# Convolution of a polynomial with itself

1. May 11, 2010

### Char. Limit

Again, in my quest to learn things I won't use in a class for at least a year, I've been looking at convolutions. Specifically, after finishing the multiple choice section of an AP Chemistry test 50 minutes early, I looked at the convolution of a polynomial with itself. I'm confused about one thing, though.

Here's what I have:

$$t^m \star t^m = t^{2m+1} \sum_{n=1}^{m+1} \frac{\binom{m}{n}}{n+m}$$

However, I don't like that sum in the solution, I don't think I computed it right, and I want it gone. Is there any way to do that?

2. May 11, 2010

### Staff: Mentor

That doesn't look like the convolution I learned quite a few years ago, which was this:
$$(f \star g)(t) = \int_{-\infty}^{\infty} f(\tau)g(t - \tau)d\tau$$

3. May 12, 2010

### Char. Limit

Hmm... I used the same formula, but the bounds I learned were from 0 to t. Using that, I came up with the OP formula for the convolution of a polynomial (really a monomial, now that I think about it, but I can't change the title) with itself.

4. May 12, 2010

### Staff: Mentor

I got the formula I posted from Wikipedia, but it's the same formula that I remember from when I studied convolution way back when. If you can show us how you got your summation formula from the definition it would be helpful.

5. May 12, 2010

### Char. Limit

It was just the pattern I noticed. For example...

For t:

$$t^3(\frac{1}{2}-\frac{1}{3})$$

For t^2:

$$t^5(\frac{1}{3}-\frac{2}{4}+\frac{1}{5})$$

For t^3:

$$t^7(\frac{1}{4}-\frac{3}{5}+\frac{3}{6}-\frac{1}{7})$$

For t^4:

$$t^9(\frac{1}{5}-\frac{4}{6}+\frac{6}{7}-\frac{4}{8}+\frac{1}{9})$$

The summation was my attempt at phrasing this pattern, which of course I forgot to put in a (-1)^(n+1). I just noticed that the denominators for t^m's convolution with itself were in consecutive integers from m+1 to 2m+1, and the numberators were each a row of Pascal's triangle, which if I remember right is also the binomial coefficients.