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| == Generating Functions ==
| | '''William Donald Hamilton''' [[Royal Society|FRS]] (1 August 1936 – 7 March 2000) was an [[English people|English]] [[evolutionary biology|evolutionary biologist]] whom [[Richard Dawkins]] praised as one of the greatest [[evolution]]ary theorists of the 20th century.<ref>[http://www.edge.org/3rd_culture/hamilton/hamilton_index.html Obituary by Richard Dawkins – ''The Independent'' – 10 March 2000]</ref> |
| In Stanley's magnificent book ''Enumerative Combinatorics'', he comments the generating function as "the most useful but most difficult to understand method (for counting)".
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| The solution to a counting problem is usually represented as some <math>a_n</math> depending a parameter <math>n</math>. Sometimes this <math>a_n</math> is called a ''counting function'' as it is a function of the parameter <math>n</math>. <math>a_n</math> can also be treated as a infinite series:
| | Hamilton became famous through his [[Theory|theoretical]] work on [[kin selection]] and [[altruism]]. He explained its [[Genetics|genetic]] basis, and this was a key part of the gene-centered view of [[evolution]]. In doing this, he became one of the forerunners of [[sociobiology]], as popularized by [[E.O. Wilson]]. Hamilton was certainly a big influence on Dawkins. He also published important work on [[sex ratio]]s and the [[evolution of sex]]. From 1984 to his death in 2000, he was the [[Royal Society]] Research Professor at [[Oxford University]]. He died of [[malaria]] contracted in the [[Democratic Republic of the Congo]]. |
| :<math>a_0,a_1,a_2,\ldots</math>
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| The '''ordinary generating function (OGF)''' defined by <math>a_n</math> is | | == Hamilton's equation == |
| :<math> | | Hamilton's equation describes whether or not a gene for altruistic behaviour will spread in a population.<ref>Hamilton W.D. 1996. ''Narrow roads of geneland: the collected papers of W.D. Hamilton'', vol 1. Freeman, Oxford.</ref> The gene will spread if '''r'''x'''b''' is greater than '''c''': |
| G(x)=\sum_{n\ge 0} a_nx^n.
| | :<math>rb > c \ </math> |
| </math> | | where: |
| | * <math>c \ </math> is the reproductive cost to the altruist, |
| | * <math>b \ </math> is the reproductive benefit to the recipient of the altruistic behavior, and |
| | * <math>r \ </math> is the probability, above the population average, of the individuals sharing an altruistic gene – the "degree of relatedness". |
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| So <math>G(x)=a_0+a_1x+a_2x^2+\cdots</math>. An expression in this form is called a [http://en.wikipedia.org/wiki/Formal_power_series '''formal power series'''], and <math>a_0,a_1,a_2,\ldots</math> is the sequence of '''coefficients'''.
| | == Collected papers == |
| | Hamilton started to publish his collected papers starting in 1996, with short essays giving each paper context. He died after the preparation of the second volume, so the commentaries for the third volume came from his coauthors. |
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| Furthermore, the generating function can be expanded as
| | * Hamilton W.D. 1996. ''Narrow roads of gene land vol. 1: Evolution of social behaviour''. Freeman, Oxford. ISBN 0-7167-4530-5 |
| :G(x)=<math>(\underbrace{1+\cdots+1}_{a_0})+(\underbrace{x+\cdots+x}_{a_1})+(\underbrace{x^2+\cdots+x^2}_{a_2})+\cdots+(\underbrace{x^n+\cdots+x^n}_{a_n})+\cdots</math> | | * Hamilton W.D. 2002. ''Narrow roads of gene land vol. 2: Evolution of sex''. Oxford University Press, Oxford. ISBN 0-19-850336-9 |
| so it indeed "generates" all the possible instances of the objects we want to count.
| | * Hamilton W.D. 2005. ''Narrow roads of gene land, vol. 3: Last words'' (with essays by coauthors, ed. M. Ridley). Oxford University Press, Oxford. ISBN 0-19-856690-5 |
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| Usually, we do not evaluate the generating function <math>GF(x)</math> on any particular value. <math>x</math> remains as a '''formal variable''' without assuming any value. The numbers that we want to count are the coefficients carried by the terms in the formal power series. So far the generating function is just another way to represent the sequence
| | == References == |
| :<math>(a_0,a_1,a_2,\ldots\ldots)</math>.
