(b)False position or Regula Falsi method:

Bisection method converges slowly. Here while defining the new interval $[a_{n}, b_{n}]$ the only utilization of the function $'f(x)'$ is in checking whether $f(a_{n})f(b_{n})<0$ but not in actually calculating the end point of the interval. False position or Regular Falsi method uses $f(x)$ not only in deciding the new interval $[a_{n}, b_{n}]$ as in bisection method but also in calculating one of the end points of the new interval. Here one of end points of $[a_{n}, b_{n}]$ say $w$ is calculated as a weighted average defined on previous interval $[a_{n-1},\quad b_{n-1}]$ as:

$$w=\frac{\vert f(b_{n-1})\vert.a_{n-1}+\vert f(a_{n-1})\vert.b_{n-1}}{\vert f(b_{n-1})\vert+\vert f(a_{n-1})\vert}$$

$$=\frac{f(b_{n-1})a_{n-1}-f(a_{n-1})b_{n-1}}{f(b_{n-1})-f(a_{n-1})}$$

( $\because f(a_{n-1}),f(b_{n-1})$ have opposite signs).
The algorithm for computing the root of function $f(x)$ by thus method is given below.

Algorithm:
Given a $f(x)$ continuous on an interval $[a,b]$ satisfying the criteria $f(a)f(b)<0$, carry out the following steps to find the root $\xi$ of $f(x)$ in $[a,b]$:
(1) Set $a_{0}=a,\quad b_{0}=b$
(2) For n = 0,1,2.... until convergence criteria is satisfied do:
(a) Compute      $${w=\frac{[f(b_{n})a_{n}-f(a_{n})b_{n}]}{[{f(b_{n})-f(a_{n})}]}}$$
(b) If $(f(a_{n})f(w) \leq 0)$    Then

set $a_{n+1}=a_{n}\quad ;\quad b_{n+1}=w$
otherwise
set $a_{n+1}=w \quad;\quad b_{n+1}=b$
Note:
Use any one of the convergence criteria discussed earlier under bisection method. For the sake of carrying out a comparative we will stick both to the same convergence criteria as before i.e. $\vert f(e_{sn})\vert=\vert f(w)\vert<\epsilon=10^{-6}$(say) and to the example problems.

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

Solution: Since $f(1)\, f(2)=-33<0$, we go ahead in finding the root of given function f(x) in [1,2].
Set $a_{0}=1,\quad b_{0}=2$.

$$w=\frac{f(b_{0})a_{0}-f(a_{0})b_{0}}{f(b_{0})-f(a_{0})}=\frac{f(2).1-f(1).2}{f(2)-f(1)}=1.47826087$$

$$f(w)=f(1.47826.087)=-2.23489761$$

$$\because f(a_{0})\quad f(w)=(-5.5)\times(-2.23489761)>0$$

set $a_{1}=w=1.47826087\quad;\quad b_{1}=b_{0}=2$
$\because \quad \vert f(w)\vert>\varepsilon=10^{-6}$ proceed with iteration.
Iteration details are provide below in a tabular form:

		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	1.6198574305	-0.5488323569
2	1.6198574305	2.0000000000	1.6517157555	-0.1169833690
3	1.6517157555	2.0000000000	1.6583764553	-0.0241659321
4	1.6583764553	2.0000000000	1.6597468853	-0.0049594725
5	1.6597468853	2.0000000000	1.6600278616	-0.0010169938
6	1.6600278616	2.0000000000	1.6600854397	-0.0002089010
7	1.6600854397	2.0000000000	1.6600972414	-0.0000432589
8	1.6600972414	2.0000000000	1.6600997448	-0.0000081223

Note : One may note that Regular Falsi method has converged faster than the Bisection method.

Geometric Interpretation of Regula Falsi Method:

Let us plot the polynomial considered in the above example and trace $w$, its movement and new intervals $[a_{n}, b_{n}]$ with iteration. From the figure() one can verify that the weighted average

$$w=\frac{f(b_{n})a_{n}-f(a_{n})b_{n}}{f(b_{n})-f(a_{n})}$$

is the point of intersection of the secant to $f(x)$, passing through points $(a_{n},f(a_{n}))$ and $(b_{n},f(b_{n}))$ with the x-axis. Since here $f(x)$ is concave upward and increasing the secant is always above $f(x)$. Hence, $w$ always lies to the left of the zero. If $f(x)$ were to be concave downward and increasing, $w$ would always lie to the right of the zero.

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

		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	0.7032044530	0.0251714624
3	0.5000000000	0.7032044530	0.7015219927	0.0021148270
4	0.5000000000	0.7015219927	0.7013807297	0.0001767781
5	0.5000000000	0.7013807297	0.7013689280	0.0000148928
6	0.5000000000	0.7013689280	0.7013679147	0.0000009526