11
$\begingroup$

Can anyone please help me with the following proof:

Prove that there exists infinitely many positive integers, $n$, such that $$\sin^2{(na)}+\sin^2{(nb)}\le \dfrac{2\pi^2}{n}\quad a,b\in \Bbb R$$

$\endgroup$
  • $\begingroup$ Look, I'm sure that this task belongs to the Diophantine approximations. "Middle-school" methods may help you to prove that "there exist solution for every $a,b$", but the passage about "infinitely many integers" points me to the Dirichlet theorem. And you don't need constructive proof there, as the thorem doesn't provide it. I understand that the proof that uses some exercise taken elsewhere looks not pleasant, but the good way is to check your reading for some form of Dirichet's theorem that was given to you and try to use it the similar way I did. Sorry, that's my opinion. $\endgroup$ – Slowpoke Aug 19 '15 at 20:44
  • $\begingroup$ The formula I used is Corrolary 1B at Ch2, paragraph 1 in Schmidt diophantine approximation I think it's credible source enough. $\endgroup$ – Slowpoke Aug 19 '15 at 20:57
  • 2
    $\begingroup$ Hint: Use Jordan inequality $$\frac{2}{\pi}x \le \sin x \le x$$ for all $x \in [0,\frac{\pi}{2}]$. Caution: For $x=0$, one may see that the inequality does not hold, but it is a special case by defining the function $f(x)=\frac{\sin x}{x}$ for $x\ne 0$ and $f(x)=1$ for $x=0$ so that $\frac{2}{\pi}\le 1$. $\endgroup$ – Mohammad W. Alomari Aug 20 '15 at 22:35
  • $\begingroup$ then can you post full solution? $\endgroup$ – user246688 Aug 21 '15 at 11:42
8
$\begingroup$

Here is my idea (if I'm wrong, I will delete my answer):

According to the Dirichlet's theorem for any given irrational $\alpha$ there are infinitely many integers $p,q$ such that $$|\alpha - \frac{p}{q}|<\frac{1}{q^2} $$

Let's take $\alpha = a/\pi, a \in \mathbb{R}; \ q = n, p = k$ $$|an-\pi k|<\frac{\pi}{n}$$ Thus there are infinitely many positive integers $n, k$ such that $$|\sin(an)|=|\sin(an)-\sin( \pi k)| < \frac{\pi}{n}<\frac{\pi}{\sqrt{n}}$$

Thus inequality $\sin^2(an) < \frac{\pi^2}{n}$ is strict.

EDIT: There is a problem when $a \neq b$ then the sets of $n$'s for them may not intersect, so I was searching for some generalizations of Dirichlet's theorem and it seems I found one (Corrolary 1B at Ch2, paragraph 1 in Schmidt - Diophantine approximation):

For any given irrational $\alpha_1, \alpha_2$, the system of inequalities $$|\alpha_j - \frac{p_j}{q}|<\frac{1}{q\sqrt{q}}, \ j=1,2 $$ has infinite number of solutions $p_1,p_2,q \in \mathbb{Z}$.

Then by taking (irrational) $\alpha_1=a/\pi, \alpha_2 = b/\pi$, and repeating steps above we see that there exist infinite number integer solutions $n,k_1,k_2$ for which $$|an - \pi k_1|<\frac{\pi}{\sqrt{n}}, \ |bn - \pi k_2|<\frac{\pi}{\sqrt{n}} $$ which gives us desired inequality. So the last thing left is to prove it when $a,b$ are such that $\alpha_1,\alpha_2$ are rational.

In this case $a=\pi\frac{p}{q}$, so $\sin(na)=0$ for $n=qk, \ k \in \mathbb{Z}$; the same holds for $b$ with another coefficient: $n=lk$; Thus $n=qlk, \ k \in \mathbb{Z}$ will do it.

