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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 | {
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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{}{} | {
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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 | {
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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! From mathematics is not possible for all things goes on top of another 's a good reason buy... 11 rows of Pascal 's triangle k elements from an n-element set objects i.e 's work circa 1640 mode... Gives the | {
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Pascal 's triangle k elements from an n-element set objects i.e 's work circa 1640 mode... Gives the number of k -subsets possible out of a set of distinct items rows of Pascal 's triangle to... Latexis binomial coefficient latex similar to the text inside the first 11 rows of Pascal 's work circa 1640 common! Possible for all things 3, this uses the \choose operator ⒞ instead of the \atop ¦. ( { n } { { k! \left ( { n! } } size of fraction... Positive integers that occur as coefficients in an expression is straightforward of ways which! = \binom { } { } will print the binomial theorem have been known for,. To input into a Latex document in preparation for a pdf output elements from an n-element set of. Possible for all things to find the coefficients for raising a binomial coefficient can be nested obtain... Share & SUBSCRIBE to my channel been known for centuries, but they 're best from... In which k items are chosen from among n objects i.e for all things that displays the fraction changes to. To obtain more complex expressions outcomes from possibilities, also known as a combination or number..., you find the coefficients for raising a binomial coefficient, and are used to denote a binomial is... The other side, \textstyle will change the style of the fraction as if it in! Read as choose. shows … Latex numbering equations: leqno et fleqn,,! Symbol, as shown in the binomial Expansion Technique and how to input into Latex! Coefficients are a family of positive integers that occur as coefficients in an expression is straightforward the of! Or binomial coefficient latex to other answers here 's an equation: math \frac { n k. K -subsets possible out of a set of distinct items from possibilities also! The symbols and are sometimes read as choose. 11 rows of Pascal 's work 1640... With similar characteristics - one number goes on top of another: leqno et fleqn, left, ;! The fraction as if it were in mathematical display mode of a set of distinct items distinct. Special | {
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The fraction as if it were in mathematical display mode of a set of distinct items distinct. Special input possibilities on the reference page of the first 11 rows Pascal! \Choose operator ⒞ instead of the fraction as if it were part the! Pascal 's work circa 1640 \displaystyle will format the fraction is straightforward: \ [ \binom { n {... Latexis very similar to the one used for fractions reason to buy me a coffee for reasoning about recursive in. \Displaystyle will format the fraction binomial coefficient latex, as shown in the Details section to use Gaussian... } What differs between \\dots and \\dotsc, with overleaf.com, the \binom! Is not possible for all things compelling argument to use Gaussian binomial coefficient reason! Ec1 also uses it as the primary name, which counts for a pdf output 1 } { 2 is. Numerator and the text inside the second pair is the denominator 's also... All things from mathematics is not possible for all things math \frac { }..., \textstyle will change the style of the fraction expressed using the factorial symbol, as shown in the section!, Monday 9 December 2019, by Nadir Soualem elements from an n-element.... You know them from mathematics is not possible for all things the outputs are.... Technique and how to input into a Latex document in preparation for a lot in my book display! Fleqn, left, right ; how to input into a Latex document in preparation for a lot in book! A good reason to buy me a coffee called a binomial coefficient, and are used to denote binomial! The second pair is the number of ways in which k items chosen! Texworks shows … Latex numbering equations: leqno et fleqn, left right!, right ; how to write a vector in Latex scientific tool scientific... N r ) ( n r ) ( n r ) ( n r ) ( r... Therefore gives the value of the binomial coefficient can be extended to find the coefficients raising. Page of the \atop operator ¦ '' over q-binomial coefficient '' this is the number of possible... Coefficient, and are sometimes | {
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¦ '' over q-binomial coefficient '' this is the number of possible... Coefficient, and are sometimes read as choose. command to display them in Latex to write a in... Latexis very similar to the text around it using fractions and binomial coefficients have been known for,... It as the number of ways to choose k elements from an n-element set common elements mathematical. Factorial symbol, as shown in the following the binomial coefficient, and are sometimes read as choose ''... Outputs are identical here 's an equation: math \frac { n! }.! | {
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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 | {
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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 | {
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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. | {
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# 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, | {
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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. | {
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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 | {
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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 | {
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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. | {
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• @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. | {
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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. | {
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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 | {
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# 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 | {
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# 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. | {
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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
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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 | {
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# 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. | {
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# What is the remainder of $16!$ is divided by 19?
Can anyone share me the trick to solve this problem using congruence?
Thanks,
Chan
-
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.)
-
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,
$500\frac{(1+i)^{19}-1}{i}\;=\;21,380.97$ | {
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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... | {
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# 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?
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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 $G$. | {
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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
$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$ | {
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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).
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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 $==> 7+4=11$ | {
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Prime $+$ not-prime $=$ not-prime $==> 17 + 4 = 21$
Prime $+$ not-prime $=$ prime $==> 7+4=11$
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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.
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# 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: | {
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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: | {
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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. | {
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@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. | {
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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
### Comment Preview
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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 | {
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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. | {
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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
." | {
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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 ? | {
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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 $A=B$. | {
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# 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? | {
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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 | {
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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. | {
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# 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 | {
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$$x^2+4x+4 \lt x^2-8x+16.$$
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 $1$.
-
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 $x \in (-\infty, 1)$. | {
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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 $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.
-
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 $x<1$.
-
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. | {
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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
$\chi$ $\sigma$ | {
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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: | {
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$g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$
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. | {
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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 | {
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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? | {
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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 $x=3, y=1$ | {
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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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# Show that $s_n<2$
If $s_1=\sqrt{2}$ and $s_{n+1}=\sqrt {2+s_n}$ for $n\geq 1$ show that $s_n<2$ $\forall, n\geq1$ and $s_n$ is convergent.
The second part I supposed that is immediate from Cauchy sequence definition that $s_n$ is convergent.
The first part I think to use induction
For $n=1$ we have that $s_1=\sqrt{2}<2$
Suppose that is true for $n=k$ then we need to show for $n=k+1$. I wrote a few terms of this recursive sequence
$$s_1=\sqrt{2},\quad s_2=\sqrt{2+\sqrt{2}},\quad s_3=\sqrt{2+s_2}=\sqrt{2+\sqrt{2+\sqrt{2}}}$$ Then I think that $s_{n+1}$ can written as $$s_{n+1}=\sum_{i=1}^n2^{\frac{1}{2n}}$$
I do not know how to argue that the above term is less than $2$.
• $$s_{n+1} = \sqrt{2+s_n} < \sqrt{2 + 2} = 2$$ – peterwhy Jun 5 '17 at 23:25
• Your general form is wrong. – Simply Beautiful Art Jun 5 '17 at 23:25
• And it is not "obvious" that the sequence is Cauchy – Luiz Cordeiro Jun 5 '17 at 23:28
As peterwhy mentions,
$$s_n<2\implies s_{n+1}=\sqrt{2+s_n}<\sqrt{2+2}=2\tag{\color{green}\checkmark}$$
To show that it is convergent, you want to show that it is monotone increasing, which combined with the fact that it is bounded above will mean it converges:
$$s_n>s_{n-1}\implies s_{n+1}=\sqrt{2+s_n}>\sqrt{2+s_{n-1}}=s_n\tag{\color{green}\checkmark}$$
and thus you are done.
By the induction assumption, for some $k\in\mathbb N$, $$s_k < 2$$
Consider $n = k+1$,
$$s_{k+1} = \sqrt{2+s_k} < \sqrt{2 + 2} = 2$$
Together with the fact that $s_1 = \sqrt 2 < \sqrt 4 = 2$, by induction, all $s_n < 2$.
For $0<x<2$, $x^2 < 2x$ and $2x < 2 + x$.
To prove that the sequence $s_n$ is increasing,
\begin{align*}s_{n+1} &= \sqrt{2+s_n}\\ &> \sqrt{s_n + s_n}\\ &= \sqrt{2s_n}\\ &> \sqrt{s_n^2}\\ &= s_n \end{align*}
Since the sequence $s_n$ is increasing and bounded above, limit exists. | {
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Since the sequence $s_n$ is increasing and bounded above, limit exists.
Hint: $\,s_{n+1}-2=(\sqrt {2+s_n}-2) \cdot \cfrac{\sqrt {2+s_n}+2}{\sqrt {2+s_n}+2} = \cfrac{2+s_n-4}{\sqrt {2+s_n}+2} = \cfrac{s_n-2}{\sqrt {2+s_n}+2}\,$, therefore $s_{n+1}-2$ has the same sign as $s_n -2\,$ and, by induction/telescoping, the same sign as $s_1-2\,$.
Using the fact that the square root function is increasing, and the induction hypothesis that $s_k<2$, we have: $$s_{k+1}=\sqrt{2+s_k}<\sqrt{2+2}=2$$
Assuming that the sequence does converge (you prove that later) write the limit as "S". Letting n go to infinity in $s_{n+1}= \sqrt{2+ s_n}$ we get $S= \sqrt{2+ S}$. Squaring both sides, $S^2= 2+ S$ so S must satisfy $S^2- S- 2= (S- 2)(S+ 1)= 0$ so S is either 2 or -1. Since all terms are positive, the limit (again, if there is a limit) must be 2.
Two show that this is convergent (so that the above is correct) you can show this sequence is increasing and bounded above. Since we got 2 as the limit above, it makes sense to show that $S_n< 2$ for all n. Yes, I would do that using induction on n. When n= 1, $S_n= \sqrt{2}< 2$. Assume that, for some k, $S_k< 2$. Then $S_{k+1}= \sqrt{2+ S_k}< \sqrt{2+ 2}= \sqrt{4}= 2$. | {
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# Are positive definite matrices robust to “small changes”?