| | {{Reflist}} |
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| The true power of generating functions comes from the various algebraic operations that we can perform on these generating functions. We use an example to demonstrate this.
| | {{DEFAULTSORT:Hamilton, William Donald}} |
| | | [[Category:1936 births]] |
| === Combinations ===
| | [[Category:2000 deaths]] |
| Suppose we wish to enumerate all subsets of an <math>n</math>-set. To construct a subset, we specifies for every element of the <math>n</math>-set whether the element is chosen or not. Let us denote the choice to omit an element by <math>x_0</math>, and the choice to include it by <math>x_1</math>. Using "<math>+</math>" to represent "OR", and using the multiplication to denote "AND", the choices of subsets of the <math>n</math>-set are expressed as
| | [[Category:English mathematicians]] |
| :<math>\underbrace{(x_0+x_1)(x_0+x_1)\cdots (x_0+x_1)}_{n\mbox{ elements}}=(x_0+x_1)^n</math>.
| | [[Category:Geneticists]] |
| | | [[Category:English evolutionary biologists]] |
| For example, when <math>n=3</math>, we have
| | [[Category:Fellows of the Royal Society]] |
| :<math>\begin{align}
| |
| (x_0+x_1)^3
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| &=x_0x_0x_0+x_0x_0x_1+x_0x_1x_0+x_0x_1x_1\\
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| &\quad +x_1x_0x_0+x_1x_0x_1+x_1x_1x_0+x_1x_1x_1
| |
| \end{align}</math>.
| |
| | |
| So it "generate" all subsets of the 3-set. Writing <math>1</math> for <math>x_0</math> and <math>x</math> for <math>x_1</math>, we have <math>(1+x)^3=1+3x+3x^2+x^3</math>. The coefficient of <math>x^k</math> is the number of <math>k</math>-subsets of a 3-element set.
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| In general, <math>(1+x)^n</math> has the coefficients which are the number of subsets of fixed sizes of an <math>n</math>-element set.
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| -----
| |
| | |
| Suppose that we have twelve balls: <font color="red">3 red</font>, <font color="blue">4 blue</font>, and <font color="green">5 green</font>. Balls with the same color are indistinguishable.
| |
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| We want to determine the number of ways to select <math>k</math> balls from these twelve balls, for some <math>0\le k\le 12</math>.
| |
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| The generating function of this sequence is
| |
| :<math>\begin{align}
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| &\quad {\color{Red}(1+x+x^2+x^3)}{\color{Blue}(1+x+x^2+x^3+x^4)}{\color{OliveGreen}(1+x+x^2+x^3+x^4+x^5)}\\
| |
| &=1+3x+6x^2+10x^3+14x^4+17x^5+18x^6+17x^7+14x^8+10x^9+6x^{10}+3x^{11}+x^{12}.
| |
| \end{align}</math>
| |
| The coefficient of <math>x^k</math> gives the number of ways to select <math>k</math> balls.
| |
| | |
| === Fibonacci numbers ===
| |
| Consider the following counting problems.
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| * Count the number of ways that the nonnegative integer <math>n</math> can be written as a sum of ones and twos (in order).
| |
| : The problem asks for the number of compositions of <math>n</math> with summands from <math>\{1,2\}</math>. Formally, we are counting the number of tuples <math>(x_1,x_2,\ldots,x_k)</math> for some <math>k\le n</math> such that <math>x_i\in\{1,2\}</math> and <math>x_1+x_2+\cdots+x_k=n</math>. | |
| : Let <math>F_n</math> be the solution. We observe that a composition either starts with a 1, in which case the rest is a composition of <math>n-1</math>; or starts with a 2, in which case the rest is a composition of <math>n-2</math>. So we have the recursion for <math>F_n</math> that
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| ::<math>F_n=F_{n-1}+F_{n-2}</math>.
| |
| * Count the ways to completely cover a <math>2\times n</math> rectangle with <math>2\times 1</math> dominos without any overlaps.
| |
| : Dominos are identical <math>2\times 1</math> rectangles, so that only their orientations --- vertical or horizontal matter.
| |
| : Let <math>F_n</math> be the solution. It also holds that <math>F_n=F_{n-1}+F_{n-2}</math>. The proof is left as an exercise.
