# Limit of sin(n)^n and working set

How to prove that the limit $$\lim_{n\to\infty}sin(n)^n$$ n integer towards infinity does not exist ?

If n is a real then it's obvious since we can take n=Pi/2*k k being an integer.

But if n is a integer then sin(n) is always smaller than 1, hence the power n should tend towards 0. I know this reasoning is wrong.

So is it not important the working set and we could use the reasoning on real set for n ?

Thanks.

arildno
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I do not immediately see why the limit should NOT exist, but rather that when we work with integers, the limit does, indeed, exist, and equals 0.
This is in no contradiction with the result for the reals.

Subsequences may perfectly well have well-defined limits, even though the sequence itself doesn't have such a limit.

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But, again:
I do not know whether the limit exists or not, I might be wrong, but I think it does exist, equalling 0.

I thought that way too until i tried to evaluate numerically the series with that general term. I then saw it does not converge and checked for a necessary criterion which is precisely that limit.
Numerically we obtain a lot of 0s but even for n large as 5000 result near 1 and -1 are obtained ?!
The reasoning with multiples of pi/2 is not a greater set than the integers it is just scaled.

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It's not obvious whether it should exist. Although we're taking very high powers of numbers from $(-1,1)$, those numbers are sometimes very close to $\pm 1$. Indeed, for every $\epsilon>0$, infinitely many members of the sequence appear in each of $(-1,-1+\epsilon)$ and $(1-\epsilon, 1)$. This isn't enough to draw a conclusion about whether the limit exists. I'm just saying it's not obvious.

arildno
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Then it certainly is a lot trickier. Interesting problem!
If it does NOT exist, it must mean that there is no UNIFORM convergence towards zero in its subsequences, i.e for every fixed L, you can always find N's greater than L, such as for sin(N)^N, you may be arbitrarily close to 1.

My guess is that the sequence has $1,-1,0$ all as limit points (and so doesn't converge), but that for any $\epsilon\in(0,1)$, the terms that lie outside $(-\epsilon,\epsilon)$ become rarer and rarer. This is because the sequence $(sin(n))_{n=1}^\infty$ "fills the space" in $(0,1)$ in a predictable way.

In more detail:
The sequence $(sin(n),cos(n))_{n=1}^\infty$ of vectors on the unit circle is in some sense uniformly distributed over the circle. This means it's very rarely near the right and left edge, but still always goes arbitrarily near there eventually.

arildno
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My guess is that the sequence has $1,-1,0$ all as limit points (and so doesn't converge), but that for any $\epsilon\in(0,1)$, the terms that lie outside $(-\epsilon,\epsilon)$ become rarer and rarer. This is because the sequence $(sin(n))_{n=1}^\infty$ "fills the space" in $(0,1)$ in a predictable way.

In more detail:
The sequence $(sin(n),cos(n))_{n=1}^\infty$ of vectors on the unit circle is in some sense uniformly distributed over the circle. This means it's very rarely near the right and left edge, but still always goes arbitrarily near there eventually.
This would essentially agree with my view.
We may find subsequences of sin(n)^n that are actually always INCREASING towards 1, as n goes towards infinity

We may find subsequences of sin(n)^n that are actually always INCREASING towards 1, as n goes towards infinity

Yeah. With 1 being an unattained upper bound on the sequence, this would have to be possible if 1 were a limit point.

The sequence $(sin(n),cos(n))_{n=1}^\infty$ of vectors on the unit circle is in some sense uniformly distributed over the circle.

Isnt the sequence {sin(n)} filling the interval [-1;1] with a beta distribution ?

Coming back to the original post, if we put n=PI/2*k, k a integer we get 3 possibilities : 1^n, 0^n, (-1)^n, the latter gives a complex number : (-1)^Pi=(exp(iPI)^PI)=exp(iPI^2)=cos(PI^2)+isin(PI^2)

This let me think about another question : what is 1^Pi ? if we take 1=exp(0) we get 1, if we take 1=exp(2*i*PI) we get a complex number ?

. Since this gives 0, 1 and other numbers, we conclude the limit does not exist.

Since the limit n->Infinity and PI/2*n->Infinity is the same infinity, is it a proof that the limit over integers does not exist ?

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add : it is weird but apparently numerically $$\lim_{n\to\infty}sin(n)^{n^2}$$ seems to exist and is 0.

arildno
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Interesting, but I wouldn't say necessarily weird.

If, say, for every $\epsilon>0$, we may find a subsequence of N's so that
we have $$sin(N)\approx(1-\epsilon)^{\frac{1}{N^{p}}}, p\geq{1}$$, then such an N-sequence of sin(N)^N should converge towards 1.
But, if p<2, then those subsequences will go towards 0 in your new case with n^2 as your exponent.

