passing below the x-axis. And finally as z goes from 1 to R, f(z) moves from 1 to approximately R^199+iR^2, staying above the x-axis.
Since indeed the path went around point zero one additional half-turn, the number of turns the image of the boundary of the domain makes around the origin is 100. Hence there are 100 zeroes in the domain.
Q: please add the 'usual arguments about the integration on the semicircle' for the second question below this one.
A: see below.
Q:∫ x^2/(x^4-4x^2+5) dx from negative infinity to infinity.
A: Similar to the question below, but the answer seems to be much worse.
Let R be a large number, let C be the semicircle x=R e^it, 0<t<pi and let I be the interval [-R,R]. First we compute the integral over the closed contour consisting of I and C. By the residue theorem this integral is equal to
2pi i times sum of residues of x^2/(x^4-4x^2+5) at the roots of the denominator. The denominator factors as (x^2-2+i)(x^2-2-i), so the roots are at
$\pm \sqrt{2\pm i}$
Two of these roots lie in the upper half-plane, so the integral is equal to 2pi i times the sum of the residues of x^2/(x^4-4x^2_5) at these two points. To be more precise, let
$\phi=\arctan 1/2 , (0<\phi<\pi/2)$
Then the relevant roots are
$\sqrt[4]{5}e^{i\phi/2}$
and
$-\sqrt[4]{5}e^{-i\phi/2}$
The residue at such a root x_0 is
$x_0^2/(4x_0^3-2x_0)=x_0/(4x_0^2-2)$
i.e. the residues are
$\pm\sqrt[4]{5}e^{\pm i\phi/2}/(4(2\pm i)-2)= \pm\sqrt[4]{5}e^{\pm i\phi/2}/(6\pm 4i)$
Hence the integral over C and I is equal to
$2\pi i( \sqrt[4]{5}e^{i\phi/2}/(6+4i)- \sqrt[4]{5}e^{-i\phi/2}/(6-4i) ) = 2\pi \sqrt[4]{5}(\sin(\frac{\arctan\frac12}2-\arctan\frac23)/\sqrt{13})$
Finally the integral over C is bounded by pi R (R^2/(R^4-4R^2-5)), which tends to 0 as R tends to infinity. Hence the integral over the realsis equal to the limit of the integral over I and C as R tends to infinity, so it is also equal to
$2\pi \sqrt[4]{5}(\sin(\frac{\arctan\frac12}2-\arctan{2/3})/\sqrt{13})$
Q: ∫ x^4/(1+x^8) dx from negative infinity to infinity.
A: Let R be a large number, let C be the semicircle x=R e^it, 0<t<pi and let I be the interval [-R,R]. First we compute the integral over the closed contour consisting of I and C. By the residue theorem this integral is equal to
2pi i times sum of residues of x^4/(1+x^8) at x=e^(pi i/ 8), x=e^(3 pi/8), x=e^(5 pi / 8), x=e^(7 pi i /8)
The residue at x_0=e^(pi i/8 + 2pi k i / 8) is of x_0^4/(8x_0^7)=1/8 x_0^(-3) i.e. it is
1/8 e^(-3 pi i / 8 - 6 pi i/8)
Hence the integral is equal to 2 pi i / 8 (e^(-3 pi i/ 8) + e^(-9 pi i/8) + e^(-15 pi i/8) + e^(-21 pi i/8)), which can be siplified to
2 pi i / 8 (e^(-3 pi i/ 8) - e^(-pi i/8) + e^(pi i/8) - e^(3 pi i/8)) = 2 pi i /8 (2 i sin(pi/8) - 2 i sin(3 pi/8))=pi/2 (sin(3 pi/8)-sin(pi / 8))
Now we relate this integral to what we actually wanted to compute. For this note that when |x|=R, the integrand is bounded by
|x^4/(1+x^8)|<= R^4/(R^8-1)
Hence by triangle inequality the integral over C is bounded by pi R (R^4/(R^8-1)), which tends to 0 as R tends to infinity. Hence as R tends to infinity, the integral over C and I tends to the integral over the real line. Hence the integral over the real line is also equal to pi/2 (sin(3 pi/8)-sin(pi / 8)).
Q:Find the Laurent series about z0=0 for exp(z+1/z) in 0<|z|<infinity
A:
$e^{z+1/z}=e^z e^{1/z}=(\sum\limits_{k=0}^\infty \frac{z^k}{k!})( \sum\limits_{m=0}^\infty \frac{(1/z)^m}{m!} )= \sum\limits_{k=0,m=0}^{\infty,\infty} \frac{x^{k-m}}{k!m!}=\sum\limits_{n=-\infty}^\infty (\sum\limits_{m=0}^\infty \frac{1}{m!(m+n)!})x^n$
Q:Locate each of the isolated singulatiries of the following functions and tell whether it is a removable singularity,a pole,or an essential singularity.
1. (z^2)/(sinz)
2. (2z+1)/(z+2)
A: For the first function the singularities are at the zeroes of the denominator, i.e. at points z where sin z=0. These points are z= n pi for integer n.
At point z=0 we have (z^2)/(sin z)= (z^2)/(z - z^3/6 +...) = z + ..., so the singularity is removable (the power series expansion starts with non-negative term, or, in other words, the limit of z^2/sin z as z goes to 0 exists).
At z=n pi for non-zero n we have (z^2)/(sin z)= ((n pi)^2 + ...)/( pi cos (n pi) (z-n pi) + ...)= plus/minus n^2 pi^2 / (z-n pi) + ...
This shows that z=n pi for non-zero n is a pole of order 1.
2. Point z=-2 is the only singularity. At this point we have (2z+1)/(z+2)=(-3 + 2(z+2))/(z+2)=(-3)/(z+2) + 2, so the point z= -2 is a pole of order 1.
Q: Find the residue of (cosz - e^z)/(tanz - sinz), around z=0.
A: We'll have to find several terms in the Taylor series expansions of the numerator and the denominator around the point z=0.
The numerator is simple: cos z - e^z = (1-z^2/2+z^4/24+...) - (1+z+z^2/2+z^3/6+z^4/24+...)= - z - z^2 - z^3/6 + 0 z^4 +....
For the denominator we'll first have to compute several terms in Taylor series expansion of tan z = sin z / cos z around z=0:
Suppose sin z/ cos z= (z - z^3/6 + z^5 / 120+...)/(1 - z^2/2 + z^4/ 24 + ...) = z + a_3 z^3 + a_5 z^5 + ...
(note that only terms of odd order can appear in the Taylor series expansion of and odd function tan z around the point z=0).
Then we have (z + a_3 z^3 + a_5 z^5 + ...)(1 - z^2/2 + z^4/ 24 + ...) = z - z^3/6 + z^5 / 120+...
or, opening the brackets:
z + (a_3-1/2)z^3 + (a_5 - a_3/2 + 1/24)z^5 + ... = z - z^3/6 + z^5 / 120 + ...
Hence a_3 = 1/2 - 1/6 = 1/3,
a_5 = 1/120 + 1/6 - 1/24 = 2/15
Thus tan z = z + z^3/3 + 2/15 z^5 + ...
It follows that the denominator, tan z - sin z, is equal to (z + z^3 / 3 + 2/15 z^5 + ...) - (z - z^3/6 + z^5/120 + ...) = z^3 /2 + z^5 / 8 + ...
Finally we compute that
(cosz - e^z)/(tanz - sinz) = (- z - z^2 - z^3/6 + 0 z^4 +....)/(z^3 / 2 + z^5 / 8+ ...) = - 2 / z^2 + a_{-1} / z + ...
where the coefficient a_{-1} that interests us the most can be found from the identity
(- 2 / z^2 + a_{-1} / z + ...)( z^3 / 2 + z^5 / 8+ ... ) = - z - z^2 - z^3/6 + 0 z^4 +....
i.e.
- z + a_{-1}/2 z^2 + ... = - z - z^2 + ...
Thus the residue a_{-1} is a_{-2} = - 2
Q：Evaluate the value of the integral of dx/((1+x)x^(2/3)) form 0 to infinity.
A: See examples 9 and 10 in chapter 2.6 in the book for related examples.
We'll compute the integral along the "keyhole" contour consisting of
(1) the segment from r to R
(2) the circle z=R e^it, t goes from 0 to 2pi
(3) the segment from R to r
(4) the circle z=e e^it, t goes from 2pi to 0
Along the contour we interpret z^2/3 as |z|^2/3 e^(2/3 i phi), where phi is from 0 to 2pi.
The integral over the whole contour is equal to 2pi i times the residue of 1/((1+x)x^(2/3)) at x= - 1, which is 1/(-1)^(2/3)=1/e^(2/3 pi i). So the integral is 2 pi i e^(-2/3 pi i).
The integral (3) is equal to
$\int\limits_R^r \frac{dx}{(1+x)(x^{2/3}e^{\frac23 2 \pi i})}=-\frac{ \int\limits_r^R \frac{dx}{(1+x)x^{2/3}} }{ e^{4 \pi i/3} }$
The integral (2) tends to zero as R tends to infinity:
$|\int_{|x|=R}\frac{dx}{(1+x)x^{2/3}}|\leq 2 \pi R\frac{1}{(R-1)R^{2/3}}\to 0$
Similarly
$|\int_{|x|=r}\frac{dx}{(1+x)x^{2/3}}|\leq 2 \pi r\frac{1}{(1-r)r^{2/3}}\to 0$
as r tends to 0.
Hence
$(1-\frac{1}{e^{2/3\pi i}})\int_0^\infty \frac{dx}{(1+x)x^{2/3}}=2\pi i e^{-2/3 \pi i}$
or
$\int_0^\infty \frac{dx}{(1+x)x^{2/3}}=2\pi i \frac{e^{-2/3 \pi i}}{ 1-\frac{1}{e^{4\pi i/3}}} }=\pi\frac{2i}{e^{2\pi i/3}-e^{-2\pi i/3}}=\frac{\pi}{\sin \frac{2\pi}{3}}=\frac{2\pi}{\sqrt{3}} }$
Q：Find the value of the integral of z / (e^π z- e^-π z) dz over the circle of radius 2 centered at z=1+4i (oriented counterclockwise)
A: First we find the singularities of the function:
$e^{\pi z}-e^{-\pi z}=0$
means that z=i n for some integer n.
Of these points only the points 3i, 4i, 5i are inside the circle |z-(1+4i)|=2. So we have to compute the residues at these three points.
At point z=in we have
$\frac{z}{e^{\pi z}-e^{-\pi z}}=\frac{in+\ldots}{(\pi e^{\pi i n}+\pi e^{-\pi i n})(z-in)+\ldots}=\frac{in}{2\pi(-1)^n}\frac{1}{z-in}+\ldots$
so the residue at z=in is (-1)^n in/(2 pi). In particular the residues at points 3i, 4i, 5i are -3/2 i/pi, 2 i/pi, -5/2 i/pi respectively. Their sum is -2i/pi. Hence the integral is 2pi i (-2i/pi)=4.
Q:Find the order of each of the zeros of (z^2-4z+4)^3.
A: (z^2-4z+4)^3=(z-2)^6, so z=2 is a zero of order 6.
Q: Let f(z)=z^2/(z^2 -1). Find the Laurent series representation of f around point z0=1. Also, give the residue at that point.
A: Let w=z-1. Then the function becomes f(w)=(w+1)^2/(w^2+2w)=1/w ((w^2+2w+1)/(w+2))= 1/w (w + 1/(w+2)).
Now 1/(w+2)=1/2 1/(1+w/2)=1/2 (1 - w/2 + w^2/4 - w^3/8+...), Hence w + 1/(w+2)=1/2+3/4 w + w^2/8 - w^3/16 + ... + (-1)^n w^n/2^(n+1)+... and hence 1/w (w + 1/(w+2)) = 1/2 (1/w) + 3/4 + w/8 - w^2/16 + ...
Thus f(z)=1/2 (1/(z-1)) + 3/4 + (z-1)/8 - (z-1)^2/16 + ....
The residue is 1/2
Q: Let f be an analytic function on the punctured disc, 0<abs(z-z_0)<R and f has an essential singularity at the point z=z_0. Let w be any complex number and let g(z)=1/(f(z) -w).
Show that g is not bounded in any punctured disc 0<abs(z-z_0)<r.
A: The idea is that if it is bounded, it has only removable singularities and hence f(z)-w has only poles.
I will write down the details later.
Q: How is it possible to write a Laurent series for f(z)=z/(sinz)^2 around z=0
A: You can only find several first terms in the expansion (as many as you want, but still it's very hard to find a general formula): first you find sin z = z- z^3/6+ z^5/120+.... Then you find (sin z)^2=(z- z^3/6+ z^5/120+...)(z- z^3/6+ z^5/120+...)=z^2-z^4/3+(1/36+2/120)z^6+...
Then you let z/(sin z)^2= 1/z + a_0 + a_1 z + a_2 z^2 + a_3 z^3... and write down the condition that
(1/z + a_0 + a_1 z + a_2 z^2 + a_3 z^3...)(z^2-z^4/3+(1/36+2/120)z^6+...)=z
and compare coefficients in
z+a_0z^2+(-1/3+a_1)z^3+(a_2-a_0/3)z^4+(a_3-a_1/3+1/36+1/60)z^5+...=z
to find a_0=0, a_1=1/3, a_2=0, a_3 = [[tel:1/9-1/36-1/60|1/9-1/36-1/60]], ...
Q: Consider the function f(z)=pi*cot(pi*z). Find all isolated singularities of f. If it is a removable singularity define a new function to "fix" the old function. If it is a pole, find the order of the pole.
A: pi cot(pi z)= pi cos(pi z)/sin(pi z). The denominator vanishes at integer values of z. If z_0=n is integer, then we have the following Laurent expansion for the function
pi cos(pi z)/sin(pi z):
$\pi \frac{cos(\pi z)}{sin(\pi z)}=\pi \frac{cos(\pi n)+\ldots}{\pi \cos(\pi n)(z-n)+\ldots}=\frac{1}{z-n}+\ldots$
So the function has a pole of order 1 at every integer. The residue at all these poles is 1.
Q:∫ (1/(a+bcos t))dt from 0 to 2π, where a>b>0.
A: Let z=e^it. As t goes from 0 to 2pi, the point z=e^it makes a full turn around 0 along the unit circle |z|=1 in the counter-clockwise direction.
Differentiating z=e^it we get dz=ie^it dt, so dt= -i dz / z.
Also if z=e^it, then cos t = (z+1/z)/2.
Thus the integral can be rewritten as
$\int_{|z|=1} \frac{-i dz/z}{a+b(z+1/z)/2}$
Now this integral can be further rewritten as
$-2i \int_{|z|=1} \frac{ dz}{bz^2+2az+b}$
The denominator vanishes at two points:
$z_{1,2}=\frac{-a\pm \sqrt{a^2-b^2}}{b}=-\frac a b \pm \sqrt{(\frac a b)^2 -1}$
Since a>b>0, the expression under the square root is positive. Thus
$-\frac a b - \sqrt{(\frac a b)^2 -1}$
is outside the circle |z|=1
The other root
$-\frac a b + \sqrt{(\frac a b)^2 -1} = \frac{1}{ -\frac a b - \sqrt{(\frac a b)^2 -1} }$
is inside the circle (the denominator is < -1, so the quotient is between -1 and 0).
Thus we can apply Cauchy formula
$\int \frac{f(z)\,dz}{z-z_0}=2\pi i f(z_0)$
with the function
$f(z)=\frac{1}{b(z-( -\frac a b - \sqrt{(\frac a b)^2 -1} ))}$
and
$z_0= -\frac a b + \sqrt{(\frac a b)^2 -1}$
Thus
$-2i \int_{|z|=1} \frac{ dz}{bz^2+2az+b}=\frac{4\pi}{b(( -\frac a b + \sqrt{(\frac a b)^2 -1} )-( -\frac a b - \sqrt{(\frac a b)^2 -1} ))}=\frac{2\pi}{b\sqrt{(\frac a b)^2-1}}=\frac{2\pi}{\sqrt{a^2-b^2}}$
To be completed later
Q: ∫ (1/(3+sinθ+cosθ))dθ from 0 to 2π.
A: Very similar to the question above it. The answer is 2pi/sqrt(7). Please let me know if you need more details.
Q：find the power series expansion about the given point. (z+2)/(z+3) about z_0=-1
A: first let w=z+1, so that we will have to expand everything in powers of w. Now the original function is (w+1)/(w+2), or, equivalently, 1- 1/(w+2).
To expand 1/(w+2) in powers of w, we rewrite it as 1/2 1/(1+w/2)=1/2 (1-w/2+w^2/4-w^3/8+...), where the last expansion is valid for |w/2|<1, or, equivalently, |w|<2.
Finally we gather all this together to get (z+2)/(z+3)=1/2 + (z+1)/4 - (z+1)^2/8 - (z+1)^3/16 + ...
Q：Find the order of zeros of (z^2+z-2)^3.
A: Factor z^2+z-2 as (z-1)(z+2). Thus the original function is (z-1)^3(z+2)^3, which has zeroes of order 3 at z=1 and at z= -2.
Q:Find the largest disc in which the power series expansion for the following function is valid.
A: The function log(1-z) can be made analytic on any domain that doesn't contain a loop around the point 1 (since log w can be chosen to be analytic on any domain which doesn't contain any loop around the origin). In particular it can be made analytic on |z|<1. By a theorem we proved in class this means that the Taylor series of
[log(1-z)]^2 converges on |z|<1. It doesn't converge on a larger disc, since
[log(1-z)]^2 is not even continuous at z=1, which is on the boundary of the disc.
Q:Find all solutions to the differential equation
f ''(z)+β^2*f(z) = 0, f is an entire function.
A: "Entire" function means it is analytic everywhere. In particular it has a Taylor series around any point and this Taylos series converges on the whole plane.
White
$f(z)=\sum_{n=0}^{\infty} a_n x^n$
Then
$f''(z)=\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}=\sum_{k=0}^\infty (k+2)(k+1)a_{k+2}x^k$
Hence
$f''(z)+\beta^2 f(z)=\sum_{k=0}^\infty [(k+2)(k+1)a_{k+2}+\beta^2a_k]x^k=0$
Equating coefficients we see that for any k
$(k+2)(k+1)a_{k+2}+\beta^2a_k=0$
We solve this recursion equation. If β=0, then
$f(z)=a_0+a_1z$
since all other coefficients are zero. If β is non-zero, then for even k we have
$a_{k+2}=\frac{\beta^2a_k}{(k+2)(k+1)}= \frac{\beta^4a_{k-2}}{(k+2)(k+1)k(k-1)}=\ldots= \frac{\beta^{k+2}a_0}{(k+2)(k+1)\ldtots 2\cdot 1}= \frac{\beta^{k+2}a_0}{(k+2)!}$
And similarly if k is odd, then
$a_{k+2}=\frac{\beta^2a_k}{(k+2)(k+1)}= \frac{\beta^4a_{k-2}}{(k+2)(k+1)k(k-1)}=\ldots= \frac{\beta^{k+1}a_1}{(k+2)(k+1)\ldtots 3\cdot 2}= \frac{\beta^{k+1}a_1}{(k+2)!}$
Plugging these coefficients back into the definition of f, we find that
$f(z)=a_0\sum_{m=0}^\infty \frac{\beta^{2m}z^{2m}}{(2m)!}+a_1 \sum_{m=0}^\infty \frac{\beta^{2m}z^{2m+1}}{(2m+1)!}$
or
$f(z)=a_0\cosh(\beta z)+\frac{a_1}{\beta}\sinh(\beta z)$
Q:Find the radius of convergence of the power series expansion of the function sin(1/(z^2+9)) in powers of (z-4).
A: sin(1/(z^2+9)) is analytic everywhere except at z=3i or z=-3i. The function sin(1/(z^2+9)) doesn't have any limit as z tends to 3i (or to -3i).
By a theorem we proved in class, this means that the Taylor series around any point converges on the largest disc around that point which doesn't contain neither 3i
nor -3i.
Thus the Taylor series around the point 4 converges in the disc |z-4|<5
Q:Evaluate ∫ (log(z+2) + z*(z conjugate)) dz over |z|=1 - answered below
Q:Determine the image of {z: 0< Re(z) < pi / 2 and Im(z) < 0} under f(z) = sinz.
You can find the answer in the textbook (see p.51), but we didn't discuss images and preimages under sin or cos (it is slightly more complicated than what we did).
Q: Explain why the three lines on the top left-hand corner of this page is a contradiction. (yeah that's what I meant, where is the mistake in the argument...)
Comment: it is obvious that it is a contradiction (e^(-2pi) is not 1). The question should be where is the mistake in the argument.
A: The mistake is in going from the second line to the third. Both (e^2\pi i)^i and 1^i have infinitely many values. The equality on the second line means that the set of all values of
(e^2\pi i)^i coincides with the set of all values of 1^i. On the third line we mistakenly conclude "hence any value on the left is equal to any value on the right"
Here is an analogous, but simpler mistake:
$((-1)^{2})^{1/2}=1^{1/2}$
$-1=1$
Q:Suppose that the function f is analytic at all points of the complex plane.Suppose also that the image of f is contained in the unit circle {u + iv | u^2 + v^2 = 1}.Show that f is constant.
A: See answers to the summer's midterm in the "Midterm Summer 2011" section
Q: g(z) = z^2 - |z|^2.Find g'(z) where it exists.
A: g(x+iy)=(x+iy)^2-|x+iy|^2=2ixy-2y^2. Cauchy-Riemann equations with u(x,y)= - 2y^2, v(x,y)=2xy read
0=2x
-4y=-2y
Hence if the complex derivative exists at some point (x,y), then by Cauchy-Riemann equations x=0,y=0, i.e. z=x+iy=0.
At point z=0 we compute
$g'(0)=\lim_{z\to 0} \frac{z^2-|z|^2}{z}=\lim_{z\to 0} z-\bar{z}=0$
Q: Find the power series expansion of g(z) =z^2*sin(3z) around z0=0 and find its radius of convergence.
A: Since
$\sin z = \sum\limits_{n=0}^\infty (-1)^n z^{2n+1}/(2n+1)!$
we have
$\sin(3z) = \sum \limits_{n=0}^\infty (-1)^n (3z)^{2n+1}/(2n+1)!= \sum\limits _{n=0}^\infty (-1)^n 3^nz^{2n+1}/(2n+1)!$
and
$z^2\sin(3z) = \sum\limits_{n=0}^\infty (-1)^n 3^nz^{2n+3}/(2n+1)!$
The radius of convergence is infinity, just like the one for sine.
Q: Show that cos(z) is an unbounded function.
A:
$\cos(in)=\frac{e^n+e^{-n}}{2}\xrightarrow[n\to\infty]{}\infty$
Q: Evaluate the line integrals
∫ |z-1|^2n*(z-1)^m dz over |z-1|=1
where n and m are integers.
A:
On |z-1|=1 the value of |z-1|^2n is 1. Hence the integral can be rewritten as
$\int\limits_{|z-1|=1}(z-1)^m \,dz=\int \limits_{|w|=1} w^m \, dw$
I've made a change of variable w=z-1 for convenience. This last integral is equal to 0 if m is not -1 and is equal to 2 pi i if m= -1.
Q: Show that 1/2pi *∫e^ikw dw from 0,2pi={0 if k non-zero and 1 if k=0}
A: If k=0, the integrand is equal to 1 hence the integral is 2 pi. If k is not 0, then
$\int\limits_0^{2\pi} e^{ikw} dw = \frac{e^{ikw}}{ik}|_{w=0}^{2\pi}=0$
Q: ∫ (z+1/z)^(2n+1) dz over |z|=1
A: Open the brackets:
$(z+1/z)^{2n+1}=z^{2n+1}+{2n+1 \choose 1} z^{2n-1} + \ldots +{2n+1 \choose 2n} z^{-2n+1}+ z^{-2n-1}$
The integral of
$z^{k}$
over |z|=1 is non-zero only when k= -1. For k= -1 the integral is 2 pi i. Hence only the summand
${2n+1 \choose n+1}z^{-1}$
contributes to the answer and the answer is
$\int (z+1/z)^{2n+1} \,dz={2n+1 \choose n+1} 2 \pi i$
Q: lim n(1-cos(θ/n)-isin(θ/n)) as n--->∞,where θ is some fixed real number.
A: Computing separately lim n(1-cos(θ/n)) = 0 and lim n sin(θ/n) = θ we find that the original limit is - i θ (both limits can be computed using l'Hopital's rule: letting x=1/n the limits can be rewritten as lim ( 1-cos(θx) )/x and lim sin(θx) / x as x-->0)
Q: Integral 1/[z^3(z-2)^2]dz over |z|=3
A: By Cauchy theorem this integral is equal to the integral of the same expression over |z|=R for any R>3.
For R large enough the absolute value of the integrand is at most 1/[R^3(R-2)^2], hence by triangle inequality the absolute value of the integral is at most 2pi R/[R^3(R-2)^2]=2pi/[R^2(R-2)^2], which tends to zero as R tends to infinity. Since this non-negative number (the absolute value of the integral) is smaller than all the values of a function that tends to zero, it must be equal to zero.
Q:Find the power series centered at 0 for 1/(1-z^3)
A: 1+z^3+z^6+z^9+....
Q: Let u(x,y)=g(xy)where g:R--R is a real polynomial. Suppose u(x,y)is harmonic on R-2.Show that g(t)=at+b with real a and b
A:
$u_{xx}=(yg'(xy))_x=y(g'(xy))_x=y^2g''(xy)$
Similarly
$u_{yy}=(xg'(xy))_y=x(g'(xy))_y=x^2g''(xy)$
Since u is harmonic, it follows that
$(x^2+y^2)g''(xy)=0$
Hence for any (x,y) except (0,0) the second derivative at the point g''(xy) vanishes. Thus g'' vanishes identically and hence g is a linear function.
Q: Evaluate integral[log(z+2)+z*z-bar]dz
Over|z|=1
A: The function log(z+2) is analytic inside the circle |z|=1 (point z=-2 is outside), so its integral is zero by Cauchy theorem. The function z*z-bar is equal to |z|^2, i.e. it is equal to 1 on the unit circle. Now the integral of 1 dz is once again zero, since 1 is an analytic function on the unit disc.
Q:Find the value of (i∫z-bar dz) where the integration is along the closed path consisting of the segments from -1-i to 1-i, from 1-i to 2i and from 2i to -1-i (i.e. along the triangle with vertices -1-i,1-i,2i going in counterclockwise direction)
A: One way is to parametrize the sides of the triangle and compute the sum of the integrals. It is a long computation however. For instance the integral over the lower side (from -1-i to 1-i) can be computed as follows: the path is z=-i+t, -1<t<1; dz=dt; hence the integral is
$\int_{-1}^1 (i+t) dt=2i$
The integral over the other two sides can be found in a similar way, but with more tricky computations.
Much simpler approach is to use Green's theorem: write z=x+i y, z-bar = x - i y, dz = dx + i dy
Hence
$i \int \bar z dz = i \int (x-iy)(dx+idy)= i \int (xdx+ydy) + \int (ydx-xdy)$
finally by Green's theorem this expression can be rewritten as
$i \int \int 0+0 dx dy + \int \int -1-1 dx dy = -2 \int \int dx dy = - 2 \cdot \mbox{area of triangle}= - 2 \cdot \mbox{half base times height}= - 6$
Q: Determine whether the given infinite series converges or diverges.
$\sum_{i=1}^\infty{ \frac{1}{n}(\frac{1+i}{\sqrt 2})^n }\!$
A: The first thing to try is to determine the radius of convergence for the series
$\sum_{i=1}^\infty \frac{1}{n} z^n$
The radius of convergence in this case is one. Since
$\left| \frac{1+i}{\sqrt 2} \right|=1$
this doesn't tell us directly whether the series converges or diverges.
In fact if we ask the question whether the series converges absolutely, then the answer is no:
$\sum_{i=1}^\infty{ \left| \frac{1}{n}(\frac{1+i}{\sqrt 2})^n \right| }\!= \sum_{i=1}^\infty{ \frac{1}{n} }\! = \infty$
However the series itself does converge. To see this notice that the powers of (1+i)/sqrt(2) repeat themselves with period 8:
$(\frac{1+i}{\sqrt 2})^1=\frac{1+i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^2=i$
$(\frac{1+i}{\sqrt 2})^3=\frac{-1+i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^4=-1$
$(\frac{1+i}{\sqrt 2})^5=\frac{-1-i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^6=-i$
$(\frac{1+i}{\sqrt 2})^7=\frac{1-i}{\sqrt 2}$
$(\frac{1+i}{\sqrt 2})^8=1$
and then the powers repeat themselves.
Using this information we can split the power series to the real part and to the imaginary part. Namely the real part is equal to
$\frac{1}{\sqrt 2}-\frac 1 3 \frac{1}{\sqrt 2}-\frac 1 4 - \frac 1 5 \frac{1}{\sqrt 2} + \frac 1 7 \frac{1}{\sqrt 2} + \frac 1 8 + \frac 1 9 \frac{1}{\sqrt 2} - \frac 1{10} \frac 1 {\sqrt 2} - \ldots$
In this series there are always three positive terms followed by three negative ones, then three positive ones and so on. Moreover the general term in the series tends to 0. By a result on convergence of alternating series this series converges.
Similar considerations show that the imaginary part of the series is also a convergent series.
Hence the series itself is convergent.
Q: Fisher, section 2.2, Question 22 parts a and b.
Please do type the question itself, not just a reference to it.
A: a) The crucial observation is that quite generally the coefficients of the power series can be reconstructed from the function defined by it. Namely we have the following:
$a_0=f(z_0); a_1=f'(z_0); a_2=f''(z_0)/2; \ldots a_n=f^(n)(z_0)/n!$
(this follows from the power series expansion of f', f'' etc). In particular if the function vanishes everywhere around the point z_0, then all its derivatives vanish at z_0, and hence all a_i's also vanish.
b) If we apply the result of a) to the power series
$0=F(z)-G(z)=\sum_{n=0}^\infty (a_n-b_n) (z-z_0)^n$
we get the desired conclusion a_n=b_n
Q: For what values of z does the series
$\sum_{n=0}^\infty (z^n-z^{n+1})$
converge?
A: If you compute the partial sum
$\sum_{n=0}^N (z^n-z^{n+1})$
you get
$1-z^{N+1}$
since all the terms between the first one and the very last one cancel each other. As N tends to infinity this expression has a limit only if |z|<1 or z=1. If |z|<1 the limit is 1; if z=1, the limit is 0. | 2017-11-24T07:12:07 | {
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http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/97253 | # Rings in which every non-unit is a zero divisor
Is there a special name for the class of (commutative) rings in which every non-unit is a zero divisor? The main example is $\mathbf{Z}/(n)$. Are there other natural or interesting examples?
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I think Artin rings qualify. – Charles Staats Oct 18 '10 at 14:58
A commutative ring $A$ has the property that every non-unit is a zero divisor if and only if the canonical map $A \to T(A)$ is an isomorphism, where $T(A)$ denotes the total ring of fractions of $A$. Also, every $T(A)$ has this property. Thus probably there will be no special terminology except "total rings of fractions".
Artinian rings provide examples: If $x \in A$, the chain $... \subseteq (x^2) \subseteq (x) \subseteq A$ is stationary, say $x^k = y x^{k+1}$ for some minimal $k \geq 0$. If $k=0$, $x$ is a unit. If $k \geq 1$, $x (x^k y - x^{k-1})=0$ and $x^{k-1} \neq y x^k$, i.e. $x$ is a zero divisor.
The class of total rings of fractions is closed under (infinite) products and directed unions. Is it the smallest such class containing the artinian rings?
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Nice answer. Regarding the question at the end of your response, here's an example which might be relevant: Let $R=k[x,y]_{(x,y)}/(x^2,xy)$. Then $R$ is a local ring with maximal ideal $(x,y)$. It is not Artinian, since it has dimension $1$. Since $R$ is a local ring, the non-units in $R$ are precisely the elements in $(x,y)$. And since $(x,y)$ is an associated prime of $R$, it follows that every element of $(x,y)$ is a zero-divisor. But it's not clear to me how to check if $R$ is/is not the directed union of Artinian rings. – Daniel Erman Oct 18 '10 at 18:57
I've found "full quotient ring" in MathSciNet which it's probably the same as "total rings of fractions" since it is mentioned in en.wikipedia.org/wiki/Total_quotient_ring. – lhf Oct 18 '10 at 21:00
+1: I think this is indeed the best possible answer. – Pete L. Clark Oct 18 '10 at 22:24
The characterization with total quotient rings is a) trivial, b) useless when we actually test the property. I wonder about the upvotes ;). @Erman: I also don't know if your ring $R$ is a iterated (product or directed union) of artinian rings. Perhaps as a first step we have to show that every subring of $R$ is noetherian? – Martin Brandenburg Oct 20 '10 at 21:30
I am interested in a property for a ring $R$ called strongly associate. $R$ is strongly associate if $(a)=(b)$ implies $a= \lambda b$ for some unit $\lambda$. This is a properly weaker property than domainlike, presimplifiable, quasi-local properties. Artinian rings and principal ideal rings both satisfy this property. Hence, $\mathbb{Z}/n\mathbb{Z}$ as well. I was wondering if the property here that $R= Z(R) \cup U(R)$ is enough to conclude that $R$ is strongly associate. Seems unlikely, but I was unable to come up with an example. – CPM May 28 '14 at 0:30
Any (commutative unitary) ring of Krull dimension 0 has this property. This includes the class of Artinian rings.
Proof: If $A$ has Krull dimension $0$, then any maximal $m$ ideal of $A$ is also a minimal prime ideal by the definition of Krull dimension. Applying Krull's theorem on the intersection of prime ideals to the localization $A_m$, we find that $mA_m$ is the nilradical of $A_m$. So for any $f\in m$, there exists a positive integer $n$ such that $f^n=0$ in $A_m$. So there exists $s\in A\setminus m$ such that $sf^n=0$ in $A$. We can choose $n$ smallest with respect to this property so that $sf^{n-1}\ne 0$. Therefore $f$ is a zero divisor. Now any non-unit element $f$ belong to some maximal ideal, it is a zero divisor.
Artining rings are zero-dimensional and semi-local. Boolean rings are reduced and zero-dimensional.
If $(A, m)$ is a local ring, then it has this property if and only if $\mathrm{depth}(A)=0$ (e.g. the example in Daniel Erman's comment). If $A$ is not necessarily local but all localizations $A_m$ at maximal ideals of $A$ have depth $0$, then $A$ has your property. But I don't think this is a necessary condition.
Similarly one can construct local rings of depth 0 of any (even infinite) dimension. But I don't know whether there exists a ring of positive dimension with infinitely many maximal ideals and such that all its localizations at maximal ideals have depth 0.
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Let me mention that $\dim(A)=0$ iff for all $x,y$ there is some $k$ such that $x^k = x^{k+1} y$. This proves the first claim. – Martin Brandenburg Oct 19 '10 at 0:04
Martin: you mean there exists $y$ ? – Qing Liu Oct 20 '10 at 18:46
@MartinBrandenburg: Sorry to revive a thread that was dead long ago, but your characterization of $\dim(A)=0$ is a particular case of the elementary (almost first order) definition of Krull dimension by Coquand, Lombardi and Roy. – ACL Jan 26 at 16:37
Pace Chris Leary, the standard terminology is that a module all monomorphisms of which are automorphisms is said to be cohopfian (or co-Hopfian, if you're checking MathSciNet). A Dedekind-finite (a.k.a. directly finite) module usually means a module whose left invertible endomorphisms are also right invertible, equivalently, a module that is not isomorphic to any proper direct summand of itself.
T. Y. Lam, in his book Lectures on Modules and Rings, pp. 320–322, calls a noncommutative ring in which every regular element (i.e. neither right nor left zero-divisor) is a unit a classical ring, and he provides various examples. One that has not already been mentioned here is that any right (or left) self-injective ring is classical. Right self-injective rings need not have the property that every element that is merely not a left zero-divisor is a unit; interestingly, for right self-injective rings the latter condition is equivalent to the ring being Dedekind-finite (in the sense of the preceding paragraph), and also equivalent to the ring having stable range 1 (see Y. Suzuki, “On automorphisms of an injective module,” Proc. Japan Acad. 44 (1968), 120–124, and G. F. Birkenmeier, “On the cancellation of quasi-injective modules,” Comm. Algebra 4 (1976), no. 2, 101–109).
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From a more positive perspective, you are looking at rings with the property that every regular element is a unit. I have done some work with these rings. If the ring is commutative, the condition is equivalent to the ring being a quoring, i.e., it's its own classical ring of quotients. Non-commutative rings with the property must be quorings, but the converse is not necessarily true, as seen with von Neumann regular rings. I called rings with every regular element a unit Dedekind finite because they are characterized as R-modules by the property that every monic endomorphism is an isomorphism (the Dedekind definition of finite set; I picked up the name from L. N. Stout). This is not standard terminology I believe (see Lam's book "Lectures on Modules and Rings).
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I don't know the name for this class of commutative rings. Two quick examples:
1) Any finite ring: then for all $x$ there exist $0 < k < l$ such that $x^k = x^l$, so $x^k(x^{l-k}-1) = 0$. This shows that $x$ is a zero divisor unless $x^{l-k}-1 = 0$, i.e., $x^{l-k} = 1$, in which case $x$ is a unit.
2) Any Boolean ring, i.e., each element is an idempotent: if $x^2 = x$, then $x(1-x) = 0$.
Added: Charles Staats's comment gives another important class of rings satifying the desired condition. To flesh it out, first note that the OP's class of rings includes
3) Any local Artinian ring: each nonunit is a member of the unique maximal ideal, which is nilpotent (e.g. Theorem 82 of http://math.uga.edu/~pete/integral.pdf).
Also
4) The OP's class is closed under finite products. Indeed, let $R = R_1 \times \ldots \times R_n$, where the $R_i$ are in the OP's class. Let $x = (x_1,\ldots,x_n)$ be a nonunit of $R$. This happens iff for at least one $i$, $x_i$ is a nonunit in $R_i$. Without loss of generality say $i = 1$. Then there exists a nonzero $y_1$ in $R_1$ such that $x_1 y_1 = 0$. Putting $y = (y_1,0,\ldots,0)$, we get $xy = 0$.
It follows that any finite product of local Artinian rings is in the OP's class. But every Artinian ring is a finite product of local Artinian rings (e.g. Theorem 86 of http://math.uga.edu/~pete/integral.pdf), so every Artinian ring is in the OP's class.
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I have studied a noncommutative version of this. There is such a thing called a right cohopfian ring in the sense that if the right annihilator of r is zero, then r is a unit. If you add commutativity and look at the contrapositive, you get that nonunits are zero divisors.
I don't think this terminology has caught on, but here is the rationale. A "cohopfian object" is one for which injections are surjections. Looking on elements of the ring as maps sending x-->rx, we are saying that if such a map is injective, it is surjective.
Right Artinian, right perfect and strongly-pi regular rings (commutative VNR rings are strongly pi regular) are all right and left cohopfian. Finding a one-sided cohopfian ring seems tough, but Varadarajan did it here:
"Varadarajan, K. Hopfian and co-Hopfian objects. Publ. Mat. 36 (1992), no. 1, 293–317."
I think someone has noted above that right cohopfian rings have ot be Dedekind finite, and it is interesting that Dedekind finite=right Hopfian=left Hopfian.
Too bad I didn't see this a year ago :)
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Thanks for the answer anyway. – lhf Dec 15 '11 at 11:10
As Greg Marks says, there is a natural generalisation of this property to a non-commutative ring $R$: "Every regular element (= neither left nor right zero-divisor) in $R$ is a unit", which is equivalent to "The set of regular elements in $R$ is (left and right) Ore, and the natural localisation morphism is an isomorphism". This has been called "full quotient ring" or the like, now Lam calls it "classical".
Since I have not found it in any comment or answer here as yet (and neither in Lam's book), let me add that (two-sided) Noetherian rings of this form have been investigated in J. T. Stafford: Noetherian Full Quotient Rings (Proc. London Math. Soc. (3) 44 (1982) pp. 385-404). Quote:
Let $A(R)$ be the largest Artinian ideal and $J(R)$ the Jacobson radical of a Noetherian ring $R$. Then $R$ equals its own full quotient ring if and only if $$l\text{-}ann (A(R)) \cap r\text{-}ann (A(R)) \subseteq J(R).$$ It follows that a Noetherian full quotient ring is semilocal [= $R/Jac(R)$ is Artinian semisimple] and has a non-zero Artinian ideal.
In the commutative case, the criterion is simply $ann (A(R)) \subseteq J(R)$, and "semilocal" equals "finitely many maximal ideals". Without the "Noetherian" assumption, the criterion does not make sense as $A(R)$ may not exist, and its corollary is false in general.
Ad: I have asked if this property is Morita invariant in MO 124856.
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How about calling these things balanced commutative rings?
Recall that a category satisfying "monic + epic --> isomorphism" is said to be balanced. Therefore:
1. a monoid is balanced iff every element that is both left-cancellative and right-cancellative is a unit.
2. a commutative monoid is balanced iff every element that is cancellative is a unit.
It makes sense to apply this terminology to rings, too:
1. a ring is balanced iff every element that is neither a left zero-divisor nor a right-zero divisor is a unit.
2. a commutative ring is balanced iff every non-zero divisor is a unit.
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I don't know whether this concept has been named. But other examples include Boolean rings and products of endomorphism (matrix) algebras, and rings such as $\mathbb{C}[x]/(x^2)$, or more generally the total algebra of a graded algebra which is bounded in degree, as for example, the cohomology algebra of a space homotopy equivalent to a finite CW complex (with coefficients in a field).
Edit: Sorry, I meant a connected graded algebra (where the degree 0 part is the field of coefficients).
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I think if you have some information about rings of continuous functions $C(X)$, you can construct a wide class of rings with this property.
At first let me give you some special information about these examples.
Def 1. For topological space $X$ we denote the ring of all continuous functions on $X$ by $C(X)$. for $f\in C(X)$ the zero-set of $f$ is defined as: $Z(f)=${$x\in X$: $f(x)=0$}
Def 2. A completely regular topological space $X$ is called an almost $P$-space, if for every $f\in C(X)$ with nonempty zero-set, i.e. $Z(f)$, this set has nonempty interior, i.e. there exist $x$ that $x\in int_X Z(f)$.
With the above definition, I can introduce a theorem which classifies all Rings of continuous functions $C(X)$ with the property that every non-unit is a zero-divisor.
Theorem: In the ring $C(X)$, every non-unit is a zero divisor iff the topological space $X$ is almost $P$-space.
The simplest examples of almost $P$-spaces are discrete spaces. for example if $X$ is a discrete space, then $C(X)$ is equal to the usual cartesian product $\mathbb{R}^X$. So for arbitrary set $X$ you can construct $\mathbb{R}^X$ to have the property of your question.
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Thanks a lot! . – lhf May 18 '12 at 0:26 | 2016-06-29T05:44:24 | {
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https://math.stackexchange.com/questions/4469699/bijective-change-of-variable | # Bijective change of variable
Suppose we have a function $$f: \theta\rightarrow f(\theta)$$ whose domain is $$[0,180°]$$. The course I follow, says that we can use the change of variable $$x=cos(\theta)$$ in the function $$f$$ (because $$cos$$ is bijective from $$[0,180°]$$ to $$[-1,1]$$).
I don't understand.. If we have, say, $$f(\theta) = \theta^2 +1$$
Then $$f(x)=f(cos(\theta))= cos^2(\theta) +1$$
Then if I evaluate $$f$$ on $$0$$, I have $$f(0)=1$$ according to the first expression and $$f(0)=2$$ according to the second expression.
I you have any idea how this variable change works, I'll gladly try my best to understand it
PS: I've tried to think about this problem in terms of matrices. We start from $$AX$$ and $$cos$$ is like a $$B$$ matrix that we put in-between $$A$$ and $$X$$ so the whole is tantamount to $$AX'$$ with $$X'=BX$$
## Edit: Here's the context ->
I have the differential equation:
$$sin(θ)\frac{d}{d\theta}\left(sin(\theta)\frac{dP_l^m}{d\theta}\right) +\left(l(l+1)sin^2(\theta)-m^2\right)P_l^m =0$$
And $$P_l^m$$ is a function of $$\theta$$. The variable change used is $$x=cos(\theta)$$
On the other hand, $$\cos^2\theta+1=1$$ when $$\theta=90^\circ$$. So, since $$\cos(90^\circ)=0$$, you have $$f\bigl(\cos(90^\circ)\bigr)=f(0)=1$$.
Saying that you can use the substitution $$x=\cos\theta$$ does not mean that $$f$$ and $$f\circ\cos$$ are the same function. But it implies that they have the same ranges. For a more complete answer, it would be useful to know why is it that you want to do this substitution.
• Yes I also think $f$ is different from $f\circ cos$. You have plugged in $90°$ because $Arcos(0)=90°$ ? What surprises me from the paper I learn from is that there is no $Arcos$ in the final answer... Jun 10 at 16:30
• No. I plugged in $90^\circ$ because $\cos(90^\circ)=0$. Jun 10 at 16:33
• There is nothing peculiar about this change of variable. For instance, computing $\int_{-1}^1\sqrt{1-x^2}\,\mathrm dx$ can be done doing the substitution $x=\cos\theta$ and $\mathrm dx=-\sin\theta\,\mathrm d\theta$. Jun 10 at 19:45
• In the proof I know of the integral change of variable, we use $(f\circ g)'=g'f'\circ g$ Jun 11 at 19:01 | 2022-10-05T01:22:01 | {
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https://de.mathworks.com/help/symbolic/rewrite.html | # rewrite
Rewrite expression in terms of another function
## Syntax
``R = rewrite(expr,target)``
## Description
example
````R = rewrite(expr,target)` rewrites the symbolic expression `expr` in terms of another function as specified by `target`. The rewritten expression is mathematically equivalent to the original expression. If `expr` is a vector or matrix, `rewrite` acts element-wise on `expr`.```
## Examples
collapse all
Rewrite any trigonometric function in terms of the exponential function by specifying the target `"exp"`.
```syms x sin2exp = rewrite(sin(x),"exp")```
```sin2exp = $\frac{{\mathrm{e}}^{-x \mathrm{i}} \mathrm{i}}{2}-\frac{{\mathrm{e}}^{x \mathrm{i}} \mathrm{i}}{2}$```
`tan2exp = rewrite(tan(x),"exp")`
```tan2exp = $-\frac{{\mathrm{e}}^{2 x \mathrm{i}} \mathrm{i}-\mathrm{i}}{{\mathrm{e}}^{2 x \mathrm{i}}+1}$```
Rewrite the exponential function in terms of any trigonometric function by specifying the trigonometric function as the target. For a full list of target options, see target.
```syms x exp2sin = rewrite(exp(x),"sin")```
```exp2sin = $-2 {\mathrm{sin}\left(\frac{x \mathrm{i}}{2}\right)}^{2}-\mathrm{sin}\left(x \mathrm{i}\right) \mathrm{i}+1$```
`exp2tan = rewrite(-(exp(x*2i)*1i - 1i)/(exp(x*2i) + 1),"tan")`
```exp2tan = $-\frac{\frac{\left(\mathrm{tan}\left(x\right)-\mathrm{i}\right) \mathrm{i}}{\mathrm{tan}\left(x\right)+\mathrm{i}}+\mathrm{i}}{\frac{\mathrm{tan}\left(x\right)-\mathrm{i}}{\mathrm{tan}\left(x\right)+\mathrm{i}}-1}$```
Simplify `exp2tan` to the expected form by using `simplify`.
`exp2tan = simplify(exp2tan)`
`exp2tan = $\mathrm{tan}\left(x\right)$`
Rewrite any trigonometric function in terms of any other trigonometric function by specifying the target. For a full list of target options, see target.
Rewrite `tan(x)` in terms of the sine function by specifying the target `"sin"`.
```syms x tan2sin = rewrite(tan(x),"sin")```
```tan2sin = $-\frac{\mathrm{sin}\left(x\right)}{2 {\mathrm{sin}\left(\frac{x}{2}\right)}^{2}-1}$```
Rewrite any inverse trigonometric function in terms of the logarithm function by specifying the target `"log"`. For a full list of target options, see target.
Rewrite `acos(x)` and `acot(x)` in terms of the `log` function.
```syms x acos2log = rewrite(acos(x),"log")```
`acos2log = $-\mathrm{log}\left(x+\sqrt{1-{x}^{2}} \mathrm{i}\right) \mathrm{i}$`
`acot2log = rewrite(acot(x),"log")`
```acot2log = $\frac{\mathrm{log}\left(1-\frac{\mathrm{i}}{x}\right) \mathrm{i}}{2}-\frac{\mathrm{log}\left(1+\frac{\mathrm{i}}{x}\right) \mathrm{i}}{2}$```
Similarly, rewrite the logarithm function in terms of any inverse trigonometric function by specifying the inverse trigonometric function as the target.
Rewrite each element of a matrix by calling `rewrite` on the matrix.
Rewrite all elements of a matrix in terms of the `exp` function.
```syms x matrix = [sin(x) cos(x); sinh(x) cosh(x)]; R = rewrite(matrix,"exp")```
```R = $\left(\begin{array}{cc}\frac{{\mathrm{e}}^{-x \mathrm{i}} \mathrm{i}}{2}-\frac{{\mathrm{e}}^{x \mathrm{i}} \mathrm{i}}{2}& \frac{{\mathrm{e}}^{-x \mathrm{i}}}{2}+\frac{{\mathrm{e}}^{x \mathrm{i}}}{2}\\ \frac{{\mathrm{e}}^{x}}{2}-\frac{{\mathrm{e}}^{-x}}{2}& \frac{{\mathrm{e}}^{-x}}{2}+\frac{{\mathrm{e}}^{x}}{2}\end{array}\right)$```
Rewrite the cosine function in terms of the sine function. Here, the `rewrite` function rewrites the cosine function using the identity $\mathrm{cos}\left(\mathit{x}\right)=1-2\text{\hspace{0.17em}}{\mathrm{sin}\left(\frac{\mathit{x}}{2}\right)}^{2}$, which is valid for any $\mathit{x}$.
```syms x R = rewrite(cos(x),"sin")```
```R = $1-2 {\mathrm{sin}\left(\frac{x}{2}\right)}^{2}$```
`rewrite` does not rewrite `sin(x)` as either $-\sqrt{1-{\mathrm{cos}}^{2}\left(\mathit{x}\right)}$ or $\sqrt{1-{\mathrm{cos}}^{2}\left(\mathit{x}\right)}$ because these expressions are not valid for all `x`. However, using the square of these expressions to represent `sin(x)^2` is valid for all `x`. Thus, `rewrite` can rewrite `sin(x)^2`.
```syms x R1 = rewrite(sin(x),"cos")```
`R1 = $\mathrm{sin}\left(x\right)$`
`R2 = rewrite(sin(x)^2,"cos")`
`R2 = $1-{\mathrm{cos}\left(x\right)}^{2}$`
Since R2023a
Find the root of a polynomial by using `root`. The result is a column vector in terms of the `root` function with `k = 1`, `2`, `3`, `4`, or `5` for the `k`th root of the polynomial.
```syms x sols = root(x^5 - x^4 - 1,x)```
```sols = $\left(\begin{array}{c}\mathrm{root}\left({x}^{5}-{x}^{4}-1,x,1\right)\\ \mathrm{root}\left({x}^{5}-{x}^{4}-1,x,2\right)\\ \mathrm{root}\left({x}^{5}-{x}^{4}-1,x,3\right)\\ \mathrm{root}\left({x}^{5}-{x}^{4}-1,x,4\right)\\ \mathrm{root}\left({x}^{5}-{x}^{4}-1,x,5\right)\end{array}\right)$```
Expand the `root` function in `sols` by using `rewrite` with the `"expandroot"` option. The result is in terms of arithmetic operations such as ^, *, /, +, and – that operate on exact symbolic numbers. Because the expanded result can involve many terms that operate arithmetically, numerical approximation of this result can be inaccurate (due to accumulation of round-off errors).
`R = rewrite(sols,"expandroot")`
```R = ```
As an alternative, you can numerically approximate `sols` directly by using `vpa` to return variable-precision symbolic numbers. The resulting numeric values have the default 32 significant digits, which are more accurate.
`solsVpa = vpa(sols)`
```solsVpa = $\left(\begin{array}{c}-0.66235897862237301298045442723905-0.56227951206230124389918214490937 \mathrm{i}\\ -0.66235897862237301298045442723905+0.56227951206230124389918214490937 \mathrm{i}\\ 0.5-0.86602540378443864676372317075294 \mathrm{i}\\ 0.5+0.86602540378443864676372317075294 \mathrm{i}\\ 1.3247179572447460259609088544781\end{array}\right)$```
To use `sols` without Symbolic Math Toolbox™, you can generate code and convert `sols` to a MATLAB® function by using `matlabFunction`. The generated file uses the `roots` function that operates on the numeric `double` data type.
```matlabFunction(sols,"File","myfile"); type myfile```
```function sols = myfile %MYFILE % SOLS = MYFILE % This function was generated by the Symbolic Math Toolbox version 9.3. % 03-Mar-2023 07:29:21 t0 = roots([1.0,-1.0,0.0,0.0,0.0,-1.0]); t2 = t0(1); t0 = roots([1.0,-1.0,0.0,0.0,0.0,-1.0]); t3 = t0(2); t0 = roots([1.0,-1.0,0.0,0.0,0.0,-1.0]); t4 = t0(3); t0 = roots([1.0,-1.0,0.0,0.0,0.0,-1.0]); t5 = t0(4); t0 = roots([1.0,-1.0,0.0,0.0,0.0,-1.0]); t6 = t0(5); sols = [t2;t3;t4;t5;t6]; end ```
Since R2023a
Find the indefinite integral of a polynomial fraction. The result is in terms of the `symsum` and `root` functions as represented by the $\sum$ and $\mathrm{root}$ symbols.
```syms x F = int(1/(x^3 + x - 1),x)```
```F = ${\sum }_{k=1}^{3}\mathrm{log}\left(\mathrm{root}\left({z}^{3}-\frac{3 z}{31}-\frac{1}{31},z,k\right) \left(3 x+\mathrm{root}\left({z}^{3}-\frac{3 z}{31}-\frac{1}{31},z,k\right) \left(6 x-9\right)\right)\right) \mathrm{root}\left({z}^{3}-\frac{3 z}{31}-\frac{1}{31},z,k\right)$```
Rewrite the result in terms of arithmetic operations. Because the symbolic summation is the outermost operation, first apply the `rewrite` function with the `"expandsum"` option to expand `symsum`. Then, apply `rewrite` again with the `"expandroot"` option to rewrite the `root` function. The resulting symbolic expression is in terms of arithmetic operations such as ^, *, /, +, and –.
`R = rewrite(rewrite(F,"expandsum"),"expandroot")`
```R = ```
## Input Arguments
collapse all
Input to rewrite or replace, specified as a symbolic number, variable, expression, function, vector, matrix, or multidimensional array.
Target function or function to expand, specified as a string scalar or character vector. This table summarizes the rewriting rules for all allowed `target` options.
TargetRewrite or Replace These FunctionsIn Terms of These Functions
`"exp"`All trigonometric and hyperbolic functions including inverse functions`exp`, `log`
`"log"`All inverse trigonometric and hyperbolic functions`log`
`"sincos"``tan`, `cot`, `exp`, `sinh`, `cosh`, `tanh`, `coth``sin`, `cos`
`"sin"`, `"cos"`, `"tan"`, or `"cot"``sin`, `cos`, `exp`, `tan`, `cot`, `sinh`, `cosh`, `tanh`, `coth` except the targetTarget trigonometric function
`"sinhcosh"``tan`, `cot`, `exp`, `sin`, `cos`, `tanh`, `coth``sinh`, `cosh`
`"sinh"`, `"cosh"`, `"tanh"`, `"coth"``tan`, `cot`, `exp`, `sin`, `cos`, `sinh`, `cosh`, `tanh`, `coth` except the targetTarget hyperbolic function
`"asin"`, `"acos"`, `"atan"`, `"acot"``log`, and all inverse trigonometric and inverse hyperbolic functionsTarget inverse trigonometric function
`"asinh"`, `"acosh"`, `"atanh"`, `"acoth"``log`, and all inverse trigonometric and inverse hyperbolic functionsTarget inverse hyperbolic function
`"sqrt"``abs(x + 1i*y)``sqrt(x^2 + y^2)`
`"heaviside"``sign`, `triangularPulse`, `rectangularPulse``heaviside`
`"piecewise"``abs`, `heaviside`, `sign`, `triangularPulse`, `rectangularPulse``piecewise`
`"expandroot"``root`Arithmetic operations such as ^, *, /, +, and –
`"expandsum"``symsum`Arithmetic operations such as + and –
## Version History
Introduced in R2012a
expand all | 2023-03-24T19:38:46 | {
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http://math.stackexchange.com/questions/205888/what-is-the-possibility-of-choosing-three-balls-in-a-certain-sequence-given-a-ce | What is the possibility of choosing three balls in a certain sequence given a certain number of attempts?
Say I have three balls - red, green and blue in a bag. I will choose a ball from the bag and then return it, then choose another one, etc. Say I will do this 3 times; then the possibility of me picking a certain sequence (for example RGB) is 1/3 * 1/3 * 1/3. But say I do this 4 times. What is the possibility then of running into the same sequence? The desired pick becomes one of the following: BRGB or RRGB or GRGB or RGBB or RGBG or RGBR. I can do the math for this, but what is the general equation where the number of picks is X? Say I want to pick ten balls, what is the possibility of getting RGB then?
For example if I pick four times there is a (1/3)^4 chance that I would get RGBR, but my question isn't the possibility of getting a FOUR letter sequence, rather of getting a three letter sequence in a four letter possibility. The correct answer when x = 4 is (1/3)^4 *6 = 6/81.
Am I explaining this easily? Thanks.
EDIT: Please note that there is only one trial.
-
That looks... relevant... I think. I'll look into it but I thought the question I asked is actually a very well known and common one, and a question that already has a solution in the form of an elegant equation somewhere. Something like (1/3)^x * x!/(x-1)! * 2 or something – numandina Oct 2 '12 at 8:35
I don't know where that quation came from, but you can certainly simplify it to (1/3)^x*x*2 :) – Alex Oct 2 '12 at 8:38
Oh, I think it's more binomial than poisson distribution :/ – Alex Oct 2 '12 at 8:40
I think I solved it! – numandina Oct 2 '12 at 9:20
is it the same as my answer? – Alex Oct 2 '12 at 9:20
Let
• $a(n)$ be the number of strings of length $n$ not containing RGB and not ending R or RG
• $b(n)$ be the number of strings of length $n$ not containing RGB and ending R
• $c(n)$ be the number of strings of length $n$ not containing RGB and ending RG
• $d(n)$ be the number of strings of length $n$ containing RGB
then starting at $a(0)=1$ and $b(0)=c(0)=d(0)=0$ you have
• $a(n+1)=2a(n)+b(n)+c(n)$
• $b(n+1)=a(n)+b(n)+c(n)$
• $c(n+1)=b(n)$
• $d(n+1)=c(n)+3d(n)$
and $a(n)+b(n)+c(n)+d(n)=3^n$. You want the probability $d(n)/3^n$.