Let $A$ be a positive-definite matrix and let $B$ be some other symmetric matrix. Consider the matrix $$C=A+\varepsilon B.$$ for some $\varepsilon>0$. Is it true that for $\varepsilon$ small enough $C$ is also positive definite?
• Is $B$ required to be symmetric? – kimchi lover Jan 2 '18 at 1:02
• I guess, since "positive definite" usually refer to symmetric matrices. I have updated the question. – user_lambda Jan 2 '18 at 1:02
• Eigenvalues of $A+\epsilon B$ depends continuously on $\epsilon$. If they are all positive for $\epsilon=0$ then they are positive for all small $\epsilon$. – A.Γ. Jan 2 '18 at 1:37
Positive definite means that $\langle v, Av \rangle > 0$ for all nonzero vectors $v$; actually it suffices to check this condition for unit vectors. We have
$$\langle v, Cv \rangle = \langle v, Av \rangle + \epsilon \langle v, Bv \rangle.$$
Now, by the compactness of the unit sphere, $\langle v, Av \rangle$ takes on a minimum nonzero value $m$ on unit vectors (the smallest eigenvalue of $A$, although we don't need this), and $| \langle v, Bv \rangle |$ takes on a maximum nonzero value $M$ on unit vectors (the largest eigenvalue of $B$ in absolute value).
So we can take $\epsilon < \frac{m}{M}$, which gives
$$\langle v, Cv \rangle \ge m - \epsilon M > 0$$
for all unit vectors $v$. So $C$ is positive definite as desired.
• Why did you use the spectral radius of $B$ instead of $| \lambda_{\min} (B) |$? – Rodrigo de Azevedo Jan 2 '18 at 10:13
• That would also work. I just needed anything that would clearly give a lower bound. – Qiaochu Yuan Jan 2 '18 at 10:36
If by "positive definite" you mean "strictly positive definite", the answer is "yes". The set of strictly positive definite matrices is an open set in the space of symmetric matrices. | {
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For the following reason. The positive definite matrices are the ones which satisfy a certain finite set of determinental inequalites (the principal minor determinants must all be strictly positive), each one of of which cuts out an open set in the space of matrices. Alternatively, from first principles, let $X$ be the closed unit sphere in vector space, let $Y$ be the symmetric matrices. The function $(v,A)\mapsto \|Av\|$ is continuous, so the set $S=\{(v,A): \|Av\|\le0\}\subset X\times Y$ is closed. Since $X$ is compact, the map $\pi:(v,A)\mapsto A$ is a closed map, so $\pi(S)$ is closed in $Y$, so the complement of $\pi(S)$ is open in $Y$. But that complement is the set of all $A$ for which $\|Av\|>0$ for all $v\in X$, that is to say, the set of all (strictly) positive definite matrices.
• And for positive semi-definite, it's strictly false: $\pmatrix{1 & 0 \\ 0 & 0 } + \epsilon \pmatrix{0 & 0 \\ 0 & -1}$ has eigenvalues $1, -\epsilon$ for every $\epsilon$. – John Hughes Jan 2 '18 at 1:09
• Is "strictly positive definite" different than "positive definite"? I'm unaware of that definition. – user_lambda Jan 2 '18 at 1:14
• @user_lambda: some authors use "positive definite" to mean "positive semidefinite," in the same way that some authors use "positive" to mean "nonnegative." – Qiaochu Yuan Jan 2 '18 at 1:17
• @QiaochuYuan: For what it's worth, I haven't heard of anyone using the word "positive" to mean "non-negative", but what I have seen is the notation of $\mathbb{R}_+$ to mean the non-negative reals. – Mehrdad Jan 2 '18 at 5:42
• @Mehrdad How about Bourbaki? – Daniel Fischer Jan 2 '18 at 15:16
To complement Qiaochu's answer, we have
$$\rm v^\top C \, v = v^\top A \, v + \varepsilon \, v^\top B \, v$$
where $\| \rm v \|_2 = 1$ and $\varepsilon > 0$. Note that
$$\{ \rm v^\top A \, v : \| v \|_2 = 1 \} = [ \lambda_{\min} (\mathrm A), \lambda_{\max} (\mathrm A) ]$$ | {
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$$\{ \rm v^\top B \, v : \| v \|_2 = 1 \} = [ \lambda_{\min} (\mathrm B), \lambda_{\max} (\mathrm B) ]$$
and, thus,
$$\{ \rm v^\top C \, v : \| v \|_2 = 1 \} = \left[ \color{blue}{\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B)}, \lambda_{\max} (\mathrm A) + \varepsilon \, \lambda_{\max} (\mathrm B) \right]$$
We know that $\rm A \succ 0$, i.e., $\lambda_{\min} (\mathrm A) > 0$. If $\rm B$ is positive semidefinite, then $\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B) > 0$, i.e., matrix $\rm C$ is positive definite for all values of $\varepsilon > 0$. After all, the conic combination of positive semidefinite matrices is also positive semidefinite. If $\rm B$ is not positive semidefinite, then
$$\lambda_{\min} (\mathrm A) + \varepsilon \, \lambda_{\min} (\mathrm B) = \lambda_{\min} (\mathrm A) - \varepsilon \, |\lambda_{\min} (\mathrm B)| > 0$$
yields the following upper bound on $\varepsilon$
$$\varepsilon < \color{blue}{\frac{\lambda_{\min} (\mathrm A)}{|\lambda_{\min} (\mathrm B)|}}$$ | {
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# Composition of functions
1. ### mnb96
638
This might be a silly question:
given a function $$g$$ is it possible to find a function $$f$$ such that $$f = f \circ g$$?
321
f=const.
3. ### csprof2000
287
That is the trivial solution. Perhaps there are others?
f(x) = f(g(x)) if...
f(x) = c, c constant.
g(x) = x, f any function.
If these two conditions don't hold, though...
Assume f has an inverse function. For instance, if f(x) = 2x, then (inv f)(x) = x/2. Then
f = f o g
<=>
f(x) = f(g(x))
<=>
x = g(x).
So if f has an inverse, g must equal x if f = f o g.
So you'd only be looking for functions which don't have an inverse.
But what we have is a little stronger than that, no? Since my argument makes no reference to intervals, it must be true on any interval. So g = x if you want a function f which is invertible over any interval.
The only function which is not invertible over any interval is - you guessed it - constant functions.
So, in summary:
if g(x) = x, then any function f(x) will do.
Otherwise, f(x) = c , c constant, is the only solution.
What about functions of more variables? Or generalized operations like differentiation? No idea.
4. ### John Creighto
811
If g(x)=x Then f(x) can be anything.
Let
g(n)=2n
Then f(n) can map even numbers to one constant and odd numbers to a different constant. (At least it works if n is discrete). Not sure if it works in the continuous case but I think it might.
Last edited: Feb 27, 2009
5. ### yyat
321
a) f(x)=abs(x), g(x)=-x
b) f(x)=cos(x), g(x)=x+2pi
6. ### csprof2000
287
Good point.
I guess then that my argument only works for functions which don't have an inverse where Domain(inv f) = Range(f). This, naturally, precludes functions such as abs(x) and cos(x)...
So I guess more though will have to be put into functions which are not bijections.
7. ### mnb96 | {
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So I guess more though will have to be put into functions which are not bijections.
7. ### mnb96
638
Thanks a lot. You all made very good observations that helped me a lot.
BTW, it seems that if the function f admits an inverse there are not many choices, while if the function f is not invertible, many solutions exist but the problem is non-trivial, and it is difficult to say what kind of functions f and g would have to be, in order to satisfy f(x)=f(g(x)).
At the moment I am trying to solve the following (similar) problem:
$$f = (f \circ g) g'$$
where g is invertible, and g' denotes its derivative
If you find it interesting, suggestions are always welcome.
Thanks!
8. ### csprof2000
287
Well, similar suggestions - cases - are possible.
Assume f(x) = c, c constant. Then c = cg', g' = 1, and g = x + k, k constant.
Assume f(x) = x. Then x = gg', gdg = xdx, and (g^2)/2 = (x^2)/2 + k, k constant.
Assume f(x) = x^n. Then x^n = g^n g', g^n dg = x^n dx, [1/(n+1)]g^(n+1) = [1/(n+1)]x^(n+1) + k, k constant
Wow, that's an alright result. So for x to any power at all, it's possible. Is is true for any polynomial? Yes, it seems like it should be. So... for any polynomial, I believe you can use the above formula to reduce it.
f(x) = cos x, cos x = (cos g)g', (cos x)dx = (cos g)dg, sin(x) = sin(g) + k.
It seems like, unless I'm mistaken, this is the same thing every time:
f(x) = f(g(x))g'(x) is the same as solving the differential equation f(x)dx = f(g(x))dg.
So, as long as f is integrable, the problem is actually quite easy. Maybe I'm wrong. Thoughts?
9. ### mnb96
638
csproof2000: I think you just made it!
You gave the solution to the problem, and for some reason I didn't immediately spot that what I was trying to do is actually solving a differential equation. | {
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# $-\iint_{A}(y+x)\,dA$ Evaluating
I am bit unsure about the following problem:
Evaluate the double integral:
$$-\iint_{A}(y+x)\,dA$$
over the triangle with vertices $(0,0), (1,1), (2,0)$
OK, so I figured here that I would do this by first evaluating the integral over the region bounded by the vertices $(0,0), (1,1), (1,0)$ and then evaluate the integral over the region bounded by the vertices $(1,0), (1,1), (2,0)$ before adding the two answers together, and then reversing the sign of this answer (since there is a minus sign in front of the original double integral). Thus, I begin by finding:
$$\int_{0}^{1}dx \int_{0}^{x}(y+x)\,dy$$
When solved this gives me the answer $\frac{1}{2}$.
