What kind of expansion is $\sqrt{1+u} = 1+ \frac12u + \frac18u^2+\dotso$. What kind of expansion  is this $$ \sqrt{1+u} = 1+ \frac12u - \frac18u^2+\dotso$$
Where the 8 came from ?
Thank you
 A: We know that
$$(1+x)^a=1+ax+\frac {a (a-1)}{2}x^2+\frac {a (a-1)(a-2)}{3!}x^3+.... $$
if $a=\frac {1}{2} $,  the third  term gives $\frac {-1}{8} x^2$.
A: $f(u)=\sqrt{1+u}$ is an analytic function in a neighbourhood of the origin and a solution of $f(0)=1$, $f(u)=2(1+u)\,f'(u)$. In particular, by assuming 
$$ f(u) = 1+\sum_{n\geq 1} c_n u^n \tag{A}$$
for any $u$ sufficiently close to the origin, we have
$$ 2(1+u)f'(u)=2(1+u)\sum_{n\geq 1}n c_n u^{n-1} = \sum_{n\geq 1} 2n c_n (u^n+u^{n-1})\\=2c_1+\sum_{n\geq 1}\left(2nc_n+2(n+1)c_{n+1}\right)u^n\tag{B} $$ 
hence by comparing $(A)$ and $(B)$ we get $c_1=\frac{1}{2}$ and
$$ c_n = 2n c_n + 2(n+1) c_{n+1}, \qquad c_{n+1}=-\frac{2n-1}{2n+2}c_n\tag{C}$$
so $c_2=-\frac{1}{4}c_1 = -\frac{1}{8}$ and in general $c_n=\frac{(2n-1)!!\cdot(-1)^{n+1}}{(2n-1)\cdot(2n)!!}$.
A: With $f(u)=\sqrt{1+u}$, a term in the Taylor series of $f$ at $0$ is
$$\frac{u^2}{2!} f''(0) = \frac{u^2}{2} \left[\frac{d}{du} \frac{1}{2\sqrt{1+u}}\right]_{u=0} = - \frac{u^2}{8} \left[(1+u)^{-3/2}\right]_{u=0}$$
A: It is a polynomial expansion:
$$\sqrt{1+x}=(1+x)^{1/2}=\color{blue}{a_0}+\color{red}{a_1}x+\color{green}{a_2}x^2+\color{magenta}{a_3}x^3+\cdots$$
Plug $x=0$ to find $a_0$: 
$$(1+0)^{1/2}=a_0+0+0+0+\cdots \Rightarrow \color{blue}{a_0=1}.$$
Take derivative of both sides and plug $x=0$ to find $a_1$: $$\frac12(1+0)^{-1/2}=a_1 \Rightarrow \color{red}{a_1=\frac12}.$$
Take derivative of both sides twice and plug $x=0$ to find $a_2$: 
$$-\frac14(1+0)^{-3/2}=2a_2 \Rightarrow \color{green}{a_2=-\frac18}.$$
Take derivative of both sides three times and plug $x=0$ to find $a_3$: 
$$\frac38(1+0)^{-5/2}=6a_3 \Rightarrow \color{magenta}{a_3=\frac{1}{16}}.$$
