Secant Method

Like Newton with a secant approximation to $f'(x)$. Newton's Method requires $f'(x)$
Replace $f'(x_n)\approx \displaystyle \frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}$
since $f'(x)=\displaystyle \lim_{n\to x}\displaystyle \frac{f(h)-f(x)}{h-x}$

Tangent replaced by secant.

$$% latex2html id marker 8360\therefore p_{n+1}=p_{n}-f(p_n) \left[\frac{p_n-p_{n-1}}{f(p_n)-f(p_{n-1})}\right]\,n\ge 1$$

So we need 2 points

Rate of Convergence:

$$e_{n+1}=\mathcal{C} e_n e_{n-1}\sim A\vert e_n\vert^{(1+\sqrt{5})/2}$$

Sine the golden mean $$\frac{(1+\sqrt{5})}{2}\approx 1.62 <2 \Rightarrow$$ superlinear, i.e. better than linear but not as good as quadratic.

$\Box$

The algorithm for the Secant method is depicted in Figure 17.

Figure 17: Secant Method Algorithm

Algorithm-Secant

Find $f(x)=0$ given $p_0,\, p_1$

input $p_0, p_1$ tolerance, $N_0$.
output $p$, or failure message

Step 1 Set $i=2$
$q_0=f(p_0)$
$q_1=f(p_1)$
Step 2 While $i\le N_0$ do 3 - 6:

Step 3 Set $p=p_1-q_1(p_1-p_0)\bigg/ (q_1-q_0)$     % compute $p_i$

Step 4 if $\vert p-p_1\vert<$ TOL then
output $p$     % done
Stop
Step 5 $i=i+1$
Step 6 Set $p_0=p_1$    % update $p_0, q_0, p_1, q_1$
$q_0=q_1$
$p_1=p$
$q_1=f(p)$

Step 7 Output (`Method failed after $N_0$ iterates).
STOP $\Box$