Talk:Wallis product

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    !!! I believe the sums should start at n,m=1, not at n,m=0! Please check and correct, whoever wrote this!

    [edit] Wrong "proof"

    This "proof" is seriously lacking in many ways.

    When someone writes:

    \frac{\sin(x)}{x} = k \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots
    where k is a constant.
    • First show that the right hand side of the equality is a holomorfic function (just like the left hand side). This should not be hard.
    • Most importantly the assertion that k is constant is not necessarily true. Assuming the previous point, you have 2 holomorphic functions that share the same roots with the same order, and thus all you can say is that their quotient is another holomorphic function with no zeros. You haven't proved that this function is constant. In this particular case it is but the justification is lacking.

    If you have any doubts, apply the same reasoning to e^z \frac{\sin(z)}{z} and reach the conclusion that both e^z \frac{\sin(z)}{z} and \frac{\sin(z)}{z} have the same infinite product expansion and are thus equal, resulting in e^z = 1, \forall_{z\in\mathbb{C}} which is absurd.

    Cláudio Valente 12:54, 27 April 2007 (UTC)

    In fact, your first equation is incorrect. Or rather, it is ambiguous. The equation
    \sin(x) = x\prod_{n = 1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right)
    is true, but you cannot factor those terms unless you keep them together; that is, the product
    x\prod_{n = -\infty}^\infty\left(1 - \frac{x}{n\pi}\right)
    does not converge. You need a "correction factor", which is part of the theory of Weierstrass products. However, a proof of this identity is unnecessary in this article, since the result is true and part of a different topic. Given it, the rest of the "wrong proof" here is actually right. Ryan Reich (talk) 16:51, 8 December 2007 (UTC)