Shortcut with x0 for taylor expansion I am confused about the point $x_0$ in Taylor Series Expansion.
$f(x) = 1/(1-x)$ and $\sum_n^\infty x^n$ at $x_0=0$ 
so I thought that if $x_0$=2 I don't need to go through all the process solving for $f^n(x_0)$
and I have a shortcut for $x_0=2$  like :
$$f(x) = 1/(1-(x-2))$$ and series for it is$$\sum_n^\infty (x-2)^n$$ but it is not true.
My exercise was to find $f(x)$ for $\sum_n^\infty((x-1)/2)^n$. So I see here that $x_0=-1$ and
immediately I thought of this:
$$1/(1-(x-1)/2) = \sum_n^\infty((x-1)/2)^n$$
so I just got lucky with this example.
Question : Isn't there a shortcut to find the series of a function like $1/(1-x)$ where $x_0=const$ without going through all the pain of finding the derivative for $f^n(x_0)$ ?
Thanks
 A: Consider the general case where you need the Taylor expansion of $$f(x)=\frac{1}{1-x}$$ built at $x=a$. The derivatives are easy to compute and you arrive at $$f(x)=\sum_{n=0}^{\infty}\frac{(x-a)^n}{(1-a)^{n+1}}$$
A: You didn't get lucky in the second example, you just didn't do it properly in the first.
Let's see. 
You have $f(x)=\frac{1}{1-x}=\sum x^n$. You want to obtain $f(x)=\sum a_n(x-2)^n$. 
Then $\sum x^n=\sum a_n(x-2)^n$. Put  $x=y+2$. 
So $f(y+2)=\sum (y+2)^n=\sum a_ny^n$. 
Therefore, to obtain the coefficients $a_n$ we just need to expand $f(y+2)$ at $y_0=0$.
But $f(y+2)=\frac{1}{1-(y+2)}=-\frac{1}{1-(-y)}=-\sum (-y)^n=\sum(-1)^{n+1}y^n$. Therefore $a_n=(-1)^{n+1}$. Therefore
$$f(x)=\sum (-1)^{n+1}(x-2)^n.$$
Summarizing: To expand $f(x)$ at $x_0$ (assuming that what we know is to expand things at $0$) is to expand $$f(x+x_0)=\sum a_nx^n$$ at $0$ and take those coefficients to get 
$$f(x)=\sum a_n(x-x_0)^n.$$
A: Why would you think it is a pain?
Set $x_0=0$
$f^1(x)=-(1-x)^{-2}$
$f^2(x)=2(1-x)^{-3}$
$f^3(x)=-2*3(1-x)^{-4}$
So, as you can see, $f^n(0)=(-1)^{n}*n!$
And even if you don't want to set $x_0=0$, for any $x_0$, you'd have $f^n(x_0)=(-1)^n*n!*(1-x_0)^{n-1}$
For other cases, I am not an expert at this, but there are some helpful "tricks".
Assume that the expression for the derivative is easy to obtain, so that we have the coefficients $c_0',c_1',\ldots$ then we have $(f')^{(k)}(0)=c_k'*k!$ and thus $f^{k+1}(0)=c'_k*k!$ which would make it easy to get the coefficients for $f$
After simplification, you'd get $f(x)=f(0)+\frac{c_0'}{1}x+\frac{c_1'}{2}x^2+\ldots$
This would be helpful in the case of getting the taylor series for $arctan(x)$ for example.
