}}{{k!\left( {n - k} \right)!}}. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? One can drop one of the numbers in the bottom list and infer it from the fact that sum … binomial The combination (n r) (n r) is called a binomial coefficient. }}{{k!\left( {n - k} \right)!}} The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. C — All combinations of v matrix. = \binom {n} {k}  This is the binomial coefficient. The binomial coefficient $\binom{n}{k}$ can be interpreted as the number of ways to choose k elements from an n-element set. where A is the permutation, A_n^k = \frac{n!}{(n-k)! In latex mode we must use \binom fonction as follows: \frac {n!} Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Then it's a good reason to buy me a coffee. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. k-combinations of n-element set. Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. Gerhard "Ask Me About System Design" Paseman, 2010.03.27 \endgroup – Gerhard Paseman Mar 27 '10 at 17:00 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … This article explains how to typeset them in LaTeX. The binomial coefficient is defined by the next expression: \binom {n}{k} = \frac {n ! }}{{k!\left( {n - k} \right)!}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. Latex binomial coefficient Definition. If your equation requires specific numbers in place of the "n" or "k," click on a letter to select it, press "Delete" and enter a number in its place. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … The binomial coefficient is defined by the next expression: \ [ \binom{n} {k} = \frac{n!} (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. Knowledge base dedicated to Linux and applied mathematics. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. In latex mode we must use \binom fonction as follows: \frac{n!}{k! The order of selection of items not considered. (adsbygoogle = window.adsbygoogle || []).push({}); Thank you ! The possibility to insert operators and functions as you know them from mathematics is not possible for all things. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. \boxed, How to write table in Latex ? Using fractions and binomial coefficients in an expression is straightforward. The Texworks shows … Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. ( n - k )! Here's an equation: math \frac {n!} (n−k)! Mathematical Equations in LaTeX. = \binom{n}{k} This will give more accuracy at the cost of computing small sums of binomial coefficients. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". n! This same array could be expressed using the factorial symbol, as shown in the following. And of course this command can be included in the normal text flow \ (\binom{n} {k}\). All combinations of v, returned as a matrix of the same type as v. Specially useful for continued fractions. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Binomial Coefficient. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Then it's a good reason to buy me a coffee. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. binomial coefficient Latex. LaTeX forum ⇒ Math & Science ⇒ Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. Asking for help, clarification, or responding to other answers. }{k ! therefore gives the number of k -subsets possible out of a set of distinct items. It is especially useful for reasoning about recursive methods in programming. All combinations of v, returned as a matrix of the same type as v. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Open an example in Overleaf (n - k)!} formulas, graphs). So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… Binomial coefficient, returned as a nonnegative scalar value. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. } ... Pascal’s triangle. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. Identifying Binomial Coefficients. It will give me the energy and motivation to continue this development. Ak n = n! The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. (n-k)!} I agree. The binomial coefficient is defined by the next expression: \[\binom {n}{k} = \ frac {n!}{k!(n-k)! Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In Counting Principles, we studied combinations.In the shortcut to finding ${\left(x+y\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. On the other side, \textstyle will change the style of the fraction as if it were part of the text. Don't forget to LIKE, COMMENT, SHARE & SUBSCRIBE to my channel. Usually, you find the special input possibilities on the reference page of the function in the Details section. How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? For these commands to work you must import the package amsmath by adding the next line to the preamble of your file In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. \vec,\overrightarrow; Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Also, the text size of the fraction changes according to the text around it. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. Stanley's EC1 also uses it as the primary name, which counts for a lot in my book. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. The text inside the first pair of braces is the numerator and the text inside the second pair is the denominator. The command \displaystyle will format the fraction as if it were in mathematical display mode. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is Latex Binomial coefficient, returned as a nonnegative scalar value. begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? (n - k)!} {k! = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\frac{n!}{k! (n - k)!} This website was useful to you? {k! In Counting Principles, we studied combinations. k-combinations of n-element set. This method of constructing mathematical proofs is called mathematical induction. samedi 11 juillet 2020, par Nadir Soualem. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? Since binomial coefficients are quite common, TeX has the \choose control word for them. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. In this video, you will learn how to write binomial coefficients in a LaTeX document. I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an n-element set. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … Latex k parmi n - coefficient binomial. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. For example, … The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. therefore gives the number of k-subsets possible out of a set of distinct items. You can set this manually if you want. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file How to write it in Latex ? C — All combinations of v matrix. The usage of fractions is quite flexible, they can be nested to obtain more complex expressions. coefficient are the different ordered arrangements of a k-element subset of an n-set, $$\binom{n}{k} = \binom{n-1}{k-1} +\binom{n-1}{k}$$. {k! Using fractions and binomial coefficients in an expression is straightforward. (−)!. . Below is a construction of the first 11 rows of Pascal's triangle. 2019, by Nadir Soualem { n! } { 2 } is the that! Second pair is the one that displays the fraction as if it were in mathematical,! Triangle can be nested to obtain more complex expressions fleqn, left, right ; how typeset! Top of another combination ( n r ) is called a binomial coefficient using the symbol. On top of another top of another are common mathematical elements with similar characteristics - number! Further and say  q-binomial coefficient '' over  q-binomial coefficient '' clear there! Will print the binomial coefficient raising a binomial to any whole number exponent in which items! 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## YumYum247 one year ago how do i plot a velocity time graph from an average velocity of an object?

1. YumYum247

if a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....

2. anonymous

You will plot velocity on the vertical (y) axis and the time on the horizontal (x) axis. All you have to do is plot the values and connect all of them by a line. Make sure velocity is given in meters per second (m/s) and time in seconds (s) unless instructed otherwise.

3. YumYum247

can you please show me how do i make the data table so i can plot the points neatly...

4. YumYum247

|dw:1438454610769:dw|

5. YumYum247

how do i use the average velocity given in the question to plot the points on the graph.

6. anonymous

Velocity (in m/s) is [distance in metres] divided by [time in seconds.]

7. YumYum247

so v = d/t

8. YumYum247

but how's that gunna help me?

9. YumYum247

if i want to know how far did the runner get in 7 sec, how do i figure that out???

10. YumYum247

@oldrin.bataku

11. anonymous

if the acceleration is constant $$a$$ then the total change in velocity from the start until time $$t$$ is just $$v-v_0=at$$; in our case, we starta t rest so $$v_0=0$$

12. YumYum247

yes the accelaeration is constant.

13. anonymous

so the velocity is given by $$v=at$$, so in terms of a velocity vs time plot we see that the graph is a simple line with slope $$a$$ passing through the origin

14. YumYum247

i just need to plot how much distance he covered from rest to 20sec.

15. anonymous

well, if we have a velocity vs time plot, then the distance covered between $$t=t_1$$ and $$t=t_2$$ is just the total area bounded by our velocity graph between those points:

16. anonymous

|dw:1438456604986:dw|

17. YumYum247

can you please just do one for me, i'll do the rest myself.... can you figure out how much distance the runner covered in let's say 7 seconds????

18. anonymous

|dw:1438456642313:dw|

19. anonymous

so the distance they go between $$t_1,t_2$$ is just the area under the graph between $$t_1,t_2$$.

20. YumYum247

the area under the triangle is the displacement of the object....i get that...but

21. YumYum247

did you use Tf - Ti = a X t ?

22. YumYum247

because i need to make a data table to record all the points until 20 second and plot them on the velocity time graph.

23. anonymous

okay, so here we have $$t_1=0$$, $$t_2=7$$, and our equation is again $$v(t)=0.5t$$. so our total displacement is: $$A=\frac12 bh$$here our base is $$t_2=7$$ since $$t_1=0$$ and our height is $$v(t_2)=v(7)=3.5$$ so: $$A=\frac12(7)(3.5)=12.25$$

24. anonymous

you could also have done that work using an integral instead of manually finding teh area of a triangle

25. YumYum247

it doesn't look right because i have the answer key and the maximum displacement the runner makes under 20 sec is 10 meters.... i'm willing to show you the answer tho but i don't know how they got those points.....in the first place.

26. anonymous

i dont know what points youre talking about

27. YumYum247

|dw:1438457170341:dw|

28. YumYum247

this is how it looks......

29. anonymous

the kinematic equation gives $$x=x_0+v_0 t+\frac12 at^2$$, here $$a=0.5,t=7$$ and $$x_0=v_0=0$$ so $$x=\frac12\cdot0.5\cdot7^2=12.25$$

30. anonymous

actually, no it doesn't; you've misread it. that's the *velocity* against time, so it's going 10 meters *per second* in velocity after 20 seconds of time

31. YumYum247

i'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec

32. anonymous

yes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.

33. YumYum247

ok

34. anonymous

his speed after 7 seconds is only 3.5 m/s as I stated before

35. anonymous

with an average speed of $$(3.5-0)/2=1.75$$ m/s, which is indeed correct: $$\frac{12.25}7=1.75$$ m/s

36. YumYum247

ok what formula did you use plase?????? :(

37. YumYum247

ok so i need to make a data table of this situation as drawn above, so what equation do i use to plot the speed of the runner every second????

38. anonymous

since the velocity vs time graph is a line, the average velocity between $$t_1=0,t_2=7$$ s is just the midpoint, hence $$(3.5-0)/2=1.75$$ m/s. this matches the alternative calculation of the average velocity as the total displacement $$12.25$$ m divided by the time elapsed $$t_2-t_1=7$$ s, since $$12.25/7=1.75$$ m/s as well

39. anonymous

the equation for the velocity at time $$t$$ is just $$v=0.5t$$, since the acceleration is constant. this is the straight line we've been graphing

40. YumYum247

ok let me try on my own and i'll ask you to check my work. Thank you!!! :")

41. YumYum247

omg i see how you got 10m....LOL

42. YumYum247

ok one more thing, if the average velocity of an object is 0.5m/sec2 how quick or how fast will it accelerate every second.


