Calculating the Lyapunov exponent of the times-m map, $E_{m}$.

Question

I'm trying to compute the Lyapunov exponents for $E_{m}$, where $E_{m}:S^{1}\to S^{1}$, $x\mapsto mx\mod 1$.

The Lyapunov exponent is given by $$\chi(x,v)=\overline{\lim_{n\to\infty}}\frac{1}{n}\log\|df^{n}(x)v\|$$

I take $f^{n}(x)=E^{n}_{m}x=(mx\mod 1)\underbrace{\circ...\circ}_{n\text{ times}}(mx\mod 1)$

Then $$\chi(x,v)=\overline{\lim_{n\to\infty}}\frac{1}{n}\log\|dE^{n}_{m}xv\|$$

$E_{m}$, $m>1$, has positive exponents at all points. So there is a sequence $n_{j}\to\infty$ such that for all $\eta>0$, $$\|dE^{n}_{m}xv\|\ge e^{(\chi-\eta)n_{j}}\|v\|$$

To be honest I don't really know where I am going with this. I have never calculated Lyapunov exponents before and I could not find any similar examples.

Answer 1

What stops you from using the definition directly? The derivative of $E_m$ is everywhere equal to $m$. It is enough.