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 $f^n(x_0)$ ?