# If $\gcd(a, b) = 1$ and if $ab = x^2$, prove that $a, b$ must also be perfect squares; where $a,b,x$ are in the set of natural numbers

Problem:

If $\gcd(a, b) = 1$ and If $ab = x^2$ ,prove that $a$, $b$ must also be perfect squares; where $a$,$b$,$x$ are in the set of natural numbers

I've come to the conclusion that $a \ne b$ and $a \ne x$ and $b \ne x$ but I guess that won't really help me.. I understand that if the $\gcd$ between two numbers if $1$ then they obviously have no common divisors but where do I go from this point?

Any tips at tackling this would be great. It looks quite easy though I'm still trying to get my hand around these proofs! Any pointers in the right direction would be great.

• You just wrote "$b\ne b$." I really hope that's not what you meant. – apnorton Feb 14 '14 at 3:51
• I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. – user61527 Feb 14 '14 at 3:52
• Hi Tyler! I really appreciate the heads up. I'm working away at an assignment right now. I've gotten most of the problems down but I still have only a few left. Thanks so much again! – A A Feb 14 '14 at 4:08

Fundamental theorem of arithmetic says that every number has a unique prime factorization.

If gcd(a,b) = 1, then all of these factors are unique (no prime factor is shared between a and b). What does this say about $x^2$?

Hint 2:

Let $a_i$ be a prime factor of a and $b_i$ be a factor of b. Then,

$$ab = \prod {a_{i}^{m_i}}\prod {b_{i}^{n_i}}$$

But $x$ has to have a unique factorization in the form,

$$x = \prod {x_{i}^{e_i}}$$, where $m, n, e$ are integer exponents. Keep in mind it is unique and we can order these factors in any way we please. What does this say about $x^2$ compared to $ab$?

• Hmm, x^2 must then: 1) have unique prime factors – A A Feb 14 '14 at 4:14
• I'll edit my post with another hint. – Chantry Cargill Feb 14 '14 at 4:17
• I think I may have it: x^2 = (p1^ip2^i2...pn^i2)^2 = (p1^2i2*p2^2i3*...pn^2in) = ab a and b must be factors of x^2, and all the factors of x^2 are squares, therfore a and b must be squares. Hmm, Does this sound correct to you? – A A Feb 14 '14 at 4:54
• Yes, more or less! Just make sure that it's clear that you can order the factors of $x^2$ such that the primes match up with a and b. Then the root will just be ab. – Chantry Cargill Feb 14 '14 at 5:11

Below is an approach employing universal gcd laws (associative, commutative, distributive), some of which you may need to (simply) prove before you can use this method. But once you do so, you will gain great power. Below we explicitly show that $\rm\:a,b\:$ are squares by taking gcds. Namely

Lemma $\rm\ \ \color{#0a0}{(a,b,c) = 1},\,\ \color{#c00}{c^2 = ab}\ \Rightarrow\ a = (a,c)^2,\,\ b = (b,c)^2\$ for $\rm\:a,b,c\in \mathbb N$

Proof $\rm\ \ (c,b)^2 = (\color{#c00}{c^2},b^2,bc) = (\color{#c00}{ab},b^2,bc) = b\color{#0a0}{(a,b,c)} = b.\$ Similarly for $\,\rm(c,a)^2.\ \$ QED

Yours is the special case $\rm\:(a,b) = 1\ (\Rightarrow\ (a,b,c) = 1)$.

Generally $\rm\: \color{#c00}{ab = cd}\: \Rightarrow\: (a,c)(a,d) = (aa,\color{#c00}{cd},ac,ad) = a\: (a,\color{#c00}b,c,d) = a\:$ if $\rm\:(a,b,c,d) = 1.\:$ For more on this and closely related topics such as Euler's four number theorem (Vierzahlensatz), Riesz interpolation, or Schreier refinement see this post and this post.

Compare the following Bezout-based proof (this is a simplified form of the proof in Rob's answer). For comparison, I append an ideal-theoretic version of the proof of the more complex direction.

Note that $\ 1=\overbrace{a{\rm u}+b\,{\rm v}}^{\large (a,b)}\,\ \overset{\large \times\,a}\Rightarrow\ a = \color{#c00}{a^2}{\rm u}+\!\!\overbrace{ab}^{\Large\ \ \color{#c00}{c^2}}{\rm v} \ \,$ so $\,\ d=(a,c)\mid a,c\,\Rightarrow\, d^2\!\mid \color{#c00}{a^2,c^2}\,\Rightarrow\, d^2\!\mid a$

Conversely $\ d = (a,c)= au+cv\,\ \Rightarrow\,\ d^2=\,\color{#c00}{a^2}u^2+2\color{#c00}acuv+\color{#c00}{c^2}v^2\ \$ thus $\ \ \color{#c00}{a\mid c^2}\ \Rightarrow\,\ \color{#c00}a\mid d^2$

$\quad\ \ {\rm i.e.}\quad (d)= (a,c)\ \ \Rightarrow\ \ (d^2) \subseteq\, (a,c^2)\,\subseteq\, (a)\ \$ by $\ \ a\mid c^2\,\ \$ [simpler ideal form of prior]

Notice how the ideal version eliminates the obfuscatory Bezout coefficients $\,u,v$.

• Hi Bill! Thank you for the response! Your way ahead of me, I'm not too familiar with the notation used/ – A A Feb 14 '14 at 4:53
• Bezouted.$\ \$ – robjohn Feb 14 '14 at 10:16
• @AA I use the standard notation $\ (a,b,\,\ldots)\, =\, \gcd(a,b,\,\ldots).\,$ I added another Bezout-based proof. $\$ – Bill Dubuque Feb 15 '14 at 4:50

Here is a proof using Bezout's Identity.

Let $x^2=ab$ and $\gcd(a,b)=1$, where $a,b\gt0$.

There are $u,v$ so that $$au+bv=1\tag{1}$$ Let $s_a=\gcd(x,a)$. Rewriting $(1)$, we have \begin{align} s_a\left(\dfrac{a}{s_a}\right)u+bv=1 &\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2=1-b(2v-bv^2)\tag{2}\\ &\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2a+ab(2v-bv^2)=a\tag{3}\\ &\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2a+s_a^2\left(\dfrac{x}{s_a}\right)^2(2v-bv^2)=a\tag{4}\\[8pt] &\implies s_a^2\mid a\tag{5} \end{align} Justification:
$(2)$: Move $bv$ to the right side and square
$(3)$: Move $b(2v-bv^2)$ to the left side and multiply by $a$
$(4)$: $x^2=ab$
$(5)$: $s_a^2$ divides each term on the left side

There are $u_a,v_a$ so that \begin{align} xu_a+av_a=s_a &\implies\frac{x}{s_a}u_a+\frac{a}{s_a}v_a=1\tag{6}\\ &\implies\frac{x^2}{s_a^2}u_a^2=1-\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)\tag{7}\\ &\implies\frac{ab}{s_a^2}u_a^2+\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)=1\tag{8}\\ &\implies abu_a^2+as_a\left(2v_a-\frac{a}{s_a}v_a^2\right)=s_a^2\tag{9}\\[7pt] &\implies a\mid s_a^2\tag{10} \end{align} Justification:
$\ \:(6)$: Divide by $s_a$
$\ \:(7)$: Move $\frac{a}{s_a}v_a$ to the right side and square
$\ \:(8)$: Move $\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)$ to the left side, $x^2=ab$
$\ \:(9)$: Multiply by $s_a^2$
$(10)$: $a$ divides each term on the left side

Combining $(5)$ and $(10)$ yields $a=\gcd(x,a)^2$. Symmetry yields, $b=\gcd(x,b)^2$.

• I simplified it a bit - see my answer. In $\,(2\!-\!5)\,$ it suffices to use the Bezout identity itself (vs. its square). In $(6\!-\!10)$ it's simpler to immediately square the Bezout identity (vs. rearrange it first). – Bill Dubuque Feb 15 '14 at 4:47


Isometry group of a norm is always contained in some Isometry group of an inner product?

Does there always exist some inner product $\<,\>$ on $V$ such that $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ ?

Update:

As pointed by Qiaochu Yuan the answer is positive.

This raises the question of uniqueness of the inner product $\<,\>$ which satisfies $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$.

Is it unique (up to scalar multiple)?

Remarks:

1) Determining $\<,\>$ (up to scalar multiple) is equivalent to determining $\text{ISO}(\<,\>)$.

Clearly if we know the inner product we know all its isometries. The other direction follows as a corollary from an argument given here which shows which inner products are preserved by a given automorphism.

2) Since there are "rigid" norms (whose only isometries are $\pm Id$ ) the uniqueness certainly doesn't hold in general.

One could hope for that in the case of "rich enough norms" (norms with many isometries, see this question) the subset $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ will be large enough to determine $\text{ISO}(\<,\>)$.

(which by remark 1) determines $(\<,\>)$).

Yes. This is because an isometry group is always compact (with respect to the topology on $\text{End}(V)$ induced by the operator norm: this is a consequence of the Heine-Borel theorem). Hence you can average an inner product over it with respect to Haar measure.

• @Asaf: that is not the suggestion. The suggestion is the following: pick an arbitrary inner product $\langle v, w \rangle$. Construct the new inner product $\langle v, w \rangle_G = \int_G \langle gv, gw \rangle \, dg$. (There is some work necessary to prove that this is still an inner product but it's not so bad.) By construction, this new inner product is $G$-invariant, so every $g \in G$ is an isometry with respect to the corresponding norm. This is a standard maneuver in representation theory called "Weyl's unitary trick"; you can try to google that term for references. – Qiaochu Yuan Jun 14 '15 at 19:38

For completeness, I am writing more detailes of the solution suggested by Qiaochu:

Denote by $G$ the isometry group of $(V,\|\cdot\|)$. $G\subseteq \text{End}(V)$. On $\text{End}(V)$ we have the operator norm $\| \|_{op}$ (w.r.t the given norm $\|\cdot\|$), which induces a topology on $\text{End}(V)$.

