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.

Let $x=(x_1,\ldots,x_n)\in[-1,1]^n$, $k=(k_1,\ldots,k_n)$ and $\mathbb{N}^n_0=\mathbb{N}^n \cup\{0,\ldots,0\}$. Define $ x^k$ by $$x^k=\prod_{i=1}^n x_i^{k_i}$$

Show that $$\sum_{k\in\mathbb{N}^n_0}x^k=\prod_{i=1}^n (1-x_i)^{-1}$$

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

The factors $(1-x_i)^{-1}$ can be expanded as convergent geometric series $1+x_i+x_i^2+\cdots,$ since $x_i\in[-1,1].$ Thus, $\prod_{i=1}^n(1-x_i)^{-1}=(\sum_k x_1^k)\cdots(\sum_k x_n^k).$ Expanding we get as summands all possible products $x_1^{k_1}\cdots x_n^{k_n}$ with $(k_1,\ldots, k_n)\in\Bbb N^n,$ where $\Bbb N$ is nonnegative integers.

There is some justification to be made about why expanding is equal to the original product of series.

share|improve this answer
add comment

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.