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?

$$f(x)=\sum_{n=0}^\infty b_n(x-c_2)^n$$


particularly without directly applying Taylor's theorem everywhere.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.