POTW Infinite Sequences of Sines

[Petek]

$k, n$ depend on $\epsilon$. If the same $k,n$ suffice for all $\epsilon$, then we must have $k\pi = n$, which is impossible.

Perhaps I should flesh out my argument, since I wasn't assuming that k and n were fixed. Your lemma states, in effect, that given any $\epsilon > 0$, there exist integers k and n (depending on $\epsilon$) such that $|k\pi - n| < \epsilon$. I argue that, if this is true, then there exist sequences of integers $k_1, k_2, k_3, \cdots$ and $n_1, n_2, n_3, \cdots$, such that

$|k_1\pi - n_1|= 0. 0a_1 a_2 a_3 \dots$
$|k_2\pi - n_2| = 0. 00b_1 b_2 b_3 \dots$
$|k_3\pi - n_3| = 0. 000c_1 c_2 c_3 \dots$
{writing $k_1$ and $n_1$ corresponding to $\epsilon_1$ (as defined below) and so on.

As in my prior post, we may assume that $\epsilon < 1$. Then, by assumption, there exist integers $k_{\epsilon}$ and $n_{\epsilon}$ such that $|k_{\epsilon}\pi - n_{\epsilon}| < \epsilon < 1$. Since the difference between $k_{\epsilon}\pi$ and $n_{\epsilon}$ is less than one, we must have that $n_{\epsilon}$ is the integer nearest to $k_{\epsilon}\pi$ and so $|k_{\epsilon}\pi - n_{\epsilon}|$ is the decimal part of $k_{\epsilon}\pi$. Setting $\epsilon_1 = 0.1, \epsilon_2 = 0.01$, and so on, we get the sequence shown above. It is unknown whether such a sequence, corresponding to arbitrarily many zeros at the beginning of $k_i\pi$'s decimal expansion, exists. (I originally included the part about base k to simplify the argument, but it's not needed.)

 

[PeroK]

I don't follow that argument. I'm pretty sure that lemma must hold for all irrationals.

 

[PeroK]

Here's an outline proof. Let $x$ be a positive irrational and consider the infimum of the set $S = \{k' = kx -[kx]: k \in \mathbb N \}$. By choosing $\epsilon < inf S$ we can find $k_1, k_2$ such that $k'_1 - inf S = \epsilon$ and $infS < k'_2$ and $k'_1 - k'_2 < \epsilon$.

Then each $k_2 -k_1$ steps we reduce $k'$ by a fixed quantity less than $inf S$, which leads to some $k' < inf S$. Unless, of course, $\inf S = 0$.

 

[Petek]

Here's an outline proof. Let $x$ be a positive irrational and consider the infimum of the set $S = \{k' = kx -[kx]: k \in \mathbb N \}$. By choosing $\epsilon < inf S$ we can find $k_1, k_2$ such that $k'_1 - inf S = \epsilon$ and $infS < k'_2$ and $k'_1 - k'_2 < \epsilon$.

Then each $k_2 -k_1$ steps we reduce $k'$ by a fixed quantity less than $inf S$, which leads to some $k' < inf S$. Unless, of course, $\inf S = 0$.

What if $x = 0.101001000100001000001 \cdots$? (with the number of zeros between ones increasing by one each time). That number is clearly irrational, but the set S contains numbers of the form $0.10 \cdots, 0.0010 \cdots, 0.00010 \cdots$ (letting k equal a suitable power of 10) and so on. 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?

 

[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.

 

[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 <br /> 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.