Infinite Sequences of Sines

[PeroK]

Therefore, Inf(S) = 0. There are plenty of other irrationals such that Inf(S) = 0. You recognize that's an issue, but how do you handle it? Can you even prove that $S = \{k' = k\pi -[k\pi]: k \in \mathbb N \}$ doesn't have Inf (S)= 0?

The whole point is that $\inf(S) = 0$ for all irrationals; as I've just proved. And that proves the original lemma.

 

[bob012345]

First, we need to prove that for any irrational number $x$ and $\epsilon > 0$, there exist integers $k, n$ such that $|kx - n| < \epsilon$. I thought this would follow from the density of the rationals, but I haven't found a proof yet.

Let's assume that holds. And, in particular, applies to $\pi$.
...

But, we still need a proof of the initial lemma.

I don't understand why you needed to prove it for any $x$ instead of just $\pi$ which is what you are interested in using. If you can show you can construct the sequence around $\pi$ then its existence is proved isn't it?

 

[PeroK]

I don't understand why you needed to prove it for any $x$ instead of just $\pi$ which is what you are interested in using. If you can show you can construct the sequence around $\pi$ then its existence is proved isn't it?

It amounted to the same thing. The irrationality of $\pi$ was the key property, so it seemed logical to generalise.

 

[bob012345]

I realize the proof does not require it but I'm still interested in seeing an actual algorithm or method to generate the $m_i's$. I tried but ran into a mental block.

 

[pasmith]

Alternative method:

Define $f: \mathbb{N} \to S^1 : m \mapsto (\cos m , \sin m)$. Then by sequential compactness of $S^1$, $(f(m))_{m\in\mathbb{N}}$ has a convergent subsequence $(f(m_n))_{n\in\mathbb{N}}$. Then for positive integer $\ell$ let $$P_\ell: S^1 \to S^1 : (\cos \theta, \sin \theta) \mapsto (\cos \ell \theta, \sin \ell \theta)$$ which is a polynomial of degree $\ell$ and so continuous. Hence $(P_{\ell}\circ f)(m_n) = (\cos(\ell m_n), \sin(\ell m_n))$ converges.

I realize the proof does not require it but I'm still interested in seeing an actual algorithm or method to generate the $m_i's$. I tried but ran into a mental block.

EDIT: I got confused with a different thread concerning convergence of $\sin^2(a_n)$, which has period $\pi$; however the principle is clear.

import numpy as np
a = np.zeros(6,dtype=int)
a[0] = 1
xprev = a[0] % np.pi
# Constructing a sequence such that x_{m_n} tends to a multiple of pi:
for n in range(1,6):
    m = a[n-1] + 1
    while True:
        x = m % np.pi
        if min(x, np.pi-x) < min(xprev, np.pi-xprev):
            a[n] = m
            xprev = x
            break
        m = m + 1

        
a
array([     1,      3,     22,    333,    355, 103993])
np.sin(a)**2
array([7.08073418e-01, 1.99148567e-02, 7.83456762e-05, 7.78129716e-05,
       9.08682039e-10, 3.65931487e-10])

 

[PeroK]

Here's a full proof:

Let $x$ be a positive irrational and consider the infimum of the set $S = \{k' = kx -[kx]: k \in \mathbb N \}$. We need to show that $s = \inf(S) = 0$.

Assume $0 < s < 1$. We can find $k_1$ such that$$k_1x = n_1 + s + \epsilon$$where $\epsilon < s$ and $s + \epsilon < 1$.

Now, we can find $k_2 > k_1$ such that $$k_2x = n_2 + s + \delta$$where $\delta < \frac \epsilon 2$. Note to ensure that $k_2 > k_1$, we look for $\delta < \min\{\frac \epsilon 2, 1'-s, 2'-s \dots k'_1-s\}$.

Moreover:$$(2k_2 - k_1)x = 2n_2 + 2s + 2\delta - n_1 - s - \epsilon = 2n_2 - n_1 + s + 2\delta - \epsilon$$This is a contradiction, as$$0 < s + 2\delta - \epsilon < s$$and hence we have a member of $S$ less than the infimum. The conclusion is that $\inf(S) = 0$, as required.

There is a little more work to cover the case where $s \in S$. First, $s \ne x-[x]$, since $x- [x], 2x -[2x], 3x-[3x] \dots$ cannot increase indefinitely and when it overflows $1$ and decreases we must have $nx - {nx} < x - [x]$, because $x- [x]$ is effectively the increment each time.

And, if $s = nx - [nx]$, then we take $nx$ as our starting irrational with the same $s$, which is impossible, by the above argument.

 

[PeroK]

Actually, the above tidy-up gives an idea for a simpler elementary proof. In outline, we start with $x - [x]$, then find a sequence $n_k$ such that $n_{k+1}x - [n_{k+1}x] < n_kx - [n_kx]$. This sequence is bounded below, hence converges and hence the difference in terms in the sequence must tend to zero. That gives the required sequence.

 

[paige turner]

I assume you mean radians? If it were interpreted as degrees it would have the trivial solution $sin(l m_i ) = 0$ for all $l, m_i =360 i$ with $i= 1 ,2,3...∞$ converging to zero.

OP, pls answer this. I was all excited that I found an easy solution.

 

[Euge]

OP, pls answer this. I was all excited that I found an easy solution.

Hi @paige turner, please read post #12.

 

[Euge]

I didn't expect that this POTW would invite so much attention and discussion. Thanks to all for participating!

Spoiler

Since the sequence $\{\sin n\}_{n = 1}^\infty$ is bounded, by the Bolzano Weierstrass theorem it has a convergent subsequence $\{\sin n_1(k)\}_{k = 1}^\infty$. The sequence $\{\sin 2n_1(k)\}_{k = 1}^\infty$ is bounded, so it admits a further subsequence $\{\sin 2n_2(k)\}_{k = 1}^\infty$. Continuing the process, we extract a sequence of sequences $\{n_1(k)\}_{k = 1}^\infty, \{n_2(k)\}_{k = 1}^\infty,\ldots$ such that for every positive integer $\ell$ the sequence $\{\sin \ell n_\ell(k)\}_{k = 1}^\infty$ converges. Define $m_k = n_k(k)$ for $k = 1, 2, 3,\ldots$. Then $m_1 < m_2 < \cdots$, and given a positive integer $\ell$, the sequence $\{\sin \ell m_k\}_{k = 1}^\infty$ is eventually a subsequence of the convergent sequence $\{\sin \ell n_\ell(k)\}_{k = 1}^\infty$, so it converges.