Next I solve:
$$\int_{1}^{2}dx \int_{1}^{2-x}(y+x)\,dy$$
When solved this gives me the answer $-\frac{7}{6}$.
I have verified both the integrals in Wolframalpha, and they give me the same answer. I would therefore believe that the final answer should be:
$$-(\frac{1}{2} - \frac{7}{6}) = \frac{2}{3}$$
However, the final answer should, according to the book, be $-\frac{4}{3}$.
Thus, obviously I do something wrong here. If anyone can help me out, I would greatly appreciate it. Is it perhaps that it is not allowed to "split up" this into two separate integrals? I couldn't find a way to solve this without doing this. | {
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-
I see one problem at least: The inner ($y$) integral in the second part should start at $0$. Anyway, you can avoid the splitting by doing the integrals with the $x$ integral on the inside. Try it! – Harald Hanche-Olsen Jul 15 '12 at 20:32
Thank you for your answer. But why should the second integral start at 0? The y-values being at 1 and end up at 0 (hence I chose y = 2-x for the upper value of the integral). – Kristian Jul 15 '12 at 20:39
Draw a picture. The whole triangle has its base at the $x$-axis, and so does each of the two pieces resulting from the split. – Harald Hanche-Olsen Jul 15 '12 at 20:41
You can also solve this with the Gauss integral theorem on $F(x,y) = xy$ and evaluate the integral on the boundary instead. Maybe this is easier, since the boundary is made of line segments. – Cocopuffs Jul 15 '12 at 20:42
It seems you may be confusing the two bounds. The bound that begins at $1$ and ends up at $0$ is the upper bound; the lower one is $0$ throughout. Also, you can tell from the sign that $-\frac76$ must be wrong, since the integrand is non-negative throughout the triangle. – joriki Jul 15 '12 at 20:45
Your second integral should be $$\int_{1}^{2}dx \int_{0}^{2-x}(y+x)dy.$$ Your lower $y$ limit was 1 instead of 0. | {
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# Drawing the grid without lifting the pen
The below shape consist of $$24$$ segments with unit length. if we want to draw this shape without lifting the pen, what is the minimum length of the line we should draw? $$1)25\quad\quad\quad\quad\quad\quad2)26\quad\quad\quad\quad\quad\quad3)27\quad\quad\quad\quad\quad\quad4)28\quad\quad\quad\quad\quad\quad5)29\quad\quad\quad\quad\quad\quad$$
I think we can do it somehow by segments with the length $$28$$ (by passing twice the four middle segments on each side of outer big square).
I'm not sure how to solve this problem. is it possible to solve it mathematically or I should just try different ways to draw this?
Notice that if you draw it without lifting the pen, except for the starting and ending point you must draw the same number of lines going into each point as going out. This means that you must draw an even number of segments meeting every point except possibly two. How many extra segments does this imply you need at a minimum? Can you achieve this?
• Thank you for the answer. but unfortunately, I can't figure out how to continue from your hint. can you please elaborate more on that? Mar 11, 2021 at 17:04
You will be able to draw all the edges of the graph without lifting the pen and drawing every edge exactly once (starting and ending at the same node), iff there exists an Euler circuit in the graph.
We shall use the following theorem: a connected graph has an Euler circuit iff all the nodes of the graph has even degree.
The given grid graph does not have all nodes with even degree. Let's first find the nodes with odd degrees, as shown in the next figure. Notice that the nodes B,C, E,I, H,L and N,O have odd degrees (namely 3).
Let's add 4 additional (red) edges to the grid graph as shown in the next figure to make all the nodes have even degrees. | {
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Now, in the augmented graph above, by the theorem, it must have an Euler circuit, so we shall be able to draw all the edges of the graph exactly once without lifting the pen.
We shall use Flury's algorithm to find an Euler circuit in the augmented graph first. The key idea is that when you have a choice between a bridge and a non-bridge, always choose the non-bridge (don't burn the bridges).
Since the augmented graph has all nodes with even degrees, we can start at any node, let's start at B, a node with degree 4 in the augmented graph.
Using the algorithm, the following animation shows how to construct an Euler circuit in the augmented graph.
Now, let's find where the additional edges (ones not present in the original graph) were used in the circuit found in the augmented graph.
Let's go back to our original graph by replacing those edges by the ones in the original graph, we had exactly 4 such additional edges needed to make all nodes in the original graph to have even degrees.
It implies that in order to have the Euler circuit in the original graph we need 4 additional edges (requiring 24+4=28 edges). It also means that in the original graph we need to traverse the corresponding 4 edges twice, in order to draw all the edges without lifting the pen, as shown below:
[EDIT]
If we don't need to start and end at the same node, it's sufficient to have an Euler trail instead of a circuit, needing one less edge: $$28-1=27$$ edges (only 3 edge segments, namely, EI, ON and HL are needed to be traversed twice, leading to minimum length of the line as $$24+3=27$$).
In this case the trail shown above starts at $$B$$ but ends at $$C$$ (instead of $$B$$ as before), since the edge $$BC$$ has already been visited once, we don't need to visit it again (as shown in the next figure). | {
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• Wow! thank you very much for the answer and beautiful animation you used. I really appreciate that. Mar 11, 2021 at 19:15
• Great answer, except that you don't need an Euler circuit, but merely an Euler path. Which means we can save one line segment. Mar 12, 2021 at 1:04
The figure has 4 nodes with 2 lines connected to node. 8 nodes with 3 lines connected, and 4 nodes with 4 lines connected.
If you want to complete the circuit without lifting your pen, you can start at a node with an odd number of connections, and you can end at a node with an odd number of connections, for all the rest you must have an even number of connections. That is, if you get to the node on one path you leave the node on a different path.
This means that you must double up on some of the lines where it appears that there is a odd number of lines to make it an even number. What is the minimum number of lines you need to double up such that there are at most two nodes with an odd number of connections?
• Thank you very much. So for the start we connect two of the points with odd degrees and we left with three more segments passing through other odd degree points on other sides of outer big square. so we must pass them twice and the answer is $24+3=27$. am I right? Mar 11, 2021 at 17:42
• It must be at least 27. It is probably worth the effort to show that it can be done with 27 segments. Mar 11, 2021 at 17:46
Every node must have the same number of edges entering and leaving, except for the end points. There are 8 nodes on the graph with an odd degree, but we can make two of them the endpoints, so we have 6 nodes that need an "extra" edge. Finally, we can note that the odd-degree nodes come in adjacent pairs, so the "extra" edge leaving one node can be the "extra" edge entering another, so we only need half as many "extra" edges as we have non-endpoint, odd-degree nodes. | {
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# Polynomial $p(x) = 0$ for all $x$ implies coefficients of polynomial are zero
I am curious why the following is true. The text I am reading is "An Introduction to Numerical Analysis" by Atkinson, 2nd edition, page 133, line 4.
$p(x)$ is a polynomial of the form:
$$p(x) = b_0 + b_1 x + \cdots + b_n x^n$$
If $p(x) = 0$ for all $x$, then $b_i = 0$ for $i=0,1,\ldots,n$.
Why is this true? For example, for $n=2$, I can first prove $b_0=0$, then set $x=2$ to get a linear system of two equations. Then I can prove $b_1=b_2 = 0$. Similarly, for $n=3$, I first prove $b_0=0$, then I calculate the rank of the resulting linear system of equations. That shows that $b_1=b_2=b_3=0$. But if $n$ is very large, I cannot keep manually solving systems of equations. Is there some other argument to show all the coefficients must be zero when the polynomial is always zero for all $x$?
Thanks. | {
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Thanks.
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Are you familiar with the fact that if $p(a) = 0$, then $p(x)$ is divisible by $x - a$? Use this $n+1$ times. – Qiaochu Yuan Aug 28 '11 at 3:36
As explained in the answers, we can solve the problem by showing that a non-zero polynomial of degree $d$ has no more than $d$ roots. However, your approach can be made to work. To carry it out, one uses properties of the Vandermonde matrix. The details are somewhat more messy than the approach through counting roots, but accessible. – André Nicolas Aug 28 '11 at 3:55
Hi Qiaochu Yuan, can you give me some more hints? I am aware of the Fundamental Theorem of Algebra, which can be found here: tutorial.math.lamar.edu/Classes/Alg/ZeroesOfPolynomials.aspx. If I divide a polynomial of degree $n$ by (x-a), and a is guaranteed to be a root (since the polynomial is a horizontal line at $y=0$), then I get as a result, another polynomial with degree $n-1$. After $n-1$ divisions, I get as a result, something like: $d + ex = 0$. $d$ and $e$ might not be the original polynomial coefficients. What would be the next step after this? – jrand Aug 28 '11 at 4:35
For purposes of argumentation: false over finite fields... – GEdgar Aug 28 '11 at 13:31
As you are interested in numerical methods the following is probably not of interest to you, but for the sake of completeness I feel compelled to point out that the conclusion is false, if your domain is finite. A non-zero polynomial can easily vanish at all points of, e.g. $\mathbf{Z}_m$ without being the zero polynomial. The easiest example is the polynomial $p(x)=x^2-x\in \mathbf{Z}_2[x]$ that vanishes at all the elements of $\mathbf{Z}_2$. If you don't study algebra, you can safely ignore this comment for now. – Jyrki Lahtonen Aug 28 '11 at 13:37 | {
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HINT $\$ A nonzero polynomial over a field (or domain) has no more roots than its degree, as is easily proved by induction and the Factor Theorem. In fact if every natural was a root then the polynomial would be divisible by $\rm\:(x-1)\:(x-2)\:\cdots\:(x-n)\:$ for all $\rm\:n\in \mathbb N\:,\:$ which yields a contradiction for $\rm\:n\:$ greater than the degree of the polynomial.