Lemma 1: $G$ is compact in $\text{End}(V)$

Proof: $\text{End}(V)$ is a finite dimensional normed space, hence it is linearly homeomorphic to $\mathbb{R}^n$ (This is in fact true for every finite dimensioanl real topological vector space). So, the Heine-Borel theorem aplies. (Every closed and bounded subset is compact).

$G$ is bounded since for every isometry $g\in G, \|g\|_{op}=1$, hence $G$ is contained in the unit sphere of $(\text{End}(V),\| \|_{op})$ .

$G$ is closed: Assume $g_n \rightarrow g,g_n\in G$. Fix some $v\in V$. $\|g_n(v)-g(v)\|_V\leq \|g_n-g\|_{op}\cdot\|v\|_V \xrightarrow{n\to\infty} 0$. So $g_n(v)\xrightarrow{n\to\infty}g(v)$. Now by the continuity of the norm $\| \cdot \|$ (w.r.t to the topology it induces on $V$) we get that: $\|v\| \stackrel{g_n isometry}{=} \|g_n(v)\|\xrightarrow{n\to\infty}\|g(v)\|$. This forces $g$ to be an isometry.

Corollary1: $G$ is locally compact Hausdorff topological group.

Proof: $\text{End}(V)$ is Hausdorff (Any metric space is...), and every subspace of Hausdorff is also Hausdorff. It is sdandard fact that $GL(V)$ is a topological group (t.g), and any subgroup of a t.g is a t.g.

Now, there exists a left-invariant measure $\mu$ on the Borel $\sigma$-algebra of $G$ such that $\mu(G)>0$. (This is the Haar measure which can be constructed on any locally compact Hausdorff topological group).

Now take any inner product $\<,\>$ on $V$. Fix $v,w \in V$. Define $f_{v,w}:G\rightarrow \mathbb{R},f_{v,w}(g)=\<gv,gw\>$.

Lemma 2: $f_{v,w}$ is continuous

Proof: Since $G$ is a metric space (a subspace of the normed space $\text{End}(V)$) it is enough to check sequential continuity. Take
$g_n \rightarrow g,g_n\in G$. We already showed this implies $g_n(v)\xrightarrow{n\to\infty}g(v)$ so by the continuity of the inner product $f_{v,w}(g_n)=\<g_nv,g_nw\> \xrightarrow{n\to\infty} f_{v,w}(g)$.

In particular $f_{v,w}$ is measurable, so we can integrate it. (Compactness of $G$ implies $f_{v,w}$ is bounded, and $G$ being a finite measure space guarantees the integral will be finite).

So we define: $\<v, w \>' = \int_G f_{v,w} \, d\mu = \int_G \< gv, gw \> \, d\mu$.

Now all is left is to show $\<, \>'$ is an inner product on $V$ that is presrved by each $h\in G$. (since this means $G=\text{ISO}(V,\|\cdot\|) \subseteq \text{ISO}(V,\<, \>')$ as required.

Lemma 3: $\<,\>'$ is an inner product.

The only non-trivial thing is positive-definiteness. (The rest follows from the linearity of $g\in G$ and the integral, and the bilinearity of $\<,\>$). But this follows from standard measure theory:

Fix $v\neq 0$. $f_{v,v} > 0$ on $G$ (since each $g\in G$ is injective and the original inner prodcut is positive). But this forces $\<v,v\>'>0$ as required.

Lemma 4: $\<, \>'$ is $G$-invariant. But this follows from another standard proposition in measure theory: (See "Real Analaysis" by H.L.Royden) chapter 22 pg 488 proposition 10).

• Still not enough langles and rangles... – Qiaochu Yuan Jun 16 '15 at 3:19
• You are right... I thought I already fixed it but I was wrong. I hope now its OK. – Asaf Shachar Jun 16 '15 at 7:10
• I really like that you accepted Qiaochu's answer and then added the details in your own. Nice! – Joachim Jun 21 '15 at 21:00


# Solving $\left(1/3\right)^k n = 1$ for $k$

The goal is to show that $$\left(\frac{1}{3}\right)^kn=1 \Rightarrow k = \log_3 n\,.$$

So I started with $$\left(\frac{1}{3}\right)^kn=1 \Leftrightarrow \left(\frac{1}{3}\right)^k=\frac{1}{n}$$ in order to use the identity $$y=a^x \Leftrightarrow x=\log_a y$$, which then yields $$k=\log_{1/3} \frac{1}{n}$$ which using $$\log \frac{1}{x}=-\log a$$ can be written as $$k = -\log_{1/3} n\,.$$ But that is not what I wanted to show, as $$\log_3 n \neq -\log_{1/3} n$$.

I don't know where the mistake is.

Note that $$-\log_{1/3} n = \frac{\log_{1/3} n}{\log_{1/3}3} = \log_3 n$$

• Thank you! That was fast. I will accept your answer as soon as it will let me. – user500664 Feb 25 '19 at 15:45
• Sorry, I've got another question: the identity you used is $\log_b (a)=\frac{\log_c a}{\log_c b}$, right? In your answer, $c=\frac{1}{3}$, but couldn't it be any arbitrary number as $c$ bears no relation to either $a$ or $b$? – user500664 Feb 25 '19 at 15:55
• @ThomasFlinkow Yes, but keep in mind that $0<c \neq 1$ for the logarithms to be defined. – Haris Gušić Feb 25 '19 at 15:58

Alternatively,

$$\left( \frac{1}{3^k}\right)n=1$$

Multiplying $$3^k$$ on both sides, $$n=3^k$$

Hence $$k = \log_3 n$$

• Thank you very much. This is even more elegant. I'm afraid I already accepted an answer – user500664 Feb 25 '19 at 16:12
• Don't worry about reputations. I just see a question and just lift a few fingersl ;) The other solution teach you why $-\log_{\frac13} n = \log_3 n$. – Siong Thye Goh Feb 25 '19 at 16:16


# expected payoff of dice game (paying for every next roll)

I am thinking of a variation of the dice game, where

one has the option to throw a die unlimited number of times.The first throw is free and every next throw costs 1 dollar. One will earn the face value of the die and has the option to stop after each throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?

If I calculate the expected value as

for 2 rolls as  (1/6)(3+4+5+6)  +   (1/3) (3.5-1) = 3.83$for 3 rolls as (1/6)(3+4+5+6) + (1/3) (3.83-1) = 3.94$
for 4 rolls as  (1/6)(3+4+5+6) +   (1/3) (3.94-1) = 3.98$for 5 rolls as (1/6)(3+4+5+6) + (1/3) (3.98-1) = 3.99$

it asymptotically tends to 4

Is this approach correct?
Shoudn't the player go bankrupt after 7 rolls (negative expectation)?

• I don't follow your method here. Can you explain your reasoning? – Matthew Conroy Oct 27 '16 at 7:36
• @ Matthew Conroy let's take 2 rolls. The expectation of a single trow is 3.5, but it costs 1, therefore the expectation is 2.5. Thus, if one gets 1 or 2 in the first try then one keeps rolling because the expectation is higher (2.5). If 3,4,5, or 6 comes one stops and takes the payoff, so $$(\frac{1}{6} 3+\frac{1}{6} 4+\frac{1}{6} 5+\frac{1}{6} 6) + \frac{2}{6}(3.5-1) = 3.83$$ – Michal Oct 28 '16 at 2:41
• Sometimes they will lose money, but on average they will net $4$. See my answer below. (There is no "bankrupt" here, since you have not specified how much money the player has to begin with.) – Matthew Conroy Oct 29 '16 at 20:39

I'd do it this way.

Let $E_n$ be the expected net if we roll until we get $n$ or greater.

Suppose $n=6$. We roll once. If it's a $6$, we are done. If not, we are in exactly the situation we started in, except that our "first" roll isn't free: it cost $1$ dollar. Hence $$E_6 = \frac{1}{6}(6) + \frac{5}{6}(E_6-1).$$ Solving, we find $E_6=1$. (Alternatively: on average it takes $6$ rolls to get a $6$, and all but the first roll is free, so $E_6=6-5=1$.)

Similarly, \begin{align} E_5 &= \frac{2}{6}\left(\frac{5+6}{2} \right) + \frac{4}{6}(E_5-1) & \text{ so } &E_5=\frac{7}{2}=3.5 \\ E_4 &= \frac{3}{6} \left( \frac{4+5+6}{3} \right) + \frac{3}{6}(E_4-1) & \text{ so } &E_4=4 \\ E_3 &= \frac{4}{6} \left( \frac{3+4+5+6}{4} \right) + \frac{2}{6}(E_3-1) & \text{ so} &E_3=4 \\ E_2 &= \frac{5}{6} \left( \frac{2+3+4+5+6}{5} \right) +\frac{1}{6}(E_2-1) & \text{ so } &E_2=\frac{19}{5} = 3.8 \\ \end{align} and, of course, $E_1=3.5$.

So the best strategy appears to be "roll until $4$ or more" with an expected net of $4$. Sometimes you will roll, say, 10 times with this strategy and lose money, but the average net is $4$.

Here is the result of $10^6$ simulated plays with the $4$ strategy:

net/number of occurrences

-21 1

-20 0

-19 0

-18 0

-17 0

-16 1

-15 0

-14 1

-13 5

-12 6

-11 8

-10 19

-9 32

-8 80

-7 146

-6 293

-5 579

-4 1122

-3 2272

-2 4517

-1 9147

0 18256

1 36474

2 72912

3 145747

4 291215

5 250332

6 166835

with an average net of $4.000515$.

