<s>[INST] Taylor series for $\sqrt{1+x^2}$ I want to expand 
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$ 
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\frac{1}{2}x^2+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{x^4}{2!}+\frac{1}{2} \left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\frac{x^6}{3!}$$
Then
$$\sqrt{1+x^2}=\sqrt{1+(x-2)^2-4+4x}=\sqrt{(x-2)^2-3+4x}=\sqrt{(x-2)^2-3+4(x-2)+8}$$
I could not complete , what can we do then ?
(I know that we can differentiate the function and substitute in the Taylor formula , but I want a shorter way)
for example : 
if we want to expand $$g(x)=\frac{1}{1-x}$$ around $x=2$ we can start by
\begin{align}
g(x) & =\frac{1}{1-x}=\frac{1}{1-(x-2)-2}=\frac{-1}{1+(x+2)} \\[10pt]
& =-[1-(x-2)+(x-2)^2-(x-2)^3+\cdots]
\end{align}
So  I want to get convert $f(x)$ to a form that we can write its expansion without getting derivatives, like I did with $g(x)$ above
 [/INST] To give the general term expansion let $y=x-2$, then
\begin{align}
\sqrt{1+x^2}
&= \sqrt{1+ (y+2)^2}
= \sqrt{5 + 4y + y^2}
= \sqrt{5} \sqrt{1 + \frac{y}{5}(4+y)}\\
&= \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \left[\frac{y}{5}(4+y)\right]^n
= \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \frac{y^n}{5^n}\sum_{k=0}^n\binom{n}{k}4^k y^{n-k}
\end{align}
Extracting the $m$th coefficient, $[y^m]\sqrt{1+(y+2)^2}$, which may be found by considering which multiples of $y^n$ and $y^{n-k}$ equal $y^m$, i.e., when $n+(n-k) = 2n-k = m$, so we only take the $k=2n-m$ term from the inner summation above. Hence the $m$th coefficient is
\begin{align}
[y^m] \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \frac{y^n}{5^n}\sum_{k=0}^n\binom{n}{k}4^k y^{n-k}
&= \sqrt{5} \sum_{n=0}^{\infty} \binom{1/2}{n} \frac{1}{5^n} \binom{n}{2n-m}4^{2n-m}\\
&= \frac{\sqrt{5}}{4^m} \sum_{n=0}^{\infty} \binom{1/2}{n} \binom{n}{2n-m} \left(\frac{16}{5}\right)^n
\end{align}
Moreover, the binomial coefficient $\binom{n}{2n-m}$ is $0$ for $2n-m<0$ and for $n < 2n-m$, so we need only consider $n \geq m/2$ and $n\leq m$. Removing zero terms and a bit of algebra gives the Taylor series
$$\sqrt{1+x^2} = \sum_{m=0}^{\infty} c_m y^m = \sum_{m=0}^{\infty}c_m (x-2)^m,\quad\text{where}\;
c_m \equiv\frac{\sqrt{5}}{4^m}\sum_{n=\lceil m/2\rceil }^m\binom{1/2}{n}  \binom{n}{2n-m} \left(\frac{16}{5}\right)^n$$
</s>