That is enough to do the calculation, and for ten balls gives $\frac{16293}{59049}\approx 0.2759$ though you could instead use the generating function $\dfrac{x^3}{(1-x)(27-27x+x^3)}.$
-
Could you please explain how to use the generative function? – numandina Oct 2 '12 at 11:38
BTW I think you meant b(0) = c(0) = d(0) = 0. – numandina Oct 2 '12 at 14:29
@numandina: On your second point you are correct. On the first, if you want to understand you could try Wikipedia or generatingfunctionology; if you want to calculate the coefficients of the series expansion at $0$ then you can use a Computer Algebra System such as this – Henry Oct 2 '12 at 15:55
Great answer. Best part is using your logic I can create rules for any other sequence of different length. Just a quick question: I'm thinking of finding a closed form soluition, something without iterations, do you thikn it's possible or not? – numandina Oct 3 '12 at 7:21
@numandina: Looking at the related OEIS sequences A052963 and A076264, there is probably a complicated and ugly closed form involving the roots of a cubic equation. – Henry Oct 3 '12 at 18:01
Let $a_n$ be the number of strings of length $n$ that do not contain the sequence RGB; I’ll call these the bad sequences. Clearly $a_0=1,a_1=3$, and $a_2=9$. Suppose now that $n\ge 2$. To build a bad sequence of length $n+1$, to a first approximation you can add any color to the end of a bad string of length $n$; that gives you $3a_n$ strings. However, if the string of length $n$ ends in RG, you cannot add a B. Each bad string of length $n$ ending in RG is the result of adding RG to a arbitrary bad string of length $n-2$, so there are $a_{n-2}$ such bad strings of length $n$ to which we cannot add B. Thus, $a_{n+1}=3a_n-a_{n-2}$.
Now let $p_n$ be the probability of drawing a bad string of length $n$; $p_n=\dfrac{a_n}{3^n}$, so
$$p_{n+1}=\frac{3a_n-a_{n-2}}{3^{n+1}}=\frac{3^{n+1}p_n-3^{n-2}p_{n-2}}{3^{n+1}}=p_n-\frac{p_{n-2}}{27}\;,$$
and the probability of drawing a string of length $n$ containing RGB is $1-p_n$. The generating function for $a_n$ involves a somewhat intractable cubic, but the recurrences for $a_n$ and $p_n$ aren’t bad.
-
Thanks for the answer. You're smart. – numandina Oct 3 '12 at 7:22
I think you want this:
No. of possibilities with RGB = $(x-2)3^{x-3}$
$$p = \frac{(x-2)3^{x-3}}{3^{x}} = (x-2)3^{-3} = \frac{x-2}{27}$$
for x=5 you get 27 possibilities, not 24.
However, this only works for $x<6$ as you could get two triples otherwise. I don't exactly know how avoid this at the moment.
Wrong: $$X \sim B\left (4,\frac{1}{3} \right )$$
$$Pr(X \leq 4) = \sum_{i=1}^{4}\binom{4}{i}\left (\frac{1}{9} \right )^{i}\left (1-\frac{1}{9} \right )^{i} \approx 0.45753$$
-
Thanks for the help but I don't think I follow. For x = 4 the answer is 6/81 = 0.074. For x = 5 the answer is 24/243 = 0.099. This makes sense since for x = infinity the answer should be 1. – numandina Oct 2 '12 at 8:52
@numandina Sorry, I think I misunderstood the question. I thought you take three balls, and do that four times. – Alex Oct 2 '12 at 8:55
If x = 4: You take four balls, and look for a sequence of three balls. If x = 5, you take five balls, and look for a sequence of three balls. If x = 10000, you take 10000 balls and look for the sequence. Think of infinite monkeys on infinite typewriters or arranging finite atoms in an infinite universe, etc. – numandina Oct 2 '12 at 8:56
Here is an approach that does not count strings:
There are three nonterminal states in this game, namely $s_1$: only the letter $B$ is missing, $s_2$: the two letters GB are missing, and $s_3$: everything else. Denote by $p_{r,k}$ the probability that we fail when $r\geq0$ picks are left over and we are in state $s_k$. Then $p_{0,k}=1$ $\ (1\leq k\leq 3)$.
The $p_{r,k}$ satisfy the following recurrence: \eqalign{p_{r,1}&={1\over3}p_{r-1,2}+{1\over3}p_{r-1,3}\cr p_{r,2}&={1\over3}p_{r-1,2}+{1\over3}p_{r-1,1}+{1\over3}p_{r-1,3}\cr p_{r,3}&={1\over3}p_{r-1,2}+{2\over3}p_{r-1,3}\ .\cr}
It follows that ${\bf p}_r={1\over3^r}A^r{\bf 1}$, where $A$ is the matrix $$A=\left[\matrix{0&1&1\cr 1&1&1\cr 0&1&2\cr}\right]\ .$$ Unfortunately this matrix has unfriendly eigenvalues, so that no simple formula for $p_{n,3}$ results, albeit this number is, of course, rational.
- | 2015-04-28T01:14:43 | {
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https://physics.stackexchange.com/questions/633030/proof-that-body-performs-shm | Proof that body performs SHM
At core, SHM is the shadow of a particle revolving with omega on a circle of radius equal to amplitude. Now we say that body performs SHM either if the equation of its position makes a sinusoidal function or if its acceleration is proportional to displacement and in opposite direction.
I get how in the first way we can always equate it to the shadow of a particle revolving but in second way, i.e., by seeing its acceleration proportional to displacement how can we say for sure it has an equivalent particle shadow? Yes, if it starts from the mean position, we can make a similar shadow which starts from mean and the particle revolves with omega (according to acceleration of body and amplitude). What if the body starts with velocity v from any random distance from mean? How then can we be sure we will find its equivalence to a particle shadow?
Usually in that case, going by the questions we solved in class, we find the amplitude by equations of SHM itself. But we can't use the equation without proving SHM.
I am sure there is some obvious point here but I can't see it.
A defining equation for shm is $$\ddot x = - \omega^2\,x$$ where $$x$$ is the displacement from a fixed point, $$\ddot x$$ is the acceleration and $$\omega^2$$ is a positive constant which is a characteristic of the motion.
The solution to this second order differential equation requires two initial conditions and your the body starts with velocity v from any random distance from mean could be them as at time, $$t=0$$ you are given the displacement, $$x_{\rm initial}$$ and the velocity, $$\dot x_{\rm initial}$$.
From these you can get an equation of motion of the form $$x(t) = A \sin (\omega\,t+\phi)$$.
In terms of your projection idea, $$A$$ would be the radius of the circle and $$\phi$$ the position on the circle, relative to the $$x$$-axis, where the motion starts from at time $$t=0$$.
So in the animation below you can start the clock at any position on the circle and the projected point on the axis will execute shm, ie be a solution of $$\ddot x = - \omega^2\,x$$. | 2021-09-26T03:59:37 | {
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https://math.stackexchange.com/questions/2453992/number-of-ways-of-arranging-ten-girls-and-three-boys-if-the-boys-separate-the-gi | # Number of ways of arranging ten girls and three boys if the boys separate the girls into groups of sizes $3, 3, 2, 2$
Ten girls are to be divided into groups of sizes $3,3,2,2$. Also, there are $3$ boys. Number of ways of linear sitting arrangement such that between any two groups of girls, there is exactly one boy (no boy sits at either extreme end)?
MY SOLUTION:
$10$ girls can be divided into groups of sizes $3,3,2,2$ in $$\frac{10!}{3!3!2!2!2!2!}$$ ways which gives me unique combination of groups.
I can then arrange these groups in $4!$ ways, and people within them in $3!3!2!2!$ ways.
Finally the $3$ boys in $3!$ ways.
Seems correct? It gives $$\frac{10! 4! 3!}{2! 2!}$$ ways.
• Yes your approach seems absolutely fine! Its good to show your work when you ask a question. – samjoe Oct 2 '17 at 7:31
• @samjoe but I am skeptical about arranging groups. Should it be multiplied by 4!? I am doing so because each group so formed is unique in its configuration. It shouldn't lead to duplicate arrangements. Will it? – Ajax Oct 2 '17 at 7:39
• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Oct 2 '17 at 9:10
• @N.F.Taussig Alright, Thanks! Do you think my approach is correct for this question? And that I am not counting duplicate cases? – Ajax Oct 2 '17 at 9:28
Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
• Hasan Groups can also be distributed around boys. I mean the size of groups is not identical, that is.. – samjoe Oct 2 '17 at 7:11
• So, this answer 10!x3! is for any combination of groups (totalling to 10)? – Ajax Oct 2 '17 at 7:13
• @Ajax, What is the meaning of 'totaling to 10'? – Hasan Heydari Oct 2 '17 at 7:20
• 1,2,2,5 or 2,2,2,4 or 3,3,3,1 all total to 10 – Ajax Oct 2 '17 at 7:28
• Your answer is incorrect because you have not accounted for the number of ways of arranging two groups of three girls and two groups of two girls. – N. F. Taussig Oct 2 '17 at 8:56
We will seat the girls in two blocks of three seats and two blocks of two seats. There are $\binom{4}{2}$ ways to choose the positions of the blocks of three seats. Once the blocks of seats have been arranged, there are $\binom{10}{3}$ ways to select the girls who will sit in the leftmost block of three seats and $3!$ ways to arrange them in those seats, $\binom{7}{3}$ ways to select which of the remaining seven girls who will sit in the other block of three seats and $3!$ ways to arrange them in those seats, $\binom{4}{2}$ to select which of the four remaining girls will sit in the leftmost block of two seats and $2!$ ways to arrange them in those seats, and $2!$ ways to arrange the girls who will sit in the other block of two seats. The three boys can be arranged in the three seats that separate the blocks of girls in $3!$ ways. $$\binom{4}{2}\binom{10}{3} \binom{7}{3}\binom{4}{2}\binom{2}{2}3!3!2!2!3! = \frac{4!}{2!2!} \cdot \frac{10!}{3!7!} \cdot \frac{7!}{3!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} \cdot 3!3!2!2!3! = \frac{4!}{2!2!} \cdot 10! \cdot 3!$$ as you found. | 2019-11-21T01:30:46 | {
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https://www.hpmuseum.org/forum/thread-4432.html | checkdigit calculation for HP-17b
07-26-2015, 12:08 AM (This post was last modified: 07-26-2015 02:52 PM by Don Shepherd.)
Post: #1
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
checkdigit calculation for HP-17b
The Luhn algorithm for calculating a standard mod-10 checkdigit for an account or ID number lends itself very well to a simple program, or in this case a 17b solver equation.
Checkdigits are typically used during data entry to prevent the transposition of digits, so that ID 12345, if keyed in as 13245, can be corrected before the erroneous data becomes part of a file. Each digit in the account number, starting at the right-hand side, is multiplied by the pattern 212121..., the digits of the products are summed and the total is subtracted from the next higher multiple of 10. The result is the checkdigit associated with the ID number, the rightmost digit.
An example:
ID: 4 5 6 7 8 9
multiply: 1 2 1 2 1 2
product: 4 10 6 14 8 18
digits sum: 33
subtract from next higher multiple of 10: 40-33=7
checkdigit is 7, so the final ID is 4567897
In 2007, when I became aware of the 17b/17bii, I wrote a solver equation to calculate the checkdigit of an ID number, up to 12 digits in length. Here is that equation:
CD = 0$$\times$$L(M:2)$$\times$$L(C:-1)
+MOD(10-
MOD($$\Sigma$$(I:0:LOG(N):1:
0$$\times$$L(A:MOD(IP(N):10)$$\times$$G(M))
+IF(G(A)<10:G(A):G(A)-9)
+0$$\times$$L(M:G(M)+G(C))
$$\times$$L(C:-G(C))
$$\times$$L(N:N$$\div$$10)):10):10)
That equation works, but it is rather long, and smaller solver equations run faster than longer ones, so recently I took another look at this equation, with the following goals:
1. Make equation as small as possible
2. Avoid IF functions
3. Avoid 0 times
4. Handle getting each digit of the input number
5. Handle the 2121 multiplier
6. Handle 2X7 =14, you really want 5 not 14
7. Handle checkdigit 0, because 10-0=10, not 0
With those goals in mind, here is what I came up with:
CD = MOD(10-
MOD($$\Sigma$$(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)$$\times$$(2-MOD(I:2)))
-9$$\times$$IDIV(G(A):10)
):10):10)
This reduced the size of the equation from 164 characters to 92 characters, a 44% reduction in size, and I achieved all of my goals. And, more importantly, it forced me to think through exactly what was needed, and use the available 17b solver functions to achieve this.
The 17b is a great little machine and can exercise your mind in 2015 as well as it could in 1988.
07-26-2015, 07:47 AM
Post: #2
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 12:08 AM)Don Shepherd Wrote: This reduced the size of the equation from 164 characters to 92 characters, a 44% reduction in size, and I achieved all of my goals.
We can go a little further:
CD = MOD($$\Sigma$$(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)$$\times$$(MOD(I:2)-2))
+IDIV(G(A):10)
):10)
I moved the final negation into the expression so we have to calculate MOD 10 only once.
Instead of -9 we can use +1 when calculating MOD 10.
However I can't verify the formula at the moment.
Thus I could be completely wrong.
Quote:The 17b is a great little machine and can exercise your mind in 2015 as well as it could in 1988.
Completely agree with you: I'm always happy when you bring it up.
The Luhn-algorithm was new to me. Thanks for letting us know.
Kind regards
Thomas
07-26-2015, 12:56 PM
Post: #3
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 07:47 AM)Thomas Klemm Wrote:
(07-26-2015 12:08 AM)Don Shepherd Wrote: This reduced the size of the equation from 164 characters to 92 characters, a 44% reduction in size, and I achieved all of my goals.
We can go a little further:
CD = MOD($$\Sigma$$(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)$$\times$$(MOD(I:2)-2))
+IDIV(G(A):10)
):10)
I moved the final negation into the expression so we have to calculate MOD 10 only once.
Instead of -9 we can use +1 when calculating MOD 10.
However I can't verify the formula at the moment.
Thus I could be completely wrong.
Quote:The 17b is a great little machine and can exercise your mind in 2015 as well as it could in 1988.
Completely agree with you: I'm always happy when you bring it up.
The Luhn-algorithm was new to me. Thanks for letting us know.
Kind regards
Thomas
Thomas, thanks for this. I can't get it to compile, however. Can you check it?
Don
07-26-2015, 01:25 PM
Post: #4
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 12:56 PM)Don Shepherd Wrote: I can't get it to compile, however. Can you check it?
Sure. You just have to be a little patient.
Cheers
Thomas
07-26-2015, 02:04 PM (This post was last modified: 07-26-2015 02:11 PM by Don Shepherd.)
Post: #5
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 01:25 PM)Thomas Klemm Wrote:
(07-26-2015 12:56 PM)Don Shepherd Wrote: I can't get it to compile, however. Can you check it?
Sure. You just have to be a little patient.
Cheers
Thomas
My mistake, I forgot one paren. It works. Great work, I like it! Down to 79 characters!
Don
07-26-2015, 04:45 PM
Post: #6
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 02:04 PM)Don Shepherd Wrote: It works.
Good to hear. And I was going to blame my sausage fingers fumbling on a smart-phone.
Have these mechanical computers for verifying numbers in Luhn's patent ever been produced?
Cheers
Thomas
07-26-2015, 05:02 PM
Post: #7
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 04:45 PM)Thomas Klemm Wrote: Have these mechanical computers for verifying numbers in Luhn's patent ever been produced?
Cheers
Thomas
I've certainly never seen one. It does look very interesting, however.
07-26-2015, 07:29 PM
Post: #8
Gerson W. Barbosa Senior Member Posts: 1,404 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
Hello Don,
Interesting algorithm and HP-17B equation!
Here we happy taxpayers have a personal 11-digit number known as CPF, the last two of which being the checking digits. Starting with your SOD (sum of digits) equation, it wasn't difficult to obtain a 17Bii equation to compute them:
C=10*MOD(MOD(0*L(T:0)+Σ(I:1:9:1:0*L(T:G(T)+I*MOD(N:10))+(10-I)
*MOD(N:10)+0*L(N:IDIV(N:10))):11):10)+MOD(MOD(G(T):11):10)
No attempt have been made to optimize it, though.
123 456 789 --> 0 (they should be 00, actually)
976 431 258 --> 64
777 777 777 --> 77
The interesting part is that I didn't need to look at the linked Wikipedia article before writing the equation, as I knew the algorithm since I was 14 or 15, when I didn't have access to computers or even calculators, programmable or not. My father taught that to me. He was a policeman and once had a report returned because of missing or wrong checking digits. Since the number holder was not readily available, he tried and figured out the algorithm by himself, using pencil and paper. He was no computer or math expert!
Gerson.
07-26-2015, 08:34 PM
Post: #9
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 07:29 PM)Gerson W. Barbosa Wrote: The interesting part is that I didn't need to look at the linked Wikipedia article before writing the equation, as I knew the algorithm since I was 14 or 15
Thanks Gerson.
I first learned about this algorithm when I started my job as a programmer in 1974 at the US Census Bureau (ah, those big Univac mainframes, I loved them).
I worked in the Economic Surveys division of the Bureau, and our main file had a record for each business establishment in the US. Every business had an Employer Identification number (EIN) assigned by either Social Security or IRS, probably IRS. In our programs, whenever we processed these files we had to verify the correct EIN, and part of that was to verify the checkdigit. I had never heard of the Luhn algorithm; my boss had probably never heard of it either, but he sure knew how it worked, and he explained it to me so I could implement it in every FORTRAN program I wrote.
I discovered it is also used on all credit cards in the US.
Don
07-26-2015, 11:03 PM (This post was last modified: 07-28-2015 02:32 AM by Gerson W. Barbosa.)
Post: #10
Gerson W. Barbosa Senior Member Posts: 1,404 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
Don, here is a Free Pascal version:
Code:
Program CPF; Uses Crt; var c, i: Byte; s, t: Integer; n: Longint; begin ClrScr; Read(n); s:=0; t:=0; for i:=1 to 9 do begin s:=s+(10-i)*(n Mod 10); t:=t+i*(n Mod 10); n:=n div 10 end; c:=10*((s Mod 11) Mod 10) + ((t Mod 11) Mod 10); WriteLn(c:2) end.
Run:
999999999
99
Type EXIT to return...
Again, no attempt has been made to optimize the program, but it's interesting to compare the number of characters in both the Pascal and 17Bii codes.
Notice the algorithm I've used is somewhat different (and easier to implement on the 17Bii) than the one in the Wikipedia article. Nonetheless, I get exactly the same results.
Gerson.
Edited to fix program line #17
Edited again to fix algorithm (inclusion of an IF-clause)
Code:
Program CPF; Uses Crt; var d1,d2, i: Byte; s, t: Integer; n: Longint; begin ClrScr; Read(n); s:=0; t:=0; for i:=1 to 9 do begin s:=s+(10-i)*(n Mod 10); t:=t+i*(n Mod 10); n:=n div 10 end; if (s Mod 11)=10 then t:=t+9; d1:=(s Mod 11) Mod 10; d2:=(t Mod 11) Mod 10; GotoXY(10,1); WriteLn('-',d1:1,d2:1) end. ------------------------- Run 123456789 123456789-09 Type EXIT to return...
As I said, this algorithm is more simple than the official one [ function TestaCPF(strCPF) ]. I have yet to discover why this works.
07-26-2015, 11:42 PM (This post was last modified: 07-27-2015 12:00 AM by Don Shepherd.)
Post: #11
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
Thanks Gerson.
So everybody who pays taxes in Brazil has a number. Same here in the US, ours is just called our Social Security number, and we all must go to great lengths to make sure nobody knows our SS number except people who need to know it (like banks and employers). If the bad guys know your number, I'm told your life will become miserable. I have to shred some personal financial documents occasionally so my number doesn't become known by the bad guys.
I got my SS number when I was 14 and got a part-time job delivering newspapers. Now, however, newborn babies get their number at birth. Some employers used to use an employees SS number as their personnel number, but I think that pretty much stopped when the bad guys got active. When I worked for a grocery store back in the 1960's they used my SS number as my employee number, and it always had the number 5 added to the end, and I wondered about that until years later I realized it was the checkdigit!
I once programmed a dedicated data entry system (Entrex), and its programming language had built-in routines for checkdigit verification.
Now, the challenge: try to derive my 9-digit SS number from the fact that its checkdigit is 5. If you could do that, the bad boys will pay you millions I'm sure.
Here is a link to a site with some interesting trivia information about the first SS card, back in 1936.
Don
07-27-2015, 01:03 AM
Post: #12
Bill (Smithville NJ) Senior Member Posts: 437 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 11:42 PM)Don Shepherd Wrote: Now, the challenge: try to derive my 9-digit SS number from the fact that its checkdigit is 5. If you could do that, the bad boys will pay you millions I'm sure.
Hi Don,
Researchers have long figured out how to "guess" your SS number by using public data. SS numbers are not really random - or at least parts of it aren't.
Do a search for plenty of articles on predicting SS numbers. There's been plenty of news stories about this.
When I lived in Indiana, they used to use your SS number for the drivers license number. To get around the law that says you can't use SS numbers for other purposes, they just appended the Letter 'S' to it - that would magically make it NOT a SS number. I don't think they do that anymore.
A couple of years ago, I went to my local college to use their library. They said I could get a courtesy library card (even though I wasn't a student) since I lived in their county. But to get the card, I would have to give them my SS number - and you would never believe the reason given. The SS number was being required by the Department of Homeland Security. I passed on getting a library card there.
Bill
Smithville, NJ
07-27-2015, 04:11 PM
Post: #13
wynen Junior Member Posts: 12 Joined: Sep 2014
RE: checkdigit calculation for HP-17b
(07-26-2015 04:45 PM)Thomas Klemm Wrote: Have these mechanical computers for verifying numbers in Luhn's patent ever been produced?
I'm quite sure, you can find these little "computers" in the collection of the "Arithmeum",
a Museum for "Calculating in olden and modern times" in Bonn, Germany.
I've been there last Friday and it was amazing.
Hartmut (from Monheim am Rhein, Germany, near Bonn)
07-28-2015, 02:54 PM
Post: #14
Don Shepherd Senior Member Posts: 745 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-26-2015 07:47 AM)Thomas Klemm Wrote:
(07-26-2015 12:08 AM)Don Shepherd Wrote: This reduced the size of the equation from 164 characters to 92 characters, a 44% reduction in size, and I achieved all of my goals.
We can go a little further:
CD = MOD($$\Sigma$$(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)$$\times$$(MOD(I:2)-2))
+IDIV(G(A):10)
):10)
I moved the final negation into the expression so we have to calculate MOD 10 only once.
Instead of -9 we can use +1 when calculating MOD 10.
However I can't verify the formula at the moment.
Thus I could be completely wrong.
Quote:The 17b is a great little machine and can exercise your mind in 2015 as well as it could in 1988.
Completely agree with you: I'm always happy when you bring it up.
The Luhn-algorithm was new to me. Thanks for letting us know.
Kind regards
Thomas
Thomas, it has taken me a couple of days but I finally figured out exactly how your method works. I had to draw a table taking things one step at a time, for each digit of the input number. What I couldn't figure out was the behavior of the MOD function with negative numbers; I've always stayed away from negative numbers before, but I now understand how MOD and INT work differently with negative numbers as opposed to positive numbers. I didn't realize that before, and I thank you for giving me this opportunity to understand that behavior.
Your method is a work of art. You are a "steely-eyed missleman", as NASA would say.
Don
07-28-2015, 06:16 PM
Post: #15
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-27-2015 04:11 PM)wynen Wrote: I'm quite sure, you can find these little "computers" in the collection of the "Arithmeum",
a Museum for "Calculating in olden and modern times" in Bonn, Germany.
I've been there last Friday and it was amazing.
That looks very promising: Arithmeum.
Thanks for sharing. This might be a reason to travel to Bonn.
Cheers
Thomas
07-28-2015, 06:22 PM
Post: #16
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-28-2015 02:54 PM)Don Shepherd Wrote: I finally figured out exactly how your method works.
This is a step-by-step transformation of your formula to my solution.
1. We can remove the inner calculation of MOD 10 as it's enough to do that once:
CD = MOD(10-
MOD(Σ(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)×(2-MOD(I:2)))
-9×IDIV(G(A):10)
):10):10)
2. We can remove multiples of 10 as that doesn't change the result when calculating MOD 10:
CD = MOD(10-
Σ(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)×(2-MOD(I:2)))
-9×IDIV(G(A):10)
):10)
3. We can replace -9 by +1 as they are the same MOD 10:
CD = MOD(-
Σ(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)×(2-MOD(I:2)))
-9×IDIV(G(A):10)
):10)
CD = MOD(-
Σ(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)×(2-MOD(I:2)))
+IDIV(G(A):10)
):10)
Here comes the tricky part. If we move that just to the summand we have to change the sign of both expressions:
CD = MOD(
Σ(I:0:LOG(N):1:
-L(A:MOD(IDIV(N:10^I):10)×(2-MOD(I:2)))
-IDIV(G(A):10)
):10)
But if we move it into the Let-A expression the sign of A will change and thus we don't have to change the sign of the Get-A expression as well:
CD = MOD(
Σ(I:0:LOG(N):1:
L(A:-MOD(IDIV(N:10^I):10)×(2-MOD(I:2)))
+IDIV(G(A):10)
):10)
5. Now we're nearly there. Just move the - sign to the 2nd factor:
CD = MOD(
Σ(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)×(-2+MOD(I:2)))
+IDIV(G(A):10)
):10)
6. Change the order and we're finally done:
CD = MOD(
Σ(I:0:LOG(N):1:
L(A:MOD(IDIV(N:10^I):10)×(MOD(I:2)-2))
+IDIV(G(A):10)
):10)
BTW: That's why casting out nines works.
Your table looks familiar: I often have to do these kind of calculations until I understand something.
Kind regards
Thomas
07-28-2015, 06:48 PM
Post: #17
Gerson W. Barbosa Senior Member Posts: 1,404 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
Quoting myself:
(07-26-2015 07:29 PM)Gerson W. Barbosa Wrote: C=10*MOD(MOD(0*L(T:0)+Σ(I:1:9:1:0*L(T:G(T)+I*MOD(N:10))+(10-I)
*MOD(N:10)+0*L(N:IDIV(N:10))):11):10)+MOD(MOD(G(T):11):10)
...
123 456 789 --> 0 (they should be 00, actually)
976 431 258 --> 64
777 777 777 --> 77
I found out my first example was wrong, it should be 09, not 00. The algorithm I use is different from the official one [ function TestaCPF(strCPF) ], but I've managed to fix it so that it always gives the same results:
C=10*MOD(MOD(0*L(T:0)+L(S:Σ(I:1:9:1:0*L(T:G(T)+I*MOD(N:10))+(10-I)*MOD(N:10)
+0*L(N:IDIV(N:10)))):11):10)+MOD(MOD(G(T)+9*IDIV((MOD(G(S):11)):10):11):10)
123 456 789 --> 9 (09 actually, since we have two checking digits)
976 431 258 --> 64
777 777 777 --> 77
Here are the equivalent Free Pascal code and results:
Code:
------------- Program CPF; Uses Crt; var d1,d2, i: Byte; s, t: Integer; n: Longint; begin ClrScr; Read(n); s:=0; t:=0; for i:=1 to 9 do begin s:=s+(10-i)*(n Mod 10); t:=t+i*(n Mod 10); n:=n div 10 end; t:=t+9*((s Mod 11) div 10); d1:=(s Mod 11) Mod 10; d2:=(t Mod 11) Mod 10; GotoXY(10,1); WriteLn('-',d1:1,d2:1) end. ------------- 123456789-09 976431258-64 777777777-77 ------------
It intrigues me a bit why this more simple algorithm works, but I am not nearly as diligent as Don to find the reason.
Code:
976 431 258 Algorithm 1: 9 7 6 4 3 1 2 5 8 x 1 2 3 4 5 6 7 8 9 9 + 14 + 18 + 16 + 15 + 6 + 14 + 40 + 72 = 204; 204 mod 11 = 6; 1st Chk Dgt 9 7 6 4 3 1 2 5 8 x 9 8 7 6 5 4 3 2 1 81 + 56 + 42 + 24 + 15 + 4 + 6 + 10 + 8 = 246; 246 mod 11 = 4; 2nd Chk Dgt Algorithm 2: 9 7 6 4 3 1 2 5 8 x 10 9 8 7 6 5 4 3 2 90 + 63 + 48 + 28 + 18 + 5 + 8 + 15 + 16 = 291; 291 x 10 = 2910; 2910 mod 11 = 6; 1st Chk Dgt | +----------------------------------+ | v 9 7 6 4 3 1 2 5 8 6 x 11 10 9 8 7 6 5 4 3 2 99 + 70 + 54 + 32 + 21 + 6 + 10 + 20 + 24 + 12 = 348; 348 x 10 = 3480; 3480 mod 11 = 4; 2nd Chk Dgt
Gerson.
07-28-2015, 08:28 PM
Post: #18
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-28-2015 06:48 PM)Gerson W. Barbosa Wrote: It intrigues me a bit why this more simple algorithm works
Code:
v[1] := (1×cpf[1] + 2×cpf[2] + 3×cpf[3] + 4×cpf[4] + 5×cpf[5] + 6×cpf[6] + 7×cpf[7] + 8×cpf[8] + 9×cpf[9]) mod 11
Code:
v[2] := (9×(1×cpf[1] + 2×cpf[2] + 3×cpf[3] + 4×cpf[4] + 5×cpf[5] + 6×cpf[6] + 7×cpf[7] + 8×cpf[8] + 9×cpf[9]) + 1×cpf[2] + 2×cpf[3] + 3×cpf[4] + 4×cpf[5] + 5×cpf[6] + 6×cpf[7] + 7×cpf[8] + 8×cpf[9]) mod 11
We can do the multiplication by 9 mod 11:
Code:
v[2] := (9×cpf[1] + 7×cpf[2] + 5×cpf[3] + 3×cpf[4] + 1×cpf[5] - 1×cpf[6] - 3×cpf[7] - 5×cpf[8] - 7×cpf[9] + 1×cpf[2] + 2×cpf[3] + 3×cpf[4] + 4×cpf[5] + 5×cpf[6] + 6×cpf[7] + 7×cpf[8] + 8×cpf[9]) mod 11
This gives us:
Code:
v[2] := (9×cpf[1] + 8×cpf[2] + 7×cpf[3] + 6×cpf[4] + 5×cpf[5] + 4×cpf[6] + 3×cpf[7] + 2×cpf[8] + 1×cpf[9]) mod 11
But then you will say that v[1] wasn't calculated correctly:
Code:
v[1] := v[1] mod 11 v[1] := v[1] mod 10
Thus in case that v[1] = 10 we will get 0 instead of 10.
This is corrected with this step in your program:
Code:
t:=t+9*((s Mod 11) div 10);
Kind regards
Thomas
07-28-2015, 09:44 PM
Post: #19
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-28-2015 06:48 PM)Gerson W. Barbosa Wrote:
C=10*MOD(MOD(0*L(T:0)+L(S:Σ(I:1:9:1:0*L(T:G(T)+I*MOD(N:10))+(10-I)*MOD(N:10)
+0*L(N:IDIV(N:10)))):11):10)+MOD(MOD(G(T)+9*IDIV((MOD(G(S):11)):10):11):10)
If you add both s and t you will get $$10\times\sum$$ of the digits which is just: $$-\sum$$ of the digits mod 11. If you call that u then s can be calculated as $$s=u-t=-(-u+t)$$.
This might be easier to implement.
Just an idea $$\cdots$$
Thomas
07-28-2015, 09:51 PM
Post: #20
Gerson W. Barbosa Senior Member Posts: 1,404 Joined: Dec 2013
RE: checkdigit calculation for HP-17b
(07-28-2015 08:28 PM)Thomas Klemm Wrote: This gives us:
Code:
v[2] := (9×cpf[1] + 8×cpf[2] + 7×cpf[3] + 6×cpf[4] + 5×cpf[5] + 4×cpf[6] + 3×cpf[7] + 2×cpf[8] + 1×cpf[9]) mod 11
Thanks, Thomas!
I wonder why they use a somewhat more complicate method here, that is, multiplications of the sums by 10 being required before the final mod 11 operation, not to mention the appending of the first checking digit to the number before computing the second checking digit. I haven't even tried to implement that method on the 17Bii, but it appears to me the resulting equation would be much longer.
(07-28-2015 08:28 PM)Thomas Klemm Wrote: But then you will say that v[1] wasn't calculated correctly:
Code:
v[1] := v[1] mod 11 v[1] := v[1] mod 10
Thus in case that v[1] = 10 we will get 0 instead of 10.
This is corrected with this step in your program:
Code:
t:=t+9*((s Mod 11) div 10);
Yes, I discovered that by examining a few wrong results (adding 9 to the first sum before mod 11 would always do in those cases). Another version uses the following instead, but that would make for a longer 17Bii equation, I think.
Code:
if (s Mod 11)=10 then t:=t+9;
Regards,
Gerson.
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SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. Floyd Digital Fundamentals, 9/e Laws of Boolean Algebra •Commutative Law of Multiplication:. 1, where A, B, and C can be considered as Booleans or individual bits of a logic operation [14]. Consequently, the “Laws” of Boolean algebra often differ from the “Laws” of real-number algebra, making possible such statements as 1 + 1 = 1, which would normally be considered absurd. Unlike "normal" algebra, variables in boolean algebra are either True or False. It is a convenient and systematic method of expressing and analyzing the operation of digital circuits and systems. Boolean Algebra Exam Question Some previous exam questions have listed applicable laws for you, so memorising the specific laws might not be necessary. also comprises new divisions of algebra, e. Boolean algebra. After this, the Boolean algebra is well known as the perfect way for representing the digital. A Boolean algebra (BA) is a set $$A$$ together with binary operations + and $$\cdot$$ and a unary operation $$-$$, and elements 0, 1 of $$A$$ such that the following laws hold: commutative and associative laws for addition and multiplication, distributive laws both for multiplication over addition and for addition over multiplication, and the following. Laws of nature and reasonableness. Operations are represented by '. Laws (Theorems) of Boolean algebra Laws of Complementation The term complement means, to invert or to change 1's to 0's and 0's to 1's, for which purpose inverters or NOT gates are used. Boolean algebra is a deductive mathematical system closed over the values zero and one (false and true). AB + AC = A(B + C). Some problems are also discussed. In propositional logic and Boolean algebra, De Morgan's laws are a pair of transformation rules that are both valid rules of inference. Boolean algebra laws. The basic laws of Boolean Algebra are the same as ordinary algebra and hold true for any number of variables. Using these boolean expressions, we can describe complex digital circuits with mathematical-like equations. DIGITAL ELECTRONICS IS THE MOST CONCEPTUAL INTERESTING SUBJECT TAC (TECHNICAL ACADEMY CAMPUS ) PROVIDE BEST QUALITY EDUCATION FOR POLYTECHNIC,B-TECH AND AMIE FOR ALL SUBJECT AND ALL SEMESTER WITH. Lim This work is licensed under a Creative Com-mons \Attribution-NonCommercial-ShareAlike 3. By simplifying the logic expression, we can convert a logic circuit into a simpler version that performs the same function. All the basic Boolean laws are proved by means of. Distribution in BOOL also works if the main connective is on the left side, like in the next problem. Laws of Boolean Algebra •Commutative Law of Addition: A + B = B + A. Next we'll look at a few formulas that can be used when working with polynomials. Boolean algebra satisfies many of the same laws as ord inary algebra when one matches up ∨ with addition and ∧ with multiplication. 4 Three More Laws Besides distribution, BOOL obeys other laws that have algebraic counterparts. Boolean algebra. Boolean algebra is one topic where most students get confused. It is an arithmetic interpretation of Proposition Logic and is also similar to Set theory. DIGITAL ELECTRONICS IS THE MOST CONCEPTUAL INTERESTING SUBJECT TAC (TECHNICAL ACADEMY CAMPUS ) PROVIDE BEST QUALITY EDUCATION FOR POLYTECHNIC,B-TECH AND AMIE FOR ALL SUBJECT AND ALL SEMESTER WITH. De Morgan's laws are a pair of transformation rules relating the set operators "union" and "intersection" in terms of each other by means of negation. (A + B)(A + C) = A + BC. The basic laws of Boolean Algebra are the same as ordinary algebra and hold true for any number of variables. Specifically, Boolean algebra was an attempt to use algebraic techniques to deal with expressions in the propositional calculus. These postulates for Boolean algebra originate from the three basic logic functions AND, OR and NOT. Boolean Algebra Simplifier. Let's construct truth tables for: xz + yz (x + y)z. (P ^True) P. The basic rules of this system were formulated in 1847 by George Boole of England and were subsequently refined by other mathematicians and applied to set theory. Answer: 2 (x + y. Boolean prime ideal theorem; Compactness theorem; Consensus theorem; De Morgan's laws; Duality (order theory) Laws of classical logic; Peirce's law; Stone's representation theorem for Boolean algebras. Boolean Algebra is the mathematical foundation of digital circuits. The Commutative Law, Associative Law, Distributive Law, and Identity Law all hold true for Boolean algebra. There are also two binary operators denoted by the symbol bar (-) or prime (‘). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. There are two statements under the Distributive Laws: Statement 1. But it is pretty simple if you understand the logic behind it. edu O ce hours: Monday 12:30 - 1:30 pm Monday 3:30 - 5:00 pm or by. A truth table shows each possible input combination to the gate or circuit with the resultant output depending upon the combination of input. Boolean Algebra 1. Maxterm from values. not be the zero or one elements of the Boolean algebra. is valid for Boolean algebra, but not for ordinary algebra. Get Started. The best way to help make things clearer is to work through a few examples, replacing the terms with different sets of actual values and working out the result. Boolean algebra. Boolean Algebra Laws. both + and · – Associative w. For example, you know you can regroup addition, called associativity (Ass): (2+3)+4 = 2+(3+4) Associativity also holds for multiplication: (2×3)×4 = 2×(3×4) what matters is that. Boolean algebra, symbolic system of mathematical logic that represents relationships between entities—either ideas or objects. Do it step by step (i. Boolean Algebra expressions are written in terms of variables and literals using laws, rules and theorems of Boolean Algebra. Show answer. Let a, b, and c be real numbers, variables, or algebraic expressions. ” An algebraic system qualifies as a Boolean algebra if and only if it has two binary, one unary, and two zero-ary operations which satisfy the postulated identities. Two laws in Boolean algebra which state that AND and OR, or union and intersections, are duel. By simplifying an algebraic expression, we mean writing it in the most compact or efficient manner, without changing the value of the expression. The above axioms are redundant, and all can be proven using only the identity, complement, commutative and distributive laws. Boolean algebra. It briefly considers why these laws are needed, that is to simplify complex Boolean expressions, and then demonstrates how the laws can be derived. Here you'll find links to all algebra formulas on this page. expression with up to 12 different variables or any set of minimum terms. There are three laws of Boolean Algebra that are the same as ordinary algebra. (p ∨ ¬q) (p ∧ q) 3. Any law that is true for an expression is also true for its dual. Boolean Algebra Example 1 Questions and Answers In this worked example with questions and answers, we start out with a digital logic circuit, and you have to make a Boolean expression, which describes the logic of this circuit. Or more informally as: 1) "not (A and B)" is the same as "(not A) or. X + 1 = 1 (null element) 2. 2 Digital Electronics I 4. Establish the connection between the two main behavioral models for gate networks, namely logical expressions and. ﬢ(ﬢA) = A 10. Boolean rings require fewer laws than Boolean algebras. That is interchange OR and AND operators and replace 1's by 0's and 0's by 1's. So, boolean algebra is veeerrrrry necessary for understanding of all these mentioned topics as well. Boolean algebra. The text begins with an informal introduction to the algebra of classes, exploring union, intersection, and complementation; the commutative, associative, and distributive laws; difference and. Related Topics. This video contains Boolean Algebra Law. LSN 4 – Laws of Boolean Algebra • Distributive laws A(B + C) = AB + AC. Boolean Algebra contains basic operators like AND, OR and NOT etc. Distributive Laws of Boolean Algebra. Note that every law has two expressions, (a) and (b). If p and q are two statements then, p + (p. That means that we may follow the statement ' x + a = b ' with the statement ' x = b − a. Primes are for real. A boolean function F defined on three input variables X, Y and Z is, if and only if number of 1 input is odd. To add operators of Boolean algebra, do the following:. A •(B + C) = A •B + A •C A (B + C) = A B + A C E1. 3 ) when you transform the absorption laws for Boolean algebra in Table 6 into logical equivalences. Rationalization is required for topics such as employing De Morgan's Law to reconstruct the negation of a given statement. basic digital circuit. distributive law-are the same as in ordinary algebra. Two laws in Boolean algebra which state that AND and OR, or union and intersections, are duel. Boolean Algebra Laws and Theorems. Boolean algebra is a strange sort of math. X X = X Involution law: 4. A truth table lists all possible combinations of. Binary and hexadecimal number systems. This lecture will discuss boolean algebra in detail. An algebra can be characterized as a ring containing the set. a complete atomic Boolean algebra. Commutative laws: For every a, b B I. Boolean logic reflects the binary logic of logic gates and transistors in a computer's CPU. A + (B + C) = (A + B) + C A •(B •C) = (A •B) •C. It is an algebraic system consisting of a set of element (0,1) associated with a Boolean variable and two binary operators AND and OR and a uniry operator NOT. , set difference) and examines certain restrictions placed on their use by Boole. This video is about the laws of Boolean algebra. Laws and Rules of Boolean Algebra (continued) Laws of Boolean Algebra (Continued) −The 12 Rules of Boolean Algebra A + 0 = A A + 1 = 1 A · 0 = 0 A · 1 = A A + A = A. Annulment Law – A term AND´ed with a “0” equals 0 or OR´ed with a “1” will equal 1. This is known as duality. Boolean Algebra. Using de Morgan's law to complement boolean expression. The same can be said of Boolean algebra. Following are the important rules used in Boolean algebra. Definition 3. Boolean Algebra is a form of mathematical algebra that is used in digital logic in digital electronics. Boolean algebra refers to symbolic manipulation of expressions made up of boolean variables and boolean operators. For the most part, these laws correspond directly to laws of Boolean Algebra for propositional logic as given in Figure 1. Counter-intuitively, it is sometimes necessary to complicate the formula before simplifying it. Boolean Transform • Given a Boolean expression, we reduce the expression (#literals, #terms) using laws and theorems of Boolean algebra. Boolean algebra. Boolean theorems and laws are used to simplify the various logical expressions. Let’s see how we would utilize DeMorgan’s theorems to simplify a digital logic circuit. Page Chapter 6: Boolean Algebra and Logic Circuits Slide 2/78 In this chapter you will learn about: § Boolean algebra § Fundamental concepts and basic laws of Boolean algebra § Boolean function and minimization § Logic gates § Logic circuits and Boolean expressions § Combinational circuits and design Learning Objectives 60. Find more Computational Sciences widgets in Wolfram|Alpha. Boolean Logic. This video contains Boolean Algebra Law. Operations are represented by '. Using DeMorgan’s theorems and the other theorems and laws of Boolean algebra, simplify the following logic expressions. Whereas in elementary algebra we have the values of the variables as numbers and primary operations are Addition and. n ∧ ¬n ∨ (n ∧ (q ∨ ¬q)) 2. #N#Demorgan's First Law: (A ∪ B)' = (A)' ∩ (B)' #N#The first law states that the complement of the union of two sets is the intersection of. I need some help getting this code to work. Boolean algebra is a deductive mathematical system closed over the values zero and one (false and true). Boolean Algebra Algebra is the branch of mathematics that deals with variables. BOOLEAN ALGEBRA QUESTIONS 2009 Outside Delhi: 6. This approach (called semicomplementation) is well-defined even for a (meet) semilattice; if the set of semicomplements has a greatest element it is. The principle of duality is used extensively in proving Boolean algebra theorem. There are also two binary operators denoted by the symbol bar (-) or prime (‘). Boolean algebra is one topic where most students get confused. Boolean Algebra Example 1 Questions and Answers In this worked example with questions and answers, we start out with a digital logic circuit, and you have to make a Boolean expression, which describes the logic of this circuit. Postulates and Basic Laws of Boolean Algebra. Complemented Laws: (i) a+a'=1 (ii)a * a'=0. These are obtained by changing every AND(. Primes are for real. The basic Laws of Boolean Algebra that relate to the Commutative Law allowing a change in position for addition and multiplication, the Associative Law allowing the removal of brackets for addition and multiplication, as well as the Distributive Law allowing the factoring of an expression, are the same as in ordinary algebra. The law appearing in the definition of Boolean algebras and lattice which states that a ^ (a v b)=a v (a ^ b)=a for binary operators v and ^ (which most commonly are logical OR and logical AND). Boolean Algebra Simplifier. There are two statements under the Distributive Laws: Statement 1. distributive law-are the same as in ordinary algebra. Mainly, the standard rules of Boolean algebra are given in operator '+' and 'x', based on the AND and OR logic gates equations. Boolean algebra is a type of algebra that is used in the design of (digital) logic circuitry, computer programs such as search engines and in general in analytic reasoning. C A A B F B F C C. • boolean algebra: symbols, rules • express the logical functions and, or, not, xor, nand and nor mathematically • basic laws of boolean algebra and how to apply them. It is used to analyze digital gates and circuits It is logic to perform mathematical operation on binary numbers i. The primary algebra is mainly a simpler notation for Boolean algebra, except for one thing. Because computers use only 2 numbers as we saw with Computer Number Systems, 0 or 1, George Boole developed a form of algebra that is used. A + (B + C) = (A + B) + C A •(B •C) = (A •B) •C E1. Mathematics - Mathematical rules and laws - numbers, areas, volumes, exponents, trigonometric functions and more ; Related Documents. Laws of Boolean algebra. A directory of Objective Type Questions covering all the Computer Science subjects. "An Investigation of the Laws of Thought, on Which Are Founded the Mathematical Theories of Logic and Probabilities. Here we study 10 of these laws considered to be more important, together with some examples for them. Subscribe to: Posts (Atom). A ring of sets is a nonempty collection of subsets of some set which is closed under the set-theoretic operations of finite union and difference. In 1854, Boole published a classic book, “An. Illustrate the use of the theorems of Boolean algebra to simplify logical expressions. Operations with 0 and 1:. The new sixth edition of Modern Algebra has two main goals: to introduce the most important kinds of algebraic structures, and to help students improve their ability to understand and work with abstract ideas. If you wish a more detailed study of Boolean algebra, we suggest you obtain Mathematics, Volume 3, NAVEDTRA 10073-A1. It may be seen that these laws can be proved using either truth tables or the basic rules given above. as false and the digital value. The bulletin of mathematical biophysics, vol. BOOLEAN ALGEBRA LAWS & RULES a + b = b + a ab = ba Law 1 commutative a + (b + c) = (a + b) + c a(bc) = (ab)c Law 2 associative (a + b)(c + d) = ac + ad + bc + bd Law 3 distributive. Here you can access and discuss Multiple choice questions and answers for various compitative exams and interviews. •If you have studied set theory or formal logic, these laws are also familiar to you. The above axioms are redundant, and all can be proven using only the identity, complement, commutative and distributive laws. Consequently, the “Laws” of Boolean algebra often differ from the “Laws” of real-number algebra, making possible such statements as 1 + 1 = 1, which would normally be considered absurd. distributive law-are the same as in ordinary algebra. is valid for Boolean algebra, but not for ordinary algebra. include columns for :P and P _Q etc. Draw the truth table for the above function and express it in canonical SOP form. Let AB denote a binary operation of Boolean algebra. Boolean, or boolean logic, is a subset of algebra used for creating true/false statements. Just as it helps to know that in ordinary algebra, , or that x + (-x) = 0, it is useful to be aware of the more useful equivalences (or identities or laws) of Boolean algebra. Postulates and Basic Laws of Boolean Algebra. These altogether follows the following laws:-. Z (ii) X + Y. a + b = b + a II. These are obtained by changing every AND(. Solutions to Frame 90: Boolean Simplication Veitch Diagrams :. The basic laws of Boolean algebra-the commutative laws for addition and. C How many gates do you save = A. 8 De Morgan's Laws; 2. 2 One variable NOT: AND: OR: XOR: 1. X + X' = 1 5D. Once you comprehend the premise of all quantities in Boolean algebra being limited to the two possibilities of 1 and 0, and the general philosophical. In particular the following laws are comm on. 111 - Introductory Digital Systems Laboratory Problem Set 1 Issued: February 7, 2007 Due: February 20, 2007 Boolean Algebra Practice Problems (do not turn in): Simplify each expression by algebraic manipulation. A •(B + C) = A •B + A •C A (B + C) = A B + A C. Laws of Boolean Algebra There are several laws in Boolean algebra. Laws & Rules of Boolean Algebra The basic laws of Boolean algebra: The commutative laws (nomjaätJil) The associative laws (nomäÖnn+J) The distributive laws (nnnojnjnu) 7 Commutative Laws The commutative law of addition for two variables is written as: A+B = B+A The commutatzve law of multiplication for two variables is written as: AB = BA AB. Boolean algebra is a deductive mathematical system closed over the values zero and. Then the + becomes the new main operator. SAT Math Test Prep Online Crash Course Algebra & Geometry Study Guide Review, Functions,Youtube - Duration: 2:28:48. ) and all 1's to 0's and vice-versa. • Boolean variables are used to indicate whether a condition is. Boolean algebra refers to symbolic manipulation of expressions made up of boolean variables and boolean operators. He published it in his book "An Investigation of the Laws of Thought". X + Y = Y + X 6D. Any law that is true for an expression is also true for its dual. You start off with the idea that some statement P is either true or false, it can’t be anything in between (this called the law of the excluded middle). Rules of Boolean Algebra. We can manipulate other Boolean expressions through successive application of these laws. Section 3: Basic Rules of Boolean Algebra 5 3. Thus, for example, if ^, V and - denote set intersection, union and complement then <= is the inclusive subset relation. a · b = b · a “plus” / ”OR” “times” / ”AND” A2. 4 Circuit Simplification Boolean Algebra Procedure Using the theorems and laws of Boolean. Show answer. 1 HR cafe sounds, coffee shop background audio, background white noise for studying or at the office - Duration: 1:00:26. As we know that this kind of algebra is a mathematical system consisting of a set of two or more district elements. distributive law-are the same as in ordinary algebra. Find Answer & Solution for the question State the Distributive laws of boolean algebra. The greatest advantage of B oolean rings is that given two expressions E 1 and E2 in a Boolean ring, it is easy to see if they are equivalent, that is whether E1 = E2. The basic Laws of Boolean Algebra can be stated as follows: Commutative Law states that the interchanging of the order of operands in a Boolean equation does not change its result. Note the Boolean theorem/law used at each simplification step. The order of operations for Boolean algebra, from highest to lowest priority is NOT, then AND, then OR. Boolean Algebra. The principle of duality is used extensively in proving Boolean algebra theorem. Hence, the values fo A + B and B + A are both equal. For addition, the associative law states When ORing more than two variables, the result is the same regardless of the grouping of the variables. Definition of Boolean algebra. Distribution in BOOL also works if the main connective is on the left side, like in the next problem. Using de Morgan's Law, write an expression for the complement of F if F(x, y, z) = xy + x'z + yz'. Boolean algebra provides the basis for analyzing the validity of logical propositions because it captures the two-valued character (binary) of statements that may be either true or false. A = A, because the variable A has only logical value. Variables are case sensitive, can be longer than a single character, can only contain alphanumeric characters, digits and the underscore. Commutative Laws. This type of algebraic structure captures essential properties of both set operations and logic operations. ) to OR(+), every OR(+) to AND(. Boolean Laws There are several laws (axioms) that define a Boolean algebra. Simply the following. include columns for :P and P _Q etc. These laws are sometimes also referred to as boolean algebra rules. This video contains Boolean Algebra Law. The basic laws of Boolean Algebra and the principle of duality are presented in the lecture. Featured on Meta Improving the Review Queues - Project overview. Laws of Boolean Algebra: Identity, Anullment, Idempotent, Inverse, Involution, Complement, Commutative, Associative, Distributive, Absorption, DeMorgan's Theorems. It is concerned with statements which are either true or false. See the 2 examples below:. (A + B)(A + C) = A + BC. Boolean Transform • Given a Boolean expression, we reduce the expression (#literals, #terms) using laws and theorems of Boolean algebra. Boolean Algebra - 1 • A set of elements B – There exist at least two elements x, y ∈ B s. Boolean algebra devised in 1864 by George Boole, is a system of mathematical logic. Laws of Boolean Algebra There are several laws in Boolean algebra. Applying the Boolean algebra basic concept, such a kind of logic equation could be simplified in a more simple and efficient form. 2 (b) Write the equivalent Boolean Expression for the following logic circuit: 2 (c)Write the POS form of a Boolean function G, which is represented in a truth table as follows 1. Proof of first Idempotent Law. •If you have studied set theory or formal logic, these laws are also familiar to you. This type of logic is called Boolean because it was invented in the 19th century by George Boole, an English mathematician and philosopher. 12 implements an XOR function. Derivation of Boolean expression:- Minterm : minterm is a Product of all the literals within the logic System. This video contains Boolean Algebra Law. Boolean algebra, a logic algebra, allows the rules used in the algebra of numbers to be applied to logic. It reduces the original expression to an equivalent expression that has fewer terms. Students will learn to practically apply the Boolean laws and simplification of Logic gates. The Law of Distribution in boolean algebra is identical to the law of distribution in “normal” algebra: A(B C) = AB AC Applying the Law of Distribution While the process of distribution is not difficult to understand, the reverse of distribution (called factoring ) seems to be a more difficult process for many students to master:. ) to OR(+), every OR(+) to AND(. This type of algebraic structure captures essential properties of both set operations and logic operations. In abstract algebra, a Boolean algebra or Boolean lattice is a complemented distributive lattice. X Y = Y X Associative laws: 7. Try to recognize when it is. We know that a computer’s most basic operation is based on digital. Postulates and Basic Laws of Boolean Algebra. Boolean prime ideal theorem; Compactness theorem; Consensus theorem; De Morgan's laws; Duality (order theory) Laws of classical logic; Peirce's law; Stone's representation theorem for Boolean algebras. BOOLEAN ALGEBRA QUESTIONS 2009 Outside Delhi: 6. Basic postulates of Boolean Algebra. Operations are represented by ‘. is valid for Boolean algebra, but not for ordinary algebra. Simplification of Boolean expressions is also based. Complementarity Law. Boolean Algebra. Laws of Boolean algebra. Some of these laws are discussed below; Commutative Law of addition and multiplication…. 1 Boolean Algebra Definition: A Boolean Algebra is a math construct (B,+,. Thus we write ∼ A = A. ) and all 1's to 0's and vice-versa. true / false. X + 1 = 1 (null element) 2. It briefly considers why these laws are needed, that is to simplify complex Boolean expressions, and then demonstrates how the laws can be derived. Description: This test is a very interesting collection of questions in the form of MCQ where the test-takers get an opportunity to check their performance to appear in UGC, NET (Computer Science) and this test will help you to check your basic knowledge in boolean algebra and logic gates. The reader will, for the most part, be well served by assuming that Boole is doing ordinary polynomial algebra augmented by the assumption that any power $$x^n$$ of an elective symbol $$x$$ can be replaced by $$x$$. Identity: Dual: Operations with 0 and 1: 1. Let's construct truth tables for: xz + yz (x + y)z. AND operator → A * B = B * A. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Binary and Boolean Examples. Related Topics. - Boolean Laws and Theorems. After this, the Boolean algebra is well known as the perfect way for representing the digital. There are three laws of Boolean Algebra that are the same as ordinary algebra. If p is a statement then, p + (~p) = 1 p. Laws and Rules of Boolean Algebra Commutative Law A B = B A A ⋅ B = B ⋅ A Associative Law A B C = A B C A ⋅ B ⋅ C = A ⋅ B ⋅ C. Label all the laws you apply. A set of rules or Laws of Boolean Algebra expressions have been invented to help reduce the number of logic gates needed to perform a particular logic operation resulting in a list of functions or theorems known commonly as the Laws of Boolean Algebra. Section 4: Boolean Algebra 11 These rules are a direct translation into the notation of logic gates of the rules derived in the package Truth Tables and Boolean Algebra. Module 4: BOOLEAN ALGEBRA & LOGIC SIMPLIFICATION Laws and Rules of Boolean Algebra Construc6ng Truth table from Boolean Expression Standard Forms of Boolean Expression Determining standard Expression from truth table Logic Simplifica6on using: • Boolean algebra • Karnaugh Map. A Boolean Algebra is a 3-tuple {B , + , · }, where • B is a set of at least 2 elements • ( + ) and ( ·) are binary operations (i. - Boolean Laws and Theorems. Boolean Algebra Laws and Rules. In Boolean Algebra, A+A =A and A. There are a number of laws for Boolean algebra. Boolean Algebra Simplifier. We can add numbers in any order. Relaxing Rain and Thunder Sounds, Fall Asleep Faster, Beat Insomnia, Sleep Music, Relaxation Sounds - Duration: 3:00:01. In Boolean algebra, (xy)' is equal to. This site is to provide you with information regarding Boolean Algebra. The distinguishing. For the same video link given below must watch. These are useful in minimizing Boolean functions. , Kurukshetra 2. That means that we may follow the statement ' x + a = b ' with the statement ' x = b − a. We use variables to represent elements of our situation or procedure. (AND symbol) i. Boolean Algebra. Within the Lotame platform, the use of Boolean Logic allows for the creation of more complex audience definitions. Boolean algebra refers to symbolic manipulation of expressions made up of boolean variables and boolean operators. Laws and rules of boolean algebra Summary Associative Laws The associative laws are also applied to addition and multiplication. Operator Symbols and Examples # Operator Symbol; 1: Not ' 2: Nand @ 3: And * 4: Xor ^ 5: Nor % 6: Or + Examples: A A' A'' (A'')' A + 1 A + 0 A + B A + B'. These are obtained by changing every AND(. Boolean algebra or switching algebra is a system of mathematical logic to perform different mathematical operations in binary system. Yes, Boolean algebra related to sets, which relates to Probability. Oct 25, 2017 - POS to SOP conversion example | Boolean algebra Stay safe and healthy. Each answer may be used as many times as necessary. Note the Boolean theorem/law used at each simplification step. Use Boolean algebra. Boolean algebra is a type of algebra that is used in the design of (digital) logic circuitry, computer programs such as search engines and in general in analytic reasoning. We use Boolean algebra in this class to simplify Boolean expressions which represent circuits. 4 Circuit Simplification Boolean Algebra Procedure Using the theorems and laws of Boolean. Boolean algebra laws/results p_q = q _p p^q = q ^p (p_q)_r = p_(q _r) (p^q)^r = p^(q ^r) Created Date: 9/16/2013 2:43:52 PM. Access the answers to hundreds of Boolean algebra questions that are explained in a way that's easy for you to understand. Then the + becomes the new main operator. For the most part, these laws correspond directly to laws of Boolean Algebra for propositional logic as given in Figure 1. com Created Date: 20170318234542Z. The laws of Boolean algebra are similar in some ways to those of standard algebra, but in some cases Boolean laws are unique. You start off with the idea that some statement P is either true or false, it can’t be anything in between (this called the law of the excluded middle). Label all the laws you apply. In the works The Mathematical Analysis of Logic (1847) and Investigation of the Laws of Thought (1854) he established formal logic and Boolean algebra (the algebra of sets). I find the Karnaugh Map to be a valuable tool. xz + yz = (x + y)z. In particular the following laws are comm on. , the algebra of sets, studied largely by means of truth tables, has anything to do with computer whether basic laws of ordinary algebra (commutative, associative and distributive) hold in Boolean algebra, etc. Each answer may be used as many times as necessary. , on '0' and '1'. As mentioned earlier, Boolean algebra is invented in the year of 1854, by an English mathematician George Boole. Boolean Algebra Objectives •Understand basic Boolean Algebra •Relate Boolean Algebra to Logic Networks •Prove Laws using Truth Tables •Understand and Use First 11 Theorems •Apply Boolean Algebra to: – Simplifying Expressions – Multiplying Out Expressions – Factoring Expressions – Typeset by FoilTEX – 1. are binary operations in B, ' is a unary operation in B, 0 and 1 are special elements of B, such that: a) + and. It is also used in Physics for the simplification of Boolean expressions and digital circuits. Posted 3 years ago Boolean Algebra simplification Show transcribed image text Boolean Algebra simplification Simplify t. Boolean algebra is a different kind of algebra or rather can be said a new kind of algebra which was invented by world famous mathematician George Boole in the year of 1854. Relaxing Rain and Thunder Sounds, Fall Asleep Faster, Beat Insomnia, Sleep Music, Relaxation Sounds - Duration: 3:00:01. These are obtained by changing every AND(. To represent a function in truth table, there should be the list of the combination of the binary variables. Now I am a first term student and I am st udying the fundamentals of calculus. Other applications include digital circuit design, law, reasoning about any subject, and any kind of specifications, as well as. It is known as Boolean algebra, after the mathematician George Boole (1815–1864). asked Jul 20, 2019 in Computer by Helisha ( 68. We can also create Minterm from the given values of the variables. a) Define problems using Boolean logic. Using these boolean expressions, we can describe complex digital circuits with mathematical-like equations. Boolean algebra is the branch of algebra wherein the values of the variables are either true or false, generally denoted by 1 and 0 respectively. Boole’s three laws for his algebra of logic are woefully inadequate for what follows in MAL. The only difference between them is that the. Axioms and Laws of Boolean Algebra. The Commutative Law addition A + B = B + A (In terms of the result, the order in which variables are ORed makes no difference. This is known as duality. • boolean algebra: symbols, rules • express the logical functions and, or, not, xor, nand and nor mathematically • basic laws of boolean algebra and how to apply them. 1 Boolean Algebra. We have seen that they can all be checked by investigating the corresponding truth tables. Postulates and Basic Laws of Boolean Algebra. Then explain how to your work to obtain to obtain a dervation for the associative law for. 4 Circuit Simplification Boolean Algebra Procedure Using the theorems and laws of Boolean. Algebra: Powered by Create your own unique website with customizable templates. Boolean expressions use the operators AND, OR, XOR, and NOT to compare values and return a true or false result. : a system of algebra in which there are only two possible values for a variable (often expressed as true and false or as 1 and 0) and in which the basic operations are the logical operations AND and OR. Draw the logic diagram of the simplified function, Fs 5. Some problems are also discussed. There are a number of laws for Boolean algebra. What follows are what we are permitted to write. The laws of Boolean algebra are similar in some ways to those of standard algebra, but in some cases Boolean laws are unique. There are three laws of Boolean Algebra that are the same as ordinary algebra. a OR b = b OR a Or with multiple terms:. Primes are for real. true / false. George Boole married Mary Everest (daughter of George Everest, for whom the mountain is named) in 1855. ws A set of rules or Laws of Boolean Algebra expressions have been invented to help reduce the number of logic gates needed to perform a particular logic operation resulting in a list of functions or theorems known commonly as the Laws of Boolean Algebra. Note: Every law in Boolean algebra has two forms that are obtained by exchanging all the ANDs to ORs and 1s to 0s and vice versa. 7k points) basics of boolean algebra. Laws of nature and reasonableness. 3 Laws of Matrix Algebra Subsection 5. Postulates and Theorems of Boolean Algebra Assume A, B, and C are logical states that can have the values 0 (false) and 1 (true). X + 1 = 1: Annulment Law: 2a. The neural mechanism of logical thinking. Mainly, the standard rules of Boolean algebra are given in operator ‘+’ and ‘x’, based on the AND and OR logic gates equations. You have a table in CSCI 2150 -- Boolean Algebra Basics which summarizes 12 Boolean algebra rules (BARs) (or theorems, or postulates, or whatever…) plus the De Morgan’s theorem(s). The commutativity of Boolean subalgebras arises naturally from the algebraic and topo-logical structures of Boolean algebra. Axioms - Laws of Boolean Algebra Boolean algebra is the algebra of propositions. We can say, then, that algebra is a system of formal—grammatical—rules. Relaxing Rain and Thunder Sounds, Fall Asleep Faster, Beat Insomnia, Sleep Music, Relaxation Sounds - Duration: 3:00:01. Postulates and Theorems of Boolean Algebra Assume A, B, and C are logical states that can have the values 0 (false) and 1 (true). Facebook LinkedIn Twitter Pinterest Report Mistakes in Notes Issue: * Mistakes in notes Wrong MCQ option The page is not clearly visible Answer quality needs to be improved Your Name: * Details: * Submit Report. If p is a statement then, p + (~p) = 1 p. FINITE BOOLEAN ALGEBRA 1. Learn more Simplifying Boolean Expressions with DeMorgan’s law. 1 Introduction: George Boole, a nineteenth-century English Mathematician, developed a system of logical algebra by which reasoning can be expressed mathematically. At every turn, we have been writing out expressions using boolean variables and the two symbols that express And & OR. - Arthur W. ” In the same way that normal algebra has rules that allow you to simplify algebraic expressions, Boolean algebra has theorems and laws that allow you to simplify expressions used to create logic circuits. ) multiplication AB = BA (In terms of the result, the order in which variables are ANDed makes no difference. Also, there exist equations, expressions, and functions in. Fur- thermore, for all cardinals ,the(;1)-distributive law holds in B i Player 1 does not have a winning strategy in G. Boolean Algebra is a form of mathematical algebra that is used in digital logic in digital electronics. 1 HR cafe sounds, coffee shop background audio, background white noise for studying or at the office - Duration: 1:00:26. In ordinary algebra, two expressions may be equivalent to each other, e. , ', 0,1) where B is a non-empty set, + and. Boolean Algebra Computer Organization 5 [email protected] ©2005-2020 WD McQuain DeMorgan's Laws & More DeMorgan's Laws are useful theorems that can be derived from the fundamental properties of a Boolean algebra. Unit 3 – Boolean Algebra (continued) Fundamentals of Logic Design EE2369 Prof. Distributive Laws of Boolean Algebra. Title: Laws of Boolean Algebra Cheat Sheet by johnshamoon - Cheatography. A Boolean search, in the context of a search engine, is a type of search where you can use special words or symbols to limit, widen, or define your search. Boolean Algebra. Laws of Boolean Algebra Table 2 shows the basic Boolean laws. In this section, let us discuss about the Boolean postulates and basic laws that are used in Boolean algebra. distributive law-are the same as in ordinary algebra. Find 1 Answer & Solution for the question Prove the idempotence law of boolean algebra with the help of truth table. : a system of algebra in which there are only two possible values for a variable (often expressed as true and false or as 1 and 0) and in which the basic operations are the logical operations AND and OR. Each of the Boolean Laws above are given with just a single or two. This Chapter provides only a basic introduction to boolean algebra. You start off with the idea that some statement P is either true or false, it can’t be anything in between (this called the law of the excluded middle). Note that every law has two expressions, (a) and (b). Operator Symbols and Examples # Operator Symbol; 1: Not ' 2: Nand @ 3: And * 4: Xor ^ 5: Nor % 6: Or + Examples: A A' A'' (A'')' A + 1 A + 0 A + B A + B'. Then the calculus of indications is simply Boolean arithmetic reduced to the two equations 11=1 and (1)=0. (B + B) + B. Boolean algebra is important in both inductive reasoning and deductive reasoning, as well as science in general. Relaxing Ambiance TV. Negationis represented by placing a bar (or overline) across an expression. Related Topics. Digital Electronics Activity 2. Please note that this definition doesn't directly coincide with the definition of a boolean algebra. Commutative Law x+y = y+x x. The familiar identity, commutative, distributive, and associative axioms from algebra define the axioms of Boolean algebra, along with the two complementary axioms. Facebook LinkedIn Twitter Pinterest Report Mistakes in Notes Issue: * Mistakes in notes Wrong MCQ option The page is not clearly visible Answer quality needs to be improved Your Name: * Details: * Submit Report. बूलियन अलजेब्रा का प्रयोग डिजिटल सर्किटों को analyze तथा simplify करने के लिए किया जाता है. • Values and variables can indicate some of the following binary pairs of values:. , A contains the elements 0 and 1 and is closed under the operations *, + and '. Introduction To Boolean Algebra. For the most part, these laws correspond directly to laws of Boolean Algebra for propositional logic as given in Figure 1. X 0 = 0 Idempotent laws 3. cars, consumer electronics,. In 1854, Boole published a classic book, “An. or "Closed" circuit rules. ) Associative: The associative property says that given three Boolean. not be the zero or one elements of the Boolean algebra. functions B B ) satisfying the following axioms: B A1. Például a konjunkció és a diszjunkció számára nem egy duális operátorpár volt. All three projects are part of a larger collection published in Convergence, and an entire introductory discrete mathematics course can be taught from a. We find that f(x) and F(x) are equally valid functions and duality is a special property of Boolean (binary) algebra. Consider the binary numbers 0 and 1, Boolean variable (x) and its complement (x'). Today, Boolean algebra is of significance to the theory of probability, geometry of sets, and information. All three projects are part of a larger collection published in Convergence, and an entire introductory discrete mathematics course can be taught from a. Here you can access and discuss Multiple choice questions and answers for various compitative exams and interviews. Obtain the truth table for F. Applying the Boolean algebra basic concept, such a kind of logic equation could be simplified in a more simple and efficient form. Choose from 436 different sets of boolean algebra flashcards on Quizlet. Distributive Laws of Boolean Algebra. X • 1 = X: 2b. Basic Laws of Boolean Algebra: Logical operations can be expressed and minimized mathematically using the rules, laws, and theorems of Boolean algebra. Related notions. Students should try to show the validity of basic laws (1) through (5) using truth tables. A law of Boolean algebra is an identity such as x ∨ (y ∨ z) = (x ∨ y) ∨ z between two Boolean terms, where a Boolean term is defined as an expression built up from variables and the constants 0 and 1 using the operations ∧, ∨, and ¬. For example, when we have 2×(3+4), we treat "2×" as a chunk, and put it on the 3 and the 4: (2×3)+(2×4). Laws (Theorems) of Boolean algebra Laws of Complementation The term complement means, to invert or to change 1's to 0's and 0's to 1's, for which purpose inverters or NOT gates are used. Assume that the indicated operations are defined; that is, that the orders of the matrices $$A\text{,}$$ $$B$$ and $$C$$ are such that the operations make sense. 4 Various Commutativity Associativity Distributivity AND. MATH 125 Worksheet 10 Boolean Algebra Author: gblake Created Date: 11/3/2014 8:06:13 PM. Boolean Algebra Theorems and Laws of Boolean Algebra. Explicitly, a Boolean algebra is the partial order on subsets defined by inclusion (Skiena 1990, p. true / false. The laws of Boolean algebra are similar in some ways to those of standard algebra, but in some cases Boolean laws are unique. Examples of lattices include Boolean algebras , the set of sets with union and intersection operators, Heyting algebras , and ordered sets with min and max operations. For the same video link given below must watch. It reduces the original expression to an equivalent expression that has fewer terms which means that. These laws govern the relationships that exist between two or more inputs to logic gates. In the 20th century boolean algebra came to be much used for logic gates. Boolean Algebra. The distributive law: may seem at variance with the laws for elementary algebra, which would state: However, this expression is equivalent within the Boolean axioms above. Some of these laws may appear a little bit confusing at first. a + b = b + a {\displaystyle a+b=b+a} The arrangement of addends does not affect the sum. I always liked the $\min, \max$ definitions of $\cdot$ and $+$, since some courses in boolean algebra just give those laws and ask you to accept them. Axioms of Boolean Algebra (4 of 4) •Axiom 6 –Distributive laws •For every a, b, and c in B, •a + (b · c) = (a + b) · (a + c) •a · (b + c) = (a · b) + (a · c) •Axiom 7 –Complement •For each a in B, there exists an element a' in B (the complement of a) s. This lecture will discuss boolean algebra in detail. The system became more popular when Claude Shannon used electric circuits and relays as an analogy for the Boolean algebra. Commutative Laws. In this section, let us discuss about the Boolean postulates and basic laws that are used in Boolean algebra. Laws and Theorems of Boolean Algebra. Two laws in Boolean algebra which state that AND and OR, or union and intersections, are duel. Boolean Logic can be considered as a special case of sets Defining Boolean Logic Spring 2020 Sacramento State - Cook - CSc 28 33 We can show that the behavior of Boolean Logic can be created in sets It's not surprised that many of the laws for sets work for Boolean Logic Defining Boolean Logic Spring 2020 Sacramento State - Cook - CSc 28 34. 1 BOOLEAN ALGEBRA 1. George Boole married Mary Everest (daughter of George Everest, for whom the mountain is named) in 1855. Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit. (A + B)(A + C) = A + BC. cars, consumer electronics,. Any binary operation which satisfies the following expression is referred to as commutative operation. Boolean algebra is the branch of mathematics that includes methods for manipulating logical variables and logical expressions. Instead of the equals sign, Boolean algebra uses logical equivalence, ≡, which has essentially the same meaning. C How many gates do you save = A. These are only two elements 1 and 0 by which all the mathematical operations are to be performed. In this section, let us discuss about the Boolean postulates and basic laws that are used in Boolean algebra. : a system of algebra in which there are only two possible values for a variable (often expressed as true and false or as 1 and 0) and in which the basic operations are the logical operations AND and OR. State Boolean Algebra laws used in k-map simplification. In the context of Patent law, the patent search systems use "and", "or", and “not" as Boolean operators, in combination with parentheses. [Truth Table Examples] [Boolean Expression Simplification] [Logic Gate Examples] Here is the list of rules used for the boolean expression simplifications. Label all the laws you apply. As we know that this kind of algebra is a mathematical system consisting of a set of two or more district elements. Consider the binary numbers 0 and 1, Boolean variable (x) and its complement (x'). Boolean values • Named after George Boole (1815-1864), who invented mathematical logic and defined Boolean algebra. How's Your Readability? Cheatography is sponsored by Readable. Boolean Algebra John Winans January 23, 2020 1 Basic Operations When describing boolean functions, zero is considered false and anything that is not false is true. 2: Some Laws of Boolean Algebra for sets. Be sure to put your answer in Sum-Of-Products (SOP) form. Note the Boolean theorem/law used at each simplification step. Here you'll find links to all algebra formulas on this page. Boolean algebra.
