# Simple reduction formula

1. Apr 26, 2007

### John O' Meara

Given that $$I_{m,n}=\int_0^1x^m(1-x)^ndx$$ prove that $$I_{m,n} = \frac{n}{m+n+1}I_{m,n-1}$$ and evaluate $$I_{4,4}$$

I set the integral up as follows: $$\int_0^1(x^m(1-x))(1-x)^{n-1}dx \\$$
$$= \int_0^1 x^m(1-x)^{n-1}dx - \int_0^1 x^{m+1}(1-x)^{n-1}dx\\$$
$$= \frac{m}{m-n+1}I_{m-1,n} \\$$ which is not what is requested, I would welcome help on how to get $$I_{m,n-1}$$. Thanks for the help.

2. Apr 26, 2007

### Office_Shredder

Staff Emeritus
You seem to have $I_{m,n} = I_{m,n-1} - I_{m+1,n-1}$.

But if what you posted afterwards is correct, you can transform $I_{m+1,n-1}$ into $\frac{(m+1)}{(m+1)-(n-1)+1} I_{m,n-1}$ and you should be done

Last edited: Apr 26, 2007
3. Apr 26, 2007

### Dick

So far you've got I(m,n)=I(m,n-1)-I(m+1,n-1) which is a good start. I don't know how you turned that into what you finally presented. What I did was get a second relation between the I's using integration by parts and use that to eliminate the I(m+1,n-1). DId you do something like that?

4. Apr 26, 2007

### John O' Meara

I integrated a second time to get $$I_{m,n}= -\frac{m}{n}I_{m-1,n} + \frac{m+1}{n}I_{m,n}$$, I may have made a simple error or two, I will have to redo it.

5. Apr 26, 2007

### Dick

How can you 'integrate a second time'?? I don't think what you are doing is very kosher. Try something more like this. If f(x)=x^m*(1-x)^n then do integration by parts like this: integral(f(x)*dx)=-integral(x*d(f(x)))=-integral(x*f'(x)*dx). Notice I could ignore the boundary terms since f(0)=0 and f(1)=0.

6. Apr 26, 2007

### John O' Meara

I'll do what you said do. Thanks for the help.

7. Apr 27, 2007

### John O' Meara

Yesterday I got the following: $$I_{m,n} = I_{m,n-1} - I_{m+1,n-1}$$...(i); then I was wondering what to do next. According to you What I needed to do next; was to get a second relation between the I's. So I try the following (integrating by parts):
$$I_{m+1,n-1} = \int x^{m+1}(1-x)^{n-1}dx \\$$
$$= \int(x^{m+1}(1-x))(1-x)^{n-2} dx \\$$
$$= -\frac{m+1}{n-1} \int x^m(1-x)^{n-1} dx \\$$
$$+ \frac{m+2}{n-1} \int x^{m+1}(1-x)^{n-1} dx \\$$
Therefore I get: $$I_{m+1,n-1}(n-m-3) = -(m+1)I_{m,n-1} \\$$. Substituting into equation ...(i)
$$I_{m,n}=I_{m,n-1}(\frac{-n+2}{m-n+3})$$

8. Apr 27, 2007

### Dick

You didn't really do what I said. I don't know what you did. Do you agree with me that I(m,n)=integral(f(x)*dx)=-integral(x*f'(x)*dx)? Do you see how the integration by parts works? Then just carefully write out -integral(x*f'(x)*dx) and work out what it is in terms of I's. PS you don't really 'integrate' anything. Just rearrange the parts.

9. Apr 27, 2007

### John O' Meara

Yes I agree with you about I(m,n)=integral(f(x)*dx)=-integral(x*f'(x)dx)

10. Apr 27, 2007

### Dick

So what is the RHS? Just differentiate x^m*(1-x)^n and multiply by x. I get two terms. One looks like I(m,n) and the other like I(m+1,n-1). What are the coefficients?

Last edited: Apr 27, 2007