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| In both problems, the solution is given by <math>F_n</math> which satisfies the following recursion.
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| :<math>F_n=\begin{cases}
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| F_{n-1}+F_{n-2} & \mbox{if }n\ge 2,\\
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| 1 & \mbox{if }n=1\\
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| 0 & \mbox{if }n=0.
| |
| \end{cases}</math>
| |
| | |
| <math>F_n</math> is called the [http://en.wikipedia.org/wiki/Fibonacci_number Fibonacci number].
| |
| | |
| {{Theorem|Theorem|
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| ::<math>F_n=\frac{1}{\sqrt{5}}\left(\phi^n-\hat{\phi}^n\right)</math>,
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| :where <math>\phi=\frac{1+\sqrt{5}}{2}</math> and <math>\hat{\phi}=\frac{1-\sqrt{5}}{2}</math>.
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| }}
| |
| The quantity <math>\phi=\frac{1+\sqrt{5}}{2}</math> is the so-called [http://en.wikipedia.org/wiki/Golden_ratio golden ratio], a constant with some significance in mathematics and aesthetics.
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| We now prove this theorem by using generating functions.
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| The ordinary generating function for the Fibonacci number <math>F_{n}</math> is
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| :<math>G(x)=\sum_{n\ge 0}F_n x^n</math>.
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| We have that <math>F_{n}=F_{n-1}+F_{n-2}</math> for <math>n\ge 2</math>, thus
| |
| :<math>\begin{align}
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| G(x)
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| &=
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| \sum_{n\ge 0}F_n x^n
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| &=
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| F_0+F_1x+\sum_{n\ge 2}F_n x^n
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| &=
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| x+\sum_{n\ge 2}(F_{n-1}+F_{n-2})x^n.
| |
| \end{align}
| |
| </math>
| |
| For generating functions, there are general ways to generate <math>F_{n-1}</math> and <math>F_{n-2}</math>, or the coefficients with any smaller indices.
| |
| :<math>
| |
| \begin{align}
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| xG(x)
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| &=\sum_{n\ge 0}F_n x^{n+1}=\sum_{n\ge 1}F_{n-1} x^n=\sum_{n\ge 2}F_{n-1} x^n\\
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| x^2G(x)
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| &=\sum_{n\ge 0}F_n x^{n+2}=\sum_{n\ge 2}F_{n-2} x^n.
| |
| \end{align}
| |
| </math>
| |
| So we have
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| :<math>G(x)=x+(x+x^2)G(x)\,</math>,
| |
| hence
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| :<math>G(x)=\frac{x}{1-x-x^2}</math>.
| |
| The value of <math>F_n</math> is the coefficient of <math>x^n</math> in the Taylor series for this formular, which is <math>\frac{G^{(n)}(0)}{n!}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n</math>. Although this expansion works in principle, the detailed calculus is rather painful.
| |
| | |
| ----
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| It is easier to expand the generating function by breaking it into two geometric series.
| |
| {{Theorem|Proposition|
| |
| :Let <math>\phi=\frac{1+\sqrt{5}}{2}</math> and <math>\hat{\phi}=\frac{1-\sqrt{5}}{2}</math>. It holds that
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| ::<math>\frac{x}{1-x-x^2}=\frac{1}{\sqrt{5}}\cdot\frac{1}{1-\phi x}-\frac{1}{\sqrt{5}}\cdot\frac{1}{1-\hat{\phi} x}</math>.
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| }}
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| It is easy to verify the above equation, but to deduce it, we need some (high school) calculation.
