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Math Help - An inequality from Alex Lupas

  1. #1
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    An inequality from Alex Lupas

    If possible, find two positive integers m,n such that: 8\pi - 2 < m \pi - n < e^{\pi}.
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    If possible, find two positive integers m,n such that: 8\pi - 2 < m \pi - n < e^{\pi}.
    Do you exactly want the numbers or prove existence of them?
    If you want the existence, for any \lambda\in\mathbb{R}\backslash\mathbb{Q} the set S:=\{m\lambda+n:m,n\in\mathbb{Z}\} is dense in \mathbb{R}, hence for any interval of type (p,q) with {\color{red}{p<q}}, there exists at least one r\in S such that r\in(p,q).
    Last edited by bkarpuz; December 25th 2008 at 10:02 AM. Reason: misprint pointed out by NonCommAlg is corrected.
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    Hey bkarpuz, I think he expects us to "find" m and n.

    The inequality is pretty tight. I think m = 3391, n = 10630 works. But I did that the inelegant way (using MATLAB)

    I am wondering if a "Elementary and Middle School Math" technique exists for this
    Last edited by Isomorphism; December 25th 2008 at 12:33 AM.
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    Quote Originally Posted by bkarpuz View Post
    Do you exactly want the numbers or prove existence of them?
    If you want the existence, for any \lambda\in\mathbb{R}\backslash\mathbb{Q} the set S:=\{m\lambda+n:m,n\in\mathbb{Z}\} is dense in \mathbb{R}, hence for any interval of type (p,q) with \color{red}p < q, there exists at least one r\in S such that r\in(p,q).
    that's almost the idea but note that we're looking for an element of S which is in the form m\lambda + n with m > 0 and n < 0. you might want to check the question again?
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    that's almost the idea but note that we're looking for an element of S which is in the form m\lambda + n with m > 0 and n < 0. you might want to check the question again?
    By computations, it seems hard for me , but as Isomorphism given, it may be possible to ask mathematical programming languages for such values.
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  6. #6
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    According to my little pocket calculator (which may not be quite up to the job),

    \begin{aligned}8\pi-2&\approx23.132741,\\<br />
16612\pi-52165&\approx23.137161,\\<br />
e^\pi&\approx23.140693.\end{aligned}

    If 8\pi-2<m\pi-n<e^\pi then 0<(m-8)\pi-(n-2) < e^\pi-(8\pi-2)\approx23.140693 - 23.132741 = 7.952\times10^{-3}. Therefore 0<\pi - \frac{n-2}{m-8}< \frac{7.952\times10^{-3}}{m-8}.

    So we are looking for a rational approximation to π that is less than π, but close enough that the difference is of the order of one-thousandth of the reciprocal of its denominator. Among the known rational approximations to π, the first one that seems to fit these conditions is \frac{52163}{16604}. So take n-2=52163,\;\;m-8=16604. Unless my little calculator is misleading me, that appears to work.

    Those are bigger numbers than Isomorphism gets, but I don't have MATLAB to numbercrunch for me.
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    let \{x \}=x-\lfloor x \rfloor. it's known that if \theta \notin \mathbb{Q}, then the sequence \{\{k \theta \}: \ \ k \in \mathbb{N} \} is dense in the unit interval. so there exists k \in \mathbb{N} such that: \left \{\frac{k}{\pi} \right \} > 9 - \frac{e^{\pi} + 2}{\pi}. \ \ \ \ \ (*)

    now let m=\left \lfloor \frac{k}{\pi} \right \rfloor + 9, and n=k+2. then it's easy to see that 8\pi - 2 < m \pi - n < e^{\pi}. this proves the existence of m,n. to find m,n, we need to find a solution for (*).
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