Infinite Sequences of Sines

[Euge]

Prove the existence of a strictly increasing sequence $m_1 < m_2 < m_3 < \cdots$ of integers satisfying the property that for all positive integers $\ell$, the sequence $\sin(\ell m_1), \sin(\ell m_2), \sin (\ell m_3),\ldots$ converges.

 

[anuttarasammyak]

It is just a conjecture but
$$m_1=3$$
$$m_2=31$$
$$m_3=314$$
---
$$m_n=[\pi*10^{n-1}]$$
may satisfy the statement with convergence to zero. Is it so ?

 

[PeroK]

Why would you get convergence? The fraction you are losing depends on the nth digit of the decimal expansion of pi, which doesn't converge.

 

[anuttarasammyak]

Sure, my bad. Thanks.
Another attempt is to find better rational number approximations of $\pi$ by computer,
$$\frac{3}{1}, \frac{22}{7},\frac{355}{113},\frac{3550}{1130},\frac{99733}{31746},\frac{833719}{265381},\frac{5419351}{1725033},\frac{80143857}{25510582},...$$
which are the best ones with both numerator and denominator are under $10^n$. I made series of the numerators,
$$m_1=3,m_2=22,m_3=355,m_4=3550,m_5=99733,m_6=833719,m_7=5419351,m_8=80143857,...$$

x	sinx
3​	0.141120008​
22​	-0.008851309​
355​	-3.01444E-05​
3550​	0.000301444​
99733​	-0.000380862​
833719​	2.31292E-06​
5419351​	-3.82004E-08​
80143857​	-1.47741E-08​

sin x are small but I am afraid it is not enough for the statement.

Say integers a, b are
$$\frac{b}{a}=\pi+\epsilon$$
$$\sin(bl)=\sin(al\epsilon)$$
Regard them as sequence, so that $b_n$ be what we want, we must show
$$a_n \epsilon_n \rightarrow 0$$
I hope to find a good approximation sequence satisfying it. Diophantine approximation https://en.wikipedia.org/wiki/Diophantine_approximation seems to assure the existence of such a sequence, but I have no idea how to construct it actually.

​

 

[Petek]

What about using the partial convergents of the continued fraction for $\pi$?

 

[bob012345]

Prove the existence of a strictly increasing sequence $m_1 < m_2 < m_3 < \cdots$ of integers satisfying the property that for all positive integers $\ell$, the sequence $\sin(\ell m_1), \sin(\ell m_2), \sin (\ell m_3),\ldots$ converges.

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.

 

[bob012345]

Does strictly increasing mean just each term larger or each term larger according to some algorithm? Can the integers be arbitrary as long as they are larger?

 

[PeroK]

Does strictly increasing mean just each term larger or each term larger according to some algorithm? Can the integers be arbitrary as long as they are larger?

A strictly increasing sequence is a sequence where each term is larger that the previous term. Nothing more than that.

 

[pasmith]

My idea is to exploit the periodicity of the sine function by considering $x_m = \ell m \mod 2\pi$ as a sequence in $[0,2\pi]$ and using Bolzano-Weierstrass to extract a convergent subsequence $x_{m_n} \to x \in [0,2\pi]$. Then by continuity of the sine function, $\sin(\ell m_n) = \sin(x_{m_n}) \to \sin x$.

Unfortunately that shows "for every positive integer $\ell$ there exists a sequence of strictly inreasing integers $m_n$" and we require the reverse, so there is more work to be done.

 

[PeroK]

A sequence can be obtained using the density of the rationals.

 

[PeroK]

Find a sequence first for $l =1$, then show the same sequence works for any $l$.

 

[Euge]

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.

Yes, it is in radians.

 

[Euge]

Does strictly increasing mean just each term larger or each term larger according to some algorithm? Can the integers be arbitrary as long as they are larger?

The strictly increasing sequence is expressed in the problem statement: $m_1 < m_2 < m_3 < \cdots$. Thus $m_2$ is larger than $m_1$, $m_3$ is larger than $m_2$, and so on.

