show that $|e_{n+1}|\le M |e_n||e{n-1}| $ for some constant $M$

Question

Let $f\in C^2(a, b)$ assume that $|f'(x)|\ge \delta > 0$ for all $x \in [a, b], f(p) = 0$ and that the secant method defines a sequence $\{p_n\}$ converging to p.

One can show that the absolute error satisfy: $|e_{n+1}| = |\frac{f''(\alpha_1)}{f'(\alpha_2)}e_ne_{n-1}|$ for every $n\ge 1$ and for some $\alpha_1,\alpha_2 \in [a, b]$.

Using this fact, show that $|e_{n+1}| \le M|e_n||e_{n-1}|$ for some constant $M$.

Answer 1

Hint: you almost proved it $\left| \frac{f''(a_1)}{f'(a_2)}e_n e_{n-1} \right|=\left|\frac{f''(a_1)}{f'(a_2)}\right|\left|e_n\right|\left|e_{n-1} \right|\leq \frac{\left|f''(a_1)\right|}{\delta}\left|e_n\right|\left|e_{n-1} \right| $ and use Extreme value theorem for the 2nd derivative.