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Math Help - Reduction Formulae

  1. #1
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    Reduction Formulae

    I need some help on this reduction formulae question.

    The question is:
    Given that  \<br />
I_n  = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx<br />
\
    Show that
    \<br />
I_{2n}  = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }}<br />
\

    I know that the reduction formulae is
    \<br />
I_n  = \frac{{n - 1}}{n}I_{n - 2} <br />
\
    But I don't know how manipulate it into the above form. Could someone help please?
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by free_to_fly View Post
    I need some help on this reduction formulae question.

    The question is:
    Given that  \<br />
I_n  = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx<br />
\
    Show that
    \<br />
I_{2n}  = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }}<br />
\

    I know that the reduction formulae is
    \<br />
I_n  = \frac{{n - 1}}{n}I_{n - 2} <br />
\
    But I don't know how manipulate it into the above form. Could someone help please?
    I_0 = \int_0^{\pi /2} sin^0(x)dx = \int_0^{\pi /2} dx = [x]_0^{\pi /2} = \pi /2

    I_1 = \int_0^{\pi /2} sin^1(x)dx = \int_0^{\pi /2} sin(x) dx = [-cos(x)]_0^{\pi /2} = 1

    -----
    n=1
    I_2 = \frac 12 * \frac{\pi}2

    -----
    n=2
    I_4 = \frac 34 * \frac 12 * \frac{\pi}2

    -----
    n=3
    I_6 = \frac 56 *\frac 34 * \frac 12 * \frac{\pi}2

    -----
    n=4
    I_8 = \frac 78 * \frac 56 *\frac 34 * \frac 12 * \frac{\pi}2



    By now you should be noticing the pattern.

    Lets look at the numerator:
    the leading term is always 2n-1 and it skips every other number. so we can write it like this:

    7*5*3*1*\pi

    = \frac{8*7*6*5*4*2*1}{8*6*4*2}\pi

    =\frac{8!}{4(2)*3(2)*2(2)*1(2)}\pi

    = \frac{8!}{(2*2*2*2)(4*3*2*1)}\pi

    = \frac{8!}{2^4*4!}\pi

    = \frac{(2*4)!}{2^4*4!}\pi

    And of course you notice that this n=4 and that this really means the numerator is

    = \frac{(2*n)!}{2^n*n!}\pi

    -----

    And now we look at the denominator
    8*6*4*2*2 And we notice that the leading term is always 2n. So we evaluate it a bit.

    8*6*4*2*2

    =4(2)*3(2)*2(2)*2(1)*2

    =(2*2*2*2)(4*3*2*1)*2

    =2^4*4!*2

    =(2*2^4)*4!

    =2^{4+1}*4!

    And of course you notice that n=4 and that this holds for any iteration. So the denominator becomes

    =2^{n+1}*n!


    -----

    Then you consolidate the function for the numerator and denominator
    \frac{\frac{(2*n)!}{2^n*n!}\pi}{2^{n+1}*n!}

    =\frac{(2*n)!}{2^n*n!}\pi * \frac 1{2^{n+1}*n!}

    =\frac{(2n)!\pi}{2^n*n!*2^{n+1}*n!}

    =\frac{(2n)!\pi}{(2^n*2^{n+1})(n!*n!)}

    =\frac{(2n)!\pi}{2^{n+1+n}(n!)^2}

    =\frac{(2n)!\pi}{2^{2n+1}(n!)^2}
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  3. #3
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    Quote Originally Posted by free_to_fly View Post
    I need some help on this reduction formulae question.

    The question is:
    Given that  \<br />
I_n  = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx<br />
\
    Show that
    \<br />
I_{2n}  = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }}<br />
\

    I know that the reduction formulae is
    \<br />
I_n  = \frac{{n - 1}}{n}I_{n - 2} <br />
\
    But I don't know how manipulate it into the above form. Could someone help please?
    You could also read this.
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    You could also read this.
    Interestingly enough, 4.5 months ago I did this: http://www.mathhelpforum.com/math-he...443-post2.html

    When I become a cyborg, I won't forget things like this.
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