# The Unapologetic Mathematician

## Products of Power Series

Formally, we defined the product of two power series to be the series you get when you multiply out all the terms and collect terms of the same degree. Specifically, consider the series $\sum\limits_{n=0}^\infty a_nz^n$ and $\sum\limits_{n=0}^\infty b_nz^n$. Their product will be the series $\sum\limits_{n=0}^\infty c_nz^n$, where the coefficients are defined by

$\displaystyle c_n=\sum\limits_{k+l=n}a_kb_l=\sum\limits_{k=0}^na_kb_{n-k}$

Now if the series converge within radii $R_a$ and $R_b$, respectively, it wouldn’t make sense for the product of the functions to be anything but whatever this converges to. But in what sense is this the case?

Like when we translated power series, I’m going to sort of wave my hands here, motivating it by the fact that absolute convergence makes things nice.

Let’s take a point $z_1$ inside both of the radii of convergence. Then we know that the series $\sum\limits_{n=0}^\infty a_nz_1^n$ and $\sum\limits_{n=0}^\infty b_nz_1^n$ both converge absolutely. We want to consider the product of these limits

$\displaystyle\left(\sum\limits_{k=0}^\infty a_kz_1^k\right)\left(\sum\limits_{l=0}^\infty b_lz_1^l\right)$

Since the limit of the first sequence converges, we’ll just take it as a constant and distribute it over the other:

$\displaystyle\sum\limits_{l=0}^\infty\left(\sum\limits_{k=0}^\infty a_kz_1^k\right)b_lz_1^l$

And now we’ll just distribute each $b_lz_1^l$ across the sum it appears with:

$\displaystyle\sum\limits_{l=0}^\infty\left(\sum\limits_{k=0}^\infty a_kb_lz_1^{k+l}\right)$

And now we’ll use the fact that all the series in sight converge absolutely to rearrange this sum, adding up all the terms of the same total degree at once, and pull out factors of $z_1^n$:

$\displaystyle\sum\limits_{n=0}^\infty\left(\sum\limits_{k+l=n}a_kb_lz_1^n\right)=\sum\limits_{n=0}^\infty\left(\sum\limits_{k+l=n}a_kb_l\right)z_1^n$

As a special case, we can work out powers of power series. Say that $f(z)=\sum\limits_{n=0}^\infty a_n(z-z_0)^n$ within a radius of $R$. Then within the same radius of $R$ we have

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

where the coefficients are defined by

$\displaystyle c_n(p)=\sum\limits_{m_1+m_2+...+m_p=n}a_{m_1}a_{m_2}...a_{m_p}$