infinite product

[jet]

Does anyone know how to compute product
a1*a2*...
where ai=1-x^i, x in (0,1)

Thank you in advance

[Janitor]

If I ever knew how to do that sort of thing, I guess I've forgotten. I did a quickie spreadsheet and found these approximate results:

x=.25, product=0.688537537

x=.333333, product=0.56012661

x=.5, product=0.288788095

x=.666666, product=0.069273004

x=.75, product=0.015549724

I eyeballed the result for x=.5 in particular, and cannot think of any neato type of number that it corresponds to.

[Janitor]

I once came across something called "the tower of exponents." Take a number and raise it to a power equal to the number. Now instead, take a number and raise it to a power equal to that number raised to the power of itself. Continue this indefinitely. The tower converges over some range of arguments. I can't remember the range, but its endpoints were based on Euler's number. Something along the lines of 1/e to 1/(e-1), that sort of thing.

[Zurtex]

I've never ever done this so I am probably wrong, but is this what you want to know:

(1-x)(1-x^2)(1-x^3)...

Where 0 < x < 1. Well correct me if I am wrong but does that not mean where n \geq 1, |1 - x^n| < 1. And thus an infinite product would equal 0? :confused:

[matt grime]

the product exists and is non-zero exactly when the sum of the a_is exists and none of the individual terms 1-a_i is zero. at least that is the theory of products.

in general the product is the limit of

1-\sum_{i=1}^{n}a_1 + \sum_{i<j\ i,j=1}^n a_ia_j + \ldots

as n tends to infinity

i think you might be able to work it out int this case though i don't know what the answer is.

[Hurkyl]

Zurtex, consider this product:

3/4 * 8/9 * 15/16 * 24/25 * ...

Note that the partial products are:

3/4
2/3
5/8
3/5
7/12
...

these partial products approach 1/2, despite each term being less than 1.

[Zurtex]

Oh right thanks :smile:

[TALewis]

I've never formally done infinite products, but isn't the following true:

\lim_{i\rightarrow\infty} 1-x^i = 1\, ,\quad 0<x<1

Eventually you're just multiplying by 1, so the infinite product should converge somewhere. Is that line of reasoning valid?

[Janitor]

Eventually you're just multiplying by 1, so the infinite product should converge somewhere. Is that line of reasoning valid?- TALewis

When I first read the thread-starter post, my worry was that all such products might converge trivially to zero. But then my spreadsheet and Hurkyl's post convinced me otherwise.

[Muzza]

TALewis, I don't think you can split up the limit of a product that way (assuming I'm reading your mind correctly). Consider \lim_{x\rightarrow\infty} \frac{1}{x}x, it's not equal to \lim_{x\rightarrow\infty} \frac{1}{x} * \lim_{x\rightarrow\infty} x, is it?

[TALewis]

Well, I'm not saying this:


\lim_{i\rightarrow\infty} 1-x^i = \lim_{n\rightarrow\infty}\prod_{i=1}^n1-x^i

That's clearly untrue. I think I was trying to say that a_i approaches 1, so that a1*a2*a3... isn't going to be zero. But that could be an invalid leap in reasoning; I'm not sure. This isn't my strongest area.

[Muzza]

That's not what I thought you were saying either. ;) But you can probably ignore my post anyway, I confused myself by using x (which was a constant in the original post). It would seem as if your thinking is correct, since the factors converges to 1, the product should converge too.

[matt grime]

it is no more true, in fact is exactly as false as saying that an infinite sum converges if its terms got to zero as to say that a product converges if its terms go to 1. go back and read my post about 5 previously where I explained that.

[TALewis]

I see. I guess it's similar to the divergence test for infinite series:

\sum_{i=1}^\infty a_i \mathrel{\mbox{is divergent if}} \lim_{i\rightarrow\infty}a_i \neq 0$

However, this test cannot be used to show the convergence of a series, even though it might appear to "make sense," just as my previous proposition sounded correct to me at first. I stand corrected.

[Muzza]

Sorry Matt, I must've missed your post. Do you have an example of a product which diverges, but where the factors go to 1? (Not that I doubt your word or anything ;)).

[matt grime]

Firstly let's be clear than when we say does not converge we mean, in this case, that the 'limit' is zero. after all it is not possible to have non-zero things multiplying to give zero. that's the rule of thumb we adopt (of course it may also wander off to infinity).

since the rule i gave was that 1-a_r has convergent product iff a_r has convergent sum the first thing you should think is: do i know any sums that misbehave?

how about a_r=1/r?

the sum diverges, yet the terms go to zoer, similarly even though the terms 1-a_r tend to 1, the product is the limit

1/2*2/3*3/4....*(n-1)/n = 1/n which tends to zero. and that is your canonical non-example.

if you want something that is more convincing consider a_r=-1/r, where the product does tend to infinity.

[TALewis]

Muzza, here's a real simple one I looked up:


\begin{align*}
\lim_{k\rightarrow\infty}1+\frac{1}{k} &= 1\mbox{, whereas}\\
\prod_{k=1}^\infty 1+\frac{1}{k} &= \infty
\end{align*}