According to Taylor series expansion at x = 0, we have sin(x) = x -x^3/3! + x^5/

Question

According to Taylor series expansion at x = 0, we have sin(x) = x -x^3/3! + x^5/5! - x^7/7! + x^9/9! - x^11/11! + ... = sigma_k=0^infinity (-1)^k x^2k+1/(2 k + 1)! and cos (x) = 1 - x^2/2! + x^4/4! - x^+/6! + x^8/8! - x^10/10! + ... = sigma_k=0^infinity (-1)^k x^2k/(2k)! where x is in radian. Using these two equations, we can approximately evaluate the values of sin and cos functions for a given argument x. Write two functions g(x) = sin(x) and h(x) = cos(x) using the series above to obtain accuracy to 5 decimal places. Write a C++ program that uses the functions above to calculate f(n) for integer n = 0, 1, 2, ..., 6, where f(n) = 5g(n) * h (4000 pi n + pi/3 = 5 sin(n) * cos (4000 pi n + pi/3). The program also display the result in the following table format: n 5 sin(n) cos(4000 pi n + pi/3) f(n)

Explanation / Answer

#include<iostream>
#include<cmath>
#include <iomanip>
using namespace std;

double g(double x);
double h(double x);
long int factorial(int m);
double f(int n);

const double PIE= 3.14159;

int main()
{
   //x is in radians
   double x;
   int n;

   cout<<"Enter x in radians = ";
   cin>>x;

   cout << setprecision(5);
   //find g(x) and h(x) by calling those functions
   cout<<"Value of g(x) = "<<g(x) << " and h(x) = "<<h(x)<<endl;

   //find f(n)
   cout<<"Enter value of n = ";
   cin>>n;

cout<<"Value of f(n)= "<< f(n)<<endl;

}

double g(double x)
{
int n;
double val=0;
   //for approximation to 5 places
for (n=0;n<5;n++)
{
double p = pow((long double)-1,n);
double px = pow(x,2*n+1);
long fac = factorial(2*n+1);
val += p * px / fac;
}
return val;
}

double h(double x)
{
int n;
double val=0;
   //for approximation to 5 places
for (n=0;n<5;n++)
{
double p = pow((long double)-1,n);
double px = pow(x,2*n);
long fac = factorial(2*n);
val += p * px / fac;
}
return val;
}

long int factorial(int m)
{
if (m==0 || m==1) return (1);
else return (m*factorial(m-1));
}

double f(int n)
{
   double value;
   //f(n) = 5sin(n) * cos(4000pien+pie/3
   value = 5*g(n) * cos((4000*PIE)+PIE/3);

   return value;
}

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