Convolution of a polynomial with itself

Click For Summary

Homework Help Overview

The discussion revolves around the convolution of a polynomial with itself, specifically focusing on the original poster's (OP) expression and their concerns about a summation term in the result. The subject area includes convolutions, polynomial functions, and potentially combinatorial aspects related to binomial coefficients.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The OP presents a convolution formula and expresses confusion regarding a summation term. Other participants share their understanding of the convolution definition and discuss variations in bounds used in the integral. Some participants question the derivation of the OP's summation formula and suggest demonstrating the connection to the convolution definition.

Discussion Status

The discussion is active, with participants exploring different interpretations of the convolution process and sharing their recollections of the definitions. There is an ongoing inquiry into the derivation of the OP's formula and the nature of the summation, indicating a productive exchange of ideas without a clear consensus yet.

Contextual Notes

Participants reference different bounds for the convolution integral, which may indicate variations in their educational backgrounds or the specific contexts in which they learned about convolutions. The OP's desire to simplify their expression suggests a focus on clarity and correctness in mathematical representation.

Char. Limit
Gold Member
Messages
1,222
Reaction score
23
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?
 
Physics news on Phys.org
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
 
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.
 
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.
 
Mark44 said:
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.

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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K