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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.

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  • $\begingroup$ Write $c_1=c_2+\Delta$ (where $\Delta=c_1-c_2$) and run the calcs with binomial theorem. The new coefficients will be power series in $\Delta$, so you want $\lvert\Delta\rvert$ to be smaller than the radius of convergence of the original series. $\endgroup$
    – user228113
    Dec 7 '16 at 17:29
  • $\begingroup$ @G.Sassatelli could you explain more as an example? $\endgroup$ Dec 7 '16 at 17:41
  • $\begingroup$ It is basically a convolution. $\endgroup$
    – copper.hat
    Dec 7 '16 at 17:51
  • $\begingroup$ @copper.hat I'm not familiar with convolutions. $\endgroup$ Dec 7 '16 at 21:10
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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.

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  • $\begingroup$ Ah, thank you :D I guess I could've figured that out with some ingenuitive thinking. $\endgroup$ Dec 7 '16 at 21:29
  • $\begingroup$ What happens when $|c_2|>r$? Does it simply fail to converge or something weird like that? $\endgroup$ Dec 8 '16 at 1:33
  • $\begingroup$ It fails to converge, at least locally around $c_2$. $\endgroup$ Dec 8 '16 at 6:54
  • $\begingroup$ How does $|z-c_2| < r - |c_2|$ imply $|z|<r$? And how is $|z|<r$ used in the following equations? $\endgroup$ Jan 10 '17 at 9:28
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    $\begingroup$ @MusséRedi $|z|=|z-c_2+c_2|\leq |z-c_2|+|c_2|<r-|c_2|+|c_2|=r$ $\endgroup$ Jan 10 '17 at 9:37

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