- #1

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Does anyone know how to compute product

a1*a2*...

where ai=1-x^i, x in (0,1)

Thank you in advance

a1*a2*...

where ai=1-x^i, x in (0,1)

Thank you in advance

- Thread starter jet
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- #1

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Does anyone know how to compute product

a1*a2*...

where ai=1-x^i, x in (0,1)

Thank you in advance

a1*a2*...

where ai=1-x^i, x in (0,1)

Thank you in advance

- #2

Janitor

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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.

- #3

Janitor

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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.

- #4

Zurtex

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

[tex](1-x)(1-x^2)(1-x^3)...[/tex]

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

[tex](1-x)(1-x^2)(1-x^3)...[/tex]

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

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- #5

matt grime

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in general the product is the limit of

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

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.

- #6

Hurkyl

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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.

- #7

Zurtex

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Oh right thanks

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

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

- #9

Janitor

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

- #10

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

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- #11

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[tex]

\lim_{i\rightarrow\infty} 1-x^i = \lim_{n\rightarrow\infty}\prod_{i=1}^n1-x^i[/tex]

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.

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- #13

matt grime

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- #14

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

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

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.

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

However, this test cannot be used to show the

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- #16

matt grime

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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.

- #17

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[tex]

\begin{align*}

\lim_{k\rightarrow\infty}1+\frac{1}{k} &= 1\mbox{, whereas}\\

\prod_{k=1}^\infty 1+\frac{1}{k} &= \infty

\end{align*}

[/tex]

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