Well I wish to prove that in case I may choose $x_0,x_1$ close enough to the root $a$ of $f(x)$, then the secant method $x_{n+1} = x_n -\frac{x_n -x_{n-1}}{f(x_n)-f(x_{n-1})}f(x_n)$ converges to the root, mean $\lim_{n\rightarrow\infty} x_n-a=0$.

So - I tried to show that $e_{n+1}<e_n$ when $e_n =x_n-a$ . One may choose $x_0,x_1$ such that $e_1<e_0$, now let us assume that $e_n<e_{n-1}$ by induction.

$x_{n+1}-a = x_n-a - \frac{x_n - x_{n-1}}{f(x_n)-f(x_{n-1}) }f(x_n)$ using $e_n$ notation we get: $e_{n+1} = e_n- \frac{e_n-e_{n-1}}{f(x_n)-f(x_{n-1})}f(x_n) = \frac{e_{n-1}f(x_n)-e_{n}f(x_{n-1})}{f(x_n)-f(x_{n-1})}$

The last result seems odd, casue $e_{n-1} >e_n$ thus $\frac{e_{n-1}f(x_n)-e_{n}f(x_{n-1})}{f(x_n)-f(x_{n-1})} >\frac{e_{n}f(x_n)-e_{n}f(x_{n-1})}{f(x_n)-f(x_{n-1})} = e_n$ , so it leads to $e_{n+1} > e_n$ which is completely the opposite to what I wish to show.

Anyway I'll demonstrate the result I got using Taylor series:

$e_{n+1} = \frac{e_{n-1}f(x_n)-e_{n}f(x_{n-1})}{f(x_n)-f(x_{n-1})}$ $ = \frac{e_{n-1}[f'(a)e_n+(1/2)f''(a)e_n^2 +(1/6)f'''(d_1)e_n^3]-e_n[f'(a)e_{n-1}+(1/2)f''(a)e_{n-1}^2 + (1/6)f'''(d_2)e_{n-1}^3]}{f'(a)e_n+(1/2)f''(c_1)e_n^2-f'(a)e_{n-1}+(1/2)f''(c_2)e_{n-1}^2}$ $ = e_ne_{n-1} \frac{(1/2)f''(a)(e_n-e_{n-1}) +(1/6)[f'''(d_1)e_{n-1}^2-f'''(d_2)e_n^2]}{f'(a)(e_n-e_{n-1})+(1/2)[f''(c_1)e_n^2-f''(c_2)e_{n-1}^2]} $

If I could show that $\lim_{n\rightarrow \infty} \frac{e_{n+1}}{e_n} <1$ then by the ration test $lim_{n\rightarrow \infty} e_n = 0$ as needed. Yet I didn't succeed to show it. (The previous comment, before Taylor usage makes me think it may even be wrong)


Write $$ e_{n+1}=e_ne_{n-1}\,\frac{\frac{f(x_n)}{x_n-a}-\frac{f(x_{n-1})}{x_{n-1}-a}}{f(x_n)-f(x_{n-1})} $$ and apply the extended mean value theorem to the last factor (note that $\frac{f(x)}{x-a}$ is a smooth function if $a$ is a regular root), $$ e_{n+1}=e_ne_{n-1}\,\frac{\frac{f'(c)}{c-a}-\frac{f(c)}{(c-a)^2}}{f'(c)} $$ Now insert the Taylor expansion at $a$ to get $$ e_{n+1}=e_ne_{n-1}\,\frac{-\frac12f''(a)-\frac16f'''(a)(c-a)-...}{f'(a)+\frac12f''(a)(c-a)+...} $$ So if you get that $|Me_{n-1}|\ll 1$ and $|Me_n|\ll 1$ with $M=-\frac{f''(a)}{2f'(a)}$, then you can also conclude that $|Me_{n+1}|\ll 1$.

For a more precise result you need to bound how fast the first and second derivatives of $f$ grow around $a$ to define a neighborhood where for instance $|f'(c)|>\frac12|f'(a)|$ and $|f''(c)|<2|f''(a)|$ so that $|Me_k|\ll1$ can be replaced by $|Me_k|<\frac14$. Then you get $|4Me_k|\le (4M\max(|e_0,e_1|)^{F_{k+1}}$ with $(F_k)$ the Fibonacci sequence $(0,1,1,2,3,5,8,...)$.


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