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Simple reduction formula

  1. Apr 26, 2007 #1
    Given that [tex] I_{m,n}=\int_0^1x^m(1-x)^ndx [/tex] prove that [tex] I_{m,n} = \frac{n}{m+n+1}I_{m,n-1} [/tex] and evaluate [tex] I_{4,4} [/tex]

    I set the integral up as follows: [tex]\int_0^1(x^m(1-x))(1-x)^{n-1}dx \\[/tex]
    [tex]= \int_0^1 x^m(1-x)^{n-1}dx - \int_0^1 x^{m+1}(1-x)^{n-1}dx\\[/tex]
    [tex]= \frac{m}{m-n+1}I_{m-1,n} \\ [/tex] which is not what is requested, I would welcome help on how to get [tex] I_{m,n-1} [/tex]. Thanks for the help.
  2. jcsd
  3. Apr 26, 2007 #2


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    You seem to have [itex]I_{m,n} = I_{m,n-1} - I_{m+1,n-1}[/itex].

    But if what you posted afterwards is correct, you can transform [itex]I_{m+1,n-1}[/itex] into [itex]\frac{(m+1)}{(m+1)-(n-1)+1} I_{m,n-1}[/itex] and you should be done
    Last edited: Apr 26, 2007
  4. Apr 26, 2007 #3


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    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?
  5. Apr 26, 2007 #4
    I integrated a second time to get [tex]I_{m,n}= -\frac{m}{n}I_{m-1,n} + \frac{m+1}{n}I_{m,n}[/tex], I may have made a simple error or two, I will have to redo it.
  6. Apr 26, 2007 #5


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    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.
  7. Apr 26, 2007 #6
    I'll do what you said do. Thanks for the help.
  8. Apr 27, 2007 #7
    Yesterday I got the following: [tex] I_{m,n} = I_{m,n-1} - I_{m+1,n-1} [/tex]...(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):
    [tex] I_{m+1,n-1} = \int x^{m+1}(1-x)^{n-1}dx \\ [/tex]
    [tex] = \int(x^{m+1}(1-x))(1-x)^{n-2} dx \\ [/tex]
    [tex] = -\frac{m+1}{n-1} \int x^m(1-x)^{n-1} dx \\[/tex]
    [tex]+ \frac{m+2}{n-1} \int x^{m+1}(1-x)^{n-1} dx \\ [/tex]
    Therefore I get: [tex] I_{m+1,n-1}(n-m-3) = -(m+1)I_{m,n-1} \\[/tex]. Substituting into equation ...(i)
    [tex] I_{m,n}=I_{m,n-1}(\frac{-n+2}{m-n+3})[/tex]
  9. Apr 27, 2007 #8


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    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.
  10. Apr 27, 2007 #9
    Yes I agree with you about I(m,n)=integral(f(x)*dx)=-integral(x*f'(x)dx)
  11. Apr 27, 2007 #10


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    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
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