Using the fact that, a succession a_n has limit if and only if any sub-succession has, then you can see that if you pick the subsuccession n=Pi/2 r, r natural, then you have non convergence.

By the way, for the same reason also sin(x)^x^2 doesn't exist, numerically it gives a result probably because the algorythm stops iteracting and approssimates.

arildno
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Using the fact that, a succession a_n has limit if and only if any sub-succession has, then you can see that if you pick the subsuccession n=Pi/2 r, r natural, then you have non convergence.

This is true for the set of REALS.
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But, the problem in THIS thread is to limit ourselves to the set of INTEGER values of the sine argument, and pi/2*r is NEVER an integer, for r integer.

I see, i misunderstood the point then.

Maybe this:

The function sequence $f_n (x)=sin(x)^{n}$ converges to the function:
$f(x)=$ 1, if x=$\frac{n\pi}{2}$
$f(x)=$ -1, if x= $\frac{3n\pi}{2}$
$f(x)=$ 0, elsewhere

Then, if $n\in N$, f(n)=0, $\forall n$, so the limit of f(n), should be zero(not sure if this last step holds though)

I believe we need to first show that the values $|\sin(n)|$ are dense in $(0,1)$. If so then I believe we can make a claim that:

$$\exists\; n : 1-\epsilon<|sin(n)|<1$$ for any $\epsilon>0$ such that $1-\delta<|\sin(n)|^n<1$ for some $\delta>0$.

But that means there exists an n such that

$$1-2\delta<|sin(n)|^n<1-\delta$$

for any $\delta>0$ and in general, there exists an n, such that:

$$1-n\delta<|sin(n)|^n<1-(n-1)\delta$$

and that would suggest the set $\{|\sin(n)^n|\}$ forever oscillates within a boundary of some type of wave between 0 and 1 suggesting $\{\sin(n)^n\}$ does so between -1 and 1 with most of the values being near zero but never all going to zero if the set $\{\sin(n)\}$ is indeed dense in (-1,1) for all natural numbers.

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Does that claim holds though? Because (0,1) and $$N$$ have not the same cardinality..

Office_Shredder
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This is essentially a number theory question. I'm going to answer a slightly different question because it's easier to work with (and then explain why I can't answer this one but I suspect it has the same answer).

Consider
$$\lim_{n\to \infty} \cos(n)^n$$

If there is a large value of n (call it N) for which this is close to 1 or -1, then cos(N) must be close to 1 or -1. This means that N is close to $M \pi$ for some M. Equivalently,
$$\frac{N}{M} \approx \pi$$
OK, now we are in business, because we are asking how well can we approximate an irrational number by a fraction. Hurwitz's theorem says that
http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory [Broken])

we can find N and M such that
$$\left| \frac{N}{M} - \pi \right| \leq \frac{1}{M^2}$$

In this case, $$\left|N-M\pi \right| \leq \frac{1}{M}$$
and
$$\cos(N)^N \approx \left(\pm 1-\frac{1}{2M^2} \right)^{N}$$
from the Taylor series expansion of cosine (the plus or minus comes from whether M is odd or even, let's assume it's even for now, otherwise multiply N and M by two and we still get something that is good enough) and noting that M and N differ by a constant ( since N is approximately pi*M)
$$\approx \left(1-\frac{1}{2M^2} \right)^{\pi M}$$

For really large M this is going to be approximately $e^{-\pi/(2M)}$ which is going to converge to 1 as M goes to infinity. So there are values of N such that $\cos(N)^N$ is close to 1.

The problem with trying to apply this to sin(n)n is that we want to find N and M such that
$$N \approx \pi/2 + M\pi$$
and this doesn't quite fit Hurwit'z theorem but maybe someone can see how to do it or derive the sin(n)n result from the cos(n)n one.

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arildno
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The phase shift of sine and cosine ought not matter, but I agree Office Shredder, we are deep into number theoretical subtleties here..

So if we take Hurwitz theorem, we can find m,n such that |m/n-Pi/2|<1/n^2
hence |m-pi/2*n|<1/n.

hence sin(m)=sin(Pi/2*n-1/n)~1-1/2/n^2

so that sin(m)^m~(1-1/2/n^2)^(pi/2*n) which makes a subsequence that tends to 1

but then sin(m)^(m^2) tends to exp(-Pi^2/8) but this is not 0 as we could expect ?

Following this result only sin(m)^(m^3) tends towards 0, accepting that Hurwitz theorem gives the best approximation for pi/2*n.

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$sin^2n=1-cos^2 n = 1-\frac{1}{1+tan^2 n}=\frac{tan^2 n}{1 + tan^2 n}=\frac{1}{1+\frac{1}{λ}}$
Where $λ=tan (n)$
And $$\lim_{n\to\infty}sin(n)^n=\lim_{n\to\infty}(\frac{1}{1+\frac{1}{λ}})^{n/2}$$