Talk:Wallis product

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

Wrong "proof"

This "proof" is seriously lacking in many ways.

When someone writes:

${\displaystyle {\frac {\sin(x)}{x}}=k\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots }$
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 ${\displaystyle e^{z}{\frac {\sin(z)}{z}}}$ and reach the conclusion that both ${\displaystyle e^{z}{\frac {\sin(z)}{z}}}$ and ${\displaystyle {\frac {\sin(z)}{z}}}$ have the same infinite product expansion and are thus equal, resulting in ${\displaystyle e^{z}=1,\forall _{z\in \mathbb {C} }}$ 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
${\displaystyle \sin(x)=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right)}$
is true, but you cannot factor those terms unless you keep them together; that is, the product
${\displaystyle x\prod _{n=-\infty }^{\infty }\left(1-{\frac {x}{n\pi }}\right)}$
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)

In view of the serious flaws that the "proof" given in this article, should not it be considered for deletion/overhaul? 210.212.55.3 (talk) 18:40, 28 September 2008 (UTC)

Order of Factors

Suppose we write

${\displaystyle W_{n}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdots {\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}}$

Then

{\displaystyle {\begin{aligned}\ln(W_{n})&=\ln \left({\frac {2}{1}}\right)+\ln \left({\frac {2}{3}}\right)+\ln \left({\frac {4}{3}}\right)+\ln \left({\frac {4}{5}}\right)+\ln \left({\frac {6}{5}}\right)+\dots +\ln \left({\frac {2n}{2n-1}}\right)+\ln \left({\frac {2n}{2n+1}}\right)\\&=\ln \left({\frac {2}{1}}\right)-\ln \left({\frac {3}{2}}\right)+\ln \left({\frac {4}{3}}\right)-\ln \left({\frac {5}{4}}\right)+\ln \left({\frac {6}{5}}\right)-\dots +\ln \left({\frac {2n}{2n-1}}\right)-\ln \left({\frac {2n+1}{2n}}\right)\\\end{aligned}}}

where, in the last line, all the logs are positive.

For the sum of the magnitudes of these terms, we have

${\displaystyle \sum _{r=1}^{2n}\ln \left({\frac {r+1}{r}}\right)>\int _{1}^{2n}\ln \left({\frac {x+1}{x}}\right)\,dx=2n\ln \left({\frac {2n+1}{2n}}\right)+\ln \left({\frac {2n+1}{4}}\right)}$

Both of the terms in the result of the integration are positive and the second clearly diverges as ${\displaystyle n\rightarrow \infty }$; so, therefore, must our sum of magnitudes. ${\displaystyle \ln(W_{n})}$ is thus conditionally convergent. By suitably ordering its terms, we can therefore make it converge to any real value and it follows that, by ordering its factors, ${\displaystyle W_{n}}$ can be made to converge to any positive real number. Any proof of the Wallis formula which fails to take into explicit account the order of the factors would thus appear to be spurious. IanHH (talk) 16:40, 25 July 2008 (UTC)

Infinity Factorial

Using zeta function regularization,
${\displaystyle -\zeta '(s)=\sum _{n=1}^{\infty }{\frac {\ln n}{n^{s}}}}$
Plugging in ${\displaystyle s=0}$ gives:
${\displaystyle -\zeta '(0)={\frac {1}{2}}\ln \left(2\pi \right)=\sum _{n=1}^{\infty }\ln n=\ln \prod _{n=1}^{\infty }n=\ln(\infty !)}$
${\displaystyle \Rightarrow \infty !={\sqrt {2\pi }}}$

This is wrong. It is wrong because the series definition of ${\displaystyle \zeta '(s)}$ is divergent at 0. Therefore, it cannot be used to calculate the value of ${\displaystyle \zeta '(0)}$ and is not equivalent to its actual value, ${\displaystyle {\frac {1}{2}}\ln \left(2\pi \right)}$. In other words, ${\displaystyle {\frac {1}{2}}\ln \left(2\pi \right)\neq \sum _{n=1}^{\infty }\ln n}$. — Preceding unsigned comment added by 173.80.198.94 (talk) 01:36, 3 January 2012 (UTC)