Question

HI, ANY PROGRAMMING LANGUAGE WOULD WORK, FROM PYTHON TO C++, MOST PREFERABLY OR ELSE ANY FOR YOUR CONVENIENCE.

Let f(x) = x⁶ + 7x⁵− 15x⁴− 70x³ + 75x² + 175x − 125.

a.) Write a program that carries out the Secant Method on f(x). You do not need to make your program take arbitrary

input (i.e. you can tailor it to this specific f(x)). Your program should take as input the appropriate number of initial

guesses, an interval [a, b] and a desired error tolerance TOL. It should output a root r that is within the desired

tolerance and it should output N, the number of iterations it took to get to within the error tolerance. Make it clear

what procedure you are using to compute the error of your approximations.

b.)  Write a program that carries out Steffensen′s Method on f(x). You do not need to make your program take arbitrary

input (i.e. you can tailor it to this specific f(x)). Your program should take as input the appropriate number of

initial guesses, an interval [a, b] and a desired error tolerance TOL. It should output a root r that is within the

desired tolerance and it should output N, the number of iterations it took to get to within the error tolerance. Make

it clear what procedure you are using as your underlying recursion, and how you are computing the error of your

approximations.

c.) Write a program that carries out Aitken′s Method on f(x). You do not need to make your program take arbitrary input

(i.e. you can tailor it to this specific f(x)). Your program should take as input the appropriate number of initial guesses,

an interval [a, b] and a desired error tolerance TOL. It should output a root r that is within the desired tolerance and

it should output N, the number of iterations it took to get to within the error tolerance. Make it clear what procedure

you are using as your underlying recursion, and how you are computing the error of your approximations.

d.) Write a program that carries out Newton′s Method on f(x). You do not need to make your program take arbitrary

input (i.e. you can tailor it to this specific f(x)). Your program should take as input the appropriate number of initial

guesses, an interval [a, b], and a desired error tolerance TOL. It should output a root r that is within the desired

tolerance and it should output N, the number of iterations it took to get to within the error tolerance. Make it clear

what procedure you are using to compute the error of your approximations

Answer 1

a)Program to find root of an equations using secant method

Last Updated: 04-01-2019

The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates x1 and x2 for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if the difference between two intermediate values is less than convergence factor.

Examples :

Input : equation = x³ + x - 1

        x1 = 0, x2 = 1, E = 0.0001

Output : Root of the given equation = 0.682326

         No. of iteration=5

Algorithm

Initialize: x1, x2, E, n         // E = convergence indicator

calculate f(x1),f(x2)

if(f(x1) * f(x2) = E); //repeat the loop until the convergence

    print 'x0' //value of the root

    print 'n' //number of iteration

}

else

    print "can not found a root in the given interval"

// C++ Program to find root of an

// equations using secant method

#include <bits/stdc++.h>

using namespace std;

// function takes value of x and returns f(x)

float f(float x)

{

    // we are taking equation as x^3+x-1

    float f = pow(x, 3) + x - 1;

    return f;

}

  

void secant(float x1, float x2, float E)

{

    float n = 0, xm, x0, c;

    if (f(x1) * f(x2) < 0) {

        do {

            // calculate the intermediate value

            x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));

  

            // check if x0 is root of equation or not

            c = f(x1) * f(x0);

  

            // update the value of interval

            x1 = x2;

            x2 = x0;

  

            // update number of iteration

            n++;

  

            // if x0 is the root of equation then break the loop

            if (c == 0)

                break;

            xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));

        } while (fabs(xm - x0) >= E); // repeat the loop

                                // until the convergence

  

        cout << "Root of the given equation=" << x0 << endl;

        cout << "No. of iterations = " << n << endl;

    } else

        cout << "Can not find a root in the given inteval";

}

  

// Driver code

int main()

{

    // initializing the values

    float x1 = 0, x2 = 1, E = 0.0001;

    secant(x1, x2, E);

    return 0;

}

Output :

Root of the given equation = 0.682326

No. of iterations = 5

Time Complexity = O(1)