Infintite product true or not

  • Thread starter Werg22
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  • #1
Werg22
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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:

Answers and Replies

  • #2
TD
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It doesn't seem right to me, try x = 2pi.
 
  • #3
Werg22
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Sorry I did a mistake. I meant sin(x).
 
  • #4
TD
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Try x = pi.
 
  • #5
Werg22
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I'm affraid I do not see what you mean... the serie is infinite...
 
  • #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]
 
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  • #7
TD
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Werg22 said:
I'm affraid I do not see what you mean... the serie is infinite...
I was trying to point out that sin(pi) = 0 but your serie would never go to 0.
 
  • #8
Werg22
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Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.
 
  • #9
shmoe
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Werg22 said:
Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.

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

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