# derivation of generating function for the reciprocal central binomial coefficients

According to the article, the ordinary generating function for $\dbinom{2n}{n}^{-1}$ is

 $\frac{4\left(\sqrt{4-x}+\sqrt{x}\arcsin\left(\frac{\sqrt{x}}{2}\right)\right)}% {(4-x)^{3/2}}$

To see this, let $C_{n}=\binom{2n}{n}^{-1}$, and $C(x)=\sum_{n\geq 0}C_{n}x^{n}$ its ordinary generating function. Then

 $\displaystyle C_{n+1}$ $\displaystyle=\dbinom{2n+2}{n+1}^{-1}=\frac{(n+1)!(n+1)!}{(2n+2)!}$ $\displaystyle=\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\frac{n!n!}{(2n)!}$ $\displaystyle=\frac{n+1}{2(2n+1)}\cdot C_{n}$

Thus

 $(4n+2)C_{n+1}=(n+1)C_{n}$

so that

 $\sum_{n\geq 0}(4n+2)C_{n+1}x^{n}=\sum_{n\geq 0}(n+1)C_{n}x^{n}$

A little algebra gives

 $4\sum_{n\geq 0}(n+1)C_{n+1}x^{n}-2\sum_{n\geq 0}C_{n+1}x^{n}=\sum_{n\geq 0}nC_% {n}x^{n}+\sum_{n\geq 0}C_{n}x^{n}$

so that

 $4C^{\prime}(x)-\frac{2}{x}(C(x)-1)=xC^{\prime}(x)+C(x)$

and, collecting terms,

 $(4x-x^{2})C^{\prime}(x)=(x+2)C(x)-2$

We now have a first-order linear ODE to solve. Put it in the form

 $C^{\prime}(x)+\frac{-x-2}{x(4-x)}C(x)=\frac{-2}{4x-x^{2}}$

and we must now integrate the coefficient of $C(x)$. Expand by partial fractions and integrate to get

 $\int\frac{-x-2}{x(4-x)}dx=\ln\left(\frac{(4-x)^{3/2}}{\sqrt{x}}\right)$

Thus the solution to the equation is

 $\displaystyle C(x)$ $\displaystyle=\frac{\sqrt{x}}{(4-x)^{3/2}}\left(k+\int\frac{(4-x)^{3/2}}{\sqrt% {x}}\cdot\frac{-2}{x(4-x)}dx\right)$ $\displaystyle=\frac{k\sqrt{x}}{(4-x)^{3/2}}-\frac{2\sqrt{x}}{(4-x)^{3/2}}\int% \frac{\sqrt{4-x}}{x^{3/2}}dx$ $\displaystyle=\frac{k\sqrt{x}}{(4-x)^{3/2}}-\frac{2\sqrt{x}}{(4-x)^{3/2}}\left% (\frac{-2(4-x)}{\sqrt{x(4-x)}}-\arcsin\left(\frac{x}{2}-1\right)\right)$ $\displaystyle=\frac{4}{4-x}+\frac{\sqrt{x}}{(4-x)^{3/2}}\left(k+2\arcsin\left(% \frac{x}{2}-1\right)\right)$

To determine the constant $k$, note that we should have $C^{\prime}(x)\big{\lvert}_{x=0}=\frac{1}{2}$; looking at $\lim_{x\to 0}C^{\prime}(x)$ we see that for $k=\pi$ this equation holds. Thus

 $C(x)=\frac{4}{4-x}+\frac{\sqrt{x}}{(4-x)^{3/2}}\left(\pi+2\arcsin\left(\frac{x% }{2}-1\right)\right)$

We show below that the following is an identity:

 $\sqrt{\frac{z+1}{2}}=\sin\left(\frac{\pi}{4}+\frac{1}{2}\arcsin(z)\right)$

Assuming that result, substitute $\frac{x}{2}-1$ for $z$ and simplify to get

 $\frac{\sqrt{x}}{2}=\sin\left(\frac{\pi}{4}+\frac{1}{2}\arcsin\left(\frac{x}{2}% -1\right)\right)$

so that

 $4\arcsin\left(\frac{\sqrt{x}}{2}\right)=\pi+2\arcsin\left(\frac{x}{2}-1\right)$

and then

 $\displaystyle C(x)$ $\displaystyle=\frac{4}{4-x}+\frac{\sqrt{x}}{(4-x)^{3/2}}\left(4\arcsin\left(% \frac{\sqrt{x}}{2}\right)\right)$ $\displaystyle=\frac{4\left(\sqrt{4-x}+\sqrt{x}\arcsin\left(\frac{\sqrt{x}}{2}% \right)\right)}{(4-x)^{3/2}}$

as desired.

Finally, to prove the identity, first expand the right-hand using the formula for $\sin(a+b)$, and then apply the half-angle formulas:

 $\displaystyle\sin\left(\frac{\pi}{4}+\frac{1}{2}\arcsin(z)\right)$ $\displaystyle=\frac{\sqrt{2}}{2}\left(\cos\left(\frac{1}{2}\arcsin(z)\right)+% \sin\left(\frac{1}{2}\arcsin(z)\right)\right)$ $\displaystyle=\frac{\sqrt{2}}{2}\left(\sqrt{\frac{1+\cos(\arcsin(z))}{2}}+% \sqrt{\frac{1-\cos(\arcsin(z))}{2}}\right)$ $\displaystyle=\frac{\sqrt{2}}{2}\left(\sqrt{\frac{1+\sqrt{1-z^{2}}}{2}}+\sqrt{% \frac{1-\sqrt{1-z^{2}}}{2}}\right)$ $\displaystyle=\frac{1}{2}\left(\sqrt{1+\sqrt{1-z^{2}}}+\sqrt{1-\sqrt{1-z^{2}}}\right)$

Now square this expression to get

 $\frac{1}{4}\left(2+2\sqrt{1-1+z^{2}}\right)=\frac{\lvert z\rvert+1}{2}$

Thus the identity holds for $0\leq z\leq 1$; an almost identical computation using $-z$ in of $z$ shows that it also holds for $-1\leq z\leq 0$.

Title derivation of generating function  for the reciprocal central binomial coefficients  DerivationOfGeneratingFunctionForTheReciprocalCentralBinomialCoefficients 2013-03-22 19:04:58 2013-03-22 19:04:58 rm50 (10146) rm50 (10146) 4 rm50 (10146) Result msc 05A10 msc 05A15 msc 05A19 msc 11B65