# If the initial points for secant iteration method are sufficiently close to the root, the iteration converges to the root

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 $$e_{n+1} when $$e_n =x_n-a$$ . One may choose $$x_0,x_1$$ such that $$e_1, now let us assume that $$e_n 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 $$(0,1,1,2,3,5,8,...)$$.