 

[bob012345]

Find a sequence first for $l =1$, then show the same sequence works for any $l$.

I think one must derive a solution for any $\ell$ up front. There are an infinite number of sets of $m_i's$ but in general they will not work for any $\ell$ so the way the set of $m_i's$ are generated is the key in my mind. So far I can only think of ways that are not scalable with $\ell$.

 

[bob012345]

The strictly increasing sequence is expressed in the problem statement: $m_1 < m_2 < m_3 < \cdots$. Thus $m_2$ is larger than $m_1$, $m_3$ is larger than $m_2$, and so on.

Thanks. The word sequence made me think they must be some consistent algorithm but I understand now it just means list and $m_{i+1}$ need have no relation to $m_i$ except $m_{i+1}>m_i$.

 

[PeroK]

I think one must derive a solution for any $\ell$ up front.

Not necessarily. Get a solution for $l =1$ first.

 

[PeroK]

I worked out a proof in my head, but something I thought was equivalent to the density of the rationals turns out to be quite difficult to prove!

 

[pasmith]

My idea is to exploit the periodicity of the sine function by considering $x_m = \ell m \mod 2\pi$ as a sequence in $[0,2\pi]$ and using Bolzano-Weierstrass to extract a convergent subsequence $x_{m_n} \to x \in [0,2\pi]$. Then by continuity of the sine function, $\sin(\ell m_n) = \sin(x_{m_n}) \to \sin x$.

Unfortunately that shows "for every positive integer $\ell$ there exists a sequence of strictly inreasing integers $m_n$" and we require the reverse, so there is more work to be done.

For positive integer $\ell$, $\sin(\ell m_n)$ is a polynomial of degree $\ell$ in $\sin(m_n)$. and will therefore converge to a limit if $\sin(m_n)$ does. I think this competes the proof.

 

[bob012345]

For positive integer $\ell$, $\sin(\ell m_n)$ is a polynomial of degree $\ell$ in $\sin(m_n)$. and will therefore converge to a limit if $\sin(m_n)$ does. I think this completes the proof.

What is the polynomial?

 

[PeroK]

For positive integer $\ell$, $\sin(\ell m_n)$ is a polynomial of degree $\ell$ in $\sin(m_n)$. and will therefore converge to a limit if $\sin(m_n)$ does. I think this competes the proof.

I think this only works if you choose a potentialy different sequence $m_n$ for each $l$.

Or, perhaps I got confused by the previous post?

 

[PeroK]

Your idea seems to work simply by:

$\forall m > 0$, let $x_m \in (0, 2\pi)$, where $m = 2k\pi + x_m$ for some $k \ge 0$.

By B-W there is a convergent subsequence $x_{m_n} \to x$, and $\sin(m_n) = \sin(x_{m_n})$ converges by continuity of the sine function. Likewise $\cos(m_n)$ converges.

Finally, $\forall l$, $\sin(lx)$ can be written as a continuous function of $\sin(x), \cos(x)$.

 

[PeroK]

What is the polynomial?

Use the Euler formula:$$\sin(lx) = Im[e^{ilx}] = Im[(\cos x + i\sin x)^l]$$

 

[bob012345]

Is it necessary in this problem that convergence means convergence to zero?

 

[Guineapigs_181]

Sorry if I'm misunderstanding, but since each term has to be bigger than the last, wouldn't
m₁=2π
m₂=4π
m₃=6π
where
m_n=[2nπ]
work?

 

[bob012345]

Sorry if I'm misunderstanding, but since each term has to be bigger than the last, wouldn't
m_1=2pi
m_2=4pi
m_3=6pi
---
m_n=[2npi]
work?

The $m_i's$ must be integers.

 

[bob012345]

For positive integer $\ell$, $\sin(\ell m_n)$ is a polynomial of degree $\ell$ in $\sin(m_n)$. and will therefore converge to a limit if $\sin(m_n)$ does. I think this completes the proof.

