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Math Help - evaluating an infinite product

  1. #1
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    evaluating an infinite product

    How do I evaluate \prod_{i=1}^\infty \frac{10^i - 1}{10^i} = .9\cdot .99 \cdot .99 \cdot \ldots? Note that it's equivalent to solving the recurrence relation p_{n+1} = p_n - \frac{p_n}{10^{n+1}} for p_1 = \frac{9}{10}.

    I know how to prove it's between .8 and 1, but I can't figure out an exact answer.
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  2. #2
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    Quote Originally Posted by rn443 View Post
    How do I evaluate \prod_{i=1}^\infty \frac{10^i - 1}{10^i} = .9\cdot .99 \cdot .99 \cdot \ldots? Note that it's equivalent to solving the recurrence relation p_{n+1} = p_n - \frac{p_n}{10^{n+1}} for p_1 = \frac{9}{10}.

    I know how to prove it's between .8 and 1, but I can't figure out an exact answer.
    I don't think there is a closed formula. There is a famous formula by Euler that goes as follow : \prod_{n=1}^\infty (1-q^n) = \frac{1}{\sum_{n=0}^\infty p(n) q^n} where p(n) is the number of partitions of n. But no explicit formula is usually given; that says it all.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    I don't think there is a closed formula. There is a famous formula by Euler that goes as follow : \prod_{n=1}^\infty (1-q^n) = \frac{1}{\sum_{n=0}^\infty p(n) q^n} where p(n) is the number of partitions of n. But no explicit formula is usually given; that says it all.
    There is some more information about this function here. That link gives the expansion of \prod_{n=1}^\infty (1-x^n) as 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} - x^{35} - x^{40} + x^{51} + \ldots. It's not stated in the link, but it looks very much as though the nonzero coefficients are all \pm1 (with a pattern of two 1s followed by two +1s). If so then \prod_{n=1}^\infty (1-10^{-n}) = 0.8900101 - 10^{-12} - 10^{-15} with an error of less than 10^{-21}.

    In fact, that pattern of pairs of 1s looks extremely regular, with the k'th pair being the coefficients of x^{p_k} and x^{q_k}, where p_k=\tfrac12k(3k-1) and q_k=\tfrac12k(3k+1). Again, that's just based on observations from the list of the first 200 coefficients in the Encyclopedia of Integer Sequences. I don't have any proof for it.
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    Quote Originally Posted by Opalg View Post
    There is some more information about this function here. That link gives the expansion of \prod_{n=1}^\infty (1-x^n) as 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} - x^{35} - x^{40} + x^{51} + \ldots. It's not stated in the link, but it looks very much as though the nonzero coefficients are all \pm1 (with a pattern of two 1s followed by two +1s). If so then \prod_{n=1}^\infty (1-10^{-n}) = 0.8900101 - 10^{-12} - 10^{-15} with an error of less than 10^{-21}.

    In fact, that pattern of pairs of 1s looks extremely regular, with the k'th pair being the coefficients of x^{p_k} and x^{q_k}, where p_k=\tfrac12k(3k-1) and q_k=\tfrac12k(3k+1). Again, that's just based on observations from the list of the first 200 coefficients in the Encyclopedia of Integer Sequences. I don't have any proof for it.
    You seem to be correct. It is called Euler's pentagonal theorem and two proofs are given here.
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  5. #5
    MHF Contributor chisigma's Avatar
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    A reasonable alternative to the 'direct computation' of the 'infinite product' is the computation of its logarithm...

    L(x)= \ln \prod_{n=1}^{\infty} (1-x^{n}) = \sum_{n=1}^{\infty} \ln (1-x^{n}) (1)

    Using the standard Taylor expansion of \ln (1-x^{n}) we obtain...

    L(x)= -\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{x^{kn}}{k} = \sum_{k=1}^{\infty}\frac {x^{k}}{k\cdot (x^{k}-1)} (2)

    Both the 'infinite product' and the 'infinite sum' are convergent for |x|<1. The computation using the first three hundred terms of the 'product' and the first three hundred terms of the 'sum' leads us to the same result approximated in twelve digits...

     \prod_{n=1}^{\infty} (1-.1^{n}) \approx .890010099999...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    A reasonable alternative to the 'direct computation' of the 'infinite product' is the computation of its logarithm...

    L(x)= \ln \prod_{n=1}^{\infty} (1-x^{n}) = \sum_{n=1}^{\infty} \ln (1-x^{n}) (1)

    Using the standard Taylor expansion of \ln (1-x^{n}) we obtain...

    L(x)= -\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{x^{kn}}{k} = \sum_{k=1}^{\infty}\frac {x^{k}}{k\cdot (x^{k}-1)} (2)

    Both the 'infinite product' and the 'infinite sum' are convergent for |x|<1. The computation using the first three hundred terms of the 'product' and the first three hundred terms of the 'sum' leads us to the same result approximated in twelve digits...

     \prod_{n=1}^{\infty} (1-.1^{n}) \approx .890010099999...

    Kind regards

    \chi \sigma
    A few related question please -
    1. Is there a easy way to show if the result is rational or irrational?
    2. @rn443: Can you please share how you proved that the result is greater than 0.8?

    Thanks
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  7. #7
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    Quote Originally Posted by aman_cc View Post
    A few related question please -
    1. Is there a easy way to show if the result is rational or irrational?
    Yes.

    Because of the formula

    \prod_{n=1}^\infty (1+q^n)=1+\sum_{n=1}^\infty (-1)^n (q^{\frac{n(3n-1)}{2}}+q^{\frac{n(3n+1)}{2}}),

    we have \alpha=\prod_{n=1}^\infty \left(1+\frac{1}{10^n}\right) = 1.00001010000000000000010001\ldots - 0.11000000000100100\ldots (only ones and zeros).

    Note as well that 0.111111\cdots = \frac{1}{9}. Therefore,

    \alpha + \frac{1}{9} = 1.0011212111101101111112111211\cdots.

    There may be mistakes, but you see the idea: \alpha+\frac{1}{9} has a decimal expansion with only 0,1,2's. And because the exponents in the first formula are quadratic (not linear), this expansion is not eventually periodic.

    We conclude that \alpha+\frac{1}{9} is irrational, and thus so is \alpha.
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