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Context:


We know that $\cos(x)$ if taken recursively on itself, converges to the Dottie number, which is the function's stable fixed point then.

On the other hand, for a function like $f(x)=3x$, keeps diverging if taken on itself.

On the other hand, we know that calculating Lyapunov's exponent $\lambda$ helps to find out about such behavior of functions.

Meaning of exponent's sign:

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enter image description here

$\lambda$ for 2nd picture should be positive, mistake in picture...


Question:

  • Can one analytically calculate and show that for example $\lambda$ for $\cos(x)$ is negative, and $\lambda$ is positive for $3x$ , as expected?
  • Source of image: http://hypertextbook.com/chaos/43.shtml –  Oct 21 '14 at 18:00
  • Are you aware that simply the definition implies that the Lyapunov exponent for the $3x$ function is $\log3\gt0$? And a similarly direct computation is available for the cosine case... – Did Oct 21 '14 at 18:46
  • @Phonon ?? Anything not zero. – Did Oct 21 '14 at 20:28
  • @did haha right ok :) – Ellie Oct 21 '14 at 20:38
  • @Did thanks for the comment, but I don't see how you get it... for either case using formula on wiki, I'm stuck at these limits (putting $\partial Z_0=1$): $\lim_{t \to \infty} \frac{\ln(-sint dt)}{t}$ and for the other one $\lim_{t \to \infty} \frac{\ln(3 dt)}{t}$ ... how to proceed? thanks for your help. –  Oct 22 '14 at 17:11

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for either case using formula on wiki, I'm stuck at these limits (putting $\partial Z_0=1$): $\lim_{t \to \infty} \frac{\ln(-sint dt)}{t}$ and for the other one $\lim_{t \to \infty} \frac{\ln(3 dt)}{t}$

Sorry? What is, in the second case, say, $3dt$ when $t\to\infty$? And why put $\partial Z_0=1$, as you say, when the limits are considered for $\delta Z_0\to0$?

What does the WP page you link to, actually say? One considers a dynamical system, that is, in continuous time $Z'(t)=f(Z(t))$ or in discrete time $Z_{n+1}=g(Z_n)$, and one is interested in

the speed at which trajectories starting from two different points diverge one from the another when the time is large.

To make this precise in the discrete setting, one considers two starting points $x$ and $y$ and the systems $Z^x_{n+1}=g(Z^x_n)$, $Z^x_0=x$, and $Z^y_{n+1}=g(Z^y_n)$, $Z^y_0=y$. If $Z^x_n-Z^y_n$ behaves roughly as $\mathrm e^{n\lambda}$, then one calls $\lambda$ the Lyapunov exponent.

Of course, more care is needed to define this rigorously: one actually fixes some $x$ and one considers the limit $y\to x$, then, if $|Z^y_n-Z^x_n|/|y-x|\to k_n(x)$ when $y\to x$ and if $k_n(x)=\mathrm e^{n\lambda(x)+o(n)}$ when $n\to\infty$, then $\lambda(x)$ is the Lyapunov exponent at $x$. Thus, $$\lambda(x)=\lim_{n\to\infty}\frac1n\lim_{y\to x}\log\left(\frac{|Z^y_n-Z^x_n|}{|y-x|}\right).$$ This may seem abstruse but the idea is simple, so let us see what happens when $g:z\mapsto3z$ (your second case). Then $Z^x_n=3^nx$ for every $x$ and every $n$ hence $$\frac{|Z^y_n-Z^x_n|}{|y-x|}=3^n,$$ for every $y\ne x$, in particular $k_n(x)=3^n$ for every $n$ and indeed $\lambda(x)$ exists for every $x$ and $\lambda(x)=3$.

Likewise, if $g:z\mapsto\cos(z)$ and $x=\cos(x)$, that is, $x\approx0.739$ is the Dottie number, then, for every $y$, $Z^y_n=\cos^{\circ n}(y)$ hence $k_n(x)=|(\cos^{\circ n})'(x)|$. For every $z$, $$(\cos^{\circ n})'(z)=(-\sin z)(-\sin(\cos z))\cdots(-\sin(\cos^{\circ (n-1)} z)),$$ in particular $k_n(x)=(\sin x)^n$ hence $\lambda(x)=\log\sin x\approx-0.395$.

Did
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