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Math Help - Conjecture

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Conjecture

    I posted this on another forum a while back but nobody was able to prove or disprove this conjecture of mine. I have empirical reasons to believe that it is true.

    Let A \subset \mathbb{N} be an infinite subset of the positive integers. Define

    f(x) = \sum_{a \in A}\frac{x^a}{a!}.

    Prove that, as x \rightarrow \infty, f(x)^{1/x} \rightarrow e, or supply a counter-example.
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  2. #2
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    Quote Originally Posted by Bruno J. View Post
    Prove that, as x \rightarrow \infty, f(x)^{1/x} \rightarrow e, or supply a counter-example.
    Neither a proof nor a counter-example, just first thoughts...

    Since f(x)\leq e^x, we have the right upper bound.

    There is a matching lower bound: for a\in A, f(a)\geq \frac{a^a}{a!}=g(a) and, using \log n!=n\log n-n+o(n) (consequence of Stirling formula), we obtain g(a)^{1/a}\rightarrow e. Therefore, with the upper bound:

    \lim_{a\to\infty, a\in A} f(a)^{1/a}=e.

    However, I suspect that f(x)^{1/x} can fluctuate when x is between two successive points of A that would be "far away from each other". Allowing for arbitrary A seems too weak an assumption to me: it can be an incredibly "hollow" sequence...

    Can you prove the theorem for A=\{n^2|n\in\mathbb{N}\} ? Or for any other sequence that grows faster than linearly? (I would be curious of what method you would use)
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  3. #3
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    Quote Originally Posted by Laurent View Post
    Neither a proof nor a counter-example, just first thoughts...
    Now I have a proof that fluctuations between points of A can break the conjecture. I shall provide a family of examples where \liminf_{x\to\infty} f(x)^{1/x}\leq 1.

    Note that f(x)^{1/x}\to e is equivalent to \frac{\log f(x)}{x}\to 1.

    Suppose x is such that m<x<M where m,M\in A and A\cap(m,M)=\emptyset (thus (m,M) is a gap in A, thought of as a large gap). More precise conditions will come later.

    We have the following upper bound that consists in taking all terms of the series outside the gap:

    f(x)\leq \sum_{a=0}^m \frac{x^a}{a!}+\sum_{a=M}^\infty \frac{x^a}{a!}.

    For the first sum, note that the largest term is the last one (each term is \frac{x}{a}(>1) times the previous one), so that the sum is less than (m+1)\frac{x^m}{m!}.

    For the second sum, I factorize by the first term and use a geometric series to bound the second factor:

    \sum_{a=M}^\infty \frac{x^a}{a!}=\frac{x^M}{M!}\left(1+\frac{x}{M+1}  +\frac{x^2}{(M+1)(M+2)}+\cdots\right) \leq \frac{x^M}{M!}(1+\frac{x}{M}+\frac{x^2}{M^2}+\cdot  s)=\frac{x^M}{M!}\frac{1}{1-\frac{x}{M}}.

    Summarizing, we have f(x)\leq (m+1)\frac{x^m}{m!}+\frac{x^M}{M!}\frac{1}{1-\frac{x}{M}}.

    I shall now give conditions that ensure \frac{log T}{x}\to 0 or -\infty where T is either of the two terms of the upper bound, and the limit is taken along some sequence x_n\to\infty (corresponding to some intervals [m_n,M_n]).

    First one: \frac{1}{x}\log\frac{x^m}{m!}=\frac{m\log x-\log m!}{x}=\frac{m}{x}(\log\tfrac{x}{m}+O(1)) =\frac{\log\frac{x}{m}}{\frac{x}{m}}+O(\frac{m}{x}  ). (Using \log m!=m\log m+O(m)), hence \frac{1}{x}\log\frac{x^m}{m!}\to 0 as soon as \frac{m}{x}\to 0 and m\to\infty. (Indeed, \frac{\log x}{x}\to_{x\to\infty} 0) * I forgot \frac{1}{x}\log(m+1), which also tends to 0 (faster) under this condition.

    Second one: \frac{1}{x}\log(\frac{x^M}{M!}\frac{1}{1-\frac{x}{M}})=\frac{1}{x}(M\log x-\log M!-\log(1-\frac{x}{M})) =\frac{M\log x-M\log M+M-\frac{1}{2}\log M+O(1)}{x}+\frac{1}{M}+o(\frac{1}{M}) if \frac{x}{M}\to 0 and M\to\infty (using Stirling estimate: \log n!=n\log n-n+\frac{1}{2}\log n+O(1) and \log(1+x)=x+o_{x\to 0}(x)), hence this term equals \frac{M}{x}(-\log\frac{M}{x}+1-\frac{1}{2}\frac{\log M}{M}x)+O(\frac{1}{x})+\frac{1}{M}+o(\frac{1}{M}), so that it tends to -\infty as soon as \frac{x}{M}\to 0 and M\to\infty.

    Since \log(a+b)\leq \max(\log (2a),\log (2b))=\log 2+\max(\log a,\log b), we deduce that \liminf f(x)^{1/x}\leq 1 where the bound corresponds to a limit taken along sequences of x that are in gaps (m,M) of A such that m<\!\!< x<\!\!<M (" <\!\!<" meaning "negligible compared to"). Such sequences of x's exist when A grows very fast (we must have m<\!\!<M where m and M are successive terms in the sequence): for instance, n!<\!\!< n!\sqrt{n}<\!\!<(n+1)! so A=\{n!|n\in\mathbb{N}\} and x_n=n!\sqrt{n} is such an example.

    --

    However, I would still be interested in knowing how you would deal with A=\{n^2|n\in\mathbb{N}\} if you know that.
    Last edited by Laurent; June 29th 2009 at 05:12 AM. Reason: Mistake corrected
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    MHF Contributor Bruno J.'s Avatar
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    Marvellous! You're a champ. I wouldn't have known how to do this myself.

    I do not know how to deal with the specific case you give. I only had some empirical reasons to believe the conjecture was true - I did some tests with Mathematica and it seemed to hold with some very sparse sets...

    Thanks and congrats again!
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  5. #5
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Bruno J. View Post
    I posted this on another forum a while back but nobody was able to prove or disprove this conjecture of mine. I have empirical reasons to believe that it is true.

    Let A \subset \mathbb{N} be an infinite subset of the positive integers. Define

    f(x) = \sum_{a \in A}\frac{x^a}{a!}.

    Prove that, as x \rightarrow \infty, f(x)^{1/x} \rightarrow e, or supply a counter-example.
    Wouldn't  A = \mathbb{N} if  A is an infinite subset of  \mathbb{N} ?
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  6. #6
    Moo
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    Quote Originally Posted by Sampras View Post
    Wouldn't  A = \mathbb{N} if  A is an infinite subset of  \mathbb{N} ?
    Consider the set of odd natural integers, which is an infinite subset of \mathbb{N}

    A would be isomorphic to \mathbb{N} (if I'm not mistaking )
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