Coefficient of Taylor Series of $\sqrt{1+x}$ The coefficient of $x^3$ in the Taylor series of the function $f(x) = \sqrt{1+x}$ about the point $a = 0$ is $$1\over 16$$ 
Can someone show me how to get this value?
 A: Let $f$ be an $n$ times differentiable function at $a$. Then, for any $0 \leq k \leq n$, the $k^{\text{th}}$ Taylor polynomial of $f$ at $a$ is $$P_k(x) = \sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i.$$ In your situation, you have $a = 0$; we sometimes refer to this as a Maclaurin polynomial: $$P_k(x) = \sum_{i=0}^k\frac{f^{(i)}(0)}{i!}x^i.$$ So for $k \geq 3$ there is an $x^3$ term, in fact, only one, and it has coefficient $\displaystyle\frac{f^{(3)}(0)}{3!}$ which is $\dfrac{1}{16}$ (details of which can be found below by putting your mouse over the grey box).

 $f(x) = (1+x)^{\frac{1}{2}}$, $f'(x) = \frac{1}{2}(1+x)^{-\frac{1}{2}}$, $f''(x) = -\frac{1}{4}(1+x)^{-\frac{3}{2}}$, $f'''(x) = \frac{3}{8}(1+x)^{-\frac{5}{2}}$ so $f^{(3)}(0) = f'''(0) = \frac{3}{8}$. Therefore $$\displaystyle\frac{f^{(3)}(0)}{3!} = \frac{\frac{3}{8}}{6} = \frac{1}{16}.$$

A: Another way to see this is to expand in Binomial series:
$$
(1+x)^{\frac{1}{2}} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k
$$
Although $\binom{\frac{1}{2}}{k}$ looks intimidating, it's nothing when you replace $\frac{1}{2} = \alpha \Rightarrow \binom{\alpha}{k} = \alpha\cdot (\alpha-1)\cdots(\alpha-k+1) \cdot \frac{1}{k!} = \frac{1}{2} \cdots (-\frac{1}{2}) \cdots \frac{1}{2}-(k-1)\cdot \frac{1}{k!} = \frac{\sqrt{\pi}}{k! 2 \Gamma(\frac{3}{2}-k)}$
This is easy since you only need to plug in $k=3$ and $\Gamma(-\frac{3}{2})=\frac{4 \sqrt{\pi}}{3}$ and therefore $\binom{\frac{1}{2}}{3} = \frac{1}{16}$