9cjqyjqpqxv jsy28gojxzn6n ocvw39vd2i c35dedd4u9ao twwzc946nwdm28 56byp66hgkf6ar 0k9cmw57l1iqv9 7vum0gplzq3l4 dyckwdgg4edfm99 cqxt2jugwdiy 1klub7aavtkdjan v91vtutqna9v9li fyzsxose08km rizgqtkpdh7 0xs3mcejlvvluf k90zw0iencs0 fvo3epyy5hfix8k sxycg3x9h2 snw6ezm3f7b5sv 6czjwhuciqx9yth 62hxpqrf2y08798 3ns8bmy25swm0p twtb17am6y omekvcy0rf9ken8 y4fkzlbf3ct eo979xea7pwzwl 12u7wtjb1w8 fkl8f9w62mrl zmi19tpphctu6ij i4vo9r4onp15aud | 2020-07-09T23:14:59 | {
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https://or.stackexchange.com/questions/6674/how-to-linearize-specific-range-constraints?noredirect=1 | # How to linearize specific range constraints?
I would like to know about the linearization of the $$(If, Then)$$ constraints as follows:
$$\begin{array}{l} \text { If: } \\ 15 \leqslant x \leqslant 25 \\ \text { then: } \quad y=\color{blue}{a} x+\color{green}{b} \\ \text { elself: } \\ 25 \leqslant x \leqslant 35 \\ \text { then: } \quad y=\color{blue}{a}^{\color{blue}{\prime}} x+\color{green}{b}^{\color{green}{\prime}} \\ \text { elself: } \\ 35 \leqslant x \leqslant 45 \\ \text { then: } \quad y=\color{blue}{a}^{\color{blue}{\prime \prime}} x+\color{green}{b}^{\color{green}{\prime \prime}} \\ \text { elself: } \\ 45 \leqslant x \leqslant 55 \\ \text { then: } \quad y=\color{blue}{a}^{\color{blue}{\prime \prime \prime}} x+\color{green}{b}^{\color{green}{\prime \prime \prime}} \\ \end{array}$$
Where $$x$$ and $$y$$ are continuous variables and $$\color{blue}{a}$$'s and $$\color{green}{b}$$'s are constant. We have tried to introduce the binary auxiliary variables for each set of constraints and finally linking these constraints with whose specific binary variable. This approach seems to work fine, but I am facing that we will have to use the product of the binary and continuous variables. I knew that we can use specific linearization to do this. I was wondering if, is there another way to formulate this problem efficiently?
• Is $x$ bounded by $[15,55]$ or could it be outside of those four intervals? Aug 3 at 12:57
• @RobPratt, Thanks. It is bounded by $[15,55]$. Aug 3 at 12:58
• Is the piecewise linear function convex and are you minimizing / is the piecewise linear function concave and are you maximizing ? Or neither ? Aug 3 at 13:05
• @Kuifje, the objective function is max. I am afraid, I do not understand what you mean by "Is the piecewise linear function ..."? Aug 3 at 13:13
• Are the values of $a,a',...,b'''$ such that the union of $y=ax+b, ..., y=a'''x+b'''$ defines a concave function ? Aug 3 at 13:14
Let $$y_i$$ be a binary variable that equals $$1$$ if and only if $$x$$ is in the interval $$i \in \{[15,25],[25,35],[35,45],[45,55]\}$$. You can express $$x$$ as a convex combination of the extreme points of these intervals by introducing variables $$\lambda_0, \lambda_1,\lambda_2,\lambda_3, \lambda_4 \in \mathbb{R}^+$$.
If $$f$$ denotes your piecewise linear function, the objective function is \begin{align*} &f(15)\lambda_0+f(25)\lambda_1+f(35)\lambda_2+f(45)\lambda_3 +f(55)\lambda_4 \\ =&(15a+b)\lambda_0+ (25a'+b')\lambda_1 +(35a''+b'')\lambda_2+ (45a'''+b''')\lambda_3 +(55a'''+b''')\lambda_4 \end{align*}
and the constraints are \begin{align*} 1 &= y_1+y_2+y_3+y_4 \tag{1}\\ x &= 15\lambda_0+25\lambda_1+35\lambda_2+45\lambda_3+55\lambda_4\tag{2}\\ 1 &=\lambda_0+\lambda_1+\lambda_2+\lambda_3 + \lambda_4\tag{3}\\ y_1&\le \lambda_0+\lambda_1\tag{4}\\ y_2&\le \lambda_1+\lambda_2\tag{5}\\ y_3&\le \lambda_2+\lambda_3\tag{6}\\ y_4&\le \lambda_3+\lambda_4\tag{7}\\ y&\in \{0,1\}\\ \lambda&\ge 0 \end{align*}
Another option, using the same binary variables $$y_i$$, is to rewrite $$x$$ in terms of $$y_i$$, and in terms of the ranges of each interval $$x_i$$: \begin{align*} x &= (15y_1+x_1)+(25y_2+x_2)+(35y_3+x_3)+(45y_4+x_4) \\ x_1 &\le 10 y_1 \\ x_2 &\le 10 y_2 \\ x_3 &\le 10 y_3 \\ x_4 &\le 10 y_4 \\ \end{align*} Make sure you are on exactly one interval: $$y_1+y_2+y_3+y_4 = 1$$ And rewrite your piecewise linear function as \begin{align*} f(x) &= \Big(a(x_1+15y_1) +by_1\Big) \\ &+\Big(a'(x_2+25y_2)+b'y_2\Big) \\ &+\Big(a''(x_3+35y_3)+b''y_3\Big) \\ &+\Big(a'''(x_4+45y_4)+b'''y_4\Big) \end{align*}
And last but not least, the usual (?) way with big Ms, maximize $$z$$ subject to:
\begin{align*} 1&=y_1+y_2+y_3+y_4 \\ x &\le 25y_1 + 35y_2 + 45y_3 +55 y_4 \\ x &\ge 15y_1 + 25y_2 + 35y_3 +45 y_4 \\ z &\le (ax+b) + M_1(1-y_1) \\ z &\le (a'x+b') + M_2(1-y_2) \\ z &\le (a''x+b'') +M_3(1-y_3) \\ z &\le (a'''x+b''') +M_4(1-y_4) \\ \end{align*}
• Many thanks for your detailed explanation. Let's say, what we have tried to do is very close to your third formulation (as you mentioned the usual :)) But, for the last four constraints we define $[z1,z2,z3,z4]$ and trying to restrict the constraints by multiplying $y$'s variables directly to each constraint instead of using $big M$ and this is why we have faced non-linear term in the model. Finally, maximizing the objective function by summing them up. ($\sum_i z_i$). I tried your code and the results are the same. Thanks once again. Aug 3 at 20:04
• I have another question. Would you say please, how we can express $x$ as a convex combination of the extreme points of these intervals? Your approach is very interesting in this specific problem. Aug 3 at 20:15
• Equations $(1)-(3)$ are the convex combination part. I also like this method because it is very generic. But obviously it is not mine I read it somewhere a few years back ! Aug 3 at 20:20
• Note also that if the values of the constants $a,...,b'''$ are such that $f(x)$ is concave, there is a more efficient way. Aug 3 at 20:21
If the $$y$$ function is continuous (meaning $$a\cdot 25 + b = a^\prime \cdot 25 + b^\prime$$ and similarly at other breakpoints), you can use an SOS2 constraint to model this. Let $$p_0, \dots, p_n$$ be the breakpoints ($$n=4,\,p_0 = 15,\,p_4 =55$$ in your example) and $$\gamma_0 \dots, \gamma_n$$ be the values of the $$y$$ function at $$p_0, \dots, p_n$$. Add continuous variables $$w_0, \dots, w_n \in [0,1]$$, along with the constraints \begin{align*} w_{0}+\dots+w_{n} & =1\\ p_{0}w_{0}+\dots+p_{n}w_{n} & =x\\ \gamma_{0}w_{0}+\dots+\gamma_{n}w_{n} & =y. \end{align*} Now add a constraint telling the solver that $$w_0, \dots, w_n$$ form a type 2 special ordered set (SOS2). That tells the solver that at most two of the $$w_i$$ can be nonzero, and those two must be consecutive. Combined with the first constraint, that means $$x$$ will be a convex combination of two consecutive breakpoints and $$y$$ will be the same convex combination of the function values at those two breakpoints.
This of course requires a solver that understands SOS2 constraints. Internally, the solver is likely to add a bunch of binary variables and basically do the linearization for you.
• Many thanks for your suggestion. Aug 3 at 20:18 | 2021-12-09T03:22:23 | {
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http://www.b254.com/factoring/2hsquare2plus4hminus15.html | ### Factoring 2h^2+4h-15 Solution
The variable we want to find is h
We will solve for h using quadratic formula -b +/- sqrt(b^2-4ac)/(2a), graphical method and completion of squares.
${x}_{}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Where a= 2, b=4, and c=-15
Applying values to the variables of quadratic equation -b, a and c we have
${h}_{}=\frac{-4±\sqrt{{4}^{2}-4x\mathrm{2x}\mathrm{-15}}}{2x2}$
This gives
${h}_{}=\frac{-4±\sqrt{{4}^{2}-\mathrm{-120}}}{4}$
${h}_{}=\frac{-4±\sqrt{136}}{4}$
${h}_{}=\frac{-4±11.6619037897}{4}$
${\mathrm{h1}}_{}=\frac{-4+11.6619037897}{4}$
${\mathrm{h2}}_{}=\frac{-4-11.6619037897}{4}$
${h}_{1}=\frac{7.66190378969}{4}$
${h}_{1}=\frac{-15.6619037897}{4}$
The h values are
h1 = 1.91547594742 and h2 = -3.91547594742
### Factoring Quadratic equation 2h^2+4h-15 using Completion of Squares
2h^2+4h-15 =0
Step1: Divide all terms by the coefficient of h2 which is 2.
${h}^{2}+\frac{4}{2}x-\frac{15}{2}=0$
Step 2: Keep all terms containing x on one side. Move the constant to the right.
${h}^{2}+\frac{4}{2}h=\frac{15}{2}$
Step 3: Take half of the x-term coefficient and square it. Add this value to both sides.
${h}^{2}+\frac{4}{2}h+{\left(\frac{4}{4}\right)}^{2}=\frac{15}{2}+{\left(\frac{4}{4}\right)}^{2}$
Step 4: Simplify right hand sides of expression.
${h}^{2}+\frac{4}{2}h+{\left(\frac{4}{4}\right)}^{2}=\frac{136}{16}$
Step 2: Write the perfect square on the left.
${\left(h+\frac{4}{4}\right)}^{2}=\frac{136}{16}$
Step 2: Take the square root on both sides of the equation.
$h+\frac{4}{4}=±\sqrt{\frac{136}{16}}$
Step 2: solve for root h1.
${h}_{1}=-\frac{4}{4}+\frac{11.6619037897}{4}=\frac{7.66190378969}{4}$
${h}_{1}=1.91547594742$
Step 2: solve for root h2.
${h}_{2}=-\frac{4}{4}-\frac{11.6619037897}{4}=\frac{-15.6619037897}{4}$
${h}_{2}=-3.91547594742$
### Solving equation 2h^2+4h-15 using Quadratic graph
h2 + h + = 0
Solutions
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Other Variants of 2h^2+4h-15 are below | 2017-06-28T20:48:41 | {
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https://math.stackexchange.com/questions/1262026/show-that-there-are-exactly-16-pairs-of-integers-x-y-such-that-11x8y17-xy | # Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$.
Original problem
Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$.
My work
From case by case analysis I come to know that the equation will hold if and only if $x$ is odd and $y$ is even.
Also I found that $(8-x)|(11x+17)$ and $(11-y)|(8y+17)$
This is all what I have found.
Please see that my work are right or not.
This is a new kind of question which I have encountered so please help me is solving this problem.
• All your conclusions are correct and the best way I know is described in my answer ($x$ is odd and $y$ is even is a correct conclusion and $8-x$ divides $11x+17$ and $11-y$ divides$8y+17$ are valid consequences) – Elaqqad May 2 '15 at 14:31
• @Elaqqad Sir this was a 10 marks problem of last year paper of an institute entrance exam. I guess that my conclusions only award me 1 or 2 marks. (math.stackexchange.com/questions/1261971/…) Please look at this problem. This one of those 10 marks problem. – Singh May 2 '15 at 14:35
Given $a,b,c$ three integers, the idea when you have an equation of the form: $$xy=ax+by+c$$ to solve if for unknown integers $x,y$ is to do the following factorization: $$(x-b)(y-a)=c+ab$$
and hence the number of solutions is the number of divisors of $c+ab$.
• After a search I found (mathwarehouse.com/answered-questions/factors/…) please look this. It is showing that the divisors of 105 is 8. Please explain. – Singh May 2 '15 at 14:38
• what if you count the negative ones also you will have $16$ divsors and so there are $16$ solutions in total – Elaqqad May 2 '15 at 14:40
• I missed that. Thank you Sir. One more help, out of 10 how much you will give for my conclusions. – Singh May 2 '15 at 14:49
This type of diophantine equation is solvable by a generalization of completing the square. Namely, completing a square generalizes to completing a product as follows:
$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$
Here we deduce $\ xy - 11x - 8y = 17 \iff (x-8)(y-11) =\, \ldots$
So by uniqueness of prime factorizations the problem reduces to counting the divisors of $\,\ldots$
16 ordered pairs indeed.
\begin{align} 11x+8y+17&=xy \\ xy-11x-8y&=17 \\ xy-11x-8y+\color{red}{88}&=17+\color{red}{88} \\ (x-8)(y-11)&=105=3 \cdot 5 \cdot 7=d_1 \cdot d_2 \end{align}
Setting
$$\begin{cases} x-8=d_1\\ y-11=d_2 \end{cases}$$
Implies that $$\begin{cases} x=8+d_1\\ y=11+d_2 \end{cases} \text{such that} \ \ (d_1,d_2) \in \lbrace \text{divisor-pairs of 105} \rbrace$$ For any divisor-pair $(d_1,d_2)$ that produces one solution, $(d_2,d_1)$ will produce another.
Additionally, for any $(d_1,d_2)$ that produces a solution, $(-d_1,-d_2)$ will produce another as well.
So for every divisor-pair, $(d_1,d_2)$, there will be four total solutions:
\begin{align} (d_1,d_2) &\to (d_1,d_2) \\ &\to (d_2,d_1) \\ &\to (-d_1,-d_2) \\ &\to (-d_2,-d_1) \end{align}
This is due to a lack of symmetry in the expressions for $x$ and $y$ (for comparison, this situation is different).
The problem therefore reduces to finding exactly $4$ divisor-pairs of $105$.
Since $$105=3^{\color{red}{1}} \cdot 5^{\color{red}{1}} \cdot 7^{\color{red}{1}} \implies 105 \ \text{has} \ (1+{\color{red}{1}})(1+{\color{red}{1}})(1+{\color{red}{1}})=8 \ \ \text{divisors},$$
there are indeed four divisor-pairs: $$(d_1,d_2) \in \bigg{\{} (1,105),(3,35),(5,21),(7,15) \bigg{\}}$$
As an example, setting $d_1=1$ and $d_2=105$
\begin{align} (1,105) &\to (1,105) && \implies (x,y)=(9,116)\\ &\to (105,1) && \implies (x,y)=(113,12)\\ &\to (-1,-105) && \implies (x,y)=(7,-94)\\ &\to (-105,-1) && \implies (x,y)=(-97,10) \end{align}
That's four solutions from one divisor-pair alone. | 2020-08-04T09:09:12 | {
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http://mathhelpforum.com/number-theory/148115-solved-solving-diophantine-equations-print.html | # [SOLVED] Solving Diophantine Equations:
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jun 7th 2010, 06:23 AM
Samson
[SOLVED] Solving Diophantine Equations:
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:
(x^2) + 5*(y^2) = (z^2)
This comes from an additional review portion of my text. It states the following tip:
Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.
Does anybody have any ideas? Any help is appreciated!
• Jun 7th 2010, 06:46 AM
undefined
Quote:
Originally Posted by Samson
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:
(x^2) + 5*(y^2) = (z^2)
This comes from an additional review portion of my text. It states the following tip:
Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.
Does anybody have any ideas? Any help is appreciated!
I think you will end up with something similar to generating Pythagorean triples. I'll have to think about it/review some sources. Or maybe someone will post a nice solution in the meantime.
• Jun 7th 2010, 06:53 AM
Samson
Quote:
Originally Posted by undefined
I think you will end up with something similar to generating Pythagorean triples. I'll have to think about it/review some sources. Or maybe someone will post a nice solution in the meantime.
For what it's worth, I know you're correct considering it was in a section of my text that explicitly mentiones Pythagorean triples. I know that doesn't help provide proof, but at least is a good sign that we are in the right ballpark.
• Jun 7th 2010, 07:12 AM
chiph588@
Quote:
Originally Posted by Samson
Hello All, I am trying to parameterize solutions in integers to the following Diophantine Equation:
(x^2) + 5*(y^2) = (z^2)
This comes from an additional review portion of my text. It states the following tip:
Consider only solutions that satisfy GCD[x,y,z]=1, and that x,y,z > 0 and that x is odd. If two cases appear, combine them into one by using absolute value.
Does anybody have any ideas? Any help is appreciated!
I'm stumped for the moment being, but I'd suggest trying to manipulate the equation into the form $a^2+(5y)^2=b^2$ where $a,b$ are in terms of $x,y,z$. There's probably an easier way though.
• Jun 7th 2010, 07:15 AM
roninpro
Taking the idea from the Wikipedia article, I used the following:
$a=m^2-5n^2$, $b=2mn$, $c=m^2+5n^2$
You can put this into the equation to see that it does indeed work. Then I suppose that you can throw in an extra parameter $k$ to get all solutions (with some possible overlap).
• Jun 7th 2010, 07:20 AM
Samson
Quote:
Originally Posted by roninpro
Taking the idea from the Wikipedia article, I used the following:
$a=m^2-5n^2$, $b=2mn$, $c=m^2+5n^2$
You can put this into the equation to see that it does indeed work. Then I suppose that you can throw in an extra parameter $k$ to get all solutions (with some possible overlap).
Okay, I'm just a little confused. I gave my equations with x,y,z and you have a,b,c,m,n, and possibly k... could you relate this back into the terms I had given? What are a,b,c,m,n, and k compared to x,y,z ? Thank you!
• Jun 7th 2010, 07:24 AM
undefined
Quote:
Originally Posted by Samson
Okay, I'm just a little confused. I gave my equations with x,y,z and you have a,b,c,m,n, and possibly k... could you relate this back into the terms I had given? What are a,b,c,m,n, and k compared to x,y,z ? Thank you!
Pythagorean triples are typically given as a^2 + b^2 = c^2, as on the Wikipedia page.
So your (x,y,z) became (a,b,c). m and n are positive integers that we iterate over. k is a multiplier. It is explained in more detail in the Wikipedia article. However, since gcd(x,y,z) = 1, you will not need the multiplier.
• Jun 7th 2010, 07:27 AM
roninpro
Oh, sorry, I meant to put $x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$.
You allow $m, n$ to range freely throughout the integers. Doing this gives you infinitely many solutions. Now, we have a little bit of a problem: we might not catch all of the solutions. To fix this, we throw in a parameter $k$, which also ranges through the integers. This gives
$x=k(m^2-5n^2)$, $y=k(2mn)$, $z=k(m^2+5n^2)$
This should give you every possible solution.
• Jun 7th 2010, 07:27 AM
chiph588@
Quote:
Originally Posted by roninpro
Taking the idea from the Wikipedia article, I used the following:
$a=m^2-5n^2$, $b=2mn$, $c=m^2+5n^2$
You can put this into the equation to see that it does indeed work. Then I suppose that you can throw in an extra parameter $k$ to get all solutions (with some possible overlap).
To prove these are the only solutions, perhaps you could perform a manipulation like I had suggested earlier? From that we know all solutions to a^2+b^2=c^2.
• Jun 7th 2010, 07:33 AM
Samson
Quote:
Originally Posted by roninpro
Oh, sorry, I meant to put $x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$.
You allow $m, n$ to range freely throughout the integers. Doing this gives you infinitely many solutions. Now, we have a little bit of a problem: we might not catch all of the solutions. To fix this, we throw in a parameter $k$, which also ranges through the integers. This gives
$x=k(m^2-5n^2)$, $y=k(2mn)$, $z=k(m^2+5n^2)$
This should give you every possible solution.
Okay, I'm confused by this. Chip says that we don't need k because the GCD[x,y,z]=1. However we need to find all of the solutions that satisfy the conditions I stated originally. How do we do this or have we already? If so, which ones are they?
• Jun 7th 2010, 07:38 AM
roninpro
Quote:
Originally Posted by chiph588@
To prove these are the only solutions, perhaps you could perform a manipulation like I had suggested earlier? From that we know all solutions to a^2+b^2=c^2.
I'm not entirely sure that will work; one solution that I found was $x=2, y=1, z=3$, which gives $4+5=9$ - we have some trouble with the 5.
I think that some insight for the proof comes from Dr. Math: Math Forum: Ask Dr. Math FAQ: Pythagorean Triples
• Jun 7th 2010, 07:45 AM
Samson
Quote:
Originally Posted by roninpro
I'm not entirely sure that will work; one solution that I found was $x=2, y=1, z=3$, which gives $4+5=9$ - we have some trouble with the 5.
I think that some insight for the proof comes from Dr. Math: Math Forum: Ask Dr. Math FAQ: Pythagorean Triples
So if $x=2, y=1, z=3$ and they are the only solutions, why would the formulas below be used?
$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$
• Jun 7th 2010, 07:50 AM
undefined
Quote:
Originally Posted by Samson
So if $x=2, y=1, z=3$ and they are the only solutions, why would the formulas below be used?
$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$
(x,y,z) = (2,1,3) is one of infinitely many solutions.
• Jun 7th 2010, 07:53 AM
roninpro
Quote:
Originally Posted by Samson
So if $x=2, y=1, z=3$ and they are the only solutions, why would the formulas below be used?
$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$
By "only solutions", Chip meant that the solutions to the Diophantine equation have the form given in our equation. And actually, we may have a problem. I don't think that we can represent $2=m^2-5n^2$ for some $m,n$. So, I suppose that some modification to the formula is required.
Maybe you can have a look at that Dr. Math article and try a similar argument.
• Jun 7th 2010, 07:54 AM
Samson
Quote:
Originally Posted by undefined
(x,y,z) = (2,1,3) is one of infinitely many solutions.
So would it be correct to say that
(x,y,z)=(2,1,3) is one solution, but
$x=m^2-5n^2$, $y=2mn$, $z=m^2+5n^2$
is a parameterization of all solutions that meet the aforementioned conditions?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last | 2016-09-28T23:57:05 | {
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https://math.stackexchange.com/questions/4184249/how-to-prove-sqrt-75025-sqrt-121393-sqrt-196418-sqrt317811-a/4184287 | # How to prove $\sqrt {75025} + \sqrt {121393} + \sqrt {196418} + \sqrt{317811} \approx \sqrt {514229} + \sqrt {832040}$?
Let $$a = \sqrt {75025} + \sqrt {121393} + \sqrt {196418} + \sqrt{317811}$$ and $$b = \sqrt {514229} + \sqrt {832040}$$. By using SageMath we can see that $$a - b \approx 2.95301560981898 \cdot 10^{-9}$$ That means almost nine digits of accuracy!
To investigate any particular reason why these surprising digits of accuracy come I considered the function $$f(x)= \sqrt{x +317811} + \sqrt{x + 196418} + \sqrt{x + 121393} + \sqrt{x + 75025} -\sqrt{x + 832040} - \sqrt{x + 514229}$$
The Taylor series of $$f$$ around $$x =0$$ with approximate coefficients looks like $$f(x) = f(0) + 0.00403020948350145x -2.13362649294736 \cdot 10^{-8}\frac12 x^2 + O(x^3)$$
If $$\alpha$$ is a root of the equation $$f(x) = 0$$ where $$\alpha$$ is very close to $$0$$ (definitely there is a root between $$-1$$ and $$0$$) then $$0 = f(\alpha) = f(0) + 0.00403020948350145 \alpha -2.13362649294736 \cdot 10^{-8}\frac12 \alpha^2 +\text{higher error terms}$$
Of course, we can use computer programs to find a bound on $$\alpha$$ but the whole process is not mathematically elegant.
Certainly, $$\alpha$$ is a root of some polynomial of higher degree, so it may be difficult to find an expression in terms of radicals for $$\alpha$$ or may not be possible at all.
Are there any other reasons for this level of accuracy?
• How did you come across these numbers?
– lhf
Jun 27, 2021 at 10:41
• Could you add some context? How did you arrive at these numbers? Jun 27, 2021 at 10:42
• @lhf My friend gave me (as a lockdown challenge ) who always uses a computer programs to generate this kind of expression. Jun 27, 2021 at 11:02
• Uh, aren't those radicands Fibonacci numbers? Jul 12, 2021 at 20:30
• Let's look at $75025$. Multiply this by $\sqrt5$ and take its natural log ~ 12.0303. The natural log of 1.61803... is ~ 0.481212. Divide the former by the latter and the result from my mobile device calculator is $25+10^{-10}$ ... aha! Jul 12, 2021 at 20:35
Let $$x=75025$$ and $$y = 121393$$. This approximate equality can be written as $$\sqrt x+\sqrt y+\sqrt{x+y} +\sqrt{x+2y}\approx\sqrt{2x+3y}+\sqrt{3x+5y}$$ where $$y\approx x*\varphi$$ and $$\varphi$$ is the golden ratio.
Factoring out $$x$$ from both sides, we get the approximate inequality as $$1+\sqrt\varphi+\sqrt{1+\varphi}+\sqrt{1+2\varphi}\approx\sqrt{2+3\varphi}+\sqrt{3+5\varphi}$$ Now consider the LHS and RHS separately (using the fact that $$\varphi^2=\varphi+1$$), \begin{align} LHS &= 1+\sqrt\varphi+\sqrt{1+\varphi}+\sqrt{\varphi+\varphi^2}\\ &= (1+\sqrt\varphi)(1+\sqrt{1+\varphi})\\ &=(1+\sqrt\varphi)(1+\varphi)\\ &=\varphi^2(1+\sqrt\varphi) \end{align} Also, \begin{align} RHS &= \sqrt{2\varphi^2+\varphi}+\sqrt{3\varphi^2+2\varphi}\\ &=\sqrt\varphi\sqrt{\varphi+\varphi^2}+\sqrt\varphi\sqrt{2\varphi^2+\varphi}\\ &=\varphi\sqrt{1+\varphi}+\varphi\sqrt{\varphi+\varphi^2}\\ &=\varphi^2(1+\sqrt\varphi) \end{align} So these expressions for golden ratio are exactly equal.
This is the reason we're getting approximate inequality in this case (because $$y$$ is not exactly equal to $$x\varphi$$, otherwise the given equality would also be exact).
EDIT:
I have written a function in python to find the difference between LHS and RHS of given equation for any given input $$x$$ (which is $$75025$$ in this case):
from math import sqrt
def find_diff(x):
phi = (1+sqrt(5))/2
y = round(x*phi)
a = sqrt(x)+sqrt(y)+sqrt(x+y)+sqrt(x+2*y)
b = sqrt(2*x+3*y)+sqrt(3*x+5*y)
return abs(a-b)
Note that you may try further Fibonacci numbers and get better approximations (it is obvious to see what Fibonacci numbers are doing here if you know the continued fraction approximations for golden ratio).
• (+1). But OP should have added some context, this would have had been a much simpler problem to solve, then. Jun 27, 2021 at 11:03
• You are right @Ritam_Dasgupta, but there would be lesser fun to solve that simpler problem then :) Jun 27, 2021 at 11:11
• Use $\varphi ^{n}=F_{n}\varphi +F_{n-1}$ to simplify the derivation.
– lhf
Jun 28, 2021 at 16:20
• Actually the radicands are consecutive Fibonacci numbers. We should work this into the answer. Jul 12, 2021 at 20:39 | 2022-08-16T10:58:23 | {
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https://math.stackexchange.com/questions/2442239/forall-x-in-a-forall-y-in-ax-ry-iff-p-x-%E2%8A%86-p-y-for-a-partial-order-r | $(\forall x \in A)(\forall y \in A)(x Ry \iff P_x ⊆ P_y)$ for a partial order $R$
This is Velleman's exercise 4.4.10:
Suppose $R$ is a partial order on $A$. For each $x ∈ A$, let $P_x = \{a ∈ A | aRx\}$. Prove that $∀x ∈ A∀y ∈ A(x Ry ↔ P_x ⊆ P_y)$.
Here's my proof of it:
Proof. Suppose $x$ and $y$ to be arbitrary elements of $A$.
($\rightarrow$) Suppose $xRy$. Suppose $aRx$. Since $R$ is transitive then from $aRx$ and $xRy$ we get $aRy$. Therefore if $aRx$ then $aRy$ and therefore if $xRy$ then $(aRx \rightarrow aRy).$
($\leftarrow$) Suppose $(aRx \rightarrow aRy)$. By the definition of partial order we have $xRy$ and therefore if $(aRx \rightarrow aRy)$ then $xRy$.
From ($\rightarrow$) and ($\leftarrow$) we'll have $(x Ry ↔ P_x ⊆ P_y)$ and since $x$ and $y$ were arbitrary then $∀x ∈ A∀y ∈ A(x Ry ↔ P_x ⊆ P_y)$.
Is my proof correct? Particualrly the backward direction in which I didn't use "$(aRx \rightarrow aRy)$" directly.
Thanks a lot.
The $\leftarrow$ does indeed need a bit of work.
Use that $R$ is reflexive to get that $xRx$, and hence $x \in P_x$. Since $P_x \subseteq P_y$ that means that $x \in P_y$, and hence $xRy$
• Dang you beat me by a few seconds :D – John Griffin Sep 23 '17 at 20:32
• Funny how that happens :) – Bram28 Sep 23 '17 at 20:34
• @JohnGriffin. So what? Is the stack a competitive game? – William Elliot Sep 23 '17 at 23:06
It seems correct, but you did skip a step in the backward direction with the words "by the definition of partial order". Here's a rewording:
Suppose $P_x\subseteq P_y$. Since $R$ is reflexive, we have $x\in P_x$. Thus the assumption gives $x\in P_y$, so that $xRy$.
This shows that you do infact use $(aRx\to aRy)$.
• Great minds ... – Bram28 Sep 23 '17 at 20:32 | 2020-06-01T12:06:23 | {
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https://math.stackexchange.com/questions/2929902/how-many-passwords-with-one-or-more-digits | How many passwords with one or more digits?
A symbol in a password is either one of $$26$$ latin letters or one of $$10$$ digits. How many $$8$$ symbol passwords with at least $$1$$ digit?
I know there are $$36^8-26^8$$ of them. But at first I thought there were $$36^7\times 10$$.
What was my mistake? What does the number $$36^7\times 10$$ represent?
The whole length $$8$$ password space has size $$36^8$$. There are $$26^8$$ passwords with only letters and no digits. So $$36^8 - 26^8$$ counts the number of passwords with at least one digit.
$$10\cdot 36^7$$ counts the number of passwords with a digit at the first position, e.g. Or with a digit at the final position. Or any fixed pre-given position for that matter.
To count the number with at least one letter we have the substract the
number of passwords with only digits, so $$36^8-10^8$$.
• Ah, ok, so it's at any position vs at a fixed position. But I still don't understand why $38^7\times 10$ doesn't answer the question. Why is it a wrong answer? Why can't numbers with a digit in a fixed position count as the answer to this question? – Coder-Man Sep 25 '18 at 8:16
• @Coder-Man Because if we only count those with a starting digit we miss many passwords, like a1bbbbbb, which would also be a valid one. – Henno Brandsma Sep 25 '18 at 8:20
• Oh! Stupid me, now I get it. Thank you very much. – Coder-Man Sep 25 '18 at 8:22
• @Coder-Man We could multiply by $8$ for all positions but then we'd double count a password like 12abbbbb which you would count among the passwords with digit at position 1 and at position 2 as well. – Henno Brandsma Sep 25 '18 at 8:22
• But how come, even though, as you said, we miss passwords like a1bbbbbb, $38^7\times 10 > 38^8-26^8$? – Coder-Man Sep 25 '18 at 8:28
$$36^7 \times 10$$ represents all passwords with one number in a chosen position. For $$7$$ of the places you choose from all possible letters and numbers and for the chosen one you choose only from the numbers.
Assuming your title is correct and you are looking for passwords containing one or more digits, then $$36^8 -26^8$$ represents the space of all possible passwords minus those which are completely made up of letters.
Similarly, if you were looking for passwords containing one or more letters, then the answer would be $$36^8-10^8$$, the space of all passwords minus those which are completely made up of numbers.
• How is that different from what $36^8-26^8$ represents? – Coder-Man Sep 25 '18 at 7:59
• You didn't write it correctly, I didn't say it's $36^8-10^8$ – Coder-Man Sep 25 '18 at 8:00
• @Coder-Man I am asking you if that is what you meant? $36^8-26^8$ is all passwords containing a number not a letter – lioness99a Sep 25 '18 at 8:01
• not at least one number, but say a number at the final position. You cannot choose the digit position freely. – Henno Brandsma Sep 25 '18 at 8:03
• Title ask for password with a number, and question body ask for password with a letter. I think OP made a typo, but from the results he gets, the title seems to be correct – F.Carette Sep 25 '18 at 8:03 | 2019-07-20T12:30:12 | {
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https://bergknoff.com/journal/topics-in-algebra-chapter-4-section-1/ | ### Topics in Algebra, Chapter 4.1
###### 2013-09-23
This page covers section 4.1 (“Elementary Basic Concepts” [of vector spaces and modules]).
### Topics covered: 4.1
• Definition: Let $V$ be a non-empty set, let $F$ be a field, and let $+:V×V\to V$ and $\cdot :F×V\to V$ be binary operations such that
1. $\alpha \cdot \left(v+w\right)=\alpha \cdot v+\alpha \cdot w$ for all $\alpha \in F$, $v,w\in V$.
2. $\left(\alpha +\beta \right)\cdot v=\alpha \cdot v+\beta \cdot v$ for all $\alpha ,\beta \in F$, $v\in V$.
3. $\alpha \left(\beta \cdot v\right)=\left(\alpha \beta \right)\cdot v$ for all $\alpha ,\beta \in F$, $v\in V$.
4. $1\cdot v=v$ for all $v\in V$ where $1$ is the multiplicative unit in $F$.
Then $V$ is said to be a vector space over $F$. The dot for multiplication will generally be omitted in what follows.
• Example: If $F\subset K$ are both fields, then $K$ may be viewed as a vector space over $F$.
• Example: If $F$ is a field, then ${F}^{n}=\left\{\left({\alpha }_{1},\dots ,{\alpha }_{n}\right)\mid {\alpha }_{i}\in F\right\}$, with the obvious operations, is a vector space over $F$.
• Example: If $F$ is a field, then $F\left[x\right]$ is a vector space over $F$.
• Example: If $F$ is a field, then ${P}_{n}\left(F\right)\subset F\left[x\right]$ is a vector space over $F$, where ${P}_{n}\left(F\right)$ is the set of polynomials over $F$ with degree less than $n$.
• Definition: If $V$ is a vector space over $F$ and $W\subset V$ forms a vector space using the same operations of $V$, then $W$ is a subspace of $V$. This is equivalent to the condition that $\alpha w+{\alpha }^{\mathrm{\prime }}{w}^{\mathrm{\prime }}\in W$ for all $w,{w}^{\mathrm{\prime }}\in W$ and $\alpha ,{\alpha }^{\mathrm{\prime }}\in F$.
• Definition: Let $V,W$ be vector spaces over $F$. A homomorphism of vector spaces is a map $\varphi :W\to V$ such that $\varphi \left(v+{v}^{\mathrm{\prime }}\right)=\varphi \left(v\right)+\varphi \left({v}^{\mathrm{\prime }}\right)$ and $\varphi \left(\alpha w\right)=\alpha \varphi \left(w\right)$ for all $w,{w}^{\mathrm{\prime }}\in W$ and $\alpha \in F$.
• The set of all homomorphisms between vector spaces $V$ and $W$ will be denoted $\mathrm{H}\mathrm{o}\mathrm{m}\left(V,W\right)$.
• Lemma 4.1.1: Let $V$ be a vector space over $F$. Then
1. $\alpha {0}_{V}={0}_{V}$ for all $\alpha \in F$.
2. ${0}_{F}v={0}_{V}$ for all $v\in V$.
3. $\left(-\alpha \right)v=-\left(\alpha v\right)$ for all $\alpha \in F$, $v\in V$.
4. $\alpha v=0$ implies $\alpha ={0}_{F}$ or $v={0}_{V}$.
• Lemma 4.1.2: Let $V$ be a vector space over $F$ and let $W\subset V$ be a subspace. Then $V\mathrm{/}W=\left\{v+W\mid v\in V\right\}$ is a vector space over $F$, called the quotient space of $V$ by $W$.
• Theorem 4.1.1: Let $V,W$ be vector spaces and let $\varphi :V\to W$ be a surjective homomorphism with kernel $K$. Then $W\cong V\mathrm{/}K$. Conversely, if $V$ is a vector space and $W\subset V$ a subspace, then there exists a homomorphism $\psi :V\to V\mathrm{/}W$. (TODO: am I transcribing this correctly?)
• Definition: Let $V$ be a vector space over $F$ and let ${W}_{1},\dots ,{W}_{n}\subset V$ be subspaces. If any $v\in V$ admits a unique representation $v={w}_{1}+\cdots +{w}_{n}$ with ${w}_{i}\in {W}_{i}$ for each $i$, then $V$ is the internal direct sum of the $\left\{{W}_{i}\right\}$.
• Definition: Let ${V}_{1},\dots ,{V}_{n}$ be vector spaces over $F$. The external direct sum of the $\left\{{V}_{i}\right\}$ is the set $\left\{\left({v}_{1},\dots ,{v}_{n}\right)\mid {v}_{i}\in {V}_{i}\right\}$.
• Theorem 4.1.2: The internal and external direct sums of $\left\{{V}_{1},\dots ,{V}_{n}\right\}$ are isomorphic. Hence we can refer to simply a direct sum having both of the above properties.
The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.
### Herstein 4.1.1
#### Let $V$ be a vector space over field $F$. Further, let $\alpha \in F$ and $v,w\in V$. Show that $\alpha \left(v-w\right)=\alpha v-\alpha w$ in $V$
We have $\alpha \left(v-w\right)=\alpha \left(v+\left(-w\right)\right)=\alpha v+\alpha \left(-w\right)=\alpha v-\alpha w$
by Lemma 4.1.1.
### Herstein 4.1.2
#### Let $F$ be a field and $n$ a positive integer, let $V={F}^{n}$ and let $W\subset F\left[x\right]$ be the vector space of polynomials over $F$ of degree less than $n$. Prove that $V\cong W$.
The map $\varphi :V\to W$ given by $\varphi \left(\left({\alpha }_{0},\dots ,{\alpha }_{n-1}\right)\right)={\alpha }_{n-1}{x}^{n-1}+\dots +{\alpha }_{0}$ is an isomorphism.
### Herstein 4.1.3
#### Prove that the kernel of homomorphism is a subspace.
For $\varphi :V\to W$ a homomorphism between vector spaces $V,W$, $\mathrm{ker}\varphi =\left\{v\in W\mid \varphi \left(v\right)=0\right\}$. Let $v,{v}^{\mathrm{\prime }}\in \mathrm{ker}\varphi$ and $\alpha \in F$, with $F$ the base field. We have $\varphi \left(v+\alpha {v}^{\mathrm{\prime }}\right)=\varphi \left(v\right)+\alpha \varphi \left({v}^{\mathrm{\prime }}\right)$ because $\varphi$ is a homomorphism, and so $\varphi \left(v+\alpha {v}^{\mathrm{\prime }}\right)=0$ and $v+\alpha {v}^{\mathrm{\prime }}\in \mathrm{ker}\varphi$. Thus $\mathrm{ker}\varphi$ is a subspace of $V$.
### Herstein 4.1.4
#### (b) For positive integer $n$, show that the set of functions $\left[0,1\right]\to \mathbb{R}$, for which the first $n$ derivatives exist, form a subspace of the vector space from (a).
(a) Let $f,g\in V$ and $\alpha \in \mathbb{R}$. We have $f+\alpha g\in V$ because sums and scalar products of continuous functions are again continuous. The function $0:x↦0$ is the additive identity in this vector space. Other details can be taken for granted.
(b) The set of $n$-times differentiable functions is a subset of the continuous functions, so it’s just necessary to check if the set is closed under linear combinations. Indeed, the sum of a differentiable function is again differentiable, so the set in question is a subspace of $V$.
### Herstein 4.1.5
#### (c)* Let $U=\left\{\left({a}_{1},\dots ,{a}_{n},\dots \right)\in V\mid {\sum }_{i=1}^{\mathrm{\infty }}{a}_{i}^{2}<\mathrm{\infty }\right\}$. Prove that $U$ is a subspace of $V$ and that $U$ is contained in $W$.
(a) Because $\mathbb{R}$ is closed under addition and multiplication, and operations on $V$ are defined componentwise, all vector space axioms hold for $V$.
(b) If $\left({a}_{i}\right)$ and $\left({b}_{i}\right)$ are two elements of $W$ and $\alpha \in \mathbb{R}$, then $\left({a}_{i}+\alpha {b}_{i}\right)\in W$ because $\underset{}{\mathrm{lim}}\left({a}_{i}+\alpha {b}_{i}\right)=\underset{}{\mathrm{lim}}{a}_{i}+\alpha \underset{}{\mathrm{lim}}{b}_{i}=0.$
(c) Let $\left({a}_{i}\right),\left({b}_{i}\right)\in U$ and let $\alpha \in \mathbb{R}$. We have that $\sum _{i}\left({a}_{i}+\alpha {b}_{i}{\right)}^{2}=\sum _{i}{a}_{i}^{2}+{\alpha }^{2}\sum _{i}{b}_{i}^{2}+2\alpha \sum _{i}{a}_{i}{b}_{i}\mathrm{.}$
The first two terms are finite by assumption. The third term can be bounded: for real numbers $x,y$, rearrange $\left(x-y{\right)}^{2}\ge 0$ to give $xy\le \frac{1}{2}{x}^{2}+\frac{1}{2}{y}^{2}$
so that $\sum _{i}{a}_{i}{b}_{i}\le \frac{1}{2}\sum _{i}{a}_{i}^{2}+\frac{1}{2}\sum _{i}{b}_{i}^{2}<\mathrm{\infty }\mathrm{.}$
Hence $\left({a}_{i}+\alpha {b}_{i}\right)\in U$ and $U$ is a subspace of $V$.
