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Infinite product

  1. Jun 6, 2004 #1


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

    Thank you in advance
  2. jcsd
  3. Jun 6, 2004 #2


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    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.
  4. Jun 6, 2004 #3


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    Also worth mentioning-

    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.
  5. Jun 6, 2004 #4


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


    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? :confused:
    Last edited: Jun 6, 2004
  6. Jun 6, 2004 #5

    matt grime

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    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

    [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.
  7. Jun 6, 2004 #6


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    Zurtex, consider this product:

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

    Note that the partial products are:


    these partial products approach 1/2, despite each term being less than 1.
  8. Jun 6, 2004 #7


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    Oh right thanks :smile:
  9. Jun 6, 2004 #8
    I've never formally done infinite products, but isn't the following true:

    [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?
  10. Jun 6, 2004 #9


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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.
  11. Jun 6, 2004 #10
    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?
    Last edited: Jun 6, 2004
  12. Jun 6, 2004 #11
    Well, I'm not saying this:

    \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.
  13. Jun 6, 2004 #12
    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.
  14. Jun 6, 2004 #13

    matt grime

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    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.
  15. Jun 6, 2004 #14
    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.
    Last edited: Jun 6, 2004
  16. Jun 6, 2004 #15
    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 ;)).
  17. Jun 6, 2004 #16

    matt grime

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    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.
  18. Jun 6, 2004 #17
    Muzza, here's a real simple one I looked up:

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