help with convolution proof

Apr 2009
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Hey MHF i need some help with this proof.

If\(\displaystyle f(g)= t^m\) and \(\displaystyle g(t)= t^n\), where m and n are positive integers, show that:

\(\displaystyle (f*g)(t)= t^{m+n+1} \int_0^1 \! u^m(1-u)^n \, du.\)



show : \(\displaystyle \int_0^1 \! u^m(1-u)^n \, du = \frac{m!n!}{(m+n+1)!}\)
 

Prove It

MHF Helper
Aug 2008
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show : \(\displaystyle \int_0^1 \! u^m(1-u)^n \, du = \frac{m!n!}{(m+n+1)!}\)
The Binomial Theorem and the fact you can switch integral and summation signs should help you for this one.

Are \(\displaystyle m\) and \(\displaystyle n\) positive integers? I assume so since your answer involves factorials...
 
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Drexel28

MHF Hall of Honor
Nov 2009
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Hey MHF i need some help with this proof.

If\(\displaystyle f(g)= t^m\) and \(\displaystyle g(t)= t^n\), where m and n are positive integers, show that:

\(\displaystyle (f*g)(t)= t^{m+n+1} \int_0^1 \! u^m(1-u)^n \, du.\)



show : \(\displaystyle \int_0^1 \! u^m(1-u)^n \, du = \frac{m!n!}{(m+n+1)!}\)
I suppose you don't know that \(\displaystyle \int_0^1 u^m(1-u)^ndu=\beta(m+1,n+1)?\).

Beta function - Wikipedia, the free encyclopedia
 
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simplependulum

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Jan 2009
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Hey MHF i need some help with this proof.

If\(\displaystyle f(g)= t^m\) and \(\displaystyle g(t)= t^n\), where m and n are positive integers, show that:

\(\displaystyle (f*g)(t)= t^{m+n+1} \int_0^1 \! u^m(1-u)^n \, du.\)


show : \(\displaystyle \int_0^1 \! u^m(1-u)^n \, du = \frac{m!n!}{(m+n+1)!}\)


Use integration by parts \(\displaystyle n \) times ,

\(\displaystyle \int_0^1 \! u^m(1-u)^n \, du \)

\(\displaystyle = \left[ \frac{ u^{m+1} }{m+1} (1-u)^n \right]_0^1 + \frac{n}{m+1} \int_0^1 u^{m+1} (1-u)^{n-1}~du\)

\(\displaystyle = \frac{n}{m+1} \int_0^1 u^{m+1} (1-u)^{n-1}~du \)

\(\displaystyle = \frac{n(n-1)}{(m+1)(m+2)} \int_0^1 u^{m+2} (1-u)^{n-2}~du\)


\(\displaystyle = ... = \frac{n(n-1)...1}{(m+1)(m+2)(m+3)...(m+n)} \int_0^1 u^{m+n}~du \)

\(\displaystyle = \frac{n!}{(m+1)(m+2)(m+3)...(m+n)(m+n+1)} \)

\(\displaystyle = \frac{m! n!}{m! (m+1)(m+2)(m+3)...(m+n)(m+n+1)} \)

\(\displaystyle = \frac{m! n!}{(m+n+1)!} \)
 
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May 2008
138
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May 2008
138
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It really looks like the first part of the proof is something with the Gamma function, but I can't pinpoint it.

Been staring at it for 3 days now :/

Edit: is the explanation of the beta function i commented on correct?

For the first part:

\(\displaystyle
(f*g)(t)= \int_0^t \! \tau^m(t-\tau)^n \, d\tau.
\)

I have been playing with the fact that \(\displaystyle \Gamma \left( x +1\right) = x\Gamma \left( x \right)\) and using that with an integral proof I found, but to no avail.

I am also not sure how the \(\displaystyle \tau \) and integrals limits get changed between the two steps. I know \(\displaystyle u \) and \(\displaystyle \tau \) are just symbols...so maybe they are just changed for the sake of being changed. Throwing me off because my teacher has never used u for those problems
 
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Apr 2009
85
0
To prove (a) can we use the result that



with \(\displaystyle u= \tau\) and\(\displaystyle t=1\) ?
 

Drexel28

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Nov 2009
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Berkeley, California
It really looks like the first part of the proof is something with the Gamma function, but I can't pinpoint it.

Been staring at it for 3 days now :/

Edit: is the explanation of the beta function i commented on correct?

For the first part:

\(\displaystyle
(f*g)(t)= \int_0^t \! \tau^m(t-\tau)^n \, d\tau.
\)

I have been playing with the fact that \(\displaystyle \Gamma \left( x +1\right) = x\Gamma \left( x \right)\) and using that with an integral proof I found, but to no avail.

I am also not sure how the \(\displaystyle \tau \) and integrals limits get changed between the two steps. I know \(\displaystyle u \) and \(\displaystyle \tau \) are just symbols...so maybe they are just changed for the sake of being changed. Throwing me off because my teacher has never used u for those problems
[It is the gamma function! Like I said your integral is equal to \(\displaystyle B(m+1,n+1)=\frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+1+m+1)}=\frac{m!n!}{(m+n+1)!}\)
 
May 2008
138
0
[It is the gamma function! Like I said your integral is equal to \(\displaystyle B(m+1,n+1)=\frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+1+m+1)}=\frac{m!n!}{(m+n+1)!}\)
I'm sorry Drexel. I was referring to the first part of the problem. It's not very easy to see due to how the post was formatted. Thanks for your help with part A though!(Rofl)

Part A:

show that:

\(\displaystyle (f*g)(t)= t^{m+n+1} \int_0^1 \! u^m(1-u)^n \, du\)
 
Last edited:
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