The iterated Sine

I am a bit surprised that none of my readers like to solve problems. OK, then here is a proof. I adopt the notation $$sin_k$$ for the iterated sine.

It is a simple problem to show that $$\sin_k(pi/2)$$ goes to 0 decreasing. So there is a k such that

$$\sin_{k+1}(\pi/2) \le x \le \sin_k(\pi/2)$$

Since all

$$\sin_n$$

are monotone by induction, we get

$$\dfrac{\sin_{k+n+1}(\pi/2)}{\sin_n(\pi/2)} \le \dfrac{\sin_{n}(x)}{\sin_n(\pi/2)} \le \dfrac{\sin_{k+n}(\pi/2)}{\sin_n(\pi/2)}$$

Now, all we need to show is for all k

$$\lim_{t to 0} \dfrac{\sin_k(t)}{t} = 1$$

But this is the same as

$$\sin’_k(0)=1$$,

which easily follows by induction on k.

The same can be done with any other monone  function of the form

$$0 < f(x) < x, \quad f'(0)=1$$

For the iterated function, we always get

$$\lim_{n \to \infty} \dfrac{f_k(x)}{f_k(y)} = 1$$

for $$0 < x < y$$. An example of such a function, which is easy to compute is

$$f(x) = x-x^2$$

Since you are so fond of my problems, here is another one, I cannot solve yet myself. Numerical experiments show that

$$\lim_{n \to \infty} \sqrt{n} \sin_n(\pi/2)$$

exists. Can you prove that? And what is that limit?

2 Gedanken zu „The iterated Sine“

1. watchmath

You said that there is k such that

$latex sin_{k+1}(pi/2)leq xleq sin_k(pi/2) &s=-1$.

How if $latex x=2 &s=-1$?

2. mga010 Beitragsautor

That’s right! I overlooked this. You have to define

$sin_0(x) = dfrac(pi}{2} &s=-1$

to make the proof work.

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