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Lim sup sin(n)

  • Thread starter mmmboh
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  • #1
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I figured out limn->∞ sup sin(n)=1, but I can only prove it using Diophantine approximation, and this question is for analysis. My teacher said we can do this, but it was obvious he prefer us do it another way that doesn't involve theorems we haven't learned, but I can't figure out another way to do this.

I was thinking of somehow showing that for any e>0, there are infinitely many terms of sin(n) such that sin(n)>1-e, but I am not even sure this is possible without other theorems not learned in analysis. I've also tried bounding it between things that converge to 1 at some point, but that was unsuccessful.

Can someone help?
Thanks.
 

Answers and Replies

  • #2
sin(n) is bounded thus, any subsequencial limit must in the interval [-1, 1], why ? Prove it or use it if you have already proved it in class.

It is easy to construct s subsequence of sin(n) with a limit of 1.
 
  • #3
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I need nk->π/2(4m+1) for some m, but I'm not seeing what the sequence is :confused:. n is an integer by the way, if it wasn't clear.
 
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  • #4
I need nk->π/2(4m+1) for some m, but I'm not seeing what the sequence is :confused:. n is an integer by the way, if it wasn't clear.
I don't see the problem.

The sequence you need is what you just stated :confused:.

The sequence is simply [tex] sin \frac{ \pi(4n+1)}{2}[/tex] where [tex] n_k = 4n+1[/tex].

Thus, [tex] sin(n_k) \rightarrow 1[/tex].
 
  • #5
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But [tex] \frac{ \pi(4n+1)}{2}[/tex] isn't an integer, so I can't use that as a sequence..the argument of sine I use must be an integer.
 
  • #6
But [tex] \frac{ \pi(4n+1)}{2}[/tex] isn't an integer, so I can't use that as a sequence..
:facepalm I am sorry, I replied in a hurry.

We will have to use other theorems for this.


Do you know the proof of bolzano weierstrass ? The one about peaks ?

You will have to modify it to create a monotone increasing subsequence (not explicitly though) and then use the monotone convergence theorem to conclude that the bounded subsequence converges to the suprema.

However, this is almost a circular argument since you will have to prove that 1 is the suprema of sin(n).

Do you have any ideas, as to, how to prove 1 is the suprema of sin(n) ?
 
  • #7
Dick
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I'm kind of guessing mmmboh already knows one proof. Since pi is irrational it's pretty easy to show that you can find integers k such that pi/2+k*pi is arbitrarily close to being an integer. If that's the Diophantine approximation thing referred to. I'm not sure I know any more 'analysis' type proof.
 
  • #8
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How can you show pi/2+k*pi can be made arbitrarily close to an integer? I know the typical epsilon arguments for showing things like this, but I don't seen how you would apply it to this situation.
 
  • #9
I just found a proof online.


The idea was to show that you can construct a subsequence such that [tex]|sin(n_k) -1 | < \frac{1}{k+1}[/tex].

Which is possible to do with the archimedean property.

The online proof was a bit different with some geometry( lol I suck at geometry).
 
  • #10
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Can you give me a hint on how to construct the sequence?
 
  • #11
LeonhardEuler
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How can you show pi/2+k*pi can be made arbitrarily close to an integer? I know the typical epsilon arguments for showing things like this, but I don't seen how you would apply it to this situation.
The sin([itex]\theta[/itex]) is the y-component of the position on a unit circle when the angle is [itex]\theta[/itex]. If we start out at some point on the circle and repeatedly add the number 'a_1' and move that arc length counterclockwise along the circle (increase the angle by 'a_1' each step) we will cross the top of the circle at some point. Suppose we just crossed the top of the circle and call that step 1. Right before that we were at step 0.

We proceed more and more steps until after N steps we cross our position at step 0. Can we be at the starting point of step 1 if 'a_1' is rational?

We are in a sector of circle with arc length a_1. Take the minimum of the distance between our current point and one of the end points. It can be at most a_1/2. Call this a_2. It is the distance moved along the circle in either N or N-1 steps. Now start making compound moves of N or N-1 moves until you cross the circle again. Do you see where this is going?
 
  • #12
Dick
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How can you show pi/2+k*pi can be made arbitrarily close to an integer? I know the typical epsilon arguments for showing things like this, but I don't seen how you would apply it to this situation.
Do you know that if r is irrational, you can find integers k such that k*r is arbitrarily close to an integer? Split the unit interval into subintervals of size 1/N. Now pick N+1 different values of k and take k*r mod 1. One of the intervals must contain two different kr values. It's the pigeonhole principle. Your problem is basically the same thing.
 
  • #13
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Dick: Edit: yes I do.

LeonardEuler: I believe it gets arbitrarily close to pi/2, although I don't think I totally get it, but I'm reading it over. Is there a way to do this without mentioning geometry? Maybe how ╔(σ_σ)╝ was thinking?

Edit: I wonder if it would be easier to use the density of rationals somehow.
 
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  • #14
Dick: Edit: yes I do.

