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Infintite product true or not

  1. Sep 19, 2005 #1
    Is this true?

    f(x)=sin(x)=x(x-2pi)(x-4pi)(x-6pi)...

    edit:

    f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...
     
    Last edited: Sep 19, 2005
  2. jcsd
  3. Sep 19, 2005 #2

    TD

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    It doesn't seem right to me, try x = 2pi.
     
  4. Sep 19, 2005 #3
    Sorry I did a mistake. I meant sin(x).
     
  5. Sep 19, 2005 #4

    TD

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    Try x = pi.
     
  6. Sep 19, 2005 #5
    I'm affraid I do not see what you mean... the serie is infinite...
     
  7. Sep 19, 2005 #6

    shmoe

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    No, your infintite product does not even converge.

    Infinite product form for sine:

    [tex]\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right)[/tex]
     
    Last edited: Sep 19, 2005
  8. Sep 19, 2005 #7

    TD

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    I was trying to point out that sin(pi) = 0 but your serie would never go to 0.
     
  9. Sep 19, 2005 #8
    Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

    Really sorry.
     
  10. Sep 19, 2005 #9

    shmoe

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    Still doesn't converge, the absolute value of the terms is growing without bound.
     
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