Multiplying power formal series when reindexing is required?

Question

Normally, when I multiply the two following series:

$$\sum_{n\ge0}a_nx^n$$

$$\sum_{n\ge0}b_nx^n$$

I have:

$$\sum_{n\ge0}(\sum_{i=0}^na_ib_{n-i})x^n$$

Which is nice and intuitive. However, what if the exponents of x, or the indices don't match up? For example, what if I wanted to multiply these two:

$$\sum_{n\ge0}a_nx^{100n}$$

$$\sum_{n\ge0}b_nx^n$$

If I were to try to do this I'd reindex the second series to:

$$\sum_{\frac{n}{100}\ge0}b_nx^{100n}$$

But then the indices wouldn't line up. Am I missing something?

Answer 1

We can nicely see the similarities when adding an intermediate step.

• On the one hand we have \begin{align*} \sum_{k=0}^\infty a_kx^k\sum_{l=0}^\infty b_lx^l &=\sum_{n=0}^\infty\left(\sum_{\color{blue}{{k+l=n}\atop{k,l\geq 0}}}a_kb_l\right)x^n =\sum_{n=0}^\infty \sum_{\color{blue}{k=0}}^{\color{blue}{n}} a_kb_{n-k} x^n\\ \end{align*}

• On the other hand we have \begin{align*} \sum_{k=0}^\infty a_kx^{100k}\sum_{l=0}^\infty b_lx^l &=\sum_{n=0}^\infty\left(\sum_{\color{blue}{{100k+l=n}\atop{k,l\geq 0}}}a_kb_l\right)x^n =\sum_{n=0}^\infty \sum_{\color{blue}{k=0}}^{\color{blue}{\left\lfloor\frac{n}{100}\right\rfloor}}a_kb_{n-100k} x^n\\ \end{align*}

Comment:

• In the middle sums we have rearranged the terms according to increasing powers of $x^n$. Consequently the conditions are $k+l=n$ resp. $100k+l=n$ besides $k,l\geq 0$.

• In the right-hand sums we eliminate $l$ by substituting $l=n-k$ resp. $l=n-100k$. Consequently is the upper limit of the inner sum $n$ resp. $\left\lfloor\frac{n}{100}\right\rfloor$.

Answer 2

We have \begin{eqnarray*} \left( \sum_{n=1}^{ \infty} a_n x^{100n} \right) \left(\sum_{ m=1}^{\infty }b_m x^m\right) = \sum \sum a_n b_m x^{ \color{red}{100n+m}}. \end{eqnarray*} So to collect "like" $x$ terms, we need to list solutions of $100n+m=N$ ... this will give \begin{eqnarray*} \sum_{N=1}^{\infty} \left(\sum_{m \mid m+100n=N} a_n b_m \right) x^{ \color{red}{N}}. \end{eqnarray*}

Answer 3

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\begin{align} &\left.\begin{array}{r} \ds{\sum_{n = 0}^{\infty}a_{n}x^{\mu n}} \\[2mm] \ds{\sum_{n = 0}^{\infty}b_{n}x^{\nu n}} \end{array}\right\} \implies \left\{\begin{array}{rcl} \ds{\sum_{j = 0}^{\infty}a_{j}\, x^{\mu j}\sum_{k = 0}^{\infty}b_{k}\,x^{\nu k}} & \ds{=} & \ds{\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty}a_{j}\, b_{k}\sum_{n = 0}^{\infty} \delta_{n,\mu j + \nu k}\,\,\,x^{n}} \\[2mm] & \ds{=} & \ds{\sum_{n = 0}^{\infty}\pars{\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty}a_{j}\,b_{k}\, \delta_{n,\mu j + \nu k}}}x^{n} \\[2mm] & \ds{=} & \bbx{\ds{\sum_{n = 0}^{\infty}\pars{\sum_{j = 0}^{\infty}a_{j} \,b_{\pars{n - \mu j}/\nu}\,\,\, \bracks{{n - \mu j \over \nu} \in \mathbb{N}_{\geq 0}}}}x^{n}} \\ && \end{array}\right. \end{align}