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.

• 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.
– user228113
Dec 7 '16 at 17:29
• @G.Sassatelli could you explain more as an example? Dec 7 '16 at 17:41
• It is basically a convolution. Dec 7 '16 at 17:51
• @copper.hat I'm not familiar with convolutions. Dec 7 '16 at 21:10

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.

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