Lecture 6: False Position Method

False Position Method

The method of False Position generates approximations in the same manner as the Secant method, but it includes a test to ensure that the root is
always bracketed between successive iterations.
First choose initial approximations p0 and p1 with f ( p0) ? f ( p1) < 0. The approximation p2 is chosen in the same manner as in the Secant method, as the x-intercept of the line joining ( p0, f ( p0)) and ( p1, f ( p1)). To decide which secant line to use to compute p3, consider f ( p2) ? f ( p1);
• If f ( p2) ? f ( p1) < 0, then p1 and p2 bracket a root. Choose p3 as the x-intercept of the line joining ( p1, f ( p1)) and ( p2, f ( p2)).
• If not, choose p3 as the x-intercept of the line joining ( p0, f ( p0)) and ( p2, f ( p2)).
In a similar manner compute p4, and so on.


Example1: Find the root of xln(x)-1=0 using False Position in the interval [1, 2] and ?=0.001
Solution:
f(x1)=-1, f(x2)=0.3863
x=(1*(0.3863)-2*(-1))/(0.3863+1)=1.7213
|x_1-x|=|1-1.7213|=0.7213> ?
f(x)= - 0.0652
x1 = x; y1 = y;
x=(1.7213*(0.3863)-2*(-0.0652))/(0.3863-(-0.0652))=1.7615
|x_1-x|=|1.7213-1.7615|=0.0402> ?
f(x) =- 0.0027
x1 = x; y1 = y;
x=(1.7615*0.3863-2*(-0.0027))/(0.3863-(-0.0027))=1.7632
|x_1-x|=|1.7615-1.7632|=0.0017>?
f(x)=-0.00004
x2 = x; y2 = y;

Example2:
Use the method of False Position to find a solution to x = cos x. Compare the results with fixed point, Newton and Secant methods.
Solution: To make a reasonable comparison we will use the same initial approximations as in the Secant method, that is, p0 = 0.5 and p1 = ?/4. In the table below, notice that the False Position and Secant approximations agree through p3 and that the method of False Position requires an additional iteration to obtain the same accuracy as the Secant method.






Exercises:
Q\ Use Secant, Newton and false position methods to find solutions to within 10?7 for the following problems.
a) x2 ? 4x + 4 ? ln x = 0 for 1 ? x ? 2 and for 2 ? x ? 4
b) x + 1 ? 2 sin ?x = 0 for 0 ? x ? 1/2 and for 1/2 ? x ? 1