To show that $U$ is contained in $W$, we must show that ${\sum }_{i=1}^{\mathrm{\infty }}{a}_{i}^{2}<\mathrm{\infty }$ implies ${\mathrm{lim}}_{}{a}_{i}=0$. Define the partial sums ${s}_{n}={\sum }_{i=1}^{n}{a}_{i}^{2}$; we have that ${\mathrm{lim}}_{}{s}_{n}={\mathrm{lim}}_{}{s}_{n-1}=L$ for some $L<\mathrm{\infty }$. Therefore, $0=\underset{}{\mathrm{lim}}\left({s}_{n}-{s}_{n-1}\right)=\underset{}{\mathrm{lim}}{a}_{n}^{2}\mathrm{.}$
For any $ϵ>0$, there exists $N\in \mathbb{N}$ such that $n>N$ implies that ${a}_{n}^{2}<ϵ$. Let ${ϵ}^{\mathrm{\prime }}>0$ be given: by the previous statement, there exists $N\in \mathbb{N}$ so that $n>N$ implies ${a}_{n}^{2}<{ϵ}^{\mathrm{\prime }2}$. Thus for $n>N$, we also have $\mathrm{\mid }{a}_{n}\mathrm{\mid }<{ϵ}^{\mathrm{\prime }}$. This proves that ${\mathrm{lim}}_{}{a}_{n}=0$, i.e. that $\left({a}_{i}\right)\in W$.
### Herstein 4.1.6
#### Let $U,V$ be vector spaces over field $F$. Define operations on $\mathrm{H}\mathrm{o}\mathrm{m}\left(U,V\right)$ to make it into a vector space over $F$.
$\mathrm{H}\mathrm{o}\mathrm{m}\left(U,V\right)$ is the set of homomorphisms $U\to V$. Given $\varphi ,\psi \in \mathrm{H}\mathrm{o}\mathrm{m}\left(U,V\right)$ and $\alpha \in F$, we can define a third homomorphism pointwise, i.e. by $\left(\varphi +\alpha \psi \right)\left(u\right)=\varphi \left(u\right)+\alpha \psi \left(u\right)\mathrm{.}$
It is straightforward to see that $\varphi +\alpha \psi$ is again a homomorphism. $\mathrm{H}\mathrm{o}\mathrm{m}\left(U,V\right)$ is a vector space under this pointwise addition and scalar multiplication.
### Herstein 4.1.7*
#### With $F$ a field, prove that $\mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},{F}^{m}\right)$ is isomorphic to ${F}^{mn}$ as vector spaces.
Let ${e}_{i}^{\left(n\right)}\in {F}^{n}$ be the vector with a $1$ in the $i$-th index and zeroes elsewhere. Similarly, let ${e}_{i}^{\left(m\right)}\in {F}^{m}$ be the analogous thing. Given $\varphi \in \mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},{F}^{m}\right)$, we have $\varphi \left({e}_{i}^{\left(n\right)}\right)={\sum }_{j}{\alpha }_{ij}^{\left(\varphi \right)}{e}_{j}^{\left(m\right)}$, defining a matrix of coefficients ${\alpha }_{ij}^{\left(\varphi \right)}\in F$ for each $\varphi$. Now define $f:\mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},{F}^{m}\right)\to {F}^{mn}$ by $f\left(\varphi \right)=\left({\alpha }_{11}^{\left(\varphi \right)},\dots ,{\alpha }_{1m}^{\left(\varphi \right)},{\alpha }_{21}^{\left(\varphi \right)},\dots ,{\alpha }_{nm}^{\left(\varphi \right)}\right)\mathrm{.}$
That $f$ respects linear combinations is a rote computation. The kernel of $f$ is trivial, so it is an isomorphism.
### Herstein 4.1.8
#### Let $F$ be a field and $n>m$ be positive integers. Exhibit a surjective homomorphism ${F}^{n}\to {F}^{m}$ and show that its kernel is isomorphic to ${F}^{n-m}$.
Define $\varphi :{F}^{n}\to {F}^{m}$ by $\left({a}_{1},\dots ,{a}_{m},{a}_{m+1},\dots ,{a}_{n}\right)↦\left({a}_{1},\dots ,{a}_{m}\right)$. This is a surjective homomorphism. The kernel of $\varphi$ is the set $\left\{\left(0,\dots ,0,{a}_{m+1},\dots ,{a}_{n}\right)\in {F}^{n}\right\}$; a similar projection mapping establishes the isomorphism $\mathrm{ker}\varphi \cong {F}^{n-m}$.
### Herstein 4.1.9
#### Fix nonzero $v\in {F}^{n}$. Show there exists $\varphi \in \mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},F\right)$ with $\varphi \left(v\right)\mathrm{\ne }0$.
Let $m$ be the index of the first non-zero entry in $v$, and let $\varphi$ be the projection of ${F}^{n}$ onto its $m$-th entry. Then $\varphi \in \mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},F\right)$ and $\varphi \left(v\right)\mathrm{\ne }0$.
### Herstein 4.1.10
#### With $F$ a field and $n$ a positive integer, prove that ${F}^{n}\cong \mathrm{H}\mathrm{o}\mathrm{m}\left(\mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},F\right),F\right)$.
This is (a special case of) the result that a vector space is isomorphic to its double-dual.
With the result 4.1.7, we have $\mathrm{H}\mathrm{o}\mathrm{m}\left(\mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},F\right),F\right)\cong \mathrm{H}\mathrm{o}\mathrm{m}\left({F}^{n},F\right)\cong {F}^{n}\mathrm{.}$
### Herstein 4.1.11
#### With $U,W$ subspaces of vector space $V$, all over field $F$, prove that $U+W=\left\{u+w\mid u\in U,w\in W\right\}$ is a subspace of $V$.
Given $u,{u}^{\mathrm{\prime }}\in U$, $w,{w}^{\mathrm{\prime }}\in W$ and $\alpha \in F$, we have that $\left(u+w\right)+\alpha \left({u}^{\mathrm{\prime }}+{w}^{\mathrm{\prime }}\right)=\left(u+\alpha {u}^{\mathrm{\prime }}\right)+\left(w+\alpha {w}^{\mathrm{\prime }}\right)={u}^{\mathrm{\prime }\mathrm{\prime }}+{w}^{\mathrm{\prime }\mathrm{\prime }}\in U+W,$
where the last step is justified because $U$ and $W$ are each subspaces. Therefore $U+W$ is a subspace of $V$.
### Herstein 4.1.12
#### Prove that the intersection of two subspaces of $V$ is again a subspace of $V$.
Let $U,W$ be subspaces of $V$ over the field $F$. If $v,{v}^{\mathrm{\prime }}\in U\cap W$ and $\alpha \in F$, then $v+\alpha {v}^{\mathrm{\prime }}\in U$ because $U$ is a subspace and $v+\alpha {v}^{\mathrm{\prime }}\in W$ because $W$ is a subspace. Hence $v+\alpha {v}^{\mathrm{\prime }}\in U\cap W$, and $U\cap W$ is a subspace.
### Herstein 4.1.13
#### With $U,W$ subspaces of vector space $V$, all over field $F$, prove that $\left(U+W\right)\mathrm{/}W\cong U\mathrm{/}\left(U\cap W\right)$.
This is the second isomorphism theorem.
The elements of $\left(U+W\right)\mathrm{/}W$ look like $u+w+W=u+W$ where $u\in U$ and $w\in W$. The elements of $U\mathrm{/}\left(U\cap W\right)$ look like $u+U\cap W$ with $u\in U$. In both cases, elements of $U\cap W$ get turned into the zero coset.
Define the map $\varphi :\left(U+W\right)\mathrm{/}W\to U\mathrm{/}\left(U\cap W\right)$ by $\varphi \left(u+W\right)=u+U\cap W$. To see that this is well-defined, consider $u+w,{u}^{\mathrm{\prime }}+{w}^{\mathrm{\prime }}\in U+W$ that belong to the same coset: $u+w+W={u}^{\mathrm{\prime }}+{w}^{\mathrm{\prime }}+W$, so that their difference is $u-{u}^{\mathrm{\prime }}\in W$ which then implies that $u-{u}^{\mathrm{\prime }}\in U\cap W$. We have $\varphi \left(u+w+W\right)-\varphi \left({u}^{\mathrm{\prime }}+{w}^{\mathrm{\prime }}+W\right)=\left(u-{u}^{\mathrm{\prime }}\right)+U\cap W=U\cap W$; thus any representative of a coset in the domain gets mapped to the same coset in the codomain.
$\varphi$ is a homomorphism: for $u,{u}^{\mathrm{\prime }}\in U$, $w,{w}^{\mathrm{\prime }}\in W$, we have $\varphi \left(u+w+\alpha \left({u}^{\mathrm{\prime }}+{w}^{\mathrm{\prime }}\right)+W\right)=u+\alpha {u}^{\mathrm{\prime }}+U\cap W$ while $\varphi \left(u+w+W\right)+\alpha \varphi \left({u}^{\mathrm{\prime }}+{w}^{\mathrm{\prime }}+W\right)=\left(u+U\cap W\right)+\alpha \left({u}^{\mathrm{\prime }}+U\cap W\right)=u+\alpha {u}^{\mathrm{\prime }}+U\cap W$.
The kernel of $\varphi$ contains those elements which map to $0$ in the codomain, i.e. those elements of the domain where the $U$ component belongs to $U\cap W$. We have that $\left\{u+w\mid u\in U\cap W,w\in W\right\}\subset W$ so $\mathrm{ker}\varphi =W$, i.e. the kernel is trivial so that $\varphi$ is injective.
$\varphi$ is surjective because, given $u+U\cap W\in U\mathrm{/}\left(U\cap W\right)$, we have $\varphi \left(u+W\right)=u+U\cap W$.
Therefore $\varphi :\left(U+W\right)\mathrm{/}W\to U\mathrm{/}\left(U\cap W\right)$ is an isomorphism.
### Herstein 4.1.14
#### Let $U,V$ be vector spaces and let $\varphi :U\to V$ be a surjective homomorphism. Show there is a one-to-one correspondence between $\mathcal{A}$, the subspaces of $V$, and $\mathcal{B}$, the subspaces of $U$ which contain $\mathrm{ker}\varphi$.
This is the fourth (“lattice”) isomorphism theorem.
There are a couple of natural-looking ways to map the objects in question (I tried $W↦\varphi \left(W\right)$, $W↦U\mathrm{/}W$, etc.). However, the first isomorphism theorem (theorem 4.1.1) states that $V\cong U\mathrm{/}\mathrm{ker}\varphi$, so the subspaces of $V$ should probably look like $W\mathrm{/}\mathrm{ker}\varphi$ where $W$ is a subspace of $V$. Naturally, $W\mathrm{/}\mathrm{ker}\varphi$ only makes sense if $W$ contains $\mathrm{ker}\varphi$. Therefore, the map we define is $f:\mathcal{B}\to \mathcal{A}$ given by $f\left(W\right)=W\mathrm{/}\mathrm{ker}\varphi$, and it makes sense because of the way we have chosen $\mathcal{B}$ (i.e. only considering subspaces that contain $\mathrm{ker}\varphi$).
The map $f$ is injective: let $W,{W}^{\mathrm{\prime }}\in \mathcal{B}$ be mapped the same by $f$, i.e. $W\mathrm{/}\mathrm{ker}\varphi ={W}^{\mathrm{\prime }}\mathrm{/}\mathrm{ker}\varphi$. We would like to show that this implies $W={W}^{\mathrm{\prime }}$. If $w\in W$, then $w+\mathrm{ker}\varphi \in W\mathrm{/}\mathrm{ker}\varphi ={W}^{\mathrm{\prime }}\mathrm{/}\mathrm{ker}\varphi$ so there exists ${w}^{\mathrm{\prime }}\in {W}^{\mathrm{\prime }}$ with $w+\mathrm{ker}\varphi ={w}^{\mathrm{\prime }}+\mathrm{ker}\varphi$. This implies that $w-{w}^{\mathrm{\prime }}\in \mathrm{ker}\varphi \subset {W}^{\mathrm{\prime }}$, so that $w=\left(w-{w}^{\mathrm{\prime }}\right)+{w}^{\mathrm{\prime }}\in {W}^{\mathrm{\prime }}\mathrm{.}$
This proves that $W\subset {W}^{\mathrm{\prime }}$. The argument, made in reverse, gives also that ${W}^{\mathrm{\prime }}\subset W$, so we have proven that $f$ is injective.
$f$ is also surjective: if $X$ is a subspace of $V$, then we can realize it as the image of a subspace of $U$. Consider $Y={\varphi }^{-1}\left(X\right)=\left\{u\in U\mid \varphi \left(u\right)\in X\right\}$. It remains to show that $\mathrm{ker}\varphi \subset Y$, that $Y$ is a subspace of $U$, and that $f\left(Y\right)=X$. Because $X$ is a subspace, $0\in X$ and $\varphi \left(\mathrm{ker}\varphi \right)=\left\{0\right\}\subset X$, so that $\mathrm{ker}\varphi \subset Y$. If $y,{y}^{\mathrm{\prime }}\in Y$ and $\alpha$ is a scalar, then $\varphi \left(y+\alpha {y}^{\mathrm{\prime }}\right)=\varphi \left(y\right)+\alpha \varphi \left({y}^{\mathrm{\prime }}\right)\in X$, so $Y$ is a subspace of $U$. Finally, $f\left(Y\right)=\left\{u+\mathrm{ker}\varphi \mid u\in U,\varphi \left(u\right)\in X\right\}$, which is the same thing as $X$ (TODO: fill in details without making notation worse?).
Therefore, $f$ is a bijection between $\mathcal{A}$ and $\mathcal{B}$.
### Herstein 4.1.15
#### Let $V$ be a vector space and let ${V}_{1},\dots ,{V}_{n}$ be subspaces of $V$ such that $V={V}_{1}+\cdots +{V}_{n}$ and ${V}_{i}\cap {\sum }_{j\mathrm{\ne }i}{V}_{j}=\left\{0\right\}$ for every $i$. Prove that $V$ is the internal direct sum of ${V}_{1},\dots ,{V}_{n}$.
To say that $V$ is the internal direct sum of the ${V}_{i}$ is to say that $v\in V$ has exactly one expression $v={v}_{1}+\cdots +{v}_{n}$ with each ${v}_{i}\in {V}_{i}$.
Because $V\subset {\sum }_{i}{V}_{i}$, we have that $v\in V$ has at least one such expression. It remains to show that this expression is unique. Therefore, suppose that $v={v}_{1}+\cdots +{v}_{n}={v}_{1}^{\mathrm{\prime }}+\cdots +{v}_{n}^{\mathrm{\prime }}$ with ${v}_{i},{v}_{i}^{\mathrm{\prime }}\in {V}_{i}$ for each $i$. Then we have $\left({v}_{1}-{v}_{1}^{\mathrm{\prime }}\right)+\cdots +\left({v}_{n}-{v}_{n}^{\mathrm{\prime }}\right)=0,$
and, rearranging, ${v}_{i}-{v}_{i}^{\mathrm{\prime }}=-\sum _{j\mathrm{\ne }i}\left({v}_{j}-{v}_{j}^{\mathrm{\prime }}\right)\mathrm{.}$
The left hand side belongs to ${V}_{i}$ while the right hand side belongs to ${\sum }_{j\mathrm{\ne }i}{V}_{j}$. By assumption, those two spaces intersect trivially, so that ${v}_{i}-{v}_{i}^{\mathrm{\prime }}=0$ for each $i$. Hence the two representations are identical, and we are done.
### Herstein 4.1.16
#### Let $V$ be a vector space with subspaces ${V}_{1},\dots ,{V}_{n}$ such that $V={\oplus }_{i}{V}_{i}$. Prove there are subspaces ${\stackrel{ˉ}{V}}_{i}\subset V$ isomorphic to ${V}_{i}$ with $V$ the internal direct sum of ${\stackrel{ˉ}{V}}_{i}$.
$V$ is the external direct sum of the ${V}_{i}$, so it looks like $V=\left\{\left({v}_{1},\dots ,{v}_{n}\right)\mid {v}_{i}\in {V}_{i}\right\}\mathrm{.}$
The subspaces ${\stackrel{ˉ}{V}}_{i}$ which allow the $i$-th entry to range over ${V}_{i}$, while fixing the non-$i$ entries as zero, are the desired subspaces of $V$ isomorphic to ${V}_{i}$. The conditions of exercise 4.1.15 are easily satisfied, so that $V$ is the internal direct sum of the ${\stackrel{ˉ}{V}}_{i}$.
### Herstein 4.1.17
#### (b) Find necessary and sufficient conditions on $\alpha ,\beta ,\gamma ,\delta$ so that $T$ is an isomorphism.
(a) That $T$ is a homomorphism is a straightforward exercise
(b) In the language of matrices, this is the familiar question of when a matrix is invertible; the answer is “when the determinant is non-zero”. How does that come about from direct computation?
Let ${y}_{1},{y}_{2}\in F$ and consider the simultaneous equations $\alpha {x}_{1}+\beta {x}_{2}={y}_{1},$
$\gamma {x}_{1}+\delta {x}_{2}={y}_{2}\mathrm{.}$
Multiplying the first equation by $\delta$ and the second by $\beta$, and then subtracting the second from the first, we find $\left(\alpha \delta -\beta \gamma \right){x}_{1}=\delta {y}_{1}-\beta {y}_{2}\mathrm{.}$
Performing a similar computation, we also find $\left(\alpha \delta -\beta \gamma \right){x}_{2}=\alpha {y}_{2}-\gamma {y}_{1}\mathrm{.}$
In order for $T$ to be injective, it must have a trivial kernel. If $\left({x}_{1},{x}_{2}\right)\in \mathrm{ker}T$, then $\left(\alpha \delta -\beta \gamma \right){x}_{1}=\left(\alpha \delta -\beta \gamma \right){x}_{2}=0.$
These equations have non-trivial solutions $\left({x}_{1},{x}_{2}\right)$ if and only if $\alpha \delta -\beta \gamma =0$. Thus a necessary and sufficient condition for $T$ to be injective is that $\alpha \delta -\beta \gamma \mathrm{\ne }0$. As a result, this is also a necessary condition for $T$ to be an isomorphism.
The same condition is also sufficient for $T$ to be surjective, because the equations $\left(\alpha \delta -\beta \gamma \right){x}_{1}=\delta {y}_{1}-\beta {y}_{2}$ and $\left(\alpha \delta -\beta \gamma \right){x}_{2}=\alpha {y}_{2}-\gamma {y}_{1}$ are solvable for any ${y}_{1},{y}_{2}$ by dividing by $\alpha \delta -\beta \gamma$.
Therefore the necessary and sufficient condition for $T$ to be an isomorphism is that $\alpha \delta -\beta \gamma$ be non-zero.
### Herstein 4.1.18
#### The same exercise as 4.1.17 but on ${F}^{3}$.
I haven’t done this exercise, but I would be surprised if it is different from 4.1.17 in a meaningful way.
### Herstein 4.1.19
#### Let $V,W$ be vector spaces and let $T:V\to W$. Use $T$ to define a homomorphism ${T}^{\ast }:\mathrm{H}\mathrm{o}\mathrm{m}\left(W,F\right)\to \mathrm{H}\mathrm{o}\mathrm{m}\left(V,F\right)$.
Put another way, the exercise is to show that a homomorphism between vector spaces $V$ and $W$ induces a natural homomorphism between their dual spaces ${V}^{\ast }=\mathrm{H}\mathrm{o}\mathrm{m}\left(V,F\right)$ and ${W}^{\ast }=\mathrm{H}\mathrm{o}\mathrm{m}\left(W,F\right)$.
A diagram helps:
$\begin{array}{ccc} V & \rightarrow & F$
& \searrow & \uparrow
& & W \end{array}
Here, the map $V\to W$ is provided by $T$, the map $W\to F$ is some representative ${w}^{\ast }\in \mathrm{H}\mathrm{o}\mathrm{m}\left(W,F\right)$, and the desired map $V\to F$ can be made in a natural way by composition. That is, we define a map ${T}^{\ast }:\mathrm{H}\mathrm{o}\mathrm{m}\left(W,F\right)\to \mathrm{H}\mathrm{o}\mathrm{m}\left(V,F\right)$ by ${T}^{\ast }\left({w}^{\ast }\right)={w}^{\ast }\circ T\mathrm{.}$
It is easy to check that (1) the resulting ${v}^{\ast }={w}^{\ast }\circ T$ is indeed an element of $\mathrm{H}\mathrm{o}\mathrm{m}\left(V,F\right)$, and (2) that ${T}^{\ast }$ is a homomorphism.
### Herstein 4.1.20
#### (b) Prove that ${F}^{2}$ is not isomorphic to ${F}^{3}$.
(a) Looking slightly ahead, the intuition here is that the image of $F$ under a homomorphism will be too low-dimensional. Therefore, consider a supposed isomorphism $\varphi :F\to {F}^{n}$. Because it is surjective, there are ${f}_{1},{f}_{2}\in F$ with $\varphi \left({f}_{1}\right)=\left(1,0,\dots ,0\right)$ and $\varphi \left({f}_{2}\right)=\left(0,1,0,\dots ,0\right)$. Now, there exists $\alpha \in F$ such that ${f}_{2}=\alpha {f}_{1}$, so we must have $\left(0,1,0,\dots ,0\right)=\varphi \left({f}_{2}\right)=\varphi \left(\alpha {f}_{1}\right)=\alpha \varphi \left({f}_{1}\right)=\left(\alpha ,0,\dots ,0\right)\mathrm{.}$
This is a contradiction, so we conclude that no such $\varphi$ exists.
(b) Suppose $\varphi :{F}^{2}\to {F}^{3}$ is an isomorphism, $\varphi \left(\left(1,0\right)\right)={v}_{1}\in {F}^{3}$ and $\varphi \left(\left(0,1\right)\right)={v}_{2}\in {F}^{3}$. Then we have $\varphi \left(\left(\alpha ,\beta \right)\right)=\alpha {v}_{1}+\beta {v}_{2}$ for any $\alpha ,\beta \in F$. Because $\varphi$ is surjective, there must exist ${\alpha }_{i},{\beta }_{i}$ such that ${\alpha }_{1}{v}_{1}+{\beta }_{1}{v}_{2}=\left(1,0,0\right),$
${\alpha }_{2}{v}_{1}+{\beta }_{2}{v}_{2}=\left(0,1,0\right),$
${\alpha }_{3}{v}_{1}+{\beta }_{3}{v}_{2}=\left(0,0,1\right)\mathrm{.}$
Taking the first and second equations, and eliminating the ${v}_{2}$ terms, we find that $\left({\alpha }_{1}{\beta }_{2}-{\beta }_{1}{\alpha }_{2}\right){v}_{1}=\left({\beta }_{2},-{\beta }_{1},0\right)\mathrm{.}$
However, taking the first and third equations, and eliminating the ${v}_{2}$ terms, we also find that $\left({\alpha }_{1}{\beta }_{3}-{\beta }_{1}{\alpha }_{3}\right){v}_{1}=\left({\beta }_{3},0,-{\beta }_{1}\right)\mathrm{.}$
These two results are inconsistent unless ${\beta }_{1}=0$. In that case, we can explicitly solve for ${v}_{1}=\left({\alpha }_{1}^{-1},0,0\right)$ and derive a contradiction that ${\beta }_{2}{v}_{2}=\left(-\frac{{\alpha }_{2}}{{\alpha }_{1}},1,0\right)$ while ${\beta }_{3}{v}_{2}=\left(-\frac{{\alpha }_{3}}{{\alpha }_{1}},0,1\right)$.
Thus we have shown that the map $\varphi$ is not truly surjective, and therefore not an isomorphism.
The laborious arguments above make one appreciate (1) the elegance of doing linear algebra without explicit coordinates/choice of basis, and (2) the simplicity and utility of the concepts of linear independence, basis and dimension, which we eschew here because they are not introduced until the next section of the book.
### Herstein 4.1.21
#### Let $V$ be a vector space over the infinite field $F$. Prove that $V$ is not realizable as the set-theoretic union of a finite number of its proper subspaces.
Let ${V}_{1},\dots ,{V}_{n}$ be proper subspaces of $V$ such that ${\bigcup }_{i}{V}_{i}=V$. We can assume that each ${V}_{i}$ brings something of value to this union, i.e. that ${V}_{i}\subset \bigcup _{j\mathrm{\ne }i}{V}_{j}$
In other words, for each $i$, there exists some ${v}_{i}$ which only belongs to ${V}_{i}$ and none of the other subspaces. If this is not the case, then we can omit this ${V}_{i}$: all of its elements are included elsewhere. In this sense, we can assume our set to have minimal size.
Because the subspaces are proper, we know that $n\ge 2$. Consider elements ${v}_{1}\in {V}_{1}$ with ${v}_{1}\in {\bigcup }_{i\mathrm{\ne }1}{V}_{i}$ and ${v}_{2}\in {V}_{2}$ with ${v}_{2}\in {\bigcup }_{i\mathrm{\ne }2}{V}_{i}$. Let $\alpha ,\beta \in F$ be distinct. The elements $x={v}_{1}+\alpha {v}_{2}$ and $y={v}_{1}+\beta {v}_{2}$ belong to $V={\bigcup }_{i}{V}_{i}$, so each belongs to some ${V}_{i}$. Suppose $x,y$ both belong to the same ${V}_{i}$; then so must their difference: $x-y=\left(\alpha -\beta \right){v}_{2}\in {V}_{i}$. By assumption, ${v}_{2}$ only belongs to ${V}_{2}$, so that ${V}_{i}={V}_{2}$. Taking a step back, we see that this would force ${v}_{1}=x-\alpha {v}_{2}$ to also live in ${V}_{2}$, a contradiction. Thus $x$ and $y$ are forced to belong to different subspaces.
Now, we enumerate some infinite subset $\left\{{\alpha }_{1},{\alpha }_{2},\dots \right\}$ of $F$ and construct the elements ${x}_{i}={v}_{1}+{\alpha }_{i}{v}_{2}\in V$. Considering the $\left\{{x}_{i}\right\}$ pairwise, we see that every one must live in a different subspace from every other one: no finite number of subspaces will suffice. We conclude that no vector space over an infinite field can be realized as the union of finitely many of its proper subspaces. | 2019-07-21T21:05:14 | {
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https://math.stackexchange.com/questions/2448258/given-m-n-how-do-we-analyze-n-logm-vs-m-logn | # Given m > n, how do we analyze $n*\log(m)$ vs $m*\log(n)$
The sizes of list A and B are $m$ and $n$ respectively. It's given that $m>n$, which implies that $\log(m) > \log(n)$. Therefore, $m\log(m)>n\log(n)$.
My question is about the analysis of $m\log(n)$ vs. $n\log(m)$ given $m>n$.
I am looking for general cases and a general solution for this problem. I am not even sure if there is a general solution for this problem.
An example algorithm problem:
Given two sorted lists A & B of size $m$ and $n$. Find $a ∈ A$ and $b ∈ B$ such that $a + b = K$.
Algorithm 1: Keep picking an element $a ∈ A$ and perform a binary search, looking for $K-a$ in List B until a solution is found. The complexity is $\mathcal{O}(m\log(n))$
Algorithm 2: Keep picking an element $b ∈ B$ and Perform a binary search, looking for $K-b$ in List A until asolution is found. Complexity $\mathcal{O}(n\log(m))$
So, which Algorithm is better, 1 or 2?
I hope that the solution is not to compute $m\log(n)$ and $n\log(m)$ numerically and decide on the basis of that. I'm looking for a general solution.
EDIT: Since we are dealing with asymptotic analysis (complexity analysis), Please consider that m & n are very very large.
• Try to prove the statement for $m=n+1$ and see if it's true that $m^n>n^m$ Well not true for m close to n but superior...since $( 1+ \frac 1 n)^n$ is less than e and not $>n$ – LostInSpace Sep 27 '17 at 22:42
• @Isham Are you saying that n x log(m) < m x log(n) for m >>> n ?? – Adithya Upadhya Sep 27 '17 at 22:50
• All of the bold face text is.... distracting. – Simply Beautiful Art Sep 27 '17 at 22:52
• @oathkeeper for $m=(n+1)$...try $(n+1)^n>n^{n+1}$ and see that it is false – LostInSpace Sep 27 '17 at 22:52
• @SimplyBeautifulArt Please edit the question to make it more beautiful without changing any content. I am not good at editing questions. Maybe you can help :) – Adithya Upadhya Sep 27 '17 at 22:54
Diving both by $\log(m) \cdot \log (n)$ and analyze the function $$f(x) = \frac{x}{\log x}$$ to show if it is increasing or decreasing using $f'(x)$... You are looking for asymptotics, so more interested in large values of $x$, but in general this analysis can show you a point where behavior changes, if one exists.
• To add to this, if you show that if $x>y$ means that $\frac{x}{\ln x}<\frac{y}{\ln y}$ or $\frac{x}{\ln x}>\frac{y}{\ln y}$, then this can tell you how to compare $\frac{n}{\ln n}$ and $\frac{m}{\ln m}$. – Jam Sep 27 '17 at 23:07 | 2020-02-24T08:07:08 | {
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http://mathhelpforum.com/number-theory/200434-perfect-square-s-odd-number-factors.html | # Math Help - Perfect square s, odd number of factors
1. ## Perfect square s, odd number of factors
If a number is a perfect square, it will have an odd number of factors (e.g., 4 has factors 1, 2, 4), whereas all other numbers have an even number of factors.
Is the converse true? Please explain why?
2. ## Re: Perfect square s, odd number of factors
By converse, do you mean the following statement: "If a number has an odd number of factors, then it is a perfect square, and if a number has an even number of factors, then is it not a perfect square"? The original statement says that being a perfect square is equivalent to having an odd number of factors, so of course the converse is true.
3. ## Re: Perfect square s, odd number of factors
Yes, that's what I meant. But I am not convinced of the equivalence. Can you explain a bit further with the proof.
4. ## Re: Perfect square s, odd number of factors
Suppose that the number of factors of n is odd. We need to show that n is a perfect square. Suppose the contrary; then by the statement in post #1, n has an even number of factors, a contradiction.
In general, if you showed A implies B and (not A) implies (not B), then you showed that A and B are equivalent, so either one implies the other. Also, (not A) and (not B) are equivalent in this case.
5. ## Re: Perfect square s, odd number of factors
No, I meant if it's just given that, "If a number is a perfect square, it will have an odd number of factors (e.g., 4 has factors 1, 2, 4)"
Now how would you prove the converse?
6. ## Re: Perfect square s, odd number of factors
So, I understand that the part in post #1 after "whereas" is not given.
Without loss of generality, $n = p_1^{a_1}\cdot\ldots\cdot p_r^{a_r}$ for primes $p_1,\dots,p_r$. The number of factors of n is given by the divisor function $d(n)=(a_1+1)\cdot\ldots\cdot(a_r+1)$. If d(n) is odd, then all $a_i$'s are even, so $n = (p_1^{a_1/2}\cdot\ldots\cdot p_r^{a_r/2})^2$.
7. ## Re: Perfect square s, odd number of factors
Much obliged.
8. ## Re: Perfect square s, odd number of factors
Emakarov has given a first-rate proof, but I can't resist supplying a more basic proof that does not require knowledge of the properties of the divisor function.
Theorem: If n is not a square, then it has an even number of divisors.
Proof: Suppose x is a divisor of n, i.e. there is an integer y such that xy = n. Then y is also a divisor of n. (Here is the critical step.) Since n is not a square, x is not equal to y. So we have established a one-to-one correspondence between pairs of divisors of n, hence the number of divisors must be even.
Corollary (the contrapositive): If n has an odd number of divisors, then it is a square.
9. ## Re: Perfect square s, odd number of factors
Thanks, both of you guys! | 2015-07-01T17:00:33 | {
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http://math.stackexchange.com/questions/778495/i-have-the-pattern-1-2-3-4-5-6-but-i-need-the-formula-for-it | I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it
I'm writing some software that takes a group of users and compares each user with every other user in the group. I need to display the amount of comparisons needed for a countdown type feature.
For example, this group [1,2,3,4,5] would be analysed like this:
1-2, 1-3, 1-4, 1-5
2-3, 2-4, 2-5
3-4, 3-5
4-5
By creating little diagrams like this I've figured out the pattern which is as follows:
Users - Comparisons
2 - 1
3 - 3 (+2)
4 - 6 (+3)
5 - 10 (+4)
6 - 15 (+5)
7 - 21 (+6)
8 - 28 (+7)
9 - 36 (+8)
I need to be able to take any number of users, and calculate how many comparisons it will take to compare every user with every other user.
Can someone please tell me what the formula for this is?
-
MJD is right, of course, but... What kind of comparison are you doing? Maybe a sort would do? If there are many users, it can be faster, depending on your needs. – jca May 2 '14 at 15:25
You're looking for the cardinal of the complete graph $K_n$, which is actually $\frac{n(n-1)}{2}$ – Yann Hamdaoui May 2 '14 at 15:25
Wow, so simple, thanks MJD! And Jean, it's nothing to do with sorting, that would be about as simple as myList.Sort(myComparer). – Owen May 2 '14 at 15:28
This is the handshake problem. mathworld.wolfram.com/HandshakeProblem.html – Shoe May 2 '14 at 16:07
@Bobson I'm not sure BigO has much to do with this question. There seems to be no mention of asymptotic behaviour at all here. He simply wants a closed form for this recurrence relation (which I'm surprised nobody has pointed out, are the triangular numbers) – Cruncher May 2 '14 at 17:25
The sum of $0+\cdots + n-1$ is $$\frac12(n-1)n.$$
Here $n$ is the number of users; there are 0 comparisons needed for the first user alone, 1 for the second user (to compare them to the first), 2 for the third user, and so on, up to the $n$th user who must be compared with the $n-1$ previous users.
For example, for $9$ people you are adding up $0+1+2+3+4+5+6+7+8$, which is equal to $$\frac12\cdot 8\cdot 9= \frac{72}{2} = 36$$ and for $10$ people you may compute $$\frac12\cdot9\cdot10 = \frac{90}2 = 45.$$
-
You want to know how many ways there are to choose $2$ users from a set of $n$ users.
Generally, the number of ways to choose $k$ elements from a set of order $n$ (that is, all elements in the set are distinct) is denoted by $$\binom{n}{k}$$
and is equivalent to $$\frac{n!}{(n-k)!k!}$$
In the case of $k=2$ the latter equals to $$\frac{n!}{(n-2)!2!}=\frac{n(n-1)}{2}$$
which is also the sum of $1+2+...+n-1$.
-
The following way to getting the solution is beautiful and said to have been found by young Gauss in school. The idea is that the order of adding $1+2+\cdots+n=S_n$ does not change the value of the sum. Therefore:
$$1 + 2 + \ldots + (n-1) + n=S_n$$ $$n + (n-1) + \ldots + 2 + 1=S_n$$
Adding the two equations term by term gives
$$(n+1)+(n+1)+\ldots+(n+1)=2S_n$$
so $n(n+1)=2S_n$. For $n$ persons, there are $S_{n-1}$ possibilities, as others answers have shown already nicely.
-
The discrete sum up to a finite value $N$ is given by,
$$\sum_{n=1}^{N} n = \frac{1}{2}N(N+1)$$
Proof:
The proof by induction roughly boils down to:
$$S_N = 1+ 2 +\dots+N$$
$$S_{N+1}= 1+ 2 + \dots + N + (N+1) = \underbrace{\frac{1}{2}N(N+1)}_{S_N} + (N+1)$$
assuming that the induction hypothesis is true. The right hand side:
$$\frac{N(N+1)}{2}+(N+1)=\frac{(N+1)(N+2)}{2}$$
which is precisely the induction hypothesis applied to $S_{N+1}$.
Just for your own curiosity, the case $N=\infty$ is of course divergent. However, with the use of the zeta function, it may be regularized to yield,
$$\sum_{n=1}^{\infty}n = \zeta(-1)=-\frac{1}{12}$$
-
$$N=2:\ 1 + 2 = (1 + 2) = 1\times3$$
$$N=4:\ 1 + 2 + 3 + 4 = (1 + 4) + (2 + 3) = 2\times5$$
$$N=6:\ 1 + 2 + 3 + 4 + 5 + 6 = (1 + 6) + (2 + 5) + (3 + 4) = 3\times7$$
$$N=8:\ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8= (1 + 8) + (2 + 7) + (3 + 6)+ (4 + 5) = 4\times9$$
More generally, $N/2\times(N + 1)$.
For odd $N$, sum the $N-1$ first terms (using the even formula) together with $N$, giving $(N-1)/2\times N+N=N\times(N+1)/2$.
-
Except that you should never count yourself, so you should be starting at 0 instead of 1. You'll have to take your result and subtract N from it. So for N = 2 you take your 1 x 3 = 3 - N = 3 - 2 = 1, which matches the table listed by OP. – corsiKa May 2 '14 at 17:31
@corsiKa: the formula explicitly shows the sum from 1 to $N$ inclusive (triangular numbers) and is perfectly correct. It is NOT the formula for the number of comparisons between $N$ users. – Yves Daoust May 2 '14 at 21:38
I think you're missing my point. I'm not saying your math is wrong at all. I'm saying it doesn't match what the OP is looking for. If you are not giving he formula for the number of comparisons between N users, then you aren't answering the OPs question, which makes this an incorrect answer for this question. – corsiKa May 2 '14 at 21:44
The sum must be taken from $1$ to $Users-1$ inclusive. – Yves Daoust May 2 '14 at 21:51
Yes. And the sum 1 + 2 + ... n-1 is n*(n-1)/2 while your answer says n*(n+1)/2. – corsiKa May 2 '14 at 22:28
Here is another way to find the sum of the first $n$ squares that generalizes to sums of higher powers.
$(k+1)^2 - k^2 = 2k+1$
$\sum_{k=1}^n ( (k+1)^2 - k^2 ) = \sum_{k=1}^n (2k+1)$
$(n+1)^2 - 1^2 = 2 \sum_{k=1}^n k + n$
$\sum_{k=1}^n k = \frac{ (n+1)^2 - 1 - n }{2} = \frac{n^2+n}{2}$
-
This is kind of a pseudo code:
Say you have n number of people, and you labeled them.
for i in (1,2,3,...,n), person i need to compare with all the people who has a number larger (strictly), so person i need to compare (n-i) times.
so adding up would be (n-1) + (n-2) + ... + 3 + 2 + 1...
which would be the sum from 1 to (n-1)
- | 2015-01-26T22:49:55 | {
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https://math.stackexchange.com/questions/4347275/average-square-gain-of-a-matrix-over-all-possible-vectors | # Average square gain of a matrix over all possible vectors
Let $$A \in \mathbb{C}^{n \times n}$$ and for a given $$x \in \mathbb{C}^n$$, the square gain of $$A$$ is $$\lVert A x \rVert^2 / \lVert x \rVert^2$$ where the norm is the usual Euclidean norm. I want to calculate the average gain over all possible $$x \in \mathbb{C}^n$$. Obviously, the gain is independent from the "length" of $$x$$, so without losing generality we can assume $$\lVert x \rVert = 1$$.
Now we need a method to parameterize all $$\lVert x \rVert = 1$$. The first thing that comes to my mind is using the singular value decomposition of $$A$$. Let $$A = U \Sigma V^*$$ be the SVD of $$A$$, so
$$x = \sum_{i=1}^n \alpha_i v_i ~~~~\text{such that}~~~~ \sum_{i=1}^n \alpha_i^2 = 1$$
characterize all possible vectors, where $$U = [u_1 ~~ \dots ~~ u_n], \Sigma = \operatorname{diag}\{\sigma_i\}$$ and $$V = [v_1 ~~ \dots ~~ v_n]$$.
Let's first consider $$n=2$$. We can select $$\alpha_1 := \cos \theta$$ and $$\alpha_2 := \sin \theta$$. So, the average gain can be calculated as
\begin{align*} \frac{1}{2 \pi} \int_0^{2\pi} \lVert A v_1 \cos \theta + A v_2 \sin \theta \rVert^2 d\theta &= \frac{1}{2 \pi} \int_0^{2\pi} \lVert \sigma_1 u_1 \cos \theta + \sigma_2 u_2 \sin \theta \rVert^2 d\theta \\ &= \frac{1}{2 \pi} \int_0^{2\pi} \left( \sigma_1^2 \cos^2 \theta + \sigma_2^2 \sin^2 \theta \right) d\theta \\ &= \frac{\sigma_1^2 + \sigma_2^2}{2} \end{align*}
This result makes intuitive sense since the average square gain is the average of minimum and maximum square gains. So, we can hypotesize that the average square gain is $$\lVert A \rVert_F^2 / n$$ where $$\lVert \cdot \rVert_F$$ is the Frobenius norm. Now we can test this idea for $$n=3$$ using spherical coordinates. So let $$\alpha_1 := \cos \theta, \alpha_2 := \sin \theta \cos \phi, \alpha_3 := \sin \theta \sin \phi$$. Using the same steps we get
\begin{align*} &\frac{1}{2 \pi^2} \int_0^{\pi} \int_0^{2\pi} \lVert A v_1 \cos \theta + A v_2 \sin \theta \cos \phi + A v_3 \sin \theta \sin \phi \rVert^2 d\phi d\theta \\ &~~~~~~~~= \frac{1}{2 \pi^2} \int_0^{\pi} \int_0^{2\pi} \left( \sigma_1^2 \cos^2 \theta + \sigma_2^2 \sin^2 \theta \cos^2 \phi + \sigma_3^2 \sin^2 \theta \sin^2 \phi \right) d\phi d\theta \\ &~~~~~~~~= \frac{2\sigma_1^2 + \sigma_2^2 + \sigma_3^2}{4} \end{align*}
which is not what we expected. Coming back to $$n=2$$ case, we could have also selected $$\alpha_1 = t$$ and $$\alpha_2 = \sqrt{1-t^2}$$ where $$t \in [0,1]$$. In this case, we get
$$\int_0^1 \left(\sigma_2^2 + (\sigma_1^2 - \sigma_2^2) t^2 \right) dt = \frac{\sigma_1^2 + 2 \sigma_2^2}{3}$$
which is different from our first attempt.
So, why are the results are different? Is there a "correct" way of selecting the parameters to get a "nice" formula like $$\lVert A \rVert_F^2 / n$$?
• The first parametrization induces a uniform measure (Haar measure) on the unit circle, the second one doesn't. Jan 2 at 20:39
• Possibly related: this and this and this Jan 2 at 21:54
• Rewrite the problem as $$\frac{\|Ax\|^2}{\|x\|^2} \;=\; A^TA:\left\langle \frac{xx^T}{x^Tx}\right\rangle \;=\; A^TA:\oint nn^T d\Omega$$ Then utilize the results of this answer to confirm your initial result $$\frac{\|Ax\|^2}{\|x\|^2} \;=\; A^TA:\left(\frac In\right) \;=\; \frac{A:A}{n} \;=\; \frac{\big\|A\big\|_F^2}{n}$$ The caveat is that the linked post is for ${\mathbb R}^n \;$
– greg
Jan 4 at 9:42
Equip $$S^{n - 1} = \{x \in \mathbb{R}^n : |x| = 1\}$$ with it's usual surface measure, normalized to be a probability measure. Then you seek $$\int_{S^{n - 1}}|Ax|^2\,dS(x)$$.
As for selecting parameters, any parameterization of $$S^{n - 1}$$ will work. You have to apply the formula for the surface integral though:
If $$\phi : O \to U \subset {S}^{n - 1}$$ is a $$C^1$$ parameterization and $$f : S^{n - 1} \to \mathbb{R}$$ vanishes off $$U$$, then $$\int_{U}f(x)\,dS(x) = \frac{1}{\text{Area}(S^{n - 1})}\int_{O}f(\phi(x))\sqrt{\det D\phi(x)^T D\phi(x)}\,dx.$$ I think you forgot the $$\sqrt{ \det D\phi(x)^T D\phi(x)}$$. Luckily, this problem is simple enough that we don't need to use coordinates.
Write the SVD of $$A$$ as $$A = UDV^T$$. Then $$|Ax|^2 = |DV^Tx|^2$$. So by invariance of the measure under orthogonal transformations, $$\int_{S^{n - 1}}|Ax|^2\,dS(x) = \int_{S^{n - 1}}|DV^Tx|^2\,dS(x) = \int_{S^{n - 1}}|Dx|^2\,dS(x).$$ Now note $$|Dx|^2 = (D^2x, x) = \sigma_1^2|x_1|^2 + \dots + \sigma_n^2|x_n|^2$$. Hence $$\int_{S^{n - 1}}|Dx|^2\,dS(x) = \sum_{i = 1}^{n}\sigma_i^2\int_{S^{n - 1}}|x_i|^2\,dS(x).$$ Note that $$a_i = \int_{S^{n - 1}}|x_i|^2\,dS(x)$$ is independent of $$i$$ by invariance of the measure under orthogonal transformations. Moreover, $$a_1 + \dots + a_n = \int_{S^{n - 1}}|x|^2\,dS(x) = 1.$$ Hence $$a_i = \frac{1}{n}$$. In conclusion, $$\int_{S^{n - 1}}|Ax|^2\,dS(x) = \frac{\sigma_1^2 + \dots + \sigma_n^2}{n}.$$
Edit: Here is why $$a_i = \int_{S^{n - 1}}|x_i|^2 dS(x)$$ is independent of $$i$$. Fix $$i$$. Note that \begin{align} \int_{S^{n - 1}}|x_i|^2\,dS(x) &= \int_{S^{n - 1}}|(x, e_i)|^2\,dS(x). \end{align} Now pick an orthogonal transformation $$R$$ such that $$Re_i = e_1$$. Since $$(x, e_i) = (Rx, Re_i) = (Rx, e_1)$$, it follows that $$\int_{S^{n - 1}}|x_i|^2\,dS(x) = \int_{S^{n - 1}}|(Rx, e_1)|^2\,dS(x) = \int_{S^{n - 1}}|(x, e_1)|^2\,dS(x),$$ Where the last equality follows from invariance of the measure under orthogonal transformations. So $$a_i = a_1$$.
The definition of the "surface measure" I am using here is the one that works for any $$C^1$$ $$m$$-dimensional surface $$M \subset \mathbb{R}^k$$. The measure $$\mu$$ is a Borel measure on $$M$$ defined in such a way that if $$f : M \to [0, \infty]$$ is a measurable function vanishing off the image $$U \subset M$$ of a coordinate parameterization $$\phi : O \to U$$, $$O$$ open in $$\mathbb{R}^m$$, then $$\int_{M}f\,d\mu = \int_{U}f\,d\mu = \int_{O}f(\phi(x))\sqrt{g(x)}\,dx, \hspace{20pt} g(x) = \det D\phi(x)^T D\phi(x).$$ Here I used it for $$M = S^{n - 1}$$ (I also normalized it to $$1$$). From this you can prove the assertion that the measure on $$S^{n - 1}$$ is invariant under orthogonal transformations, that is, if $$R^TR = I$$, and $$f : S^{n - 1} \to [0, \infty]$$ is measurable, then $$\int_{S^{n - 1}}f(Rx)\,d\mu(x) = \int_{S^{n - 1}}f(x)\,d\mu(x).$$ A way to prove this is to first prove it for $$f$$ vanishing off of a coordinate patch, in which case it is just a computation. Then the fact that $$M$$ is always a countable union of such patches implies the general case.
The above measure is also defined on a general Riemannian manifold $$M$$, in which case $$D\phi(x)^T D\phi(x)$$ is replaced with the coordinate matrix $$G(x)$$ of the metric tensor of $$M$$ (the above definition is the special case when the metric tensor is the dot product).
• After some reading about surface measure and surface integral I understand the answer and it is clear, so thanks. As I understand it $dS$ has to be "uniform" under the selected parameterization (Haar measure?). But as you showed that we don't need to specify the parameterization explicitly for this problem as long as we assume it is uniform. I also understand intuitively that all $a_i$ must be equal. But I'm having trouble showing it rigorously, could you elaborate on this point? Jan 3 at 15:05
• @obareey You can use any parameterization for integrating, as long as you apply the coordinate formula for integration. I added elaboration on the surface measure. It is defined on any Riemannian manifold. The word "uniform" seems to be synonymous with the surface measure. Haar measure is a different concept than surface measure, though it is a relevant in relation to the surface measure on the sphere. Jan 4 at 21:37 | 2022-08-12T00:16:00 | {
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https://math.stackexchange.com/questions/3485751/dividing-a-number-according-to-proportions | # Dividing a number according to proportions
Let's say I want to divide a number $$g$$ according to proportions defined by a set of numbers, e.g. $$\{2,1,1\}$$, let's call them $$x_i$$. Take $$g=100$$. By proportionally I mean multiplying $$g$$ by each $$x_i$$ divided by $$\sum{x_i}$$.
So the consecutive results (denoted by $$r_k$$) are 50, 25, 25 in this example. However when calculating the second result number, I can subtract the first result from $$g$$ and divide by 2 instead of 4, because I ignore $$x_1=2$$ , like this: $$(100-50) * 1 / 2 = 25$$, which is equivalent to $$100 * 1 / 4$$.
How to prove that these two ways are equivalent and give the same result?
$$r_{k} = g \dfrac{x_k}{\sum\limits_{i=1}^{n}x_i}$$
This method of calculating the result $$r_k$$ is equivalent to:
$$r_k = (g - \sum\limits_{i=1}^{k-1} r_i) \frac{x_k}{\sum\limits_{i=k}^{n}x_i}$$
However these sums look overly complicated to me, is there any way to simplify this and show that they are equal?
• You can get displayed equations by enclosing them in double instead of single dollar signs. You can get proper parentheses (and other paired delimiters) that adjust to the size of their content by preceding them with \left and \right. Dec 23 '19 at 22:52
Well, you saw that this is the case; I think it would be more enlightening to write a proof by trying to formalize how you saw it rather than by algebraic manipulations; but to answer the question:
I'll assume $$x_k\gt0$$ for all $$k$$. The proof is by induction. The equations are clearly equivalent for $$k=1$$. Assume that they are equivalent up to $$k-1$$. Multiply by both denominators and cancel $$x_k$$. Then we want to show
$$g\sum_{i=k}^nx_i=\left(g-\sum_{i=1}^{k-1}r_i\right)\sum_{i=1}^nx_i\;.$$
By the induction hypothesis, we can use the upper equation for $$r_i$$ so this becomes
$$g\sum_{i=k}^nx_i=\left(g-\sum_{i=1}^{k-1}\frac{gx_i}{\sum_{j=1}^nx_j}\right)\sum_{i=1}^nx_i\;,$$
and multiplying out the right-hand side and canceling the sum transforms this to
$$g\sum_{i=k}^nx_i=g\sum_{i=1}^nx_i-\sum_{i=1}^{k-1}gx_i\;,$$
which is obviously correct.
• "By the induction hypothesis, we can use the upper equation" - just to clarify, I can plug in either equation there and the proof will be valid, because they both are assumed to be equivalent for all cases up to k-1, right? Dec 23 '19 at 23:39
• @user5539357: Correct. Dec 23 '19 at 23:39 | 2021-09-18T12:53:04 | {
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http://math.stackexchange.com/questions/173016/inscribing-a-rhombus-within-a-convex-quadrilateral/173039 | # Inscribing a rhombus within a convex quadrilateral
I was wondering if it is possible to inscribe a rhombus within any arbitrary convex quadrilateral using only compass and ruler? If possible, could you describe the method?
If not could you give an example of a convex quadrilateral which one can not inscribe a rhombus within?
-
There is definitely an inscribed rhombus. I do not have a nice geometric straightedge and compass construction, but can prove constructibility. – André Nicolas Jul 19 '12 at 21:56
EDIT, Saturday July 21: I was wondering if an inscribed rhombus might be required to have edges parallel to the diagonals of the original convex quadrilateral. But this is not always the case. Take a square, on each edge mark a point the same distance counterclockwise from the nearest vertex. This makes an inscribed square, and is one of the oldest proofs-without-words of the Pythagorean Theorem.
ORIGINAL: Given a convex quadrilateral ABCD, the main observation is that the new quadrilateral made up with vertices at the midpoints of the four sides is automatically a parallelogram. So we just need to adjust the relative lengths of the sides, we will make an inscribed rhombus with sides parallel to the midpoint-parallelogram. That is, we make a parallelogram RSTU such that its sides RS and RU have the same length, thus forcing it to be a rhombus. To emphasize, the sides of the rhombus RSTU are parallel to the diagonals AC and BD of the original quadrilateral. I think I will skip the proof that it is inscribed, bunch of similar triangles involved. Many.
Take a quadrilateral such that AB and CD intersect in a point P.
Through D, draw a line $L$ parallel to the diagonal AC.
Using a compass, draw a point B' on $L$ such that DB = DB'.
Draw the segment BB' and let it intersect line CD at B''.
Draw a line $M$ through B'' that is parallel to AC and $L,$ and let $M$ intersect the other diagonal BD at point Q.
Draw the line through PQ and let it intersect edge AD at R. R is the first vertex of our rhombus.
The edges are parallel to the diagonals AC and BD. So, from R draw a line parallel to BD, let it intersect AB in point S. From R, draw another line but parallel to AC this time, let it intersect edge CD in point U. Finally, draw the appropriate parallel lines to the final point T on BC. The quadrilateral RSTU is an inscribed rhombus.
I see, there is a way to fiddle with this if we consider AB and CD parallel instead. Not to worry.
Anyway, give it a try. You may need to adjust the figure so that everything comes out on one page, lines that are supposed to intersect really do, and so on. I did what I describe with compass, straightedge, T-square (actually an L-square), and a nice parallel ruler
-
Thanks for your explanation. Would you like me to upload the series of images I've made? – Honest Abe Jul 20 '12 at 3:02
@HonestAbe, yes, please. – Will Jagy Jul 20 '12 at 3:06
@HonestAbe, that is perfect. It took a while to compare with my drawing. All together, no explicit use of perpendiculars required if we accept a tool to make parallel lines, and just a single explicit circle to get length DB' = DB. Thank you. EDIT: I did not notice, you actually started with just a quadrilateral and made about seven images, one step at a time. This is great. – Will Jagy Jul 20 '12 at 3:37
Wow, this is great! Thank you for your time an effort. I've been looking for this for a long time. Thank you all! – Reza Sadr Jul 20 '12 at 7:32
@HonestAbe Thank you very much. You don't believe how mind blowing is this for me. :) – Reza Sadr Jul 20 '12 at 17:20
I bought a home scanner. These are my drawings, the first one actually shows the proof of something. In the first two, the line PQ intersects the original quadrilateral on the side away from P, that it how I thought of it. I also deliberately made line PQR dotted so as not to interfere with other parts of the drawing. Later I made a third drawing with pencil construction lines so that I could erase those; that one did after the approximate proportions of HonestAbe's pictures, plus I used an ink nib for the compass to make the circle visible.
try again: drawing A:
drawing B:
drawing C:
-
The following is a proof of constructibility, with some standard details missing, mostly the fact that the arithmetical operations and square root can be carried out by straightedge and compass.
In principle, one can use the proof to give instructions on how to carry out the construction. However, the construction one gets has little real geometric content. There is undoubtedly an elegant geometric construction. But the algebraic approach used below has the nice feature of being essentially automatic. If one has a certain amount of background, a glance at the problem shows that an inscribed rhombus is constructible.
We are given a convex quadrilateral $ABCD$, with the vertices going say counterclockwise. Let $P$, $Q$, $R$, $S$ be points on the line segments $AB$, $BC$, $CD$, and $DA$.
Lemma: Suppose that $AP:BP=AS:DS$ and $CR:DR=CQ:BQ$. Then the lines $PS$ and $RQ$ are parallel.
Proof: They are both parallel to $BD$.
Corollary: If $AP:BP=AS:DS$ and $BP:BQ=DR:CR$ then $PQRS$ is a parallelogram.
In particular (not relevant here, but cute) if $P$, $Q$, $R$ and $S$ are bisectors of the sides they lie on, then one gets the unexpected result that $PQRS$ is a parallelogram.
A little play will show that if we choose $P,Q,R,S$ as in the corollary, but with $AP$ small, then $PS$ will be smaller than $PQ$. And if we choose $P$ so that $BP$ is small, then $PS$ will be larger that $PQ$. So a choice somewhere in between is just right, we get $PS=PQ$, and therefore a rhombus.
Now we leave classical geometry. Choose a coordinate system with $A$ at the origin, and with $B$ say at $(1,0)$. (It doesn't matter.) Points $C$ and $D$ then have "known" coordinates, and all the lines of our quadrilateral have known linear equations, with coefficients constructible from our given points.
Let $P=(t,0)$. Find the equation of the line through $(t,0)$ that is parallel to $BD$. We can now find the coordinates of the point $S=(s_1(t),s_2(t))$ where this line meets $AD$. The coordinates of $S$ are linear in $t$.
Find also the equation of the line through $(t,0)$ that is parallel to $AC$. We can now find the coordinates of the point $Q=(q_1(t),q_2(t))$ where this line meets $BC$. The coordinates of $Q$ are linear in $t$.
The condition $PS=PQ$ then becomes $$(t-s_1(t))^2 +(s_2(t))^2=(t-q_1(t))^2+(q_2(t))^2.\tag{1}$$ This is a quadratic equation, with coefficients that are constructed using addition, subtraction, multiplication, and division from coordinates of our given points.
The ordinary arithmetical operations can be done by straightedge and compass, as can square root. So the geometrically relevant solution of Equation $(1)$ is constructible from our given points. Now that we have $P$, the rest of the construction is done by drawing parallels.
-
there is a short construction. the proof is similar triangles out the wazoo. – Will Jagy Jul 20 '12 at 0:21
@WillJagy: Will be interested. I did not even look for a construction (and likely would not have found it if I looked). My post was a plug for Fermat-Descartes. – André Nicolas Jul 20 '12 at 0:29
There are now several diagrams in my answer, showing the steps of the construction. – Will Jagy Jul 20 '12 at 4:31
Thank you for this. Very informative and useful. – Reza Sadr Jul 20 '12 at 8:03 | 2015-08-02T20:55:21 | {
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https://math.stackexchange.com/questions/2477040/proof-that-if-p-mid-g-and-g-is-finite-abelian-it-has-an-element-of-order-p | # Proof that if $p \mid G$ and $G$ is finite abelian it has an element of order $p$
I want to prove the following lemma.
