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Homework Help: Induction to prove an expanded integral

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\int_0^1{x^n (1-x)^r}\,dx = \frac{n!}{(r+1)(r+2)...(r+n+1)}[/itex]

    2. Relevant equations
    n≥0, r≥0

    3. The attempt at a solution

    i solved it for the case n=0, where RHS and LHS are the same. however, im having trouble integrating the LHS for the induction step where n=k+1.

    i'm trying to do it by parts, but since the integral keeps expanding i am having trouble with it. any help would be greatly appreciated. thanks!
     
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  3. Jan 20, 2012 #2

    SammyS

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    I suspect you will have to use integration by parts.

    Let u = xk+1 , dv = (1-x)r dx .

    Then do some algebra.
     
  4. Jan 20, 2012 #3
    i think this is what i'm having trouble with. how do i deal with expanded integral?
     
  5. Jan 20, 2012 #4

    SammyS

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    What do you get after applying integration by parts?
     
  6. Jan 20, 2012 #5
    [itex] \frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^k[/itex]


    CORRECTION:
    restriction on r as stated in problem is,
    [itex]r>-1[/itex]
     
  7. Jan 20, 2012 #6

    SammyS

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    Use " \displaystyle " in LaTeX to make the fractions & ∫ sign larger.
    [itex] \displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx[/itex]​
    I also inserted the dx .

    The following equation has been edited.
    That integral can be broken up as follows:
    [itex]\displaystyle\int(1-x)(1-x)^{r} x^k dx=\int(1-x)^{r} x^kdx-\int\,x(1-x)^{r} x^{k}dx[/itex]​

    And what you're trying to evaluate is the integral [itex]\displaystyle\int(1-x)^{r} x^{k+1}dx\,,[/itex] which is the same as the second integral on the right hand side of the equation immediately above..

    To make the algebra a bit easier, define Ik as [itex]\displaystyle I_k=\int(1-x)^{r} x^{k}dx\,.[/itex]

    Solve for Ik+1 .

    Added in Edit:
    Fixed typo above. I messed that all up with cutting & pasting !
     
    Last edited: Jan 20, 2012
  8. Jan 20, 2012 #7
    I've applied integration by parts to Ik+1

    [itex]\int(1-x)^{r} x^{(k+1)}dx[/itex]

    by setting u=xk+1 and dv=(1-x)r and the result was,

    [itex] \displaystyle\frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\int(1-x)^{(r+1)} x^kdx[/itex]​


    im confused about where to proceed from here.
     
  9. Jan 20, 2012 #8

    SammyS

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    I had some cut/paste errors in my previous post. :redface: Sorry about that !!! :redface:

    So, my previous post basically says ... The integral you get in the integration by parts is [itex]\displaystyle \int(1-x)^{(r+1)} x^kdx = I_k - I_{k+1}.[/itex]

    Rewrite your integration by parts result as:
    [itex] \displaystyle I_{k+1}= \frac{-x^{(k+1)}(1-x)^{(r+1)}}{(r+1)} + \frac{k+1}{r+1}\left(I_k - I_{k+1}\right)[/itex]​
    Solve that for Ik+1. (The first term is zero when evaluated at x=0 and x=1.)
     
  10. Jan 20, 2012 #9
    i still feel as if im integrating in circles and i don't see the bigger picture.

    EDIT:
    just saw your most recent post. i'll try your suggestion.
    thanks!
     
  11. Jan 20, 2012 #10
    Do you mean solving for Ik+1 and then evaluating the integral between x=0 and x=1?

    i'm ending up with Ik+1 = 0
     
  12. Jan 20, 2012 #11

    SammyS

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    Yes, solve for Ik+1. It doesn't give Ik+1 = 0 .

    Remember that [itex]\displaystyle I_k=\frac{k\,!}{(r+1)(r+2)...(r+k+1)}[/itex] when evaluated from x=0 to x=1.
     
  13. Jan 20, 2012 #12
    EDIT:
    this post was confusing. sorry!
    got it now though.
     
    Last edited: Jan 20, 2012
  14. Jan 20, 2012 #13
    Hey!
    I actually just got it.

    Thank you so much!
     
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