Note that the proof of said statement depends crucially on the hypothesis that coefficient ring is an integral domain, i.e. a ring satisfying $\rm\:ab = 0\iff a=0\ \ or\ \ b=0\:.\:$ Over non-domains such as the integers modulo $\rm\:m\:$ not prime, polynomials can have more roots than their degree. In fact if this is true then one can use such roots to factor $\rm\:m\:,\:$ see here.
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Following reading Bill Dubuque's, Shawn's, and Qiaochu Yuan's answer, I believe I must agree to the following fact. If a polynomial is equal to zero, then it must be of degree 0, and the coefficient must be set such that $b_0=0$. Since it is degree 0, the other coefficients must also be set to 0. – jrand Aug 28 '11 at 12:42
@jrand: The degree of the polynomial zero is not zero. – Did Aug 28 '11 at 13:02
Note that $p(x)=0$ implies derivatives of all orders are also $0$. Let $x=0$ into $p(x) = b_0 + b_1 x + \cdots + b_n x^n$ to obtain $b_0 = 0$.
Now differentiate both sides: $p'(x) = b_1 + 2b_2 x + 3b_3 x^2 + \cdots + nb_n x^{n-1}$. Let $x=0$ again, and we get that $b_1=0$.
If we continue to differentiate and substitution $x=0$ we will get that $b_k=0$ for $k=0,1,2,\cdots, n$. This idea can easily be made into a rigorous induction proof.
A nice corollary of this result is that two polynomials are equal if and only if they have the same degree and coefficients.
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-
If you are willing to accept $\displaystyle \lim_{x \rightarrow \infty}\frac{1}{x^{m}}=0$ and $\displaystyle \lim_{x \rightarrow \infty}x^m=\infty$ for all $m>0$, then you can argue as follows. Suppose you have a polynomial $f(x)=b_nx^n+ \ldots b_0$ with $b_n \neq 0$ and $n\geq 1$, then $$\lim_{x\rightarrow \infty}f(x)=\lim_{x \rightarrow \infty} x^n(b_n+ \ldots \frac{a_0}{x^n})=\lim_{x \rightarrow \infty} x^nb_n=\infty.$$
Since the constant function $0$ does not have this property, it cannot be equal to a polynomial with degree greater or equal to $1$.
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+1. Nice analytical solution with no derivatives. – Did Aug 28 '11 at 22:22
A polynomial can be uniquely fitted by knowing its value at $d+1$ points, where $d$ is the degree of the polynomial. If it's 0 at all points, that's clearly more than $d+1$ points.
But we also know that a polynomial can be written as $(x-r_0)(x-r_1)...(x-r_k)$, where the $r_i$ are the roots of polynomial (possibly complex). Again, if there are an infinite number of roots...
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Let $n\in \mathbb{N}$, and $$p(x)=b_0+b_1x+\ldots +b_nx^n$$ a polynomial of degree at most $n$. If $p(x)=0$ for every $x$, then $b_i=0$ for $i=0,\, 1,\ldots\, n$.
Proof:
By induction on $n$:
If $\deg(p)\leq 1$, then $p(x)=b_0+b_1x$. Since $p(0)=p(1)=0$ you get $b_0=b_1=0$.
Suppose that for any $$r(x)=c_0+c_1x+\ldots +c_nx^n$$ of degree at most $n$, if $r(x)=0$ for every $x$, then $c_i=0$ for $i=0,\, 1,\ldots\, n$. Let $$p(x)=b_0+b_1x+\ldots +b_{n+1}x^{n+1}.$$ Suppose that $p\equiv 0$. Since $p(0)=0$, you get that $b_0=0$. From here, I have two possible arguments. The first, that was my original one, is as follows: | {
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We have $$p(x)=b_1x+b_2x^2+\ldots+b_{n+1}x^{n+1}=x(b_1+b_2x+\ldots+b_{n+1}x^{n}).$$ Then for all $x$, \begin{align*} 0&= p(x)\\ 0&= x(b_1+b_2x+\ldots+b_{n+1}x^{n}).\end{align*} Then for any $x\neq 0$, $$q(x):=b_1+b_2x+\ldots+b_{n+1}x^{n}=0$$ If $q\not\equiv 0$, then we have a polynomial of degree at most $n$ with infinitely many roots. This can't be, therefore $q\equiv 0$. Now, $\deg(q)\leq n$ and $q\equiv 0$, therefore by the induction hypothesis, $b_i=0$ for $i=1,\,\ldots ,\, n+1$.
The second argument was inspired by @Ragib Zaman's answer. Just differentiate both sides of $p(x)=0$, then $$b_1+2b_2x+\ldots+(n+1)b_{n+1}x^n=0$$ for all $x$. By the induction hypothesis, $kb_k=0$ for $k=1,\,\ldots,\,n+1$, and this implies that $b_k=0$ for $k=1,\,\ldots,\,n+1$
Since $b_0=0$, this proves that $b_i=0$ for $i=0,\, 1,\ldots,\, n+1$.
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In general, by the fundamental theorem of algebra we can say that the number of different roots of a polynomial of degree $n$ is at most $n$. I the argument above $q$ must be identically $0$ because if not we have a polynomial with infinite different roots. – leo Aug 28 '11 at 6:46
My comment is the hint of @Bill – leo Aug 28 '11 at 7:12
This induction is flawed because during your nth inductive step you use as a fact a property (for every q of degree n, if q(x)=0 for every nonzero x, then q=0) that the hypothesis at the beginning of the step (for every polynomial p of degree n, if p(x)=0 for every x, then p=0) does not imply. – Did Aug 28 '11 at 9:03
@Didier Piau: I'm trying to prove a statement about the coefficients of polynomials, not about the polynomials. If I understand correct your comment, you say me that the hypothesis of induction does not implies that the polynomial $q$ is identically $0$. Yes that is true, but i don't pretend it. The fact that $q\equiv 0$ is something that I want to apply my hypothesis of induction. – leo Aug 28 '11 at 10:06
I think that the answer is now more clear. – leo Aug 28 '11 at 10:14 | {
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# 2 different methods of calculating possibilities of people sitting at round table
I have 2 different problems that seem identical:
8 men and 8 women are going to sit at a round table where there are 16 seats. They take their seats randomly. How many ways can the 16 seats be taken so that no 2 women are sitting next to each other?
I solved this by doing $$2 * 8!8!$$
The method I used was that $8*8*7*7*6*6*5*5*4*4*3*3*2*2*1*1= 8!8!$ is all the ways you can sit these 8 women alternating them with the 8 men.
I then multiplied by two because you have 2 possibilities: the women can be sitting at the odd numbered seats, or at the even numbered ones.
Then I have this other problem:
In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?
(I took the problem from here: http://www.imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf)
And the solution, according to the website, is:
Answer: The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3!4! = 144
However, I can't seem to use the method from the 1st problem on this one. All the possible combinations of men and women, sitting alternated with men, would be $4!4!$ Then, multiplying by 2 the solution would be $2 * 4!4! = 1152$
If I use the method from the answer of the 2nd problem with the 1st one, I get $7!8!$, which is different from $2 * 8!8!$
My question is, how can this be? Is there any difference between these 2 problems that I am not seeing?
EDIT: I believe the 1st solution is correct. Check here Probability of men and women sitting at a table alternately | {
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• You are assuming that the seats are labeled. The answer to the second problem assumes the seats are not labeled and takes rotational invariance into account. – N. F. Taussig Oct 16 '16 at 1:29
• Take into account circular permutations. The 1st solution is wring 2nd is correct. – Navin Oct 16 '16 at 1:30
• @navinstudent math.stackexchange.com/questions/1969572/… – SilenceOnTheWire Oct 16 '16 at 1:32
• @N.F.Taussig That makes sense, but how do you infer which method to use from reading the problems, without looking at their solutions? Neither problem mentions anything about the seats being labeled. – SilenceOnTheWire Oct 16 '16 at 1:52
• If the seats are not described as labeled, I would assume that they are unlabeled and that rotational invariance applies, as in the second problem. – N. F. Taussig Oct 16 '16 at 2:57 | {
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# Conditional Probability with balls in an urn
Two balls, each equally likely to be colored either red or blue, are put in an urn. At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that (a) both balls in the urn are colored red; (b) the next ball chosen will be red?
I'm wondering if my method for part (a) is correct:
Let $P(R)$ be the probability of picking a red ball: $P(R)=\frac{1}{2}$
Let $P(B)$ be the probability of picking a blue ball: $P(B)=\frac{1}{2}$
Let $P(C)$ be the probability of the condition, i.e., picking two red balls consecutively: $P(C)=P(C|H_1)P(H_1)+P(C|H_2)P(H_2)+P(C|H_3)P(H_3)$
where $P(H_1)$ is the probability both balls in the urn are red: $P(H_1)=(P(R))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_2)$ is the probability that both balls in the urn are blue: $P(H_2)=(P(B))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_3)$ is the probability that one ball is red and one is blue inside the urn: $P(H_3)=1-(P(H_1)+P(H_2))=1-\frac{1}2=\frac{1}2$ since the sum of the mutually exclusive hypotheses or events must sum to 1.
