How to use the generalized binomial theorem to produce the power series of $(1-x)^{1/2}$

Question

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.

also related: http://math.stackexchange.com/q/124293/ http://math.stackexchange.com/q/135894/ — Grigory M, Jul 05 '16 at 14:09

score 2 · Accepted Answer · answered Jul 05 '16 at 14:31

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!

(Thank you) Factorial – Antoni Parellada Jul 05 '16 at 15:25 — Antoni Parellada, Jul 05 '16 at 15:25
You're welcome Toni. My pleasure. -Mark – Mark Viola Jul 05 '16 at 15:29 — Mark Viola, Jul 05 '16 at 15:29

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

1

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.

edited Apr 13 '17 at 12:21

Community

1

answered Jul 05 '16 at 14:09

pancini

19,216

How to use the generalized binomial theorem to produce the power series of $(1-x)^{1/2}$

2 Answers2