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Math Help - VERY HARD! Reduction Formula Question

  1. #1
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    VERY HARD! Reduction Formula Question

    The question is as follows:

    Given that I_{n} = \int_0^\frac{\pi}{2} \sin^n (x) dx where n \epsilon N (used capital N to show the sign for natural number or normal number, not too sure what it is)

    a) show that I_{2n} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2}

    b) show further that I_{2n-1} I_{2n} = \frac{\pi}{4n}

    really need help with this one, its bugging me so much that i cant get it
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Stylis10 View Post
    The question is as follows:

    Given that I_{n} = \int_0^\frac{\pi}{2} \sin^n (x) dx where n \epsilon N (used capital N to show the sign for natural number or normal number, not too sure what it is)

    a) show that I_{2n} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2}
    So write I_{2n} :

    I_{2n}=\int_0^{\pi/2} \sin^{2n} (x) ~dx

    But \sin^{2n}(x)=\sin^{2n-2}(x) \sin^2(x)=\sin^{2n-2}(x)-\cos^2(x) \sin^{2n-2}(x)

    So I_{2n}=I_{2n-2}-\int_0^{\pi/2} \cos^2(x) \sin^{2n-2}(x) ~dx \qquad {\color{red}\star}

    Now let's study J=\int_0^{\pi/2} \cos^2(x) \sin^{2n-2}(x) ~dx
    Integrate by parts with u=\cos(x) \Rightarrow du=-\sin(x) and dv=\cos(x)\sin^{2n-2}(x) \Rightarrow v=\frac{1}{2n-1} \cdot \sin^{2n-1}(x)

    Hence we have :
    J=\left. \frac{1}{2n-1} \cos(x) \sin^{2n-1}(x) \right|_0^{\pi/2}+\frac{1}{2n-1} \int_0^{\pi/2} \sin^{2n}(x) ~dx
    \boxed{J=\frac{1}{2n-1} \cdot I_{2n}}

    Now substitute in {\color{red}\star} :

    I_{2n}=I_{2n-2}-\frac{1}{2n-1} \cdot I_{2n}

    Hence I_{2n}=\frac{2n-1}{2n} \cdot I_{2(n-1)} (recursive formula)



    Now it should be easy... And I voluntarily skipped steps so that you'll have to think a little
    Last edited by Moo; February 21st 2009 at 08:51 AM.
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  3. #3
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    Thanxs Moo, but i got all of what you've done an upto that point, i however don't understand the factorails bit, its just eluding me
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  4. #4
    Moo
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    Quote Originally Posted by Stylis10 View Post
    Thanxs Moo, but i got all of what you've done an upto that point, i however don't understand the factorails bit, its just eluding me
    You could have shown your work... I wouldn't have bothered explaining these things


    So for the factorial thing, you can do it by induction.
    If n=0, then I_0=\pi/2 but \left. \frac{(2n)!\pi}{2^{2n+1}(n!)^2} \right|_{n=0}=\pi/2. So it is ok.
    Now assume I_{2n}=\frac{(2n)!\pi}{2^{2n+1}(n!)^2}

    I_{2(n+1)}=\frac{2(n+1)-1}{2(n+1)} \cdot I_{2n} (by the recursive formula)
    =\frac{2n+1}{2(n+1)} \cdot \frac{(2n)!\pi}{2^{2n+1}(n!)^2}
    =\frac{(2n+1)(2n+2)}{2^2 (n+1)^2} \cdot \frac{(2n)!\pi}{2^{2n+1}(n!)^2}

    It's easy to prove that this equals \frac{(2(n+1))! \pi}{2^{2(n+1)+1} [(n+1)!]^2}=\frac{(2n+2)!\pi}{2^{2n+3} [(n+1)!]^2}, which is what you are looking for.




    Question 2. > similarly, find a recursive formula for I_{2n-1}
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  5. #5
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    WOW!, than you so much!, i never once thought to use induction, ive spent near enough 5 hours trying to show it n i jus coudlnt get it
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