How to use the generalized binomial theorem to produce the power series of $(1-x)^{1/2}$ I am trying to see how to get from $\sqrt{1-x}$ to the power series $\displaystyle\sum_{m=0}^\infty\frac{-1}{2m-1}\,{2m \choose m}\,\frac{x^m}{4^m}$, ideally using the generalized binomial theorem.
I see that the Taylor expansion evaluated at $0$ is,
$f(x)=(1-x)^{1/2}$; $f(0)=1$; and the first coefficient of the power series is $1$.
$f^1(x)= - \frac{1}{2}(1-x)^{-1/2}$; $f^1(0)=-1/2$; and the second coefficient, $-\frac{x}{2}$.
$f^2(x) = -\frac{1}{4}(1-x)^{-3/2}$; $f^2(0)=-1/4$; and the third coefficient, $-\frac{x^2}{8}$.
$f^3(x)=-\frac{3}{8}(1-x)^{-5/2}$; $f^3(0)=-3/8$; and the fourth coefficient, $-\frac{3\cdot x^3}{8\cdot3!}=-\frac{x^3}{16}$.
It is also likely that ${2m\choose m}=\frac{2m!}{m!m!}$ may be part of the derivation, but I don't see a straightforward link.
 A: The binomial theorem states
$$(1+x)^r=\sum_{k=0}^\infty {r\choose k}x^k$$
and thus
$$(1-x)^{1/2}=\sum_{k=0}^\infty (-1)^k{1/2 \choose k}x^k.$$
Your question is really, how to simplify $1/2$ choose $k$, which can be found here.
A: Let $f(x)=(1-x)^{1/2}$.  Then, differentiating we have
$$\begin{align}
f^{(1)}(x)&=(-1)\left(\frac12\right) (1-x)^{-1/2}\\\\
f^{(2)}(x)&=(-1)^2\left(\frac12\right) \left(-\frac12\right) (1-x)^{-3/2} \\\\
f^{(3)}(x)&=(-1)^3\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right) (1-x)^{-5/2}\\\\
f^{(4)}(x)&=(-1)^4\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right)\left(-\frac52\right) (1-x)^{-7/2}\\\\
\vdots\\\\
f^{(n)}(x)&=-\frac{(2n-3)!!}{2^n}(1-x)^{-(2n-1)/2} \\\\
\end{align}$$
Therefore, $f(x)$ has the series expansion
$$f(x)=1-\sum_{n=1}^\infty \frac{(2n-3)!!}{2^n\,n!}\,x^n \tag 1$$
We can express the double factorial term in terms of single factorials by writing
$$\begin{align}
(2n-3)!!&=(2n-3)(2n-5)(2n-7)\cdots(5)(3)(1)\\\\
&=\frac{(2n-3)!}{(2n-4)(2n-6)(2n-8)\cdots (6)(4)(2)}\\\\
&=\frac{(2n-3)!}{2^{n-2}(n-2)!}\\\\
&=\frac{(2n)!}{(2n-1)2^n\,n!}\\\\
&=\frac{n!}{(2n-1)2^n}\binom{2n}{n} \tag 2
\end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align}
f(x)&=1-\sum_{n=1}^\infty \frac{1}{4^n(2n-1)}\binom{2n}{n}\,x^n\\\\
&=-\sum_{n=1}^\infty \frac{1}{4^n(2n-1)}\binom{2n}{n}\,x^n
\end{align}$$
as was to be shown!
