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I understand the Cauchy product for power series, but we have slightly different notation here. Suppose we have the following power series:

$$f(x) = \sum_{n=0}^\infty a_nx^n$$ $$g(x) = \sum_{m=0}^\infty b_mx^m$$

Their product is written as follows: $$g(x) \times f(x) = \sum_{n=0}^\infty \sum_{m=0}^\infty a_n b_{m}x^{n+m}$$ $$= \sum_{k=0}^\infty x^k (\sum_{n=0}^k a_n b_{k-n})$$

where, $k=n+m, m=k-n, 0\le n \le k$.

Could you please tell me how we got $$g(x) \times f(x) = \sum_{k=0}^\infty x^k (\sum_{n=0}^k a_n b_{k-n})$$

Avv
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    The swap is just a notation, noting really changed – Measure me Dec 02 '20 at 01:12
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    This is called the Cauchy product; convergence aside, it's what you get when you multiply two series together. Try it for polynomials, say $A(x) = a_0 + a_1 x+ a_2 x^2$ and $B(x) = b_0+b_1 x+b_2 x^2$ to see how it works. – Integrand Dec 02 '20 at 01:36
  • @Integrand. I did that but I found some terms not found in the original product before I post the question. For example, when we multiply $a_2x^2$ with $ (b_0+b_1x+b_2x^2)$, we get only one product of power $4$ which is $a_2b_2x^4$, but the formula starts from $n=0$ up to $k=4$ for $x^4$, which gives terms not found in the product of the terms you gave as the formula will give $x^4(a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4_b_0)$ – Avv Dec 02 '20 at 01:43
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    Avra, glad you multiplied it out. Yes, if you truncate it, you will only get accuracy up to the order of the truncation. The purpose of doing it in the finite case is to observe, say, the coefficient of $x^2$ is $a_0b_2+a_1b_1+a_2b_0$ and try to generalize that. – Integrand Dec 02 '20 at 01:47
  • @Integrand. So the formula does not give the exact number? This what confused me because there are extra terms in the formula for power of $x$. – Avv Dec 02 '20 at 01:50
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    What I think Integrand means is that they are equivalent mathematically, but if you interrupt the "infinite" process of summation in the 2 different expressions, you get two different results becouse they "pick" terms differently, in a different order. – Measure me Dec 02 '20 at 02:09

2 Answers2

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If you have $$g(x) \times f(x) = \sum_{n=0}^\infty \sum_{m=0}^{\infty} a_n b_{m}x^{n+m}$$ then if you set $k=n+m$, the coefficient for the $k$ degree term is $\sum_{n=0}^ka_nb_{k-n}$ becouse you have to go trough all elements of the sums that combined give $k$.

So you have that with one sum it is the seires that has that as coefficient to the $k$ term, which means: $$g(x) \times f(x) = \sum_{k=0}^\infty(\sum_{n=0}^k a_n b_{k-n}) x^k $$ which is what you wrote as the answer.

Measure me
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    Ask if any part is unclear! – Measure me Dec 02 '20 at 01:36
  • How $k$ goes for the outer sum and $n$ goes for the inner sum? And why the subscript $n$ in $a_n$ does not change although we changed $k$ to be $n+m$ and $m=k-n$? – Avv Dec 02 '20 at 01:58
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    I can't find any additional terms; you were talking about $x^4(a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)$, but for example you get the first coefficient for $n=0$ and $m=4$, the secondo for $n=1$ and $m=3$, and so on. – Measure me Dec 02 '20 at 02:05
  • Measure, try to multiple $A(x)$ and $B(x)$ that integrand gave for $a_2x^2$ with $(b_0+b_1x+b_2x^2)$, you will get one term with power $x^4$, which is $a_2b_2x^4$. But the formula give more than one term associated with $x^4$:

    $x^{4} (a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)$

     – Avv Dec 02 '20 at 02:09
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    yes but if you multiply $a_3x^3$ with $(b_0+b_1x+b_2x^2)$ you also get $a_3b_1x^4$, and so on... – Measure me Dec 02 '20 at 02:11
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    So you get all of the coefficient of the degree $4$ terms, simply you have to be carefull to search them all. – Measure me Dec 02 '20 at 02:11
  • There is no $a_3x^3$ in integrand equations as we are taking it up to $a_2x^2$, but the formula still gives additional terms. – Avv Dec 02 '20 at 02:13
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    That's what i am saying. Infinite series and a finite sums are different. As I wrote in the other comment section, the two expressions give different results if you just get it over a finite sum. The series are really the same, the sums over finite index are not. – Measure me Dec 02 '20 at 02:16
  • I got that, so the series has to be infinite. So what about my last question measure, can you please let me know how $k$ goes for the outer sum and $n$ goes for the inner sum? And why the subscript $n$ in an does not change although we changed $k$ to be $n+m$ and $m=k−n$? – Avv Dec 02 '20 at 02:18
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    It's not that the $n$ went on the inner sum, but if you follow the argument i wrote as the answer: (1) i first state that the double series can be written as a unique series becouse the multiplication of terms gives other terms (i say a series with index $k$); (2) i try to find the coefficient of the term of degree $k$ and find it to be $\sum_{n=0}^ka_nb_{k-n}$; (3) so now I just place it as the coefficient of $x^k$. And that's it. – Measure me Dec 02 '20 at 02:23
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Here is an algebraic derivation.

We obtain \begin{align*} g(x)\times f(x)&=\sum_{n=0}^\infty\sum_{m=0}^\infty a_nb_mx^{n+m}\\ &=\sum_{k=0}^\infty\left(\sum_{{n+m=k}\atop{n,m\geq 0}} a_nb_m\right)x^k\tag{1}\\ &=\sum_{k=0}^\infty\left(\sum_{n=0}^ka_nb_{k-n}\right)x^k\tag{2} \end{align*}

Comment:

  • In (1) we reorder the sums by increasing powers of $x$ (provided the reordering is feasible, e.g. due to absolute convergence of the series).

  • In (2) we eliminate the variable $m$ by using $m=k-n$.

Markus Scheuer
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