An inequality from Alex Lupas

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December 24th 2008, 03:50 PM
NonCommAlg

An inequality from Alex Lupas

If possible, find two positive integers $m,n$ such that: $8\pi - 2 < m \pi - n < e^{\pi}.$
December 24th 2008, 11:33 PM
bkarpuz

Quote:

Originally Posted by NonCommAlg

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 $r\in(p,q)$ .
December 25th 2008, 12:13 AM
Isomorphism

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) (Crying)

I am wondering if a "Elementary and Middle School Math" technique exists for this (Thinking)
December 25th 2008, 12:15 AM
NonCommAlg

Quote:

Originally Posted by bkarpuz

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? (Smile)
December 25th 2008, 05:59 AM
bkarpuz

Quote:

Originally Posted by NonCommAlg

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? (Smile)

By computations, it seems hard for me (Doh), but as Isomorphism given, it may be possible to ask mathematical programming languages for such values. (Thinking)
December 26th 2008, 10:08 AM
Opalg

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.
December 26th 2008, 12:35 PM
NonCommAlg

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 $(*).$ (Wondering)