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| {|border="2" width="100%" cellspacing="4" cellpadding="3" rules="all" style="margin:1em 1em 1em 0; border:solid 1px #AAAAAA; border-collapse:collapse;empty-cells:show;"
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| :{|
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| <math>1-x-x^2</math> has two roots <math>\frac{-1\pm\sqrt{5}}{2}</math>.
| |
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| Denote that <math>\phi=\frac{2}{-1+\sqrt{5}}=\frac{1+\sqrt{5}}{2}</math> and <math>\hat{\phi}=\frac{2}{-1-\sqrt{5}}=\frac{1-\sqrt{5}}{2}</math>.
| |
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| Then <math>(1-x-x^2)=(1-\phi x)(1-\hat{\phi}x)</math>, so we can write
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| :<math>
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| \begin{align}
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| \frac{x}{1-x-x^2}
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| &=\frac{x}{(1-\phi x)(1-\hat{\phi} x)}\\
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| &=\frac{\alpha}{(1-\phi x)}+\frac{\beta}{(1-\hat{\phi} x)},
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| \end{align}
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| </math>
| |
| where <math>\alpha</math> and <math>\beta</math> satisfying that
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| :<math>\begin{cases}
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| \alpha+\beta=0\\
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| \alpha\phi+\beta\hat{\phi}= -1.
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| \end{cases}</math>
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| Solving this we have that <math>\alpha=\frac{1}{\sqrt{5}}</math> and <math>\beta=-\frac{1}{\sqrt{5}}</math>. Thus,
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| :<math>G(x)=\frac{x}{1-x-x^2}=\frac{1}{\sqrt{5}}\cdot\frac{1}{1-\phi x}-\frac{1}{\sqrt{5}}\cdot\frac{1}{1-\hat{\phi} x}</math>.
| |
| |}
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| :<math>\square</math>
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| |}
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| Note that the expression <math>\frac{1}{1-z}</math> has a well known geometric expansion:
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| :<math>\frac{1}{1-z}=\sum_{n\ge 0}z^n</math>.
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| Therefore, <math>G(x)</math> can be expanded as
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| :<math>
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| \begin{align}
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| G(x)
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| &=\frac{1}{\sqrt{5}}\cdot\frac{1}{1-\phi x}-\frac{1}{\sqrt{5}}\cdot\frac{1}{1-\hat{\phi} x}\\
| |
| &=\frac{1}{\sqrt{5}}\sum_{n\ge 0}(\phi x)^n-\frac{1}{\sqrt{5}}\sum_{n\ge 0}(\hat{\phi} x)^n\\
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| &=\sum_{n\ge 0}\frac{1}{\sqrt{5}}\left(\phi^n-\hat{\phi}^n\right)x^n.
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| \end{align}</math>
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| So the <math>n</math>th Fibonacci number is given by
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| :<math>F_n=\frac{1}{\sqrt{5}}\left(\phi^n-\hat{\phi}^n\right)=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n</math>.
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| == Solving recurrences ==
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| The following steps describe a general methodology of solving recurrences by generating functions.
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| :1. Give a recursion that computes <math>a_n</math>. In the case of Fibonacci sequence
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| ::<math>a_n=a_{n-1}+a_{n-2}</math>.
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| :2. Multiply both sides of the equation by <math>x^n</math> and sum over all <math>n</math>. This gives the generating function
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| ::<math>G(x)=\sum_{n\ge 0}a_nx^n=\sum_{n\ge 0}(a_{n-1}+a_{n-2})x^n</math>.
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| :: And manipulate the right hand side of the equation so that it becomes some other expression involving <math>G(x)</math>.
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| ::<math>G(x)=x+(x+x^2)G(x)\,</math>.
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| :3. Solve the resulting equation to derive an explicit formula for <math>G(x)</math>.
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| ::<math>G(x)=\frac{x}{1-x-x^2}</math>.
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| :4. Expand <math>G(x)</math> into a power series and read off the coefficient of <math>x^n</math>, which is a closed form for <math>a_n</math>.
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| The first step is usually established by combinatorial observations, or explicitly given by the problem. The third step is trivial.
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| The second and the forth steps need some non-trivial analytic techniques.
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| === Algebraic operations on generating functions ===
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| The second step in the above methodology is somehow tricky. It involves first applying the recurrence to the coefficients of <math>G(x)</math>, which is easy; and then manipulating the resulting formal power series to express it in terms of <math>G(x)</math>, which is more difficult (because it works backwards).
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| We can apply several natural algebraic operations on the formal power series.
