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Math Help - Double limit.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Double limit.

    Prove the following:

    \lim_{n\to\infty}\lim_{k\to\infty}\cos^{2k}(\pi n! x)=\begin{cases} 1 \text{ if  }x \in\mathbb{Q}  \\  0 \text{ if } x\in \mathbb{R}/\mathbb{Q}\end{cases} and show that:

    \lim_{n\to\infty}\lim_{k\to\infty}\cos^{2k}(\pi n! e)\neq \lim_{k\to\infty}\lim_{n\to\infty}\cos^{2k}(\pi n! e)
    Last edited by Plato; May 25th 2011 at 09:04 AM. Reason: LaTeX fix
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  2. #2
    Behold, the power of SARDINES!
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    Hint if

    x \in \mathbb{Q}

    Then there exists an N such that for all n > N

    xn! \in \mathbb{Z}

    the above fails if x is irrational. See where this takes you.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Hint if

    x \in \mathbb{Q}

    Then there exists an N such that for all n > N

    xn! \in \mathbb{Z}

    the above fails if x is irrational. See where this takes you.

    Yes, I know this part. My trouble is writing the proof or the "explanation" in a right way.

    Is the following will consider a mathematical proof:

    When x is rational the double limit can be written as:

    \{cos((2t-1)\pi)cos((2t-1)\pi)\}\cdot \{cos((2t-1)\pi)cos((2t-1)\pi)\}\cdot ...=\{(-1)(-1)\}\{(-1)(-1)\}...=1\cdot 1\cdot 1 ...=1

    or:

    \{cos(2t\pi)cos(2t\pi)\}\cdot \{cos(2t\pi)cos(2t\pi)\}\cdot ...=1\cdot 1\cdot 1\cdot ... =1

    and if x irrational:

    The double limit can be understood like that(?):

    \prod_{n = 0}^\infty x_n^2 when |x_i|<1 for all i, which is equals (the product) to 0.


    And what about:

    \lim_{k\to\infty}\lim_{n\to\infty}cos^{2k}(\pi n!e)=?


    Thank you!
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