• @ Matthew Conroy I think you have a mistake in $E_2$ (in case of 2 you keep rolling, so 2 shouldn't be in the first bracket). The expectation keeps going up, asymptotically tending to 4 as n increases. on the other hand is it not at odds with common sense? any strategy with more than 7 rolls is loosing money, the higher n the bigger loss – Michal Oct 28 '16 at 10:58
• @ Matthew Conroy secondly, why in your calculation the options you decide to pay off are decreasing with every next roll? for the 2 roll case E(x) value of the second roll is 2.5 so you pay off straight away after first roll if 3,4,5,6 come. for the 3 roll case and the next ones the E(x) is below 4 so you keep rolling in case of 1,2,3 and pay off in case of 4,5,6, correct? – Michal Oct 28 '16 at 14:10
• My $E_2$ is the expected value if we roll until we get $2$ or greater. In this case, we stop when we get a $2$. – Matthew Conroy Oct 28 '16 at 22:13
• Regarding your claim "the higher n the bigger loss", consider: if I've rolled a 1 (for example), it is always to my advantage, on average, to roll again, regardless of how many rolls I've taken already. Cheers! – Matthew Conroy Oct 29 '16 at 4:44
• If you roll $1$ six times in a row, then you will lose money. However, on average, you will lose less money if you continue to roll for a higher number. The same goes for rolling $2$s. The $E_2$ strategy is not optimal, as the calculation above shows. It is still a strategy. If you still don't agree with these calculations, I recommend you write a simulation and try out these strategies yourself. Sometimes the $4$ strategy loses money, but on average the net is $4$ dollars. Cheers! – Matthew Conroy Oct 29 '16 at 20:05


# Different results in integrating both sides of $\sin{2x}=2\cos x\sin x$

I feel like there is something I am missing here. When integrating both sides of the trigonometric identity $\sin{2x}=2\cos x\sin x$ I get different results.

The left side of course results in $-\frac{1}{2}\cos{2x}+C$.

The right side I solve with u-substitution:

$u=\cos x$

$du=-\sin x dx$

$-2\int udu=-u^2+C=-\cos^2 x+C$

While writing this question I noticed another identity $\cos^2 x=\frac{1}{2}+\frac{1}{2}\cos 2x$. So apparently the $\frac{1}{2}$ falls out because of the $+C$ resulting from indefinite integration? This is still a little confusing to me.

You are correct to recall that $\cos^2x=\frac{1+\cos 2x }{2}$.

This is an indefinite integral. So, the constant term $\frac 12$ is not relevant. That is

$$\int \sin 2x \,dx=-\frac12\cos (2x)+C_1 \tag 1$$

and

\begin{align} \int \sin 2x \,dx&=-\cos^2 x+C_2\\\\ &=-\frac12\cos 2x+(-\frac12 +C_2)\\\\ &=-\frac12\cos 2x+C_3\tag 2 \end{align}

where we absorbed the constants $-\frac12+C_2$ into a new constant and called that new constant $C_3$. Inasmuch as the integration constant is arbitrary, $(1)$ and $(2)$ are equivalent statements.

• Thanks, it's clear now. I was thrown off by the cosines having a different power. – Rubenknex Sep 3 '15 at 20:04
• You're welcome. My pleasure. Although it can be confusing, you were on the right track. – Mark Viola Sep 3 '15 at 20:30

Integration process gives you results correct up to an arbitrary constant.

Both results are essentially the same.

$$- \frac{\cos 2 x}{2} + C_1 = - \frac { 2 \cos ^2 x -1}{2} + C_1 = - \cos ^2 x + C_2$$

That is why it is better to write $C_1, C_2$ for the integration constants.


# What is the remainder of $16!$ is divided by 19?

Can anyone share me the trick to solve this problem using congruence?

Thanks,
Chan

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If you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)

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I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks. –  Chan Feb 26 '11 at 6:17
@Chan: $\rm 17 \equiv -2,\ 18\equiv -1$ –  Bill Dubuque Feb 26 '11 at 6:21
@Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\equiv -1$ and $17\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) –  Jonas Meyer Feb 26 '11 at 6:21
Many thanks, I got it now ;) –  Chan Feb 26 '11 at 6:24

HINT $\$ By Wilson's theorem $\$ mod $19\::\ -1 \ \equiv\ 18\:!\ \equiv\ 18\cdot17\cdot 16\:!\ \equiv\ (-1)\ (-2)\cdot 16\:!$

Similarly we have the Wilson reflection formula $\rm\displaystyle\ (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p)\:,\$ $\rm\:p\:$ prime

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@Bill Dubuque: Thank you! –  Chan Feb 26 '11 at 6:24
@Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides? –  Chan Feb 26 '11 at 6:42
@Chan: $\rm\ 2\:x\equiv 2\:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19)$ –  Bill Dubuque Feb 26 '11 at 6:50
@Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right? –  Chan Feb 26 '11 at 7:07
@Chan: $\rm\ n,m\$ coprime $\rm\iff a\ n + b\ m = 1\$ for some $\rm\:a,b\:\ \iff\ a\ n\equiv 1\ (mod\ m)\:,\:$ for some $\rm\:a\:$ $\rm\iff\ n$ is invertible/cancellable $\rm\:(mod\ m)$ –  Bill Dubuque Feb 26 '11 at 7:12

I almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:

$$10!= (2\cdot 10)\cdot (4\cdot5)\cdot(3\cdot6)\cdot(7\cdot8)\cdot 9\equiv 1\cdot 1\cdot (-1)\cdot(-1)\cdot 9\equiv9$$

Now we have

$$16!\equiv 9\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\equiv9\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4)\cdot(-3)$$

$$\equiv 9\cdot(4\cdot5)\cdot(6\cdot3)\cdot(7\cdot8)\equiv 9\cdot(1)\cdot(-1)\cdot(-1)\equiv 9$$

Of course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)

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1. Compound Interest Q Help!

Can you guys help me with this question? I';m so confused on how I should approach it:
Tom’s father paid $500 into an account on the day Tom was born. ?After that, he paid$500 into the account on Tom’s bday until Tom’s 18th bday. ?If the account accrued interest at 8%p.a compounded monthly, calculate how much Tom would receive on his 18th bday.

A:
20880.97

2. Just think it through, one payment at a time.

From the start, I'm not absolutely certain there was a payment ON the 18th birthday. let's assume that there was. If we get $500 too much, we'll discard this payment. Starting from the last payment and working backwards, we have: 18th:$500 and no accumulation for interest
17th: $500(1+i) -- 1 year's accumulation for interest 16th:$500(1+i)^2 -- 2 year's accumulation for interest
15th: $500(1+i)^3 -- 3 year's accumulation for interest ... Birth:$500(1+i)^18 -- 18 year's accumulation for interest

With any luck, one should notice this is a Geometric Sequence and we should be able to add them all up.

$500 +$500(1+i) + $500(1+i)^2 +$500(1+i)^3 + ... + $500(1+i)^18 =$500(1 + (1+i) + (1+i)^2 + (1+i)^3 + ... + (1+i)^18) =

$500\frac{(1+i)^{19}-1}{i}$

Our only remaining concern is 'i'. What is it? The formula above uses 'i' as an annual effective interest rate. We need to find one of those. We are given 8% Nominal Interest and Monthly Compounding. This gives:

$\left(1 + \frac{0.08}{12}\right)^{12} - 1 = 0.082999511 = i$

Thus,

$500\frac{(1+i)^{19}-1}{i}\;=\;21,380.97$

As can be seen, $21,380.97 -$500.00 = \$20,880.97, so I guess there was not a payment made ON the 18th Birthday Anniversary. This leaves us with a bit of a dilemma. On a written exam or a homework assignment, I would state my assumptions and provide both answers, citing the ambiguity of the word "until". Anything marked wrong would get a vigorous challenge. On a multiple-choice exam, I would be prepared to find either answer. In my view, if both appear on the multiple-choice exam, the question probably should be discarded as accepting either answer will not necessarily provide any information about a student's knowledge. One may simply have done it badly. In any case, questions should be clear. If you have discussed the word "until" in class, and it has been defined clearly to mean "NOT on the end date", then you can be expected to get the unique value. There may also be a diagram explaining the intent. It is a very hard thing to write perfectly clear questions. It is up to the student to explain any point of ambiguity. The exam writer cannot be expected to think of every possible translation, but I'm sure the exam writer tries to do that.

Well, enough of exam philosophy...


# What's the name of this algebraic property?

I'm looking for a name of a property of which I have a few examples:

$(1) \quad\color{green}{\text{even number}}+\color{red}{\text{odd number}}=\color{red}{\text{odd number}}$

$(2) \quad \color{green}{\text{rational number}}+\color{red}{\text{irrational number}}=\color{red}{\text{irrational number}}$

$(3) \quad\color{green}{\text{algebraic number}}+\color{red}{\text{transcendental number}}=\color{red}{\text{transcendental number}}$

$(4) \quad\color{green}{\text{real number}}+\color{red}{\text{non-real number}}=\color{red}{\text{non-real number}}$

If I were to generalise, this, I'd say that if we partition a set $X$ into two subsets $S$ and $S^c=X\setminus S$, then the sum of a member of $S$ and a member of $S^c$ is always in either $S^c$ or $S$.

My question is:

"Is there a name for this property (in these four cases) and is this property true in general?"

Also, does anyone have any more examples of this property?

-
Note that you're saying more than just set here: You're requiring that the objects of that set can be added. –  Semiclassical Jul 25 at 16:50
@Semiclassical Yeah, thanks! What word should I use instead of set, then? –  alexqwx Jul 25 at 16:51
This can be written as: $S$ is closed under subtraction. That is, if $a,b\in S$ and $a-b$ exists, then $a-b\in S$. –  Thomas Andrews Jul 25 at 16:52
@Semiclassical I don't see how this contradicts $(2).$ –  alexqwx Jul 25 at 16:54
@Semiclassical No, you misread my comment. The rationals are closed under subtraction. That is what the above says... –  Thomas Andrews Jul 25 at 16:55

I think this comes from the fact that if you have a group $G$ and $H$ a subgroup of $G$ then if $h\in H$ and $x\not\in H$ we get $xh\not\in H$.

