How can I find a generating function for the following mathematical term?
$$ a_r = \left(\matrix{2r \\ r}\right) $$
Is it the $\dfrac{r!}{2r(2r-r)!} = \dfrac{(2r-1)\cdot(2r-2)\cdot\ldots\cdot 1}{r\cdot(r-1)\cdot(r-2)\cdot\ldots\cdot1}$?
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You can get an answer by simple manipulation on the generating function for the Catalan numbers as given at http://en.wikipedia.org/wiki/Catalan_number – Gerry Myerson Apr 27 '14 at 11:50
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This is just $(1 - 4 z)^{- 1/2}$. You can check that: \begin{align} \binom{-1/2}{k} &= \frac{(-1/2) (-1/2 - 1) \ldots (-1/2 - k + 1)}{k!} \\ &= (-1)^k \frac{3 \cdot 5 \dotsm (2 k - 1)}{2^k k!} \\ &= (-1)^k \frac{1}{2^{2 k}} \frac{(2 k)!}{k! k!} \\ &= (-1)^k \frac{1}{2^{2 k}} \binom{2 k}{k} \end{align} Use this to expand the power given.
vonbrand
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Let the generating function be $G(x)$ for the following series. Now in $xG(x)'$ its coefficient of $x^r$ is $\frac{2(2r+1)2r!}{(r+1)!(r+1)!}$ which can be written as $\frac{2(2r+2)2r!}{(r)!(r)!}-\frac{2(1)2r!}{(r)!(r)!}$. Second term is $2G(x)$ and the first one is differentiation of $xG(x)$ multiplied by 2. This will give a differential equation for the generating function.
happymath
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@V-Moy the generating function for each term should not be different i.e. it should not depend on r. Just by looking at $x^r$ coefficient i should be able to tell the value for more details check the wiki page – happymath Apr 27 '14 at 09:18
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I still don't get it. Where I can learn this stuff from the basic? – Anastasiya-Romanova 秀 Apr 27 '14 at 09:21
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@V-Moy tris this http://www.cut-the-knot.org/blue/GeneratingFunctions.shtml – happymath Apr 27 '14 at 09:25
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