# Closed form of infinite product

1. Jan 8, 2007

### quark80

Hi,

Have come across the following infinite product:

$\prod_{n=1}^{inf} (1+ax^n)$

where for practicality 0<x<1 and 0<a<1

After much searching through old calculus notes and 2 straight days of google searching I still can't tell if this infinite product has a closed form. I'm not asking for the actual closed form, just wandering if one actually exists? And if one does exist, a nudge in the right direction would be appreciated. One of the websites I visited showed something kind of similar based on q-analoges from combinatorial theory, however this didn't really seem like what I was after.

thanks, q :)

Last edited: Jan 8, 2007
2. Jan 9, 2007

### Gib Z

On this site we don't use \$. Rather we use start the code with [ tex ] and end it with [ /tex ] (without the spaces).

$$\prod_{n=1}^{inf} (1+ax^n)$$

3. Jan 9, 2007

### theperthvan

I would say yes, at least for x>1

4. Jan 9, 2007

### quark80

Thanks :)

But say I can modify it to:

$$\sum_{n=1}^{inf} ln(1+ax^{n})$$

where 0<x<1 and 0<a<1

Would that help determine if it has a closed form or not?

I'm basically just trying to figure out whether I'm wasting my time looking for a solution or not...

Last edited: Jan 9, 2007
5. Jan 10, 2007

### StatusX

It sort of does. I was able to get show it's a series:

$$c_0 + c_1 a+ c_2 a^2 +...$$

Where the coefficients are functions of x that start off:

$$c_0=1$$

$$c_1 = \frac{1}{1-x}$$

$$c_2 = \frac{1}{2} \left( \frac{1}{1-x^2} - \frac{1}{(1-x)^2} \right)$$

And in general there is a recursive formula for the cn. I've done it really sloppily, so that might not be right. I don't see an obvious way to simplify it further, but maybe if I went back and was more careful something would show up. But I'll let you work on that if you want.

6. Jan 11, 2007

### quark80

Thanks for the help

Maybe I'm having a brainfart and doing something wrong here. But the issue seems to be that the limit of the series as n -> infinity is 0. It's making me go around in circles every time I think I'm close to finding a closed form. I just can't seem to eliminate n from the equation.

After differentiating twice (which is how I understand this problem should be addressed?) and making some substitutions I ended up with:
$$\frac{bn^{2}(a+1)}{x(b+1)^{2}}$$

so for the life of me I can't get rid of n because the other side of that equation comes out to equal 0.

Looking at it in another way leads me to the same issue. I can't eliminate n because the other side of the equation always comes out to be 0. Does this simply mean that both the infinite product and infinite sum don't have a closed form? This issue with the other side being 0 is the reason why I posed the question in my original post.

7. Jan 11, 2007

### StatusX

You want to differentiate with respect to a. I'm not sure what that expression is supposed to be. Maybe you're differentiating wrong. The formula for differntiating a product can be written:

$$\frac{d}{dx} \left( \prod_{k=1}^n f_k(x) \right) = \left( \prod_{k=1}^n f_k(x)\right) \left( \sum_{k=1}^n \frac{f_k'(x)}{f_k(x)} \right)$$

which can be derived from the product rule.

8. Jan 11, 2007

### quark80

Doh! Was differentiating wrt x. That'd be the problem :P

9. Jan 13, 2007

### quark80

Deleted. Think I have it sorted :) Cheers

Last edited: Jan 14, 2007
10. Jan 14, 2007

### benorin

Euler has solved the case of $$a=-1$$ for you, here is a link to his paper entitled "The expansion of the infinite product $$(1-x)(1-xx)(1-x^3)(1-x^4)(1-x^5)(1-x^6)\cdots$$ into a single series".

11. Jan 14, 2007

### quark80

Legend

I don't suppose you know of an existing proof for when a is not -1? I'm trying to derive it from this Euler paper, but the minor modification turns out to not be as simple as I thought it might be.

12. Jan 15, 2007

### benorin

One handy forumla is the conversion from an infinite product to an infinite series, namely

$$\prod_{k=1}^{\infty}(1+ a_k) = 1+a_{1} + \sum_{k=2}^{\infty} (1+a_{1})(1+a_{2})\cdots (1+a_{k-1}) a_{k}$$

and this is that formula Euler had used at the end of the first line when he said "it clearly follows that: <result>" where said <result> is the righthand side of the above formula applied to the given product (this post derives the same formula with slightly different terms/indices).

Last edited: Jan 15, 2007
13. Jan 15, 2007

### benorin

I have fiddled with Euler's calulations to get his result in this form:

$$\prod_{n=1}^{\infty}(1-x^n) =\sum_{n=1}^{\infty}(-1)^{n}x^{\frac{1}{2}n(3n-1)}(1-x^{1-2n})$$​

14. Jan 15, 2007

### quark80

Yeah I got that far yesterday.

It can be modified to get:
$$\prod_{n=1}^{\infty} (1-ax^{n})=\sum_{n=0}^{\infty}\frac{(-1)^{n} a^{2n} x^{n(n+1)/2}}{\prod_{k=1}^{n} (1-x^{k})}$$

Which is (from what I've since learned) a modified version of Euler's Pentagonal Number Theorem. But this leads back to my original issue. I have to find a closed form for this, if one exists. However I think I'm beginning to realise that it's nowhere near as simple as I thought it was going to be. It might just simply be that I can't simplify this problem any further and may just need to leave it as it is...

One thing that popped up in my search for an answer was Cauchy's Binomial Theorem...Which would make sense because ultimately this product is the result of a cascade problem I'm working on. But yeah, still doesn't help me simplify things.

Last edited: Jan 15, 2007