###### Question 1783 – Nielit Scientist-B 17-12-2017

November 5, 2023###### Question 11185 – Algorithms

November 5, 2023# Question 1788 – Nielit Scientist-B 17-12-2017

Using bisection method, one root of X^{4}-X-1 lies between 1 and 2. After second iteration the root may lie in interval:

Correct Answer: B

Question 3 Explanation:

Given data.

root= X

Root lies Between 1 and 2,

After second iteration=?

Using bisection method.

f(1)=X

=1-1-1

= -1

f(2)=X

= 2

=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)

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

C3. By testing the condition | f (c

root= X

^{4}-X-1.Root lies Between 1 and 2,

After second iteration=?

Using bisection method.

f(1)=X

^{4}-X-1=1-1-1

= -1

f(2)=X

^{4}-X-1= 2

^{4}-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

(1.25,1.5)

(1,1.25)

(1,1.5)

None of the options

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