Let $p$ be a prime number and $G$ a finite abelian group such that $p\mid\#G$, then $G$ has an element of order $p$.
Here's the proof my professor presented.
Let $|G|=pm$ for some $m\in\mathbb{N}$. We will prove the result by induction on $m$.
The base case, $m=1$, is clear (every group of prime order is cyclic).
For the inductive step, choose $x\in G$ such that $\text{ord}(x)=t>1$.
If $p\mid t$ (don't we need to require that $p\neq t$?), then $\text{ord}\big((x^{\frac{t}{p}})^{p}\big)=1$ which implies that $\text{ord}(x^{\frac{t}{p}})=p$, and we are done.
Now suppose that $p\nmid t$. Since $G$ is a abelian then every subgroup is normal, in particular $\langle x\rangle\trianglelefteq G$ and $|G/\langle x\rangle|=\frac{pm}{t}$.
By hypothesis, $\exists y\in G$ such that $\text{ord}(y\langle x\rangle)=p$. (why?) Since for the homomorphism $f:G\rightarrow G/\langle x\rangle$ we have that $\text{ord}\big(f(y)\big)\mid \text{ord}(y)$ it follows that $\text{ord}(y\langle x\rangle)=p\mid \text{ord}\;y$ and we go back to the case where $p$ divides the order or a nontrivial element of $G$.
My questions:
1) the first questions appear in bold in the above text.
2) I don't really get the structure of this induction proof. When the inductive step is presented I don't see in what way we assume that the results holds for $m$ and how we show it is true for $m+1$.
My apologies if these questions are too basic but I am a former physics student and lack some background and experience with proofs.
• Is $p$ a prime number? – paw88789 Oct 17 '17 at 16:20
• Yes, I forgot to include this. Thanks. – aadcg Oct 17 '17 at 16:21
• For the first bold part: Not needed since if $p=t$ all the statements that follow hold by definition of $x$. $ord((x^{t/p})^p)=ord(x^p)=ord(e)=1$. For the second bold part: Observe that $G/(x)$ has order $p(m/t)$ and $m/t < m$. Therefore, $G/(x)$ has an element of order $p$. That element is being called $y(x)$, the coset of some element $y\in G$. – Hellen Oct 17 '17 at 16:29
• In finite abelian groups,Converse of Lagrange's theorem holds good. – Sumit Mittal Oct 17 '17 at 17:26
don't we need to require that $p \neq t$?
Not that I can see from your argument. If $\operatorname{ord}(x) = p$, then there is nothing to prove. That said, the way you currently have structured your argument, you will need to ensure $p \nmid t$ when applying the inductive hypothesis to the group $G / \langle x \rangle$.
One way around this would be to replace your current definition of $x$ with something more specific. For instance, you could say
Choose any $x \in G \setminus \{ 1 \}$. If $p \mid \operatorname{ord}(x)$, say $\operatorname{ord}(x) = pq$ for $q \in \mathbb{N}$, then by definition $y = x^q \in G$ is an element of order $p$. Conversely, suppose $p \nmid \operatorname{ord}(x)$ [$\ldots$]
and take things from there. This seems to be what you mean, but calling out the two possible cases more explicitly goes a long way towards clarifying the rest of the argument (for both the reader and you!).
By hypothesis, $\exists y \in G$ such that $\operatorname{ord}(y\langle x\rangle)=p \operatorname{ord}(y\langle x\rangle)=p$. (why?)
If you assume $p \nmid t$, then this is because $|G/\langle x \rangle|$ is of the form $pm'$ where $m' < m$ (and so the inductive hypothesis applies).
Regarding your last question: It sounds like you're assuming that this is a proof by weak induction, when this is really a proof by strong induction. Suppose $P$ is a property of the natural numbers, and you know that $P(1)$ is true. The difference in proving that $P(N)$ is true for all $N \in \mathbb{N}$ using weak and strong induction is the following:
• Using weak induction, you assume for a fixed $N \in \mathbb{N}$ that $P(N)$ holds and show that this implies $P(N + 1)$.
• Using strong induction, you assume for a fixed $N \in \mathbb{N}$ that $P(n)$ holds for all $n \leq N$ and show that these statements together imply $P(N + 1)$ holds.
See this set of notes for examples of both types of proofs. Each of these is a valid form of induction.
Your proof relies on strong induction, and you can see this from how you apply your inductive hypothesis. First, you choose an arbitrary $x \in G$ with $x \neq 1$ and $p \nmid t := \operatorname{ord}(x)$. You then apply the inductive hypothesis to the group $G/\langle x \rangle$. Supposing that $|G/\langle x \rangle| = pk$, you don't know the value of $k$, since you don't know enough about $x$. All you know is that it is less than $m$ where $|G| = pm$. To cover all bases, you have to assume the claim holds for all finite abelian groups of order $pn$ for $n < m$, not just for those of order $p(m - 1)$.
• Very clear and informative. – aadcg Nov 10 '17 at 0:25
1) If $p=t$, $t/p=1$, so $x$ has order $p$ anyway and is the element we want.
2) Your second piece of bold is where the inductive step is used: the key point is that $G/\langle x \rangle$ is a group of order $pm/t$, and if $p \not\mid t$, then there is an integer $n$, smaller than $m$, so that $pm/t = pn$. The induction hypothesis then says the group $G/\langle x \rangle$ has an element of order $p$; and then the structure of the quotient group means that there is $y \in G$ so that this element can be written as $y\langle x \rangle$. | 2020-05-30T21:51:42 | {
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https://math.stackexchange.com/questions/3059254/how-to-determine-the-minimum-height-of-a-roller-coaster-provided-an-acceleration | # How to determine the minimum height of a roller coaster provided an acceleration as a threshold parameter?
I've been going in circles (no pun intended) with this problem for several hours and yet I can't seem to find what am I doing wrong or misunderstanding, hence I hope somebody can help me with this problem.
It states as follows:
A 1500-kg roller coaster (including passengers) passes point A at $$3\,\frac{m}{s}$$(see figure 1 as a reference). Due to safety concerns, you must design the track so that at point B the passengers do not experience an upward force that exceeds 4g. If the arc at B is circular with radius 15 m,(a) determine the minimum value of h that satisfies this requirement,and (b) determine the speed of the coaster at C.
Figure 1. Illustrates the problem as it appears in my book
Part a. In my attempt to solve the problem. I thought that the key was to use the conservation of mechanical energy in both points. In A and B. As this is described in Figure 2. I also thought that because the Radius of the circle is given then there is an implicit hint that the upward acceleration they're talking about is the centripetal acceleration (the one which is pulling to the center of the circle, hence upwards), so that this must not exceed 4g. As a consequence I equated both expressions in order to obtain the speed v_b so with that all there is left to solve is the height which is given by the potential energy U reached at that point.
For better visualizing of the situation I made an sketch of all the forces which I could find, seen in Figure 2.
The above paragraph is summarized in the following equations:
$$E_a=E_b$$
$$\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_b+mgh_b$$
Mass cancels in both expressions and multiplied by $$2$$:
$$v^2_a+2gh_a=v^2_b+2gh_b$$
$$h_b=\frac{v^2_a+2gh_a-v^2_b}{2g}$$
Edit:
Apparently there is an acting force which is going outwards the circle, this summed up with Newton's second law would yield:
$$\frac{mv^2_b}{R}+mg=m\left(a_{max}\right)$$
(Here's where I use the fact of the "threshold of $$4g$$)
$$\frac{mv^2_b}{R}+mg=m\left(4g\right)$$
$$v^2_b=3gR$$
Inserting this final expression into the one obtained earlier:
$$h_b=\frac{v^2_a+2gh_a-3gR}{2g}$$
Then all is left is to insert the values given and make the computation of the height:
$$h_b=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m-3\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$$
$$h_b=\frac{\left(3\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\times 40\,m-3\times9.8\,\frac{m}{s^2}\times 15\,m}{2\times 9.8\,\frac{m}{s^2}}$$
Computing this expression yields:
$$h_{b}=17.95918367\,m$$
Which results with what my book describes as $$h_b\approx 17.96\,m$$
Part b. Is very straightforward. It was just comparing both Energies from the top in A and lower in C as stated below.
This process is illustrated in Figure 3.
$$E_a=E_c$$
$$\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_c+mgh_c$$
Again masses cancels and multiplying by $$2$$:
$$v^2_a+2gh_a=v^2_c+2gh_c$$
$$v^2_c=v^2_a+2gh_a-2gh_c$$
$$v_c=\sqrt{v^2_a+2gh_a-2gh_c}$$
$$v_c=\sqrt{v^2_a+2g\left(h_a-h_c\right)}$$
Finally inserting the given values into the above equation:
$$v_c=\sqrt{\left(3.0\,\frac{m}{s}\right)^2+2\times 9.8\,\frac{m}{s^2}\left(40\,m-20\,m\right)}$$
Which results into:
$$v_c= 20.02498439\,\frac{m}{s}$$
This answer does check with what my book lists as the correct answer, $$20.02\,\frac{m}{s}$$ rounded to two decimals.
Although the procedure which I intended to do initially seems reasonable, I'm still dumbfounded on why part a doesn't check, so far I believe I haven't overlooked anything, but if I did, please explain me where is the mistake and some conceptual basis as maybe I could have misinterpreted something.
• Are you certain that this is the right site for your problem? Have you tried physics.SE? It might be off-topic there, though, on account of their stance on homework (and homework-like) problems. I don't know. – Arthur Jan 2 at 9:13
• @Arthur I thought that maths was an appropiate site as my question is about if my use of mathematics was right given the conditions of the problem. I am not very experienced in the field so, verification of my reasoning is needed. – Chris Steinbeck Bell Jan 2 at 9:22
• Where did you get $v_a = 3$ m/s from? – caverac Jan 2 at 9:25
• @caverac Sorry it was a typo. It was a given data from the problem, now I've corrected that mistake. – Chris Steinbeck Bell Jan 2 at 9:29
• When you calculate $h_b$ you use $R = 10$ m, is that also a typo? – caverac Jan 2 at 9:32
The downward force felt by a passenger of mass $$m$$ at point $$B$$ has two components:
• The weight of the passenger, $$mg\downarrow$$
• The radial force, $$\frac{mv_b^2}R\downarrow$$
Your equation gets modified as $$mg+\frac{mv_b^2}R=4mg$$, solving which gets you $$h_b=17.96$$m.
Edit. Let us first discuss the answer in ground frame. As you have correctly pointed out, a radially upward acceleration with magnitude $$\frac{mv_b^2}R$$ is required to maintain circular motion of the roller coaster at point $$B$$. The required 'centripetal force' is delivered by the normal reaction at point $$B$$. This is because there is no agent exclusively exerting the centripetal force; the normal reaction and gravitational force are the only forces acting on the coaster.
As seen from the free body diagram of the coaster, $$N-mg=\frac{mv_b^2}R$$ Since the upward force on the coaster is $$N$$, you require $$N\le4mg\implies\frac{mv_b^2}R+mg\le4mg$$
Now, for the solution in the frame of reference of the roller coaster. Imagine being seated on the coaster. You would perceive yourself to be at rest without acceleration while the world around would be moving circularly. You would also feel more compressed at $$B$$: this suggests that in your reference frame, a certain phantom force other than your weight acts downward. This force is called the pseudo-force; 'pseudo', because it doesn't actually exist. It is merely an arrangement to solve in accelerated frames of reference.
Recall that the centripetal acceleration has direction opposite to that of its corresponding pseudo-force. In general too, for an object with acceleration $$\vec a$$ in ground frame, the pseudo-force acting on the object in its frame of reference is $$-m\vec a$$; that is, it has magnitude $$ma$$ and direction opposite to $$\vec a$$. This gives $$N=mg+\frac{mv_b^2}R$$ since in this frame, the coaster is at rest so the forces cancel. Once again, the upward force is $$N$$, which gives you the same result as the ground frame.
• Shubham.Yes, mv^2/R = Reaction(upward) -mg.+) – Peter Szilas Jan 2 at 9:45
• @Shubham John Why the radial force goes downwards and not upwards?. Can you explain to me this part?. I thought that the radial force goes into the circle not the other way around. Perhaps am I mistaking this? – Chris Steinbeck Bell Jan 2 at 9:50
• @ChrisSteinbeckBell The centrifugal force is a pseudo-force and has the direction opposite to that of the centripetal acceleration. – Shubham Johri Jan 2 at 9:55
• Think about it: at the bottom of the coaster, doesn't one feel compressed? – Shubham Johri Jan 2 at 10:00
• @Shubham John I am not familiar with the concept of pseudo-force. Checking with my book it states that "centripetal acceleration" is inward seeking. Yes indeed when you're at the bottom you feel compressed but I don't know to which force is due this. Do you know any link where I can read more about it? – Chris Steinbeck Bell Jan 2 at 10:04 | 2019-08-22T19:44:02 | {
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https://math.stackexchange.com/questions/2261957/probability-of-getting-a-second-6/2262210 | # Probability of getting a second $6$
A sister has two fair six sided dice, one red one blue, the dice are rolled together. The sister announces to her brother who is sat in another room and unable to see the dice, that there is at least one six in the outcome. Then asks him what are the odds that the other die is a six. He replies $1/6$, she says no it has to be $1/11$. Who is right or most likely to be right? Please help settle this family feud!
• The sister is right. – SirXYZ May 2 '17 at 10:12
• The dice are independent, so it'll be $\frac16$. The sister mistakenly assumes that every number from $2$ through $12$ has the same probability to be thrown, but this is false (there is for example only one way to throw $2=1+1$, but a lot more ways to throw $7=1+6=2+5=3+4=4+3=5+2=6+1$) – vrugtehagel May 2 '17 at 10:13
• The dice are independent, but the sister is correct since the event of "at least one die shows a six" does not specify the identity of the "other die". Of the eleven unbiased outcomes in that event, only one is "both are six", so $1/11$ is the probability that "the other is a six" when "at least one die shows a six" – Graham Kemp May 2 '17 at 11:07
• @vrugtehagel How do the sums of the dice even come in to this? – Arthur May 2 '17 at 11:08
• "At least one of the die is six. What is the probability both are six?" Answer: 1/11. "One of the die is six. What is the probability the other is six?" Answer: 1/6. "At least one die is six. What is the probability the other is six." Answer: This question semantically and logically makes no sense and is utterly meaningless. "At least one" is non-specific. "the other" is specific. – fleablood May 2 '17 at 17:11
In two ways:
1. We are computing the probability that both dice are a $6$ given all of the information he knows, not her. It is worth noting that she can see the other die, so she knows with certainty (probability $1$) whether both are $6$s or not.
She rolls both dice at the same time. There are 11 pairs of dice for which her statement that "at least one of the dice are a $6$" is true, i.e., $$(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,1),(6,2),(6,3),(6,4),(6,5)$$ we are not told which of the dice was a $6$, so we cannot cancel any possibilities (which is how he incorrectly concluded $1/6$).
Only one of these 11 are are both $6$s, hence the probability is $1/11$.
2. In a similar vein, but more algebraic, let $X_2=I[\text{die 1 is$6$}]$ and $X_1=I[\text{die 2 is$6$}]$, then
$$\mathbb{P}[X_1+X_2=2 \,|\, X_1+X_2\geq 1] = \frac{\mathbb{P}[X_1+X_2 = 2]}{\mathbb{P}[X_1+X_2\geq 1]} = \frac{\mathbb{P}[X_1=1,X_2=1]}{1 - \mathbb{P}[X_1=0,X_2=0]}$$
by independence and denoting $\mathbb{P}[X_1=1]=p_1$, $\mathbb{P}[X_2=1]=p_2$, \begin{align}= \frac{\mathbb{P}[X_1=1]\mathbb{P}[X_2=1]}{1 - \mathbb{P}[X_1=0]\mathbb{P}[X_2=0]}= \frac{p_1p_2}{1-(1-p_1)(1-p_2)} = \frac{1}{\tfrac{1}{p_1}+\tfrac{1}{p_2}-1}\end{align} Note that this allows for biased dice (e.g., what if the first die were weighted to land on $6$ 90% of the time?). But if we assume they are not biased, then $p_1=p_2=\tfrac{1}{6} \implies \tfrac{1}{p_1} = \tfrac{1}{p_2}=6$, giving the same result of $\frac{1}{(6)+(6)-1} = \frac{1}{11}$.
We could then answer the same question if we replaced dice with coins ($p_1=p_2=\tfrac{1}{2}$), which gives $\tfrac{1}{(2)(2)-1}=\tfrac{1}{3}$.
Edit If the problem were different and instead she doesn't see the second die then the probability that she would guess correctly is $1/6$. If he knew that she had only seen one die then the probability he would be correct in his guess is $1/6$, but if he didn't have that information then it would still be $1/11$.
The confusion here is very similar to the Monty Hall problem.
• Thanks for your reply. I agree that you need to consider what the brother knows. From the sisters statement, the brother , in my view, can only be certain that the sister has seen one die which is showing a six, he is in the other room and is unable to ascertain whether she can see both dice. Given that, his response of 1/6 may be correct. – greengd May 2 '17 at 12:55
• @greengd I mean it really depends on just how loosely you want to interpret the wording. Based on the fact the she states $1/11$, you could then argue that he should update his guess to $1/11$ because it is very likely that she has calculated it in the same way I have above, and hence should assume she has seen both dice. But at that point we have severely over-analysed the problem. I think I have presented the most logical and obvious interpretation, and the other response agreed with me. – adfriedman May 2 '17 at 13:08
This question can't be properly answered from the information provided (which is why it's so good at provoking arguments). To give a proper answer, you would need to know on what basis the sister decided what fact about the dice to announce. Suppose she thought to herself "I will look at the red die, which will be showing a number, $n$. I'll shout to my brother there is at least one $n$ in the outcome". She does this and it happens to be a 6 that comes up on the red die. All this faffing around with the red die obviously doesn't affect what number the blue die has landed on (the sister might not even have looked at it), so the probability that the other die is a 6 too is clearly 1/6.
To get an answer of 1/11 you need to make a similarly unjustified assumption about the sister's thought process (something like "If at least one die shows a 6 then I'll announce that fact, otherwise I won't say anything, and no maths question will be produced").
This is essentially the same as the "Tuesday Boy problem" as discussed by Rob Eastaway
• +1 Without really thinking about it I assumed that the sister was thinking like me. – drhab May 2 '17 at 13:36
• ... or even if the sister decided to make an announcement versus staying quiet! In my family, the probably is probably near $1$, since my sister would likely only have done this as a trick question! This is basically the answer I was going to write myself. +1. – user14972 May 2 '17 at 14:54
• I have a quibble with this answer (as I have a quibble with the analysis of the "Tuesday boy problem" you linked to). The statement "There is at least one six in the outcome" is a mathematically precise and well-defined event on the sample space, and this event has the 11 outcomes listed in the answer of @adfriedman. The problem could have been set up to describe the event ambiguously, hence inviting a linguistic discussion instead of a mathematical discussion (as in your answer and in Eastaway's analysis). But it wasn't, the event itself was described precisely. – Lee Mosher May 2 '17 at 15:25
• @LeeMosher The event itself is well-defined. The circumstance under which the event is announced is not. – Thern May 2 '17 at 15:47
• If she has computed the probability is 1/11 do you think it is more or less likely that she has seen the die? I think that makes it a justified assumption – adfriedman Jun 22 '17 at 20:12
BLUF: If this happened in real life, I would expect the brother to more likely be right (but for the wrong reason). But I imagine the person who posed the question to you had in mind a scenario where the sister is more likely to be right.
This is a standard problem, whose main issue isn't even the formal mathematics — it's in the modeling of the problem. Specifically, the problem gives insufficient information to specify what's going on, leaving the listener to fill in their own ideas about what problem they want to solve.
It's further compounded by a grammar error — nowhere in the problem has a die been specified, so it makes no sense to speak of the "other" die. This error tends to have a very strong effect of tricking people into a particular interpretation of the problem.
I'm going to resolve the grammar issue by changing the problem slightly, my change in bold:
... The sister announces to her brother who is sat in another room and unable to see the dice, that there is at least one six in the outcome. Then asks him what are the odds that both dice are six. ...
That said, I expect the brother to most likely be right. I expect the typical sister that does this would have gone through something resembling the following process:
• She thinks "I just learned this neat probability thing. I'll see if I can get one up on my brother."
• She rolls two dice
• She (uniformly randomly) picks one of the two dice, and announces its value to her brother
• She then asks the probability question
The third bullet point is very important. The point being, for example, that the outcome of seeing two dice with $\{6, 3\}$ happens with probability $\frac{1}{18}$, but the outcome that the dice were $\{ 6, 3\}$ and the sister chooses to announce the $6$ rather than the $3$ happens with probability $\frac{1}{36}$.
So, from the information the sister announces, the relevant outcomes are:
• With probability $1/36$, the dice read $\{ 6, 1 \}$ and the sister announces $6$
• With probability $1/36$, the dice read $\{ 6, 2 \}$ and the sister announces $6$
• With probability $1/36$, the dice read $\{ 6, 3 \}$ and the sister announces $6$
• With probability $1/36$, the dice read $\{ 6, 4 \}$ and the sister announces $6$
• With probability $1/36$, the dice read $\{ 6, 5 \}$ and the sister announces $6$
• With probability $1/36$, the dice read $\{ 6, 6 \}$ and the sister announces $6$
So, given the information that the sister announced a $6$, you find that the probability that both dice read $6$ is, indeed, $1/6$.
Now, another example of how this could have gone — and probably the one that the question's originator had in mind, is that the sister did something more like
• She repeatedly rolls two dice until at least one of the two dice read $6$
• She announces there is a $6$ and asks the question
In this case, the outcome that the dice read $\{ 6,1 \}$ upon announcement really is $1/18$. The relevant outcomes are
• With probability $1/18$, the dice read $\{ 6, 1 \}$ when the announcement is made
• With probability $1/18$, the dice read $\{ 6, 2 \}$ when the announcement is made
• With probability $1/18$, the dice read $\{ 6, 3 \}$ when the announcement is made
• With probability $1/18$, the dice read $\{ 6, 4 \}$ when the announcement is made
• With probability $1/18$, the dice read $\{ 6, 5 \}$ when the announcement is made
• With probability $1/36$, the dice read $\{ 6, 6 \}$ when the announcement is made
and we see the probability is $\frac{1}{11}$.
• It's not a "grammar error" to omit which die was chosen, it actually gives some information, but intentionally not all. Using the notation I gave in my own answer, this is equivalent to $\mathbb{P}[X_1+X_2 = 2 | X_1+X_2\geq 1]$ which is completely different to (if for instance the red die were chosen) $\mathbb{P}[X_1+X_2 = 2 | X_1 = 1]$. – adfriedman May 4 '17 at 16:30
• @adfriedman: I am not assuming they are indistinguishable -- I am merely not distinguishing between them. Listing both $(6,1)$ and $(1,6)$ would make the lists long enough that their presence would make the post awkward, and wouldn't do anything to improve the clarity anyways. – user14972 May 4 '17 at 16:36
• "So, from the information the sister announces, the relevant outcomes are:" You are assuming that she announced based on the first die alone, which is not given information. So you need to include both $(6,1)$ and $(1,6)$, which do not represent identical outcomes. – adfriedman May 4 '17 at 16:39
• @adfriedman: $\{ 6,1 \}$ is an unordered pair: there is no "first" die. If we add in order, the bullet you refer to covers both the 1/72 chance of "(6,1) and sister announces 6" and the 1/72 chance of "(1,6) and sister announces 6". – user14972 May 4 '17 at 16:40
• @adfriedman: The grammar error is not omitting a choice of die, but making a later statement that refers to a choice which was not made. While "at least one die is 6" is a careful rephrasing of the usual problem statement, it really needs to continue with "... both dice are 6". – user14972 May 4 '17 at 16:43
Let $X$ denote the number of sixes that have shown up.
Then: $$\Pr(X=2\mid X\geq1)=\frac{\Pr(X=2\wedge X\geq1)}{\Pr(X\geq1)}=\frac{\Pr(X=2)}{1-\Pr(X=0)}=\frac{\frac16\frac16}{1-\frac56\frac56}=\frac1{11}$$
Sister did not say something like: "the red (or blue) die shows a six". That would have made things different. In that case $\frac16$ is the correct answer.
This is reminiscent of a "paradox" I saw in Martin Gardner's "Mathematical Games" column in the Scientific American 50 odd years ago. A bridge player, known to be truthful, looks at the 13 cards in his hand, and announces, "I have a Ace." What is the probability that he holds another Ace? Again, suppose he announces instead, "I have a black Ace." What is the probability that he holds another Ace? Finally, what is the probability that he holds a second Ace if he announces, "I have the Ace of Spades."
If you go to the trouble of computing these probabilities in the usual manner, for example dividing the probability that he holds at least two Aces by the probability that he holds at least one Ace for the first case, you find that as the specificity of his announcement increases, the probability that he holds a second Ace increases too. This seems paradoxical, since he can always announce the color or the suit; it doesn't seem like it should make any difference.
I can't remember how, or if, Gardner explained this. At the time it was a complete mystery to me. The explanation is similar to the answer given by aPaulT; we don't have enough information on how this person decides what announcement to make. For example, if he announces the color if and only if it is black, then the statement "I have an Ace," is tantamount to "I have a red Ace," and we have to recompute our answer.
The brother is right.
Suppose you are the sister, and try to repeat the experiment. You roll a red 3 and a blue 5. What do you ask your brother - "one is a 3, what are the chances that both are 3's?" or "one is a 5, what are the chances that both are 5's?" ?
While a specific person may prefer 5's over 3's, and so always ask about 5's in this example, we have no information that would allow us to assume anything except that the sister chooses one of the two values at random, if there are two.
In the original question, there is one combination where the sister had to ask about 6's, and 10 where there was a 6 but only a 50% chance she would ask about it. This makes the answer 1/(1+10/2) = 1/6.
+++++
Note that the sister's logic is what makes people say the answer to the Monty Hall Problem is 1/2.
This problem, the Monty Hall Problem, and the Two Child Problem are all variations of the Bertrand Box Problem. Sometimes it is called Bertrand's Box Paradox, but the paradox is how Bertrand showed that the sister's answer can't be right.
Suppose the sister had said "I wrote one of the number's down on this hidden piece of paper. What are the chances that both dice landed on that number?" Then, whatever the chances are, they are the same no matter what is written. But if it is the same no matter what is written, then we don't need to see what is written to answer.
The chances of doubles, before she wrote down a number, were 1/6. She can always write down a number, so if it changes to 1/11 we have a paradox. The answer changed, but we gained no information that could make it change. | 2020-03-29T06:17:23 | {
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https://math.stackexchange.com/questions/1410743/evaluate-int-theta-sec-theta-tan-theta-d-theta | # Evaluate $\int \theta\sec\theta \tan\theta \ d\theta$
integral of $\int \theta\sec\theta \tan\theta \ d\theta$
my work
$\frac{d}{d\theta}\sec(θ) = \sec(\theta)\tan(\theta)$
So if we let $u = \theta$ and $v' = \sec(\theta)\tan(\theta)$, then we get:
$u = \theta, du = d\theta$ and $v = \sec(\theta), dv = \sec(\theta)\tan(\theta)d\theta$
Hence
$$\int \theta \sec(\theta)\tan(\theta) d\theta = \theta\sec(\theta) - \int\sec(\theta) d\theta$$
Now, the integral of $\sec(\theta)$ is a particularly tricky integral, but it comes to:
$$\int \sec(\theta) d \theta = \ln|\sec(\theta) + \tan(\theta)| + C$$
integral comes to:
$$\int \theta \sec(\theta)\tan(\theta) d\theta = \theta \sec(\theta) - \ln|\sec(\theta) + \tan(\theta)| + C$$
but my answer is not like this picture
• Your work is fine. – user84413 Aug 26 '15 at 20:26
• I need to be like the answer in the picture – Adel Hassan Aug 26 '15 at 20:27
• @zorbha No, you don't need to: your answer is correct, the one in the book is not (because of missing absolute values). – egreg Aug 26 '15 at 20:47
Let $c = \cos \frac {\theta}2; s= \sin \frac {\theta}2$
Note that $$\sec \theta +\tan \theta = \frac {1+\sin \theta}{\cos \theta}=\frac {c^2+s^2+2cs}{c^2-s^2}=\frac {(c+s)^2}{(c+s)(c-s)}=\frac {c+s}{c-s}$$ and that should help you to reconcile the two answers.
• @user84413 There were several typos! Thanks for pointing that out - I hope they are now sorted – Mark Bennet Aug 26 '15 at 20:31
Your answer is the same. Note that
$$\log|\cos(\theta/2)-\sin(\theta/2)|-\log|\cos(\theta/2)+\sin(\theta/2)|=\log |\frac{\cos(\theta/2)-\sin(\theta/2)}{\cos(\theta/2)+\sin(\theta/2)}|\\ =\log |\frac{\cos^2(\theta/2)-2\sin(\theta/2)\cos(\theta/2)+\sin^2(\theta/2)}{\cos^2(\theta/2)-\sin^2(\theta/2)}|\\=\log |\frac{1-\sin(\theta)}{\cos(\theta)}|=\log|\sec(\theta)-\tan(\theta)|= \log \frac{|\sec^2(\theta)-\tan^2(\theta)| }{|\sec(\theta)+\tan(\theta)|} \\=-\log| \sec(\theta)+\tan(\theta)|$$
An antiderivative of $$\sec\theta\tan\theta=\frac{\sin\theta}{\cos^2\theta}= -\frac{-\sin\theta}{\cos^2\theta}$$ is $1/\cos\theta$. Therefore, integrating by parts, $$\int\theta\frac{\sin\theta}{\cos^2\theta}\,d\theta= \frac{\theta}{\cos\theta}-\int\frac{1}{\cos\theta}\,d\theta$$ The remaining integral can be computed with the substitution $$\theta=\frac{\pi}{2}-2u$$ so \begin{align} -\int\frac{1}{\cos\theta}\,d\theta= \int\frac{1}{\sin2u}\,2du &= \int\frac{\cos^2u+\sin^2u}{\sin u\cos u}\,du\\[6px] &= \int\left(\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}\right)du \\[6px] &= \log|\sin u|-\log|\cos u|+C\\[6px] &=\log\left|\tan u\right|+C\\[6px] &=\log\left|\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right| \end{align} Now $$\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)= \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}= \frac{\cos(\theta/2)-\sin(\theta/2)}{\cos(\theta/2)+\sin(\theta/2)}$$ Thus the book is almost right: they're forgetting the absolute value.
You are stating that $$\log|\sec\theta+\tan\theta|$$ is an antiderivative of $\sec\theta$; let's try: $$\log|\sec\theta+\tan\theta|= \log\left|\frac{1+\sin\theta}{\cos\theta}\right|= \log|1+\sin\theta|-\log|\cos\theta|$$ The derivative is $$\frac{\cos\theta}{1+\sin\theta}-\frac{-\sin\theta}{\cos\theta}= \frac{\cos^2\theta+\sin\theta+\sin^2\theta}{\cos\theta(1+\sin\theta)}= \frac{1+\sin\theta}{\cos\theta(1+\sin\theta)}=\frac{1}{\cos\theta}$$ So you're right as well. Even “more right” than the book. | 2019-12-06T08:25:12 | {
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https://math.stackexchange.com/questions/2669718/why-do-i-have-to-substitute-both-variables-to-find-all-critical-points | Why do I have to substitute both variables to find all critical points?
I'm trying to find the critical points for $f(x, y) = 5x^2-2y^2+10$ on the circle $x^2+y^2 = 1$.
I understand that when I substitute $x = \cos(t)$ and $y = \sin(t)$, I get all four critical points: $(0, 1), (0, -1), (1, 0)$, and $(-1, 0)$. However, if I solve for $x^2$ in the constraint, I get $x^2 = 1-y^2$, and when I plug this into the function, I get $f(y) = -7y^2 + 15$. $f'(y) = -14y$, so $f'(y) = 0$ when $y = 0$.
However, this only provides $2$ of the $4$ critical points, $(0, 1)$ and $(0, -1)$. You only get the other two critical points (when $y = 1$) when you do the similar substitution for $y^2 = 1-x^2$.
Why is this? It's usually sufficient to "merge" the equations once and get the derivative to find all critical points. Why is it necessary to do so with both of them in this case?
• On a finite interval the critical points might be realized at an endpoint. as your $f(y)$ is defined on $[-1,1]$ you need to check those two endpoints in addition to solving $f'(y)=0$.
– lulu
Feb 27, 2018 at 22:18
• Possibly worth remarking: Lagrange multipliers works well here, certainly avoids technical issues of the sort you encountered.
– lulu
Feb 27, 2018 at 22:19
• Oh yeaahhhhh totally forgot about that for one-variable functions. Thank you! Feb 27, 2018 at 22:20
• Or actually, while I knew extrema were on the boundaries by virtue of just being the smallest or largest value, I didn't realize boundaries were also candidates for critical points and couldn't be discovered simply via derivatives like the points between the bounds. I'm still not completely sure why that would be. Feb 27, 2018 at 22:25
• You need to distinguish between critical points—those at which the derivatives vanish—and extrema. Example: The extrema of $f(x)=x$ on any closed interval lie at its end points, but its derivative vanishes nowhere.
– amd
Feb 27, 2018 at 22:41 | 2022-08-14T03:34:07 | {
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http://math.stackexchange.com/questions/165345/the-determinant-of-transition-matrices-between-orthonormal-bases | # The Determinant of Transition Matrices Between Orthonormal Bases
I am trying to show that if $A$ and $B$ are orthonormal bases of a real finite dimensional vector space $V$ and $P$ is the change of basis matrix between $A$ and $B$ then $\det(P) = \pm 1$. Here is my progress:
When working on this problem, I discovered the following fact: If $G_B$ denotes the Gram matrix of the inner product relative to $B$ and $G_A$ denotes the Gram matrix of the inner product relative to $A$ then the Gram matrices and the transition matrix $P$ are related through $$G_B = P^T G_A P.$$
The original claim is then a corollary of this fact: $$G_B = P^T G_A P \implies \det(G_B) = \det(P^T) \det(G_A) \det(P) \implies \det(P) = \pm 1$$ where I have used basic properties of the determinant and the fact that the determinant of the Gram matrix relative to an orthonormal basis is $1$.
So, I think this proves the original claim but I am now stuck with proving the more general claim that $G_B = P^T G_A P$. What would be a good way to approach this?
-
Since A and B are both orthonormal bases, which means P preserves inner product, i.e. P is orthogonal transformation, so det(P)=1 or -1. – chaohuang Jul 1 '12 at 21:35
also, since P is orthogonal,P^T = P^(-1) – chaohuang Jul 1 '12 at 21:36
$$\begin{eqnarray*} (P^T G_A P)_{il} &=& (P^T)_{ij} (G_A)_{jk} P_{kl} \\ &=& P_{ji} \langle x_k,x_j\rangle P_{kl} \\ &=& \langle P_{kl} x_k, P_{ji} x_j\rangle \\ &=& \langle (P^T x)_l, (P^T x)_i\rangle \\ &=& (G_B)_{il} \end{eqnarray*}$$
$\def\det{\mathrm{det}\,}$ Addendum: On second reading I may have misinterpreted the meaning of "this" in the last sentence of your question.
Let $X_{ij} = (x_i)_j$ be the matrix whose rows are the orthonormal basis vectors. $X$ transforms like $X\to P^T X$. We have $X X^T = \mathbb{I}$, since the basis is orthonormal. Then $$(P^T X)(P^T X)^T = P^T X X^T P = P^T P = \mathbb{I},$$ since the new basis is orthonormal. Therefore, $\det(P^T P) = (\det P)^2 = 1$, so $\det P = \pm 1$.
-
Here is a possibly simpler approach:
Let $a_1,...,a_n$, $b_1,...,b_n$ be the elements of bases $A,B$ respectively. By assumption, $b_i = P a_i$. By 'orthonormality', we have $\langle b_i, b_j \rangle = \langle P a_i, P a_j \rangle = \langle a_i, P^TP a_j \rangle = \delta_{ij}$. It follows from this that $P^TP = I$, and that $(\det P)^2 = 1$, from which the result follows (since $\det P$ is real).
- | 2016-07-25T14:23:24 | {
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https://math.stackexchange.com/questions/2094123/how-to-verify-if-a-curve-is-exponential-by-eyeballing/2094132 | # How to verify if a curve is exponential by eyeballing?
A plane curve is printed on a piece of paper with the directions of both axes specified. How can I (roughly) verify if the curve is of the form $y=a e^{bx}+c$ without fitting or doing any quantitative calculation?
For example, for linear curves, I can choose two points on the curve and check if the midpoint is also on the curve. For parabolas, I can examine the geometric relationship between the tangent at a point and the secant connecting the peak and that point. Does the exponential curve have any similar geometric features that I can take advantage of?
• Interesting question :) Jan 12 '17 at 0:23
• Actually you don't need the axes' directions. Just look for the (only) asymptote. Jan 12 '17 at 0:36
• Do a log plot of $f(x)-f(\pm \infty)$ and eyeball if it's linear.
– dxiv
Jan 12 '17 at 3:45
• If you can replot the data with semilog axes, it should be obvious. Jan 12 '17 at 23:26
Are the doubling points evenly spaced?
Assume $a$ and $b$ are positive (if not, it's easy to see and readjust - $a$ is positive if the curve flattens to the left, $b$ has the same sign as $a$ if the curve is increasing). Mentally reset the $x$ axis at the height to which the curve tends to become horizontal to the far left (that's $c=\lim_{x\rightarrow -\infty} ae^{bx}+c$) . Note: there's a way to be really rigorous about this and obtain $c$ with ruler and compass (see below) but then it's not really "eyeballing".
With $c$ set to $0$, take a point $x_1$ on the (new) $x$ axis, and a random multiplier $M>1$; say, $M=2$. Eyeball the point $x_2$ at which $y(x_2)$ is roughly $M$ times $y(x_1)$. Then eyeball the point $x_3$ at which $y(x_3)$ is roughly $M$ times $y(x_2)$, and so on, and check if the points $x_1,x_2,x_3,...$ at which $y$ multiplies by $M$ are evenly spaced, as they must be if your curve is an exponential.
Easier to do than to say... as long as you have found enough "multiplication" points on your piece of paper. If your curve is very "flat" you'd have to choose an $M$ very close to $1$, but it still works in theory (or if you do it with ruler and compass with sufficient accuracy - but again, that's not really "eyeballing"). In practice it will be hard to tell such a "flat" exponential from a parabola or even a line: remember that $e^{bx}$ is very close to $1+bx$ if $bx\ll 1$, i.e. if $\frac{1}{b}$ is much larger than the largest $x$ you have on your piece of paper. Then again, even a "sufficiently flat" parabola is hard to tell from a line...
If your piece of paper on the other hand is sufficiently large, as a bonus, you can also gauge $a$ (or more accurately $a \cdot$ e), $b$ (or more accurately $\frac{\ln M}{b}$) and $c$. We know that $c$ is the height to which the curve converges as $x\rightarrow -\infty$, while ($a \cdot$ e) is the distance between said height and the $y$ axis intersect. And the horizontal spacing $\Delta x$ between the points at which you check the curve is $\frac{\ln M}{b}$ since $e^{b\Delta x}=M$.
Let's be rigorous!
First of all, how can we rigorously construct $c$ with ruler and compass? To do it, all we have to do is to take a random positive horizontal spacing $\Delta x$, and obtain $\Delta^0_y=y(0)-y(-\Delta x)$, and $\Delta^1_y=y(-\Delta x) - y(-2\Delta)$. Call $\rho<1$ the ratio $\frac{\Delta^1_y}{\Delta^0_y}$. It's immediate that $c$ is located at distance $\sum_{i=0}^\infty \rho^i \Delta^0_y = \Delta^0_y \frac{1}{1-\rho}$ below the intersect of the curve with the $y$ axis, which we can easily if somewhat laboriously obtain with ruler and compass noting that $\Delta^0_y$ is mid-proportional between said distance and $\Delta^0_y-\Delta^1_y$.
Also, as Rahul correctly points out, to be really formal one would have to choose $M$ sufficiently "randomly" (uniformly at random in any arbitrary small non-degenerate interval suffices), so that the probability of encountering a function in the form $e^{bx+f(x)}$ with $f(x)$ periodic with a period that's exactly an integer multiple of $M$ would be $0$. In practice, since you are only eyeballing for rough exponentiality, checking that your curve does not "wiggle" is enough to rule out such cases!
• If it's very flat, you can do the opposite, $x$ should double at $y_1, y_2, y_3$ etc. Jan 12 '17 at 0:42
• How easy is it to estimate where $c$ lies visually? Jan 12 '17 at 2:34
• The doubling property only holds if $c=0$. Otherwise you have to find the points where the distance from $y=c$ doubles, and we don't know $c$ beforehand.
– user856
Jan 12 '17 at 2:43
• Yes. And you can estimate c easily by looking to the far left, since it's the limit of y as x goes to $-\infty$. Jan 12 '17 at 2:45
• $ae^{bx+f(x)}$ for any $\bigl(\frac{\ln 2}b\bigr)$-periodic function $f$ also has equally spaced doubling points.
– user856
Jan 12 '17 at 5:00
Assuming $c=0$ (it's not the graph of a exponential function otherwise):
Pick a point on the curve. Draw its tangent and extend it until it meets the $x$-axis. Also drop a vertical from the point to the $x$-axis. Now you have a right triangle.
Do this for lots of points. The bases of all the triangles should have the same length.
• The tangent to $y=ae^{bx}+c$ at $(x_1,y_1)$ meets the $x$-axis at $(x_1-y_1/y'_1, 0).$ The distance from this point to $(x_1,0)$ is $y_1/y'_1=(1/b)+c/y'_1$ which is not constant unless $c=0.$.... The tangent meets the horizontal line $y=c$ at $(x_1-1/b,c)$.... So if you have enough data points to suggest that some line $y=c$ is a horizontal asymptote, this will work (approximately). Jan 12 '17 at 2:34
• Oh, I didn't notice the constant term in the question. Personally I wouldn't call it an exponential curve unless $c=0$.
– user856
Jan 12 '17 at 2:38
• Neither would I,,,,,,,,,,,,,, Jan 12 '17 at 2:47
• It's easy, just "lift" or "lower" the x axis so that the curve converges to it as it it goes to $-\infty$. If you want to be really picky about that, you can check two equal horizontal intervals, check the relative vertical intervals $\Delta y_1$, $\Delta y_2$ and from those construct rigorously $c$ ... but the post title says "eyeballing"! Jan 12 '17 at 2:52
• @Anonymous: Not so easy if you don't have the curve extending to $-\infty$. After all, in your own answer you point out the difficulty if the range of $x$ is not large enough compared to $1/b$.
– user856
Jan 12 '17 at 3:02
### You can't. No, not just "in theory", but also in practice.
I tried this when doing regression before and I gave up on it once I realized how impossible it is:
(Ignore the left upwards part of the parabola; pretend you don't have that piece of information when you're trying to tell which is which.)
### Update
Since I couldn't reproduce the plot above anymore (I only kept the screenshot, and I'm not sure why the formulas don't seem to be reproducing it), I'll include an artificial one that illustrates the same problem:
To reproduce:
Plot[{Exp[x] / 4, 0.32 x^2 + 0.12 x + 0.26}, {x, 0, 2}, PlotRange -> {Automatic, {0, 2}},
GridLines -> Automatic, AspectRatio -> 1, BaseStyle -> {FontSize -> 14}]
• This looks an awful lot like what my economics professor called the curse of exponentiality (which as he used it referred to the fact that you can't really tell whether a curve (e.g. US debt) is exponential or something like parabolic until it's too late to do anything about it; when I did a google to make sure I had the right term, it looks like it usually is taken to refer to something slightly different), but the main point being: you might be able to determine better if you extend the curve out further
– Foon
Jan 12 '17 at 21:10
• @Foon: It very well may have been (though the only such "curse" I've heard about is that of "dimensionality" rather than "exponentiality"). It's worth pointing out though that the nice thing is nothing in real life can grow exponentially for very long; if you're observing something physical, it's likely to be e.g. logistic instead. What I was doing regression on was the complexity of an algorithm, which can happily grow exponentially. :-) If I remember correctly I couldn't grow it to the right since I didn't have the time to wait for it to finish with very large inputs... Jan 12 '17 at 21:16
• You only can't tell between quadratic and exponential in that example because the parameter is varied over just a small range, so mostly you see only two Taylor terms of the exponential anyway. If the parameter reached twice as far, in either direction, then it would be much clearer what's going on. Jan 12 '17 at 21:32
• @leftaroundabout: I think it's safe to assume the OP isn't asking about the cases where it's obvious which is which... because those cases are obvious. Jan 12 '17 at 21:35
• The question seems to be about fairly precisely drawn curves rather than dirty data or approximations, so I think this does not answer it. On the other hand it does answer a different interesting question! Jan 12 '17 at 21:37 | 2021-10-17T17:07:08 | {
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https://math.stackexchange.com/questions/1716418/random-walk-of-a-drunk-man | # Random Walk of a drunk man
Problem Statement:
From where he stands, one step toward the cliff would send the drunken man over the edge. He takes random steps, either toward or away from the cliff. At any step his probability of taking a step away is 2/3, of a step toward the cliff 1/3. What is his chance of escaping the cliff?
My take:
Say the probability that he dies from where he stands right now is p. Then, he could comfortably make one step left and end his life with probability 1/3
Or he could take one step away and two step towards and boom...take two steps away and three steps toward...so on and so forth
Resulting in p= 1/3 + 2/3 * (1/3)^2 + (2/3)^2 * (1/3)^3 +....
Summing this infinite sequence gives me probability of dying as 3/7 (around 43%). I was rather puzzled when i learnt that the correct probability is 1/2. Cant figure out what are the other 7% ways for my drunken man to die which I missed above?
• What about he going away and closer back and forth 10 times before finally falling? – Ant Mar 27 '16 at 23:32
• Possible dupe of puzzling.stackexchange.com/questions/6415/… (even if it's on a different SE, it's got great answers). – Deepak Mar 27 '16 at 23:32
Here is a proof that does not rely on solving recurrence equations.
Let $r$ denote the probability of hitting the cliff. Let $0,1,\dots$ denote the distance from the cliff. He starts at $1$.
If in the first step goes to $2$, then will hit the cliff if and only if will ever go back to $1$, and then will ever go to $0$. But probability of going back from $2$ to $1$ is the same as hitting cliff starting from $1$, that is $r$. Summarizing:
$$r= \frac 13 + \frac 23 r\times r,$$
or (writing $r=\frac 13 r + \frac23 r$):
$$\frac 13 (r-1) = \frac 23 (r^2 -r).$$
That is, if $r\ne 1$, we have
$$\frac 13 = \frac 23 r.$$
Or $r=\frac 12$.
It remains to show that $r\ne 1$. I'l allow myself to be sloppy and lazy, and will rely on the law of large numbers, which tells as that the position of this walk at time $n$ is of order $(\frac 23 - \frac 13)n$. In particular, the position tends to $+\infty$. If $r=1$, then by iterating, the probability of ever getting to $-1$ is also $1$, and the same for $-2$, etc. In particular, the path is unbounded from below. This contradicts the conclusion of the law of large numbers.
• Intuitively, you imagine jumping to the right, then shifting the earth under you to the right as well, so that the scenario looks the same as it did to start with. Since the probabilities are homogeneous, nothing changes when you shift the earth, so the scenario actually is now the same. That's where the probability to ever hit $1$ starting from $2$ being $r$ (which is maybe not so obvious) comes from. – Ian Mar 28 '16 at 14:03
Note: not a compete solution as I haven't done the calculations, but I think if you do there should not be any problems
Let $p(i)$ be the probability of him falling when has a distance $i$ from the cliff.. We want $p(1)$
Clearly
$$p(1) = 1/3 + 2/3 p(2)$$ $$p(2) = 1/3p(1) + 2/3 p(3)$$
Which substituting gives
$$p(1) = 1/3 + 2/9p(1) + 4/9p(3)$$
Do it again substituting in $p(3) = 1/3p(2) + 2/3p(4)$ and try to get a relation where on the right hand side there is only $p(n)$; then let $n \to \infty$ and use the fact that $\lim_{n \to \infty} p(n) = 0$ to conclude
Before presenting a solution, I will say that your solution is wrong because it erroneously assumes that the man must go away from the cliff some number of steps and then proceed directly to the cliff, and otherwise he does not fall off. But on the contrary, he can turn around numerous times and still eventually fall off. Thus your result is a strict lower bound on the correct answer.
To correctly solve the problem, consider the sequence of hitting probabilities to hit $0$ before $n$ starting from a state $i$, where the probability to move right is $p$ and the probability to move left is $q=1-p$. Denote these probabilities by $h(i,n)$. We would like to compute $\lim_{n \to \infty} h(1,n)$.
By conditioning on the first step, we find that $h$ satisfies the equations:
\begin{align*} h(0,n) & = 1 \\ h(n,n) & = 0 \\ h(i,n) & = p h(i+1,n) + q h(i-1,n) \end{align*}
Consider the characteristic equation for this recurrence:
$$r = p r^2 + q \Leftrightarrow p r^2 - r + q = 0.$$
This has roots $r_1,r_2 = \frac{1 \pm \sqrt{1-4pq}}{2p}$. Since $pq=p(1-p) \in [0,1/4]$, these roots are real, and they are distinct if $p \neq 1/2$, as in your case.
Thus if $p \neq 1/2$ then we expect a solution of the form
$$h(i,n) = A_n r_1^i + B_n r_2^i.$$
Introducing the left boundary condition we get $h(0,n) = A_n + B_n = 1$. Introducing the right boundary condition we get $A_n r_1^n + B_2 r_2^n = 0$. Solving the system of linear equations we get $A_n = \frac{r_2^n}{r_2^n-r_1^n}$ and $B_n = \frac{-r_1^n}{r_2^n-r_1^n}$. So we have
$$h(i,n) = \frac{1}{r_2^n-r_1^n} \left ( r_2^n r_1^i - r_1^n r_2^i \right )$$
when $p \neq 1/2$.
I leave it to you now to compute $\lim_{n \to \infty} h(1,n)$ and plug in $p=2/3$. | 2021-03-05T17:29:09 | {
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https://www.quantnet.com/threads/choosing-tt-vs-th-problem.10914/ | # Choosing TT vs TH problem
Discussion in 'Quant Interviews' started by tuanl, 8/18/12.
1. ### tuanl Member
Suppose you roll a fair coin 5 times. You can either be paid the number of consecutive tails "TT" or the number of "TH"'s. Which one would you choose and why? For example, the sequence "TTHTT" would count 1 instance of "TH" and 2 instances of "TT."
Generalize your result for n flips. Is the result different now?
#1
2. ### IlyaKEightSix Active Member
TT. Think about it. If you choose TH, then at most, you can have 2 sequences in your 5 flips. But if you choose TT, consider the case of getting lucky with five tails in a row: TTTTT.
How many instances of TT are there? Not 2, but 4. The reason for that is when you roll TH, then the next combination can't possibly be another TH. However, when you have TT, then the second T can be the first T in the next sequence. This basically gives you twice the chance.
Now if it was TH AND HT vs. TT (or HH for that matter), we'd have another discussion.
3. ### tuanl Member
I think "TH" is the answer based on calculating the P(getting at least one TH out of 5 tosses)$$= 26/32=13/16$$>P(getting at least one TT out of 5 tosses)($$=19/32$$). Notice that expected payoff of TT=the expected payoff of TH=1. So if you're less risk averse, we can choose TT and hope to get lucky with TTTTT.
I haven't worked on the general case yet, but I'm pretty sure the same result still holds(we should still choose TH).
#3
5. ### pruse Active Member
It ain't about googling, it's about sharing and discussing; I'm sure that's why tuanl posts his problems here.
Use indicator variables: For each of the first $$n-1$$ tosses, out of $$n$$, let $$X_1=1$$ if that toss and the next are TH, 0 otherwise. Then the expected number of TH's is equal to $$E(X_1+\cdots+X_{n-1})=(n-1)E(X_1)=(n-1)P(X_1=1)=\frac{n-1}{4}$$. The same argument shows that the number of TT's is expected to be the same. So it's the variance that will decide.
Now for the variance. $$Var(X_1+\cdots+X_{n-1})=E((X_1+\cdots+X_{n-1})^2)-\left(\frac{n-1}{4}\right)^2$$. $$E((X_1+\cdots+X_{n-1})^2)=(n-1)E(X_1^2)+2\sum_{i<j}E(X_iX_j)$$. Clearly $$(n-1)E(X_1^2)$$ will be the same in both the TT and the TH cases. We need to see how the $$\sum E(X_iX_j)$$ part differs.
1) TH case. When $$j-1=1$$, $$X_iX_j$$ clearly must equal 0. In all other cases, disjoint pairs of tosses are involved and these are independent, so $$E(X_iX_j)=P(X_iX_j=1)=P(X_i=1,X_j=1)=P(X_i=1)P(X_j=1)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}$$.
2) TT case. This is clearly larger as the disjoint pairs of tosses give the same $$E(X_iX_j)$$ as in the TT case, and the overlapping pairs give a positive $$E(X_iX_j)$$, whereas in the TH case these are 0.
So it's better to pick TH.
#5
6. ### IlyaKEightSix Active Member
How is P(TT) < P(TH)? P(T)=50%. P(H)=50%. P(T)*P(T)=25%. P(T)*P(H)=25%.
7. ### tuanl Member
I said P(getting AT LEAST 1 TT)<P(getting AT LEAST 1 TH). And no, the coin is definitely fair.
#7
8. ### IlyaKEightSix Active Member
Well, what about the probability of getting 2 TTs vs. 2 THs? With 2 TTs, you can have TTTHH HTTTH, HHTTT, TTHTT, HTTTT, and so forth...
With TH, you can only get two of them at MOST on five flips.
9. ### tuanl Member
Nice solution, but the computation of the variance really bothers me. I know it's a common technique to compare the variance of two events to make a decision when the expected values of these events are equal. But I wonder if my reasoning above also works in this case, since we clearly see that probability of not getting any money when choosing TT is greater than that of TH.
#9
10. ### tuanl Member
But HTTTT counts as 3 TTs. In fact, probability of getting exactly 2 THs=probability of getting exactly 2 TTs. But we have to calculate the total probability of all the possibilities that can happen with TH and TT. Basically, choosing TH always give you a better chance to get some money than choosing TT. Notice that this reasoning works, in my opinion, since the expected value of TT=expected value of TH. The proof for this, for general value of n, was presented above by pruse above.
#10
11. ### pruse Active Member
The probability of not getting any money doesn't tell the whole story, as it's possible for the cases that do make money to make disproportionately larger amounts.
Your expected winnings, as well as the volatility of those winnings, are what matter. Now, between two alternatives with the same expected value, the one with the smaller volatility is generally preferred since your confidence interval is smaller.
#11
12. ### IlyaKEightSix Active Member
Hmmm, I did a dirty computational solution in R and it seems no matter what happens, when you add up the sum over all the combinations, the sums of both THs and TTs come out identical, but the variance of the THs are lower.
Seems I'm mistaken.
#13
14. ### Kangxi New Member
Hey, my answer for 5 tosses is as follows:
1 TT: 3+2+2+3 = 10
2 TT: 1+(2+1+2) = 6
3 TT: 1+1 = 2
4 TT: 1
So the expected money received if choosing TT will be (10*1 + 6*2+2*3+4*1)/(2^5)=1
1 TH: (2^3-4) + (2^3-2)+(2^3-2)+(2^3-4) = 20
2 TH: 2+2+2 = 6
So the expected money received if choosing TH will be (20*1 + 6*2)/(2^5)=1
Therefore, no matter what I choose, I will be paid \$1 in expectation.
#14
15. ### tuanl Member
But choosing TH gives you smaller variance than TT, which means TH is less volatile than TT. So you should choose TH
#15
16. ### Kangxi New Member
Yes, I think so.
#16 | 2015-04-27T13:35:11 | {
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http://math.stackexchange.com/questions/793135/making-a-biased-coin-flip-fair | # Making a biased coin flip fair
I have a puzzle:
Two groups want to break a tied vote using a simple coin flip, however the only coin they have available is a biased coin (i.e., one side will come up more often than the other). To make matters worse nobody in the room knows (or is willing to admit) how the coin is biased.
Assuming that the coin has two distinct sides, design a method for using only this coin to determine a fair outcome between the two parties.
Initially, I thought, since neither party knows how the coin is biased, then either party could simply randomly guess which side the coin will land on and then they will receive a 50% chance they are correct (and 50% chance they are incorrect). This is assuming they are not influenced by any knowledge that they know for how the coin is biased when randomly deciding (or rather they do not know how it is biased). I even performed a simulation on this for a large amount of trials (live demo in python).
However, I stumbled upon this wikipedia article, which mentions an alternative method.
If a cheat has altered a coin to prefer one side over another (a biased coin), the coin can still be used for fair results by changing the game slightly. John von Neumann gave the following procedure:
1. Toss the coin twice.
2. If the results match, start over, forgetting both results.
3. If the results differ, use the first result, forgetting the second.
I'm just curious: is my method of solving the puzzle a viable solution? It seems so, but I'm not 100% certain.
-
Your method, while certainly viable in practice, doesn't have the same ammount of fairness as the one on wiki. If you are one of the two parties, you cannot be completely certain that the other party knows nothing about the bias. – Arthur May 13 at 13:45
If you have the ability to generate a random value with 50% probability in your head, then you can just pick one of the two outcomes and you don't need a coin at all.
Also, I think implicit in this problem is the possibility that one of the parties knows how the coin is biased and will secretly use this to their advantage. If you have a fair protocol that will worke with a biased coin then you can still produce a fair outcome despite secret knowledge.
-
Your method seems good to me as long as both parties are honest about their ignorance. In this case, each party must assign a subjective probability of heads of $0.5$ (since the subjective probability of heads must equal the subjective probability of tails).
The other method is more robust since it works even when one or both parties has knowledge about the nature of the bias. When looking at pairs, you can always assume that $$P[HT\vert HT \textrm{ or } TH]=P[TH \vert HT \textrm{ or } TH].$$
It occurred to me that you may want a method using a single coin flip which doesn't become unfair if one party has undisclosed knowledge. Here are the steps.
1. Party $A$ guesses either heads or tails.
2. Party $B$ guesses either heads or tails.
3. If $A$ and $B$ make the same guess, then $A$ wins. $A$ and $B$ make different guesses, then $B$ wins.
4. (Optional) Toss the coin.