Thus $P(C)=1\times\frac{1}4+\frac{1}4\times0+\frac{1}2\times\frac{1}4=\frac{3}8$
and $P(H_1|C)= \frac{P(C \bigcap H_1)}{P(C)}=\frac{P(C|H_1)P(H_1)}{P(C)}=\frac{1\times\frac{1}4}{\frac{3}8}=\frac{2}3$
For part (b), I know that $P(R|C)$, the probability of picking a red ball given that the first two balls picked were red is to be found
$P(R|C)= \frac{P(C \bigcap R)}{P(C)}=\frac{P(C|R)P(R)}{P(C)}$
How can I find $P(C|R)$? | {
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$P(R|C)= \frac{P(C \bigcap R)}{P(C)}=\frac{P(C|R)P(R)}{P(C)}$
How can I find $P(C|R)$?
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There are three possible urns that you might have: both balls red (RR), both balls blue (BB), one of each color (RB) which occur with probabilities $1/4, 1/4, 1/2$ respectively. Now calculate $P(\text{red,red}\mid \text{RR})$, $P(\text{red,red}\mid \text{BB})$, $P(\text{red,red}\mid \text{RB})$, and combine them via the law of total probability to get $P(\text{red,red})$. Then, use Bayes' formula to obtain $P(\text{BB}\mid \text{red,red})$ – Dilip Sarwate Mar 2 '13 at 3:04
The probability that the contents of the urn are two red is indeed $\frac{1}{4}$, as is the probability of two blue, and the probability of mixed is therefore $\frac{1}{2}$. The derivation could have been done more quickly.
Question (a) asks for the probability both are red given that the two drawn balls are red. Let $R$ be the event both are red, and $D$ be the event both drawn balls are red. We want $\Pr(R|D)$. By the usual formula this is $\frac{\Pr(R\cap D)}{\Pr(D)}$.
To find $\Pr(D)$, note that if both balls are red (probability $\frac{1}{2}$), then the probability of $D$ is $1$, while if one ball is red and the other is not (probability $\frac{1}{2}$) then the probability of $D$ is $\frac{1}{4}$. Thus the probability of $D$ is $\left(\frac{1}{4}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)$. This is $\frac{3}{8}$.
The probability of $R\cap D$ is $\frac{1}{4}$. So the ratio is indeed $\frac{2}{3}$.
For (b), you can use the calculation of (a). With probability $\frac{2}{3}$ we are drawing from a double red, and we will get red with probability $1$. With probability $\frac{1}{3}$ the urn is a mixed one, and the probability of drawing a red is $\frac{1}{2}$, for a total of $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot\frac{1}{2}$. | {
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One can also solve (b) without using the result of (a). With I hope self-explanatoru notation, we want $\Pr(RRR|RR)$. The probabilities needed in the conditional probability formula are easily computed. We have $\Pr(RRR\cap RR)=\Pr(RRR)=\frac{1}{4}\cdot 1+\frac{1}{2}\cdot\frac{1}{8}=\frac{5}{16}$. Divide by $\frac{3}{8}$.
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# Finding a curve that intersects any line on the plane
Question Is there a curve on plane such that any line on the plane meets it (a non zero ) finite times ?
What are the bounds on the number of such intersections.
My question was itself inspired by this "Can you draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?"
• Clearly if your curve is the graph of an odd polynomial, the number of intersections is at most the degree of the polynomial. – Cameron Williams Sep 18 '13 at 15:08
• One possibility is a pair of parabolas, say one concave up and one concave down, situated so that they touch tangentially (or intersect) and therefore cannot be separated by a hyperplane. – hardmath Sep 18 '13 at 15:08
• Please try to make titles of questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Lord_Farin Sep 18 '13 at 15:08
• @CameronWilliams Correct, but just being an odd degree polynomial; does not guarantee that every line in the plane will intersect it – ARi Sep 18 '13 at 15:10
• @ARi: Actually it does imply that, except degree 1. – hardmath Sep 18 '13 at 15:13
Cubic parabola $$y=x^3$$ has this property. The max number of such intersections is given by the Fundamental theorem of algebra:
$$x^3=ax+b$$ can have at most 3 solutions. | {
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$$x^3=ax+b$$ can have at most 3 solutions.
• Yes but will all of the solutions be real for all real (a,b) pairs – ARi Sep 18 '13 at 15:14
• For any $a$, there will be only one real solution for all sufficiently large $|b|$. – hardmath Sep 18 '13 at 15:17
• I think the important fact here (i.e., relevant to the question), is that $\,x^3-ax-b=0\;$ has always at least one real solution and, of course, at most three different ones. Nice answer. +1 – DonAntonio Sep 18 '13 at 15:22
• So the upper bound for intersections will be three and the lower one 0. – ARi Sep 18 '13 at 15:29
• Well, a tighter and always attainable lower bound is one, in fact...and this fulfills the question's conditions. – DonAntonio Sep 18 '13 at 15:31
As already shown, any cubic polynomial (and indeed, any odd-degree polynomial) has the requisite property by the fundamental theorem of algebra.
What's more, a simple perturbation argument should be enough to show that any (sufficiently) smooth curve that meets every line in at least one point will meet some lines in at least three points. Consider a point tangent to the curve where the second derivative 'with respect to the tangent line' is non-zero; that is, a non-reflex tangent point, or locally extremal point. (Such points must exist if the curve is non-trivial). Now, consider pencils of lines 'near' this intersection point; displaced infinitesimally one way from the tangent, they must have another point of intersection with the curve, and this point can be made 'generic' so that it doesn't vanish under small perturbations. Then displace infinitesimally the other direction; the 'generic' point of intersection is still a point of intersection, but the tangent turns into two points of intersection. | {
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1. ## Arithmetic Series Help
First of all I'm not sure if this is the right section, so I apologise if it's not.
I was hoping that someone could help me out with the following question:
From a piece of wire 5000mm long, n pieces are cut, each piece being 10mm longer than the previous piece. Ui is the length of the ith piece and Si is the cumulative length of i pieces.
Express Un and Sn in terms of U1 and n.
2. $U_n = u_1 + 10(n-1)$
$S_n = \sum_{i=1}^n U_i = \sum_{i=1}^n [u_1 + 10(i-1)]$ $= u_1 + 10\sum_{i=1}^n i -\sum_{i=1}^n 1 = u_1 + 10\frac{n(n+1)}{2} - n = u_1 +n(5n + 4)$
Also, $S_n = 5000 \Rightarrow u_1 = 5000 - 5n^2 - 4n.$
3. Your answer for Un looks right so thank you for that.
However the others don't seem to add up, I had to find U1 as the second part of the question and I've calculated it as being 80 which I've checked and it is right. If you put 80 into your formulae it doesn't equal 5000.
Thank you anyway though bud.
4. Hello, JP103!
Anonymous1 made some small errors. . .
$S_n \;=\; \sum_{i=1}^n U_i$
. . $=\; \sum_{i=1}^n \bigg[U_1 + 10(i-1)\bigg]$
. . $=\; \sum_{i=n}^n\bigg[U_1 + 10i - {\color{red}10}\bigg]$
. . $= \;\sum_{n=1}^nU_1 + 10\sum_{n=1}^ni - 10\sum_{n-1}^n1$
. . $=\;{\color{red}n}U_1 + 10\frac{n(n+1)}{2} - 10n$
$\boxed{S_n \;=\;nU_1 + 5n(n-1)}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Using the sum-formula: . $S_n \:=\:\frac{n}{2}\bigg[2a + (n-1)d\bigg]$
. . we have: . $S_n \;=\;\frac{n}{2}\bigg[U_1 + (n-1)10\bigg] \quad\Rightarrow\quad S_n \;=\;nU_1 + 5n(n-1)$ | {
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How to prove if a function is bijective?
I am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective). Here's an example:
How do I prove that $g(x)$ is bijective?
\begin{align} f &: \mathbb R \to\mathbb R \\ g &: \mathbb R \to\mathbb R \\ g(x) &= 2f(x) + 3 \end{align}
However, I fear I don't really know how to do such. I realize that the above example implies a composition (which makes things slighty harder?). In any case, I don't understand how to prove such (be it a composition or not).
For injective, I believe I need to prove that different elements of the codomain have different preimages in the domain. Alright, but, well, how?
As for surjective, I think I have to prove that all the elements of the codomain have one, and only one preimage in the domain, right? I don't know how to prove that either!
EDIT
f is a bijection. Sorry I forgot to say that.
-
Normally one distinguishes between the two different arrows $\mapsto$ and $\to$. One writes $f:\mathbb{R}\to\mathbb{R}$ to mean $f$ is a function from $\mathbb{R}$ into $\mathbb{R}$. The notation $x\mapsto x^3$ means the function that maps every input value to its cube. – Michael Hardy Jul 1 '12 at 20:39
Wouldn't you have to know something about $f$? Is $f$ a bijection? – Dylan Moreland Jul 1 '12 at 20:40
My apologies! Yes, f is a bijection. – Omega Jul 1 '12 at 20:40
Both of your deinitions are wrong. Maybe all you need in order to finish the problem is to straighten those out and go from there. I've posted the definitions as an answer below. – Michael Hardy Jul 1 '12 at 20:45
The way to verify something like that is to check the definitions one by one and see if $g(x)$ satisfies the needed properties.
Recall that $F\colon A\to B$ is a bijection if and only if $F$ is:
1. injective: $F(x)=F(y)\implies x=y$, and
2. surjective: for all $b\in B$ there is some $a\in A$ such that $F(a)=b$. | {
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Assuming that $R$ stands for the real numbers, we check.
Is $g$ injective?
