derivation of generating function for the reciprocal central binomial coefficients

According to the article, the ordinary generating function for ${\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)}^{-1}$ 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 $C_n={\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^{-1}$, and $C(x)={\sum }_{n\ge 0}{C}_n{x}^n$ its ordinary generating function. Then

$C_{n+1}$ $={\left({\displaystyle \genfrac{}{}{0pt}{}{2n+2}{n+1}}\right)}^{-1}={\displaystyle \frac{(n+1)!(n+1)!}{(2n+2)!}}$

$={\displaystyle \frac{(n+1)(n+1)}{(2n+2)(2n+1)}}\cdot {\displaystyle \frac{n!n!}{(2n)!}}$

$={\displaystyle \frac{n+1}{2(2n+1)}}\cdot C_n$

Thus

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

so that

$$\sum _{n\ge 0}(4n+2)C_{n+1}{x}^n=\sum _{n\ge 0}(n+1)C_n{x}^n$$

A little algebra gives

$$4\sum _{n\ge 0}(n+1)C_{n+1}{x}^n-2\sum _{n\ge 0}{C}_{n+1}{x}^n=\sum _{n\ge 0}n{C}_n{x}^n+\sum _{n\ge 0}{C}_n{x}^n$$

so that

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

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

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

$={\displaystyle \frac{k\sqrt{x}}{{(4-x)}^{3/2}}}-{\displaystyle \frac{2\sqrt{x}}{{(4-x)}^{3/2}}}{\displaystyle \int \frac{\sqrt{4-x}}{x^{3/2}}𝑑x}$

$={\displaystyle \frac{k\sqrt{x}}{{(4-x)}^{3/2}}}-{\displaystyle \frac{2\sqrt{x}}{{(4-x)}^{3/2}}}\left({\displaystyle \frac{-2(4-x)}{\sqrt{x(4-x)}}}-\arcsin \left({\displaystyle \frac{x}{2}}-1\right)\right)$

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

To determine the constant $k$, note that we should have $C^{\prime }(x){|}_{x=0}=\frac{1}{2}$; 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

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

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

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:

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

$={\displaystyle \frac{\sqrt{2}}{2}}\left(\sqrt{{\displaystyle \frac{1+\cos (\arcsin (z))}{2}}}+\sqrt{{\displaystyle \frac{1-\cos (\arcsin (z))}{2}}}\right)$

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

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

Now square this expression to get

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

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

Title	derivation of generating function for the reciprocal central binomial coefficients
Canonical name	DerivationOfGeneratingFunctionForTheReciprocalCentralBinomialCoefficients
Date of creation	2013-03-22 19:04:58
Last modified on	2013-03-22 19:04:58
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	4
Author	rm50 (10146)
Entry type	Result
Classification	msc 05A10
Classification	msc 05A15
Classification	msc 05A19
Classification	msc 11B65