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Math Help - A problem in analysis

  1. #1
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    Question A problem in analysis

    Suppose e^(e^x)=an x^n, prove that an>=e(r ln n)^(-n) when n>1, r is a constant larger than e. Thanks
    Last edited by cyclic; December 31st 2008 at 05:44 PM.
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  2. #2
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    Quote Originally Posted by cyclic View Post

    Suppose e^(e^x)=an x^n, prove that an>=e(r ln n)^(-n) when n>1, r is a constant larger than e. Thanks
    the reason that you haven't got any response is that your question is not very clear. i don't know about other members in here but if a poster didn't bother to explain his/her question clearly, i.e.

    mathematically understandable, then i would just ignore the question. your question may be understood in different ways. for example, this one might be what you meant:


    suppose e^{e^x}=a_nx^n, for some x \in \mathbb{R}, \ a_n \in \mathbb{R}, \ 1 < n \in \mathbb{N}. then there exists a real constant r > e, independent from n and x, such that a_n \geq \frac{e}{(r \ln n)^n}.
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  3. #3
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    Thank you very much.

    Your understand is correct. I am very sorry that I don't know how to post mathematical formula.
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  4. #4
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    Quote Originally Posted by cyclic View Post
    Your understand is correct. I am very sorry that I don't know how to post mathematical formula.
    i think we also need to assume that x > 0 because otherwise it would be possible to choose a_n < 0, which would disprove the problem. for example e^{e^{x}}= -x has a root in the interval (-2,-1).
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  5. #5
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    You are wrong

    e^{e^x}=a_nx^n is a series
    Last edited by cyclic; January 2nd 2009 at 05:26 AM. Reason: Fixed the poor latex code
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  6. #6
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    Quote Originally Posted by cyclic View Post
    e^{e^x}=a_nx^n is a series
    wow! that totally changes the problem!! so you mean this: e^{e^x}=\sum_{n=0}^{\infty}a_nx^n.
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  7. #7
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    Yes, that is correct.

    I think that the problem should be very difficult, becauce in a Chinese math forum there is no one can solve it
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  8. #8
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    I post the problem again

    Suppose e^{e^x}=\sum_{n=0}^{\infty}a_nx^n, prove that
    a_n \geq \frac{e}{(r \ln n)^n} when n>1, r is a constant larger than e.
    Note that: NOt there exists a real constant r > e, BUT r is a constant larger than e
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