Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Check that this is actually what it works out to if you make both sums finite. –  Tobias Kildetoft Jul 22 '13 at 11:46
    
@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. –  Yam Marcovic Jul 22 '13 at 11:57
1  
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. –  Tobias Kildetoft Jul 22 '13 at 12:01

2 Answers 2

up vote 5 down vote accepted

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

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

share|improve this answer
    
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. –  Yam Marcovic Jul 22 '13 at 12:05
    
@Yam: You’re welcome! –  Brian M. Scott Jul 22 '13 at 12:09

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 $k$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.