how to multiply infinite power series

Question

I am doing an assignment for my precalculus 2 class. I am expanding two infinite power series and multiplying them together to prove that $\exp(ax)\exp(by) = \exp(ax+by)$ I'm not sure what I am supposed to do when I have $x(...)$ or $x+(...)$ should $(...)$ be the summation of all terms between the terms surrounding it? I know that that is technically what it means but I'm not sure how to operate on it.

Answer 1

If you have two power series $A(x)=\sum_{n\ge 0}A_nx^n$ and $B(x)=\sum_{n\ge 0}B_nx^n$, then their (formal) product is given by the power series $C(x)=A(x)B(x)=\sum_{n\ge 0}C_nx^n$, whereby

$C_n=\sum_{k=0}^n A_kB_{n-k}$

This is sometimes called Cauchy's product formula but you may want to verify this yourself. Hence, for the product of $\exp(ax)$ and $\exp(bx)$, this gives

$C_n = \sum_{k=0}^n \frac{a^k}{k!}\frac{b^{n-k}}{(n-k)!}$.

Multiply by $n!$ and obtain

$n!C_n = \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}=(a+b)^n$.

Therefore:

$\exp(ax)\exp(bx)=\sum_{n\ge 0}\frac{(a+b)^n}{n!}x^n = \exp((a+b)x)$.

Answer 2

Cauchy's product formula is easy to prove, if you look closely at what formulas with sums mean.

Take $A(z) = \sum_{r \ge 0} a_r z^r$ and $B(z) = \sum_{s \ge 0} b_s z^s$. Multiply: \begin{align} A(z) \cdot B(z) &= \left( \sum_{r \ge 0} a_r z^r \right) \cdot \left( \sum_{s \ge 0} b_s z^s \right) \\ &= \sum_{r \ge 0} \sum_{s \ge 0} a_r b_s z^{r +s} \\ &= \sum_{n \ge 0} z^n \sum_{0 \le r \le n}a_r b_{n - r} \end{align} The first is just that $r$ and $s$ are totally independent, so you can distribute at will, the second is just grouping terms with $n = r + s$, thus $s = n - r$. The range on $r$ is just to limit to terms that make sense, no negative indices.