# 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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