Alternative intuition behind series simplification

Question

Given functions $$ f(x)=\sum_{i=0}^\infty a_ix^i\,\,\,\text{ and }\,\,\,g(x)=\sum_{j=0}^\infty b_jx^j $$ the following simplification $$ f(x)g(x)=\left(\sum_{i=0}^\infty a_ix^i\right)\left(\sum_{j=0}^\infty b_jx^j \right)=\sum_{i=0}^\infty\sum_{j=0}^\infty a_ib_jx^{i+j} $$ is relatively intuitive. I guess a rather simple and informal proof would consist in expanding a few terms and recognize the emerging pattern. For example, $$ (a_0+a_1x+a_2x^2+\cdots)(b_0+b_1x+b_2x^2+\cdots)\\ =a_0b_0+a_0b_1x+a_0b_2x^2+\cdots+a_1b_0+a_1b_1x+a_1b_2x^2+\cdots $$ and recognize this as a double summation. However, I fear some of my students might be confused by it. Is there a more straighforward way to argue in order to justify the intuition behind such simplification?

Answer 1

There's no intuition, it's just $$\left(\sum_{i=0}^\infty a_ix^i\right)\cdot c=\sum_{i=0}^\infty a_ix^ic $$ with $c:=\sum_{j=0}^\infty b_jx^j$, i.e., $$\left(\sum_{i=0}^\infty a_ix^i\right)\cdot \left(\sum_{j=0}^\infty b_jx^j\right)=\sum_{i=0}^\infty \left(a_ix^i\sum_{j=0}^\infty b_jx^j\right) $$ followed by $$c\cdot \left(\sum_{j=0}^\infty b_jx^j\right) c=\sum_{j=0}^\infty cb_jx^j $$ with $c=a_ix^i$, i.e., $$a_ix^i\sum_{j=0}^\infty b_jx^j = \sum_{j=0}^\infty a_ib_jx^{i+j} $$ end therefore after summation $$\left(\sum_{i=0}^\infty a_ix^i\right)\cdot \left(\sum_{j=0}^\infty b_jx^j\right)= \sum_{i=0}^\infty\sum_{j=0}^\infty a_ib_jx^{i+j}$$

Answer 2

If $f(x)=\sum_{n=0}^\infty a_nx^n$, then $a_n=\frac{f^{(n)}(0)}{n!}$ and the same thing applies to $g$. But then, if $f(x)g(x)=\sum_{n=0}^\infty c_nx^n$:

\begin{align}c_0&=(f\times g)(0)=a_0b_0;\\c_1&=(f\times g)'(0)=f(0)g'(0)+f'(0)g(0)\\&=a_0b_1+a_1b_0;\\c_2&=\frac{(f\times g)''(0)}2\\&=\frac{f(0)g''(0)}2+f'(0)g'(0)+\frac{f''(0)g(0)}2\\&=f(0)\frac{g''(0)}2+f'(0)g'(0)+\frac{f''(0)}2g(0)\\&=a_0b_2+a_1b_1+a_2b_0\end{align}

and so on…