Talk:Infinite product

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This article is probably very non-ideal, but when I needed this material a while ago it was hard to find, and so I figured wikipedia would be a good place to hold it. Please feel free to change it around and make it more encyclopedia-like, even more so than usual. :)

We probably need a little caveat about negative numbers: the log formula won't work without problems in that case. Maybe use a different log which is defined on the negative reals? --AxelBoldt

A product of negative coefficients does not converge; it can at most flip. --Yecril (talk) 18:00, 20 December 2008 (UTC)

Inaccurate Statement

"Therefore, the logarithm log an will be defined for all but a finite number of n" neither makes sense nor true as I understand it. One can easily imagine a sequence which converges to 1 that is not finite. one simple example is a(n)= 1+(1/n) where n is an integer. This of course is not limited for a being real, for example think of any oscillating function which dampens to 1. this might have been a simple typo the author made, but i don't understand his/her intention, so i am timid to edit this —Preceding unsigned comment added by 129.49.61.161 (talk) 17:06, 17 August 2010 (UTC)

Maybe it is a left-over from careless editing. Some sources allow a finite number of 0 factors and still call the product convergent if the product of the non-zero factors is non-zero. Probably that is the meaning. I'll try to fix it. McKay (talk) 06:09, 31 March 2011 (UTC)

Why is ${\displaystyle \prod _{k=1}^{\infty }a_{k}=\lim _{n\rightarrow \infty }\prod _{k=1}^{n}a_{k}=0}$ an example of divergence?

The article currently indicates that if the infinite product is zero then that is an example of divergence. Why? I would think that that is an example of convergence, despite that a corresponding sum ${\displaystyle \sum _{k=1}^{\infty }\log(a_{k})}$ is not finite. —Quantling (talk | contribs) 18:35, 30 March 2011 (UTC)

It is just a convention. The reason for the convention is that allowing convergence to 0 would admit too many uninteresting series (for example, any sequence {an} with | an| < 1/2), and some desirable properties would disappear. McKay (talk) 06:15, 31 March 2011 (UTC)

How does this fit in?

If F(x) can be represented as a product:

${\displaystyle F(x)=\Pi _{n=1}^{\infty }f_{n}(x)}$

then define G(x)=1-F(x) and using the series expansion for the natural logarithm: ${\displaystyle \ln(1-G(x))=-\sum _{n=1}^{\infty }G(x)^{n}/n}$

${\displaystyle \ln(F(x))=-\sum _{n=1}^{\infty }G(x)^{n}/n=\sum _{n=1}^{\infty }\ln(f_{n}(x))}$

so that ${\displaystyle f_{n}(x)=e^{-G(x)^{n}/n}}$ and

${\displaystyle F(x)=\Pi _{n=1}^{\infty }\,e^{-{\frac {1}{n}}(1-F(x))^{n}}}$

I have checked this on Mathematica for a number of functions, and Mathematica agrees, but I have no idea what the conditions are for this formula to work. Can anyone find a reference for this and put a section into the article? Its a very useful and general expression, and much simpler than the more general Weierstrass development. PAR (talk) 17:45, 13 June 2011 (UTC)

Ok, probably after almost 6 years this comment might have been forgotten, but let me explain why I don't think is a good idea to include such observation in the article. What you say is a trivial statement: if ${\displaystyle c\in [-1,1]}$ is a fixed number, then
${\displaystyle \Pi _{n=1}^{\infty }\,e^{-{\frac {c^{n}}{n}}}=\exp(-\sum _{n=1}^{\infty }c^{n}/n)=\exp(\log(1-c))=1-c.}$
Since you are expanding ${\displaystyle \log(1-F(x))}$ , you are assuming that ${\displaystyle F(x)\in [-1,1]}$ for every ${\displaystyle x}$, and then you are just applying the above formula point-wise. Note that this is quite different from Weierstrass' formula (if you wish, the problem with your expression is that you write ${\displaystyle F(x)}$ in terms of ${\displaystyle F(x)}$, so that is not very useful, just like is not very useful to say that -for example- ${\displaystyle F(x)=[2F(x)]*\left[{\frac {F(x)}{2}}\right]}$). Lucha (talk) 16:10, 28 February 2017 (UTC)