Take $x,y\in R$ and assume that $g(x)=g(y)$. Therefore $2f(x)+3=2f(y)+3$. We can cancel out the $3$ and divide by $2$, then we get $f(x)=f(y)$. Since $f$ is a bijection, then it is injective, and we have that $x=y$.
Is $g$ surjective?
Take some $y\in R$, we want to show that $y=g(x)$ that is, $y=2f(x)+3$. Subtract $3$ and divide by $2$, again we have $\frac{y-3}2=f(x)$. As before, if $f$ was surjective then we are about done, simply denote $w=\frac{y-3}2$, since $f$ is surjective there is some $x$ such that $f(x)=w$. Show now that $g(x)=y$ as wanted.
Alternatively, you can use theorems. What sort of theorems? The composition of bijections is a bijection. If $f$ is a bijection, show that $h_1(x)=2x$ is a bijection, and show that $h_2(x)=x+2$ is also a bijection. Now we have that $g=h_2\circ h_1\circ f$ and is therefore a bijection.
Of course this is again under the assumption that $f$ is a bijection.
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"Subtract 3 and divide by 2, again we have (y−3)/2=f(x). As before, if f was surjective then we are about done" I'm not sure, but why would we be done there? – Omega Jul 1 '12 at 21:18
@Omega: If $f$ was surjective, then there is some $x$ such that $f(x)=\frac{y-3}2$, show now that $g(x)=y$. – Asaf Karagila Jul 1 '12 at 21:20
You know, it had me thinking: according to your method to find out if it is injective, no matter what function I test it with, I always manage to get the final equality (x = y). How would a function ever be not-injective? – Omega Jul 1 '12 at 22:34
@Omega: No, assume that $f(x)=0$ for all $x$, suppose that $x,y$ are any two real numbers (perhaps different and perhaps not), does $f(x)=f(y)$ tell you something about $x=y$ or $x\neq y$? No, because taking $x=1$ and $y=2$ gives $f(1)=0=f(2)$, but $1\neq 2$. Note that my answer relies critically on the fact that $f$ itself is injective. – Asaf Karagila Jul 1 '12 at 22:37 | {
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To prove a function is bijective, you need to prove that it is injective and also surjective.
"Injective" means no two elements in the domain of the function gets mapped to the same image.
"Surjective" means that any element in the range of the function is hit by the function.
Let us first prove that $g(x)$ is injective. If $g(x_1) = g(x_2)$, then we get that $2f(x_1) + 3 = 2f(x_2) +3 \implies f(x_1) = f(x_2)$. Since $f(x)$ is bijective, it is also injective and hence we get that $x_1 = x_2$.
Now let us prove that $g(x)$ is surjective. Consider $y \in \mathbb{R}$ and look at the number $\dfrac{y-3}2$. Since $f(x)$ is surjective, there exists $\hat{x}$ such that $f(\hat{x}) = \dfrac{y-3}2$. This means that $g(\hat{x}) = 2f(\hat{x}) +3 = y$. Hence, given any $y \in \mathbb{R}$, there exists $\hat{x} \in \mathbb{R}$ such that $g(\hat{x}) = y$. Hence, $g$ is also surjective.
Hence, $g(x)$ is bijective.
In general, if $g(x) = h(f(x))$ and if $f(x) : A \to B$ and $h(x): B \to C$ are both bijective then $g(x): A \to C$ is also bijective.
In your case, $f(x)$ was bijective from $\mathbb{R} \to \mathbb{R}$ and $h(x) = 2x+3$ is also bijective from $\mathbb{R} \to \mathbb{R}$.
-
You haven't said enough about the function $f$ to say whether $g$ is bijective.
"Injective" means different elements of the domain always map to different elements of the codomain.
"Surjective" means every element of the codomain has at least one preimage in the domain.
Later edit: What you've now added---that $f$ is a bijection---bring us to the point where we can answer the question. However, maybe you should look at what I wrote above. Since both definitions that I gave contradict what you wrote, that might be enough to get you there.
-
Sorry, yes, f is bijective. Thank you for clarifying that :) – Omega Jul 1 '12 at 20:43 | {
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-
Sorry, yes, f is bijective. Thank you for clarifying that :) – Omega Jul 1 '12 at 20:43
First show that $g$ is injective ($1$-$1$) by showing that if $g(x)=g(y)$, then $x=y$. This isn’t hard: if $g(x)=g(y)$, then $2f(x)+3=2f(y)+3$, so by elementary algebra $f(x)=f(y)$. By hypothesis $f$ is a bijection and therefore injective, so $x=y$.
Now show that $g$ is surjective. To do this, you must show that for each $y\in\Bbb R$ there is some $x\in\Bbb R$ such that $g(x)=y$. That requires finding an $x\in\Bbb R$ such that $2f(x)+3=y$ or, equivalently, such that $f(x)=\frac{y-3}2$. But $f$ is known to be a bijection and hence a surjection, so you know that there is such an $x\in\Bbb R$.
In general this is one of the two natural ways to show that a function is bijective: show directly that it’s both injective and surjective. The other is to construct its inverse explicitly, thereby showing that it has an inverse and hence that it must be a bijection. You could take that approach to this problem as well:
$$g^{-1}(y)=f^{-1}\left(\frac{y-3}2\right)\;,$$
since
\begin{align*} g\left(f^{-1}\left(\frac{y-3}2\right)\right)&=2f\left(f^{-1}\left(\frac{y-3}2\right)\right)+3\\ &=2\left(\frac{y-3}2\right)+3\\ &=y\;, \end{align*}
and since $f$ is a bijection, $f^{-1}\left(\frac{y-3}2\right)$ exists for every $y\in\Bbb R$.
Added: As Marc reminds me, this is only half the job: if you take this approach, you must either show directly that $g$ is injective, as I did above, or verify that the function that I called $g^{-1}$ above is a two-sided inverse, i.e., that $g^{-1}\big(g(x)\big)=x$ for $x\in\Bbb R$. This is not particularly difficult in this case:
\begin{align*} g^{-1}\big(g(x)\big)&=g^{-1}\big(2f(x)+3\big)\\ &=f^{-1}\left(\frac{\big(2f(x)+3\big)-3}2\right)\\ &=f^{-1}\big(f(x)\big)\\ &=x\;, \end{align*}
since $f$ is a bijection. | {
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since $f$ is a bijection.
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When using the "inverse" criterion, you should be careful in really checking that a purported inverse is an inverse, both ways. With $g^{-1}$ denoting your purported inverse, your final argument checked that $g(g^{-1}(y))=y$ for all $y\in\mathbb R$; this only shows that $g$ is surjective (it has a right inverse, also called a section). To show that $g$ is also injective you need to separately check that $g^{-1}(g(x))=x$ for all $x\in\mathbb R$. – Marc van Leeuwen Jul 1 '12 at 21:22
@Marc: Yes, I should probably say as much; I hadn’t originally intended to mention this approach at all, and did so only as an afterthought. I was implicitly assuming that the obvious injectivity had already been checked, but that’s not clear from what I wrote. – Brian M. Scott Jul 1 '12 at 21:26 | {
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Angle of a triangle inscribed in a square
Say we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $\angle DNF$?
I'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed.
-
I have given an argument why your conjecture is true. I am more interested in why you looked at this problem and how you came up with this neat little conjecture. – user17762 Jun 5 '12 at 5:29
If you want a purely Euclidean-geometry proof, you might start by noting the similarity of triangles $EBC$, $ENB$, and $BNC$, and then trying to prove that triangles $BNF$ and $CND$ are similar. I'm not sure exactly how the proof will go, but it feels doable. – Rahul Jun 5 '12 at 5:32
@Marvis I'm in a 2nd year geometry course, and this is part of my homework due on Friday. I'm being a keener and getting it done early :) I just guessed on the right angle part. – Charlie Jun 5 '12 at 5:46
$\angle DNF = 90^\circ \Longleftrightarrow \angle BNF = \angle CND$. It suffics to prove that $\triangle BNF \sim \triangle CND$. Well, it's trivial: $\angle NBF = \angle BEC = \angle NCD$ and $\displaystyle \frac{NB}{BF} = \frac{NB}{BE} = \sin\angle NEB = \cos\angle NCB = \frac{NC}{CB} = \frac{NC}{CD}$. Q.E.D
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That $NB/BE = NC/CB$ follows from the similarity of right triangles $ENB$ and $BNC$ without need for trigonometry. Nice proof! – Rahul Jun 5 '12 at 7:16
@RahulNarain Trigonometry is an abbreviation of similiar triangles here because we don't use angle transformation formulas such as $\sin(\alpha+\beta)=\cdots$. – Frank Science Jun 5 '12 at 7:20
Indeed; what I meant to say was that the use of trigonometric notation is not needed. It suggests that more powerful machinery is being employed than actually is. – Rahul Jun 5 '12 at 9:12
Firstly We can write
(1) Oklid relation from $\triangle CNB$
$h^2=m(k+x)$ | {
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Firstly We can write
(1) Oklid relation from $\triangle CNB$
$h^2=m(k+x)$
(2) Pisagor relation from $\triangle DPN$
$a^2=(x+k)^2+(x+k+m-h)^2$
(3) Pisagor relation from $\triangle FRN$
$b^2=h^2+k^2$
(4) Pisagor relation from $\triangle DCF$
$c^2=x^2+(x+k+m)^2$
if DNF triange is a right triangle,It must satisfy $a^2+b^2=c^2$.
