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Math Help - Show through C domain.

  1. #1
    Member courteous's Avatar
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    Question Show through C domain.

    Show that \displaystyle{  \int_0^\pi \sin^{2n}(\theta)d\theta=\pi\frac{(2n)!}{2^{2n}(n!  )^2}  }.
    From \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i} follows that 2i\sin(\theta)e^{i\theta}=e^{i2\theta}-1.

    Exponentiating by 2n gets, using  \theta and -\theta, respectively:
    1. \displaystyle{  2^{2n}i^{2n}\sin^{2n}(\theta)e^{i2n\theta}=(e^{i2\  theta}-1)^{2n}  }
    2. \displaystyle{  2^{2n}i^{2n}\sin^{2n}(\theta)e^{-i2n\theta}=(e^{-i2\theta}-1)^{2n}  }.


    Subtracting 2. from 1.:
    \displaystyle{  2^{2n}i^{2n}\sin^{2n}(\theta)\left[\underbrace{e^{in\theta}-e^{-in\theta}}_{2i\sin(2n\theta)}\right]=(e^{i2\theta}-1)^{2n}-(e^{-i2\theta}-1)^{2n}  }.


    Integrating both sides from 0 to \pi: <br />
2^{2n+1}i^{2n+1}\int_0^\pi \sin^{2n}(\theta)\sin(2n\theta)d\theta=\int_0^\pi\  left\{ (e^{i2\theta}-1)^{2n}-(e^{-i2\theta}-1)^{2n} \right\} \stackrel{?}{=}0


    What to do with \sin(2n\theta) inside integral, and, is RHS really zero? Assuming everything is OK before that.






    ?: Two leading 1's cancel and all other terms in expansion would give \displaystyle{  \int_0^\pi e^{i2k\theta}d\theta=0 \text{ ; } k=\pm 1, \pm 2, \hdots, \pm 2n }
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  2. #2
    Behold, the power of SARDINES!
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  3. #3
    Member courteous's Avatar
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    This result was known to Wallis. So, would you help me get it without Cauchy?

    But I will surely read your (nicely written out) question, as I've just recently been introduced to Cauchy's two theorems.
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  4. #4
    Behold, the power of SARDINES!
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    Here is a way it can be done without using complex analysis

    Define I_n=\int_{0}^{\pi}\sin^{2n}(x)dx

    Note that I_0=\pi

    Now if n \ge 1

    I_n=\int_{0}^{\pi}\sin^{2n-1}(x)\sin(x)dx

    Now use integration by parts with u=\sin^{2n-1}(x) \implies du=(2n-1)\sin^{2n-2}(x)\cos(x) and
    dv=\sin(x)dx \implies v=-\cos(x)

    This gives

    I_n=-\sin^{2n-1}(x)\cos(x)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)\cos^2(x)dx

    I_n=(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)\cos^2(x)dx-(2n-1)\int_{0}^{\pi}\sin^{2n}(x)dx

    Now using the relationship above we get

    I_n=(2n-1)I_{n-1}-(2n-1)I_{n} \iff (2n)I_n=(2n-1)I_{n-1} \iff I_n=\frac{2n-1}{2n}I_{n-1}

    Now with

    I_0=\pi
    The rest can be proved via induction.
    Last edited by TheEmptySet; October 24th 2010 at 12:23 PM. Reason: correct typo
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  5. #5
    Member courteous's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    This gives

    I_n=-\sin^{2n-1}(x)\cos(x)\bigg|_{0}^{\pi}+(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)\cos^2(x)dx

    I_n=(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)\cos^2(x)dx-(2n-1)\int_{0}^{\pi}\sin^{2n}(x)dx
    I think you meant
    I_n=(2n-1)\int_{0}^{\pi}\sin^{2n-2}(x)dx-(2n-1)\int_{0}^{\pi}\sin^{2n}(x)dx .

    Quote Originally Posted by TheEmptySet View Post
    Now using the relationship above we get

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

    Now with

    I_0=\pi
    The rest can be proved via induction.
    I wrote it out ...
    I_1=\frac{1}{2}I_0=\frac{1}{2}\pi

    I_2=\frac{3}{4}I_1=\frac{1\times 3}{2\times 4}\pi

    I_3=\frac{5}{6}I_2=\frac{1\times 3\times 5}{2\times 4\times 6}\pi

    I_4=\frac{7}{8}I_3=\frac{1\times 3\times 5\times 7}{2\times 4\times 6\times 8}\pi which is indeed \neq \frac{(2\times 4)!}{2^{2\times 4}(4!)^2}\pi, but I really don't know how would you "see" the general form of \frac{(2n)!}{2^{2n}(n!)^2}\pi.
    Last edited by courteous; October 28th 2010 at 12:35 AM. Reason: NOT equal
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  6. #6
    Member courteous's Avatar
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    Well, I_4=\frac{7}{8}I_3=\frac{1\times 3\times 5\times 7}{2\times 4\times 6\times 8}\pi is actually \neq \frac{(2\times 4)!}{2^{2\times 4}(4!)^2}\pi, so the general formula \frac{(2n)!}{2^{2n}(n!)^2}\pi isn't correct or I am (constantly) overlooking something.

    I mean, the formula doesn't hold even for I_1! What am I doing wrong?
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by courteous View Post
    Well, I_4=\frac{7}{8}I_3=\frac{1\times 3\times 5\times 7}{2\times 4\times 6\times 8}\pi is actually \neq \frac{(2\times 4)!}{2^{2\times 4}(4!)^2}\pi, so the general formula \frac{(2n)!}{2^{2n}(n!)^2}\pi isn't correct or I am (constantly) overlooking something.

    I mean, the formula doesn't hold even for I_1! What am I doing wrong?

    ummm..... I disagree

    I_0=\pi

    I_1=\frac{1}{2}I_0=\frac{1}{2}\pi=\frac{1\cdot 2}{2\cdot 2}\pi=\frac{2!}{2^2\cdot 1!}\pi

    As I recommended here is the proof by induction.

    Base case done

    Assume

    I_k=\pi\frac{(2k)!}{2^{2k}(k!)^2}

    Show I_{n+1}

    So we know from the relationship from before that

    I_{n+1}=\frac{2(n+1)-1}{2(n+1)}I_n=\frac{2n+1}{2(n+1)}I_n

    Now by the induction hypothesis we get

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

    Now multiply the numerator and denominator by (2n+2) to get

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

    \displaystyle \pi \frac{[2(n+1)]!}{2^{2(n+1)}[(n+1)!]^2}

    This completes the proof.
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