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Thread: 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 $\displaystyle A \subset \mathbb{N}$ be an infinite subset of the positive integers. Define

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

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

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

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

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

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

    Can you prove the theorem for $\displaystyle A=\{n^2|n\in\mathbb{N}\}$ ? Or for any other sequence that grows faster than linearly? (I would be curious of what method you would use)
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    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 $\displaystyle A$ can break the conjecture. I shall provide a family of examples where $\displaystyle \liminf_{x\to\infty} f(x)^{1/x}\leq 1$.

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

    Suppose $\displaystyle x$ is such that $\displaystyle m<x<M$ where $\displaystyle m,M\in A$ and $\displaystyle A\cap(m,M)=\emptyset$ (thus $\displaystyle (m,M)$ is a gap in $\displaystyle A$, 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:

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

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

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

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

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

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

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

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

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

    --

    However, I would still be interested in knowing how you would deal with $\displaystyle A=\{n^2|n\in\mathbb{N}\}$ if you know that.
    Last edited by Laurent; Jun 29th 2009 at 04:12 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 $\displaystyle A \subset \mathbb{N}$ be an infinite subset of the positive integers. Define

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

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

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