D: HALP GAIZ!

(Sorry if I'm not supposed to post homework here or something.)

My Analysis prof gives hard problems as extra-credit sometimes, and he gave this one during our first class of the semester yesterday.  I can't get anywhere with it. :(  It looks like it should be totally easy, but I'm not seeing it.  I just have to prove that the sequence {sin(1), sin(4), sin(9), ...} doesn't converge. The numbers 1,4,9... are the integer squares.

Anyone have any ideas?

Make use of the fact that sin(x^2) is periodic.

>>2
no it isn't

brute force it

So \frac{\pi}{2} - \left| \frac{\pi}{2} - n^2 \right| would have to approach zero as n went to infinity for it to converge, no? (Necessary, not sufficient.)

>>4
You can't brute force a convergence test, idiot.

>>5
Well n^2 gets large, so what you wrote would go to minus infinity, so I think maybe you meant something different.


Say the sequence converged to L.  Then L is between -1 and 1, and there are in general two numbers a and b between 0 and 2pi that give sin(x) = L.  So convergence would mean that for large n, the numbers n^2, (n+1)^2,...  are all very close to a or b plus some integer multiple of 2pi.

It seems pretty clear that's impossible, but I can't see how to prove it (yet).

Basically you want to show that the series n^2 is kind of independent of the series n*pi.

You want to find a infinite sequence of sqaures that are "close" to (2n+1/2)*pi and an infinite sequence that are "close" to (2n - 1/2)*pi.

By close enough I mean sufficiently close that sin of them is greater than 1/2 or less than -1/2.



I'd guess maybe you want to do it by using rational approximations to pi, and multiplying through by the denominator and numerator to find a multiple of pi "near" a square number.

Although, limitations to this, are I think the error of an approximation p/q is at best O(1/q^2) [this is a vague recollection], so you don't really get the multiple of pi very near the square number.




It's probably in fact easier to consider it in the abstract case, prove for an irrational number x and for e >0 that the sequence n*x comes with e of m^2 infinitely often, a statement which is invariably true.

Then tweak it a little so you can use (2n + 1/2)*x and (2n - 1/2)*x, then apply it to pi to get your result.


But I aint doing it for you

>>7
>But I aint doing it for you

Translation: You can't figure out how to make it work. :P

It's true that for any real \alpha there are arbitrarily large integers q such that for some integer p,  \left|q\alpha-p\right| < 1/q, so the sequence q\alpha comes within \epsilon of an integer infinitely many times.  The only proofs of this I know of, though, use continued fractions or the box principle, and I doubt they can be adapted to show that q\alpha comes close to square integers infinitely many times.  If you know how, I'd be interested to see it.

>>6
I think maybe you meant something different.
Right. "mod π" fell off somewhere between my brain and my fingers. Basically what has now been said.

>>8

It actually meant I haven't got the time to waste on it. In terms of the specific case of pi, I seem to remember there's some rather nice general continued fractions expansions of pi that follow a highly regular pattern involving square numbers.

Whilst i didn't think continued fractions would be the way to go either, it's always possible.

>>10
I think these are what you're talking about:
http://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80#Three_continued_fractions
The convergents still don't have square numerators, though.

Write

x \equiv y \pm \delta (\text{mod}\;\pi)

to mean that there is some number in [x-(y+\delta),x-(y-\delta)] which is an integral multiple of \pi.  Let L be a hypothetical limit of sin(n^2) and let c,c' be the two solutions in [0,2\pi) of sin(x)=L.  Note that c' \equiv -c (\text{mod}\;\pi).

With this notation, if the limit exists, then for any \delta > 0, there is an N such that for all n \ge N we have

n^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+1)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+2)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+3)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+4)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+5)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
...

(Note this actually includes the solutions of sin(x) = -L, but that doesn't matter).  Subtract the first of these from the second, the second from the third, and so on to get

2n+1 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+3 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+5 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+7 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+9 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
...

for some choice of the signs.  Now subtract the first of these from the rest to get:

0 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
2 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
4 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
6 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
8 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
10 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
...

Ignoring the error term, there are only 5 possibilities for the RHS of this last set of equations: \{-4c, -2c, 0, 2c, 4c\}.  Therefore two of these equations have the same RHS.  Of these two, subtract the one with the smaller LHS from the other to get an equation like

a \equiv \pm 8\delta (\text{mod}\;\pi)

for an even integer 2 \le a \le 10.  Contradiction for small enough \delta.

>>12

OMG TYTYTYTYTYTY! :DDD