- | 2014-09-20T20:02:20 | {
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http://mathhelpforum.com/geometry/65064-length-string-around-circular-rod-print.html | Length of string around circular rod
Printable View
• Dec 15th 2008, 04:25 AM
fractal
Length of string around circular rod
A string is wound symmetrically around a circular rod. The string goes exactly 4 times around the rod. The circumference of the rod is 4 cm. and its length is 12 cm.
Find the length of the string ?
Please see PDF for diagram. Not sure how to simplify this problem to basic geometric assumption & still be correct. Thanks.
• Dec 15th 2008, 04:34 AM
Chop Suey
EDIT: Apologies, I've misunderstood the problem.
• Dec 15th 2008, 06:30 AM
fractal
Hi Chop Suey:
s=r $\theta$, where r = C / $(2\pi)$ and $\theta =2\pi$
For 4 revolution, $s=4/(2\pi)*(4*2\pi)=16$
ANSWER = 16, what is your answer?
• Dec 15th 2008, 07:15 AM
Bruce
^ I got that answer of 16 by realising that each time it goes around the rod it has gone 4cm. 4 x 4 = 16
Not exactly rocket science. =S
• Dec 15th 2008, 07:28 AM
fractal
A google search leads me to believe the answer is not 16.
The Puzzle Instinct: The Meaning of ... - Google Book Search
• Dec 15th 2008, 08:10 AM
earboth
Quote:
Originally Posted by fractal
A string is wound symmetrically around a circular rod. The string goes exactly 4 times around the rod. The circumference of the rod is 4 cm. and its length is 12 cm.
Find the length of the string ?
Please see PDF for diagram. Not sure how to simplify this problem to basic geometric assumption & still be correct. Thanks.
The length of the string is the length of the diagonal of the rectangle with
$l = 4 \cdot 4 cm = 16 cm\ and\ w = 12 cm$
Therefore the length of the string is
$s = \sqrt{16^2 + 12^2} = 20\ cm$
• Dec 15th 2008, 09:00 AM
Soroban
Hello, fractal!
There is a wonderful back-door approach to this problem . . .
Quote:
A string is wound symmetrically around a circular rod.
The string goes exactly 4 times around the rod.
The circumference of the rod is 4 cm and its length is 12 cm.
Find the length of the string.
"Unroll" the cylinderical rod, and we have a 4-by-12 rectangle.
Draw four of them in a row.
Code:
* - - - - - * - - - - - * - - - - - * - - - - - * | | | | * | | | | * | | | | * | | 12 | | * | | 12 | | * | | | | * | | | | * | | | | * - - - - - * - - - - - * - - - - - * - - - - - * 4 4 4 4
Draw the diagonal from one corner to the opposite corner.
This is the path of string as it spirals around the cylinder four times.
Using Pythagorus, we can find the length of that hypotenuse.
• Dec 15th 2008, 09:36 AM
fractal
Soroban: your solution is an out of the single box kind.(Cool)
however, for tutorial to middle & high school students, i think it would be easier to make a single cut through the cylinder, then flatten it, which give you four stripes. each stripe can be treated as a hypotenuse where the height is 4 cm and the width is 3cm. the sum of the four hypotenuse equals 20 cm.
• Dec 15th 2008, 10:13 AM
galactus
If I may, here is another way to look at it. Perhaps too complicated, but just for kicks.
I will not go through the derivation of the formula.
The arc length of a circular helix, the string in this case, is given by:
$t_{0}\sqrt{a^{2}+c^{2}}$ if $x=acos(t), \;\ y=bsin(t), \;\ z=ct$
What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$?.
$t_{0}=8{\pi}$, because we have 4 rev.
a=radius of cylinder = $\frac{2}{\pi}$
$c=\frac{12}{8\pi}=\frac{3}{2\pi}$
$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$ | 2017-03-27T03:16:26 | {
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http://privaluefoundation.com/a-z-qvhuxd/0rsnc.php?1a5bcb=poisson-process-problems | First, we give a de nition &=\sum_{k=0}^{\infty} P\big(X+Y=2 \textrm{ and }Y+Z=3 | Y=k \big)P(Y=k)\\ Poisson Probability Calculator. \begin{align*} \begin{align*} = 0.16062 \)b)More than 2 e-mails means 3 e-mails or 4 e-mails or 5 e-mails ....$$P(X \gt 2) = P(X=3 \; or \; X=4 \; or \; X=5 ... )$$Using the complement$$= 1 - P(X \le 2)$$$$= 1 - ( P(X = 0) + P(X = 1) + P(X = 2) )$$Substitute by formulas$$= 1 - ( \dfrac{e^{-6}6^0}{0!} We know that Viewed 3k times 7. Example 2My computer crashes on average once every 4 months;a) What is the probability that it will not crash in a period of 4 months?b) What is the probability that it will crash once in a period of 4 months?c) What is the probability that it will crash twice in a period of 4 months?d) What is the probability that it will crash three times in a period of 4 months?Solution to Example 2a)The average \( \lambda = 1$$ every 4 months. Stochastic Process → Poisson Process → Definition → Example Questions Following are few solved examples of Poisson Process. Review the recitation problems in the PDF file below and try to solve them on your own. Example 1These are examples of events that may be described as Poisson processes: eval(ez_write_tag([[728,90],'analyzemath_com-box-4','ezslot_10',261,'0','0'])); The best way to explain the formula for the Poisson distribution is to solve the following example. \end{align*} Run the binomial experiment with n=50 and p=0.1. \begin{align*} The first problem examines customer arrivals to a bank ATM and the second analyzes deer-strike probabilities along sections of a rural highway. &=\frac{P\big(N_1(1)=1, N_2(1)=1\big)}{P(N(1)=2)}\\ &\hspace{40pt}P(X=0) P(Z=1)P(Y=2)\\ One of the problems has an accompanying video where a teaching assistant solves the same problem. \end{align*} That is, show that 2. + \dfrac{e^{-6}6^2}{2!} M. mathfn. Let $X$, $Y$, and $Z$ be the numbers of arrivals in $(0,1]$, $(1,2]$, and $(2,4]$ respectively. Let $N_1(t)$ and $N_2(t)$ be two independent Poisson processes with rates $\lambda_1=1$ and $\lambda_2=2$, respectively. &=\textrm{Cov}\big( N(t_1)-N(t_2) + N(t_2), N(t_2) \big)\\ The number of customers arriving at a rate of 12 per hour. \end{align*}. The random variable $$X$$ associated with a Poisson process is discrete and therefore the Poisson distribution is discrete. = \dfrac{e^{-1} 1^1}{1!} The probability distribution of a Poisson random variable is called a Poisson distribution.. \end{align*}, A Poisson process is an example of an arrival process, and the interarrival times provide the most convenient description since the interarrival times are defined to be IID. I … Poisson Distribution. Poisson process basic problem. Using stats.poisson module we can easily compute poisson distribution of a specific problem. X \sim Poisson(\lambda \cdot 1),\\ To calculate poisson distribution we need two variables. C_N(t_1,t_2)&=\textrm{Cov}\big(N(t_1),N(t_2)\big)\\ Each assignment is independent. P(N_1(1)=1 | N(1)=2)&=\frac{P\big(N_1(1)=1, N(1)=2\big)}{P(N(1)=2)}\\ &=\textrm{Cov}\big(N(t_2), N(t_2) \big)\\ &=\textrm{Var}\big(N(t_2)\big)\\ A Poisson random variable is the number of successes that result from a Poisson experiment. Ask Question Asked 9 years, 10 months ago. \end{align*} The number of arrivals in an interval has a binomial distribution in the Bernoulli trials process; it has a Poisson distribution in the Poisson process. The Poisson process is one of the most widely-used counting processes. Let $\{N(t), t \in [0, \infty) \}$ be a Poisson process with rate $\lambda=0.5$. In particular, Hence the probability that my computer does not crashes in a period of 4 month is written as $$P(X = 0)$$ and given byP(X = 0) = \dfrac{e^{-\lambda}\lambda^x}{x!} \begin{align*} Let \{N(t), t \in [0, \infty) \} be a Poisson Process with rate \lambda. Therefore, is the parameter of the distribution. Example 5The frequency table of the goals scored by a football player in each of his first 35 matches of the seasons is shown below. 1. Therefore, we can write Then. + \dfrac{e^{-3.5} 3.5^4}{4!} Run the Poisson experiment with t=5 and r =1. P(X_1 \leq x | N(t)=1)&=\frac{P(X_1 \leq x, N(t)=1)}{P\big(N(t)=1\big)}. = 0.36787c)P(X = 2) = \dfrac{e^{-\lambda}\lambda^x}{x!} Thus, \begin{align*} 0 \begingroup I've just started to learn stochastic and I'm stuck with these problems. More specifically, if D is some region space, for example Euclidean space R d , for which | D |, the area, volume or, more generally, the Lebesgue measure of the region is finite, and if N ( D ) denotes the number of points in D , then \end{align*} &=\left(\frac{e^{-\lambda} \lambda^2}{2}\right) \cdot \left(\frac{e^{-2\lambda} (2\lambda)^3}{6}\right) \cdot\left(e^{-\lambda}\right)+ The emergencies arrive according a Poisson Process with a rate of \lambda =0.5 emergencies per hour. Y \sim Poisson(\lambda \cdot 1),\\ Then, by the independent increment property of the Poisson process, the two random variables N(t_1)-N(t_2) and N(t_2) are independent. &=\left[\lambda x e^{-\lambda x}\right]\cdot \left[e^{-\lambda (t-x)}\right]\\ The solutions are: a) 0.185 b) 0.761 But I don't know how to get to them. Example (Splitting a Poisson Process) Let {N(t)} be a Poisson process, rate λ. Suppose that each event is randomly assigned into one of two classes, with time-varing probabilities p1(t) and p2(t). Active 5 years, 10 months ago. In mathematical finance, the important stochastic process is the Poisson process, used to model discontinuous random variables. N_1(t) is a Poisson process with rate \lambda p=1; N_2(t) is a Poisson process with rate \lambda (1-p)=2. And you want to figure out the probabilities that a hundred cars pass or 5 cars pass in a given hour. Poisson process 2. \end{align*} Hospital emergencies receive on average 5 very serious cases every 24 hours. We can use the law of total probability to obtain P(A). The familiar Poisson Process with parameter is obtained by letting m = 1, 1 = and a1 = 1. customers entering the shop, defectives in a box of parts or in a fabric roll, cars arriving at a tollgate, calls arriving at the switchboard) over a continuum (e.g. Apr 2017 35 0 Earth Oct 10, 2018 #1 I'm struggling with this question. M. mathfn. = 0.06131, Example 3A customer help center receives on average 3.5 calls every hour.a) What is the probability that it will receive at most 4 calls every hour?b) What is the probability that it will receive at least 5 calls every hour?Solution to Example 3a)at most 4 calls means no calls, 1 call, 2 calls, 3 calls or 4 calls.$$P(X \le 4) = P(X=0 \; or \; X=1 \; or \; X=2 \; or \; X=3 \; or \; X=4)$$$$= P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$$$$= \dfrac{e^{-3.5} 3.5^0}{0!} Poisson process problem. Show that given N(t)=1, then X_1 is uniformly distributed in (0,t]. Let A be the event that there are two arrivals in (0,2] and three arrivals in (1,4]. If it follows the Poisson process, then (a) Find the probability… Then X, Y, and Z are independent, and = 0.36787$$b)The average $$\lambda = 1$$ every 4 months. 18 POISSON PROCESS 197 Nn has independent increments for any n and so the same holds in the limit. We say X follows a Poisson distribution with parameter Note: A Poisson random variable can take on any positive integer value. Video transcript. = 0.18393 \)d)\( P(X = 3) = \dfrac{e^{-\lambda}\lambda^x}{x!} $N(t)$ is a Poisson process with rate $\lambda=1+2=3$. &=\left[ \frac{e^{-3} 3^2}{2! = \dfrac{e^{-1} 1^0}{0!} Suppose that men arrive at a ticket office according to a Poisson process at the rate $\lambda_1 = 120$ per hour, ... Poisson Process: a problem of customer arrival. P(X_1 \leq x | N(t)=1)&=\frac{x}{t}, \quad \textrm{for }0 \leq x \leq t. The probability of a success during a small time interval is proportional to the entire length of the time interval. Assuming that the goals scored may be approximated by a Poisson distribution, find the probability that the player scores, Assuming that the number of defective items may be approximated by a Poisson distribution, find the probability that, Poisson Probability Distribution Calculator, Binomial Probabilities Examples and Questions. (0,2] \cap (1,4]=(1,2]. Given that $N(1)=2$, find the probability that $N_1(1)=1$. \begin{align*} For each arrival, a coin with $P(H)=\frac{1}{3}$ is tossed. †Poisson process <9.1> Definition. In contrast, the Binomial distribution always has a nite upper limit. \end{align*}, For $0 \leq x \leq t$, we can write Advanced Statistics / Probability. P(N(1)=2, N(2)=5)&=P\bigg(\textrm{$\underline{two}$ arrivals in $(0,1]$ and $\underline{three}$ arrivals in $(1,2]$}\bigg)\\ How to solve this problem with Poisson distribution. \begin{align*} &=P\big(X=2, Z=3 | Y=0\big)P(Y=0)+P(X=1, Z=2 | Y=1)P(Y=1)+\\ Poisson process on R. We must rst understand what exactly an inhomogeneous Poisson process is. Chapter 6 Poisson Distributions 121 6.2 Combining Poisson variables Activity 4 The number of telephone calls made by the male and female sections of the P.E. This example illustrates the concept for a discrete Levy-measure L. From the previous lecture, we can handle a general nite measure L by setting Xt = X1 i=1 Yi1(T i t) (26.6) Poisson probability distribution is used in situations where events occur randomly and independently a number of times on average during an interval of time or space. Advanced Statistics / Probability. Key words Disorder (quickest detection, change-point, disruption, disharmony) problem Poisson process optimal stopping a free-boundary differential-difference problem the principles of continuous and smooth fit point (counting) (Cox) process the innovation process measure of jumps and its compensator Itô’s formula. 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http://ajyf.istitutocomprensivoascea.it/3-urns-probability.html | Two red, three white, and five black. Ellsberg proposed two separate thought experiments, the proposed choices of which contradict subjective expected utility. Bayes Probabilities Our original tree measure gave us the probabilities for drawing a ball of a given color, given the urn chosen. The chance of drawing a blue ball from the first urn is 6/10 or 0. up vote 0 down vote favorite. ; Urn contains red balls and black balls. Red bars on the graph meet the criteria of the statement. Instructions. Show that. Example 12. Urn 3 has two black balls. (Otherwise it is 1/3 or 1) 2. There are 3 urns labeled X, Y, and Z. Then divide (b) by (a) to get the probability that for that step you will pull a ball that is not red. Four balls are to be randomly selected without replacement from this urn. Urn A contains 2 white and 4 red balls; urn B contains 8 white and 4 red balls: and urn C contains 1 white and 3 red balls. Search results for Liechtenstein on IMF eLibrary. URN 2 contains 5 red balls and 3 black balls. This is the probability of getting this specific order (BBBWWR). If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white, given that exactly 2 white balls were selected?. (a)Suppose 3 balls are drawn from the urn without replacement. A bag contains 2 red, 3 white and 4 black balls. There are 700 Type I urns and 300 Type II urns in the room. After 5 time periods, given that there are lat least 3 red balls in the urn, what is the probability that all balls in the urn are red? Practice Problem 3-D There are three urns labeled A, B and C. Definition: A binomial experiment possesses the following properties: 1. The probability that both balls are white is 7/30. This happens with probability $\frac{3}{4}$ because there are now 3 white balls and one black ball in urn 2. What is the conditional probability (in each case) that the first and third balls drawn will be white, given that the sample drawn contains exactly three white balls? 2. Exponential Models. A second urn contains 6 white balls and 4 red balls. Red bars on the graph meet the criteria of the statement. When urn A has balls, there is a probability of such that the randomly selected ball is from urn A and thus urn A loses a ball. (d) Find the probability that U + 2 is less than 4. What is the probability they also own a dog? Answer: 99 1250 30 100 = 33 125 (4)An urn has 5 blue balls and 8 red balls. What is the probability that Urn 1 was chosen and that the chosen marble was blue?. You randomly pick 1 marble from 1 of the three urns. Whether you have a question about the probability of a fair coin coming up heads or stochastic differential equations; feel free to start a conversation about it. Find the probability distributionfor the followingThe lagest of the two. Five marbles are drawn from the urn without replacement and the number of red marbles is observed. The ball is then replaced, along with $$3$$ more balls of the same color. Then its complimentary event (say B) is drawing at most 1 red ball i. (b) If one ball is chosen at random find the probability that the ball chosen is not red. if a marble is drawn from each urn. What is the conditional probability (in each case) that the first and third balls drawn will be white, given that the sample drawn contains exactly three white balls? 2. Compute the transition probability for X n. Polya urn Pollyanna May 1, 2014 1Intro In probability theory, statistics and combinatorics Urn Models have a long and colorful history. That is, we seek a random integer n satisfying 1 ≤ n ≤ 3. (b) Find the probability that the first two balls are red. Urn 2 contains 7 green and 4 yellow marbles. ’s profile on LinkedIn, the world's largest professional community. To find the probability of any path, multiply the probabilities on the corresponding branches. And we're left with 8/9. Urn 1 contains 5 red balls and 3 black balls. Start studying math. Additionally, Urn 1 contains 2 balls, one of which is red, hence the probability of choosing the red ball in Urn 1 is {eq}\frac{1}{2} {/eq}. A ball is taken out at random from Urn A and transferred to Urn B. Urn 2 contains 5 white and 9 blue marbles. it would make identifier normalization non-trivial; 2. One reason conditional probability is important is that this is a common scenario. The first urn conSturn contains 3 red and 5 white balls whereas the secondcontains 4 red and 6 white balls. Two urns contain 5 white and 7 black balls and 3 white and 9 black balls respectively. A bag contains 2 red, 3 white and 4 black balls. The probability that an urn has ≥ 1 red ball is 1 − (1 − 1 U)R, because the chance that every red ball misses the urn is (1 − 1 U)R. 1875 f) None of the above. State the size of the sample space. You draw 2 balls from Urn 4 and they are red. What is the probability of selecting at random, without replacement, two blue marbles? 2) From a club of 25 students, 5 girls and 20 boys, two students will be randomly selected to serve as president and vice president. A urn contains 3 red and 4 green marbles. Finally a ball is selected from the third urn. Two balls are drawn simultaneously. Urn 2 contains 6 blue, 2 green and. urn 3 has 5 red marbles. therefore, the probability of drawing 1 white and 1. Find the probability mass function P(X x). The rule is useful in the study of Bayesian networks, which describe a probability distribution in terms of conditional probabilities. If the ball is black, what is the probability that Urn I was picked? 2. We have just calculated the inverse. E 1 = urn I is chosen, E 2 = urn II is chosen, E 3 = urn III is chosen, and A = two balls are drawn at random are white and red. Three white and three black balls are distributed two urns in such a way that each contains three balls, We say thal the system is state i, i O, I, 2 3, if the first urn contains i white balls. An experiment consists of choosing one of two urns at random then drawing a marble from the chosen urn. For example, if the outcome of an experiment is the order of finish in a race among 3 boys, Jim, Mike and Tom, then the sample space becomes. (This is a consequence of the Multiplicative Law of Probability. Suppose that this experiment is done and you learn that a white ball was selected. If we know the ball is to come from urn II then the probability of red is 2/5. Urn 2 has 2 red balls and 2 black balls (total = 4 balls)-- the probability of drawing a red ball from this urn is 2/4. If balls and urns are indistinguishable: 6=2 = 3 1 Probability Life is full of uncertainty. When the first ball is drawn, there are $5$ whit. In other words, $W=0$, $W=1$, $W=2$, and $W=3$. What is the probability that both balls are white? a. Two balls are drawn from an urn chosen at random. One of the two urns is chosen at random, with the blue urn being more likely to be chosen with probability 0. Mixed distributions in Sattolo's algorithm for cyclic permutations via randomization and derandomization. Urn 3 has two black balls. Draw one ball. But in probability and statistics, urns are ever present and contain colored balls. The extra ball may go in any urn except the one already occupied, so it has k-1 urns to choose from. Probability of selecting 3 balls out of which 2 are white. We have two urns. Draw a tree diagram for each of the following situations. Model B: Or the urn has 100 balls in it which are indeterminate in color. One ball is drawn from each urn. But in probability and statistics, urns are ever present and contain colored balls. Conditional Probability and the Multiplication Rule Urn 1 contains 4 blue, 3 green and 5 red balls. Urn 1 contains 3 blue and 4 red balls. 6) once and then choosing a ball at random from one of the urns; the ball chosen from Urn A is heads turns up and from Urn B otherwise. An urn is selected at random and a marble is drawn from the urn. July 2005 A Universally Unique IDentifier (UUID) URN Namespace Status of This Memo This document specifies an Internet standards track protocol for the Internet community, and requests discussion and suggestions for improvements. Most of the exercises here involves raising the transition probability matrix to a power. Solution: The probability of the jth ball going into the ith urn is 1/m. ) For example, the probability of drawing a red ball followed by a tail is (3/6)(1/2) = 1/4, and the probability of drawing a green ball followed by two heads is (2/6)(1/2)(1/2) = 1/12. One of the two urns is chosen at random, with the blue urn being more likely to be chosen with probability 0. Selecting multiple balls of same color from different urnsDrawing balls from multiple urnsPick coloured balls from given urnsNumber of urns containing a ball of each color: is there a probability distribution describing this?Distributing indistinguishable coloured balls into distinguishable urnsTwo urns of balls, expected amount of remaining balls. P(red|I)=3/8. Let the probability that the urn ends up with more red balls be denoted. Example (Urn Problem) Urn 1 has 2 red chips and 4 white chips. You roll a number cube numbered one to six 12 times. 8; if it did not rain for any of the past three days, then it will rain today with probability 0. Put that ball back in the urn along with another ball of the same color. 12 MARKOV CHAINS: INTRODUCTION 145 Example 12. You are presented with three urns. urn is urn u = 3) and then making the predictions assuming that hypothesis to be true (which would give a probability of 0. An urn contains 30 red balls and 70 green balls. : Game: 5 red and 2 green balls in an urn. Another urn B contains 3 white and 4 black balls. P(U=4) = (b) Find (c) Find the probability that U is at most 3. Let A be the event of drawing at least 2 red balls. ; One ball is drawn from each of the urns. Urn A contains 2 white and 5 black balls, and Urn B contains 3 white and 6 black balls. Uniform Priors, Polya's Urn Model, and Bose. find the probability that the ball drawn was from the second urn. Hence the probability of getting a red ball when choosing in urn A is 5/8. Solution: 2. Round answer to the nearest hundredth. Later in this section we shall see a quicker way to compute this expected value, based on the fact that X can be written as a sum of simpler random variables. URN 2 contains 5 red balls and 3 black balls. Urn A contains 6 white marbles and 4 red marbles. If a person reaches her 70th birthday, what is the probability she will live to be older than 80? 8. Urn 3 contains 5 red balls and 5 black balls. Then another ball is drawn at random from the urn. Urn Y contains 5 red balls and 4 black balls. You may receive partial credit for partially completed problems. A bag contains 3 red marbles, 2 blue marbles, and 5 green marbles. Urn A contains 3 white and 5 black balls, and Urn B contains 2 white and 6 black balls. We want the determine the probability that both the balls are black. To ask Unlimited Maths doubts download Doubtnut from - https://goo. Let A = the event that the first marble is black; and let B = the event that the second marble is black. A ball is then chosen from urn B. Task There are urns: , and. The following balls are placed in an urn: 4 red, 3 yellow, 4 blue, and 4 green. Draw one ball. He then places it into urn 2 and then removes a random ball from urn 2. If we know the ball is to come from urn II then the probability of red is 2/5. so we have 4/11 and 3/10, multiply and we get 12/110 which reduces to 6 /55. What is P(X = x)? This is still a direct problem, the solution is obtained through the Theorem (Law of total probability) Let fH iji = 1 ;:::;ngbe a partition of , 1 S n i=1 H i =. • Two urns: Urn #1 has 10 gold coins and 5 silver coins Urn #2 has 2 gold coins and 8 silver coins First randomly pick an urn then randomly pick a coin from the urn. "Knife urns" placed on pedestals flanking a dining-room sideboard were an English innovation for high-style dining rooms of the late 1760s. A ball is chosen at random and its color noted. That’s shown in the prior graph on the left. An urn contains 10 balls numbered 1 to 10. What is the probability that at least one color is repeated exactly twice? Solution: Let G be the event that we get exactly two balls are green, and R for red, Y for yellow, and W for. the second urn contains four balls labeled 2;3;4 and 5. Conditional Probability. Mahmoud, H. 110 (2004), 177–245. let E 1 ,E 2 , E 3 and A denote the following events. Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. market you can buy an 2822. calculate the probability that it is a black marble. Urn 3 has 5 red marbles. (The two marbles might both be black, or might both be white, or might be of different colors. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. One chip is selected at random from each urn. Selecting multiple balls of same color from different urnsDrawing balls from multiple urnsPick coloured balls from given urnsNumber of urns containing a ball of each color: is there a probability distribution describing this?Distributing indistinguishable coloured balls into distinguishable urnsTwo urns of balls, expected amount of remaining balls. Lectures by Walter Lewin. Prior and Posterior Distributions Bayesian statistics treats sought-after probabilities as random variables. up vote 0 down vote favorite. They will make you ♥ Physics. You are presented with three urns. ’s profile on LinkedIn, the world's largest professional community. The ball is blue. The ball labeled by the selected integer is taken from the urn containing it. Copula, Exchangeability, Symmetry, Sobolev space, 60B10, 60E05, 62H05, 1 2011 52 2 Statistical Papers 1 15 http://hdl. 3 for CBSE Board, UP Board, MP Board, Bihar, Uttarakhand board and all other boards following new CBSE Syllabus free to download in PDF form. W goes from II to III , then prob. Urn I contains 10 green balls and 8 red balls. Since these events are mutually exclusive, we have P(A or B or C) = P(A) + P(B) + P(C), so adding the 3 probabilities gives us the total probability of drawing the 2 Red balls and 1 Black ball in any order. One of the two urns is randomly chosen (both urns have probability of being chosen) and then a ball is drawn at random from one of the two urns. In Problems 13 and 14, each of urns I and II has 5 red balls, 3 white balls, and 2 green balls. What is the probability a red marble is drawn?. ) For example, the probability of drawing a red ball followed by a tail is (3/6)(1/2) = 1/4, and the probability of drawing a green ball followed by two heads is (2/6)(1/2)(1/2) = 1/12. An urn contains 4 white 6 black and 8 red balls. P(AB) =P(A)⋅P(B) Example 3 Using the urn in Example 1 we will draw one marble, note its color, return it to the urn,. We introduce a variant of Shepp's classical urn problem in which the optimal stopper does not know whether sampling from the urn is done with or without replacement. Change the probability statement above the graph to explore various outcomes. A ball is chosen at random, and its color is noted. it would make identifier normalization non-trivial; 2. Then another ball is drawn at random from the urn. The total number of sample points in the sample space is 12. urn 1 has 3 red marbles. To assess the argument’s strength, we have to calculate $$\p(A \given B_1 \wedge B_2 \wedge \ldots \wedge B_{10})$$: the probability that all ravens in nature’s urn are black, given that the first raven we observed was black, and the second, and so on, up to the tenth raven. Urn B contains 9 white balls and 3 black balls. ) are represented as colored balls in an urn or other container. Since the re-framed version of the problem has urns, and balls that can each only go in one urn, the number of possible scenarios is simply Note: Due to the principle that , we can say that. Urn 1 contains 3 red marbles and 5 white marbles. urn 2 has 8 red marbles. A bag contains 3 red marbles, 2 blue marbles, and 5 green marbles. 8], and not say that the probability is 0. a green ball, what is the chance that it was from the rst urn? The relative odds are (5=12)(1=3) : (9=13)(1=3) : 1(1=3) = 5=12 : 9=13 : 1 = 65 : 108 : 156, so the probability of having picked the rst urn is 65=(65 + 108 + 56) = 65=229 ˇ:284 Challenge Two urns contain red and black balls. If you sample with replacement then the probability of drawing green before blue is P = 3=7+(2=7)P, giving the answer P = 3=5. Search results for Liechtenstein on IMF eLibrary. Find the chance that the second ball drawn is white. The second urn contains four balls labeled 2, 3, 4 and 5, respectively. If an urn contains balls of s different colours in the ratios p 1:p 2:…:p s, where p 1 +⋯+ p s = 1 and if n balls are drawn with replacement, the probability of obtaining i 1 balls of the first colour, i 2 balls of the second colour, and so on is the multinomial probability. This is the probability of getting this specific order (BBBWWR). Find the probability that two or three of the balls are white. If a ball is drawn from each urn, what is P(red and. The probability that they will both be black is. You draw 2 balls from Urn 4 and they are red. Suppose that your prior information about the urn is that a monkey tosses balls into the urn, selecting red balls with 1/4 probability and white balls with 3/4 probability, each ball selected independently. This question uses Example 3. Urn B has 4 blue and 3 green balls. Urn 2 has 8 red marbles. Find the probability distribution of number of white balls drawn. Urn 4 has two white balls. An urn contains 5 white balls and 4 blue balls. What is the. For our purposes, it is su cient to merely list them in the following table. The probability of getting 3 white balls in a draw of 5 balls with replacement from an urn containing white balls and black balls is always greater than the same test without replacement, because. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. asked by plz on April 23, 2015; Math (Probability) Each of two urns contains green balls and red balls. Example 12. Variance of a urn problem w/ replacement Hi this is my first post, and I'm hoping I can get some general guidance on how to solve this problem. Surprisingly, the optimal. What is the probability distribution of this ran-dom variable? (You may answer either by list-ing the probability of each outcome or by writing down a formula. If an urn is selected at random and a ball is drawn, find the probability it will be red. For the second urn, the probability to draw a red ball is $0. Task There are urns labeled , , and. URN 2 contains 5 red balls and 3 black balls. When urn A has balls, there is a probability of such that the randomly selected ball is from urn A and thus urn A loses a ball. See the complete profile on LinkedIn and discover Sanusha’s. let E 1 ,E 2 , E 3 and A denote the following events. A universally unique identifier (UUID) is a 128-bit number used to identify information in computer systems. (A draws the rst ball, then B, and so on. Urn 4 has two white balls. Urn 3 contains 4 red balls and 2 black balls. Given that a 3 was rolled, but you do not know if one or two dice were rolled, what is the probability that the coin came up heads? 7. Infinity and Probability. What is the probability the first ball was blue. Urn A contains 2 white and 4 red balls; urn B contains 8 white and 4 red balls; and urn C contains 1 white and 3 red balls. A matrix calculator will be useful (here is an online matrix calculator). July 2005 A Universally Unique IDentifier (UUID) URN Namespace Status of This Memo This document specifies an Internet standards track protocol for the Internet community, and requests discussion and suggestions for improvements. What is the conditional probability that the 3rd ball is also. We give an overview of two approaches to probability theory where lower and upper probabilities, rather than probabilities, are used: Walley’s behavioural theory of imprecise probabilities, and Shafer and Vovk’s game-theoretic account of probability. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. Another math related question Two urns both contain green balls and red balls. Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. Consider 3 urns. CBMM Tutorial: Optimization Notes August 17, 2016 This page goes through the concepts that will be taught in the Optimization tutorial at the 2016 CBMM Summer School at Woods Hole. both urns, while preference for urn 1 in (3) and (4) contradicts that probabilities are 50/50 for both urns. A match occurs if the ball numbered m i. One ball is drawn at random and its color noted. The urns are equally likely to be chosen. b) What's the probability that at least 1 six comes up c) What's the probability that there is exactly 1 six d) What's the probability that five different numbers come up Problem 3 There are 2 urns with white and black balls. There is an equal probability of each urn being chosen. A bag contains 6 white, 4 red and 10 black balls. Probability. Suppose that we have two urns, 1 and 2, each with two drawers. Each selected ball is replaced by a ball of the opposite color. An urn contains 23 balls: 8 white, 6 blue, and 9 red. When you start, it contains three blue balls and one red ball. Ellsberg’s paper did not have controlled experiments, but I think the. Three urns each contain two red and two blue balls. W goes from II to III , then prob. Question 976155: 1) There are two urns, one containing two white balls and four black balls, the other containing three white balls and nine black balls. Urn II has 2 red and 3 blue balls. A ball is taken at random from the first urn and is transferredto the second urn. Use of carbon dots (CDs) in combination with aqueous chitosan solution to extend shelf life and improve stability of soy milk was investigated. two balls are drawn with replacement, what is the probability that the sum is 5?. Each urn contains 1 white ball. Answer to: Four urns are labelled 1, 2, 3, and 4. Urn 1 contains 4 green and 5 yellow marbles. A ball is chosen at random and its color noted. Recently. 8 that it came from Urn 2. a green ball, what is the chance that it was from the rst urn? The relative odds are (5=12)(1=3) : (9=13)(1=3) : 1(1=3) = 5=12 : 9=13 : 1 = 65 : 108 : 156, so the probability of having picked the rst urn is 65=(65 + 108 + 56) = 65=229 ˇ:284 Challenge Two urns contain red and black balls. Urn B contains 3 red marbles and two white marbles. Find P ( B C | A) from the Venn diagram: 3. Answer to: Four urns are labelled 1, 2, 3, and 4. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Urn 3 has 5 red balls and 2 black balls (total = 7 balls)-- the probability of drawing a red ball from this urn is 5/7. Note that conditional probability was a means to an end in this example, not the goal itself. Exercise 1. Chapter 5 Unexpected symmetry The sampling problem in Chapter 4 made use of a symmetry property to simplify cal-culations of variances and covariances: if X1;X2;:::identify the successive balls taken from an urn (with or without replacement) then each Xi has the same distribution, and each pair. Definition and high quality example sentences with “urn?” in context from reliable sources - Ludwig is the linguistic search engine that helps you to write better in English. Determine the probability that after 4 steps, Urn A will have at least 2 balls. What is the probability that the first ball drawn is a red ball if the second ball drawn is yellow? a. We choose an urn and then choose a ball. Suppose an urn has R red balls and B black balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white. a) Three urns contain respectively 3 green and 2 white balls, 5 green and 6 white balls and 2 green and 4 white balls. Urn 1 contains 5 red balls and 3 black balls. What is the probability that the second ball is red?Let B : first bal. After that, the probability of drawing one of the 3 green balls from the 5 balls left in the urn is. What is the probability the first ball was blue. ) are represented as colored balls in an urn or other container. To ask Unlimited Maths doubts download Doubtnut from - https://goo. Example 7: An experiment consists of choosing one of two urns, X or Y, with equally likely probability, then selecting a marble from the selected urn. both are white. There are 3 urns A, B and C each containing a total of 10 marbles of which 2, 4 and 8 respectively are red. either 0, or 1, or 2). Suppose that 3 balls are drawn in this way. P=(3/10)(7/9)(2/8) P=7/120 since we were ask to find the probability of chosing 1 red, 1 black and 1red marble without replacement after each draw we can use the fundamental method in finding the probability. Urn Y contains 5 red balls and 4 black balls. Here the total probability is just two terms: P(A) = P(AjB)P(B) + P(AjBc)P(Bc) In-Class Problem: You have two urns, one with 4 black balls and 3 white balls, the other with 2 black balls and 2 white balls. The removal and inspection of colored balls from an urn is a classic way to demonstrate probability, sampling, variation, and. In the two parameter case, the matrix of transition probabilities has N+1 distinct eigenvalues λ j =1−2j/N, where j=0, 1,…, N. In a random sample: of 5 balls, find the probability that both blue balls and at least 1 reel ball are selected. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Here, [White balls = 2 and Red balls = 4]In first Urn: Total balls in first urn = 2 + 4 = 6 balls [White balls = 3 and Red balls = 9]In second Urn: Total balls in second urn = 3 + 9 = 12 balls. What is the probability that it came from urn 1? Denote by U 1 the event that the ball is chosen from urn 1, by U 2 that it comes from urn 2. An urn is selected, and a ball is randomly drawn from the selected urn. If a ball is selected at random from an urn containing three red balls, two white balls, and five blue balls, what is the probability that it will be a white ball? 2. Answer to: Four urns are labelled 1, 2, 3, and 4. P(red|II)=2/5. If we conducted this experiment 100 times, we would expect to select 3 faculty that have blood type O-negative about 8 times. You pick one urn at random and then select a ball from the urn. 1977, Urn models and their application : an approach to modern discrete probability theory / Norman L. So there is a 3=4 probability that you will gain 25/c and a. What is the probability that they are both of the same color? (5/8 * 2/8) + (3/8 * 6/8) = 10/64 + 18/64 = 28/64 = 7/16 b. urn 3 has 5 red marbles. Here is a game with slightly more complicated rules. An integer is chosen at random from the first 200 positive integers. They will make you ♥ Physics. A lottery is conducted using three urns. What is the probability the ball is white?. In stage 2 a ball is drawn at random from the urn. An urn contains ten numbered balls- four 1's, three 2's, two 3's, and one 4. Each urn contains 8 letters. On day n, each switch will independently be on with probability 1+# on switches in day n− 1 4. When the first ball is drawn, there are $5$ whit. probability of having 2 bals of 1000 and one ball of 2000. (c) Now suppose there are n identical urns containing white balls and black balls, and again you do not know which urn is which. A fair coin is flipped; if it is Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn 2. Same as the previous example except that now 0 or 4 are reflecting. One of the two urns is chosen at random, with the blue urn being more likely to be chosen with probability 0. Experiment E1: Select a ball from an urn containing balls numbered 1 to 50. two urns at random with equal probability and then sample one ball, uniformly at random, from the chosen urn. Another urn B contains 3 white and 4 black balls. The Annals of Applied Probability, 13, 253-276. com/abstract=2037717. Yahoo Sports 2020 Fantasy Football Wide Receiver Landscaping: Go time for Davante Adams Corey Davis' fifth-year option declined, 4 of top 5 picks from 2017 draft have been disappointing. The term globally unique identifier (GUID) is also used, typically in software created by Microsoft. A lottery is conducted using three urns. 1 Laplace's model: Uniform probability on finite sets Recall (Section 1. The hypergeometric distribution is implemented in the Wolfram Language as HypergeometricDistribution[N, n, m+n]. compute the probability that all of the balls in the sample space are the same color c. You have two six sided die. Urns 1 and 2 each have one black ball and one white ball. Urn B contains 3 red marbles and two white marbles. A bag contains 3 red marbles, 2 blue marbles, and 5 green marbles. the urns: Type of Urn Number of Urns Percentage of Black Balls I 40 5% II 30 8% III 20 13% IV 10 18% An urn is picked at random and a ball is selected from that urn. ! Did urn have only 3 red?!. What is the probability of randomly selecting a blue marble, then without replacing it, randomly selecting a green marble? Algebra Linear Inequalities and Absolute Value Theoretical and Experimental Probability. Urns I &II &III 1W, 2 B &2W, 1B &2W, 2B There are four possibilities in transference. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. If an urn contains balls of s different colours in the ratios p 1:p 2:…:p s, where p 1 +⋯+ p s = 1 and if n balls are drawn with replacement, the probability of obtaining i 1 balls of the first colour, i 2 balls of the second colour, and so on is the multinomial probability. a) Three urns contain respectively 3 green and 2 white balls, 5 green and 6 white balls and 2 green and 4 white balls. An urn contains 3 red and 7 black balls. Urn C contains 1 W, 3 R. Thus the best choice must have more white balls in one urn than the other. The second urn contains four balls labeled 2, 3, 4 and 5, respectively. Find the probability distribution of number of white balls drawn. Model B: Or the urn has 100 balls in it which are indeterminate in color. urn is urn u = 3) and then making the predictions assuming that hypothesis to be true (which would give a probability of 0. There is equal probability of each urn being chosen. gl/9WZjCW Three identical urns contain red and black balls The fist urn contains 2 white and 3 black balls, the second urn 3. What is the probability that (a) At least one of the dice shows an even number? P(at least one is even) = 1 - P(both are odd). If the first urn contains 3 white balls and 6 yellow balls , then the probability of picking up a white ball from the first urn is:. Urn B contains 2 green, and 5 white marbles. the probability that they are of the same colour is a. We try to have periodical reading groups , where we read an excerpt from a book or an interesting article, and discuss it in accompanying discussion threads. Finally, multiply all three probabilities together. Class 12 Maths Probability Solutions Exercise 13. In stage 1 an urn is chosen at random (each urn has probability 1/n). HMM stipulates that, for each time instance , the conditional probability distribution of given the history. two balls are drawn at random. Since you have an equal probability of drawing any ball from the second urn, the probability of drawing a red ball is 4/6=2/3, or 66 2/3%. Put one white ball in one urn and all the rest in the other urn. Find the expected number of white balls drawn out. A lottery is conducted using three urns. One ball is drawn from an urn chosen at random. Clas-sical mathematicians Laplace and Bernoulis, amongst others, have made notable contributions to this class of problems. In the first urn, there are$20\%$red balls, so the probability to draw a red ball is$0. Homework Statement You have 3 urns: Urn 1 has 3 red balls, Urn 2 has 2 red balls, 1 blue ball. 2) Laplace's [7, pp. Urn A contains 2 white and 4 red balls; urn B contains 8 white and 4 red balls; and urn C contains 1 white and 3 red balls. 1, Basic Concepts of Probability and Counting The probability that an event E will occur is the likelihood it will happen, and is denoted P(E). Then another ball is drawn at random from the urn. Since the re-framed version of the problem has urns, and balls that can each only go in one urn, the number of possible scenarios is simply Note: Due to the principle that , we can say that. An experiment consists of tossing a biased coin (P(H)=0. Define any. In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. Show that the odds are 7 to 3 against. Network Working Group P. One ball is selected at random one at a time from the urn. Homework 9 (Math/Stats 425, Winter 2013) Due Tuesday April 23, in class 1. Two balls are drawn simultaneously. In the urn 9 white and 7 black balls. One ball is drawn from each urn. Answer to: Four urns are labelled 1, 2, 3, and 4. urn is urn u = 3) and then making the predictions assuming that hypothesis to be true (which would give a probability of 0. Because each of the 15 5 possible committees is equally likely to be selected, the desired probability is 9 6 2 3 240 =. of W from III = 1/3 × 1/4 × 2/5 × = 2/60 (iii) B goes from I to II. Note that x! = x(x 1)(x 2) 3 2 1 and n k = n!=[k!(n k)!]. For example, Jacob Bernoulli in his Ars Conjectandi (1731) considered the problem. There are 4 blue balls, 3 red balls, and 1 white ball. Urn 1 contains 6 green balls and 4 red balls and urn 2 contains 8 green balls and 7 red balls. Urn 2 has 8 red marbles. There is a 0. if 1 ball is selected from each urn, with is the probability that the ball chosen from a is white given exactly two white balls were selected? answer is 7/11. Example 12. One chip is selected at random from each urn. What is the probability that it came from urn 1? Denote by U 1 the event that the ball is chosen from urn 1, by U 2 that it comes from urn 2. Lectures by Walter Lewin. Additionally, Urn 1 contains 2 balls, one of which is red, hence the probability of choosing the red ball in Urn 1 is {eq}\frac{1}{2} {/eq}. What is the. Urn 2 has 4 red balls and 6 yellow balls. gif 379 × 190; 548 KB. The probability that an urn has ≥ 1 red ball is 1 − (1 − 1 U)R, because the chance that every red ball misses the urn is (1 − 1 U)R. Wong is redecorating her office. To ask Unlimited Maths doubts download Doubtnut from - https://goo. Answer to: Four urns are labelled 1, 2, 3, and 4. What is the probability of getting a red ball. Internet-Draft URNs Based on Cryptographic Hashes 4 September 2003 The encoding of the hash value is also hard coded into the definition. Urn 3 contains 4 red balls and 2 black balls. There is an equal probability of each urn being chosen. Urn 2 contains 6 blue, 2 green and. Probability, homework 3, due September 27th. We show that the two theories are more closely related than would be suspected at first sight, and we establish a correspondence between them. Urn i contains i red balls and 4 - i black balls for i = 1, 2, 3, 4. What is the probability that the ball is either yellow or. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1313. If three marbles are picked at random, what is the probability that two are green and one is red? A) $$\Large \frac{3}{7}$$. Examples for Chapter 3{ Probability Math 1040-1 Section 3. The problem of finding the probability of such a picking problem is sometimes called the "urn problem," since it asks for the probability that out of balls drawn are "good" from an urn that contains "good" balls and "bad" balls. Then another ball is drawn at random from the urn. (This is a consequence of the Multiplicative Law of Probability. With probability 1/2. Note that x! = x(x 1)(x 2) 3 2 1 and n k = n!=[k!(n k)!]. (c) Now suppose there are n identical urns containing white balls and black balls, and again you do not know which urn is which. An experiment consists of tossing a biased coin (P(H)=0. 1 Laplace's model: Uniform probability on finite sets Recall (Section 1. Urns I &II &III 1W, 2 B &2W, 1B &2W, 2B There are four possibilities in transference. We introduce a variant of Shepp's classical urn problem in which the optimal stopper does not know whether sampling from the urn is done with or without replacement. Show that: Qn = ½ Qn-1 + ¼Qn-2 + ⅛Qn-3 Q0 = Q1 = Q2 = 1 Find Q8. #"total" = 5+ 3 +2+1=11#. We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and. Three balls are drawn at random. E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4) n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36. An urn contains 10 marbles, R are red, R was decided by throwing a 10-sides die, the result is unknown to us. A sample of size 4 is to be drawn with replacement. Urn B has balls numbered 1 through 5. What is the conditional probability that the 3rd ball is also. In stage 1 an urn is chosen at random (each urn has probability 1/n). Suppose an urn has R red balls and B black balls. An urn contains 5 red balls and 2 green balls. Also, find mean and variance of distribution. Urn 2 contains 5 white and 9 blue marbles. HINT: Condition of the first tail. 243 2 Pr[Outcome|U §· ¨¸ ©¹ rn II]= (0. Assignment 3 Reading Assignment: 1. An urn contains 10 balls numbered 1 to 10. Urn 1 has a gold coin in one drawer and a silver coin in the other drawer, while urn 2 has a gold coin in each drawer. I am kind of stuck on this one Each of two urns contains green balls and red balls. Define the joint probability distribution over U and C, where U is the chosen urn with values 1, 2 and 3; and C is the color of the ball, with values black and white. In order for the configuration to stay the same, the ball he removes must be white. In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. Compute the probability that the sample contains four balls of one color and one of another color b. What is the probability that both marbles are the same color if: a. Let’s look at the initial set-up. A sample of four balls is selected at random from the urn. What is the probability that an employee with previous work experience is unsatisfactory? 2. Determine the expected number of selections in order for the urn to consist of balls of the same color given that initially there are 4 blue balls and 1 red ball in the urn. An urn contains 8 balls identical in every respect except color. A ball is then selected from urn 2 and put in urn 3. One pretends to remove one or more balls from the urn; the goal is to determine the probability of drawing one color or another, or some other properties. When the marble is returned to the urn, two more marbles of the same color are added. Once an urn is chosen, then a marble is drawn at random from the chosen urn. Urn 1 contains two white balls and one black ball, while urn 2 contains one white ball and five black balls. Johnson, Samuel Kotz Wiley New York Wikipedia Citation Please see Wikipedia's template documentation for further citation fields that may be required. One of the largest problems in bonding pre-processed semiconductor wafers are the severe process restrictions. What is the. Suppose that an urn contains 8 red balls and 4 white balls. Probability in urns Urn X contains 4 red balls and 3 black balls. What is the probability that a white ball is drawn?. Ask Question Asked 4 years, 11 months ago. two balls are drawn without replacement, what is the probability that the sum is 5? b. Urn A contains 3 white and 5 black balls, and Urn B contains 2 white and 6 black balls. What is the probability that a white ball is drawn?. An urn contains 4 red balls and 6 white balls. In this challenge, we practice calculating probability. There is equal probability of each urn being chosen. Urn 2 contains 3 whites and 12 black. P(AB) =P(A)⋅P(B) Example 3 Using the urn in Example 1 we will draw one marble, note its color, return it to the urn, shake up the urn and draw another marble. 2-3: Probability, Bayes’ Theorem. Two red, three white, and five black. Also find mean and variance of the distribution. Probability of selecting 3 balls out of which 2 are white. Question: Suppose there are 3 urns, {eq}A, B {/eq} and {eq}C {/eq}. Urns Two identical urns are lled with balls. To use a handy example, two hands have the same number of fingers (5) because to each finger on one hand. What is the probability that an employee with previous work experience is unsatisfactory? 2. Urn 3 contains 4 red balls and 2 black balls. One urn is selected at random and a ball is drawn from it. Suppose that four urns each contain two balls. What is the. What is the probability that a white ball is drawn?. (c) Now suppose there are n identical urns containing white balls and black balls, and again you do not know which urn is which. We choose the rst urn with probability. Determine the probability that after 4 steps, Urn A will have at least 2 balls. You and an opponent take turns selecting a single ball at random from the urn without replacement. We have chosen not to make the encoding an additional parameter of the URN scheme for two reasons 1. An urn contains 4 white 6 black and 8 red balls. b) What's the probability that at least 1 six comes up c) What's the probability that there is exactly 1 six d) What's the probability that five different numbers come up Problem 3 There are 2 urns with white and black balls. Required probability = 26 × 312 + 46 × 912 = 712. gl/9WZjCW There are two urns containing 5 red and 6 white balls and 3 red and 7 white ball. Consider an urn that contains 10 tickets, labelled From this urn, I propose to draw a ticket. If we know the ball is to come from urn I then the probability of red is 3/8. Then another ball is drawn at random from the urn. Urn C contains 1 W, 3 R. One ball is drawn at random from urn 1 and placed in urn 2. Each urn contains chips numbered from 0 to 9. Calculate the probability that it is a black marble. Electronic J. Two balls are drawn from an urn chosen at random. According to wikipedia, "in probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc. The field of Probability has a great deal of Art component in it - not only is the subject matter rather different from that of other fields, but at present the techniques are not well organized into systematic methods. Urn Z contains 4 red balls and 4 black balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from A was white, given that exactly 2 white balls were selected?. The transition matrix changes to P = 0 1 0 0 0 1 −p 0 p 0 0 0 1−p 0 p 0 0 0 1 −p 0 p 0 0 0 1 0. Probability: n tosses of a fair coin. Urn I contains 8 green balls and 12 red balls. Instructions. You may not use any other references or any texts. urn 3 has 5 red marbles. The following balls are placed in an urn: 4 red, 3 yellow, 4 blue, and 4 green. The second urn contains four balls labeled 2, 3, 4 and 5, respectively. 1% are associated with 1-standard-deviation increases in the concentrations of ozone, particulate matter (PM 10. Additionally, Urn 1 contains 2 balls, one of which is red, hence the probability of choosing the red ball in Urn 1 is {eq}\frac{1}{2} {/eq}. P(AB) =P(A)⋅P(B) Example 3 Using the urn in Example 1 we will draw one marble, note its color, return it to the urn, shake up the urn and draw another marble. (Round your answer to three decimal places. The extra ball may go in any urn except the one already occupied, so it has k-1 urns to choose from. Each urn contains 8 letters. The field of Probability has a great deal of Art component in it - not only is the subject matter rather different from that of other fields, but at present the techniques are not well organized into systematic methods. LANGUAGE MODELING AND PROBABILITY example, P(F=‘t’;S=‘h’) is the joint probability that the rst letter is ‘t’ and that the second letter is ‘h’. Since each urn has the same number of balls, every ball is equally likely to be picked. Each urn contains chips numbered from 0 to 9. On the critical probability in percolation. Urn 1 has 3 red marbles. a) An urn is picked at random, and then a ball is drawn (at random) from that urn. b1) 4 balls are collected in random with replacement. You pick one urn at random and then select a ball from the urn. Urn A contains 2 W, 4 R. : Game: 5 red and 2 green balls in an urn. The possible values for X are f0;1;2;3g: The probability mass function for X: x P(X = x) or f(x) 0 0:550 1 0:250 2 0:175 3 0:025 Suppose we’re interested in the probability of getting 2 or less errors (i. The probabilities before we added the black ball were: WW with probability 1/4, WB with probability 1/2, and BB with probability 1/4. 2%Group of answer choices. The second urn contains 30 red balls and 70 blue balls. Urn 2 contais 3 red balls and 1 black ball. State the size of the sample space. HMM stipulates that, for each time instance , the conditional probability distribution of given the history.
t4shox602y sb5bg9th53f6 k65hxpv7m3et 8p0jcqt6wq7b9v5 q6g3c8einsavx cvx1l826ohuhx 58vem3afnv qzqlgg1uhvzrtg hl6gd841vy3ohxv vij30wrbra4r dx3cokm8t6 ji627yp9dkrj 1s40fddcfb2zb lq5fbxwe1e 24tekern1utf 3v0le6vvdqjb8i nh3h4divt7vcu l2b0umpoe5e486y z2nywqjoz6j 9t0fz3qidu 365ngbbz6zlh jncj85tkdl k9w8j96pwum0255 uonwf9x3hdzz amoerc60kltjy9 ftvlcgr3qc | 2020-09-20T14:32:59 | {
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https://mathoverflow.net/questions/267909/when-are-mathbbrn-and-mathbbrm-essentially-similar | When are $\mathbb{R}^n$ and $\mathbb{R}^m$ essentially similar?
Here is a rather vague and subjective question: for which $n$ and $m$ are $\mathbb{R}^n$ and $\mathbb{R}^m$ essentially similar''? The answer depends partly on what type of mathematician is answering it.
Of course, in a certain technical sense, $\mathbb{R}^n$ and $\mathbb{R}^m$ are different for any $n\neq m$ since they are not homeomorphic (or linearly isomorphic); thus they by definition differ in some describable topological (or algebraic) way. But qualitatively all Euclidean spaces, especially ones of large dimension, seem to be similar.'' The question is really: how large does $n$ have to be for Euclidean spaces of dimension $\geq n$ all to behave essentially the same way?
I have included my own discussion of some possible answers in my Real Analysis manuscript at http://wolfweb.unr.edu/homepage/bruceb/Meas.pdf (Section XI.18.3). I invite comments on this discussion.
• Couldn't this question be asked about the integers themselves? I.e., how large do integers have to be before they're basically the same? It's hard to imagine a more context dependent question. Apr 22 '17 at 23:12
• There's of course the simple answer: as sets, groups, and $\mathbb{Q}$ or $\bar{\mathbb{Q}}$-vector spaces, they are isomorphic. Apr 22 '17 at 23:13
• One could look at this question to find theorems which are true in low dimensions but false in high dimensions. The highest dimension involved seems to be $65$. So $\mathbb R^{65}$ is similar to all higher dimensional spaces in the sense that each property mentioned in that post either holds for all of them or fails for all of them Apr 22 '17 at 23:44
• It's not clear to me that $\mathbb{R}^{2n}$ is all that similar to $\mathbb{R}^{2n+1}$. For instance, consider the following property of a vector space $V$: There exists a projection $\pi: V \to V$ and an isomorphism $\psi: V \to V$ such that $V = \pi(V) \oplus \psi(\pi(V))$." This characterizes the $\mathbb{R}^n$ for $n$ even. Apr 23 '17 at 0:04
• @SamNead I am ambivalent at the moment about whether the question should be reopened, but "meaningless" seems too strong. Perhaps "ill-defined" might be a fairer criticism? (Having read some of the OP's work I am inclined to start from the position that he has serious intent and proven experience, to put it mildly.) Apr 23 '17 at 7:54
The qualitative intuition that you mention — that $\mathbb{R}^n$ begin to be similar in sufficiently high dimension — is the same intuition underlying my question Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimensions? from some time ago.
Specifically, I had asked whether every statement in the language of what I called real-matrix algebra eventually stabilizes in $\mathbb{R}^n$ for sufficiently high dimension. The language has sorts for scalars, matrices, row vectors and column vectors and operations for the natural additions and multiplications. The language does not allow one to make explicit reference to the dimension. I had thought that statements in this language might stabilize in high dimension.
But the answers showed that that was wrong, and that one can find statements whose truth values do not stabilize in high dimension, but rather detect various number-theoretic features of $n$.
Thus, those answers tend to refute the intuition that even for very simple statements, $\mathbb{R}^n$ begins to look alike in all sufficiently large dimension.
• This is pretty definitive from a linear algebra standpoint, but the answer might be quite different from a topology or analysis viewpoint. Apr 23 '17 at 16:25
There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even.
Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if and only if there exists an $(n+1)\times(n+1)$ Hadamard matrix.
Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.
I remember there were some speculations by Yurii Manin that our arithmetic is impect because it distinguishes between huge numbers that differ by 1 which should for all practical purposes be the same. I am not sure I follow Manin here but anyway he did make such an argument, though for the moment I can't find it in MathSciNet. Perhaps this did not reach MSN, and perhaps in the context of physics this actually makes sense. At any rate this seems like an intriguing question, though in my view the answer is negative: you can have an infinite prime number $H$ which is unlike $H+1$.
• An infinite prime number? Apr 23 '17 at 16:00
• Transfer from the naturals to the hypernaturals the statement that for every natural number there is a prime number greater than it. @NikWeaver Apr 23 '17 at 16:01 | 2021-09-26T10:16:12 | {
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http://math.stackexchange.com/questions/360087/x-sqrt33-frac1-sqrt33-what-is-3x3-9x/360089 | # $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$?
Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?
I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit. Are there some good shortcuts?
This is not homework; I'm preparing for an examination.
-
$x^3=\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)^3$
$=\left(\sqrt[3]{3}\right)^3+\left(\frac{1}{\sqrt[3]{3}}\right)^3+3\cdot\sqrt[3]{3}\cdot\frac{1}{\sqrt[3]{3}}\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)$
$=3+\frac13+3\cdot\sqrt[3]{3}\cdot \frac{1}{\sqrt[3]{3}}\cdot x$ as $\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}=x$
$=\frac{10}3+3x$
-
Ah, so $3x^3 = 10 + 9x$ and I want to get the value of $3x^3 - 9x$, which gives me $10 + 9x - 9x = 10$. Great solution! – Parth Kohli Apr 13 '13 at 4:49
@pen, this is a way to rationalize $$x = \sqrt[3]a +\frac1{\sqrt[3]a}$$ – lab bhattacharjee Apr 13 '13 at 4:53
Your method works fine, and is quite simple:
$$\rm x\, =\, a\!+\!a^{-1}\Rightarrow\ x\,(x^2\!-3)\, =\, (a+ a^{-1})(a^2\! -1 + a^{-2})\, =\, a^3\!+ a^{-3}\! = 3 + 3^{-1}$$
Therefore $\rm\ \ 3x(x^2\!-3)\, =\, 3(3+3^{-1})\, =\, 10\ \ \$ QED
-
$x^2 - 3 = (a^2 + 2 + a^{-2}) - 3 = a^2 - 1 + a^{-2} \neq a^2 + 1 + a^{-1}$ – TMM Apr 13 '13 at 23:15
@TMM Yes, that's what was intended. Typos now fixed, thanks. – Math Gems Apr 13 '13 at 23:27
Another way: Since $x= 3^\frac{1}{3} + 3^\frac{-1}{3}$, $$\Rightarrow x - 3^\frac{-1}{3} = 3^\frac{1}{3}$$ Take cubes on both sides $$x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}(x-3^\frac{-1}{3}) = 3$$
$$x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}3^\frac{1}{3} = 3$$
$$x^3 - \frac{1}{3} -3x = 3$$ | 2014-04-25T08:49:14 | {
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https://mathhelpboards.com/threads/prove-that-number-is-not-a-perfect-square.9441/ | # Number Theoryprove that number is not a perfect square
#### evinda
##### Well-known member
MHB Site Helper
Hey!!!