The proof is by contradiction, suppose $xh=l$ with $l\in H$. Then postmultiplying by $h^{-1}$ gives $x=lh^{-1}$ which is in $H$ since $H$ is a subgroup of $G$.

-
Why have you used $xh$ instead of $x+h$ in your generalisation (does it matter?)? –  alexqwx Jul 25 at 16:58
$xh$ just means the group operator acting on the pair $(x,h)$ if the operation in the particular group is addition then $xh$ is just $x+h$. the plus sign is normally used for abelian groups. But this result generalizes to any group. –  Jorge Fernández Jul 25 at 17:03
Got it. Thanks! One final thing: why did you mention post-multiplying rather than just multiplying (since addition is commutative)? –  alexqwx Jul 25 at 17:05
well, in general groups don't need to be commutative, however this assumption is not required for the result to be true. –  Jorge Fernández Jul 25 at 17:22
Talking about this in terms of groups seems just a bit too strict. The first example given in the question works just as well, for example, if we were to only talk about positive even integers (no inverse or zero element under addition.) –  Semiclassical Jul 25 at 17:51

I call this the complementary subgroup law, because the composition law arises via the following complementary view of the Subgroup Test ($\rm\color{#c00}{ST}$),  cf. below from one of my old sci.math posts.

Theorem  Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$ Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G.$

Proof $\$ $\rm\,G\,$ is a subgroup of $\rm\,H\!\overset{\ \large \color{#c00}{\rm ST}}\iff\! G\,$ is closed under subtraction, so, complementing

$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$

Instances of this law are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here (and also on sci.math).

-
Nice! ByteErrant, in the answers below, has shown that prime +non-prime can be either prime or non-prime itself, which seems to contradict the above theorem. Could you explain? –  alexqwx Jul 25 at 21:36
@alex The set $\,\color{#0a0}G\,$ of $\rm\color{#0a0}{primes}$ is not a subgroup of $\,H =$ integers under addition, since $\,\color{#0a0}G\,$ is not closed under subtraction, e.g. $\,\color{#0e0}{11}-\color{#0a0}7 = \color{#c00}4\,$ is not $\rm\color{#0a0}{prime,\,}$ so $\, \color{#0a0}7+\color{#c00}4 = \color{#0e0}{11}\,\Rightarrow\, \color{#0a0}G+\color{#c00}{\bar G}\subseteq \color{#c00}{\bar G}\,$ fails. That agrees with the (negation of) the theorem, i.e. $\,G+\bar G\subseteq G\,$ fails $\iff$ $\,G\,$ is not a subgroup of $\,H\,$ $\iff$ $\,G\,$ is not closed under subtraction. –  Bill Dubuque Jul 25 at 23:21

This is basically just "the complement" of the statement that the sets of numbers in green are subgroups of $(\Bbb R,+)$.

Given that for a subgroup $S<\Bbb R$ $a,b\in S$ implies $a-b\in S$, you can quickly compute that if $c\notin S$, $a+c=b\in S$ implies $a-b=c\in S$, a contradiction. Therefore $a+c\notin S$ for any $a\in S$, $c\notin S$.

You could also add to your list anything like "an integer plus a noninteger is a noninteger" and "for any subring $S\subseteq \Bbb R$, an element in $S$ plus an element outside of $S$ results in an element outside of $S$." Again, you don't even really need a subring: this is true for any proper subgroup.

-

Don't prime numbers offer a counter-example to the general truth of this property?

Prime $+$ not-prime $=$ not-prime $==> 17 + 4 = 21$

Prime $+$ not-prime $=$ prime $==> 7+4=11$

-
Ah. I hadn't thought about this. Thanks! Do you (or does anyone else) have any explanation for this? –  alexqwx Jul 25 at 21:34
@alexqwx Primes don't form a group under addition. Evens do, rationals do, algebraics do, and reals do. –  Cory Jul 25 at 21:41

I would say that the behaviour of this property under addition is isomorphic to addition in $\mathbb{Z}_{2}$. E.g.: If you assign "even" to 0, "odd" to 1, then even + odd = odd is expressed in 1 + 0 = 1.

-


# Calculus Made Easy (Free book)

By Murray Bourne, 25 Apr 2009

OK, it looks old and dusty, but Calculus Made Easy [PDF] is an excellent book and I strongly recommend it to those of you who are struggling with calculus concepts. It's also great for teachers, to give you ideas on how to explain calculus so it doesn't confuse the hell out of everyone. He quite rightly points out that many math text book writers are more interested in impressing the reader with sophisticated calculus techniques than explaining the basic concepts.

One of the early pages has:

BEING A VERY-SIMPLEST INTRODUCTION TO
THOSE BEAUTIFUL METHODS OF RECKONING
WHICH ARE GENERALLY CALLED BY THE
TERRIFYING NAMES OF THE
DIFFERENTIAL CALCULUS
AND THE
INTEGRAL CALCULUS.
BY
SILVANUS P. THOMPSON

In other words, this was one of the first ever "Calculus for Dummies" books. Thompson puts great effort into explaining what is going on, rather than jumping straight into the calculations. He humbly calls himself a "fool", but doesn't treat the reader as one.

He quotes from an "ancient Simian proverb":

"What one fool can do another can."

To give you an idea of how the book is written, in Chapter 1, "To Deliver You From the Preliminary Terrors", we read:

∫ which is merely a long S, and may be called (if you like) "the sum of." Thus ∫dx means the sum of all the little bits of x; or ∫dt means the sum of all the little bits of t. Ordinary mathematicians call this symbol "the integral of".

Now any fool can see that if x is considered as made up of a lot of little bits, each of which is called dx, if you add them all up together you get the sum of all the dx's, (which is the same thing as the whole of x). The word "integral" simply means "the whole".

The book is now copyright free. Grab the PDF: Calculus Made Easy.

[Thanks to Denise at LetsPlayMath for the link.]

1. solarhene says:

thanks so much for sharing this excellent book!

best regards,
an old fool

2. Murray says:

You're welcome Solarhene. I'm glad that you find it useful.

3. ayuk federick besong says:

i believe it can work and it will work

4. Noor says:

I'm so interested in re-learning calculus that I watched many different video tutorials, but I found this book summarizing Integral Calculus in its first 2 pages!

Thank you so much!

5. Sarah says:

Thanks so much for posting this. I'm so glad I clicked on your link while viewing my friend Dana's blog. My 9 year old read it and was very excited that it made so much sense. He's currently taking Physics and some of the math problems were made to be so confusing so this will help him so much.

6. lawal says:

Good DAY SIR,i want to know how i can get this text book,calculus made easy.i am mailing from Lagos Nigeria.thank u sir.

7. Murray says:

Hi Lawal

This is an e-book in PDF form (not a physical book). The link is in the article above, in the first line.

You could print it out from the PDF if you would rather hard copy.

8. D. Pal, India. says:

Its a brilliant book. Thanks for sharing the book

9. B Atchley says:

Thanks for sharing this book! It will be a great tool for my independent study students!

10. Winstone Kapanje says:

Thanks for great job well done. Do the same with other topics such as trigonometry and complex numbers

11. Judith Kanyama Chikoti says:

the way you have started culculus is very interesting but how could one get the whole PDF paper?

12. Murray says:

Hi Judith. The link to the PDF is in the article above! Do you mean that you want a hard copy? You would need to print it yourself.

13. jared okoth says:

if every mathematics topic could be introduced the way this has, then i sure no one will think of maths as the hardest of courses.

14. eduardo says:

make me realize the math is easy

15. Kaset One says:

one of the greatest books.. many thanks for sharing with us

16. Rod says:

I was contemplating purchasing a course on dvd called "Calculus Made Clear" selling for \$60.00 US. Then I decided to google "Calculus Made Easy" and found your link to this great book. I read the intro and was amazed at how easy it was to follow. I will continue to read the whole text! by the way, I was searching for books on calculus for my son so now we can read it together. Thanks!

17. Murray says:

You're welcome, Rod. Glad you found it useful.

18. fatemeh says:

i from iran and i don't english.
ineed a study book reference and i have n't book english and i request from you that a link book free download for laplas, integral ,...
thank you very much

19. Murray says:

@Fatemeh: I'll look for such materials and post it if I find any good ones.

20. Piracha, J. L. says:

Thanks

thanks! and great work which nobody cannot be appreciated for your effort for sharing such good book.

22. Dalcde says:

http://www.gutenberg.org/ebooks/33283
which is typed in LaTeX and the file size is reduced tenfold.

23. Murray says:

@Dalcde: Thanks for the tip!

24. Nayyir says:

if u r one of those looking to understand calculas at an elementary level,then u've got it here.Calculas Made Easy is indeed very helpful a book for those struggling to understand it as a concept.so do go for it , it is certainly one of the best materials available. go for it.have a nice maths day.

25. Ian Thomson says:

Hi :
Thanks for the pdf on Calculus Made Easy.
I have always been curious and terrified at the same time of calculus.

Chapter One says it all. Getting past the fancy notation, helps a huge amount.

Kind Regards Ian Thomson

26. percival says:

Hi! Ian,My name is percy and I teach Maths in grade 12. Please foreward me the calculus doc as I also struggle on calculus section.

27. Murray says:

@Percival: The link to the PDF is at the top (and again at the bottom) of the article.