$\endgroup$
  • 1
    $\begingroup$ Given $a$ and $b$ you have show which $n$ satisfies the hypotesis. In this case you didn't this. $\endgroup$ – Euler88 ... Aug 16 '15 at 22:18
  • 1
    $\begingroup$ @Euler88... In which part exactly? If it is about the last part where $\alpha_1,\alpha_2$ are rational, then it is possible only when $a,b$ look like $\pi \frac{p}{q}$. We just need to take such $n$ that will eliminate the denominator in both cases, then $0+0<\frac{2\pi^2}{n}$. The solution is an infinite set $\{qlk, \ k \in \mathbb{Z}\}$ $\endgroup$ – Slowpoke Aug 16 '15 at 22:25
  • $\begingroup$ @Euler88... The infinite set $\{n_i\}$ of possible solutions will depend on $a,b$ in any case. $\endgroup$ – Slowpoke Aug 16 '15 at 22:35
7
+25
$\begingroup$

Using the identity $|\sin x| < |x|$ we can try to check a "linearized" version of the inequality:

$$ ||n a||_{\mathbb{R}/\mathbb{Z}}^2 + ||n b ||_{\mathbb{R}/\mathbb{Z}}^2 < \frac{2\pi^2}{n}$$

where $ ||x||_{\mathbb{R}/\mathbb{Z}} = \min \big\{ |x - n\pi | : n \in \mathbb{Z}\big\}$. Both terms are positive; maybe we can get both error terms to be small:

$$ ||n a||_{\mathbb{R}/\mathbb{Z}} < \frac{\pi}{\sqrt{n}} \hspace{0.25in}\text{and}\hspace{0.25in} ||n b ||_{\mathbb{R}/\mathbb{Z}} < \frac{\pi}{\sqrt{n}} $$

Dirichlet's approximation in $\mathbb{R}^2$ says we can do just that. Why only $\frac{1}{\sqrt{n}}$ ? This is a bigger error than the 1D Dirichlet theorem.


We can prove Dirichlet's theorem using the pigeonhole principle. Consider the square $[0,\pi]^2$ and divide it into $\sqrt{N} \times \sqrt{N}$ pieces

$$ [0, \pi]\times [0, \pi] =\bigcup_{0 \leq m,n < \sqrt{N}} \Big[ \tfrac{(m-1/2)\pi}{\sqrt{N}}, \tfrac{(m+1/2)\pi}{\sqrt{N}}\Big]\times \Big[ \tfrac{(n-1/2)\pi}{\sqrt{N}}, \tfrac{(n+1/2)\pi}{\sqrt{N}}\Big]$$

By pigeonhole, one square must contain at least $2$ - in fact infinitely many - of the values $(na, nb) \in \mathbb{R}^2 / \mathbb{Z}^2$ with $n \in \mathbb{Z}$. $$ \Big((m-n)a, (m-n)b\Big) \in \Big[ -\tfrac{\pi}{\sqrt{N}}, \tfrac{\pi}{\sqrt{N}}\Big]^2 $$

For more, see What is your favorite application of the Pigeonhole Principle?


For example $n = 100, a = \sqrt{2}, b = \sqrt{3}$. Try to get an error within $\frac{1}{10}$.

enter image description here

$\endgroup$
1
$\begingroup$

A classical problem in simultaneous approximation.

Let $a_n=(\lambda n\pmod{\pi},\mu n\pmod{\pi})\in\mathbb{T}$ and let we fix some huge $N$.

Let $Q_n\subset\mathbb{T}$ be a square centered in $a_n$ having side length $\frac{\pi}{\sqrt{N}}$.

Among $Q_1,\ldots,Q_N$, at least two squares have to overlap, since the sum of their Lebesgue measures equals $\mu(\mathbb{T})$. The linearity of the sequence $\{a_n\}_{n\geq 1}$ hence implies that there is some $M\leq N$ (for clarity, $M$ is the absolute difference of the indices of the previous overlapping squares) such that both $\lambda M\pmod{\pi}$ and $\mu M\pmod{\pi}$ are $\leq\frac{\pi}{\sqrt{N}}$.

The claim follows by considering that $\sin x$ is a Lipschitz function with Lipschitz constant $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy