Closed form of divergent infinite product? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:25:44Z http://mathoverflow.net/feeds/question/28986 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28986/closed-form-of-divergent-infinite-product Closed form of divergent infinite product? Max Muller 2010-06-21T20:06:42Z 2010-06-21T21:21:22Z <p>Okay, we know that </p> <p>$$\frac{sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\cdot\pi^2}\Big)$$ .</p> <p>Is there some known (trigonometric(?)) function that is equal to the following infinite product? </p> <p>$$\prod_{n=1}^{\infty} \Big(1-\frac{x}{n\cdot\pi}\Big)$$</p> <p>I'd be happy as well if someone could provide me with a function that is equal to a similar divergent infinite product (a function, for example, that is equal to 'my' inifite product, only $\pi=1$, or $x=x^2$, or something in that direction). </p> <p>Thanks in advance,</p> <p>Max Muller</p> http://mathoverflow.net/questions/28986/closed-form-of-divergent-infinite-product/28988#28988 Answer by coudy for Closed form of divergent infinite product? coudy 2010-06-21T20:15:09Z 2010-06-21T20:15:09Z <p>I would suggest the development of the Gamma function</p> <p>$$1/\Gamma(z) = z e^{\gamma z}\ \Pi_{n=1}^\infty\ (1+{z\over n})\ e^{-{z\over n}}$$</p> http://mathoverflow.net/questions/28986/closed-form-of-divergent-infinite-product/28990#28990 Answer by Robin Chapman for Closed form of divergent infinite product? Robin Chapman 2010-06-21T20:19:46Z 2010-06-21T20:19:46Z <p>It's a <strong>divergent</strong> infinite product. You might as well ask for the sum of $$\sum_{n=1}^\infty\frac{x}{n\pi}.$$ You can "cure" the divergence by multipliying each term by a suitable factor, so for instance $$f(x)=\prod_{n=1}^\infty e^{x/n\pi}\left(1-\frac{x}{n\pi}\right)$$ does converge (as the $n$-th term is like $\exp(x^2/2n^2\pi^2)$). You can express this in terms of the gamma function which satisfies $$\frac1{\Gamma(x)}=x e^{\gamma x}\prod_{n=1}^\infty e^{-x/n}\left(1+\frac{x}{n}\right).$$ By using the identity $$f(x)f(-x)=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)$$ one can deduce the identity $$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}.$$</p> http://mathoverflow.net/questions/28986/closed-form-of-divergent-infinite-product/28998#28998 Answer by castal for Closed form of divergent infinite product? castal 2010-06-21T21:21:22Z 2010-06-21T21:21:22Z <p>Take a look at the first dozen pages of Andrews and Askey, which you can read online - <a href="http://books.google.com/books?id=nMm13WXpLt8C&amp;lpg=PP1&amp;dq=andrews%20askey&amp;pg=PA1#v=onepage&amp;q&amp;f=false" rel="nofollow">http://books.google.com/books?id=nMm13WXpLt8C&amp;lpg=PP1&amp;dq=andrews%20askey&amp;pg=PA1#v=onepage&amp;q&amp;f=false</a></p> <p>Already on page 3, they give the product representation of 1/Gamma, which is essentially your function, modified to make it convergent.</p> <p>On page 10, they treat the reflection formula, which shows that 1/Gamma is "half of the sine function", i.e it contributes the zeros on the negative x axis.</p>