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Math Help - Limit Question

  1. #1
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    Limit Question

    Hey guys,

    I'm having a REAL problem with the following limit


    \displaystyle \lim_{p\to\infty} \biggl(\frac{{| d |}^{np+p-1} |x|^{p-1}}{(p-1)!} \{(|x| + |\alpha_1|) \ldots (|x| + |\alpha_n|) \}^p\biggr)

    I've tried the squeeze rule, but an upper bound eludes me. I've only found the lower bound, which is just 0. I've tried to group together all of the powers of p, as follows:


    \displaystyle  \lim_{p\to\infty} \biggl(\frac{{| d x|}^{p-1}}{(p-1)!} \{|d|^n(|x| + |\alpha_1|) \ldots (|x| + |\alpha_n|) \}^p\biggr)

    and maybe use the fact that

    \displaystyle \frac{{| d x|}^{p-1}}{(p-1)!}

    tends to zero as p tends to infinity, since the bottom half rises faster than the top half. However, I'm stuck because I don't know how to go about this formally, or if this is the right way to go about it. I'd be really grateful if help was provided, and possibly a solution.

    Thank you very very much in advance.

    HTale
    Last edited by HTale; October 26th 2008 at 08:14 AM.
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  2. #2
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    This is what I'd do:

    \left\{d^n(x+a_1)\cdots(x+a_n)\right\}^p\leq C^p

    for some value of C

    Likewise (dx)^{p-1}C^p\leq K^p for some K and:

    \lim_{p\to\infty}\frac{K^p}{(p-1)!}\to 0\; \forall K\in\mathbb{R}^+
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  3. #3
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    Hey Shawsend, thanks a billion for your quick reply. But I have one question regarding this part:

    \left\{d^n(x+a_1)\cdots(x+a_n)\right\}^p\leq C^p for some value of C

    Can I assume that the polynomial is an integer if it is given for some fixed x? I think also the question applies for the other part too. That is, when writing the solution out, should I state that this is for a fixed x?

    Thanks in advance
    HTale

    EDIT: Also, how would I go about proving the final limit? Just a few pointers will do. Once again, thanks in advance!
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  4. #4
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    Quote Originally Posted by HTale View Post
    Hey Shawsend, thanks a billion for your quick reply. But I have one question regarding this part:

    \left\{d^n(x+a_1)\cdots(x+a_n)\right\}^p\leq C^p for some value of C


    Can I assume that the polynomial is an integer if it is given for some fixed x? I think also the question applies for the other part too. That is, when writing the solution out, should I state that this is for a fixed x?

    Thanks in advance
    HTale

    EDIT: Also, how would I go about proving the final limit? Just a few pointers will do. Once again, thanks in advance!
    I treated it as for a fixed x,d and fixed set of a_n, I could always find a C which would make the expression true and as far as pointers, I'm sorry, I don't feel I'm good enough in analysis to be giving pointers. I just stated what I'd do. Maybe the analysis experts in here can make it more rigorous.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by HTale View Post
    Hey guys,

    I'm having a REAL problem with the following limit


    \displaystyle \lim_{p\to\infty} \biggl(\frac{{| d |}^{np+p-1} |x|^{p-1}}{(p-1)!} \{(|x| + |\alpha_1|) \ldots (|x| + |\alpha_n|) \}^p\biggr)


    I've tried the squeeze rule, but an upper bound eludes me. I've only found the lower bound, which is just 0. I've tried to group together all of the powers of p, as follows:


    \displaystyle \lim_{p\to\infty} \biggl(\frac{{| d x|}^{p-1}}{(p-1)!} \{|d|^n(|x| + |\alpha_1|) \ldots (|x| + |\alpha_n|) \}^p\biggr)


    and maybe use the fact that

    \displaystyle \frac{{| d x|}^{p-1}}{(p-1)!}

    tends to zero as p tends to infinity, since the bottom half rises faster than the top half. However, I'm stuck because I don't know how to go about this formally, or if this is the right way to go about it. I'd be really grateful if help was provided, and possibly a solution.

    Thank you very very much in advance.

    HTale
    This is very easy indeed...assuming I am understanding it correctly...by that I mean I am assuming that \left\{d,x,\alpha,\cdots,\alpha_n\right\}\in\mathb  b{R}

    If so then similarly to the previous answer let \prod_{k=1}^{n}\bigg[|x|-\alpha_k\bigg]=C\in\mathbb{R}

    [tex]\Therefore \lim_{p\to\infty}\frac{|dx|^{p-1}\prod_{k=1}^{n}\bigg[|x|-\alpha_k\bigg]}{(p-1)!}=\lim_{p\to\infty}\frac{|dx|^{p-1}C^p}{(p-1)!}=\frac{1}{dx}\lim_{p\to\infty}\frac{\left(C|dx  |\right)^p}{(p-1)!}

    Now using the fact that p!\sim\sqrt{2\pi{p}}p^pe^{-p}

    We can rewrite this as

    \frac{1}{|dx|}\lim_{p\to\infty}\frac{p\left(|dx|C\  right)^P}{\sqrt{2\pi{p}}p^pe^{-p}}=\frac{1}{|dx|\sqrt{2\pi}}\lim_{p\to\infty}\fra  c{\sqrt{p}\left(e|dx|C\right)^p}{p^p} =\frac{1}{|dx|\sqrt{2\pi}}\lim_{p\to\infty}\sqrt{p  }\left(\frac{M}{p}\right)^p \quad\text{where }M=e|dx|C From there it should be pretty apparent the answer is zero
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    This is very easy indeed...assuming I am understanding it correctly...by that I mean I am assuming that \left\{d,x,\alpha,\cdots,\alpha_n\right\}\in\mathb  b{R}

    If so then similarly to the previous answer let \prod_{k=1}^{n}\bigg[|x|-\alpha_k\bigg]=C\in\mathbb{R}

    [tex]\Therefore \lim_{p\to\infty}\frac{|dx|^{p-1}\prod_{k=1}^{n}\bigg[|x|-\alpha_k\bigg]}{(p-1)!}=\lim_{p\to\infty}\frac{|dx|^{p-1}C^p}{(p-1)!}=\frac{1}{dx}\lim_{p\to\infty}\frac{\left(C|dx  |\right)^p}{(p-1)!}

    Now using the fact that p!\sim\sqrt{2\pi{p}}p^pe^{-p}

    We can rewrite this as

    \frac{1}{|dx|}\lim_{p\to\infty}\frac{p\left(|dx|C\  right)^P}{\sqrt{2\pi{p}}p^pe^{-p}}=\frac{1}{|dx|\sqrt{2\pi}}\lim_{p\to\infty}\fra  c{\sqrt{p}\left(e|dx|C\right)^p}{p^p} =\frac{1}{|dx|\sqrt{2\pi}}\lim_{p\to\infty}\sqrt{p  }\left(\frac{M}{p}\right)^p \quad\text{where }M=e|dx|C From there it should be pretty apparent the answer is zero

    Aha, Stirlings approximation! That's quite ingenious - thanks man! But, we are dealing in the complex domain - does this change things?
    Last edited by HTale; October 26th 2008 at 11:15 AM.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by HTale View Post
    Aha, Stirlings approximation! That's quite ingenious - thanks man! But, we are dealing in the complex domain - does this change things?
    It should not.
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