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Given the following continued fraction

$$F(x) =\cfrac{1}{x+\cfrac{2^2(2^2-1)}{6x+\cfrac{3^2(3^2-1)}{12x+\cfrac{4^2(4^2-1)}{20x+\cfrac{5^2(5^2-1)}{30x+\ddots}}}}}=\frac{1}{\sqrt{x^2+4}}$$

Then $$\frac{1}{x}F\left(\frac{1}{x}\right)=\sum^{\infty}_{n=0}(-1)^{n} \binom{2n}{n} x^{2n}$$

Where $\binom{2n}{n}$ are central binomial coefficients

How do we prove that the given continued fraction is a generating function of central binomial coefficients?

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    $\begingroup$ You stated those closed forms for the continued fractions, but which of them do you already know are true, and which do you want to prove? $\endgroup$ – Bart Michels May 23 '16 at 18:30
  • $\begingroup$ So, you are effectively asking how to differentiate a continued fraction, then? $\endgroup$ – J. M. is a poor mathematician May 23 '16 at 18:34
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$F(x)$ can be rewritten as $\displaystyle\;\frac{1}{\frac{2}{P(x)} - x}$ where $\displaystyle\;\def\CF{\mathop{\LARGE\mathrm K}} P(x) = \cfrac{1\cdot 2}{1\cdot 2 x + \cfrac{ (1 \cdot 2)(2\cdot 3)}{2\cdot 3 x + \cfrac{(2\cdot 3)(3\cdot 4)}{3\cdot 4 x + \ddots} }} $.

The CF $P(x)$ has the form $ \displaystyle\; \CF_{\ell=1}^{\infty} \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_\ell} $ where $\gamma_0 = 1$ and $\displaystyle\; \begin{cases} \alpha_\ell &= \ell (\ell+1)\\ \beta_\ell &= \ell (\ell+1) x\\ \gamma_\ell &= \ell (\ell+1) \end{cases} $ for $\ell > 0$.

In general, CF of this form is invariant when we scale all $\alpha_\ell, \beta_\ell, \gamma_\ell$ by same factor for all $\ell > 0$.

If we scale $\alpha_\ell, \beta_\ell, \gamma_\ell$ by $\ell(\ell+1)$ for all $\ell > 0$, we find

$$P(x) = \CF_{\ell=1}^{\infty} \frac{1}{x} = \cfrac{1}{x + \cfrac{1}{x + \cfrac{1}{x + \ddots}}}$$

The CF at RHS is well known. It is not hard to verify its convergents have the form: $$ \CF_{\ell=1}^{n} \frac{1}{x} = \lambda_{+}\lambda_{-}\frac{\lambda_{-}^n - \lambda_{+}^n}{\lambda_{-}^{n+1} - \lambda_{+}^{n+1}} \quad\text{ where }\quad \lambda_{\pm} = \frac{-x \pm \sqrt{x^2+4}}{2}$$ When $x > 0$, we have $|\lambda_{-}| > |\lambda_{+}|$. This implies $$\begin{align} P(x) &= \lim_{n\to\infty} \CF_{\ell=1}^{n} \frac{1}{x} = \lambda_{+} = \frac{\sqrt{x^2+4} - x}{2}\\ \implies F(x) &= \frac{1}{\frac{4}{\sqrt{x^2+4}-x} - x} = \frac{1}{(\sqrt{x^2+4}+x) - x} = \frac{1}{\sqrt{x^2+4}} \end{align} $$

Recall for any $\alpha \in \mathbb{C}$ and $|z| < 1$, we have

$$\frac{1}{(1-z)^{\alpha}} = \sum_{k=0}^\infty \frac{(\alpha)_k}{k!} z^k$$ where $(\alpha)_k = \alpha(\alpha+1)\cdots(\alpha+k-1)$. When $\alpha = \frac12$, the coefficient for $z^k$ becomes $$\frac{(\frac12)_k}{k!} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(j + \frac12\right) = \frac{(2k-1)!!}{2^k k!} = \frac{(2k)!}{4^k(k!)^2} = \frac{1}{4^k} \binom{2k}{k}$$ From this, we find $$\frac{1}{\sqrt{1-4z}} = \sum_{k=0}^\infty \binom{2k}{k} z^k \quad\implies\quad \frac1x F\left(\frac1x\right) = \frac{1}{\sqrt{1+4x^2}} = \sum_{k=0}^\infty (-1)^k \binom{2k}{k} x^{2k}$$

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  • $\begingroup$ @ achille hui: would you be interested in this problem? $\endgroup$ – Nicco Aug 8 '16 at 22:34
  • $\begingroup$ @Nicco I have looked at that before ( a week ago?) but nothing seems to work... $\endgroup$ – achille hui Aug 9 '16 at 16:37

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