Multiplication of sums

Question

In this set of online math notes the following is stipulated:

$$\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right) =\sum_{n=0}^\infty (c_n) $$

Where apparently $c_n = \sum_{i=0}^n a_ib_{n-i}$. Replacing $c_n$ into above equation we get

$$\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right) =\sum_{n=0}^\infty \left(\sum_{i=0}^n a_ib_{n-i}\right) $$

I tried to expand a few terms of $c_n$ using sympy:

$$ {c}_{0} = \sum_{i=0}^{0} {a}_{i} {b}_{- i} = {a}_{0} {b}_{0} $$ $$ {c}_{1} = \sum_{i=0}^{1} {a}_{i} {b}_{- i + 1} = {a}_{0} {b}_{1} + {a}_{1} {b}_{0} $$ $$ {c}_{2} = \sum_{i=0}^{2} {a}_{i} {b}_{- i + 2} = {a}_{0} {b}_{2} + {a}_{1} {b}_{1} + {a}_{2} {b}_{0} $$

Then I ran this example on a verifiable multiplication of sums as follows:

$$(2+x+2x^2)(3-5x+x^2) = 6-7x+3x^2-9x^3+2x^4$$

By setting \begin{equation*} a_0 = 2, a_1 = x, a_2 = 2x^2 \end{equation*} and \begin{equation*} b_0 = 3, b_1 = -5x, b_2 = x^2 \end{equation*}

Following the online reference this should equal $$\sum_{n=0}^2 c_n =c_0+c_1+c_2 =a_0b_0+a_0b_1+a_1b_0+a_0b_2+a_1b_1+a_2b_0 $$

However when you substitute in the values you get $$ 3 x^{2} - 7 x + 6 $$ Which is not equal to $6-7x+3x^2-9x^3+2x^4$.

I have a feeling the notes are wrong here. Could anyone confirm this? THanks.

Answer 1

The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.

Answer 2

The first identity should read $\sum_{n\ge 0}a_n x^n\sum_{n\ge 0}b_n x^n=\sum_{n\ge 0}(\sum_{i= 0}^n a_i b_{n-i})x^n$, or even better $\sum_{j\ge 0}a_j x^j\sum_{k\ge 0}b_k x^k=\sum_{n\ge 0}(\sum_{i= 0}^n a_i b_{n-i})x^n$. The proof is simple: just find all contributions to the $x^n$ coefficient on the left-hand side. When multiplying polynomials, which have finite degrees, all sufficiently late terms in each sequence are $0$. In this case $a_n=0$ for $n>2$, $b_n=0$ for $n>2$, $c_n=0$ for $n>4$.