help with convolution proof

mmattson07

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
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...

silencecloak

Drexel28

MHF Hall of Honor
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

silencecloak

simplependulum

MHF Hall of Honor
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 ,

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

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

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

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

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

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

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

$$\displaystyle = \frac{m! n!}{(m+n+1)!}$$

silencecloak

silencecloak

What would be the trick for the first part of the problem?

silencecloak

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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mmattson07

To prove (a) can we use the result that

with $$\displaystyle u= \tau$$ and$$\displaystyle t=1$$ ?

Drexel28

MHF Hall of Honor
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 $$\displaystyle B(m+1,n+1)=\frac{\Gamma(n+1)\Gamma(m+1)}{\Gamma(n+1+m+1)}=\frac{m!n!}{(m+n+1)!}$$

silencecloak

[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:

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

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