Is this Infinite Product True or Not?

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Discussion Overview

The discussion revolves around the validity of an infinite product representation of the sine function, specifically whether the proposed forms accurately represent sin(x) and converge appropriately. The scope includes mathematical reasoning and exploration of infinite products.

Discussion Character

  • Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes an infinite product representation of sin(x) as f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...
  • Another participant questions the validity of this representation by suggesting testing specific values like x = 2pi and x = pi.
  • A later reply indicates that the proposed infinite product does not converge and provides an alternative infinite product form for sine: \sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right).
  • Some participants express confusion regarding the implications of the infinite series and its behavior at specific points, particularly noting that sin(pi) = 0, which raises concerns about the convergence of the proposed series.
  • Repeated clarifications from the original poster indicate a misunderstanding in their initial formulation of the infinite product.
  • Another participant reiterates that the proposed series does not converge, stating that the absolute value of the terms is growing without bound.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the validity of the infinite product representation, with multiple competing views regarding its convergence and correctness remaining unresolved.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about convergence and the definitions of the infinite product forms. The implications of specific values on the proposed series are also not fully explored.

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:

[tex]\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right)[/tex]
 
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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