28. Joe Kenyon says:

Dear Murray
I am surprised and delighted to see my old Dear Friend Sylvanus P. Thompson here. As a very discouraged high school dropout I found an original hardback version at the rubbish tip site where I used to spend my spare time. I picked it up and the proverb just grabbed me. I read the book it and loved it. I went back to school and excelled at the age of 25. Now after 40 years in senior positions in Telecommunications in Australia my copy is very much treasured and repaired and again being used as I transition into becoming a teacher myself! The bit that initially grabbed me was "What one fool can do another can" I ask that you put that wonderful saying up on this site. It was so powerfuly encouraging it got me off the bad path I was on and led to a wonderful career and a wonderful wife and family. Imagine, it hardly seems possible that an ancient seven word proverb, repeated in one book by a man long dead could do so much good.
I love this guy and would have loved to have met him.

Kind Regards
Joe Kenyon

29. Murray says:

Thanks for your inspiring story, Joe!

I have included the quote in the article.

30. sachin sharma says:

Hi,
I have been searching such work for past couple of months and at last I got here in the form of "Calculus Made Easy" May GOD Bless all who have made such efforts in preparing such an excellent book.

Thanks to all.....

Sachin Sharma
India

31. drcobol2000 says:

Murray,

Thanks for posting this...and for free! While I struggled with math from grade school to grad school, I'm now a bit of a math junkie and books with subjects like this really pique my interest. I wish I could get my hands on an actual copy of that book for myself.

One question comes to mind, though. I noticed there are two different dates in the front of the book that confuse me (I'm a sucker for old books). There is a date of 1914 in the preface and another of 1943 by itself on another page. Do you know which date was the actual date of publication? There isn't an actual copyright year specified and as I read through the text, the absence of that date combined with the style and choice of words lead me to believe this was a work from the early 20th century. I rely on a barely mediocre literary acumen to make that assessment, so take it with an enormous grain of salt.

Thanks and best wishes!

32. Murray says:

@drcobol2000 The publication date would be 1914, as that's when he wrote the Preface to this second edition. So your "early 20th century" language observation is right on the button.

I don't believe the "6-14-43" on the separate page is a date, since in the US, the date order is normally day then month. I suspect it's there to trigger thought.

33. kevin says:

Thanks you so much sir for this ebook

I am an engineering student, and i am struggling with calculus.

This book helped me a lot even though this book is quite old.

34. Pyae Phyo Aung says:

Thank you really really much sir. You help me a lot.

35. punith says:

thank u so much its very helpful

36. rahillah says:

hi Murry, thanks for sharing the brilliant book,hope you have a lovley day!

37. Murray says:

Glad you found it useful, Rahillah.

38. Nae says:

I'm not sure if it'll work but I'll give it a try. I'll write back at the end of the semester to let you know how it worked for me.

Thanks

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# How do we define the domain of a function with removable and nonremovable discontinuities?

For instance consider the function

$$f(x) = \frac{x-2}{(x-2)(x+2)}$$

I don't understand why some discontinuities are considered "removable". Yes, we can cancel out the $x-2$'s but isn't this invalid at $x=2$ because it's like canceling out $\frac{0}{0}$? I don't understand why we're allowed to do this.

Anyway if we "remove" that piece we get:

$$f(x) = \frac{1}{x+2}$$

Which has a discontinuity at $x=-2$ but we can't "remove" that.

So we end up with two discontinuities -- do we define the domain to include or exclude these? Is the domain all real numbers? All except $2$? All except $-2$? All except $-2, 2$?

• Because limit . – user202729 Jan 29 '18 at 14:39

The discontinuity at $x=2$ can be removed because $\lim_{x\to2}f(x)=1/4$. When 'removing' the discontinuity, we define a new function $f^\prime$ that takes on the values of $f$ at all points in the domain of $f$ and $f^\prime(2)=1/4$, thereby extending the domain.

The discontinuity at $x=-2$ cannot be removed because the limit there is improper: $\lim_{x\to-2}f(x)=\infty$.

• How do you know the limit at each point? Is there an analytic way? Is the only difference between removable and nonremovable discontinuities whether or not the limit is a number or not? – user525966 Jan 29 '18 at 14:57
• When the limit at a certain point is improper, it cannot be represented by a real number. Only proper limits can be used to remove discontinuities. – Jeroen van Riel Jan 29 '18 at 15:03
• What then would be the domain of $f$? All reals except $-2,2$? – user525966 Jan 29 '18 at 15:04
• Yes, as @57Jimmy already pointed out in his answer. – Jeroen van Riel Jan 29 '18 at 15:13

You clearly understand the algebra, and the idea behind "removable singularity". The definition of the domain is a little subtle (and usually not particularly important, given your understanding).

The expressions on the right in

$$f(x) = \frac{x-2}{(x-2)(x+2)}$$

and

$$g(x) = \frac{1}{x+2}$$

define the same function where both make sense. That function has an unremovable singularity at $x=-2$.

The domain of $g$ contains the point $x=2$; the domain of $f$ does not, so strictly speaking they are not the same function. But the limit of $f$ at $x=2$ does exist, and has value $g(2) = 1/4$. That's exactly what we mean when we say the singularity is removable.

• So a removable discontinuity is an undefined point that has a defined limit, whereas a nonremovable discontinuity has a limit that does not exist or is infinity? – user525966 Jan 29 '18 at 14:52
• @user525966 Yes. – Ethan Bolker Jan 29 '18 at 15:33

The domain of $f$ is $\mathbb{R} \backslash \{-2,2\}$, because at this two points $f$ is not defined. Having said that, some singularities are better behaved than others: $f$ goes to infinity as $x \to -2$, whereas $\underset{x \to 2}{\lim} f(x) = \frac{1}{4}$ is well-defined. This is exactly the difference between poles and removable singularities.


Cartesian Product - Proof

Yankel

Active member
Dear all,

I am trying to prove a simple thing, that if AxA = BxB then A=B.

The intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.

How do I prove it formally ?

Thank you !

Opalg

MHB Oldtimer
Staff member
Dear all,

I am trying to prove a simple thing, that if AxA = BxB then A=B.

The intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.

How do I prove it formally ?

Thank you !
This may not be as simple as you think. To start with, what do you mean by saying that two sets are equal? I think that the only way to make sense of that is to interpret "A=B" to mean that A and B have the same cardinality.

If a set $A$ is finite then its cardinality is just the number of elements it contains, denoted by $|A|$. If $|A| = m$ then $|A\times A| = m^2.$ So if $|B| = n$ and $|A\times A| = |B\times B|$ then $m^2 = n^2$, from which it follows that $m=n$. This proves that if "$A\times A = B\times B$" then "$A=B$" in the case of finite sets.

For infinite sets the situation is more complicated. There is a theorem of Zermelo that if $A$ is an infinite set then $|A\times A| = |A|$. From that it follows immediately that if $|A\times A| = |B\times B|$ then $|A| = |B|$. However, the proof of Zermelo's theorem requires the Axiom of Choice. In models of set theory that do not satisfy this axiom, it may be that your result does not hold.

HallsofIvy

Well-known member
MHB Math Helper
Opalg said "
I think that the only way to make sense of that is to interpret "A=B" to mean that A and B have the same
cardinality
."

I disagree. To say that sets A and B are equal means "$$x\in A$$ if and only if $$x\in B$$". If two sets are equal they have the same cardinality but the converse is not true. The sets A= {1, 2, 3} and B= {a, b, c} have the same cardinality but are not equal.

Opalg

MHB Oldtimer
Staff member
Opalg said "
I think that the only way to make sense of that is to interpret "A=B" to mean that A and B have the same
cardinality
."

I disagree. To say that sets A and B are equal means "$$x\in A$$ if and only if $$x\in B$$". If two sets are equal they have the same cardinality but the converse is not true. The sets A= {1, 2, 3} and B= {a, b, c} have the same cardinality but are not equal.
In that case, the result becomes trivially true. If $A\times A$ and $B\times B$ are just two different names for the same set, then the diagonal elements of $A\times A$ (those of the form $(a,a):a\in A$) are duplicates of the elements of $A$. The same holds for the diagonal elements of $B\times B$. If those diagonals are the same, it follows that the elements of $A$ are the same as the elements of $B$, so $A=B$.

HallsofIvy

Well-known member
MHB Math Helper
Yes, it is. Saying that "A= B", where A and B are sets, means that if x is in A then it is also in B and if y is in B then it is also in A.

If x is a member of A. then (x, x) is in AxA= BxB so x is in B. If y is a member of B then (y, y) is in BxB= AxA so y is in A. Therefore A= B.

It is trivial but that is the question asked.

Olinguito

Well-known member
I am trying to prove a simple thing, that if AxA = BxB then A=B.

The intuition is clear to me. If a pair (x,y) belongs to AxA it means that x is in A and y is in A. If a pair (x,y) belongs to BxB it means that x is in B and y is in B. If the sets of all pairs are equal, it means that every x in A is also in B and vice versa.

How do I prove it formally ?

Thank you !
Let $a\in A$. Then $(a,a)\in A\times A$. Since we’re assuming $A\times A=B\times B$, this means $(a,a)\in B\times B$ and thus $a\in B$. Therefore $A\subseteq B$. The same argument with $A$ and $B$ interchanged shows that $B\subseteq A$. Hence $A=B$.


# Pair in a poker, again

1. Oct 6, 2012

### Astr

Im New on this forum, so i hope this is the right place to ask this question. I've tried to solve the problem of finding how many hands are in a poker game with exactly one pair, this way:

I can choose the first card from 52 ways. for the second card I can choose it from 3 ways (to match the pair). For the third card I can choose it from 48 possibilities (for ensure I´ve got no trios in the hand). For the fourth and fifth i can choose them in 44 and 40 different ways, respectively. And because order is not important I must divide for 5! (ways of permuting 5 cards). My (wrong) answer is:

(52x3x48x44x40)/5!=109 824

This is a tenth of the correct answer. Please can you tell whats wrong in the procedure. (English isn't my first language, sorry if Ive got some mistakes)

2. Oct 7, 2012

### CWatters

That gives you too many because a QS + QD is the same pair as QD+SQ.

There are 6 ways to get a pair of (for example) Queens..

