1. Sep 19, 2005 #1

   Werg22

   

   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

   Werg22, Sep 19, 2005
2. 
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3. Sep 19, 2005 #2

   TD

   

   Homework Helper

   It doesn't seem right to me, try x = 2pi.

    

   TD, Sep 19, 2005
4. Sep 19, 2005 #3

   Werg22

   

   Sorry I did a mistake. I meant sin(x).

    

   Werg22, Sep 19, 2005
5. Sep 19, 2005 #4

   TD

   

   Homework Helper

   Try x = pi.

    

   TD, Sep 19, 2005
6. Sep 19, 2005 #5

   Werg22

   

   I'm affraid I do not see what you mean... the serie is infinite...

    

   Werg22, Sep 19, 2005
7. Sep 19, 2005 #6

   shmoe

   

   Science Advisor

   Homework Helper

   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

   shmoe, Sep 19, 2005
8. Sep 19, 2005 #7

   TD

   

   Homework Helper

   I was trying to point out that sin(pi) = 0 but your serie would never go to 0.

    

   TD, Sep 19, 2005
9. Sep 19, 2005 #8

   Werg22

   

   Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...
   
   Really sorry.

    

   Werg22, Sep 19, 2005
10. Sep 19, 2005 #9

    shmoe

    

    Science Advisor

    Homework Helper

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

     

    shmoe, Sep 19, 2005

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