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Math Help - Prof of sumfunctions by Abel's theorem

  1. #1
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    Prof of sumfunctions by Abel's theorem

    I look for help with a prof :

    Let F(x) = \int ((1-t^2)^) dt
    Let F(1) = phi/4

    Prove for the series that phi/4 = \sum (n=2 to infinity) [*3/4 .....(2n-3)( [(2n-2)*(2n+1])

    I have the idea that Abel's theorem is relevant for the prof without being able to prove it

    Full prof not just a hint.

    I am sure that Abel's theorem is useful with changes to allow sum not from n=0 but from n=2.

    Help is most wellcome
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pragma View Post
    I look for help with a prof :

    Let F(x) = \int ((1-t^2)^) dt
    Let F(1) = phi/4

    Prove for the series that phi/4 = \sum (n=2 to infinity) [*3/4 .....(2n-3)( [(2n-2)*(2n+1])

    I have the idea that Abel's theorem is relevant for the prof without being able to prove it

    Full prof not just a hint.

    I am sure that Abel's theorem is useful with changes to allow sum not from n=0 but from n=2.

    Help is most wellcome
    This is pretty much impossible to read. Could you try to retype it?
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  3. #3
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    Carifycation - Sumfunction prove results with Abel's theorem

    Let me carify wirrting it more readiable.

    Let the Taylor series in x=0 of the function

    F(x) = integral of [(1-t^2)]^
    with integral from 0 to x


    Observe : F(1) = phi/4

    Prof the formula :

    Phi/4 = 1 - SUMMATION from n=2 to infinity of (these fractions)

    1 3 (2n-2) 1
    _ * _ *** *_______
    2 4 (2n-2) (2n(2n+1)


    I have the idea that Abel's theorem is relevant for the prof without being able to prove it

    Full prof not just a hint is needed. Abel's theorem is useful with changes to allow sum not from n=0 but from n=2 to infinity.

    FULL Prof is needed to convince me.
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  4. #4
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    Im not quite sure, but wouldn't it be an idea to actually see why

    F(1)= \frac{\pi}{4} and then use that to conclude that
    the formula holds? But i think your formula is wrong, and that it should be something like  \frac{\pi}{4} = 1-\frac{1}{6}- \sum_{n=0}^{\infty} \frac {1}{2} \frac {3}{4}...\frac {2n-3}{2n-2} \frac{1}{2n(2n+1)}
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  5. #5
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    Dialogue - explain Zaph

    Zaph

    Expalin your point. It's more like a polical option you are coming up with than a math reasoning.

    Exlain your point or better give a full proff of your point.

    BR
    Pragma
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  6. #6
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    I'm sorry, i can't help you with a proof, i was just thinking aloud i guess.. Hope you get it sorted, perhaps by people smarter than me
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