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Thread: An inequality from Alex Lupas

  1. #1
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    An inequality from Alex Lupas

    If possible, find two positive integers $\displaystyle m,n$ such that: $\displaystyle 8\pi - 2 < m \pi - n < e^{\pi}.$
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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 $\displaystyle m,n$ such that: $\displaystyle 8\pi - 2 < m \pi - n < e^{\pi}.$
    Do you exactly want the numbers or prove existence of them?
    If you want the existence, for any $\displaystyle \lambda\in\mathbb{R}\backslash\mathbb{Q}$ the set $\displaystyle S:=\{m\lambda+n:m,n\in\mathbb{Z}\}$ is dense in $\displaystyle \mathbb{R}$, hence for any interval of type $\displaystyle (p,q)$ with $\displaystyle {\color{red}{p<q}}$, there exists at least one $\displaystyle r\in S$ such that $\displaystyle r\in(p,q)$.
    Last edited by bkarpuz; Dec 25th 2008 at 10:02 AM. Reason: misprint pointed out by NonCommAlg is corrected.
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  3. #3
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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; Dec 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 $\displaystyle \lambda\in\mathbb{R}\backslash\mathbb{Q}$ the set $\displaystyle S:=\{m\lambda+n:m,n\in\mathbb{Z}\}$ is dense in $\displaystyle \mathbb{R}$, hence for any interval of type $\displaystyle (p,q)$ with $\displaystyle \color{red}p < q$, there exists at least one $\displaystyle r\in S$ such that $\displaystyle r\in(p,q)$.
    that's almost the idea but note that we're looking for an element of S which is in the form $\displaystyle m\lambda + n$ with $\displaystyle m > 0 $ and $\displaystyle n < 0.$ 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 $\displaystyle m\lambda + n$ with $\displaystyle m > 0 $ and $\displaystyle n < 0.$ 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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    According to my little pocket calculator (which may not be quite up to the job),

    $\displaystyle \begin{aligned}8\pi-2&\approx23.132741,\\
    16612\pi-52165&\approx23.137161,\\
    e^\pi&\approx23.140693.\end{aligned}$

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

    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 $\displaystyle \frac{52163}{16604}$. So take $\displaystyle n-2=52163,\;\;m-8=16604$. 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 $\displaystyle \{x \}=x-\lfloor x \rfloor.$ it's known that if $\displaystyle \theta \notin \mathbb{Q},$ then the sequence $\displaystyle \{\{k \theta \}: \ \ k \in \mathbb{N} \}$ is dense in the unit interval. so there exists $\displaystyle k \in \mathbb{N}$ such that: $\displaystyle \left \{\frac{k}{\pi} \right \} > 9 - \frac{e^{\pi} + 2}{\pi}. \ \ \ \ \ (*)$

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