# The Unapologetic Mathematician

## Inverses of Power Series

Now that we know how to compose power series, we can invert them. But against expectations I’m talking about multiplicative inverses instead of compositional ones.

More specifically, say we have a power series expansion

$\displaystyle p(x)=\sum\limits_{n=0}^\infty p_nz^n$

within the radius $r$, and such that $p(0)=p_0\neq0$. Then there is some radius $\delta$ within which the reciprocal has a power series expansion

$\displaystyle\frac{1}{p(x)}=\sum\limits_{n=0}^\infty q_nz^n$

In particular, we have $q_0=\frac{1}{p_0}$.

In the proof we may assume that $p_0=1$ — we can just divide the series through by $p_0$ — and so $p(0)=1$. We can set

$\displaystyle P(z)=1+\sum\limits_{n=1}^\infty\left|p_nz^n\right|$

within the radius $h$. Since we know that $P(0)=1$, continuity tells us that there’s $\delta$ so that $|z|<\delta$ implies $|P(z)-1|<1$.

Now we set

$\displaystyle f(z)=\frac{1}{1-z}=\sum\limits_{n=0}^\infty z^n$
$\displaystyle g(z)=1-p(z)=\sum\limits_{n=0}^\infty -p_nz^n$

And then we can find a power series expansion of $f\left(g(z)\right)=\frac{1}{p(z)}$.

It’s interesting to note that you might expect a reciprocal formula to follow from the multiplication formula. Set the product of $p(z)$ and an undetermined $q(z)$ to the power series $1+0z+0z^2+...$, and get an infinite sequence of algebraic conditions determining $q_n$ in terms of the $p_i$. Showing that these can all be solved is possible, but it’s easier to come around the side like this.

September 24, 2008

## Sudocube

DO WANT

Tipped off to its existence by Alexandre Borovik.

September 24, 2008 Posted by | Rubik\'s Cube | 1 Comment

## Composition of Power Series

Now that we can take powers of functions defined by power series and define them by power series in the same radii.. well, we’re all set to compose functions defined by power series!

Let’s say we have two power series expansions about $z=0$:

$\displaystyle f(z)=\sum\limits_{n=0}^\infty a_nz^n$

within the radius $r$, and

$\displaystyle g(z)=\sum\limits_{n=0}^\infty b_nz^n$

within the radius $R$.

Now let’s take a $z_1$ with $|z| and $\sum\limits_{n=0}^\infty\left|b_nz_1^n\right|. Then we have a power series expansion for the composite:

$\displaystyle f\left(g(z)\right)=\sum\limits_{n=0}^\infty c_nz^n$.

The coefficients $c_n$ are defined as follows: first, define $b_n(k)$ to be the coefficient of $z^n$ in the expansion of $g(z)^k$, then we set

$\displaystyle c_n=\sum\limits_{k=0}^\infty a_kb_n(k)$

To show this, first note that the hypothesis on $z_1$ assures that $|g(z_1)|, so we can write

$\displaystyle f\left(g(z_1)\right)=\sum\limits_{k=0}^\infty a_kg(z_1)^k=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty a_kb_n(k)z_1^n$

If we are allowed to exchange the order of summation, then formally the result follows. To justify this (at least as well as we’ve been justifying such rearrangements recently) we need to show that

$\displaystyle\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\left|a_kb_n(k)z_1^n\right|=\sum\limits_{k=0}^\infty\left|a_k\right|\sum\limits_{n=0}^\infty\left|b_n(k)z_1^n\right|$

converges. But remember that each of the coefficients $b_n(k)$ is itself a finite sum, so we find

$\displaystyle\left|b_n(k)\right|\leq\sum\limits_{m_1+...+m_k=n}\left|b_{m_1}\right|...\left|b_{m_k}\right|$

On the other hand, in parallel with our computation last time we find that

$\displaystyle\left(\sum\limits_{n=0}^\infty\left|b_n\right|z^n\right)^n=\sum\limits_{n=0}^\infty B_n(k)z^n$

where

$\displaystyle B_n(k)=\sum\limits_{m_1+...+m_k=n}\left|b_{m_1}\right|...\left|b_{m_k}\right|$

So we find

\displaystyle\begin{aligned}\sum\limits_{k=0}^\infty\left|a_k\right|\sum\limits_{n=0}^\infty\left|b_n(k)z_1^n\right|\leq\sum\limits_{k=0}^\infty\left|a_k\right|\sum\limits_{n=0}^\infty B_n(k)\left|z_1^n\right|\\=\sum\limits_{k=0}\left|a_k\right|\left(\sum\limits_{n=0}^\infty\left|b_nz_1^n\right|\right)^k\end{aligned}

which must then converge.

Breathe!

September 24, 2008 Posted by | Analysis, Calculus, Power Series | 1 Comment