<s>[INST] Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$ I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.
Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\frac{(x-a)^j}{j!}f^{(j)}(a)$
$f(x)=\sqrt{1+x}=(1+x)^{1/2} \\ 
f^\prime (x)=\left ( \frac{1}{2} \right )(1+x)^{-1/2} \\ 
f^{(2)}(x)=\left (  \frac{1}{2}\right )\left ( \frac{-1}{2} \right )(1+x)^{-3/2} \\
f^{(3)}(x)=\left (  \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )(1+x)^{-5/2} \\
f^{(4)}(x)=\left (  \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )\left ( \frac{-5}{2} \right )(1+x)^{-7/2} \\
f^{(j)}(x)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)(1+x)^{(2j-1)/2} \\
f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$
I'm thinking I have calculated the j-th derivative correctly, however I can't seem to figure out how to simplify it and the rest of the steps into a compact summation.
Added:  It's been pointed out to me that the $(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$ above should be $\prod_{i=0}^{j-1}2i-1$
Thus $f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )\prod_{i=0}^{j-1}2i-1$.
If anyone can give me a hint on how to simplify this further it would be appreciated.
 [/INST] Hint: Using the Binomial Theorem, prove by induction that
$$
(1+x)^{-1/2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}(-x)^k
$$
Truncate as needed. Note that $n!!$ is the Double Factorial:
$$
\begin{array}{rll}
(2k-1)!!&=1\cdot3\cdot5\cdots(2k-1)&=\frac{(2k)!}{2^kk!}\\[6pt]
(2k)!!&=2\cdot4\cdot6\cdot8\cdots(2k)&={2^kk!}
\end{array}
$$
</s>