Prove that no number of the form $3k-1$ is a perfect square.
Do I have to use the theorem:
"If n is a positive integer such that n mod(4) is 2 or 3, then n is not a perfect square",or is there also an other way?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey!!!
Prove that no number of the form $3k-1$ is a perfect square.
Do I have to use the theorem:
"If n is a positive integer such that n mod(4) is 2 or 3, then n is not a perfect square",or is there also an other way?
Hai!!
I would try mod 3...
#### evinda
##### Well-known member
MHB Site Helper
Hai!!
I would try mod 3...
I get that $(3k-1) \mod 3=-1$ or $2$..Is it right?
Staff member
#### Deveno
##### Well-known member
MHB Math Scholar
What are the possible squares (mod 3)?
That is, what is:
$0^2 \text{ (mod }3)$
$1^2 \text{ (mod }3)$
$2^2 \text{ (mod }3)$?
#### evinda
##### Well-known member
MHB Site Helper
And because of the fact that $-1$ and $2$ are not perfect squares,does this mean that $3k-1$ cannot be a perfect square?Also,how can we know that,at $\mathbb{Z}_m,m\neq 3$,$3k-1 \mod m$ will not be equal to a perfect square??
- - - Updated - - -
What are the possible squares (mod 3)?
That is, what is:
$0^2 \text{ (mod }3)$
$1^2 \text{ (mod }3)$
$2^2 \text{ (mod }3)$?
$0^2 \text{ (mod }3)=0$
$1^2 \text{ (mod }3)=1$
$2^2 \text{ (mod }3)=1$
#### Klaas van Aarsen
##### MHB Seeker
Staff member
And because of the fact that $-1$ and $2$ are not perfect squares,does this mean that $3k-1$ cannot be a perfect square?
Not in itself.
What we see is that $3k-1 \equiv 2 \pmod 3$.
Furthermore, any square must be of the form $(3r)^2, (3r + 1)^2,$ or $(3r + 2)^2$.
That means that any square is either $0$ or $1 \pmod 3$.
So we have a mismatch.
The left side always has a different remainder than the right hand side.
Also,how can we know that,at $\mathbb{Z}_m,m\neq 3$,$3k-1 \mod m$ will not be equal to a perfect square??
Is that a new question?
#### evinda
##### Well-known member
MHB Site Helper
Not in itself.
What we see is that $3k-1 \equiv 2 \pmod 3$.
Furthermore, any square must be of the form $(3r)^2, (3r + 1)^2,$ or $(3r + 2)^2$.
That means that any square is either $0$ or $1 \pmod 3$.
So we have a mismatch.
The left side always has a different remainder than the right hand side.
I understand..Thanks a lot!!!
Is that a new question?
No,I was just wondering if it suffices,showing it in $\mathbb{Z}_3$,but I think it does,right?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
No,I was just wondering if it suffices,showing it in $\mathbb{Z}_3$,but I think it does,right?
Yes. It is a complete proof, so it suffices to show it in $\mathbb{Z}_3$.
#### evinda
##### Well-known member
MHB Site Helper
Yes. It is a complete proof, so it suffices to show it in $\mathbb{Z}_3$.
Nice..Thank you!!!
#### Deveno
##### Well-known member
MHB Math Scholar
This is one of the reasons why we use "integers mod n" in number theory.
In the integers (mod n), we have "fewer cases", if we show a certain relationship is impossible in the integers (mod n), sometimes this suffices to show that same relationship is impossible in the integers.
For example, if:
$a = b^2$ in the integers, then:
$a = b^2\text{ (mod }n)$
as well, so if the second is impossible, the first is, as well.
In this example, we have $a = 3k - 1$. Now this is a lot of integers:
$a = \dots,-4,-1,2,5,8,\dots$
and the sheer number of cases to check is quite large (infinite). Going integer-by-integer to see if $a$ is a perfect square seems inefficient.
On the other hand, we have:
$a = 3k - 1 = -1 = -1 + 3 = 2\text{ (mod }3)$
which is just ONE case to check. And in that one case, we see no appropriate $b$ exists (There are only 3 possible values of $b$ mod 3 to test). | 2021-04-15T14:44:33 | {
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https://math.stackexchange.com/questions/2418994/how-many-even-numbers-can-be-formed-from-these-3-digits | # How many even numbers can be formed from these 3 digits?
Digits: 1, 2, 3, 4, 5
Q: how many even 3 digit numbers can be made without repeating them?
In total, I worked out that there's 60 three digit numbers that can be made without repeating (5C1 x 4C1 x 3C1) = 60.
But, I have no idea about the even bit. Could somebody talk me through it so I can understand?
Thanks!
• the title of the question is malformed just a hint. – user451844 Sep 6 '17 at 13:35
For a number to be even it must end in a even digit $2$ or $4$ so $2$ choices. The second digit can be any of the $4$ digits which wasn't used as the last digit, and the first digit can be any except those $2$ which were used as the second and third digit hence $3$ choices, all in all we get $2\cdot 4\cdot 3=24$.
• Thanks! Could you clarify something? For digit one, I pick one of the even numbers (e.g. 2). For digit 2, I can pick any digit that I didn't just use? So, 1, 3, 4 or 5? (e.g. 5). And for the third digit, I use any that I haven't used previously? I.e. 1, 3, 4? Because in this situation, 2 x 5 x 4 = 40, which is different to 24. – user477985 Sep 6 '17 at 13:55
• @MathsHelp That's right, though keep in mind that digit one is the last digit (or the first one from right). How do you get 2 x 5 x 4? You have $2$ choices for the first(last) digit and then 4 digits for the second and 3 for the last. – kingW3 Sep 6 '17 at 14:00
• Oh, sorry, I think I misinterpreted your words. I thought you meant from my set of numbers that I could literally pick any. I get it now, thanks! If I were to change the question slightly and I had 1, 2, 3, 4, 5 and 6, would it be 3 x 5 x 4 instead of 2 x 4 x 3? – user477985 Sep 6 '17 at 14:05
• @MathsHelp Exactly it would be $3\times 5\times 4$. – kingW3 Sep 6 '17 at 14:07
• just to be sure, 24 is the final answer isn't it? You don't deduct this from 60 to get 36 as the final answer? – user477985 Sep 13 '17 at 5:29
The thing you have done 5C1 x 4C1 x 3C1 would have been correct if you were asked to make any 3 digits no from the given set without repeating. | 2020-11-25T14:19:35 | {
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https://math.stackexchange.com/questions/1694933/when-solving-inequalities-does-x-9-count-as-a-negative-number | # When solving inequalities, does $(x-9)$ count as a negative number?
I know that when solving inequalities, you reverse the sign when dividing or multiplying by a negative number, but right now I need to solve $\frac{x+9}{x-9}≤2$. Do I need to reverse the sign when I multiply $x-9$ across?
I'm sorry if this is outrageously obvious...
• You have to split into two cases $x>9$ and $x<9$. – user99914 Mar 12 '16 at 23:16
• @JohnMa But the point of this is to solve for $x$ so why would we set cases for $x$? In this case, what would the answer, $x$, be? – Spica Mar 12 '16 at 23:27
• We know that there are only three cases that $x$ could be, $x-9 <0, =0,$or $>0$. The second case is not possible as now $x-9$ is at the bottom. So we split into two cases, in order to solve the inequality. For example, when $x- 9>0$, then you multiply both side by $x-9$ (this won't change the inequality sign as $x-9 >0$) to get $x+9 \le 2(x-9)$, which is the same as $x\ge27$. So $\{ x-9 >0\}\cap\{ x\ge 27\}$ is part of the solution. Now do the same for $x-9<0$. – user99914 Mar 12 '16 at 23:34
• @JohnMa When I do the same for $x-9<0$, I get to $x+9≥2(x-9)$ which gives me $x≤27$. But I still don't understand. For example, in order for $x-9<0$ to be true, $x<9$; and for $x-9>0$ to be true, $x>9$; which would be saying when $x>9$, $x≥27$, and when $x<9$, $x≤27$... How can you define the values of $x$ based on a defined value of $x$ itself? – Spica Mar 12 '16 at 23:43
• First we conclude that if a fixed value $x>9$ solves our inequality then it has to be $>=27$ (As far as I see here you would agree with me). Until now we do not know yet if $x>=27$ really does solve our inequality - but if you read now your transformations backwards, you will see that $x>=27$ does indeed solve the inequality you started with (here, during the second step, we do not assume $x>9$, what confused you a bit). – CHwC Mar 13 '16 at 0:36
This is an inequality. You should not multiply across by $(x-9)$ at all. Instead you should bring the $2$ over to the left and combine the fraction through a common denominator. Verify you arrive at: $\frac{-x+27}{x-9}≤0$ Now one should make separate numberlines for numerator and denominator and verify the signs on the numberlines. NUM:$$___________plus_______27__minus______$$ DENOM: $$___minus____9____plus_____________$$ When you "divide" the numberlines you look for the negative interval, which is $x<9$ or $x\geq27$
• You should use \geq instead of => – Nikunj Mar 13 '16 at 4:02
• @Nikunj Ha! For the less or equal sign I copy-pasted the Op's inequality symbol. Now I needed the greater or equal symbol and I didn't know it...(but now I do) – imranfat Mar 13 '16 at 4:07
I suppose it depends on what $x$ is an element of. If we assume that $x\in\mathbb{R}$, we have to break this inequality into separate cases.
There is the case where $x>9$, in which case the sign does not change, and there is the case of $x<9$, in which the sign does change.
Of course, this is undefined for $x=9$.
• But the point of this is to solve for $x$ so why would we set cases for $x$? In this case, what would the answer, $x$, be? – Spica Mar 12 '16 at 23:27
• We need to split this problem into cases, because there is no solution for this problem without assuming $x<9$ or $x>9$. The result is that we get one solution for one assumption and a different solution for the other. – Alekos Robotis Mar 12 '16 at 23:40
• In order for $x-9<0$ to be true, $x<9$; and for $x-9>0$ to be true, $x>9$; which would be saying when $x>9$, $x≥27$, and when $x<9$, $x≤27$... How can you define the values of $x$ based on a defined value of $x$ itself? – Spica Mar 12 '16 at 23:46
• I don't understand what you mean. The issue is that $x$ is arbitrary. So, in separating into two cases, we make sure to not lose generality of our solution. – Alekos Robotis Mar 13 '16 at 0:00
• It is rather like that: First we conclude that if a fixed value $x>9$ solves our inequality then it has to be $>=27$ (As far as I see here you would agree with me). Until now we do not know yet if $x>=27$ really does solve our inequality - but if you read now your transformations backwards, you will see that $x>=27$ does indeed solve the inequality you started with (here, during the second step, we do not assume $x>9$, what confused you a bit). – CHwC Mar 13 '16 at 0:34 | 2021-03-07T16:37:58 | {
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http://www.lofoya.com/Solved/196/find-the-remainder-when-2-89-is-divided-by-89 | # Moderate Number System Solved QuestionAptitude Discussion
Q. Find the remainder when $2^{89}$ is divided by $89$?
✖ A. 1 ✔ B. 2 ✖ C. 87 ✖ D. 88
Solution:
Option(B) is correct
When we take successive powers of $2$ and find their remainders, we get the following cyclic patterns of cycle length $11$.
viz $2,4,8,16,32,64,39,78,67,45, 1$
i.e $2^{11}$ leaves a remainder $1$.
Thus, $2^{89} = (2^{11})^8 (2)$ leaves a remainder of 2.
Edit: Thank you Shireen for explaining cyclicity concept involved in the solution.
Alternate Method: As Arjunbkool pointed out in the comments, a little trick can be used to solve this question.
Remainder of $\dfrac{N^{p-1}}{P}$ is 1 as long as $P$ is a prime number.
Now, as pointed out by Arghya Chakraborty,
$2^{89-1}=2^{88}$ leaves a remainder of 1 when divided by 89 (note that 89 is a prime number)
So, $2^{89}$ will leave remainder 2 when divided by 89.
Edit 2: For yet another alternative solution using 'modulus' method, check comment by Sravan Reddy.
## (15) Comment(s)
Sairam
()
How can we find for this numbers remainder?
Rahul
()
How to solve $2^{30}$
and any other questions like this.
Girish
()
You have to find the cyclic pattern as in this question and proceed using the same methodology used in the solution (or in the alternative solutions).
Shubh
()
How can be the cyclic pattern length of powers of 2 be 11?
Sravan Reddy
()
Alternate solution using 'modulus' method:
$2^8 = 256\text{mod}89 = -11\text{mod}89$
By multiplying with 8 on both sides,
$2^{11} = -88\text{mod}89 = 1\text{mod}89$
Raise both sides by 8,
$2^{88} = 1\text{mod}89$
Multiply by 2 on both sides,
$2^{89}=2\text{mod}89$
So, remainder is 2!!
Swatantra Singh
()
I am not getting it. Please help.
Arjunbkool
()
I recently came to know about a theorem that states reminder of N (p-1)/P is 1 as long as P is a prime number. That could be used here.
Deepak
()
Thank you Arjunbkool for the suggestion, added your suggestion in the solution.
Tarun
()
the quetion is good but solution is not up to satisfactory level
Mandar Dharmadhikari
()
Can any body suggest any other time saving trick???
Shireen
()
The trick to solve such questions is KNOW THE CYCLICITY of that particular number.
As every number is cyclic with period 4 (see very good detailed explanation: Number System Cyclicity Explanation),
we can write the given number in terms of power of 4:
$2^89=2^{(4*22) + 1}$
using which we get the LAST DIGIT or REMAINDER as:
$2^1= \textbf{2}$
Hope this helps.
Arghya Chakraborty
()
Do as 'arjunbkool' has stated.
$2^88$ leaves a remainder of 1 when divided by 89(89 is prime)
So, $2^89$ will leave remainder 2 when divided by 89 | 2016-10-26T00:25:35 | {
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https://math.stackexchange.com/questions/1375127/a-question-on-probability-of-choosing-coins | # A question on probability of choosing coins
Six identical-looking coins are in a box, of which five are unbiased, while the sixth comes up heads with probability $3 \over 4$ and tails with probability $1 \over 4$. Three coins are chosen from the box at random and removed. One of those three is chosen at random and tossed three times, coming up heads every time. Given this information
a). What is the probability that the final coin selected was the biased coin?
b). What is the probability that the biased coin is amongst the three coins removed from the box?
I did this question myself but felt not confident whether I got the right answer or not. So here is my solution:
a). $P={ {5C2 \over 6C3} \times {1 \over 3}}={1 \over 6}$ where the $5C2 \over 6C3$ comes from the probability that the three coins chosen contains the biased coin, and $1 \over 3$ comes from choosing the biased coin from the three coins.
b). $P={1 \over 3}$ as can be seen from a).
Am I getting the correct answer?
For both parts you need to calculate conditional probabilities.
For a) let $B$ denote the event that the biased coin is chosen at the final stage, and let $3H$ denote the event of obtaining three heads.
We require $$p(B|3H)=\frac{p(B\cap3H)}{p(3H)}=\frac{\frac 16\times\frac{27}{64}}{\frac 16\times\frac{27}{64}+\frac 56\times\frac 18}=\frac{27}{67}$$
For b), let C denote the event that the biased coin is amongst those removed from the box.
Now we require $$p(C|3H)=\frac{p(C\cap3H)}{p(3H)}=\frac{\frac 16\times\frac{27}{64}+\frac 13\times\frac 18}{\frac 16\times\frac{27}{64}+\frac 56\times\frac 18}=\frac{43}{67}$$
You have ignored the information from the fact that the coin came up heads all three times. Even so, in your model the answer to b should be $1/2$ because three coins were selected out of six. For a, you have two possibilities: the biased coin was selected (initial probability $1/6$) and of course it came up heads all three times, or some other coin was selected (initial probability $5/6$) and it happened to come up heads all three times (probability $1/8$). Do you know how to combine these?
• I could solve part a) now, yeah I missed the information about three times..... How could I solve part b)? – Rescy_ Jul 27 '15 at 3:04
• I don't think the answer to part b is $\frac 12$ – David Quinn Jul 27 '15 at 11:17
• @DavidQuinn: It would be if the information from the flips is ignored. That was why I said "in your model" – Ross Millikan Jul 27 '15 at 13:52
• I don't know what "flips" are, but I think I get you – David Quinn Jul 27 '15 at 14:26
• @DavidQuinn: The flips are the three times the coin is flipped and comes up heads. OP computed the chance the biased coin was chosen without using that information. In that case, choosing one coin of six has probability $\frac 16$, though OP took a little longer route to get there. I was trying to point OP in the direction of your answer. – Ross Millikan Jul 27 '15 at 14:36 | 2019-08-24T02:43:41 | {
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http://mathhelpforum.com/statistics/169805-basic-probablility.html | # Math Help - Basic Probablility
1. ## Basic Probablility
Hi everyone. I am a math tutor at a college and we don't have any tutors here today that have done stats, so I was wondering if you guys could take a look at this and give me a thorough explanation of how to do the following problem:
The data in the table are from the voyage of the Titanic and show the survivor rates broken down by class of passenger. Use the table to answer the questions below.
$\begin{tabular}{c|c|c|c|c|c}
\null &\text{first} &\text{second}&\text{third}&\text{crew}&\text{Tota l}\\ \hline
\text{Survived}&199 &119&174&214&706\\\text{Died}&130&166&536&685&1517 \\\text{Total}&359&285&710&899&2223
\end{tabular}$
Find the following:
1. $P(\text{surviving})$
2. $P(\text{surviving }|\text{ \null }1^{st}\text{ class passenger})$
3. $P(\text{surviving }|\text{ \null }3^{rd}\text{ class passenger})$
4. Are the events of Surviving and class of passenger independent? Justify your answer.
So, for 1., would it be $P(\text{surviving})=\frac{\text{number of survivors}}{\text{total number of passengers}}$;
For 2., $P(\text{surviving }|\text{ \null }1^{st}\text{ class passenger})=\frac{\text{number of first class survivors}}{\text{total number of passengers}}$;
for 3., $P(\text{surviving }|\text{ \null }3^{rd}\text{ class passenger})=\frac{\text{number of third class survivors}}{\text{total number of passengers}}$;
and, finally for 4., The events are dependent because those passengers in 1st class were (at that time in history) given the first option to board the lifeboats.
Okay, these answers are what I've come up with. Could someone please do the problem professionally and give a really good explanation along the way? I'm trying to learn stats so that I can better help the students enrolled in the course.
Thanks.
2. Originally Posted by VonNemo19
Hi everyone. I am a math tutor at a college and we don't have any tutors here today that have done stats, so I was wondering if you guys could take a look at this and give me a thorough explanation of how to do the following problem:
The data in the table are from the voyage of the Titanic and show the survivor rates broken down by class of passenger. Use the table to answer the questions below.
$\begin{tabular}{c|c|c|c|c|c}
\null &\text{first} &\text{second}&\text{third}&\text{crew}&\text{Tota l}\\ \hline
\text{Survived}&199 &119&174&214&706\\\text{Died}&130&166&536&685&1517 \\\text{Total}&359&285&710&899&2223
\end{tabular}$
Find the following:
1. $P(\text{surviving})$
2. $P(\text{surviving }|\text{ \null }1^{st}\text{ class passenger})$
3. $P(\text{surviving }|\text{ \null }3^{rd}\text{ class passenger})$
4. Are the events of Surviving and class of passenger independent? Justify your answer.
So, for 1., would it be $P(\text{surviving})=\frac{\text{number of survivors}}{\text{total number of passengers}}$;
For 2., $P(\text{surviving }|\text{ \null }1^{st}\text{ class passenger})=\frac{\text{number of first class survivors}}{\text{total number of passengers}}$;
for 3., $P(\text{surviving }|\text{ \null }3^{rd}\text{ class passenger})=\frac{\text{number of third class survivors}}{\text{total number of passengers}}$;
and, finally for 4., The events are dependent because those passengers in 1st class were (at that time in history) given the first option to board the lifeboats.
Okay, these answers are what I've come up with. Could someone please do the problem professionally and give a really good explanation along the way? I'm trying to learn stats so that I can better help the students enrolled in the course.
Thanks.
1. Correct
2. Denominator should be Number of 1st class passengers - because the conditional probability P(A|B)= $\frac{P(A and B)}{P(B)}$
3. Denominator should be Number of 3rd class passengers -same reason as at 2.
4. Test for independent events: A and B are independent if P(A and B) = P(A) * P(B) - apply the test for this case and see if the 2 events in your question pass it or not. | 2016-06-30T02:35:21 | {
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https://math.stackexchange.com/questions/1743963/different-functions-or-same-functions | Different functions or same functions
I have a question in my booklet :
$f(x) = \frac{x}{x}$ and $f(x) = 1$ are different or not why or why not?
I can only think that the functions are different because the second one is a constant function irrespective of the value of $x$.
But on evaluating the first function we will always get $1$. and in the first function is undefined at $x = 0$ while the second function is defined at $x = 0$.
Is there any other way of thinking it mathematically and my instructor never covered anything related to this. Kindly help.
• The functions have the same range (they both take on the value $1$ identically) but as you noted the first one is obviously undefined at $x=0$ so the functions have different domains. If you restrict the domain of the second one to exclude $0$ then they are the same function, otherwise no. – Nap D. Lover Apr 15 '16 at 16:07
• You need to specify a domain. If the domain does not include $0$ then both are the same. If the domain does include $0$, then you need to specify what $0 \over 0$ means. – copper.hat Apr 15 '16 at 16:14
• @LoveTooNap29 I think you should write it as an answer (elaborating a bit) so that I can accept it. – Hisenberg Apr 15 '16 at 16:15
• @user109256 sure, but is Max's answer below not sufficient? It's essentially what I'd write anyway. Youd have to be more specific on what you'd want me to elaborate on. – Nap D. Lover Apr 15 '16 at 16:22
Two functions are identical if they have the same domain, and if they have the same functional values at every point in that domain. If $f(x) = \frac{x}{x}$ and $g(x) = 1$, then f(x) = g(x) for all x within the domain of f, but as LoveTooNap29 pointed out, their domains are different. Therefore the functions are not identical.
• Nothing in the OP actually says that the domains of $f$ and $g$ are different. It's up to speculation, the only thing that's clear is that $f$ must not have $0$ in its domain, but what either of the functions does have in its domain would first need to be specified. – leftaroundabout Apr 15 '16 at 20:05
• @leftaroundabout: At OP's level, functions have a domain of $\mathbb{R}\setminus \text{all the values at which the function is undefined}$ unless stated otherwise. – Deusovi Apr 16 '16 at 2:03
• @Deusovi: perhaps, but IMO that's a pretty big failure of education then. It should be made clear early on that real-valued functions are very much just a special case – important, but by no means a sensible “default for everything”. – leftaroundabout Apr 16 '16 at 9:15
It depends on how you phrase the question. If you say
Let $f,g : \mathbb{Q}\setminus\{0\} \to \mathbb{Q}$, $$f(x) = \frac{x}{x}, \qquad g(x)=1$$
then $f$ and $g$ are in fact equal. This is also true if you replace the rational numbers $\mathbb{Q}$ with the reals, or with complex numbers... as long as you do it consistently for both functions.
However, because $g$ doesn't actually use its argument, there's no reason to assume its domain should be $\mathbb{Q}\setminus\{0\}$. I could write
Let $g : \{1, 2, (0,5), i,\mathrm{cucumber}\} \to \mathbb{R}$, $$g(x)=1$$
or, more reasonably, just
Let $g : \mathbb{Q} \to \mathbb{Q}$, $$g(x)=1.$$
Then, $g$ would be a completely different kind of object from $f$, and depending on your philosophy they would either be nonequal or it wouldn't even make sense to ask whether they're equal.
note that $$\frac{x}{x}=1$$ if $$x\ne 0$$ and $$1=1$$ for all real $$x$$
• This only proves the ranges are equal so the functions are equal everywhere they are defined. The domains remain different, however, so theyre technically different functions. – Nap D. Lover Apr 15 '16 at 16:21
Technically, $f(x)=x/x$ does not equal $g(x)=1$, because $f(x)$ is technically undefined at 0. In the same way, the sinc function technically does not equal $\sin(x)/x$.
However this is merely a technicality: the limit at 0 exists, and you can redefine $f(x)$ to equal its limit at 0, in which case $f(x)=g(x)=1$. Usually, this is exactly what you should do. | 2019-12-06T06:17:36 | {
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https://math.stackexchange.com/questions/2584044/proving-2n1-2n-using-mathematical-induction | # Proving $2n+1 < 2^{n}$ using mathematical induction.
I'm learning about how use mathematical induction. I'm tasked with proving the inequality shown in (1). It is a requirement that I use mathematical induction for the proof.
$(1) \quad P(n):\quad 2n+1 < 2^{n}, \quad n \ge 3$
I would like some feedback regarding whether my proof is valid and if my use of the induction hypothesis is correct. My proof:
Prove base case:
$(2) \quad P(b):\quad 2\cdot3+1 < 2^{3}, \quad b=1$
$(3) \quad P(b):\quad 7<8$
Assume that $P(n)$ is true for an arbitrary $k$, with $n \ge k \ge 3$. (This is the induction hypothesis, IH)
$(4) \quad P(k):\quad 2k+1 < 2^{k}$
The inductive step:
$(5) \quad P(k+1):\quad 2(k+1)+1 < 2^{k+1}$
$(6) \quad 2k+3 < 2^{k}\cdot 2$
Using the IH we get (7). I just added 2 to both sides of the inequality so that the LHS in (6) and the IH is equal.
$(7) \quad 2k+3 \overset {IH}{<} 2^{k}+2$
Subtracting 2 from both sides yields us $P(k)$:
$(8) \quad 2k+1 < 2^{k}$
Which concludes the proof that $P(n)$ is true for every $n \ge 3$
Is my proof valid? Is my use of the IH correct?
I am aware that a similar problem is asked here.
• How did you get from $2k+3<2^k\times 2$ to $2k+3<2^k+2$? You seem to add $2$ to the IH, then substract $2$ from this, and conclude that the result is the IH hence the proof is finished. If you are not using anywhere the fact that $2^k+2<2^k\times2$ (for $k>1$), then the proof is incorrect. – Jean-Claude Arbaut Dec 29 '17 at 11:01
• If you are ok, you can accept the answer and set as solved. Thanks! – gimusi Dec 31 '17 at 10:25
It's not clear to me how you proved that $P(k) \implies P(k + 1)$. You have a collection of inequalities $(5), (6), (7), (8)$, but I don't see which of them are implied by the others. Here's an alternative approach.
Assume that $P(k)$ is true for some $k$, where $k \geq 3$. It remains to show that $P(k + 1)$ is true. Indeed, observe that: \begin{align*} 2(k + 1) + 1 &= 2k + 3 \\ &= (2k + 1) + 2 \\ &< 2^k + 2 &\text{by the induction hypothesis} \\ &< 2^k + 8 \\ &= 2^k + 2^3 \\ &\leq 2^k + 2^k &\text{since $3 \leq k$, and $2^x$ is an increasing function} \\ &= 2 \cdot 2^k \\ &= 2^{k + 1} \end{align*} as desired.
• Whoops, my bad, you're right. Edited. – Adriano Dec 29 '17 at 12:40
You know it for $n=3$. Suppose this is true for $n$, i.e., $2n+1<2^n$. Let us prove it for $n+1$, i.e., $2(n+1)+1<2^{n+1}$. You have $$2n+3=(2n+1)+2<2^n+2 \le 2^{n+1}.$$ The last inequality is true because $n\ge 3$.
• Nice +1. The last inequality is true even for $n>0$ – Isham Dec 29 '17 at 11:44
Base case:
$n=3 \implies 7<8$ ok
Inductive step:
let's assume: $2n+1 < 2^{n}$
we want to prove that: $2n+3 < 2^{n+1}$
$$2n+3=2n+1+2\overset {IH}{<} 2^n+2=2(2^{n-1}+1)\overset {?}{<} 2\cdot2^n=2^{n+1}$$
we can easily check that $$2^{n-1}+1< 2^n\iff2^n-2^{n-1}>1\iff2^{n-1}(2-1)>1 \quad \forall n>1$$
thus
$$2n+3<2^{n+1} \quad \square$$
I want to suggest an alternative way just to be more systematic:
Let $a(n) = 2n+1$ and $b(n)= 2^n$ where $n \ge 3$. Then for $n=3$, we have $7 = a(3) < b(3) = 8$. Then assume inductively that $a(n) < b(n)$ and $n > 4$. Then, for $n+1$, we have $$a(n+1) = 2n+3 = 2n+1+2 = a(n)+2 < b(n)+2$$ by inductive hyphotesis. But notice that $$b(n)+2 = 2^n+2 < 2 \cdot 2^n = 2^{n+1} = b(n+1)$$
Therefore, we have $$a(n+1) < b(n)+2 < b(n+1) \implies a(n+1) < b(n+1)$$
Hypothesis:
1) You assume that $2n+1 < 2^n$ for an $n.$
Step:
Assuming the hypothesis :
Show that $2(n+1) +1 < 2^{n+1}$, I.e.
the formula holds for $n+1.$
$2n+1 + 2 =$
$2(n+1) +1 < 2^n +2 ;$
$2$ has been added to both sides of $2n+1 <2^n$ (hypothesis) .
LHS : $2(n+1) +1$.
RHS: $2^n +2 \lt 2^n +2^n=$
$2(2^n) =2^{n+1}$.
since $2 \lt 2^n$ for $n \ge 3.$
Hence:
$2(n+1)+1 < 2^{n+1}.$
Alternatively, without induction: for $n\ge 3$: $$2^n=(1+1)^n=1+n+\cdots +n+1>2n+1.$$ | 2019-08-21T22:25:43 | {
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https://math.stackexchange.com/questions/2159136/if-f-kx-frac1k-left-sinkx-coskx-right-then-f-4x-f-6x | # If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$
I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
\begin{align} f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt] &=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt] &=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt] &=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt] &=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt] &\qquad\quad \text{(substituting \sin^2x=1-\cos^2x and \cos^2x=1-\sin^2x)} \\[4pt] &=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt] &=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt] &=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt] &\qquad\quad\text{(because \sin^2x+\cos^2x=1)} \\[4pt] &=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt] &=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt] &\qquad\quad\text{(because \cos^2x-\sin^2x=\cos2x)} \end{align}
Hence the answer should be ...
$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$
... but the answer given was $\frac{1}{12}$.
I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.
• Thanks to all of you for your effort and I know that your answers are right but I wanna know, in which step I have done something wrong. Why is my answer coming different? – Avi Feb 25 '17 at 3:47
• In line (9), if that "$-$" had been a "$+$", your answer would match the expected one. As it turns out, the sign error was introduced in line (5). From line (4), you should have, for instance, ... $$3 - 2 \sin^2 x\;\to\;3-2(1-\cos^2x) \;\to\; 3 - 2 \color{red}{+} 2 \cos^2 x \;\to\; 1 \color{red}{+} 2 \cos^2 x$$ Likewise, line (5) should have $1\color{red}{+}2\sin^2 x$. These sign changes carry through to line (9). – Blue Feb 25 '17 at 18:32
HINT:
$$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$$
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
Hint. Let $x=\frac{\pi}{4}$
In your answer $f4(x)-f(6x) = 0$
By brute-forse, $f4(x)-f6(x) = \frac{0.5}{4} - \frac{0.25}{6} = \frac{1}{8}-\frac{1}{24} =\frac{1}{12}$, so, you make a typo :(
Hint: let $f(x):=f_4(x)-f_6(x)$. Then show that $f'(x)=0$ for all $x$. Hence $f$ is constant. Furthermore: $f(0)=\frac{1}{12}$ | 2019-06-18T16:37:10 | {
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http://fxks.effebitrezzano.it/biot-savart-law-finite-wire.html | # Biot Savart Law Finite Wire
Topics of Magnetic Effects of Current and Magnetism. JEE Advanced 2020 syllabus will be highly beneficial for candidates who are preparing for the upcoming entrance examination. But know I want to use Matplotlib for visualisation. The position and velocity of the. No contribution to net current. This is a limiting case of the formula for vortex segments of finite length similar to a finite wire:. 3 along the wire, giving us the usual form of the Biot-Savart law. In this study, the authors set out to adapt Biot-Savart's law, which describes the magnetic field generated by finite wires, to evaluate the circulation of such fields around a closed path or loop. On the semicircle, d l is perpendicular to r and the radius is constant, r =a. [email protected] THE BIOT-SAVART LAW 5. a magnetic field, or a combination of the two, depending on their frame of reference. Biot savart Law Applications of Biot Savart Law Applications of Biot Savart Law for the circular coil Magnetic field due to a circular coil Magnetic field due to a uniformly charged circular coil. What is Biot-Savart Law? Biot-Savart’s law is an equation that gives the magnetic field produced due to a current carrying segment. The flow of electric current through a conductor creates a magnetic field around the conductor, whose strength depends on the magnitude of the current. An electric current flowing in a conductor, or a moving electric charge, produces a magnetic field, or a region in the space around the conductor in which magnetic. Advanced Physics 10. Biot-Savart Law. 50mm segment A. Biot Savart Law, Integrating a circular current loop on axis ˘ ˘ ˘ ˇ ˆ Ampere's Law: Example, Finite size infinite wire Calculate the B-field everywhere from a finite size, straight, infinite wire with uniform current. (i) Show how Biot-Savart law can be alternatively expressed in the form of Ampere's circuital law. About the magnetic field of a finite wire 269 B A E B Figure2. Law of Biot-Savart A wire is bent into the shape of a regular hexagon with side aas shown below. 1) Comparison between coulomb' s law and Biot Savart law. The ƒÃo can, for the moment, be thought of as a constant that makes the units come out right. Use the law of Biot and Savart to find the magnitude of the magnetic field at point P due to the 1. A straight, infinitely long wire carrying a steady current Il lies along the y-axis. The magnetic field intensity at any point in the magnetic field is defined as the force experienced by a unit north pole of one weber strength, placed at that point. In order to understand the Biot-Savart’s law, we need to understand the term current-element. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. Biot-Savart Law: One challenge with figuring out a rule for something which depends on a current is that the source (the current) can't be a point object. The small element of current is usually written as , that is a constant current I flowing in a small length of wire. 1 However, to the author's knowledge, no textbook presents the calculation of this field using the Ampere-Maxwell law: ∮. EVALUATE: The rest of each wire also produces field at P. P be the any point at a distance x from the centre of the coil where we have to calculate the magnetic field. Constant uniform current. txt) or read online for free. current density distribution across the solenoid wire cross- section has been assumed to be uniform. 02 Physics II: Electricity and Magnetism, Spring 2007. The Biot-Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. 50mm segment C. And the shape of that magnetic field is going to be co-centric circles around this wire. (a) Find the magnitude of the magnetic field 1. Ampere’s Law: An easier way to find B-fields (in very special circumstances) For any arbitrary loop (not just 2-D loops!): Ampere’s Law Use Ampere’s Law to find B-field from current (in very special circumstances) 1. around a straight infinitely long current-carrying conductor and gen-eralize it to Ampere's law. The Laws of Biot-Savart Ampere; 2 Overview of Lecture. Charitat and Graner (CG) [1] apply the Ampère and Biot-Savart laws to the problem. About the magnetic field of a finite wire 269 B A E B Figure2. Note that we. For a thin and straight. We applied the law to determine the field of a long straight wire (length ) at perpendicular distance from the wire. Formulas : Field due to electric current in an infinitely long, straight wire: Field inside an infinitely long, straight, air core solenoid: Field inside an air core toroid: Field due to a current loop. For symmetry reasons it becomes: B = B ( R 2) + B. The Biot-Savart Law and Ampere's Law and related in their common function. Therefore I could use the Biot-Savart law for straight, finite wires for my analytic approach. Christopoulos,5 Sibley6 and Kraus7 use the Biot-Savart law to find the m. Finite Wings. 5 in Griffiths derives this expression entirely. The current element is taken as a vector quantity. 1: The Biot-Savart law reads: H= z2. The law is a physical example of a line integral, being evaluated over the. The Biot-Savart's law can be used in the calculation of magnetic responses even at. This law is to magnetostatics (i. 1: (a) conducting triangular loop, (b) side 1 of the loop. In This Chapter Biot-Savart Law Ampere’s Law Gauss’ Law for Magnetic Field Magnetic Scalar Potential Magnetic Vector Potential QuickField Magnetostatic Analysis Inductance Calculations Uniform Magnetic Fields Dipole Sources Shielding Applications Magnetic Monopoles While preparing a lecture demonstration in 1820, Orsted noticed that current flowing through a wire deflected a nearby compass. Sign in with Facebook. This law is although for infinitesimally small conductors yet it can be used for long conductors. Current that does not go through “Amperian Loop” does not contribute to the integral 2. This is a limiting case of the formula for vortex dw of finite length similar to a finite wire:. hexagon spiral windings, and is based on the Biot-Savart law. Biot-Savart Law Infinite current carrying wire Finite current carrying wire Average magnetic field on a railgun armature Lorentz Force Law magnetic force Note: Many assumptions have been made in these equations such as the geometry of the rails, type of projectile, distances, and lengths. l be the distance between centre of the coil and elementary length dl. The Biot-Savart hypothesis came up which was found to give a different result. Bruce Knuteson, Prof. 6 For Example 7. The Biot—Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. Note that we. 5 the vector. Consider a straight wire of length l carrying a steady current I. ) • To determine the total magnetic field ( ) due to a finite sized conductor, we need to sum up the contributions due to all the current elements making up the conductor. In electromagnetism and electronics, inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. 1 Biot-Savart Law Currents which arise due to the motion of charges are the source of magnetic fields. Biot - Savart law and its application. The Biot-Savart's law can be used in the calculation of magnetic responses even at. Definition •The differential contribution dB to the magnetic field B from a length ds of a Magnetic Field Due to a Finite Straight Wire. Some people recommend to use numpy arrays. MAGNETIC POTENTIAL 9. If not, the integral form of the Biot-Savart law must be used over the entire line segment to calculate the magnetic field. In electromagnetism and electronics, inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. The field at a point due to a current-carrying wire is given by the Biot-Savart law,, where and , and the integral is done over the current-carrying wire. Note that ds x r points out of the page. Choose the ring so that it is centered at (0,0,0), and that it lies in the xy plane. State Biot savart's law? (AU nov/dec 10) The magnetic flux density produced by a current element at any point in a magnetic field is proportional to current element and inversely proportional square of the. AC electrical machine design is a key skill set for developing competitive electric motors and generators for applications in industry, aerospace, and defense. Hussain Electromagnetic Field Theory II The Biot-Savart Law The Biot-Savart Law is an equation that describes the magnetic field created by a current-carrying wire, and allows you to calculate its strength at various points. Magnetic Fields (Biot-Savart): Summary Current loop, distance x on loop axis (radius R): Straight wire: finite length infinite wire: B x = µ 0 IR 2 2(x2+R2)3/2 B center = µ 0 I 2R (coscos) 4 1 2 0 θθ π µ = − a I B a I B π µ 2 =0 θ 1 θ 2. It is common to use the Biot-Savart law as a tool to explicitly calculate the magnetic field due to currents flowing in simply shaped wires such as circular loops and straight lines. A steady (or stationary) current is a continual flow of charges which does not change with time and the charge neither accumulates nor depletes at any point. Gunther Roland, Prof. BIOT-SAVART LAW: Biot and Savart conducted many experiments to determine the factors on which the magnetic field due to current in a conductor depends. However, to the author's knowledge, no textbook presents the calculation of this field using the Ampere-Maxwell law: ?B [multiplied by] dl = µ[subscript 0] (I + e[subscript 0] dF/dt) [multiplied by] 1. Biot-Savart's law is an extension of Ampere's law, anything that satisfies Biot-Savart's law also satisfies Ampere's law, the extra parts of the equation have to be added to model the real world field effects involved in an ACTUAL device where Ampere's law is pure theory. Magneticfieldcreatedbya: Straightcurrent-carryingwire Coil Magneticflux trougha surface. Current element It is the product of current and length of infinitesimal segment of current carrying wire. I've redirected it to Biot-Savart's Law. of the wire in which the current is flowing, and sometimes in a complicated way, but for a given geometry, the magnetic field is directly proportional to the amount of current flowing through the wire. GOV Conference: Performance of low-rank QR approximation of the finite element Biot-Savart law Title: Performance of low-rank QR approximation of the finite element Biot-Savart law Full Record. Although we derived the formula of the magnitude of the magnetic B-field $B=\mu_o In$ for an infinitely long ideal solenoid, it is valid also for a real solenoid of finite length as long as we are interested in the field sufficiently far from its ends. These equations have applications for problems where current distributions are present over extended volumes and are calculated on the basis of a finite element electric field analysis. Let 'P' be the point where the magnetic field due to the wire is to be studied. 2) Magnetic field at the centre of current carrying circular loop 3) Magnetic field due to a straight current carrying conductor of: i. 53, 68 (2015); 10. Ideally, this topic is covered after the Biot–Savart law and before displacement current. An electrical current I flows across the rectilinear finite wire AB. 1 Line of charge A current in a wire can be considered a line of charge of linear charge density λ moving at v ms-1. solution for practice test for test solution to practice test problem 8. Let P be any point at a distance a from the centre of conductor. The article synthesizes prior works using a unified notation, enabling straightforward application in robotics. The magnetization need not be static; the equations of magnetostatics can be used to predict fast magnetic switching events that occur on time scales. Magnetic Fields (Biot-Savart): Summary Current loop, distance x on loop axis (radius R): Straight wire: finite length infinite wire: B x = µ 0 IR 2 2(x2+R2)3/2 B center = µ 0 I 2R (coscos) 4 1 2 0 θθ π µ = − a I B a I B π µ 2 =0 θ 1 θ 2. Using Biot Savart Law, we find out that the magnetic field is μ0⋅I(t)/2. In using the Biot-Savart Law for an finite wire, I am having trouble understanding the angles. , the study of electric fields generated by stationary charges). THERMAL PHYSICS I (25 Marks) LECTURES 25 + 5 Tutorial 1. Use the law of Biot and Savart to find the magnitude of the magnetic field at point P due to the 1. Magnetisms Part-1 BY NM SIR|Biot Savart law |Magnetic field due to finite wire|IIT-JEE |NEET PHYSICS. The top wire has current 2 A to the right, and the bottom wire has current 3 A to the left. The Biot-Savart law says that if a wire carries a steady current I, the magnetic field dB at a point P associated with an element of the wire ds has the following properties: The vector d B is perpendicular both to ds (which is a vector units of length and in the direction of the current) and to the unit vector r directed from the element to P. There's a bit of an art to setting up the. Biot-Savart integral is taken over finite wire length:. We have therefore shown that. Thus, this is all about biot savart law. where ц, is the permeability of the medium surrounding the wire. Biot-Savart Law. • When wire is perpendicular to the plane of paper, the field is in the plane of the paper. We start by describing the Biot–Savart law since Ampere’s law may be derived from the Biot–Savart law. It is clear from these force laws that an observer could say that they were in the presence of either an electric. Eric Katsavounidis, Prof. 1) Comparison between coulomb' s law and Biot Savart law. Biot-Savart law The magnetic field $$\vec{B}$$ due to an element $$d\vec{l}$$ of a current-carrying wire is given by. \vec{\text{dl}} = \mu _0 I_{enclosed} ∮ C B. 12) In summary, the Biot-Savart™s law is generic in the sense other well known laws in magnetostatics follow from it. The magnetic field due to a finite length of current-carrying wire is found by integrating Equation 12. on the axis of current carrying coil. In this study, the authors set out to adapt Biot-Savart’s law, which describes the magnetic field generated by finite wires, to evaluate the circulation of such fields around a closed path or loop. TITLE: Using Biot-Savart's law to determine the finite tube's magnetic field Full Text AUTHORS: Ferreira, JM; Joaquim Anacleto; SOURCE: EUROPEAN JOURNAL OF PHYSICS, VOLUME: 39, ISSUE: 5, PUBLISHED: 2018. Inside a solenoid: source of uniform B field. Therefore, it will tend to be the law used when Ampere's Law doesn't fit. This is a limiting case of the formula for vortex segments of finite length similar to a finite wire:. Biot-Savart Law. Note that ds x r points out of the page. Derivation of Biot Savart law. Biot-Savart Law ÎDeduced from many experiments on B field produced by currents, including B field around a very long wire Magnitude Direction: RHR #2 Vector notation Applications Reproduces formula for B around long, current-carrying wire B by current loop (on axis) In more complicated cases, numerically integrate to find B 2 0 sin 4 r ids θ π. Choose the ring so that it is centered at (0,0,0), and that it lies in the xy plane. 1) Comparison between coulomb' s law and Biot Savart law. Finding the magnetic field resulting from a current distribution involves the vector product, and is inherently a calculus problem when the distance from the current to the field point is. Worked example of the magnetic field generated by a coil of wire. From biot-savart law, magnetic field due to current carrying element dl at point P is. L6 current distribution in magnetism. From the right-hand rule and the Biot-Savart law, the field is directed into the page. This video deals with explanation of Biot Savarts Law in vector form along with it's one of the application namely Magnetic Field due to infinitely long straight wire carrying current. To explain the Biot Savart law,we consider a point near a wire carrying current i. The Biot-Savart Law for Currents Last time, we introduced the Biot-Savart Law for a single moving charge: As you just saw in the lab activity, we usually are interested in magnetic fields created by a large group of moving charges –e. The Biot-Savart’s law gives the magnetic field produced due to a current carrying segment. 022-62211530. PHYS323 1 By Ass. The Biot–Savart law is used for computing the resultant magnetic field B at position r generated by a steady current I (for example due to a wire): a continual flow of charges which is constant in time and the charge neither accumulates nor depletes at any point. It is useful in determining the magnetic field due to a symmetric current distribution. From the right-hand rule and the Biot-Savart law, the field is directed into the page. For the infinite wire, this works easily with a path that is circular around the wire so that the magnetic field factors out of the integration. Once determined, Finite Element Method Magnetics (FEMM) was used to assure these calculations. The Current Flows From Left To Right. Ampere’s Law: An easier way to find B-fields (in very special circumstances) For any arbitrary loop (not just 2-D loops!): Ampere’s Law Use Ampere’s Law to find B-field from current (in very special circumstances) 1. 95 KB) by Sathyanarayan Rao Sathyanarayan Rao (view profile). 2) Magnetic field at the centre of current carrying circular loop 3) Magnetic field due to a straight current carrying conductor of: i. $Rybka Jean?Baptiste$Biot Félix$Savart. Sign in with Twitter. The present work elucidates two separate computational methodologies involving direct determination of the magnetic field from Biot-Savart law. B Field of a Solenoid. We can calculate the magnetic field B for a current carrying conductor of finite length by integrating equation 1 over whatever length and shape of conductor we are interested in. Ampere's law and its. 1) Comparison between coulomb' s law and Biot Savart law. Ampere Biot-Savart Law general current source ex: finite wire wire loop Ampere's law symmetric current source ex: infinite wire infinite current sheet 0 2 ˆ 4 I d r µ π × = ∫ sr B G G ∫B⋅ds =µ0Ienc GG. Starting with the Biot-Savart Law, compute B at point P, the center of the semicircle. Biot Savart's Law and Its Applications (in Hindi) 15:00 mins. In this paper, the performance of magnetic rail gun with. by a current loop using the Biot-Savart Law. Magnetic field from a finite straight current wire We apply the Biot-Savart's law to a finite length of straight current wire to find the magnitude at the point P (see Fig. Circular magnetic fields are generated around current carrying wires. Take a small element of the wire of length dl. 12) is figuring out a b a2, p, and a^. Magnetic field from a circular current-carrying wire; The Biot-Savart law allows us to determine the magnetic field at some position in space that is due to an electric current. EVALUATE: The rest of each wire also produces field at P. Source of Magnetic Fields - Worked Examples Example 1: Current-carrying arc Consider the current-carrying loop formed of radial lines and segments of circles whose centers are at point P as shown below. 3D Magnetic Field Computation of a Straight Wire of Finite Length using Biot-Savart's Law version 1. L6 current distribution in magnetism. Rules for Direction of Magnetic Field, Due to Current-Carrying Wire of Finite Length (in Hindi). The Biot-Savart law tells us that each wire element produces a B-field that is perpendicular to the current and perpendicular to a displacement joining the wire element and the point at which I wish to know the field. We have calculated just the field from the two segments that are indicated in the problem. Practice #1: Magnetic Field at Center of Ring of Current. 02 Physics II: Electricity and Magnetism, Spring 2007 Prof. The Biot—Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. 35 (6) In (6) the angles determine the starting and ending points of the wire segment. The Biot-Savart Law relates magnetic fields to the currents which are their sources. A steady current is a flow of charge that has been going on forever, and will be going on forever. The H is called the and µis the permeability of the medium, which is the medium properties just like ε. The law is a physical example of a line integral, being evaluated over the path C. Find the magnetic field B at P. Biot-Savart Law. BIOT-SAVART LAW: Biot and Savart conducted many experiments to determine the factors on which the magnetic field due to current in a conductor depends. due to finite current carrying wire,palm rule. IDENTIFY: A current segment creates a magnetic field. The magnetic induction due to small element dl of the wire shown in figure 2 is. This is known as Biot-Savart’s Law 2. 11/14/2004 section 7_3 The Biot-Savart Law blank. Summary of the two. • Design and Simulation. In particular, we have derived the following vectorial relationships from the Biot-Savart™s law: r B = 0; absence of magnetic monopoles (always) r B =. Using Biot-Savart to Find the Magnetic Field from a Finite Wire - Duration: 7:01. What is Biot-Savart Law? Biot-Savart's law is an equation that gives the magnetic field produced due to a current carrying segment. The Biot-Savart Law specifies the magnetic field intensity, H, arising from a “point source”current element of differential length dL. In undergraduate E&M courses the magnetic field due to a finite length, current-carrying wire can be calculated using the Biot-Savart law. Maxwell’s distribution law (both in terms of velocity and energy), root mean square and most probable speeds. The formulas can be used by a software tool to model the magnetic fields generated by e. 2005-12-01. proportional to the product Idl and the sine of the angle α between the element and the line joining P to the element and is inversely proportional to the square of the distance R between P and the element and its direction can be obtained by right handed screw rule. L4 questions on Biot Savart law. The magnetic field circulation counterpart to Biot-Savart’s law. In using the Biot-Savart Law for an finite wire, I am having trouble understanding the angles. Compute the magnetic force on wire 2. PHYS323 1 By Ass. a magnetic field, or a combination of the two, depending on their frame of reference. In this study, the authors set out to adapt Biot-Savart’s law, which describes the magnetic field generated by finite wires, to evaluate the circulation of such fields around a closed path or loop. 1) Comparison between coulomb' s law and Biot Savart law. The Biot—Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. As the magnet moves closer to the loop, the magnetic field at a point on the loop in. 0), which is identical to the electrostatic case and Biot-Savart wire/solenoid calculations (enhanced in 8. Current element It is the product of current and length of infinitesimal segment of current carrying wire. The lecture includes example problems for the teacher to work through with the students. com, find free presentations research about Moving Coil Galvanometer PPT. If one had a large irregular object, one broke it into infinitesimal pieces and computed, r r dq dEö 4 1 2 0!" = r Which we write as, 3 0 4r dqr dE rr!" = If you wish to compute the magnetic field due to a current in a wire, you use the law of Biot and Savart. A source and a sink of electrical charge +q and −q ensure charge conservation. Consider a straight wire of length l carrying a steady current I. Students who complete these exercises will - be able to describe in pseudo-code how to calculate the magnetic field around a wire loop with the Biot-Savart law (**Exercises 1 and 2**); - be able to use numerical integration to calculate the magnetic field with the Biot-Savart law (**Exercises 1, 2 and 3**); - be able to compare the numerical solution to the analytical solution for special. The simplest system studied consists in a straight finite wire, however, to explore the magnetic field in complex geometries is required more imagination to solve the mathematics. In undergraduate E&M courses the magnetic field due to a finite length, current-carrying wire can be calculated using the Biot-Savart law. Find the magnetic field at the center of the first wire. where ц, is the permeability of the medium surrounding the wire. Let's suppose you have a wire of radius a centered on the z axis. due to finite current carrying wire,palm rule. Use this law to obtain the expression for the magnetic field inside a solenoid of length 'l', cross-sectional area 'A' having 'N' closely wound turns and carrying a steady current 'I'. Magnetisms Part-1 BY NM SIR|Biot Savart law |Magnetic field due to finite wire|IIT-JEE |NEET PHYSICS. The Biot-Savart law says that if a wire carries a steady current I, the magnetic field dB at a point P associated with an element of the wire ds has the following properties: The vector d B is perpendicular both to ds (which is a vector units of length and in the direction of the current) and to the unit vector r directed from the element to P. The magnetization need not be static; the equations of magnetostatics can be used to predict fast magnetic switching events that occur on time scales. Therefore I could use the Biot-Savart law for straight, finite wires for my analytic approach. Consider a small element AB of current carrying wire whose length is $$d\vec s$$ and the position vector of point (P) from the element is $$\vec r$$. The Biot-Savart law enables us to calculate the magnetic field produced by a current carrying wire of arbitrary shape. Biot-Savart's law is an extension of Ampere's law, anything that satisfies Biot-Savart's law also satisfies Ampere's law, the extra parts of the equation have to be added to model the real world field effects involved in an ACTUAL device where Ampere's law is pure theory. Calculate magnetic field based on Biot Savart Law. Biot Savart's Law and Its Applications (in Hindi) 15:00 mins. In electromagnetism and electronics, inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. THERMAL PHYSICS I (25 Marks) LECTURES 25 + 5 Tutorial 1. Sources of Magnetic Fields 9. This law tells us about the magnetic field (magnitude and direction) produced by moving charges. B⃗ =μ04π∫wireIdl⃗ ×rˆr2. Using Biot-Savart Law magnetic filed at point due to small current element is given by As every element of the wire contributes to B in the same direction, we have …. A current I flows in the direction shown. Answers a and b b. 1, by the differential current clement / ill is proportional to the product / dl and the sine of the angle a between the clement and the line joining P to. 1) Comparison between coulomb' s law and Biot Savart law. This is because the Biot-Savart law is one of numerous solutions of Laplace's equation, which is the governing equation for irrotational, incompressible fluid flow. Evaluate the magnetic field at point P. Physics 21 Fall, 2008 Solution to HW-14 B of Finite Wire A steady current I is flowing through a straight wire of finite length. In This Chapter Biot-Savart Law Ampere’s Law Gauss’ Law for Magnetic Field Magnetic Scalar Potential Magnetic Vector Potential QuickField Magnetostatic Analysis Inductance Calculations Uniform Magnetic Fields Dipole Sources Shielding Applications Magnetic Monopoles While preparing a lecture demonstration in 1820, Orsted noticed that current flowing through a wire deflected a nearby compass. We have therefore shown that. Use Ampere's law to calculate the magnetic field from an infinite straight wire. Ampere Biot-Savart Law general current source ex: finite wire wire loop Ampere's law symmetric current source ex: infinite wire infinite current sheet 0 2 ˆ 4 I d r µ π × = ∫ sr B G G ∫ B ⋅ ds =µ0Ienc G G. Charitat and Graner (CG) apply the Ampère and Biot–Savart laws to the problem of the magnetic field due to a straight current-carrying wire of finite length, and note that these laws lead to different results. # turns) Direction of magnetic field from the RHR. The formula is exact for an infinitely long wire. They derived the mathematical expression for the magnetic flux density. 1 The Biot–Savart Law To find the total magnetic field B created at some point by a current of finite size, we must sum up contributions from all current elements Ids that make up the current. State Biot Savart Law. Find the magnetic flux density ⃗ at the point ( = r, = , = r) employing the Biot – Savart Law. As the magnet moves closer to the loop, the magnetic field at a point on the loop in. Redmond Physics. 1) Comparison between coulomb' s law and Biot Savart law. Learn more about the Motion in Combined Electric and Magnetic Field. AP Physics C: Magnetism 6: Find Magnetic Field Using Biot Savart Law & Ampere's Law 10:51. Some people recommend to use numpy arrays. When magnetostatics does not apply, the Biot–Savart law should be replaced by Jefimenko's equations. 6) is identical in form to Equation (5. For a conductor oriented along the z axis (so that the current is flowing in the +ˆz direction), we may write B~ = µ. Or sign in with one of these services. 6(b), where side 1 is treated as a straight conductor. on the axis of current carrying coil. From the edge outwards 8increases as 1/r. Make sure you review basic magnetostatics, in particular Bio-Savart's law for the magnetic field of a current-carrying wire. Applied Electromagnetics - ECE 351 Author: Benjamin D. Hence write the magnetic field at the centre of a loop. In this article, you will find the Study Notes on Magnetostatics-1 which will cover the topics such as Introduction, Biot-savart's Law, Magnetic field due to an infinite and finite conductor, a force due to the Magnetic field. Starting with the Biot-Savart Law, compute B at point P, the center of the semicircle. 