Convergence of secant method:

Definition: Say, $x_{n}=\xi + e_{n},\qquad x_{n+1}=\xi + e_{n+1}$ where $\xi$ is the root of $f(x)=0$. $e_{n}$, $e_{n+1}$ are the errors at n$^{th}$ and (n+1)$^{th}$ iterations and $x_n,\, x_{n+1}$ are the approximations of $\xi$ at $n^{th}$, (n+1)$^{th}$, iterations. If $e_{n+1}=k\, e^{p}_{n}$ where $k$ is a constant, then the rate of convergence of the method by which $\{x_{n}\}$ is generated is p.
Claim: Secant method has super linear convergence.
Proof: The iteration scheme for the secant method is given by

$$x_{n+1}=\frac{f(x_{n})x_{n-1}-f(x_{n-1})x_{n}}{(f(x_{n})-f(x_{n-1}))}\qquad\qquad(i)$$

$$=x_{n}-\frac{f(x_{n})(x_{n}-x_{n-1})}{f(x_{n})-f(x_{n-1})}\quad\,\,\qquad\qquad(ii)$$

Say $f(\xi)=0$ and $e_{n}=(\xi-x_{n})$ i.e the error in the n$^{th}$ iteration in estimating $\xi$.

$$\left. \begin{aligned} x_{n+1}=e_{n+1}+\xi \\ x_{n}=e_{n}+\xi \\ x_{n-1}=e_{n-1}+\xi \end{aligned} \qquad\qquad\qquad\qquad\right \}(iii)$$

Using (iii) in (ii) we get

$$e_{n+1}=\frac{e_{n-1}f(x_{n})-f(x_{n-1})e_{n}}{f(x_{n})-f(x_{n-1})}\qquad(iv)$$

By Mean value Theorem, $\exists \quad a\quad\eta_{n}$ in the interval $x_{n}$ and $\xi$ s.t.

$$f'(\eta_{n})=\frac{f(x_{n})-f({\xi})}{x_{n}-\xi}$$

$$\because \qquad f(\xi)=0,\quad x_{n}-\xi=e_{n}$$

We get

$$f'(\eta_{n})=\frac{f(x_{n})}{e_{n}}$$    i.e. $$\quad f(x_{n})=e_{n}f'(\eta_{n})\qquad \qquad(v)$$

Using (iii)above, we get

$$f(x_{n-1})=e_{n-1}\, f'(\eta_{n-1})\qquad\qquad \quad\qquad(vi)$$

$\therefore$                 using (v), (vi), in (iv) we get

$$e_{n+1}=e_n\, e_{n-1}\frac{f'(\xi_n)-f'(\xi_{n-1})}{f(x_n)-f(x_{n-1})}$$

$$i.e. \qquad\qquad e_{n+1} \propto e_n \,\, e_{n-1} \qquad\qquad (vii)$$

By def of rate of convergence the method is of order p if

$$\left. \begin{aligned} \displaystyle {e_n \propto e^p_{n-1} }\\ ... ...e {e_{n+1} \propto e^p_n} \end{aligned} \qquad\qquad\qquad\qquad\right \}(viii)$$

>From (vii) and (viii) we get

$$e^p_n \,\,\propto \,\, e_n \,\,\,e_{n-1}$$

i.e.

$$e^p_n \,\,\propto \,\, e^p_{n-1}\,e_{n-1}$$

i.e.

$$\qquad\qquad \qquad\qquad e_n \,\, \propto e_{n-1} \qquad\qquad \qquad \qquad(ix)$$

From (viii), (ix) we get

$$p=(p+1)/p$$

i.e.

$$\quad p^{2}-p-1=0$$

$$\Rightarrow p=\frac{1\pm\sqrt{5}}{2}$$

$$\because \, p \, > \, 0,\quad p=1.618$$

$$\therefore \qquad e_{n+1}\,\,\,\propto\,\,\, e_{n}^{1.618}$$

Hence the convergence is superlinear. Example:
Solve $5 \sin^{2}x-8\cos^{5}x=0$ for the root in the interval [0.5,1.5] by secant method.

		Secant Method
0	0.5000000000	1.5000000000	0.8773435354	2.1035263538
1	1.5000000000	0.8773435354	0.4212051630	-4.2280626297
2	0.8773435354	0.4212051630	0.7258019447	0.3298732340
3	0.4212051630	0.7258019447	0.7037572265	0.0327354670
4	0.7258019447	0.7037572265	0.7013285756	-0.0005388701
5	0.7037572265	0.7013285756	0.7013679147	0.0000009526