## 414/514 – Multiplying power series

In lecture today we argued that under reasonable circumstances, the product of two power series is the series given by their “formal” product. I think I unnecessarily made the argument more complicated than it is, so I am writing it here so we have a clean reference.

Let $\displaystyle A(x)=\sum_{j=0}^\infty a_j x^j$ and $\displaystyle B(x)=\sum_{j=0}^\infty b_j x^j$. Suppose that both $A$ and $B$ converge absolutely and uniformly in the interval ${}[-r,r]$. We want to prove that for $x\in[-r,r]$, we also have

$\displaystyle A(x)B(x)=\sum_{j=0}^\infty\left(\sum_{i=0}^j a_i b_{j-i}\right)x^j,$

and this latter series converges absolutely and uniformly in ${}[-r,r]$.

To see this, let $\displaystyle A_N(x)=\sum_{j=0}^N a_j x^j$ and $\displaystyle B_N(x)=\sum_{j=0}^N b_j x^j$ denote the partial sums of $A$ and $B$, and denote by $\displaystyle P_N(x)=\sum_{m=0}^N\left(\sum_{j+k=m}a_j b_k\right)x^m$ the partial sums of the “formal product” series.

We want to show that $\displaystyle\lim_{N\to\infty}P_N(x)=A(x)B(x).$

Let us call $R_N(x)$ the “remainder” or “tail” series of $B(x)$, i.e., $\displaystyle R_N(x)=B(x)-B_N(x)=\sum_{j=N+1}^\infty b_j x^j.$

We have that $P_N(x)=a_0B_N(x)+a_1xB_{N-1}(x)+\dots+a_N x^N B_0(x)$ $=a_0(B(x)-R_N(x))+a_1 x (B(x)-R_{N-1}(x))+\dots$ $+a_N x^N (B(x)-R_0(x)).$ Expanding this last expression, we find that it equals $B(x)\sum_{j=0}^Na_j x^j-[a_0R_N(x)+a_1 xR_{N-1}(x)+\dots$ $+a_N x^N R_0(x)].$

Now, $B(x)\sum_{j=0}^Na_j x^j=B(x)A_N(x)$ clearly converges to $B(x)A(x)$ as $N\to\infty$, and the convergence is uniform in the interval ${}[-r,r].$

We want to show that the remaining term $a_0R_N(x)+a_1 xR_{N-1}(x)+\dots+a_N x^N R_0(x)$ converges to 0 and does so uniformly in the interval ${}[-r,r].$

First, since $B_n(x)\to B(x)$ uniformly as $n\to\infty$, then for any $\epsilon>0$ there is an $N_0$ such that if $n\ge N_0$ then ${}|R_n(x)|<\epsilon$ for all $x\in[-r,r]$.  If $N>N_0$, we can split the sum above as $S_1(x)+S_2(x)$, where

$\displaystyle S_1(x)=a_0 R_N(x)+a_1 x R_{N-1}(x) + \dots + a_{N-N_0}x^{N-N_0} R_{N_0}(x)$

and

$\displaystyle S_2(x)=\sum_{j=N-N_0+1}^N a_j x^j R_{N-j}(x).$

Note that $|S_1(x)|\le\epsilon\sum_{j=0}^{N-N_0}|a_j x^j|\le K\epsilon$, where $K$ is the constant $\sum_j |a_j|r ^j.$ This shows that $S_1(x)\to0$ uniformly for $x\in{}[-r,r].$

To bound $S_2$, note that it is a sum of a constant number of terms, namely $N_0$. The functions $R_{N_0-1},R_{N_0-2},\dots,R_0$ are continuous and bounded in the interval ${}[-r,r]$, so there is a constant $L$ that bounds all of them (and, of course, $L$ does not depend on $N$). Hence

$\displaystyle |S_2(x)|\le L\sum_{j=N-N_0+1}^N |a_jx^j|\le L\sum_{j\ge N-N_0+1} |a_j|r^j.$

Since $\sum_j|a_j|r^j$ converges, there is an $N_1$ such that if $n\ge N_1$, then $\sum_{j\ge n}|a_j|r^j<\epsilon.$

Pick $N$ so that $N-N_0>N_1$. Then $|S_2(x)|. We have now shown that $|S_1(x)+S_2(x)|\to0$ uniformly for $x\in{}[-r,r]$, and this completes the proof.

(We also claimed that the convergence is absolute. To see this, replace all the terms $a_j,b_j,x^j,\dots$ above by their absolute values. The same argument shows convergence of the relevant series, so we indeed have absolute convergence.)