Given some power series, how do I make one centered at a different point? Suppose I have a function defined as follows:
$$f(x)=\sum_{n=0}^\infty a_n(x-c_1)^n$$
How would I deduce the coefficients of the power series centered at a different point?
$$f(x)=\sum_{n=0}^\infty b_n(x-c_2)^n$$
$$b_n=???$$
particularly without directly applying Taylor's theorem everywhere.
 A: Suppose without loss of generality that $c_1=0$ and let $r$ be the radius of $\sum_n a_n z^n$. Let us prove that for any $c_2$ such that $|c_2|<r$, the power series $$\sum_n \left(\sum_{k=n}^\infty a_k \binom kn c_2^{k-n} \right)(z-c_2)^n$$ has radius greater than $r-|c_2|$ and its sum equals $f$.
Consider some $z$ such that $|z-c_2|<r-|c_2|$. That implies $|z|<r$ in particular, so that 
$\begin{align}
f(z)=  \sum_{k=0}^\infty a_k z^k 
&=\sum_{k=0}^\infty a_k (c_2 + z-c_2)^k\\
&=\sum_{k=0}^\infty \sum_{j=0}^k  a_k \binom kj c_2^{k-j} (z-c_2)^j
\end{align}$
Let $u_{k,j}=\left\{
    \begin{array}{ll}
        a_k \binom kj c_2^{k-j} (z-c_2)^j & \mbox{if } j\leq k \\
        0 & \mbox{otherwise}
    \end{array}
\right.$
Let us prove that $(u_{k,j})$ is summable. 
Indeed 
$$\begin{align}\sum_{k=0}^\infty \sum_{j=0}^\infty |u_{k,j}|
&=\sum_{k=0}^\infty \sum_{j=0}^k |u_{k,j}|\\
&=\sum_{k=0}^\infty \sum_{j=0}^k  \left| a_k \binom kj c_2^{k-j} (z-c_2)^j\right|\\
&= \sum_{k=0}^\infty |a_k| (|c_2| + |z-c_2|)^k
\end{align}$$
The power series $\sum_n |a_n|z^n$ has same radius as the original one and, by choice of $z_2$, $|c_2| + |z-c_2|< r$, hence convergence of the last series.
Summability yields $$\begin{align}f(z) = \sum_{k=0}^\infty \sum_{j=0}^\infty u_{k,j}
&=\sum_{j=0}^\infty \sum_{k=0}^\infty u_{k,j}\\
&= \sum_{j=0}^\infty \left(\sum_{k=j}^\infty a_k \binom kj c_2^{k-j} \right)(z-c_2)^j
\end{align}$$
as wanted.