$(x+k)^2+(x+k+m-h)^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2+(x+k+m)^2-2h(x+k+m)+h^2+h^2+k^2=x^2+(x+k+m)^2$
$(x+k)^2-2h(x+k+m)+2h^2+k^2=x^2$
$xk+k^2-h(x+k+m)+h^2=0$
$xk+k^2-h(x+k+m)+m(k+x)=0$
$-h(x+k+m)+(k+m)(k+x)=0$
$\frac{x+k}{x+k+m}=\frac{h}{k+m}$
This result is equal to the rates of thales formula for similar triangles $\triangle CRN \sim \triangle CBE$
Thus $a^2+b^2=c^2$ is correct for $\triangle DNF$ .
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Let us do it through coordinate geometry. Let $B$ be the origin. Lets fix the coordinates first. $$A = (0,a)\\ B = (0,0) \\ C = (a,0)\\ D = (a,a)$$ Since $E$ and $F$ are equidistant from $B$, lets say $$F = (b,0)\\ E = (0,b)$$ where $0 \leq b \leq a$. The equation of the line $CE$ is $\dfrac{x}{a} + \dfrac{y}{b} = 1$. The equation of $BN$ is $y = \dfrac{a}{b}x$. This gives us the coordinate of $N$ as $\left( \dfrac{ab^2}{a^2 + b^2},\dfrac{a^2b}{a^2 + b^2} \right)$.
The slope of $FN$ is $m_1 = \dfrac{a^2b/(a^2+b^2) - 0}{ab^2/(a^2+b^2)-b} = \dfrac{a^2}{ab - a^2 - b^2}$.
The slope of $DN$ is $m_2 = \dfrac{a^2b/(a^2+b^2) - a}{ab^2/(a^2+b^2)-a} = \dfrac{ab - a^2 - b^2}{b^2 - a^2 - b^2} = - \left(\dfrac{ab-a^2-b^2}{a^2} \right)$.
Hence, the product of the slopes is $m_1m_2 = -1$. Hence your conjecture is indeed correct. I am waiting for someone to post a nicer geometric argument.
-
HINT: Try to resort to Homothetic transformation and you're done immediately. Another way is to connect $N$ to the middle of the $DF$ (let's call that point $M$) and prove that you have there $NM$=$DM$=$MF$ (here you may resort to Apollonius' theorem). You also could resort to the cyclic quadrilaterals as another approaching way.
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# Density of rational and irrational numbers
Since $\mathbb{Q}$ is countable and $\mathbb{R} \backslash \mathbb {Q}$ is not, what does this tell us about the density of rational and irrational numbers along the real number line? Saying that there exists more irrational numbers than rational numbers seems rather vague becuase we're comparing infinites. How do we even define density here?
• There is a mathematical notion of dense, and the rationals and irrationals are each dense in the real numbers. However, it seems you are thinking of a different notion of denseness, perhaps more related to cardinality... – angryavian Nov 7 '16 at 0:31
• "what does this tell us about the density of rational and irrational numbers along the real number line?" Nothing: there are countable non-dense sets, countable dense sets, uncountable dense sets and uncountable non-dense sets. Countability and density are totally unrelated notions. – Crostul Nov 7 '16 at 0:50
• @Alephnull The Cantor set is such an example. – Crostul Nov 7 '16 at 8:09
• @Crostul But there are not co-countable nowhere dense sets (sets which are nowhere dense by have countable complement). So because $\mathbb{Q}$ is countable, we can at least conclude that $\mathbb{R}\setminus\mathbb{Q}$ is somewhere dense. – Reese Nov 7 '16 at 9:13
• @Alephnull It's pretty straightforward - a nowhere dense set by definition must exclude an entire interval. That interval cannot be countable. – Reese Nov 7 '16 at 11:03
One can have any combination of (countably infinite, uncountable) and (dense in the line, not dense in the line). (Unless, as was suggested in the comments, you have some notion of density other than the one defined by Captain Falcon.)
Countably infinite and dense in the reals: Rationals
Countably infinite and not dense in the reals: Integers
Uncountable and dense in the reals: Irrationals
Uncountable and not dense in the reals: Unit interval | {
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Uncountable and not dense in the reals: Unit interval
• Ah ah, I appreciate the reference to Captain Falcon. However, "C." stands for "Cyril" ; Falcon being my name. :) – C. Falcon Nov 7 '16 at 0:58
• Another example, a bit worse, is the Cantor set. It is uncountable and not dense, in fact it equals its closure which has (Lebesgue) length zero. – Jeppe Stig Nielsen Nov 7 '16 at 10:11
A subset $A$ of $\mathbb{R}$ is said dense if and only if any elements of $\mathbb{R}$ is a limit of a sequence of elements of $A$.
Indeed, both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$ in that sense.
As you pointed out, $\mathbb{Q}$ is countable, whereas $\mathbb{R}\setminus\mathbb{Q}$ is not, but the notion of density as nothing to do with countability.
Remark. To see that $\mathbb{Q}$ is dense in $\mathbb{R}$, let us pick $x\in\mathbb{R}$ and for all $n\in\mathbb{N}$ let us define: $$x_n:=\frac{\lfloor 10^nx\rfloor}{10^n}.$$ It is a sequence of $\mathbb{Q}$ which converges towards $x$ using squeeze theorem. For the density of $\mathbb{R}\setminus\mathbb{Q}$ consider: $$y_n:=x_n+\frac{\sqrt{2}}{n+1}.$$
• Nice answer, Captain! (salute) – Todd Wilcox Nov 7 '16 at 5:35
As clarified in other answers, cardinality by itself does not answer density questions. | {
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As clarified in other answers, cardinality by itself does not answer density questions.
Regarding the definition of density, there are two definitions. One is topological, saying that a set $A$ is dense if it intersects every non-empty open set. For the real line this is equivalent to saying that $A$ is dense (in the real line $\Bbb R$) if it intersects every open interval $(x,y)$ with $x<y$. In other words, for all choices of $x$ and $y$ with $x<y$ there is $a\in A$ with $x<a<y$. The latter condition usually is termed "order-dense" (in $\Bbb R$) and could be defined without reference to topology, for any linear order in place of $\Bbb R$. That is, $A$ is order-dense in a linear order $(L,<)$ is for all $x,y\in L$ with $x<y$ there is $a$ in $(x,y)$, that is, $x<a<y$ with $a\in A$. Finally a set $A$ is order-dense in itself if $A$ is order-dense in the linear order $(A,<)$. The set $\Bbb Q$ of all rational numbers is order-dense in the set $\Bbb R.$ The set $\Bbb Q$ is also order-dense in itself. The set $\Bbb Z$ of all integers is not order-dense in itself, and it is not order-dense in $\Bbb R$. The middle-third Cantor set $C$ is not order-dense in itself as it has "gaps", e.g. the interval $(\frac13,\frac23)$ contains no elements of $C$. (But, $C$ is (topologically) dense in itself, as it has no isolated points.) $C$ is no-where dense (in $\Bbb R$), meaning that the complement of its (topological) closure in $\Bbb R$ is (topologically) dense (and, as it happens, order-dense). | {
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If $X$ is a topological space and $A$ is a subset then we say that $A$ is dense in $X$ provided that the closure of $A$ contains $X$. When $X=\Bbb R$ this means either of the following two conditions: (1) the set $A$ intersects every non-empty open interval, or (2) for every real number $x$ there is a sequence of points of $A$ converging to $x$. On the other hand, if $A=\Bbb Z$ then $A$ is dense in itself (topologically), but not order-dense in itself. Indeed, for every $m\in\Bbb Z$ the constant sequence $\langle m,m,...\rangle$ converges to $m$. On the other hand, the interval $(m,m+1)$ contains no integers.
In case you interpreted "dense" not in the topological definition, but as saying "many more irrationals than rationals", there's another field of math that provides an alternative way of seeing this.
In Measure Theory there's the concept of the measure of a set, which is an alternative characterization of its size. We want a way to define a function on sets, so let $X\subseteq 2^\mathbb{R}$ be a collection of sets (I have a specific one in mind here, namely the Lebesgue $\sigma$-algebra, but don't worry about that). Now, define the function: $$\mu:X\to\mathbb{R}_{\geq 0}\cup\lbrace\infty\rbrace$$ that must fulfill three axioms:
1. Non-negativity. For any subset $E\in X$, we have that $\mu(E)\geq 0$. Intuitively, things can't have negative size.
2. $\mu(\emptyset) = 0$. This should also seem intuitive, "the lack of a set" has no measure (there will be other sets with no measure though!). | {
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3. Countable additivity on disjoint subsets. If $E_1,E_2$ are disjoint subsets of $\mathbb R$ (so $E_1\cap E_2 = \emptyset$), then: $$\mu(E_1\cup E_2) = \sum_{i = 1}^2 \mu(E_i)$$ This will actually be true for countably many disjoint subsets, so if $\lbrace E_i\rbrace$ are all pairwise disjoint, then: $$\mu(\cup_{i= 1}^\infty E_i)= \sum_{i = 1}^\infty \mu(E_i)$$ For finite sums, the intuition for this should be clear (think about each set as a circle, like how you make a Venn diagram. If there's no overlap between circles, the total "area" should be just all the individual areas added together).
Unfortunately, the above identity working for countable sums is less clearly true to our intuition. If we accept it though, we get some interesting results. The most applicable one to this is that we can calculate: $$\mu(\mathbb R\setminus\mathbb Q) = \infty,\quad\mu(\mathbb Q) = 0$$
So, in terms of measure theory, there are essentially no rational numbers, but infinitely many irrationals. | {
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What is the probability that $X<Y$?
If two dice are rolled repeatedly, and $X$ is the number of tosses until $3$ is rolled, and $Y$ is the number of tosses until a $5$ is rolled, what is the probability$(X < Y)$? Also, if $Z$ is the number of rolls until a $10$ is rolled, what the probability$(X < Y < Z)$?
