Lecture 5: Secant Method

Secant Method
By definition
f^ (p_(n-1) )=lim?(x?p_n )??(f(x)-f(p_(n-1)))/(x-p_(n-1) )?
If p_(n-2) is close to p_(n-1), then
f^ (p_(n-1) )?(f(p_(n-2) )-f(p_(n-1) ))/(p_(n-2)-p_(n-1) )=(f(p_(n-1) )-f(p_(n-2) ))/(p_(n-1)-p_(n-2) )
Using the approximation of f^ (p_(n-1) ) in Newton’s formula gives

p_n=p_(n-1)-(f^ (p_(n-1) )(p_(n-1)-p_(n-2)))/(f(p_(n-1) )-f(p_(n-2)))

This technique is called the Secant method. Starting with the two initial approximations p0 and p1, the approximation p2 is the x-intercept of the line joining ( p0, f ( p0)) and ( p1, f ( p1)). The approximation p3 is the x-intercept of the line joining ( p1, f ( p1)) and ( p2, f ( p2)), and so on. If x1=p1, x2=p2, y1=f(x1) and y2=f(p2), then
x=x_2-(y_2 (x_2-x_1 ))/(y_2-y_1 )=(x_1.y_2-x_2 ?.y?_1)/(y_2-y_1 )

Example:
Find the root of f(x)=xln(x)-1 using Secant Method in the interval [1,2],and ? =0.0018.
Solution:
x1=1; x2=2; f(x1)=-1; f(x2)=0.3863;
|2-1|=1>?
x=(1*0.3863-2*(-1))/(0.3863-(-1))=1.7213
f(x)=-0.0652
x1=x2=2; y1=y2=0.3863; x2=x=1.7213; y2=y=-0.0652;
|1.7213 – 2|=0.2787>?
x=(2*(-0.0652)-1.7213*0.3863)/(-0.0652-0.3863)=1.7615
f(x)= -0.0027
x1=x2=1.7213; y1=y2= -0.0652; x2=x= 1.7615; y2=y= -0.0027;
|1.7615 – 1.7213|=0.0402>?
x=(1.7213*(-0.0027)-1.7615*(-0.0652))/(-0.0027-(-0.0652))=1.7632
f(x)= -3.5784e-5;
x1=x2=1.7615; y1=y2= -0.0027; x2=x=1.7632; y2=y= -3.5784e-5;
|1.7632 - 1.715|=0.0017Then the root is x= 1.7632
Exercises:
Q\ The fourth-degree polynomial f (x) = 230x4 + 18x3 + 9x2 ? 221x ? 9
has two real zeros, one in [?1, 0] and the other in [0, 1]. Attempt to approximate these zeros to within 10?6 using the
a. Method of False Position
b. Secant method
c. Newton’s method
Use the endpoints of each interval as the initial approximations in (a) and (b) and the midpoints as the initial approximation in (c).