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Math Help - Using a Reduction Formula

  1. #1
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    Using a Reduction Formula

    I want to prove equation (1). Note that the equation (2) can be proven by the reduction formula for \int sin^n(x)dx, but I'll just write the formula down since I know it's true.

    I want to show that:
    ----------------------------------------------------
    \int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\frac{(2)(4)(6  )...(2n)}{(3)(5)(7)...(2n+1)}
    ----------------------------------------------------

    This seems rather difficult, and I'm not entirely sure where to begin. I'm an amateur, so I don't have much experience with proofs like this. I'll write the equation like this:

    (1) \int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\prod_{i=1}^n\  frac{2i}{2i+1}

    Using the following reduction formula:

    (2) \int_0^{\frac{\pi}{2}}sin^n(x)dx=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}sin^{n-2}xdx

    I can write the hypothesis for n=k

    \int_0^{\frac{\pi}{2}}sin^{2k+1}(x)dx=\frac{(2k+1)-1}{2k+1}\int_0^{\frac{\pi}{2}}sin^{(2k+1)-2}(x)dx

    So now I have:

    \frac{2k}{2k+1}\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=1}^k\frac{2i}{2i+1}

    Factoring the left side:

    \prod_{i=1}^k\frac{2i}{2i+1}=\frac{2k}{2k+1}\prod_  {i=2}^k\frac{2(i-1)}{2i-1}


    (3) \int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=2}^k\frac{2(i-1)}{2i-1}

    All I've managed to do is write the inductive hypothesis. It's really frustrating that I can't proceed. I can't figure out how to connect one side to the other. Wring the next term for n=k+1 doesn't seem to get me anywhere. Can anyone offer some insight?
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    I want to prove equation (1). Note that the equation (2) can be proven by the reduction formula for \int sin^n(x)dx, but I'll just write the formula down since I know it's true.

    I want to show that:
    ----------------------------------------------------
    \int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\frac{(2)(4)(6  )...(2n)}{(3)(5)(7)...(2n+1)}
    ----------------------------------------------------

    This seems rather difficult, and I'm not entirely sure where to begin. I'm an amateur, so I don't have much experience with proofs like this. I'll write the equation like this:

    (1) \int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\prod_{i=1}^n\  frac{2i}{2i+1}

    Using the following reduction formula:

    (2) \int_0^{\frac{\pi}{2}}sin^n(x)dx=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}sin^{n-2}xdx

    I can write the hypothesis for n=k

    \int_0^{\frac{\pi}{2}}sin^{2k+1}(x)dx=\frac{(2k+1)-1}{2k+1}\int_0^{\frac{\pi}{2}}sin^{(2k+1)-2}(x)dx

    So now I have:

    \frac{2k}{2k+1}\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=1}^k\frac{2i}{2i+1}

    Factoring the left side:

    \prod_{i=1}^k\frac{2i}{2i+1}=\frac{2k}{2k+1}\prod_  {i=2}^k\frac{2(i-1)}{2i-1}


    (3) \int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=2}^k\frac{2(i-1)}{2i-1}

    All I've managed to do is write the inductive hypothesis. It's really frustrating that I can't proceed. I can't figure out how to connect one side to the other. Wring the next term for n=k+1 doesn't seem to get me anywhere. Can anyone offer some insight?
    It seems to me like you have all the pieces. Just assemby them.

    Assume true for n = k so

    \int_0^{\frac{\pi}{2}}sin^{2k+1}x\,dx=\prod_{i=1}^  k\frac{2i}{2i+1}

    and prove true for n = k+1, i.e.

    \int_0^{\frac{\pi}{2}}sin^{2k+3}x\,dx=\prod_{i=1}^  {k+1}\frac{2i}{2i+1} .

    From your reduction formula

    \int_0^{\frac{\pi}{2}}sin^{2k+3} x\,dx = \frac{2k+2}{2k+3}\int_0^{\frac{\pi}{2}}sin^{2k+1} x\,dx (from your reduction formula)

    = \frac{2k+2}{2k+3} \prod_{i=1}^k\frac{2i}{2i+1}

    = \prod_{i=1}^{k+1}\frac{2i}{2i+1}

    as required.
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  3. #3
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    Quote Originally Posted by Danny View Post
    It seems to me like you have all the pieces. Just assemby them.

    Assume true for n = k so

    \int_0^{\frac{\pi}{2}}sin^{2k+1}x\,dx=\prod_{i=1}^  k\frac{2i}{2i+1}

    and prove true for n = k+1, i.e.

    \int_0^{\frac{\pi}{2}}sin^{2k+3}x\,dx=\prod_{i=1}^  {k+1}\frac{2i}{2i+1} .

    From your reduction formula

    \int_0^{\frac{\pi}{2}}sin^{2k+3} x\,dx = \frac{2k+2}{2k+3}\int_0^{\frac{\pi}{2}}sin^{2k+1} x\,dx (from your reduction formula)

    = \frac{2k+2}{2k+3} \prod_{i=1}^k\frac{2i}{2i+1}

    = \prod_{i=1}^{k+1}\frac{2i}{2i+1}

    as required.
    I get it now, it was much easier than I thought.
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