Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Okay, we know that

$$ \frac{sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\cdot\pi^2}\Big) $$ .

Is there some known (trigonometric(?)) function that is equal to the following infinite product?

$$ \prod_{n=1}^{\infty} \Big(1-\frac{x}{n\cdot\pi}\Big) $$

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).

Thanks in advance,

Max Muller

share|improve this question
1  
Your question is related to the Gamma function (en.wikipedia.org/wiki/Gamma_function) at $-1$; but the product is meaningless, the Gamma function has a singularity there, and this all has been known for two centuries. –  Charles Matthews Jun 21 '10 at 20:17
    
Ok, but isn't it pretty 'obvious' that the gamma function has a singularity there, as it's 'almost' equal to my divergent series, which goes into infinity for whatever x. It isn't that 'bad' that the function has a singularity there. –  Max Muller Jun 21 '10 at 20:29
2  
It's equal to zero because the sum of x/n pi is infinite. Use that 1 - y <= e^{-y} for positive y to find a proof. Also if you plug in x = pi into your product and multiply out a few terms it will be clear what's going on. This isn't really a mathoverflow kind of question IMO. –  Michael Greenblatt Jun 21 '10 at 20:34
    
Yeh, I'm sorry, I was afraid of that already, but I I couldn't look it up somewhere easily... Sorry. –  Max Muller Jun 21 '10 at 20:38
add comment

3 Answers

up vote 7 down vote accepted

It's a divergent 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 $$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}.$$

share|improve this answer
    
I should add that this is an example of a Weierstrass product en.wikipedia.org/wiki/Weierstrass_product which can be used to construct entire functions with any admissble set of zeros. –  Robin Chapman Jun 21 '10 at 20:37
    
Ok, thanks, this is very useful, mister Chapman! –  Max Muller Jun 21 '10 at 20:39
add comment

I would suggest the development of the Gamma function

$$1/\Gamma(z) = z e^{\gamma z}\ \Pi_{n=1}^\infty\ (1+{z\over n})\ e^{-{z\over n}}$$

share|improve this answer
add comment

Take a look at the first dozen pages of Andrews and Askey, which you can read online - http://books.google.com/books?id=nMm13WXpLt8C&lpg=PP1&dq=andrews%20askey&pg=PA1#v=onepage&q&f=false

Already on page 3, they give the product representation of 1/Gamma, which is essentially your function, modified to make it convergent.

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.

share|improve this answer
    
Thanks, castal, I always like references to more useful information. –  Max Muller Jun 21 '10 at 21:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.