1. ## binomial series expansion

use the binomial series to expand the function as a power series.
$\sqrt{1+x}$
I know how to start the problem but i get stuck.
using:
$(1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...$

so I rewrite $\sqrt{1+x}$ as $(1+x)^\frac{1}{2}$

therefore $(1+x)^\frac{1}{2} = \sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^n$

after that i dont know what to do

2. Originally Posted by yoman360
use the binomial series to expand the function as a power series.
$\sqrt{1+x}$
I know how to start the problem but i get stuck.
using:
$(1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...$

so I rewrite $\sqrt{1+x}$ as $(1+x)^\frac{1}{2}$

therefore $(1+x)^\frac{1}{2} = \sum_{n=0}^{\infty}\binom{\frac{1}{2}}{n}x^n$

after that i dont know what to do
Don't write it using combinations. Use the longhand version.

$(1 + x)^k = 1 + kx + \frac{k(k - 1)}{2!}x^2 + \frac{k(k - 1)(k - 2)}{3!}x^3 + \dots$

So $(1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)}{2!}x^2 + \frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)\left(\frac{1}{2} - 2\right)}{3!}x^3 + \dots$

$= 1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}x^2 + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}x^3 + \dots$

$= 1 + \frac{1}{2}x + \frac{-\frac{1}{4}}{2}x^2 + \frac{\frac{3}{8}}{6}x^3 + \dots$

$= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \dots$