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Math Help - Problem #21: Recurrence Integral

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    Problem #21: Recurrence Integral

    Show that
    \int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}

    Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

    One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as I_n. We have the recurrence relation:
    I_{k + 1} = \left ( \frac{2k + 1}{2k + 3} \right )~ I_k

    -Dan
    Last edited by topsquark; September 2nd 2014 at 01:33 PM.
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    Re: Problem #21: Recurrence Integral

    \int_0^1 (1-x)^n \, dx = \frac{1}{n+1}
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    Forum Admin topsquark's Avatar
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    Re: Problem #21: Recurrence Integral

    Quote Originally Posted by Idea View Post
    \int_0^1 (1-x)^n \, dx = \frac{1}{n+1}
    Whoops! Typo. I have fixed it in the problem statement. Thanks for the catch!

    -Dan
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    Re: Problem #21: Recurrence Integral

    I_{k+1}=\frac{2 (k+1)}{2 k+3}I_k
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    Forum Admin topsquark's Avatar
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    Re: Problem #21: Recurrence Integral



    That's the last time I post a problem on two hours of sleep. Thanks (again) for the catch.

    Quote Originally Posted by topsquark View Post
    Show that
    \int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}

    Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

    One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as I_n. We have the recurrence relation:
    I_{k + 1} = \left ( \frac{2k + 2}{2k + 3} \right )~ I_k
    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Re: Problem #21: Recurrence Integral

    No takers this time around. Let's see if I can post the solution without any typos!

    Spoiler:
    Let's look at the recurrence relation first.

    Let I_{k + 1} = \int_0^1 (1 - x^2)^{k + 1}~dx

    Do this by parts:
    \int u ~ dv = uv - \int v~du

    I am choosing u = (1 - x^2)^{k + 1} and dv = dx

    This gives
    I_{k + 1} = \left . x \cdot (1 - x^2)^{k + 1} \right |_0^1 - \int_0^1 (k +1)x(1 - x^2)^k(-2x)~dx

    The first term vanishes leaving
    I_{k + 1} = (2k + 2) \int_0^1 x^2 (1 - x^2)^k~dx

    Now for a cool little trick:
    I_{k + 1} = (2k + 2) \int_0^1 \left [1 - (1 - x^2) \right ] (1 - x^2)^k~dx

    I_{k + 1} = (2k + 2) \left [ \int_0^1 (1 - x^2)^k~dx - \int_0^1 (1 - x^2)^{k + 1}~dx \right ]

    I_{k + 1} = (2k + 2)I_k - (2k + 2)I_{k + 1}

    So finally
    I_{k + 1} = \left ( \frac{2k + 2}{2k + 3} \right ) I_k

    Now for the induction step:

    Base case: k = 0:
    \int_0^1 ( 1 - x^2)^0~dx = 1 = \frac{2^{2 \cdot 0} (0!)^2}{1!} (Check!)

    Now let's assume the theorem is true for some integer k. We need to show that the k + 1 case works as well. So:

    I_{k +1} = \left ( \frac{2k + 2}{2k + 3} \right ) I_k  = \left ( \frac{2k + 2}{2k + 3} \right ) \cdot \frac{2^{2k} ( k!)^2}{(2k + 1)!} = \frac{2(k + 1)2^{2k}(k!)^2}{(2k + 3)(2k + 1)!}

    Multiply by 1:
    I_{k + 1} = \frac{2(k + 1)}{2k + 2} \cdot \frac{2(k + 1)2^{2k}(k!)^2}{(2k + 3)(2k + 1)!}

    I_{k + 1} = \frac{ \left [ 2(k + 1) \right ] ^2 2^{2k}(k!)^2}{(2k + 3)(2k + 2)(2k + 1)!}

     I_{k + 1} = \frac{2^{2(k + 1)} \left [ (k + 1)! \right ]^2}{\left [ 2(k + 1) + 1 \right ]!}

    So we have that the kth case implies the k+1th case. Therefore etc.

    -Dan
    Last edited by topsquark; September 7th 2014 at 04:29 PM.
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