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Let $f:\mathbb Q \rightarrow \mathbb Q$ be a function such that $f(x)=\displaystyle \sum_{i=0}^n a_ix^i$ with $a_i$ being elements of a subset of rational numbers indexed by non-negative integers from interval $[0;n]$. Let $g:\mathbb Q \rightarrow \mathbb Q$ be defined as $g(x)=\displaystyle \sum_{i=0}^m b_ix^i$ with $b_i$ defined analogically to $a_i$, as a subset of $\mathbb Q$ indexed by an interval of natural numbers $[0;m]$.

We know that $f(x)g(x)= \big (\displaystyle \sum_{i=0}^n a_ix^i \big ) \big ( \displaystyle \sum_{i=0}^mb_ix^i \big )$.

How to prove that $\big (\displaystyle \sum_{i=0}^n a_ix^i \big ) \big ( \displaystyle \sum_{i=0}^mb_ix^i \big )=\displaystyle \sum_{i=0}^{n+m} \big ( \displaystyle \sum_{i=0}^{n+m} a_ib_{n-i} \big)x^i$ - in other words, to achieve the standard formula for polynomial product?

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    Well, what are your thoughts on the problem? – 2017-01-16
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    Try some particular cases (with $m$ and $n$ small) first, and that will give you a good idea of what is going on. – 2017-01-16
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    Your formula cannot be quite correct as written since it makes reference to things like $a_{n+m}$ which are not defined. – 2017-01-16

2 Answers 2

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Here is a trick I often use:

To avoid indexing problems, define $a_i = b_j = 0$ for $i > n$ and $j > m$. The sums are not really infinite, since all but a finite number of terms are zero.

Then

$\begin{array}\\ \big ( \sum_{i=0}^n a_ix^i \big ) \big ( \sum_{j=0}^mb_jx^j \big ) &=\big ( \sum_{i=0}^{\infty} a_ix^i \big ) \big ( \sum_{j=0}^{\infty}b_jx^j \big )\\ &= \sum_{i=0}^{\infty} \big ( \displaystyle \sum_{j=0}^{i} a_jb_{i-j} \big)x^i\\ &= \sum_{i=0}^{n+m} \big ( \displaystyle \sum_{j=0}^{i} a_jb_{i-j} \big)x^i\\ \end{array} $

You need to use separate indices for the $a$-s and $b$-s, or else confusion and errors follow.

Note that if $i > n+m$ then $a_j = 0$ if $j > n$ and $b_{i-j} = 0$ if $j \le n$ because $i-j > n+m-n = m $.

Therefore, if $i > n+m$, $\sum_{j=0}^{i} a_jb_{i-j} = 0$.

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    How do you get the result $\big ( \sum_{i=0}^{\infty} a_ix^i \big ) \big ( \sum_{j=0}^{\infty}b_jx^j \big )=\sum_{i=0}^{\infty} \big ( \displaystyle \sum_{j=0}^{i} a_jb_{i-j} \big)x^i$? – 2017-01-16
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    The set of terms $a_ib_j x^{i+j}$ are collected into all $i$ and $j$ with $i+j = k$ so they become $x^k\sum_{i=0}^k a_i b_{k-i}$. The way I wrote it just relabels $k$ as $i$ and $i$ with $j$. – 2017-01-16
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$$\begin{array}\\ S&=\big ( \sum_{i=0}^n a_ix^i \big ) \big ( \sum_{j=0}^mb_jx^j \big ) \\ &=\displaystyle \sum_{i=0}^{n} \displaystyle \sum_{j=0}^{m} a_ib_{j} x^{i+j} \\ &=\displaystyle \sum_{k=0}^{n+m}\displaystyle \sum_{i=0}^{n} \displaystyle \sum_{j=0}^{m} a_ib_{j} x^{i+j} \delta(k-(i+j)) \\ &=\displaystyle \sum_{k=0}^{n+m}\displaystyle \sum_{i=0}^{n}a_i \left[ \displaystyle \sum_{j=0}^{m} b_{j} x^{i+j} \delta(j-(k-i))\right] \end{array} $$

Note that ...

$$\displaystyle \sum_{j=0}^{m} b_{j} x^{i+j} \delta(j-(k-i)) = b_{k-i}x^k $$ subject to the constraint

$0 \le \ k-i \le m \implies k-m\le i \le k$

This constraint must be combined with the constraint from the original sum $ 0 \le i \le n$

together these constraints imply ...

$$ \max(0,k-m) \le i \le \min(k,n)$$

so the final formula should read ...

$$ S =\displaystyle \sum_{k=0}^{n+m}\;\;\displaystyle \sum_{i=\max(0,k-m)}^{\min(n,k)}a_i b_{k-i}x^k$$

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    What's $\delta(k-(i+j))$? – 2017-01-16
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    Kronecker delta function that takes integer arguments and returns 1 when the argument is zero and returns zero otherwise, I put it in instead of just saying "let $k=i+j$" because I wanted to keep careful track of the limits of the sum. – 2017-01-16