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I have this generating function:

$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right) \left( \,{ \frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$

and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\sqrt {1-4\,z}}{2z}$ is the generating function for the sequence $\frac {1}{n+1}\binom{2n} {n}$.

Now, I thought that I could substitute those in there, and where they multiply I'll use a summation like this:

$\frac{1}{2}\left( 1-\frac{1}{n+1}\binom{2n} {n}-\binom{2n} {n} + \sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k}\binom{2(n-k)}{n-k} \right)$

Could this be right? It doesn't seem to work when I try in Maple. What else could I do?

I already know that the end sequence will be $\binom{2n-1}{n-2}$ if this can help...

  • 0
    @Adam: I like the [Iverson bracket](http://en.wikipedia.org/wiki/Iverson_bracket) for that purpose.2011-12-05

4 Answers 4

3

Expanding the product, and after some algebra: $ g(z) = \frac{1}{2} \left(\frac{\sqrt{1-4 z}-1}{z}+\frac{1}{\sqrt{1-4 z}}+1\right) $ Using generalized binomial theorem: $ [z]^n g(z) = \frac{1}{2} \left( \delta_{n,0} + \binom{1/2}{n+1} (-4)^{n+1} + \binom{-1/2}{n} (-4)^n \right) $ Using $ \begin{eqnarray} \binom{-1/2}{n} - 4 \binom{1/2}{n+1} &=& \frac{\Gamma(1/2)}{n! \Gamma(1/2-n)} - \frac{\Gamma(3/2)}{(n+1)! \Gamma(1/2-n)} \\ &=& (-1)^n (n-1) \frac{\Gamma(1/2+n) \Gamma(n+1)}{\Gamma(1/2) (n+1)!n!} \\ &=& (-1)^n (n-1) \frac{(2n)!}{4^n (n+1)!n!} \\ &=& (-1)^n \frac{(2n-1)!}{2 \cdot 4^{n-1} (n+1)!(n-2)!} = 2 \cdot \frac{(-1)^n}{4^n} \cdot \binom{2n-1}{n+1} \end{eqnarray} $ Hence, using Iverson's bracket: $ [z]^n g(z) = \binom{2n-1}{n+1} \left[ n \ge 2 \right] $

3

It’s not necessary to use the generalized binomial theorem and the gamma function. Let $g(x)=\frac12\left(\frac1{\sqrt{1-4x}}-1\right) \left(\frac{1-\sqrt{1-4x}}{2x}-1\right)$ and $u=\sqrt{1-4x}$. Then

$\begin{align*} g(x)&= \frac12\left(\frac1u-1\right)\left(\frac{1-u}{2x}-1\right)\\ &=\frac12\left(\frac{1-u}{2xu}-\frac{1-u}{2x}-\frac1u+1\right)\\ &=\frac12\left(\frac1{2xu}-\frac1x+\frac{u}{2x}-\frac1u+1\right)\\ &=\frac12\left(\frac{1+u^2}{2xu}-\frac1x-\frac1u+1\right)\\ &=\frac12\left(\frac{1-2x}{xu}-\frac1x-\frac1u+1\right)\\ &=\frac12\left(\frac1{xu}-\frac1x-\frac3u+1\right)\\ &=\frac12\left(\frac1x\left(\frac1u-1\right)+1-\frac3u\right)\\ &=\frac12\left(\sum_{k\ge 1}\binom{2k}k x^{k-1}+1-3\sum_{k\ge 0}\binom{2k}k x^k\right)\\ &=\frac12\left(1+\sum_{k\ge 0}\left(\binom{2k+2}{k+1}-3\binom{2k}k\right)x^k\right)\\ &=\frac12\sum_{k\ge 2}\left(\binom{2k+2}{k+1}-3\binom{2k}k\right)x^k. \end{align*}$

Now

$\begin{align*} \binom{2k+2}{k+1}-3\binom{2k}k&=\binom{2k+1}k+\binom{2k+1}{k+1}-3\binom{2k}k\\ &=\binom{2k+1}k+\binom{2k}k+\binom{2k}{k-1}-3\binom{2k}k\\ &=\binom{2k+1}k+\binom{2k}{k+1}-2\binom{2k}k\\ &=\binom{2k}{k-1}+\binom{2k}k+\binom{2k}{k+1}-2\binom{2k}k\\ &=\binom{2k}{k-1}-\binom{2k}k+\binom{2k}{k+1}\\ &=\binom{2k}{k-1}-\binom{2k}k+\binom{2k-1}k+\binom{2k-1}{k+1}\\ &=\binom{2k}{k-1}-\binom{2k-1}{k-1}-\binom{2k-1}k+\binom{2k-1}k+\binom{2k-1}{k+1}\\ &=\binom{2k}{k-1}-\binom{2k-1}{k-1}+\binom{2k-1}{k+1}\\ &=\binom{2k-1}{k-2}+\binom{2k-1}{k+1}\\ &=2\binom{2k-1}{k-2}, \end{align*}$

so $\displaystyle[x^k]g(x)=\binom{2k-1}{k-2}$, as desired.

1

Expand out the product, and look at what each term is the generating function for.

1

With this type of problem Lagrange inversion is the preferred approach. Suppose we seek to extract coefficients from $Q(z) = \frac{1}{2} \left(\frac{1}{\sqrt{1-4z}}-1\right) \left(\frac{1-\sqrt{1-4z}}{2z}-1\right).$

The closed form for the coefficients is $[z^n] Q(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2} \left(\frac{1}{\sqrt{1-4z}}-1\right) \left(\frac{1-\sqrt{1-4z}}{2z}-1\right) \; dz.$

Now put $1-4z=w^2$ so that $1/4-z=1/4 \times w^2$ or $z=1/4\times(1-w^2)$ and $dz = -1/2 \times w\; dw.$ This gives for the integral $-\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{w}{4} \frac{4^{n+1}}{(1-w^2)^{n+1}} \left(\frac{1}{w}-1\right) \left(\frac{1-w}{2\times 1/4\times(1-w^2)}-1\right) \; dw \\ = -\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^n}{(1-w^2)^{n+1}} (1-w) \left(\frac{2}{1+w}-1\right) \; dw \\ = -\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^n}{(1-w)^n (1+w)^{n+1}} \left(\frac{2}{1+w}-1\right) \; dw \\ = -\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^n}{(1-w)^n (1+w)^{n+1}} \frac{1-w}{1+w} \; dw \\ = -\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^n}{(1-w)^{n-1} (1+w)^{n+2}} \; dw.$

Prepare for coefficient extraction. $-\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^n}{(1-w)^{n-1} (2+(w-1))^{n+2}} \; dw \\= -\frac{1}{2\pi i} \int_{|w-1|=\epsilon} \frac{4^n}{2^{n+2}} \frac{1}{(1-w)^{n-1} (1+(w-1)/2)^{n+2}} \; dw \\= \frac{1}{2\pi i} \int_{|w-1|=\epsilon} 2^{n-2} \frac{(-1)^n}{(w-1)^{n-1}} \sum_{q\ge 0} {q+n+1\choose n+1} (-1)^q \frac{(w-1)^q}{2^q} \; dw.$

We need the coefficient $[(w-1)^{n-2}]$ which is $2^{n-2} (-1)^n {n-2+n+1\choose n+1} (-1)^{n-2} \frac{1}{2^{n-2}} = {2n-1\choose n+1}$ or alternatively ${2n-1\choose n-2}.$