Modified Regular Falsi method:

In this method an improvement over Regular Falsi method is obtained by replacing the secant by straight lines of even-smaller slope until $w$ falls to the otherside of the zero of $f(x)$. The various steps in the method are given in the algorithm below:

Algorithm:
Given a function $f(x)$ continuous on an interval $[a,b]$ satisfying the criteria $f(a)f(b)<0$, carry out the following steps to find the root of $\xi$ of $f(x)$ in $[a,b]$:
(1) Set $a_{0}=a\quad;\quad b_{0}=b\quad;\quad F=f(a_{0})\quad;\quad G=f(b_{0}),\quad w_{0}=a_{0}$
(2) For n=0,1,2...., until convergence criteria is satisfied, do:
(a) compute $w_{n+1}=[Ga_{n}-Fb_{n}]/(G-F)$
(b) If $(f(a_{n})f(w_{n+1})\leq 0)$     then
Set $a_{n+1}=a_{n}\quad;\quad b_{n+1}=w_{n+1}\quad;\quad G=f(w_{n+1})$
Also if $(f(w_{n})f(w_{n+1})>0)$     Set     $F=F/2$
Otherwise
Set $a_{n+1}=w_{n+1},\quad F=f(w_{n+1})\quad b_{n+1}=b_{n}$
Also if $(f(w_{n})\,\, f(w_{n+1})>0)$     Set    $G=g/2$

Example:
Solve $2x^{3}-2.5x-5=0$ for the root in the interval [1,2] by modified Regula Falsi method.

Solution: Since $f(1)\,\, f(2)=-33<0,$ we go ahead with finding the root of given function f(x) in [1,2]. Setting $a_{0}=1,\quad b_{0}=2$ and following the above algorithm. Result are provided in the table below:

		Modified Regula Falsi Method
Iteration no.	$a_n$	$b_n$	$W_n$	$f(W_n)$
0	1.0000000000	2.0000000000	1.4782608747	-2.2348976135
1	1.4782608747	2.0000000000	1.7010031939	0.5908976793
2	1.4782608747	1.7010031939	1.6544258595	-0.0793241411
3	1.6544258595	1.7010031939	1.6599385738	-0.0022699926
4	1.6599385738	1.7010031939	1.6602516174	0.0021237291
5	1.6599385738	1.6602516174	1.6601003408	0.0000002435

The geometric view of the example is provided in the figure below:



Example:
Solve $5\sin x^{2}-8\cos^{5}x=0$ for the root in the interval [0.5,1.5] by modified Regula Falsi method.

		Modified Regula Falsi Method
Iteration no.	$a_n$	$b_n$	$W_n$	$f(W_n)$
0	0.5000000000	1.5000000000	0.8773435354	2.1035263538
1	0.5000000000	0.8773435354	0.7222673893	0.2828366458
2	0.5000000000	0.7222673893	0.6871531010	-0.1967970580
3	0.6871531010	0.7222673893	0.7015607357	0.0026464546
4	0.6871531010	0.7015607357	0.7013695836	0.0000239155
5	0.6871531010	0.7013695836	0.7013661265	-0.0000235377
6	0.7013661265	0.7013695836	0.7013678551	-0.0000003363

Secant Method: Like the Regula Falsi method and the Bisection method this method also requires two initial estimates $x_{-1},\quad x_{0}$ of the root of f(x)=0 but unlike those earlier methods it gives up the demand of bracketing the root. Like in the Regular Falsi method, this method too retains the use of secants throughout while tracking the root of f(x)=0. The secant joining the points $(x_{-1}, f(_{x-1})),\quad(x_{0}, f(x_{0}))$ is given by

$$y=\frac{f(x_{0})-f(x_{-1})}{(x_{0}-x_{-1})}x+\frac{f(x_{-1})x_{0}-f(x_{0})x_{-1}}{(x_{0}-x_{-1})}$$

Say it intersects with x-axis at $x=x_{1}$, then

$$x_{1}=\frac {f(x_{0})x_{-1}-f(x_{-1})x_{0}}{f(x_{0})-f(x_{-1})}$$

If $\vert f(x_{2})\vert>\epsilon=10^{-6}$ (say) then replace $(x_{-1}, f(x_{-1})), (x_{0}, f({x_{0}}))$ with $(x_{0}, f(x_{0})), (x_{1}, f(x_{1}))$ and repeat the process to get $x_{1}$ and so on . The method is algorithmically described below:

Algorithm:
Given a $f(x)$, two initial points a, b and $\varepsilon$ the required level of accuracy carry out the following steps to find the root $\xi$ of f(x)=0.
(1) Set $x_{-1}=a,\quad x_{0}=b$
(2) For n=0,1,2... until convergence criteria is satisfied, do:
Compute

$$x_{n+1}=\frac{[f(x_{n})x_{n-1}-f(x_{n-1})x_{n}]}{[f(x_{n})-f(x_{n-1})]}$$

Example:
Solve $2x^{3}-2.5x-5=0$ for the root with $a=1, b=2$ by secant method to an accuracy of $10^{-6}$.

Solution:
Set $x_{-1}=a=1\quad ; \quad x_{0}=b=2$
$${f(x_{-1})=f(1)=-5.5}$$
$${f(x_{0})=f(2)=6.0}$$

$${x_{1} = \frac{f(x_{0})x_{-1}-f(x_{-1})x_{0}}{f(x_{0})-f(x_{-1})}}$$
$${\indent = \frac{f(2).1-f(1).2}{f(2)-f(1)})}$$
$${ \indent = \frac{6.1-(-5.5).2}{6-(-5.5)}}$$
$\indent=1.4782608747$

$${\left\vert f(x_{1})\right\vert=\left\vert-2.2348976135\right\vert > 10^{-6}}$$
$\therefore$ Repeat the process with $(x_{0}, f(x_{0})),\quad(x_{1}, f(x_{1}))$ and so on till you get a $\xi=x_{i}$ s.t. $\vert f(\xi)\vert<\epsilon=10^{-6}.$ These results are tabulated below:

		Secant Method
Iteration no.	$x_{n-1}$	$x_n$	$x_{n+1}$	$f(x_{n+1})$
0	1.0000000000	2.0000000000	1.4782608747	-2.2348976135
1	2.0000000000	1.4782608747	1.6198574305	-0.5488323569
2	1.4782608747	1.6198574305	1.6659486294	0.0824255496
3	1.6198574305	1.6659486294	1.6599303484	-0.0023854144
4	1.6659486294	1.6599303484	1.6600996256	-0.0000097955

Geometrical Visualization of the root tracking procedure by secant method for the above example: