Question 1788 – Nielit Scientist-B 17-12-2017

Given data.
root= X⁴-X-1.
Root lies Between 1 and 2,
After second iteration=?
Using bisection method.
f(1)=X⁴-X-1
=1-1-1
= -1
f(2)=X⁴-X-1
= 2⁴ -2 -1
=13
Given constraint that “root lies between 1 and 2”
Iteration-1: x1=(a+b)/2
=(1+2)/2
= 1.5
f(1.5) = 2.5625
Iteration-2: x2=(a+b)/2
=(1+1.5)/2
=1.25
f(1.25)=0.19140625 >0
Root may lie in between (1, 1.25)
Algorithm – Bisection Scheme
Given a function f (x) continuous on an interval [a,b] and f (a) * f (b) < 0
Do
c=(a+b)/2
if f(a)*f(c)< 0 then b=c

else a=c
while (none of the convergence criteria C1, C2 or C3 is satisfied)
More info:
Bisection method is the simplest among all the numerical schemes to solve the transcendental equations. This scheme is based on the intermediate value theorem for continuous functions .
Consider a transcendental equation f(x)=0 which has a zero in the interval [a,b] and
f(a)*f(b)<0. Bisection scheme computes the zero, say c, by repeatedly halving the interval [a,b]. That is, starting with
c = (a+b) / 2
The interval [a,b] is replaced either with [c,b] or with [a,c] depending on the sign of f (a) * f (c) . This process is continued until the zero is obtained. Since the zero is obtained numerically the value of c may not exactly match with all the decimal places of the analytical solution of f (x) = 0 in the interval [a,b]. Hence any one of the following mechanisms can be used to stop the bisection iterations :
C1. Fixing a priori the total number of bisection iterations N i.e., the length of the interval or the maximum error after N iterations in this case is less than | b-a | / 2N.
C2. By testing the condition | c_i – c _i-1| (where i are the iteration number) less than some tolerance limit, say epsilon, fixed a priori.
C3. By testing the condition | f (c_i ) | less than some tolerance limit alpha again fixed a priori.

http://mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/bracketing%20methods/bisection/bisection.html