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| {{Theorem|Generating function manipulation|
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| :Let <math>G(x)=\sum_{n\ge 0}g_nx^n</math> and <math>F(x)=\sum_{n\ge 0}f_nx^n</math>.
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| ----
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| ::<math>
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| \begin{align}
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| x^k G(x)
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| &= \sum_{n\ge k}g_{n-k}x^n, &\qquad (\mbox{integer }k\ge 0)\\
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| \frac{G(x)-\sum_{i=0}^{k-1}g_ix^i}{x^k}
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| &=\sum_{n\ge 0}g_{n+k}x^n, &\qquad (\mbox{integer }k\ge 0)\\
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| \alpha F(x)+\beta G(x)
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| &= \sum_{n\ge 0} (\alpha f_n+\beta g_n)x^n\\
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| F(x)G(x)
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| &= \sum_{n\ge 0}\sum_{k=0}^nf_kg_{n-k}x^n\\
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| G(cx)
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| &= \sum_{n\ge 0} c^ng_n x^n\\
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| G'(x)
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| &=
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| \sum_{n\ge 0}(n+1)g_{n+1}x^n
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| \end{align}
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| </math>
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| }}
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| When manipulating generating functions, these rules are applied backwards; that is, from the right-hand-side to the left-hand-side.
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| === Expanding generating functions ===
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| The last step of solving recurrences by generating function is expanding the closed form generating function <math>G(x)</math> to evaluate its <math>n</math>-th coefficient. In principle, we can always use the [http://en.wikipedia.org/wiki/Taylor_series Taylor series]
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| :<math>G(x)=\sum_{n\ge 0}\frac{G^{(n)}(0)}{n!}x^n</math>,
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| where <math>G^{(n)}(0)</math> is the value of the <math>n</math>-th derivative of <math>G(x)</math> evaluated at <math>x=0</math>.
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| Some interesting special cases are very useful.
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| ====Geometric sequence====
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| In the example of Fibonacci numbers, we use the well known geometric series:
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| :<math>\frac{1}{1-x}=\sum_{n\ge 0}x^n</math>.
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| It is useful when we can express the generating function in the form of <math>G(x)=\frac{a_1}{1-b_1x}+\frac{a_2}{1-b_2x}+\cdots+\frac{a_k}{1-b_kx}</math>. The coefficient of <math>x^n</math> in such <math>G(x)</math> is <math>a_1b_1^n+a_2b_2^n+\cdots+a_kb_k^n</math>.
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| ====Binomial theorem====
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| The <math>n</math>-th derivative of <math>(1+x)^\alpha</math> for some real <math>\alpha</math> is
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| :<math>\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)(1+x)^{\alpha-n}</math>.
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| By Taylor series, we get a generalized version of the binomial theorem known as [http://en.wikipedia.org/wiki/Binomial_coefficient#Newton.27s_binomial_series '''Newton's formula''']:
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| {{Theorem|Newton's formular (generalized binomial theorem)|
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| If <math>|x|<1</math>, then
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| :<math>(1+x)^\alpha=\sum_{n\ge 0}{\alpha\choose n}x^{n}</math>,
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| where <math>{\alpha\choose n}</math> is the '''generalized binomial coefficient''' defined by
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| :<math>{\alpha\choose n}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}</math>.
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| }}
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| === Example: multisets ===
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| In the last lecture we gave a combinatorial proof of the number of <math>k</math>-multisets on an <math>n</math>-set. Now we give a generating function approach to the problem.
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| Let <math>S=\{x_1,x_2,\ldots,x_n\}</math> be an <math>n</math>-element set. We have
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| :<math>(1+x_1+x_1^2+\cdots)(1+x_2+x_2^2+\cdots)\cdots(1+x_n+x_n^2+\cdots)=\sum_{m:S\rightarrow\mathbb{N}} \prod_{x_i\in S}x_i^{m(x_i)}</math>,
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| where each <math>m:S\rightarrow\mathbb{N}</math> species a possible multiset on <math>S</math> with multiplicity function <math>m</math>.