LeonardEuler: I believe it gets arbitrarily close to pi/2, although I don't think I totally get it, but I'm reading it over. Is there a way to do this without mentioning geometry? Maybe how ╔(σ_σ)╝ was thinking?

Edit: I wonder if it would be easier to use the density of rationals somehow.
You need some sort of geometry to show that such a k exist.

Sure with the archimedean you can pick a k such that 1/k+1 is abitrarily close to zero but it has to be greater than sin(n_k)-1; remember n_k > k.

Proving that such a k and n_k exist without a geometric argument is daunting.

I think your only recourse is geometry.
 
  • #15
LeonhardEuler
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Dick: Edit: yes I do.

LeonardEuler: I believe it gets arbitrarily close to pi/2, although I don't think I totally get it, but I'm reading it over. Is there a way to do this without mentioning geometry? Maybe how ╔(σ_σ)╝ was thinking?

Edit: I wonder if it would be easier to use the density of rationals somehow.
The general idea is that each time you define a new composite move it decreases the distance you move by at least half. This means the closest approach on each cycle is bounded by a number that decreases to 0, since the closest approach is at most a distance of the step size from pi/2.

As far as not using geometry, yes you can avoid it. You could set up an interval from 0 to 2pi. What you care about is in some sense "the remainder when k is divided by 2pi". (although I've never seen the word remainder used in the context of an irrational denominator, but you get what I mean) You can mirror the previous argument thinking of it this way, but I think it makes everything wordier and more awkward.
 
  • #16
Dick
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Dick: Edit: yes I do.
If you know that, I think you should use it. It's what you need. LeonhardEuler's argument is basically the same thing.
 
  • #17
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Let J be the open interval
(pi/2, delta+pi/2) where delta > 0, and consider the intervals
J + n + 2 pi m = (pi/2+ n + 2 pi m, delta + pi/2+ n + 2 pi m) where n and m are
integers. There are infinitely many of them in [-2 pi, 2 pi], and
each has length delta, so some of them must overlap: say
pi/2+n + 2 pi m < pi/2+n' + 2 pi m' < delta +pi/2+ n + 2 pi m.
But then
n - n' + 2 pi (m - m') < 0 < delta + n - n' + 2 pi (m - m').
If y = n' - n + 2 pi (m' - m), we have 0 < y < delta, and sin y = sin (n'-n)

The problem is I need pi/2< y < delta+pi/2, then I can take delta arbitrarily close to 0, but I can't figure out how to get that.
 
  • #18
LeonhardEuler
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The problem with this argument is that you are basically showing that there are values of n+2pi*m that are arbitrarily close to one another, when you want to show that there are values of n-2pi*m that are arbitrarily close to pi/2. The J in your argument just hangs around for a while and then just walks away when you start looking at differences in values.

I actually don't see a way to rescue this argument. The only solutions I see are variations on the one I mentioned, though there may well be better ones.

The basic idea of the argument I was making before was that you can think of what you are doing as you increase n is going around in a circle. By taking combinations of moves, it is possible to come up with arbitrarily small moves different from 0. Because the moves are so small, they must pass within any given distance of pi/2 (or any other value). I'm not sure how clear my last post was, so let me know if there's anything you want me to clear up.

(The way of avoiding geometry is basically by imagining moving along the interval [0, 2pi] and jumping back to 0 when you cross 2pi, or vice-versa for the negative moves that may appear when a compound move is slightly less than 2pi, instead of a circle. It's just slightly less simple in my opinion, but applies to all irrational numbers)
 
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  • #19
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Yeah, I tried it this different way because I had a bit of trouble following your last post, and couldn't get very far with it. :confused: But it does sound good if I could figure it out.
 
  • #20
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In all likelihood, we are classmates, mmmboh. :P

I haven't had any success solving this either, but I understand LeonhardEuler's approach and it's essentially what I've been trying to formalize all day.
 
  • #21
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All day? I finished the rest of this assignment a week ago, I've been stuck on this one all week haha.
 
  • #22
LeonhardEuler
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Maybe it will be clearer if I show the first iteration. It's simplified a little from how I first described it, I will use 0 degrees as the starting point instead of starting close to the top of the circle:
Start at 0. We move a distance of 1 every move, so at some point on this first cycle we will be within 0.5 of pi/2 (about 0.43 at n=2 is the closest)

Now we circle back around past the 0 at n=7. The closest we get to 2pi*n other than 0 is 6. Moving forward by 6 radians is the same as moving back by about 0.28. We knew this would be at most 1/2 before starting.

Now let's start from 7 and move along the circle in multiples of 6 moves. Each compound move puts us back by only about 0.28, so if we do a complete circle, some point will be within 0.14 of pi/2 and the next crossing of the circle will be within 0.14 of 0.

Once we cross the circle again, we will know that some number N_2 of steps moves us by at most 0.14 in some direction. We then cross the circle again in multiples of N_2 and achieve an even finer mesh, etc.
 

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