Both of the questions need to be done on C++ please. 8 Question 8 In this questi

Question

Both of the questions need to be done on C++ please.

8 Question 8 In this question, we compute the following sum: N-1 $(z) = cos z sin Set N = 20 in this question. . Define the following function f(z) = 5-11 S(r) (8.2) . Consider only values >0 in this question Find brackets of size not less than 0.1 which enclose the first 6 positive roots of f(z), i.e. where/(x) = 0. Find brackets of size not less than 0.1 which enclose the first 6 discontinuities of Using the bisection algorithm and your brackets above, calculate the values of the first, third and fifth positive roots of f(a) to an accuracy of three decimal places. . Solutions which compute roots other than the first, third and fifth will receive a score of zero. to 3 . Using the bisection algorithm and your brackets above, calculate the locations fourth and sixth discontinuities of fr) (for z >0) to an accuracy of three decimal places. Terminate the iteration if If(r) 10 in the iteration. State your reason for terminating the iteration (converged in or function of the value lf(z) > 10 ”). . Solutions which compute discontinuities other than the second, fourth and sixth will receive a score o f zero. reason for terminatOI final answer in r or (f()>

Explanation / Answer

Code for finding the bracket of root(first 6):

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <cmath>
using namespace std;
#define PI 3.14159265

double val(double x)
{
double sum =0.0 ;
for( double i=0;i<20;i++)
{
sum += cos(double(x*sin(double(i*PI/20))));
}
sum = sum/20;
return sum;
}

int main()
{
int count =0;
double f = 4; // base
double s = 1; //base
double start = 0;
double end = 0.1;
while ( count < 6 && abs(f) <= 1000000 ) {
double v1 = val(start);
double v2 = val(end);
double v3 = (5-(1/v1));
double v4 = (5-(1/v2));

if((v3 >=0 && v4 <= 0) || (v3<=0 && v4 >=0)) // if S(x) have different sign in the goven bracket
{
if((v1 >=0 && v2 <= 0) || (v1<=0 && v2 >=0)) // if f(x) have different sign in the given bracket
{
cout << endl;
//cout << " infinite value attained "<< endl; // returnn the bracket of infinite value
}
else
{
count++;
cout<< count << " domain found " << start << " " << end << endl ; // return the bracket of root of function
}
}
start = end;
end = end + 0.1;
}
return 0;
}

Code for finding the discontinuities of f (first 6):

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <cmath>
using namespace std;
#define PI 3.14159265

double val(double x)
{
double sum =0.0 ;
for( double i=0;i<20;i++)
{
sum += cos(double(x*sin(double(i*PI/20))));
}
sum = sum/20;
return sum;
}

int main()
{
int count =0;
double f = 4; // base
double s = 1; //base
double start = 0;
double end = 0.1;
while ( count < 6 && abs(f) <= 1000000 ) {
double v1 = val(start);
double v2 = val(end);
double v3 = (5-(1/v1));
double v4 = (5-(1/v2));

if((v3 >=0 && v4 <= 0) || (v3<=0 && v4 >=0)) // if S(x) have different sign in the goven bracket
{
if((v1 >=0 && v2 <= 0) || (v1<=0 && v2 >=0)) // if f(x) have different sign in the given bracket
{
count++;
cout<< count << " domain found " << start << " " << end << endl ; // return the bracket of root of function
//cout << " infinite value attained "<< endl; // returnn the bracket of infinite value
}

}
start = end;
end = end + 0.1;
}
return 0;
}

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