Lyapunov exponent vs. parameter I don't understand the graph. If x[0]=1/2, then f'(x[0])=0, and ln(|f'(x[0])|)=ln(0)=- infinity. So for x[0]=1/2 Lyapunov exponent should be - infinity for all values of p. Reply
I don't understand the graph. If x[0]=1/2, then f'(x[0])=0, and ln(|f'(x[0])|)=ln(0)=- infinity. So for x[0]=1/2 Lyapunov exponent should be - infinity for all values of p.