Convolution of a polynomial with itself

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The discussion centers on the convolution of a polynomial with itself, specifically the expression t^m ∗ t^m = t^{2m+1} ∑_{n=1}^{m+1} (binom{m}{n})/(n+m). The author expresses confusion about the summation in the solution and seeks to eliminate it. They recall a different convolution formula learned previously, which had different bounds. The author also shares observations about patterns in the coefficients derived from Pascal's triangle and attempts to clarify the formulation of the summation. The conversation highlights the complexities and nuances of polynomial convolutions.
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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?
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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