Question

Use the following pseudocode for the Newton-Raphson method to write MATLAB code to approximate the cube root (a)^1/3 of a given number a with accuracy roughly within 10^-8 using x0 = a/2. Use at most 100 iterations. Explain steps by commenting on them.
Use f(x) = x3 − a. Choose a = 2 + w, where w = 3
Algorithm : Newton-Raphson Iteration
Input: f(x)=x³−a, x₀ =a/2, tolerance 10^-8, maximum number of iterations100 Output: an approximation (a)^1/3 within 10^-8 or a message of failure
set x = x₀, xold = x₀;

for i = 1 to 100 do
x = x − f(x)/f′(x);
if |x − xold| < 10^-8 then. % checking required accuracy
FoundSolution = true; % done
break;   % leave for environment
end if
xold = x; % update xold for the next iteration end for
if FoundSolution then
print x, ‘The number of iterations =’, i
else
print ‘The required accuracy is not reached in 100 iterations’
end if
For f′, define the function df(x) = 3x² by using the command

df = @(x) 3 ∗ x.²;

Answer 1

Hey,

Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries

clc
clear all
close all
a=1/3;
f=@(x) x^3-a;
g=@(x) 3*x^2;
x0=a/2;
x=x0;
xold=x0;
flag=0;
for i=1:100
x=x-f(x)/g(x);
if(abs(x-xold)<1e-8)
flag=1;
break;
end
xold=x;
end
if(flag==1)
disp(['Root is ' num2str(x) ', The number of iterations are ' num2str(i)]);
else
disp('The required accuracy is not reached in 100 iterations');
end

Kindly revert for any queries

Thanks.