I had to think of this problem graphically rather than analytically. Using a unit circle and starting with the sine of 1 radian one easily can see how an infinite number of smaller sines can be made by ever larger angles to exploit the periodicity of the sine function as @pasmith said thus generating an infinite set of strictly increasing $m_i's$ chosen such that the sines are all in the positive quadrant and $\sin(m_i) →0$ as $m_i →∞$

What threw me is the demand $\ell$ be any integer and the $m_i's$ remain the same. Graphically, for an arbitrary $\ell$ this spreads the nicely decreasing sines in the first quadrant into the full circle and the sequence jumps around between minus one and plus one. The sequence could seemingly oscillate forever yet in the limit as $\ell m_i →∞$, there is still an infinite number of terms in the positive quadrant that converge to zero so the sequence as a whole converges.

 

[PeroK]

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 $\pi$.

Note that for every $m > 0$, we have the closest integer mutiple of $\pi$ ($n_m\pi$) and the minimum distance from $m$ to multiples of $\pi$ ($d_m = |m - n_m\pi|$).

We choose $\epsilon_1 = \frac 1 {10}$ and $\epsilon_{i+1} = min\{\frac 1 {10^{i+1}}, d_1 \dots d_{m_i}\}$. Where $m_i$ is chosen so that $|m_i -n_{m_i}\pi| = d_{m_i} < \epsilon_i$.

Involving all the previous $d_i$ (all positive integers up to $m_i$) ensures we have a strictly increasing sequence $m_i$.

Then, as $\epsilon_i$ is small, we have $|\sin(m_i)| = |\sin(m_i - n_{m_i}\pi)| < |m_i - n_{m_i}\pi| < \epsilon_i \le \frac 1 {10^i}$.

And, we see that $\sin(m_i) \to 0$.

Finally, we can use the same argument with cosines and polynomial functions to show that for any $l$ we have $\sin(lm_i) \to 0$.

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

 

[pasmith]

Your idea seems to work simply by:

$\forall m > 0$, let $x_m \in (0, 2\pi)$, where $m = 2k\pi + x_m$ for some $k \ge 0$.

By B-W there is a convergent subsequence $x_{m_n} \to x$, and $\sin(m_n) = \sin(x_{m_n})$ converges by continuity of the sine function. Likewise $\cos(m_n)$ converges.

Finally, $\forall l$, $\sin(lx)$ can be written as a continuous function of $\sin(x), \cos(x)$.

This can be cleaned up further: I missed that for positive integer $\ell$ we have simply $\sin(\ell m_n) = \sin(\ell x_{m_n}) \to \sin(\ell x)$.

 

[Petek]

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.

Is this statement true? Here's a heuristic argument casting doubt:

Let $x = \pi$, for simplicity. We can assume, wlog, that $\epsilon < 1$, Then the expression becomes $|k\pi - n| < \epsilon < 1$. That implies that $n = [k\pi]$, the greatest integer less than $k\pi$. Thus, $|k\pi - [k\pi]|$ is the fractional part of $k\pi$. If we use base k, then $k\pi$ is the base-k expansion of $\pi$, shifted by one digit. Saying that this number is less than any $\epsilon > 0$ would be like saying that the base-k expansion of $\pi$, shifted by one, has a sequence with arbitrarily many zeros. That's not known.

 

[PeroK]

Is this statement true? Here's a heuristic argument casting doubt:

Let $x = \pi$, for simplicity. We can assume, wlog, that $\epsilon < 1$, Then the expression becomes $|k\pi - n| < \epsilon < 1$. That implies that $n = [k\pi]$, the greatest integer less than $k\pi$. Thus, $|k\pi - [k\pi]|$ is the fractional part of $k\pi$. If we use base k, then $k\pi$ is the base-k expansion of $\pi$, shifted by one digit. Saying that this number is less than any $\epsilon > 0$ would be like saying that the base-k expansion of $\pi$, shifted by one, has a sequence with arbitrarily many zeros. That's not known.

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

 

[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.