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.

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## 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.