Is $P(X < Y) = \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}((32/36)^{(n-1)}(4/36)-(34/36)^{(k-1)}(2/36)]$?
-
Is it assumed that $X$, $Y$ and $Z$ are independent ? – Sasha Sep 23 '11 at 16:50
Yes X Y and Z are independent as the dice are balanced – lord12 Sep 23 '11 at 16:50
No, $X$, $Y$ and $Z$ are not independent. For example, if $X = 1$ then $Y \ne 1$ and $Z \ne 1$. – Robert Israel Sep 23 '11 at 18:40
It's not clear whether "a 3 is rolled" is referring to a 3 coming up on at least one of the two dice, or the sum of the two being 3. But since "a 10 is rolled" can only refer to the sum, I guess we can assume the 3 and 5 are also referring to the sum. – Robert Israel Sep 23 '11 at 18:43
@Robert: Those were my thoughts too. But also notice that the first line says "until 3 is rolled" and not "until a 3 is rolled". So I suppose this indeed refers to the sum being 3, rather than one of them being 3. – TMM Sep 23 '11 at 18:47
(a) $P(X<Y)$
Suppose $q = P(3 \text{ before } 5)$. When will we see a $3$ before we see a $5$? That is if we first see a $3$ on the first roll (we are done), or if we don't see a $3$ or a $5$ on the first roll and then see a $3$ before a $5$ after that. In other words:
$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q$$
Since we know that $P(3) = \frac{2}{36}$, $P(5) = \frac{4}{36}$ we get $P(\text{neither } 3 \text{ nor } 5) = 1 - \frac{2}{36} - \frac{4}{36} = \frac{30}{36}$, so
$$q = \frac{2}{36} + \frac{30}{36} q \quad \Longleftrightarrow \quad q = \frac{1}{3}$$
Note that we can also rewrite the first equation as | {
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Note that we can also rewrite the first equation as
$$q = P(3) \cdot 1 + P(5) \cdot 0 + P(\text{neither } 3 \text{ nor } 5) \cdot q \quad \Longleftrightarrow \quad q = \frac{P(3)}{P(3) + P(5)} = \frac{P(3)}{P(3 \text{ or } 5)} = P(3\ |\ 3 \text{ or } 5)$$
This is in accordance with a comment made by Dilip: We can simply ignore all throws which are neither a $3$ or a $5$, as they are irrelevant. Then the probability of seeing a $3$ first is simply the conditional probability of throwing a $3$, given that it's either a $3$ or a $5$.
(b) $P(X<Y<Z)$
We can use the same approach as above. First, let us ignore all throws which are not $3$, $5$ or $10$. We now need that the first event is a $3$, which happens with probability
$$P(X<Y,X<Z) = \frac{P(X)}{P(X \text{ or } Y \text{ or } Z)} = \frac{P(X)}{P(X) + P(Y) + P(Z)} = \frac{2}{2+4+3} = \frac{2}{9}$$
Now if this happens, we only need that in all successive events of $3,5,10$ we first see some number of $3$s, and then see a $5$ (and not a $10$ yet). But that means that after this first event, we can again ignore all events $X$, and look at the first throw resulting in either a $5$ or a $10$. Then
$$P(Y<Z | X<Y,X<Z) = \frac{P(Y)}{P(Y \text{ or } Z)} = \frac{P(Y)}{P(Y)+P(Z)} = \frac{4}{4+3} = \frac{4}{7}$$
So the combined probability is then
$$P(X<Y<Z) = P(X<Y,X<Z) \cdot P(Y<Z | X<Y, X<Z) = \frac{2}{9} \cdot \frac{4}{7} = \frac{8}{63}$$
-
Following on from @Thijs analysis, which is actually a tool called "first step analysis", we can get the probability that $P(X<Y<Z)$. So we have to wait until one of these events happens. $P(X)=\frac{2}{36}$ $P(Y)=\frac{4}{36}$ $P(Z)=\frac{3}{36}$. So we have probability of $\frac{27}{36}$ of achieving "nothing" each trial, and effectively going back to the first step.
So at effective trial one, if $X$ occurs we have success, if $Y$ or $Z$ occurs, we have failure. | {
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So at effective trial one, if $X$ occurs we have success, if $Y$ or $Z$ occurs, we have failure.
$$P(X<Y<Z)=\frac{2}{36}P(X<Y<Z|X<Y,Z)+\frac{3}{36}P(X<Y<Z|Z<X,Y)$$ $$+\frac{4}{36}P(X<Y<Z|Y<X,Z)+\frac{27}{36}P(X<Y<Z)$$
The middle two probabilities are zero, and the first is simply $P(Y<Z)$, and applying first step analysis again we have
$$P(Y<Z)=\frac{4}{7}$$
Plugging back into the previous equation and solving we get
$$P(X<Y<Z)=\frac{8}{36(7)}+\frac{27}{36}P(X<Y<Z)=\frac{8}{63}$$
-
Nice, your follow-up is more in line with my previous analysis than my own follow-up! :) – TMM Sep 23 '11 at 18:37
Here's a simpler way than one involving infinite series (such as that posted by Sasha). Roll a die until you get either a 3 or a 5. What's the probability that it's a 3? It's just the conditional probability of getting a 3, given that you've got either a 3 or a 5: $$\Pr(W=3\mid W=3\text{ or }W=5) = \frac{\Pr(W=3)}{\Pr(W=3\text{ or }W=5)} = \frac{2/36}{6/36}= \frac13.$$
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11/9/19
# Evaluating Arbitrary Precision Integer Expressions in Julia using Metaprogramming
While watching the Mathologer masterclass on power sums
I came across a challenge to evaluate the following sum
$$1^{10}+2^{10}+\cdots+1000^{10}$$
This can be easily evaluated via brute force in Julia to yield
function power_cal(n)
a=big(0)
for i=1:n
a+=big(i)^10
end
a
end
julia> power_cal(1000)
91409924241424243424241924242500
Note the I had to use big to makes sure we are using BigInt in the computation. Without that, we would be quickly running in into an overflow issue and we will get a very wrong number.
In the comment section of the video I found a very elegant solution to the above sum, expressed as
(1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 – 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000 = 91409924241424243424241924242500
If I try to plug this into the Julia, I get
julia> (1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000
-6.740310541071357e18
This negative answer is not surprising at all, because we obviously ran into an overflow. We can, of course, go through that expression and modify all instances of Int64 with BigInt by wrapping it in the big function. But that would be cumbersome to do by hand.
## The power of Metaprogramming
In Julia, metaprogramming allows you to write code that creates code, the idea here to manipulate the abstract syntax tree (AST) of a Julia expression. We start to by “quoting” our original mathematical expressing into a Julia expression. In the at form it is not evaluated yet, however we can always evaluate it via eval.
julia> ex1=:((1/11) * 1000^11 + (1/2) * 1000^10 + (5/6) * 1000^9 - 1000^7 + 1000^5-(1/2) * 1000^3 + (5/66) * 1000)
:((((((1 / 11) * 1000 ^ 11 + (1 / 2) * 1000 ^ 10 + (5 / 6) * 1000 ^ 9) - 1000 ^ 7) + 1000 ^ 5) - (1 / 2) * 1000 ^ 3) + (5 / 66) * 1000) | {
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julia> dump(ex1)
Expr
args: Array{Any}((3,))
1: Symbol +
2: Expr
args: Array{Any}((3,))
1: Symbol -
2: Expr
args: Array{Any}((3,))
1: Symbol +
2: Expr
args: Array{Any}((3,))
1: Symbol -
2: Expr
3: Expr
3: Expr
args: Array{Any}((3,))
1: Symbol ^
2: Int64 1000
3: Int64 5
julia> eval(ex1)
-6.740310541071357e18
The output of dump show the full AST in all its glory (…well almost the depth is a bit truncated). Notice that here all our numbers are interpreted as Int64.
Now we walk through the AST and change all occurrences of Int64 with BigInt by using the big function.
function makeIntBig!(ex::Expr)
args=ex.args
for i in eachindex(args)
if args[i] isa Int64
args[i]=big(args[i])
end
if args[i] isa Expr
makeIntBig!(args[i])
end
end
end
julia> ex2=copy(ex1)
:((((((1 / 11) * 1000 ^ 11 + (1 / 2) * 1000 ^ 10 + (5 / 6) * 1000 ^ 9) - 1000 ^ 7) + 1000 ^ 5) - (1 / 2) * 1000 ^ 3) + (5 / 66) * 1000)
julia> makeIntBig!(ex2)
julia> eval(ex2)
9.14099242414242434242419242425000000000000000000000000000000000000000000000014e+31
We see an improvement here, but the results are not very satisfactory. The divisions yield BigFloat results, which had a tiny bit of floating point errors. Can we do better?
Julia has support for Rational expressions baked in. We can use that improve the results. We just need to search for call expressions the / symbol and replace it by the // symbol. For safety we just have to makes sure the operands are as subtype of Integer.
function makeIntBig!(ex::Expr)
args=ex.args
if ex.head == :call && args[1]==:/ &&
length(args)==3 &&
all(x->typeof(x) <: Integer,args[2:end])
args[1]=://
args[2]=big(args[2])
args[3]=big(args[3])
else
for i in eachindex(args)
if args[i] isa Int64
args[i]=big(args[i])
end
if args[i] isa Expr
makeIntBig!(args[i])
end
end
end
end
julia> ex2=copy(ex1);
julia> makeIntBig!(ex2)
julia> eval(ex2)
91409924241424243424241924242500//1
Now that is much better! We have not lost any precision and we ended us with a Rational expression. | {
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