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| | |
| Let all <math>x_i=x</math>. Then
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| :<math>
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| \begin{align}
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| (1+x+x^2+\cdots)^n
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| &=
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| \sum_{m:S\rightarrow\mathbb{N}}x^{m(x_1)+\cdots+m(x_n)}\\
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| &=
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| \sum_{\text{multiset }M\text{ on }S}x^{|M|}\\
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| &=
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| \sum_{k\ge 0}\left({n\choose k}\right)x^k.
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| \end{align}
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| </math>
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| The last equation is due to the the definition of <math>\left({n\choose k}\right)</math>. Our task is to evaluate <math>\left({n\choose k}\right)</math>.
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| Due to the geometric sequence and the Newton's formula
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| :<math>
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| (1+x+x^2+\cdots)^n=(1-x)^{-n}=\sum_{k\ge 0}{-n\choose k}(-x)^k.
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| </math>
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| So
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| :<math>
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| \left({n\choose k}\right)=(-1)^k{-n\choose k}={n+k-1\choose k}.
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| </math>
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| The last equation is due to the definition of the generalized binomial coefficient. We use an analytic (generating function) proof to get the same result of <math>\left({n\choose k}\right)</math> as the combinatorial proof.
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| == Catalan Number ==
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| We now introduce a class of counting problems, all with the same solution, called [http://en.wikipedia.org/wiki/Catalan_number '''Catalan number'''].
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| The <math>n</math>th Catalan number is denoted as <math>C_n</math>.
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| In Volume 2 of Stanley's ''Enumerative Combinatorics'', a set of exercises describe 66 different interpretations of the Catalan numbers. We give a few examples, cited from Wikipedia.
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| * ''C''<sub>''n''</sub> is the number of '''Dyck words''' of length 2''n''. A Dyck word is a string consisting of ''n'' X's and ''n'' Y's such that no initial segment of the string has more Y's than X's (see also [http://en.wikipedia.org/wiki/Dyck_language Dyck language]). For example, the following are the Dyck words of length 6:
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| <div class="center"><big> XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY.</big></div>
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| * Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, ''C''<sub>''n''</sub> counts the number of expressions containing ''n'' pairs of parentheses which are correctly matched:
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| <div class="center"><big> ((())) ()(()) ()()() (())() (()()) </big></div>
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| * ''C''<sub>''n''</sub> is the number of different ways ''n'' + 1 factors can be completely parenthesized (or the number of ways of associating ''n'' applications of a '''binary operator'''). For ''n'' = 3, for example, we have the following five different parenthesizations of four factors:
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| <div class="center"><math>((ab)c)d \quad (a(bc))d \quad(ab)(cd) \quad a((bc)d) \quad a(b(cd))</math></div>
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| * Successive applications of a binary operator can be represented in terms of a '''full binary tree'''. (A rooted binary tree is ''full'' if every vertex has either two children or no children.) It follows that ''C''<sub>''n''</sub> is the number of full binary trees with ''n'' + 1 leaves:
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| [[Image:Catalan number binary tree example.png|center]] | |
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| * ''C''<sub>''n''</sub> is the number of '''monotonic paths''' along the edges of a grid with ''n'' × ''n'' square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up". The following diagrams show the case ''n'' = 4:
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| [[Image:Catalan number 4x4 grid example.svg.png|450px|center]] | |
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| * ''C''<sub>''n''</sub> is the number of different ways a [http://en.wikipedia.org/wiki/Convex_polygon '''convex polygon'''] with ''n'' + 2 sides can be cut into '''triangles''' by connecting vertices with straight lines. The following hexagons illustrate the case ''n'' = 4:
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| [[Image:Catalan-Hexagons-example.png|400px|center]] | |
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| * ''C''<sub>''n''</sub> is the number of [http://en.wikipedia.org/wiki/Stack_(data_structure) '''stack''']-sortable permutations of {1, ..., ''n''}. A permutation ''w'' is called '''stack-sortable''' if ''S''(''w'') = (1, ..., ''n''), where ''S''(''w'') is defined recursively as follows: write ''w'' = ''unv'' where ''n'' is the largest element in ''w'' and ''u'' and ''v'' are shorter sequences, and set ''S''(''w'') = ''S''(''u'')''S''(''v'')''n'', with ''S'' being the identity for one-element sequences.
