Check out some similar questions!

Nivg · Jul 21, 2009, 06:38 AM

Hi all,

I'm trying to figure how to do that one, the only thing that cross my mind is something with sandwiched, any hint, thought or help about this one will be most grateful.

galactus · Jul 21, 2009, 03:23 PM

Is this an infinite product or nested sines? I just want to be sure what you mean.

Is it:

$\lim_{n\to\infty}\frac{sin(x)\cdot sin(x)\cdot \cdot\cdot sin(x)}{n}$

Or are they nested like so: $sin(sin(sin(sin......sin(x)))))$

I am confused by the 'n times'. Does the 'n times' represent how many times sin occurs.

Regardless, since n is the limiting factor and $n\to \infty$ , it would appear the limit approaches 0 as $n\to \infty$

Is this the exact way it is presented to you?

Just making sure.

Nivg · Jul 22, 2009, 06:53 AM

they nested sins (n represent how many times sin occurs), and I know the limit = 0, as it's easy to check this one with computer, I just don't know how to prove it.

galactus · Jul 22, 2009, 07:05 AM

It is just a matter of observation. X is not the limiting factor. As n gets larger and larger, it goes to 0 because of the n in the denominator.

It essentially the same as proving

$\lim_{n\to \infty}\frac{x}{n}$ and x is a constant. Say x=1.

$\lim_{n\to \infty}\frac{1}{n}=0$

Nivg · Jul 22, 2009, 07:17 AM

Maybe I mislead you or misunderstood you there is no "n" in the denominator, there is no denominator, it's :

lim (sin(sin(sin... sin(x)))), as n->inf

n it's the number of times you get this sin inside sin thing.

Unknown008 · Jul 22, 2009, 10:12 AM

Well, you post seemed to me too that there was a denominator 'n' in your expression... *looks up again* yes, there is a denominator.

Nivg · Jul 22, 2009, 10:45 AM

I will use another font next time, again there is no denominator, sorry for misleading you people, now when we cleared this up what with some help here :P.

* lim (sin(sin(sin... sin(x)))), as n->inf

n it's the number of times you get this sin inside sin thing.

galactus · Jul 22, 2009, 03:36 PM

This is not very rigorous, but it may suffice. By letting

$L=sin(sin(sin(sin(......sin(x)))$

We then note that:

$sin(\underbrace{sin(sin(sin(......sin(x)))}_{ \text{this is L}})$

Then we have $sin(L)=L$

The value that satisfies this is L=0

Nivg · Jul 22, 2009, 04:02 PM

you say the series [x,sinx,sin(sinx),. ] is continues and monotone decreasing in [0,pi] or monotone increasing in [-pi,0] so limXn = limXn+1=L, meaning sinL=L and that only works when L=0, right?
that what you are saying?

galactus · Jul 24, 2009, 12:24 PM

Yes. It is monotone decreasing and bounded below. You can show the limit exists by proving this if you wish.

You may want to prove the limit sin(L)=L rigorously.

Nivg · Jul 24, 2009, 11:30 PM

OK thanks for all the help I got my answer :P