QS + QH
QS + QD
QS + QC
QH + QC
QH + QD
QD + QC

but there is a choice of 13 denomination (values) so for pair it's

6 * 13

then the other cards must not have the same value but can be of any suit.

3. Oct 7, 2012

### Astr

Thank you Cwatters. But I think i avoid the problem QS+QD=QD+QS, dividing by 5!, becasuse that ensures that the order of the 5 cards does not matter.

4. Oct 10, 2012

### CWatters

I can't immediatly see why your method is wrond but continuing from above..

Then the next three cards can have one of 12 other values (3 from 12 = 220) and be from any of 4 suits (4*4*4 = 64)

220 * 16 = 14080

Now that is also 48*44*40/3! so your last part seems ok?

5. Oct 10, 2012

### Astr

Thank you again, CWatters. Well I know how to solve the problem, and this is out of discussion:
For the pair:
52x3/2!= 78 ways
For the three different cards:
48x44x40/3!=14 080 ways
then: 78x14 080= 1098240 ways to get exactly one pair.
but my problem is: Whats wrong in the procedure i post first?

6. Oct 10, 2012

### Ray Vickson

You should not divide by 5!; instead, you should divide by 2! × 3!, because the order of the cards within the pair does not matter and the order within the three non-pairs does not matter. That would give the number of hands as 1098240, as you want.

There are some subtle points here, so let me expand. When counting the hands containing exactly one pair, the person posing the question is distinguishing between hands such as PPNNN, PNNNP, PNNPN, NNNPP, etc., even though the actual cards involved are exactly the same (and so all would be regarded as a *single* hand by a poker player). Basically the question is asking for the number of different ways you could be *dealt* a hand containing a single pair, even though after being dealt the cards you would re-arrange them to your own, personal liking.

Another way to see this is to look at P{pair}, the probability of getting dealt a pair. Suppose, first that we have 2 aces and one each of 2 3 4. The probability, p, of getting that is p = C(4,2)*C(4,1)^3/C(52,5), where C(a,b) = "a choose b" = a!/[b!(a-b)!]. This holds because there are C(4,2) ways to pick 2 aces from 4 and for each of the 2 3 and 4 there are C(4,1) ways to choose the particular card, and because there are C(52,5) says of being dealt a 5-card hand althogether. Now, of course, the pair may not be aces and the non-pairs may not be 2 3 and 4. However, for any choice of ranks, the probability of the pair and the three non-pairs are all the same as the p above. Therefore, P{pair} is p times the number of ways to choose the rank of the pair (13) times the number of ways of choosing the three different ranks from the remaining 12 (which is C(12,3)). Thus, we have
$$P\{ \text{pair}\} = \frac{13 \, C(12,3) \, C(4,2) \, C(4,1)^3}{C(52,5)} = \frac{N_p}{C(52,5)},$$
where Np = number of hands containing one pair. Thus, we have
$$N_p = 13 \, C(12,3) \, C(4,2) \, C(4,1)^3 = 1098240.$$

RGV

Last edited: Oct 10, 2012
7. Oct 10, 2012

### Astr

Ray, Thank you for your answer. Can you explain me why i should not divide by 5!, because this is my whole problem. Thanks.


# Solving $|a| < |b|$

I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.

I am looking for a way to eliminate absolute value functions in $|a| < |b|$.

I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.

Thank you!

An example problem:

$$|x + 2| < |x - 4|$$

-
perhaps you could give some specific problems that trouble you. – Will Jagy Aug 11 '12 at 5:19
Good suggestion. I've added one above. – Mark V. Aug 11 '12 at 5:41
It's useful to review the definition of the absolute value function: $$\text{abs}(a) = \begin{cases} a & 0 \leq a \\ -a & a < 0 \end{cases}$$ In order to eliminate one absolute value function from an expression, you will need to consider two cases: the case where its argument is non-negative, and the case where its argument is negative. This will give you an expression with two cases with one less absolute value function. Iterating this process will give an expression without the absolute value sign. – danportin Aug 11 '12 at 7:13
Since you mentioned trying Google, maybe you should try Wolfram Alpha instead – Ben Millwood Aug 11 '12 at 11:45

We have $|a| \lt |b|\,$ if any of these is true:

(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$

(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$

(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$

(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.

Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.

Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier.

Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality

$$x^2+4x+4 \lt x^2-8x+16.$$

The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.

But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$.

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Ha! I never would of thought of this. This is brilliant! Thank you so much for your help! – Mark V. Aug 11 '12 at 18:10

There are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:

1. If $x \in (-\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives: \begin{align} -(x+2) &< -(x-4) \\ x+2 &> x-4 \\ 2 &> -4 \end{align} This is true for all $x$ in the interval.

2. If $x \in [-2,4)$, then $x+2 \geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed: \begin{align} x+2 &< -(x-4) \\ 2x+2 &< 4 \\ x &< 1 \end{align} Combining this last inequality with the assumption that $x \in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.

3. Finally, if $x \in [4,\infty)$, neither expression's sign is reversed: \begin{align} x+2 &< x-4 \\ 2 &< -4 \end{align} This is false for all $x$ in the interval.

Putting all the information together from the above three cases, we have $x \in (-\infty, 1)$.

Note: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.

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Pedantry: you probably want $x\in[-2,4)$ for step 2 and $x\in[4,\infty)$ for step 3 (or change where you say $x-4<0$, but I think my way is more natural/consistent) – Ben Millwood Aug 11 '12 at 11:03
@BenMillwood: Fair enough! I've changed it. – Théophile Aug 11 '12 at 18:13

We can see $|x-a|$ as a distance point $x$ from point $a$.

Now, the question with the above "definition" would be like:

For which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?

Clearly, by drawing it maybe, you can observe that the answer is for all $x<1$.

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We can divide by $|b|$ to get $|a/b|<1$. Let $x=a/b$ then $|x|<1$ so $-1<x<1$.

Now multiply through by $b$.

If $b>0$ then $-b<a<b$.

If $b<0$ then $-b>a>b$

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# Math Help - A problem of Riemann Integrable

1. ## A problem of Riemann Integrable

If $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\geq0$, we have $\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?
(a.e. stands for almost everywhere)

2. Yes. Note that if $\int_{a} ^{b} \ x^nf(x)dx =0$ for all $n \in \mathbb{N}$ then $\int_{a} ^{b} \ p(x)f(x)dx =0$ for any polynomial $p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $I: C^0([a,b]) \rightarrow \mathbb{R}$ $g \mapsto \int_{a} ^{b} \ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\int_{a} ^{b} \ g(x)f(x)dx =0$ for all $g(x) \in C^0([a,b])$. It is a well known result that under this conditions $f \equiv 0$.

Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.

3. Originally Posted by Xingyuan
If $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\geq0$, we have $\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?
(a.e. stands for almost everywhere)

Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.

Tonio

4. Originally Posted by tonio
Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.

Tonio
No,In the problem,I am not assuming $f(x)\geq0$

5. Originally Posted by Xingyuan
No,In the problem,I am not assuming $f(x)\geq0$

I didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.

Tonio

6. Originally Posted by tonio
Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.

Tonio
Huh? That doesn't follow.

7. Originally Posted by tonio
I didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.

Tonio
Consider $\int_{-\pi}^{\pi}\sin(x)\,dx$

It's only $0$ at the two endpoints and $x=0$, but the whole integral is $0$.

Or, for that matter, consider $\int_{-a}^a f(x)\,dx$ where $f(x)$ is any odd function.

8. Originally Posted by rn443
Huh? That doesn't follow.

Yes it does

Tonio

9. Originally Posted by tonio
Yes it does

Tonio
Are you assuming f is nonnegative?

10. Originally Posted by rn443
Are you assuming f is nonnegative?
Ok, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.

Tonio

11. Does the theorem still hold if we instead assume $\int_{-\infty}^\infty x^n f(x) \mathrm dx = 0$ for all n >= 0, provided that $\int_{-\infty}^\infty |x^n f(x)| \mathrm dx$ exists and is finite?

12. Lets suppose, without loss of generality, that is $a=-\pi$ and $b=\pi$. Since $f(*)$ is Riemann integrable in $[-\pi,\pi]$ is can be expanded in Fourier series ...

$f(x)= \frac{a_0}{2} + \sum_{k=1}^{\infty} a_{k}\cdot \cos kx + b_{k}\cdot \sin kx$ (1)

... where...

$a_{k}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cdot \cos kx\cdot dx$

$b_{k}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cdot \sin kx\cdot dx$ (2)

Now is...

$\cos kx = \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot (kx)^{2n}}{(2n)!}$

$\sin kx = \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot (kx)^{2n+1}}{(2n+1)!}$ (3)

... and $\forall n\ge 0$ is...

$\int_{-\pi}^{\pi} x^{n}\cdot f(x)\cdot dx =0$ (4)

The conclusion is that all the integrals in (2) are null and the same is for the $a_{k}$ and $b_{k}$ ...

Kind regards

$\chi$ $\sigma$

13. Originally Posted by Jose27
Yes. Note that if $\int_{a} ^{b} \ x^nf(x)dx =0$ for all $n \in \mathbb{N}$ then $\int_{a} ^{b} \ p(x)f(x)dx =0$ for any polynomial $p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $I: C^0([a,b]) \rightarrow \mathbb{R}$ $g \mapsto \int_{a} ^{b} \ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\int_{a} ^{b} \ g(x)f(x)dx =0$ for all $g(x) \in C^0([a,b])$. It is a well known result that under this conditions $f \equiv 0$.

Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.
Yes,Riemann Integrable function is continous a.e.
But Weierstrass' approximation theorem is:
Every function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$
If we denote the set $E$ of all discontinuity point,then define:

$g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$

then $g(x)$ is a continuous function on $[a,b]$,right?

then use the conclusion above,we get the conclusion $g(x)\equiv0$.
so $f(x)=0$a.e.
Is this proof right?