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| * ''C''<sub>''n''</sub> is the number of ways to tile a stairstep shape of height ''n'' with ''n'' rectangles. The following figure illustrates the case ''n'' = 4:
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| [[Image:Catalan stairsteps 4.png|400px|center]]
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| === Solving the Catalan numbers ===
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| {{Theorem|Recurrence relation for Catalan numbers|
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| :<math>C_0=1</math>, and for <math>n\ge1</math>,
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| ::<math>
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| C_n=
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| \sum_{k=0}^{n-1}C_kC_{n-1-k}</math>.
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| }}
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| Let <math>G(x)=\sum_{n\ge 0}C_nx^n</math> be the generating function. Then
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| :<math>
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| \begin{align}
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| G(x)^2
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| &=\sum_{n\ge 0}\sum_{k=0}^{n}C_kC_{n-k}x^n\\
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| xG(x)^2
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| &=\sum_{n\ge 0}\sum_{k=0}^{n}C_kC_{n-k}x^{n+1}=\sum_{n\ge 1}\sum_{k=0}^{n-1}C_kC_{n-1-k}x^n.
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| \end{align}
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| </math>
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| Due to the recurrence,
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| :<math>G(x)=\sum_{n\ge 0}C_nx^n=C_0+\sum_{n\ge 1}\sum_{k=0}^{n-1}C_kC_{n-1-k}x^n=1+xG(x)^2</math>.
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| Solving <math>xG(x)^2-G(x)+1=0</math>, we obtain
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| :<math>G(x)=\frac{1\pm(1-4x)^{1/2}}{2x}</math>.
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| Only one of these functions can be the generating function for <math>C_n</math>, and it must satisfy
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| :<math>\lim_{x\rightarrow 0}G(x)=C_0=1</math>.
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| It is easy to check that the correct function is
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| :<math>G(x)=\frac{1-(1-4x)^{1/2}}{2x}</math>.
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| Expanding <math>(1-4x)^{1/2}</math> by Newton's formula,
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| :<math>
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| \begin{align}
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| (1-4x)^{1/2}
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| &=
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| \sum_{n\ge 0}{1/2\choose n}(-4x)^n\\
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| &=
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| 1+\sum_{n\ge 1}{1/2\choose n}(-4x)^n\\
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| &=
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| 1-4x\sum_{n\ge 0}{1/2\choose n+1}(-4x)^n
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| \end{align}
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| </math>
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| Then, we have
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| :<math>
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| \begin{align}
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| G(x)
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| &=
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| \frac{1-(1-4x)^{1/2}}{2x}\\
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| &=
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| 2\sum_{n\ge 0}{1/2\choose n+1}(-4x)^n
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| \end{align}
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| </math>
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| Thus,
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| :<math>
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| \begin{align}
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| C_n
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| &=2{1/2\choose n+1}(-4)^n\\
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| &=2\cdot\left(\frac{1}{2}\cdot\frac{-1}{2}\cdot\frac{-3}{2}\cdots\frac{-(2n-1)}{2}\right)\cdot\frac{1}{(n+1)!}\cdot(-4)^n\\
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| &=\frac{2^n}{(n+1)!}\prod_{k=1}^n(2k-1)\\
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| &=\frac{2^n}{(n+1)!}\prod_{k=1}^n\frac{(2k-1)2k}{2k}\\
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| &=\frac{1}{n!(n+1)!}\prod_{k=1}^n (2k-1)2k\\
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| &=\frac{(2n)!}{n!(n+1)!}\\
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| &=\frac{1}{n+1}{2n\choose n}.
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| \end{align}
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| </math>
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| So we prove the following closed form for Catalan number.
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| {{Theorem|Theorem|
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| :<math>C_n=\frac{1}{n+1}{2n\choose n}</math>.
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| }}
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| == Analysis of Quicksort ==
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| == Reference ==
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| * ''Graham, Knuth, and Patashnik'', Concrete Mathematics: A Foundation for Computer Science, Chapter 7.
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| * ''Cameron'', Combinatorics: Topics, Techniques, Algorithms, Chapter 4.
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| * ''van Lin and Wilson'', A course in combinatorics, Chapter 14.
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