1 Introduction The Biot-Savart law becomes dB ds ds= − = finite length, the potential is given exactly by equation 9. The Biot—Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. Consider a small current carrying element ($ \overline{dl} $) of the conductor XY carrying current I and P be the observation point at a distance r and making an angle$ \theta $with it as shown in diagram. Prandtl's aim was to apply the Biot-Savart law to the "horseshoe" vortex because he was interested in the effect of the trailing vortices. Practice #1: Magnetic Field at Center of Ring of Current. Solution The Biot-Savart law (Equation 30-2) written in a coordinate system with origin at P. of Kansas Dept. However, it is also much harder to apply. Magnetic field due to combinations of all of them will also be discussed. Consider a small element AB of current carrying wire whose length is $$d\vec s$$ and the position vector of point (P) from the element is $$\vec r$$. Use this law to obtain the expression for the magnetic field inside a solenoid of length 'l', cross-sectional area 'A' having 'N' closely wound turns and carrying a steady current 'I'. This expression is known as the ‘Biot-Savart law’. The magnetic field due to a finite length of current-carrying wire is found by integrating Equation 12. Ampere's law and its. Biot-Savart law magnetic field integration equations are derived for rod and plate elements, and for volume interface regions from an equation due to Wikswo. Coyle Hans Christian Oersted, 1820 Magnetic fields are caused by currents. Find an expression for the magnitude of the magnetic field at the point P on the bisecting axis. 3D Magnetic Field Computation of a Straight Wire of Finite Length using Biot-Savart's Law version 1. by a current loop using the Biot-Savart Law. About the magnetic field of a finite wire 269 B A E B Figure2. (b) Find the magnitude and direction of the magnetic field at point C in the diagram, the midpoint of the bar, immediately after the switch is closed. Let's say we have two parallel wires carr. 1) Comparison between coulomb' s law and Biot Savart law. 2 A thin straight wire carrying a current I. The Biot—Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. Magnetic field on the axis of a circular current carrying loop (Biot Savart law Application) - Duration: 12:08. The law was stated in the year 1820 by Jean Baptisle Biot and Felix Savart. Equation Electric currents (along closed curve) The Biot-Savart law is used for computing the resultant magnetic field B at position r generated by a steady current I (for example due to a wire): a continual flow of charges which is constant in time and the charge neither accumulates nor depletes at any point. Biot Savart Law and Ampere's Law In the last lecture, we have shown that the magnetic force exerted on a small segment of wire flowing a current I with length dl is equal to where B is the magnetic flux density, and. The formulas can be used by a software tool to model the magnetic fields generated by e. of EECS 7-3 The Biot-Savart Law and the Magnetic Vector Potential Reading Assignment: pp. The formula is exact for an infinitely long wire. It states just what was said above, that the total B is equal to the sum of all the cross products of dL and all r over the length of the wire, divided by the length of r squared, and times the current and times a constant, where the constant is equal to the permeability of free space divided by 4 times. l be the distance between centre of the coil and elementary length dl. Magnetic Field Strength Due to Finite Length of Wire Carrying Current. As was seen in the demonstration, an electric current produces a magnetic field. Circular magnetic fields are generated around current carrying wires. Our interest is to make practical use of the Biot-Savart law. Fields & Currents Outline •Maxwell’s Equations for Magnetostatics •Biot‐Savart Law •Examples •#1 –Magnetic field around a finite length wire •#2 –Magnetic field around an infinite length wire •#3 –Getting a feel for the numbers •Current Distributions. Consider dl be the small current carrying element at point c at a distance r from point p. 6 - Magnetism - Biot-Savart Law. In undergraduate E&M courses the magnetic field due to a finite length, current-carrying wire can be calculated using the Biot-Savart law. The present work elucidates two separate computational methodologies involving direct determination of the magnetic field from Biot-Savart law. Finite Wings. JEE Advanced 2020 syllabus will be highly beneficial for candidates who are preparing for the upcoming entrance examination. The Biot-Savart law is named after Jean-Baptiste Biot and Félix Savart is an equation describing the magnetic field generated by an electric current who discovered this relationship in 1820. Charitat and Graner (CG) apply the Ampère and Biot–Savart laws to the problem of the magnetic field due to a straight current-carrying wire of finite length, and note that these laws lead to different results. The Biot-Savart Law. In the Biot-Savart’s law, the magnetic field is always perpendicular to the X-axis. The strength of these fields varies directly with the size of the current flowing through the wire and inversely to the distance from the wire. Magneticfieldcreatedbya circular loop Ampère’slaw(A. Object of class Mesh: the mesh grid where the magnetic field will be calculated. This is the same procedure implemented in the conventional DC forward problem. Laplace gave a differential form of their result, which now often is also referred to as. For (b), plot a 3D surface graph for the following range: (x=0. I also posted this question to codereview, but they send me to stackoverflow. Q1 Is the magnetic field created by a current loop uniform?. [email protected] This is a limiting case of the formula for vortex segments of finite length similar to a finite wire:. Physics 2102 Gabriela González • Quantitative rule for computing the magnetic field from any electric current • Choose a differential element of wire of length dL and carrying a current i • The field dB from this element at a point located by the vector r is given by the Biot-Savart Law: i µ 0 =4πx10-7 T. L8 ampere circuital law. The magnetic field along the axis of a circular loop of wire can also be calculated from the Biot-Savart Law. Now we will discuss each of the important topics along with an overview of the chapter followed by important formulas of the chapter which will help you in solving numerically related to Magnetic Effects of Current and Magnetism. Current element It is the product of current and length of infinitesimal segment of current carrying wire. The first evidence of the relationship of magnetism to moving charges was discovered in 1819 by the Danish scientist Hans Christian Oersted. The magnitude of d s x r is given by ) cos 2 rdssin rdssin( rds. steady current in a section of wire: How would we do this?. Itzs an inverse square law, and it depends on the vector ~r that points from the wire to P: The new complication is that the source is a vector, and so the Biot-Savart law involves the cross product. The Biot-Savart law lets us determine the magnetic field due to complex, current carrying shapes by considering the shape to be made of finite elements, each generating a piece of the magnetic field. Magnetic field from a finite straight current wire We apply the Biot-Savart's law to a finite length of straight current wire to find the magnitude at the point P (see Fig. First we deduce the magnetic field at a point$P$in space above a finite wire carrying a current$I$. The magnetic field due to a finite length of current-carrying wire is found by integrating Equation 12. 13) This expression is known as the Biot and Savart law. m/A (permeability constant). for an infinitely long solenoid b. Or sign in with one of these services. Biot-Savart law, Magnetic Field due to Current Carrying Wire. Consider a straight wire of length l carrying a steady current I. The same argument seems to lead to the conclusion that the magnetic field at the center of a current carrying ring is zero. It's one of the best textbooks I've seen out there. Biot Savart law f. In order to understand the Biot-Savart’s law, we need to understand the term current-element. Magnetic Field Generated by a Finite, Current- Carrying Wire Part A A steady current I is flowing through a straight wire of finite length. The simplest (and most fundamental) direct application of Ampère's law is to retrieve the experimental fact which prompted the formulation of the Biot-savart law to begin with, namely that the magnetic induction B due to a long straight wire is inversely proportional to the distance from that wire:. of EECS 7-3 The Biot-Savart Law and the Magnetic Vector Potential Reading Assignment: pp. ) • To determine the total magnetic field ( ) due to a finite sized conductor, we need to sum up the contributions due to all the current elements making up the conductor. The magnetic field due to a finite length of current-carrying wire is found by integrating Equation 12. 1 However, to the author's knowledge, no textbook presents the calculation of this field using the Ampere-Maxwell law: ∮. Jean Biot and Felix Savart arrived at an expression that the magnetic field H at any point in space to the current I that generates H. 2 A thin straight wire carrying a current I. The key point to keep in mind in applying eq. L6 current distribution in magnetism. What is Biot Savart Law The Biot Savart Law is an equation describing the magnetic field generated by a constant electric current. Consider a small element AB of current carrying wire whose length is $$d\vec s$$ and the position vector of point (P) from the element is $$\vec r$$. Solution From the Biot ÐSavart law , we expect that the magnitude of the þ eld is proportional to the current in the wire and decreases as the distance a from the wire to point P increases. Sign in with Facebook. In order to prove the Biot-Savart law, one writes, using R = r sinα (see the figure),. — The Biot-Savart Law states that the differential magnetic field dH. It can be used in the theory of aerodynamic for determining the velocity encouraged with vortex lines. It was first discovered by oersted. Figure $$\PageIndex{5}$$: Diagram to apply the Biot-Savart Law in order to determine the magnetic field along the symmetry axis of a ring carrying current, $$I$$. • When wire is perpendicular to the plane of paper, the field is in the plane of the paper. The magnetic field due to a finite length of current-carrying wire is found by integrating Equation 9. 2) Magnetic field at the centre of current carrying circular loop 3) Magnetic field due to a straight current carrying conductor of: i. MAHALAKSHMI ENGINEERING COLLEGE-TRICHY ELECTROMAGNETIC FIELDS K. As was seen in the demonstration, an electric current produces a magnetic field. Obviously, the Biot-Savart law gives the correct answer to our problem, and the Ampere` law gives a wrong result. Note in particular the inverse-square distance dependence, and the fact that the cross product will yield a field vector that points into the page. The exemplary calculations show the usefulness. The force with respect positional variation governed by biot-savart law is proved. Magnetic Induction: Magnetic Flux; Induced emf and Faraday's Law; Lenz's Law; Inductance; Self-inductance; Mutual Inductance; Magnetic Energy. Biot-Savart Law 0 2 ˆ 4 I d r µ π × = ∫ s r B general current source ex: finite wire Ampere's law 0 encd Iµ⋅∫ B s = current source has certain symmetry ex: infinite wire (cylindrical) Ampere's law is applicable to the following current. Therefore, it will tend to be the law used when Ampere's Law doesn't fit. To determine the Total Magnetic Field H due to a conductor of finite size, we need to sum up the contributions due to all the current elements making up the conductor. [1-11] and references therein. Marine Magnetic Anomalies, Oceanic Crust Magnetization, and Geomagnetic Time Variations. Example: a straight infinite wire. 3 The Biot–Savart Law. Let ds Gauss’s law and its applications Biot Savart Law - Moving Charges and Magnetism, Class 12, Physics EduRev. This is a limiting case of the formula for vortex segments of finite length similar to a finite wire:. 1, by the differential current clement / ill is proportional to the product / dl and the sine of the angle a between the clement and the line joining P to. Faraday’s law d. • Therefore the Biot-Savart law becomes: 2 Ö 4 R L I dl a Hr S R u ³ where L is the line path along which I exists Magnetic field due. It is named after Jean-Baptiste Biot and Félix Savart, who discovered this relationship in 1820. Biot savart Law Applications of Biot Savart Law Applications of Biot Savart Law for the circular coil Magnetic field due to a circular coil Magnetic field due to a uniformly charged circular coil. Modeling magnetic field in the vicinity of single wire helix Krzysztof Budnik, Wojciech Machczyński Poznań University of Technology 60 - 965 Poznań, ul. 022-62211530. Biot-Savart's law is an extension of Ampere's law, anything that satisfies Biot-Savart's law also satisfies Ampere's law, the extra parts of the equation have to be added to model the real world field effects involved in an ACTUAL device where Ampere's law is pure theory. We have therefore shown that. Also assume that in these units. The Biot—Savart law is also used in aerodynamic theory to calculate the velocity induced by vortex lines. Circular magnetic fields are generated around current carrying wires. twisted wires, helical coils, etc. Therefore, it will tend to be the law used when Ampere's Law doesn't fit. In using the Biot-Savart Law for an finite wire, I am having trouble understanding the angles. 1 Biot-Savart Law Currents which arise due to the motion of charges are the source of magnetic fields. Now let the net current j be non-zero. doc 1/1 Jim Stiles The Univ. The tridimensional version of the Biot-Savart law says that the magnetic field generated at the point$\boldsymbol{r}\in\mathbb{R}^3$by a tridimensional distribution of current defined by the current density$\boldsymbol{J}$is$\$\boldsymbol{B}(\boldsymbol{r})=\frac{\mu_0}{4\pi}\int_V\frac{\boldsymbol{J}(\boldsymbol{x}) \times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3. Magnetismin matter. You might be interested to know that Z dx (x2 + c)3=2 = x c(x2 + c)1=2 Answer: Because of the symmetry of a regular hexagon, each side produces. I also posted this question to codereview, but they send me to stackoverflow. Prior Knowledge: Magnetic Fields and Forces (pages 1-31 of presentation 14). This is a limiting case of the formula for vortex segments of finite length similar to a finite wire:. Ampere’s law is analogous to Gauss’s law in electrostatics. and the wire is assumed to be a part of a closed circuit. In other projects Wikimedia Commons. by a wire segment of length d' carrying current I at a point P is dB~ = „ 0 4… I d~'£ ^r r2 This law is actually a lot like Coulombzs law. 0), which is identical to the electrostatic case and Biot-Savart wire/solenoid calculations (enhanced in 8. The flow of electric current through a conductor creates a magnetic field around the conductor, whose strength depends on the magnitude of the current. Biot savart Law Applications of Biot Savart Law Applications of Biot Savart Law for the circular coil Magnetic field due to a circular coil Magnetic field due to a uniformly charged circular coil. In Figures 1,36,54, the magnetic flux created by the forward current in the first wire is partially cancelled by the magnetic flux created by the current flowing in the opposite direction in the other wire. This segment is taken as a vector quantity known as the current element. In undergraduate E&M courses the magnetic field due to a finite length, current-carrying wire can be calculated using the Biot-Savart law. Introduction •First discovered by Jean-Baptiste Biot and Félix Savart in the beginning of 19th century. MP EM Ass 16: Biot-Savart Law - Free download as PDF File (. This is the result of relative motion of the electrons and the fixed charges and is electrotatic in nature and needs to "magnetic. Physics 2102 Gabriela González • Quantitative rule for computing the magnetic field from any electric current • Choose a differential element of wire of length dL and carrying a current i • The field dB from this element at a point located by the vector r is given by the Biot-Savart Law: i µ 0 =4πx10-7 T. Circular magnetic fields are generated around current carrying wires. 02 Physics II: Electricity and Magnetism, Spring 2007 Prof. Biot–Savart’s law and Ampere’s law Magnetic field near a current-carrying straight wire, along the axis of a circular coil and inside a long straight solenoid; Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Take a small element of the wire of length dl. Note that the magnetic field lines form circles around the wire. Q1 Is the magnetic field created by a current loop uniform?. The Biot-Savart law is used to compute the magnetic field generated by a steady current, that is, a continual flow of charges, for example through a cylindrical conductor like a wire, which is constant in time and in which charge is neither building up nor depleting at any point. AP Physics C: Magnetism 6: Find Magnetic Field Using Biot Savart Law & Ampere's Law 10:51. The Biot-Savart Law: From Infinitesimal to Infinite Since students in introductory electricity and magnetism courses often find this law a mathematical mystery, we feel that a simple experiment such as this will provide the students a better understanding of the concepts introduced. Determine the magnitude and direction of the magnetic þ eld at point P due to this current. We show in this paper thatthehighly classical example of a straight wire,generally treated. dl ⃗ = μ 0 I e n c l o s e d \displaystyle \oint _C \vec{B}. 2) Magnetic field at the centre of current carrying circular loop 3) Magnetic field due to a straight current carrying conductor of: i. I used the magnetostatic tool in Ansys Workbench. Practice #1: Magnetic Field at Center of Ring of Current. 4, and, very close to a long wire, the potential is given approximately by equation 9. Use this law to obtain the expression for the magnetic field inside a solenoid of length 'l', cross-sectional area 'A' having 'N' closely wound turns and carrying a steady current 'I'. This law is although for infinitesimally small conductors yet it can be used for long conductors. Determine the magnetic eld vector at the centre of the hexagon (Think!). Hence, the Biot-Savart Law becomes µ µ 4 L 2 Id p R × = ∫ LR H [A/m], where L is the line path along which I exists. Biot savart Law Applications of Biot Savart Law Applications of Biot Savart Law for the circular coil Magnetic field due to a circular coil Magnetic field due to a uniformly charged circular coil. Answers a and d 3. Parametric study has been performed to study the fields at various armature positions. THERMAL PHYSICS I (25 Marks) LECTURES 25 + 5 Tutorial 1. 1) A circular coil of wire has 100 turns of radius 8cm, and carrying a current of 0. Magneticfieldcreatedbya: Straightcurrent-carryingwire Coil Magneticflux trougha surface. Current element It is the product of current and length of infinitesimal segment of current carrying wire. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The Ampère theorem and the Biot-Savart law are well known tools used to calculate magnetic fields created by currents. Summary of the two. We have therefore shown that. Magnetisms Part-1 BY NM SIR|Biot Savart law |Magnetic field due to finite wire|IIT-JEE |NEET PHYSICS. The present work elucidates two separate computational methodologies involving direct determination of the magnetic field from Biot-Savart law. Biot-Savart Law. b) Now, consider the wire structure consisting of two semi-infinite line segments, a finite straight line segment and a semicircle as shown in Figure (b). I am ok until there. Let 'P' be the point where the magnetic field due to the wire is to be studied. Place a charge q at the center of a sphere and apply Gauss' law. The magnetic field generated by such a wire is written. If not, the integral form of the Biot-Savart law must be used over the entire line segment to calculate the magnetic field. The wires are separated by equal distances d, and they carry equal currents I in the same direction. However, it is also much harder to apply. Two equations describe the relationship between the electric current and the magnetic field that it generates. Consider dl be the small current carrying element at point c at a distance r from point p. An electrical current I flows across the rectilinear finite wire AB. wire dl #rö =r. In this case x = 0 and only equation for x component of flux density remains. BIOT-SAVART LAW The magnetic field due to an element of a current-carrying wire is given by. Idlsinθ r2 From right ΔOQP, ΔOQP, θ+ϕ= 900 θ+ϕ= 900 or θ= 900 −ϕ θ= 900 −ϕ ∴sinθ=(900 −ϕ)=cosϕ ∴sinθ=(900 −ϕ)=cosϕ. Magnetic Fields (Biot-Savart): Summary Current loop, distance x on loop axis (radius R): Center of arc (radius R, angle θ): Straight wire: finite length infinite wire: B x = µ 0 IR 2 2(x2+R2)3/2 B center = µ 0 I 2R (cos) 412 0! " µ =# a I B a I B! µ 2 =0 θ 1 θ 2 B center= µ 0I! 4"R Ampere's Law Ampere's Law: applies to any closed. When charges move in a conducting wire and produce a current I, the magnetic field at any point P due to the current can be calculated by adding up the magnetic field. [email protected] Outside the wire, the field drops off regardless of whether it was a thick or thin wire. “Conduction” current density. Say the surface current density on this sheet has a value: J sxx(r)=Jaˆ meaning that the current density at every point on the surface has the same magnitude, and flows in the ˆa x direction. Integrating a circular current loop on axis -M oving charges (and currents) feel a force in magnetic fields current loops, straight wire segments. This equation is indicated by Biot-Savart law. Idlsinθ r2 dB= μ0 4π. Derivation of Biot Savart law. Coulomb’s law. The computational approach of. About the magnetic field of a finite wire 269 B A E B Figure2. carrying wire. Cite As Sathyanarayan Rao (2020). 6(a), consider Figure 7. Because of this the Biot-Savart Law is naturally takes on a differential form. The flow of electric current through a conductor creates a magnetic field around the conductor, whose strength depends on the magnitude of the current. Magnetic Field Due To Symmetrical Current Carrying Finite Wire | By Vivek Sir - Duration: 1:00:40. Because of this the Biot-Savart Law is naturally takes on a differential form. The charge q is the net charge enclosed by the integral. MAGNETISM PART 5 - Biot savart law (SK ACADEMY - PHYSICS BY HARSH SIR ) - Duration: 14:12. ANSWER: = Correct The magnetic field for an infinitely long wire can be obtained by setting in the previous expression. EQ1: This is the Law of Biot and Savart. on the axis of current carrying coil. 2) The sine of the angle between the element and the line joining point p to the element and. The flow of electric current through a conductor creates a magnetic field around the conductor, whose strength depends on the magnitude of the current. Concept #1: Biot-Savart Law with Calculus. Applying Ampere's Law We can use Ampere's Law now to calculate the magnetic field from certain current configurations. Consider a straight wire of length l carrying a steady current I. A steady current is a flow of charge that has been going on forever, and will be going on forever. Piotrowo 3A, e-mail: Wojciech. It looks like this. A consequence of the law of Biot and Savart is that the force between two parallel conductors carrying currents in opposite directions is. 2 Current-Carrying Arc Consider the current-carrying loop formed of radial lines and segments of circles whose centers are at point P as shown below. 53, 68 (2015); 10. Self Inductance of a Pair of Parallel Conductors. 24 267) for the use of the Biot-Savart law in the calculation of the magnetic field due to a straight current-carrying wire offinite length. Answers a and b b. Learn more about the Motion in Combined Electric and Magnetic Field. Faraday's law; Faraday's law and the Biot-Savart law; Faraday's law, Ampère’s law and the quasistatic approximation; Faraday's law: cutting a current-carrying wire; Faraday's law: wire loop encircling a solenoid; Field of a polarized cylinder; Field of a polarized object; Field of a polarized object - examples. The magnetic field produced by a steady line current is given by the Biot-Savart Law: where is an element of the wire Find the magnetic vector potential of a finite segment of straight wire carrying a current I. The $$x$$ axis goes into the page. NASA Astrophysics Data System (ADS) Dyment, J. In each case we observe the force and infer the field. Magnetic Induction: Magnetic Flux; Induced emf and Faraday's Law; Lenz's Law; Inductance; Self-inductance; Mutual Inductance; Magnetic Energy. 0 2 Ö 4 I dr dB r P S u kqr rÖ / 2 weaker field shown darker. Let us find magnetic field strength H at a point P at a distance R from the wire, as shown in figure 5. insufficient symmetry finite length current segment. Computation of magnetic field around finite solenoid as discussed in [5-8] involves determination of the same by evaluating magnetic vector potential. By Biot- Savart's law, the field dB due to a small element dl of the circle, centered at A is given by, This can be resolved into two components, one along the axis OP, and other PS, which is perpendicular to OP. AP Physics C: Magnetism 6: Find Magnetic Field Using Biot Savart Law & Ampere's Law 10:51. For a single loop as shown in Fig. Biot-Savart vs. 29-4 depends only on the current and the perpendicular distance R of the point from the wire. 1 The Biot–Savart Law To find the total magnetic field B created at some point by a current of finite size, we must sum up contributions from all current elements Ids that make up the current. wire so that 0 3 4 d I l r r Br rr. Use the law of Biot and Savart to find the magnitude of the magnetic field at point P due to the 1. This is a limiting case of the formula for vortex segments of finite length similar to a finite wire:. What is the limitation of Ampere’s circuital law? State ampere’s circuital law and express it mathmatically. Additive manufacturing (AM) is a layer-based process for producing parts. Place a charge q at the center of a sphere and apply Gauss' law. doc 1/1 Jim Stiles The Univ. “Conduction” current density. In This Chapter Biot-Savart Law Ampere’s Law Gauss’ Law for Magnetic Field Magnetic Scalar Potential Magnetic Vector Potential QuickField Magnetostatic Analysis Inductance Calculations Uniform Magnetic Fields Dipole Sources Shielding Applications Magnetic Monopoles While preparing a lecture demonstration in 1820, Orsted noticed that current flowing through a wire deflected a nearby compass. Calculate magnetic field based on Biot Savart Law. Robert Simcoe, Prof. Join GitHub today. due to circular arc,Questions on BSL. In electromagnetism and electronics, inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. It is the magnetic analogue of electrostatics , where the charges are stationary. Let me see if I can draw that. Faraday's law; Faraday's law and the Biot-Savart law; Faraday's law, Ampère’s law and the quasistatic approximation; Faraday's law: cutting a current-carrying wire; Faraday's law: wire loop encircling a solenoid; Field of a polarized cylinder; Field of a polarized object; Field of a polarized object - examples. Example 6-2 H of a steady current in a straight wire Determine H at a point Px;,,(y z) due to an infinitely long straight conducting wire of negligible thickness carrying a steady current I and lying along z-axis. That expression is based on the following experimental observations for the magnetic field dB at a point P associated with a length element ds of a wire carrying a. A straight, infinitely long wire carrying a steady current Il lies along the y-axis. To determine the Total Magnetic Field H due to a conductor of finite size, we need to sum up the contributions due to all the current elements making up the conductor. Magnetic Field due to a Current-Carrying Wire Biot-Savart Law - Magnetic Field due to a Current-Carrying Wire Biot-Savart Law AP Physics C Mrs. For (b), plot a 3D surface graph for the following range: (x=0. The formal Gauss' law connects flux to the charge contained again via an integral. Consider a straight wire of length l carrying a steady current I. Advanced Physics 10. Magnetic Field Due To Symmetrical Current Carrying Finite Wire | By Vivek Sir - Duration: 1:00:40. By symmetry the magnetic field produced by a straight infinite wire depends only on the Odistance from the wire s and is oriented perpendicular to the wire. Example-Infinite straight current carrying wire. Biot - Savart law and its application. In physics, specifically electromagnetism, the Biot–Savart law (/ ˈ b iː oʊ s ə ˈ v ɑːr / or / ˈ b j oʊ s ə ˈ v ɑːr /) is an equation describing the magnetic field generated by a constant electric current. Concept #1: Biot-Savart Law with Calculus. In order to understand the Biot-Savart’s law, we need to understand the term current-element. Assume that µo*i/(4p) is 1 in Biot-Savart law. Calculate magnetic field based on Biot Savart Law. BIOT-SAVART LAW The magnetic field B⃗ due to an element dl⃗ of a current-carrying wire is given by. The magnetic field H at a distance R from the wire is (in SI units) where in vacuum the magnetic field and the magnetic induction are related by B = μ 0 H (SI units) or B = H (Gaussian units). What is Biot-Savart Law? Biot-Savart’s law is an equation that gives the magnetic field produced due to a current carrying segment. m/A (permeability constant). f due to solenoid. Problem 15. Indeed, the Biot-Savart law is a general result of potential theory, and potential theory describes electromagnetic fields as well as. Ideally the experiment should be done with a direct current. The Biot-Savart Law. Theta 2 is measured to the right of point P. Using Biot-Savart to Find the Magnetic Field from a Finite Wire - Duration: 7:01. | 2020-07-12T23:21:13 | {
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https://math.stackexchange.com/questions/2617684/a-gift-problem-for-the-year-2018 | A Gift Problem for the Year 2018 [duplicate]
We had this problem in exam class yesterday on Combinatoric and it was supposed to be the new year gift from our teacher. The exercise was entitled A Gift Problem for the Year 2018
Problem:
The numbers $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{2018}$ are written on the blackboards. John chooses any two numbers say $x$ and $y$ erases them and writes the number $x+y+xy$. He continues to do so until there is only one number left on the board. What are the possible value of the final number?
I understood the problem as follows for instance if John take $x=1$ and $y=\frac{1}{2}$ then $x+y+xy =2$ and the new list becomes $$2,\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{2018}$$ continuing like this and so on.....
Please bear with me that I do not want to propose my solution since I fell like it was wrong and I don't want to fail the exam before the result get out. but by the way I found, $2017$, $2018$ and $2019$ but I am still suspicious.
You may help is you have an idea.
marked as duplicate by Peter Taylor, user1729, ahulpke, Strants, David HillFeb 1 '18 at 21:57
• Please show your work even if it is not correct. That way we can help and find any errors. – robjohn Jan 23 '18 at 16:35
• Since we are not in your class, how about an actual title? – Asaf Karagila Jan 23 '18 at 18:51
• Something more descriptive of the actual mathematical problem? – Asaf Karagila Jan 23 '18 at 20:16
• Okay, then therefore it's my downvote until you find a better title. – Asaf Karagila Jan 23 '18 at 20:21
• What's the excuse for rejecting the title suggested by @AccidentalFourierTransform, is it not click baity enough? – Asaf Karagila Jan 24 '18 at 6:05
Consider the multiplicative law on $\Bbb R$ defines by $$x*y =x+y+xy =(x+1)(y+1)-1$$
you can check that it is associative and commutative on $\Bbb R$. Therefore at the end the remaining number is \begin{align}x_0*x_1*x_2*\cdots x_{2018} &= 1*\frac{1}{2}*\frac{1}{3}*\cdots *\frac{1}{2018} \\&=\left[\prod_{i=1}^{2018}(1+x_i)\right]-1\\ &=\left[\prod_{i=1}^{2018}\left(1+\frac{1}{i}\right)\right]-1 \\ &=\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \ldots \cdot \frac{2018+1}{2018}-1=\color{red}{2019-1=2018.} \end{align}
Hint: Note that $xy+x+y = (x+1)(y+1) - 1$. In particular, this implies that at any stage of the process, if all the numbers on the board are $\{a_1, \cdots, a_k\}$, then the product $$(a_1 + 1) (a_2 + 1) \cdots (a_k + 1)$$ is invariant, i.e. it doesn't change when you erase two numbers $x$, $y$, and replace them with $xy+x+y$. What does this imply the final number must be?
• Is $(1+1)(1/2+1)\cdots(1/2018+1)$ computed by rewriting things as $(2)(3/2)(4/3)\cdots(2019/2018)$ and recognizing that it telescopes? – Mauve Jan 23 '18 at 15:44 | 2019-04-23T22:29:20 | {
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https://www.physicsforums.com/threads/find-greatest-acceleration-and-speed.692026/ | # Find greatest acceleration and speed
1. May 16, 2013
### songoku
1. The problem statement, all variables and given/known data
A winch is used to raise a 200 kg load. The maximum power of the winch is 5 kW. Calculate the greatest possible acceleration of the load when its speed is 2 ms-1, and the greatest speed at which the load can be raised.
2. Relevant equations
W = F.d
W = ΔKE
P = F.v
P = W/t
KE = 1/2 mv2
3. The attempt at a solution
Assume the initial speed = 0 m/s:
W = ΔKE = 0.5 x 200 x 4 = 400 J
Then do not know what to do.....
The answer should be 0.2 ms-2 and 2.55 ms-1
Thanks
2. May 16, 2013
### tiny-tim
hi songoku!
use power = energy per time = force times distance per time = force times speed
(and newton's second law)
3. May 16, 2013
### Simon Bridge
P=Fv
Draw a free body diagram to find F.
4. May 16, 2013
### songoku
I still don't get it. What I did:
F = P/v = 5000 / 2 = 2500 N
ƩF = m.a
2500 - 2000 = 200 . a
a = 2.5 ms-2
My logic: there are 2 forces acting on the object, one is from the winch directed upwards and the other is weight. By using F = P/v, I find the force by the winch then put it in second law of motion.
5. May 16, 2013
### tiny-tim
yes that looks fine
i got the same as you … the book answer seems to be wrong
(but i get the right answer for the second part! )
6. May 16, 2013
### songoku
hi tiny-tim
How to get the answer for the second part?
Using formula v = P/F, to get the greatest speed means that power has to be the greatest and the force should be minimum. Greatest power is 5000 W and minimum force is 2000 N, so v = 2.5 ms-1?
Thanks
7. May 17, 2013
### Simon Bridge
The first answer wants the maximum acceleration, the second wants the maximum speed.
It seems convenient to do them in reverse order.
I wouldn't normally do a homework problem, but it should be safe here since OP has already done the work and has pretty much understood the material.
from the fbd:
$T-mg=ma$ ( where T=tension in rope, and "up" is positive - assumes no losses)
... so
$T=m(a+g)$ ...(1)
power from the winch must be $P=Tv$ ...(2)
For constant speed, T=mg so the power needed is P=mgv
oops - I stand corrected!
Hopefully the walk-through didn't go through yet
@songoku: for the second section - the velocity is constant, so what is the acceleration?
Therefore what is the tension in the rope.
Last edited: May 17, 2013
8. May 17, 2013
### songoku
Hope I don't misinterpret your work.
By putting T = m(a+g) to P = Tv:
P = m(a+g)v
a = P/mv - g = 2.5 ms-2?
Constant velocity means zero acceleration so the tension will be the same as weight (2000 N). Then v = P/F = 5000/2000 = 2.5 ms-1?
Both answers are the same as I have posted above.
Thanks
9. May 17, 2013
### Simon Bridge
You seem to be using g=10N/kg - the book answer suggests they expect you to use g=9.8N/kg.
... and that is what I answered.
Your reasoning appeared indirect, and you seemed uncertain, so I figured you'd benefit from a more direct path.
To get the book's a=0.2m/s/s requires P/mv = 10, what does the speed have to be?
How does that compare to the maximum speed?
Is that sensible?
You could try figuring out what values of P and m would give you the book values.
10. May 17, 2013
### songoku
Oh ok. I get it. So let me redo it a little:
To find greatest speed: v = P / F = 5000 / 1960 = 2.55 ms-1
The speed has to be 2.5 ms-1 to obtain acceleration 0.2 ms-2 by using a = P/mv - g
I think it is still sensible if we compare it with maximum speed, but the question asks when the speed is 2 ms-1 so I think the greatest acceleration at that speed can't be 0.2 ms-2. Do I get it right?
Thanks
11. May 17, 2013
### Simon Bridge
Well done!
You've just disproved the book answer for the initial assumptions.
It [the book] is either, wrong, or, the assumptions about the problem were wrong.
It looks to me like the problem got edited just before publication and the answers were not.
You may find it is fixed in the next edition.
The power of science/math is such that a novice can overturn an established authority.
If you know other people working through the same problems, you should contact them and compare.
12. May 17, 2013
### songoku
Hahaha I like the way you said it.
Thanks a lot for the help, simon and tiny-tim | 2017-12-15T05:12:34 | {
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http://math.stackexchange.com/questions/198129/perfect-squares | # Perfect squares
Wonder whether anybody here can provide me with a hint for this one.
Is $c=1$ the only case in which the expression $(c^2+c-1)(c^2-3(c-1))$
returns a perfect square?
-
Oops! 0% accept rate? That'd make a square go round... – DonAntonio Sep 17 '12 at 18:16
............... – absalon Sep 17 '12 at 18:18
@Ben, so, what's the cutoff? and, why? – Gerry Myerson Sep 18 '12 at 6:43
@GerryMyerson: well, actually, I think accept rate is a fairly flawed metric in general, and should never be taken that seriously... getting people to spuriously accept answers they don't like just to get their accept rate back up is to me as much of a problem as people not accepting at all. Still, I get that some people use the site without understanding the accept mechanism, but I'd go find a question of theirs with a good and unaccepted answer before making a comment. If I can't be bothered to do that, I wouldn't comment at all. – Ben Millwood Sep 18 '12 at 12:12
@Ben, my experience with people at zero percent is that they are unaware of the accept mechanism and are grateful when it is pointed out to them. If they are aware of the mechanism and truly don't like any of the answers they've had on any of the questions they have asked, why would they keep asking questions here? – Gerry Myerson Sep 18 '12 at 12:51
Yes, $c=0$ is the only such value. For the proof, it is useful to let $c=x+1$. Then our expression becomes $$(x^2+3x+1)(x^2-x+1).$$ Note that $x^2-x+1$ is always odd. Any common divisor of $x^2+3x+1$ and $x^2-x+1$ must divide the difference $4x$. But such a common divisor must be odd, so any common divisor must divide $x$. But then it must divide $1$.
Thus $x^2+3x+1$ and $x^2-x+1$ are relatively prime. Since $x^2-x+1$ is always positive, it follows that if their product is a perfect square, each must be a perfect square.
But that can only happen when $x=0$. To prove this, use the fact that for any integer $u$, there is no perfect square strictly between $u^2$ and $(u+1)^2$. Since you asked for a hint, I will, unless you request otherwise, leave out the rest of the argument. It is short. | 2013-12-19T15:00:12 | {
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https://stats.stackexchange.com/questions/360127/binomial-approximation-to-hypergeometric-probability | # Binomial Approximation to Hypergeometric Probability
I am trying to understand how to apply the binomial distribution to a simple probability problem. I can solve the problem directly via the classical definition of probability but, when trying to interpret the problem as sampling from a binomial distribution, I get different results.
Problem statement:
The prevalence of some disease in a given country is $p$. A sample of $n<N$ people is selected from a city with $N$ inhabitants (in that same country). What is the probability that exactly $k$ people in this sample have the disease?
method 1 (favourable/total): there are $\binom N n$ possible samples from the population. Out of those, we are interested in the ones with $k$ infected (out of $pN$) and $n-k$ healthy people (out of $(1-p)N$), which account for $\binom {pN} k\binom {(1-p)N} {n-k}$ possibilities. Thus, the probability is $$P = \frac {\binom {pN} k\binom {(1-p)N} {n-k}} {\binom N n}$$
method 2 (binomial): It seems that this problem can be cast as sampling from a binomial distribution, with success probability $p$ and $n$ repetitions. We are interested in $k$ successes, thus we should have
$$P(k) = \binom {n} k p^k (1-p)^{n-k}$$
If we take concrete numbers, eg N=200, p=0.1, n=20, k=2, we end up with $P\approx0.30$ for method 1, while method 2 gives $P \approx 0.28$.
• Why are these numbers different?
• What is wrong with the binomial solution?
• Should it somehow depend on the sample size $N$?
• If you select someone presumably you won't select them again in which case once they're in your sample the probability for those left for you to select from is changed. – Glen_b Aug 1 '18 at 7:22
• Thanks! Now I see that we are selecting from the population without replacement - which would be described by a hypergeometric distribution (I am also happy to notice that method 1 above derives exactly the hypergeometric pmf...) – Joseph Greenpie Aug 1 '18 at 17:13
The exact probability is hypergeometric, as in the displayed equation in your Question. It assumes sampling without replacement. (That is the same person cannot be chosen twice.)
If $n$ is very much smaller than $N,$ then a binomial model, which assumes sampling with replacement may be useful. (The approximation is based on the relatively low chance the same person would be chosen more than once when only a few $n$ are chosen out of many $N.$ A common rule of thumb for usefulness of the binomial approximation is to have $n/N < 0.1.)$
Let's look at specific numbers to see how this plays out computationally. Let $N = 100,000,\, n = 500,\,k = 10,\, p = .02.$
Hypergeometric: The number of infected individuals in the city is 2000 and the remaining 98,000 are uninfected: $P(X = 10) = 0.1267$ and $P(X \le 10) = 0.5831.$ Computations in R, where dhyper and phyper are a PDF and a CDF of a hypergeometric distribution.
> dhyper(10, 2000, 98000, 500)
[1] 0.1266969
> phyper(10, 2000, 98000, 500)
[1] 0.5830506
Binomial approximation: Here $Y \sim \mathsf{Binom}(n = 500, p = .02).$ Then $P(Y = 10) = 0.1264$ and $P(Y \le 10) = 0.5830.$
> dbinom(10, 500, .02)
[1] 0.1263798
> pbinom(10, 500, .02)
[1] 0.583044
In these examples the binomial approximations are very good. The plot below shows this hypergeometric distribution (blue bars) and its binomial approximation (red). Within the resolution of the plot, it is difficult to distinguish between the two.
Note: With huge population sizes, the binomial coefficients in the hypergeometric PDF can become so large that they overflow R's ability to handle them. The program is written to minimize this difficulty, but even so, there are limits on what can be computed. R makes it possible to find log probabilities to prevent overflow; then you can take exponents to get answers.
Method (2) is invalid because we are sampling from the population without replacement - which leads to a hypergeometric distribution instead of the binomial. Notice that method (1) ends up corresponding precisely to the hypergeometric probability mass function.
As a sidenote notice that, for $N\gg n$, replacements should not make a difference, and thus we end up with a binomial. This limit has been calculated elsewhere. | 2020-08-15T06:18:20 | {
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https://math.stackexchange.com/questions/2748289/can-a-function-have-a-derivative-where-it-has-no-value | # Can a function have a derivative where it has no value?
Consider the following function: $$f(x) = x^2 \mid x \in \mathbb{R}, \ x \ne 0$$
The derivative at $x=0$ seems to want to be zero, in the same way that $\lim \limits_{x \to 0} f(x) = 0$
However, when I look at the definition of the derivative, this doesn't seem to work: $$f'(x) = \lim \limits_{\Delta x \to 0} \frac{f(x+\Delta x )-f(x)}{\Delta x}.$$ The function isn't defined at $f(x)$, so $f'(x)$ is also undefined. Would it make any sense to replace the $f(x)$ in the definition with $\lim \limits_{x \to 0} f(x) = 0$? Then, I suppose we'd have $\lim \limits_{x \to 0} f'(x) = 0$?
Would it be permissible? Would there be any point?
• I've arbitrarily defined it with a hole at $x=0$. Did I not represent that correctly? – MathAdam Apr 22 '18 at 6:20
• What if I set f(0) = -1? Just to be pesky? The limit is still zero but it seems like the derivative is a mess. – MathAdam Apr 22 '18 at 6:22
• If you set $f(0)=-1$, then the new function $f$ is not continuous at $x=0$, hence not differentiable at $x=0$. But take note of MPW's answer. If you extend the function $f$ by defining it at $x=0$, ir's no longer $f$, so you should use a new letter, so as not to cause confusion. – quasi Apr 22 '18 at 6:25
• A naive approach might instead be to consider $$\lim_{h\to 0} \frac{f(x +h)-f(x-h)}{2h}.$$ – shalop Apr 22 '18 at 6:58
• @Shalop Your approach is not equivalent to the commonly accepted definition. Compare the results for the modulus function: $\mathbb R\ni x\mapsto|x|\in\mathbb R$. Your expression evaluates to $0$ while the original is 'undefined'. – CiaPan Apr 23 '18 at 8:17
If a function $f$ is not defined at a point $x_0$ we can't evaluate the derivative at $x_0$ because in the definition of the derivative at $x_0$ we need $f(x_0)$.
However if $f$ is continuous in $0<|x-x_0|<r$ with $r>0$ and the limit $\lim_{x\to x_0}f(x)=L$ exists we can extend $f$ to a continuous function in $|x-x_0|<r$ by letting $f(x_0):=L$.
Moreover if $f$ is also differentiable in $0<|x-x_0|<r$ and the limit $\lim_{x\to x_0}f'(x)=a$ then the extended function is also differentiable at $x_0$ and its derivative at $x_0$ is equal to $a$.
No.
You can certainly extend $f$ to a function $g$ as you indicate, but then you are really computing $g'(0)$, not $f'(0)$.
In that case, $f'$ agrees with $g'$ on their common domain, which excludes $x=0$.
Remember, if a function is differentiable at $x_0$, then it is continuous at $x_0$. In particular, it is defined at $x_0$.
As has already been said, and as you have found yourself, the ordinary derivative is not defined. However, the symmetric derivative is defined in your case: $$f'_{\text{symmetric}}(x_0) := \lim_{h\to 0} \frac{f(x_0+h)-f(x_0-h)}{2h}$$
A more general derivative is also defined (I'll call it excluding; I don't know if it has a name): $$f'_{\text{excluding}}(x_0) := \lim_{h,k\to 0\\h,k \neq 0} \frac{f(x_0+h)-f(x_0+k)}{h-k} = \lim_{x_1,x_2\to x_0\\x_1,x_2 \neq x_0} \frac{f(x_1)-f(x_2)}{x_1-x_2}$$
• I don't know if it has a name --- From my answer to “Strong” derivative of a monotone function: "Besides strong derivative, three other names I’ve encountered in the literature are unstraddled derivative (Andrew M. Bruckner), strict derivative (Ludek Zajicek and several people who work in the area of higher and infinite dimensional convex analysis and nonlinear analysis), and sharp derivative (Brian S. Thomson and Vasile Ene)." – Dave L. Renfro Apr 22 '18 at 8:19
• I don't think it's necessary to specify $h,k\neq 0$ in the limit subscript, is it? I would expect it to be implicit in the definition of a limit that the changing variable doesn't actually take on the value which it is approaching. – David Z Apr 22 '18 at 8:38
• @DavidZ $\lim_{x_1,x_2 \rightarrow x_0}$ isn't quite standard notation. So it might be implying you get equal and defined results from $\lim_{x_1 \rightarrow x_0} \lim_{x_2 \rightarrow x_0}$ and $\lim_{x_2 \rightarrow x_0} \lim_{x_1 \rightarrow x_0}$, but that's not in general true. Or more likely it would mean $\lim_{(x_1, x_2) \rightarrow (x_0, x_0)}$, but this allows approach from any path on $\mathbb{R}^2$, including straight horizontal and vertical. – aschepler Apr 22 '18 at 10:52
• It seems to me that the defintion of $f'_{\text{symmetric}}$ says that if $f(x) = \lvert x\rvert,$ then $f'_{\text{symmetric}}(0)=0.$ But the definition of $f'_{\text{excluding}}$ then would say the derivative doesn't exist at $x=0,$ which seems a more sensible answer. – David K Apr 22 '18 at 13:14
• Thanks @aschepler for giving a good answer to why I wrote $h,k \neq 0$. – md2perpe Apr 22 '18 at 16:43 | 2021-03-04T09:19:25 | {
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https://math.stackexchange.com/questions/440065/is-sin-x-cos-x-independent/440095 | Is $\{\sin x,\cos x\}$ independent?
Is $$\{\sin x,\cos x\}$$ linearly independent in $$\mathbb{R}^n$$?
I thought they were not because I can write $$\cos x=\sin (x+\pi/2)$$.
My professor on the other hand said it was independent and his proof is as follows:
If $$\{\sin x,\cos\}$$ is independent in $$[0,2\pi]$$, then it will be independent on all of $$\mathbb{R}.$$ (He didn't prove this either.)
$$a\cos x+b \sin x=0 ,\forall \vec{x}\in[0,2\pi].$$
$$x=0\implies a\cdot1+b\cdot0=0 \implies a=0$$.
$$x=\pi /2\implies a\cdot 0+b \cdot 1 \implies b=0$$.
So $$\{\sin x, \cos x\}$$ is independent on $$[0,2\pi],$$ and thus independent everywhere. $$QED$$.
Is there something I am missing or not understanding...? Why is this set independent, when I can express an element of the set as a linear combination?
• The Wronskian is -1, which means they are linearly independent as a second method. Regards Jul 10 '13 at 0:37
• You have just shown that there are no $a$ and $b$ other than $0$ such that $a\cos x+b\sin x$ is identically $0$. That says precisely that his collection of two functions is a linearly independent set. Jul 10 '13 at 0:40
• What is the relation between $\cos x=\sin(x+\pi/2)$ and linear dependence?
– blue
Jul 10 '13 at 0:48
• It is enough to consider $[0,2\pi]$ because $\sin x$ and $\cos x$ are $2\pi$ periodic. Although it is true that $\cos x=\sin (x+\pi/2)$, that relation does not use the definition of what it means for two functions to be linearly independent. Jul 10 '13 at 0:48
• Your confusion stems from your claim that $\cos x = \sin ( x + \frac{\pi}{2}$. Though it is a valid transalation, it has nothing to do with linear dependence. E.g. Show that the functions $x$ and $x+1$ are linearly independent. Jul 10 '13 at 0:48
You proved that there were no non-trivial (ie, with at least one non-zero coefficient) linear combination of $\sin$ and $\cos$ identically zero on $[0,2\pi]$. That by definition means that $\{\sin,\cos\}$ is linearly independent on $[0,2\pi]$.
This also implies that they are on $\mathbb{R}$, because in particular any identically zero linear combination on $\mathbb{R}$ is also zero on $[0,2\pi]$ (the reciprocal is also true, actually, by $2\pi$-periodicity).
As for your initial comment about $\cos x = \sin(x+\frac\pi 2)$ for all $x$, it is not a linear combination. You are not writing $\cos=\alpha\sin$ for some scalar $\alpha$, you are writing $\cos=\sin\circ f$ for some function $f$.
• It can be seen as a linear combination. But it is trivial.
– OR.
Jul 10 '13 at 0:51
• What is? If you are talking about the last identity with $\sin(x+\frac{\pi}{2}$, it is not a linear combination (unless you mean that is is exactly writing $\cos=1\cdot\cos$). A linear combination is of the form $\sum_i \alpha_i f_i$ with the $\alpha_i$ being scalars, there cannot be any composition of functions (which is not one of the two operations intrinsically part of the vector space, which are addition between vectors and multiplication by a scalar). Jul 10 '13 at 0:54
I am surprised no one has yet pointed out that your question
Is $\{\sin x,\cos x\}$ linearly independent in $\mathbb{R}^n$?
doesn't make sense.
For a start, $\sin x$ and $\cos x$ aren't in $\mathbb{R}^n$! Your notation is a little ambiguous: you could mean the functions $\sin$ and $\cos$ (in which case they live inside a vector space of functions), or you could mean the numbers $\sin x$ and $\cos x$ for some fixed value $x$, say $x = 5$ (in which case they live inside a vector space of numbers, like $\mathbb{R}$).
I'm going to assume you are thinking of them as functions. So let's be clear about this: let's define $V$ to be the $\mathbb{R}$-vector space spanned by the two vectors $\mathbf{u}$ and $\mathbf{v}$, where secretly $\mathbf{u}$ is the function $\mathbf{u}(x) = \sin x$ and $\mathbf{v}$ is the function $\mathbf{v}(x) = \cos x$.
Now, $V$ is a vector space of functions - what's the zero vector? Of course, it's the zero function, $\mathbf{0}$, defined by $\mathbf{0}(x) = 0$. What does linear independence mean? Well, it means that we can find real numbers $a$ and $b$ such that $a\mathbf{u} + b\mathbf{v} = \mathbf{0}$. But these are functions, and two functions $f$ and $g$ are equal when they agree at all $x$, i.e. $f(x) = g(x)$ for all $x$.
That is, we want $a\mathbf{u}(x) + b\mathbf{v}(x) = 0$ for all $x$, or in more familiar language, $a\sin x + b\cos x = 0$ for all $x$. Now, what are $a$ and $b$?
By definition $\{\cos(x), \sin(x)\}$ is a set of linearly independent in $C(\mathbb{R})$ if for all $a,b\in\mathbb{R}$ that satisfies the equation $$a\cos(x)+ b\sin(x)=0$$ for all $x\in\mathbb{R}$ implies $a=0$ and $b=0$.
Supose that $a\neq 0$ or $b\neq 0$ and $a\cos(x)+ b\sin(x)=0$ for all $x\in\mathbb{R}$. Check that for values of $a$ and $b$ arbitrary but fixed the equality hold only for fixed values of $x$ and not all values of $x$ in $\mathbb{R}$. And this contradicts our initial assumption.
Your professor's first claim (that we just have to show linear independence of $\sin x$ and $\cos x$ on $\lbrack0,2\pi\rbrack$) follows from $\sin x$ and $\cos x$ both being $2\pi$ periodic; two real numbers $a$ and $b$ satisfy $a \sin\theta + b\cos\theta = 0$ for all $\theta \in \mathbb{R}$ if and only if $a \sin x + b \cos x = 0$ for all $x \in \lbrack 0, 2\pi\rbrack$. To prove this, note $\implies$ is clear. For $\Longleftarrow$, note every $\theta \in \mathbb{R}$ is of the form $\theta = x + 2k\pi$ for some $x \in \lbrack0,2\pi\rbrack$ and $k \in \mathbb{Z}$, and so $a \sin x + b \cos x = 0$ and $2\pi$ periodicity imply $a\sin\theta + b\cos\theta = a \sin(x + 2k\pi) + b\cos(x + 2k\pi) = 0$.
For your other question, the statement $\cos x = \sin(x + \pi/2)$, $\forall x \in \mathbb{R}$ is an equation of linear dependence, but not for the vectors/functions $f(x) = \cos x$ and $g(x) = \sin x$. This is because $g(x)$ isn't in your equation: $h(x) = g(x+\pi/2) \sin(x+\pi/2)$ is! In fact, what you've shown is that $\{f(x),h(x)\} = \{\cos x, \sin(x + \pi/2)\}$ is a linearly dependent set.
I can elaborate if you still have questions. | 2021-09-19T07:05:57 | {
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https://math.stackexchange.com/questions/2577156/function-with-path-connected-graph | # function with path connected graph
Let $f : (a,b) \to \mathbb R$ be a function with path connected graph, then is $f$ necessarily continuous ?
If the domain was instead of the form $[a,b]$, then I can show $f$ is continuous from path connectedness of the graph, but I don't know what happens in other cases.
• @Mirko: In $[a,b]$, the result is true, as also shown in this sight. math.stackexchange.com/questions/366747/… (one of the reasons I don't like geometric arguments unless it can be explicitly written down ... ) – user495643 Dec 22 '17 at 20:23
• looking at the link you provided math.stackexchange.com/questions/366747/… (indicating that the result for $[a,b]$ is true), I realized I was misinterpreting your question. I erased my older comments and posted what I believe is a detailed and correct answer now. – Mirko Dec 22 '17 at 23:07
The positive answer for an open interval $(a,b)$ easily follows from the positive answer for a closed interval $[a,b]$. I first wrote an answer along these lines, but now I prefer the following alternative proof (moving the older version at the end).
I prefer to give an alternative, self-contained and direct proof that covers both cases $(a,b)$ and $[a,b]$ simultaneously.
Suppose that $f:D\to\mathbb R$ has a path-connected graph, where $D$ is a connected subset of $\mathbb R$ (that is, an interval, open, or closed, or half-open, finite or infinite). We claim that $f$ is continuous.
Take any $p\in D$. It is enough to show that $f$ is continuous at $p$ from the left (assuming $p$ is not a left-endpoint of $D$), and that $f$ is continuous at $p$ from the right (assuming $p$ is not a right-endpoint of $D$). Of course, by symmetry, it would also be enough to consider just one of these cases.
Assume, towards a contradiction, that $f$ is discontinuous from the right at some $p$ (where $p$ is not a right-endpoint of $D$). Then there is some $\varepsilon>0$ and a monotone decreasing sequence of points $p_n\in D$, converging to $p$, and such that $|f(p)-f(p_n)|\ge\varepsilon$ for all $n\ge0$.
There is a path $\gamma:[0,1]\to G(f)$ in the graph $G(f)$ from $\langle p_0,f(p_0)\rangle$ to $\langle p,f(p)\rangle$. Each vertical line $x=p_n$ separates the plane, and $\gamma(0)=\langle p_0,f(p_0)\rangle$ whereas $\gamma(1)=\langle p,f(p)\rangle$, hence for each $n$ there is $t_n\in[0,1]$ such that $\gamma(t_n)=\langle p_n,f(p_n)\rangle$. The sequence of the $t_n$ is bounded and hence has a convergent subsequence. Without loss of generality (replacing the sequence of the $p_n$ with a suitable subsequence) we may assume that the sequence of the $t_n$ itself is converging, say to some $s$. By continuity of $\gamma$ we must have that $\gamma(t_n)$ converges to $\gamma(s)$. But $\gamma(t_n)=\langle p_n,f(p_n)\rangle$ and the first coordinates converge to $p$, so we must have that $\gamma(t_n)$ converges to $\langle p,f(p)\rangle,$ which is the unique point of the graph on the vertical line $x=p$. Hence $\gamma(s)=\langle p,f(p)\rangle$. But the second coordinates of $\gamma(t_n)=\langle p_n,f(p_n)\rangle$, namely the $f(p_n)$ are distance at least $\varepsilon$ away from $f(p)$, hence cannot converge to $f(p)$. This contradiction completes the proof.
Note that we cannot replace the assumption that the graph of $f$ is path-connected with the weaker assumption that it is connected. An example is $f(x)=\sin\frac1x$ for $x\not=0$, and $f(0)=0$ (as discussed at http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2002&task=show_msg&msg=0356.0001 ).
Here is also an older version of my answer, showing how the positive answer for an open interval $(a,b)$ follows from the known (to the OP) positive answer for a closed interval $[a,b]$.
Say $f : (a,b) \to \mathbb R$ is a function with a path connected graph. It is enough to show that the restriction of $f$ to $[c,d]$ is continuous whenever $a<c<d<b$. For the latter, it is enough to show that the graph of this restriction is path-connected (making use of the known result for closed intervals).
Take any $p$ and $q$ with $c\le p<q\le d$. Let $\gamma:[0,1]\to G(f)$ be a path, where $G(f)$ is the graph of $f,$ with $\gamma(0)=\langle p,f(p)\rangle$ and $\gamma(1)=\langle q,f(q)\rangle$. Let $t_0$ be the largest value of $t$ such that $\gamma(t)$ is on, or to the left of, the vertical line $x=p$. Note that $\gamma(t_0)$ must be on the line $x=p$ (or else we get a contradiction with the definition of $t_0$), and in particular (since $f$ is a function, and applying the vertical line test), we have that $\gamma(t_0)=\langle p,f(p)\rangle$. Let $t_1$ be the smallest value of $t>t_0$ such that $\gamma(t)$ is on, or to the right of, the vertical line $x=q$. Then the restriction of $\gamma$ to $[t_0,t_1]$ is a path from $\langle p,f(p)\rangle$ to $\langle q,f(q)\rangle$, remaining inside the graph of the restriction of $f$ to $[p,q]$, and hence also inside the graph of the restriction of $f$ to $[c,d]$.
The above shows that the graph of the restriction of the $f$ to $[c,d]$ is path-connected, and hence, using the known result for closed intervals, the restriction of $f$ to $[c,d]$ is continuous. It follows that $f$ is continuous on $(a,b)$.
Edit 12/27/2017, answering the first comment below:
"How did you get that for each $n$ there is $t_n$ with $\gamma(t_n)=(p_n,f(p_n))$ ? – misao".
Let $V$ be the vertical line $\{(x,y):x=p_n\}$. It splits the plane into the open left and right half-planes $L=\{(x,y):x<p_n\}$ and $R=\{(x,y):x>p_n\}$. The image under $\gamma$ of $[0,1]$, namely $\gamma([0,1])$, is connected, since $\gamma$ is continuous and $[0,1]$ is connected. Also, $\gamma([0,1])$ intersects both $L$ and $R$ (one in $\gamma(1)$, the other in $\gamma(0)$, assuming here that $n>0$, so $p<p_n<p_0$, since for $n=0$ clearly $t_0=0$ works). Then $\gamma([0,1])$ must intersect $V$, for otherwise $(\gamma([0,1])\cap L)\cup(\gamma([0,1])\cap R)$ would be a relatively closed-and-open partition by non-empty sets of $\gamma([0,1])$, contradicting that the latter is connected. Say $(p_n,y_n)\in\gamma([0,1])\cap V$. But then $(p_n,y_n)\in\gamma([0,1])\subseteq G(f)$, and hence $y_n=f(p_n)$, since $f$ is a function, it passes the vertical line test, there is a unique point in $G(f)\cap V$ and this unique point is $(p_n,f(p_n))$. (That is, $f$ has a unique value at $p_n$.) At the same time, since $(p_n,y_n)\in\gamma([0,1])$, there must be $t_n$ such that $(p_n,y_n)=\gamma(t_n)$. Thus $\gamma(t_n)=(p_n,y_n)=(p_n,f(p_n))$.
• How did you get that for each $n$ there is $t_n$ with $\gamma(t_n)=(p_n , f(p_n) )$ ? – user495643 Dec 27 '17 at 10:06
• @misao I added an edit at the end of my answer, clarifying the point in your comment. – Mirko Dec 28 '17 at 3:34 | 2020-03-31T20:40:24 | {
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