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?

29. Juli 2009 von mga010
Kategorien: Uncategorized | 2 Kommentare

Kommentare (2)

  1. 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. 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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