Is this Infinite Product True or Not?

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The discussion centers on the validity of an infinite product representation of the sine function. Participants argue that the proposed form, f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)..., does not converge, particularly when evaluated at specific points like x=pi. There is confusion about the correct infinite product representation, with the standard form being sin(x)=x∏(1-(x^2/(πn)^2)). Ultimately, the consensus is that the infinite product as presented does not accurately represent the sine function and diverges instead. The conversation highlights the importance of convergence in infinite products in mathematical representations.
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:
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It doesn't seem right to me, try x = 2pi.
 
Sorry I did a mistake. I meant sin(x).
 
Try x = pi.
 
I'm affraid I do not see what you mean... the serie is infinite...
 
No, your infintite product does not even converge.

Infinite product form for sine:

\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right)
 
Last edited:
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.
 
Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.
 
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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