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Working on Spivak's Calculus problems, I searched online, trying to understand the solution provided for Problem 4a of Chapter 2. I found the question I needed: Spivak's Calculus - Exercise 4.a of 2nd chapter.

However, the answer provided there started with the following equation, and then went on from it to explain other things, which I could understand. But this part, which lies at the foundation of the argument, I don't understand.

Could I get an explanation for why it is that:

$$ \begin{align} \left(\sum_{k=0}^\infty a_kx^k\right)\left(\sum_{k=0}^\infty b_kx^k\right) &=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_j\color{#C00000}{x^j}b_{k-j}\color{#C00000}{x^{k-j}}\right) \end{align} $$

Keep in mind I'm a beginner, working on an introductory Calculus book as my first exposure to the subject. I'd appreciate both intuitive and rigorous answers.

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    $\begingroup$ Check that this is actually what it works out to if you make both sums finite. $\endgroup$ Commented Jul 22, 2013 at 11:46
  • $\begingroup$ @TobiasKildetoft I can't see how to make them finite, because I don't understand the equation, and so I don't know how the summation to k is derived. $\endgroup$ Commented Jul 22, 2013 at 11:57
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    $\begingroup$ I mean set all $a_m = 0$ and all $b_m = 0$ for $m\geq N$ (for some fixed $N$), and see that the formula holds in that case. $\endgroup$ Commented Jul 22, 2013 at 12:01

2 Answers 2

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Write the product out longhand:

$$\left(a_0+a_1x+a_2x^2+a_3x^3+\ldots\right)\left(b_0+b_1x+b_2x^2+b_3x^3+\ldots\right)\;.$$

This is the sum of all possible products of the form $(a_kx^k)(b_\ell x^\ell)=a_kb_\ell x^{k+\ell}$. The $x^n$ term in the product will therefore be the sum of all of these term $a_kb_\ell x^{k+\ell}$ for which $k+\ell=n$. Let’s look specifically at $n=3$, just to get a clearer picture of what’s going on. If $k+\ell=3$, where $k$ and $\ell$ are exponents in the two factors, then clearly $\langle k,\ell\rangle$ must be one of the pairs $\langle 0,3\rangle,\langle 1,2\rangle,\langle 2,1\rangle$, or $\langle 3,0\rangle$. Thus, the $x^3$ term in the product is

$$(a_0x^0)(b_3x^3)+(a_1x^1)(b_2x^2)+(a_2x^2)(b_1x^1)+(a_3x^3)(b_0x^0)=\sum_{k=0}^3(a_kx^k)(b_{3-k}x^{3-k})\;,$$

or

$$\left(a_0b_3+a_1b_2+a_2b_1+a_3b_0\right)x^3=\left(\sum_{k=0}^3a_kb_{3-k}\right)x^3\;.$$

If you generalize this to an arbitrary power $n$ of $x$, you get

$$(a_0x^0)(b_nx^n)+(a_1x^1)(b_{n-1}x^{n-1})+\ldots+(a_{n-1}x^{n-1})(b_1x^1)+(a_nx^n)(b_0x^0)=\sum_{k=0}^n(a_kx^k)(b_{n-k}x^{n-k})\;,$$

or

$$(a_0b_n+a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0)x^n=\left(\sum_{k=0}^na_kb_{n-k}\right)x^n\;.$$

In other words, the coefficient of $x^n$ in the product series is

$$\sum_{k=0}^na_kb_{n-k}\;.$$

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  • $\begingroup$ Thanks a lot! This cleared things up perfectly, and reminded me to try breaking things apart instead of staring at them, trying to make sense of things. $\endgroup$ Commented Jul 22, 2013 at 12:05
  • $\begingroup$ @Yam: You’re welcome! $\endgroup$ Commented Jul 22, 2013 at 12:09
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Multiply $$(ax^2+bx^3)(cx^2+dx^3).$$

You use distributive law to open brackets. In what ways can you get a term with $x^5$?

The degrees of the terms you multiply from each factor should add up to $5$. In general you are applying distributive and gathering together terms of degree $5$. So, you multiply together terms from each factor with degrees that add up to $k$. If the first has degree $j$ the second must have degree $k-j$ in order to add up to $k$.

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