# The Unapologetic Mathematician

## Translating Power Series

So we know that we can have two power series expansions of the same function about different points. How are they related? An important step in this direction is given by the following theorem.

Suppose that the power series $\sum\limits_{n=0}^\infty a_n(z-z_0)^n$ converges for $|z-z_0|, and that it represents the function $f(z)$ in some open subset $S$ of this disk. Then for every point $z_1\in S$ there is some open disk around $z_1$ of radius $R$ contained in $S$, in which $f$ has a power series expansion

$\displaystyle f(z)=\sum\limits_{k=0}^\infty b_k(z-z_1)^k$

where

$\displaystyle b_k=\sum\limits_{n=k}^\infty\binom{n}{k}a_n(z_1-z_0)^{n-k}$

The proof is almost straightforward. We expand

\displaystyle\begin{aligned}f(z)=\sum\limits_{n=0}^\infty a_n(z-z_0)^n=\sum\limits_{n=0}^\infty a_n(z-z_1+z_1-z_0)^n\\=\sum\limits_{n=0}^\infty a_n\sum_{k=0}^n\binom{n}{k}(z-z_1)^k(z_1-z_0)^{n-k}\end{aligned}

Now we need to interchange the order of summation. Strictly speaking, we haven’t established a condition that will allow us to make this move. However, I hope you’ll find it plausible that if this double series converges absolutely, we can adjust the order of summations freely. Indeed, we’ve seen examples of other rearrangements that all go through as soon as the convergence is absolute.

Now we consider the absolute values

\displaystyle\begin{aligned}\sum_{n=0}^\infty|a_n|\sum\limits_{k=0}^n\binom{n}{k}|z-z_1|^k|z_1-z_0|^{n-k}=\sum_{n=0}^\infty|a_n|(|z-z_1|+|z_1-z_0|)^n\\=\sum_{n=0}^\infty|a_n|(z_2-z_0)^n\end{aligned}

Where we set $z_2=z_0+|z-z_1|+|z_1-z_0|$. But then $|z_2-z_0|, where the last inequality holds because the disk around $z_1$ of radius $R$ fits within $S$, which fits within the disk of radius $r$ around $z_0$. And so this series of absolute values must converge, and we’ll take it on faith for the moment (to be shored up when we attack double series more thoroughly) that we can now interchange the order of summations.

$\displaystyle f(z)=\sum\limits_{k=0}^\infty\left(\sum_{n=k}^\infty a_n\binom{n}{k}(z_1-z_0)^{n-k}\right)(z-z_1)^k$

This result allows us to recenter our power series expansions, but it only assures that the resulting series will converge in a disk which is contained within the original disk of convergence, so we haven’t necessarily gotten anything new. Yet.

September 16, 2008 - Posted by | Analysis, Calculus, Power Series

1. […] a point within of , we can expand as a power series about […]

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2. […] of functions around various points, and within various radii of convergence. We even have formulas to relate expansions about nearby points. But when we move from one point to a nearby point, the […]

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3. […] when we translated power series, I’m going to sort of wave my hands here, motivating it by the fact that absolute convergence […]

Pingback by Products of Power Series « The Unapologetic Mathematician | September 22, 2008 | Reply

4. Would you explain how you obtained the equation for b_k?

Comment by Bernd | September 23, 2008 | Reply

5. Bernd, look at the last formula for $f(z)$ and compare to the first one.

Comment by John Armstrong | September 23, 2008 | Reply