Compact Form of the Taylor Series
Determine the Taylor Series $\frac{1}{\sqrt{1-x}}$ at $x=0$
I ended up with this:
$1 + \frac{1}{2}x+\frac{3}{4}x^2\frac{1}{2!}+\frac{15}{8}x^3\frac{1}{3!}+\frac{105}{16}x^4\frac{1}{4!}$
I am having trouble going from here to its compact form.
$\endgroup$ 02 Answers
$\begingroup$Catalan numbers$$ C_n = \frac{1}{n+1}\binom{2n}{n} $$ obey the recurrence relation: $$ C_{n+1} = \sum_{i=0}^{n}C_i C_{n-1} $$ hence if we set $f(z) = \sum_{n\geq 0} C_n z^n $ we get $f(z)=1+z\cdot f(z)^2$, from which: $$ \sum_{n\geq 0}\frac{(2n)!}{(n+1)!n!} z^{n} = \frac{1-\sqrt{1-4z}}{2z} $$ and: $$ \sum_{n\geq 0}\frac{(2n)!}{4^n (n+1)! n!} z^{n+1} = 2\left(1-\sqrt{1-z}\right) $$ so by differentiating both sides with respect to $z$:
$$ \frac{1}{\sqrt{1-z}} = \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} z^n.$$
That can be seen as a special case of the binomial theorem: we know that
$$ (1-z)^a = \sum_{n\geq 0}\binom{a}{n}(-1)^n z^n = \sum_{n\geq 0}\frac{\Gamma(a+1)}{\Gamma(n+1)\Gamma(a-n+1)}(-1)^n z^n $$ hence if we take $a=-\frac{1}{2}$ we get: $$ \frac{1}{\sqrt{1-z}} = \sum_{n\geq 0}\frac{\sqrt{\pi}}{\Gamma(n+1)\Gamma\left(\frac{1}{2}-n\right)}(-1)^n z^n. $$
$\endgroup$ 1 $\begingroup$Hint:
Write $\frac{1}{\sqrt{1-x}}=(1-x)^{-\frac{1}{2}}$.
Use the rule $\frac{d}{dx}f(1-x) = - f'(1-x)$ (chain rule) that can be generalized to $\frac{d^n}{d^nx}f(1-x) = (-1)^n f^{(n)}(1-x)$ and the relation $-\frac{1}{2} * -\frac{3}{2} * -\frac{5}{2} * \dots = (-1)^{n+1} \frac{(2(n+1)+1)!!}{2^{n+1}}$ where $n+1$ is the number of factors. Here, $!!$ is the double factorial defined as $b!! = b*(b-2)*(b-4)*\dots*1$ (even numbers b), $b!! = b*(b-2)*\dots*1$ (odd numbers b).
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