14. Originally Posted by Xingyuan
Yes,Riemann Integrable function is continous a.e.
But Weierstrass' approximation theorem is:
Every function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$
If we denote the set $E$ of all discontinuity point,then define:

$g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$

then $g(x)$ is a continuous function on $[a,b]$,right?

then use the conclusion above,we get the conclusion $g(x)\equiv0$.
so $f(x)=0$a.e.
Is this proof right?
Notice that I'm using Weierstrass to expand the set of functions $g$ such that $\int_{a} ^{b} g(x)f(x)dx =0$, I'm not saying anything about $f$.

Another thing, the $g$ you defined could be not well defined if f has a jump discontinuity.

Ok, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.
This doesn't contradict the claim since $\int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$

15. Originally Posted by Jose27
This doesn't contradict the claim since $\int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$
I was only contradicting the case $n=0$.

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# How to choose a control input so that the nonlinear system stays at a fixed point?

The dynamics of a magnetically suspended steel ball can be described by

$$m \ddot{h} = mg - c \frac{u^2}{h^2}$$

where $$m$$ is the mass of the ball, $$g$$ is gravitational acceleration, $$c$$ is a positive constant, and $$h$$ represents the vertical position of the ball. The input $$u$$ is the current supplied to the electromagnet.

(a) Write down a nonlinear state space model using $$x_1 = h$$ and $$x_2 = \dot{h}$$.

(b) Determine the equilibrium control input $$u_e$$ that has to be applied to suspend the ball at some position $$h=h_0 > 0$$.

For part (a), the system is \begin{align} \dot{x}_1 &= x_2 \tag{1} \\ \dot{x}_2 &= g - \frac{c}{m} \frac{u^2}{x^2_1} \tag{2} \end{align}

For part (b), the equilibrium point $$x_e=(h_0,0)$$ which is not at the origin, so we need first to make sure the equilibrium point $$x_e$$ is transformed to the origin by introducing new variables: let $$y=x-x_e$$, we get:

\begin{align} y_1 &= x_1 - h_0 &\implies \dot{x}_1 = \dot{y}_1 \tag{3}\\ y_2 &= x_2 - 0 &\implies \dot{x}_2 = \dot{y}_2 \tag{4} \end{align} From (1),(2),(3), and (4), the new system is \begin{align} \dot{y}_1 &= y_2 \\ \dot{y}_2 &= g - \frac{c}{m} \frac{u^2}{(y_1 + h_0)^2} \end{align} The equilibrium point $$y_e = (\sqrt{\frac{c}{gm}} u - h_0, 0)$$. The control input $$u_e$$ that makes $$y_e$$ is zero is $$u_e = \frac{ h_0}{\sqrt{\frac{c}{gm}}}$$, therefore, the equilibrium control input $$u_e$$ that has to be applied to suspend the ball at some position $$h=h_0 > 0$$ is $$u_e = \frac{ h_0}{\sqrt{\frac{c}{gm}}}$$

Is this correct? if not, any suggestions how to tackle this problem.

Looks correct, but you can always check this yourself by plugging your $$u_e$$ into the expression for either $$\dot{x}_2$$ or $$\dot{y}_2$$. Also note that the input only appears as $$u^2$$, so using $$-u$$ should have the same effect as $$u$$.
Lastly it is also possible to solve for $$u_e$$ by setting $$\dot{x}_2=0$$ while using $$x_1=h_0$$. However, you might also have to linearize the system around this equilibrium point, so doing this coordinate translation would probably have to be done anyway in a later stage.
• I've fixed that, thanks. But when I run the simulation via Simulink Matlab, it doesn't seem the the state $x_1 \rightarrow h_0$. I'm not sure if the input choice is correct. – CroCo Jul 25 at 0:42
• @CroCo Having a desired equilibrium does not mean that the system converges to it. For that you also need that that equilibrium is stable. This probably isn't the case. In order to achieve this you need to let $u$ deviate from $u_e$ using some sort of feedback. – Kwin van der Veen Jul 25 at 0:47
• The Jacobain matrix $A$ is $$A =\frac{\partial f}{ \partial x} = \begin{bmatrix} 0 & 1 \\ 2gh^2_0x^{-3}_1 & 0 \end{bmatrix}_{x_e=(h_0,0),u_e=\sqrt{\frac{gm}{c}} h_0} = \begin{bmatrix} 0 & 1 \\ 2gh^{-1}_0 & 0 \end{bmatrix}$$ The eigenvalues of $A$ are $\lambda_{1,2}=\pm \sqrt{\frac{2g}{h_0}}$. They are real and sign opposite, hence, the system is unstable for this input. Is this correct? – CroCo Jul 25 at 2:55


ekiim's blog

## Exercises

1. For the eeauations $x + y = 4, 2x - 2y = 4$, draw the row picture (two intersecting lines) and the column picture (a combination of two columns, equatl to the column vector $(4, 4)$ on the right side).

2. Solve the non singular triangular system: \begin{aligned} u + v + w &= b_1 \\ v + w &= b_2 \\ w &= b_3 \end{aligned} Show that your solution gives a combination of columns that equal the column on the write

3. Describe the intersection of three planes $u + v + w + z = 6$ and $u + w + z = 4$ and $u + w = 2$ (all in 4-dimensional space). Is it a line or a point or an empty set? What is the intersection of the fourth plane u = -1 is excluded?

4. Sketch the lines \begin{aligned} x + 2y &= 2\\ x - y &= 2\\ y &= 1 \\ \end{aligned} Can the three equations be solved simultaneously? What happens to the ficutre if all right hand sides zero? Is there any nonzero choice of right hand sides which allows the three lines to intersect at the same point and the three equations to have a solution?

5. Find two points on the line intersection of three plances $t=0$, and $z=0$ and $x+y+z+y=1$ in four-dimensional space.

6. When $b=(2,5,7)$, find a solution $(u, v w)$ to equation (4) other than the solutions $(1, 0 1)$ and $(4, -1, -1)$, metnioned in the text.

• Equation 4: $u \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} v \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} w \begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix} = b$
7. Give two more righthand sides in addition to $b=(2,5,7)$ for which equation (4) can be solved. Give two more right hand sides in addition to $b=(2,5,6)$ for which it cannot be solved.

8. Explain why the system: \begin{aligned} u + w + w &= 2\\ u + 2v + 3w &= 1\\ v + 2w &= 0 \\ \end{aligned} if singular, by finding a combination of three equations that adds up $0 = 1$. What value should replace the last zero on the rightside to allow the equation to have solutions — and what is one of the solutions?

9. The column picture for the prevous exercises is $u \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} v \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} w \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix} = b$ Show that the three columns on the left lie in the same plane by expressing the third column as a combination of the first two. What are the solutions $(u, v, w)$, if b is the zero vector $(0, 0, 0)$?

10. Under what condiction on $y_1, y_2, y_3$, do the points $(0, y_1), $(1, y_2)$, (2, y_3)$ lie on a straight line?

11. The equations \begin{aligned} ax + 2y &= 0\\ 2x + ay &= 0 \\ \end{aligned} are certain to have the solution $x=y=0$. For which values of $a$, is there a whole line of solutions?

12. Sketch the plane $x+y+z=1$, or the part of the plane that is in the positivie octant where $x\geq0, y\geq0, z\geq0$. Do the same for $x+y+z=2$ in the same figure. What vector is Perpendicular to does planes?

13. Starting with the line $x+4y=7$, fin d the equation for the parallel line thruough $x=0, y=0$. Find the equation of another line that meets the first at $x=3, y=1$


Question

# A $$\triangle{ABC}$$ is drawn to circumscribe a circle of radius $$3$$ cm. such that the segments $$BD$$ and $$DC$$ are respectively of length $$6$$ cm and $$9$$ cm. If the area of  $$\triangle{ABC}$$ is $$54{cm}^{2}$$, then find the lengths of sides $$AB$$ and $$AC$$

Solution

## Area of $$\triangle{ABC}=$$area of $$\triangle{OBC}+$$area of $$\triangle{OAC}+$$area of $$\triangle{OAB}$$We have $$BD=6$$cm, $$BE=6$$cm(equal tangents)$$DC=9$$cm, $$CF=9$$cm(equal tangents)Area of $$\triangle{OBC}=\dfrac{1}{2}\times b\times h=\dfrac{1}{2}\times 15\times 3=\dfrac{45}{2}{cm}^{2}$$Area of $$\triangle{OAC}=\dfrac{1}{2}\times \left(x+9\right)\times 3=\dfrac{3\left(x+9\right)}{2}{cm}^{2}$$Area of $$\triangle{OAB}=\dfrac{1}{2}\times \left(x+6\right)\times 3=\dfrac{3\left(x+6\right)}{2}{cm}^{2}$$By Heron's formula,area of $$\triangle{ABC}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{cm}^{2}$$$$s=\dfrac{x+9+x+6+15}{2}=\dfrac{2x+30}{2}=x+15$$$$\therefore \Delta=\sqrt{\left(x+15\right)\left(x+15-15\right)\left(x-15-x-4\right)\left(x+15-x-6\right)}=\sqrt{x\left(x+15\right)\times 6\times 9}$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3\left(x+9\right)}{2}+\dfrac{3\left(x+16\right)}{2}+\dfrac{45}{2}$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3}{2}\left[x+9+x+6+15\right]$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3}{2}\left(2x+30\right)$$$$\Rightarrow \sqrt{54x\left(x+15\right)}=3x+15$$Squaring on both sides we get$$\Rightarrow 54x\left(x+15\right)=9{\left(x+15\right)}^{2}$$$$\Rightarrow 6x=x+15$$$$\Rightarrow 6x-x=15$$$$\Rightarrow 5x=15$$$$\therefore x=\dfrac{15}{5}=3$$cmHence the sides are $$15$$cm, $$12$$cm and $$9$$cmMathematics

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