// This program will use three Taylor Series expansions for arctan(x), cos(x), and exp(x) to approximate the values of arctan, cos, and e to the x functions after being given x. // Amanda Macoritto and Kreepa Upadhyay. ECE 206/L. Apirl 17th, 2022. #include <iostream> #include <cmath> #include <math.h> using namespace std; double pi = 3.14159265359; //best value of pi int numbers = 6; //total terms of taylor series to be used int fact(int n){ if (n == 0) return 1; int i, ans = 1; for (i = 1; i <= n; i++) ans *= i; return ans; } //calculate arctan double arctan(double x){ // if x is greater than 1 if (x > 1) x = 1.0 / x; double ans = 0.0; int i; for (i = 1; i <= numbers; i++){ double temp = pow(x, 2 * i - 1) / (double)(2 * i - 1); // if the current term is odd term if (i % 2 == 0) ans -= temp; // if the current term is even term else ans += temp; } // if x is greater than 1 if (x > 1) ans = pi - ans; return ans; } //calculates cos double cos(double x){ double cosx = 0.0; int i; for (i = 0; i < numbers; i++) { // calculate current element double temp = pow(x, 2 * i) / (double)fact(2 * i); // if current term is odd term if (i % 2 != 0) { temp = -temp; cosx += temp; } return cosx; } } //calculate exp double exp(double x){ double ans = 0; int i; for (i = 0; i < numbers; i++) ans += pow(x, i) / (double)fact(i); return ans; } void f(double x){ cout << "n\t\t10arctan(n)\tcos(6000pi*n + pi/6)\texp(-n/2)\tf(n)" << endl; cout << "---------------------------------------------------" << endl; double a = 10 * arctan(x); double b = cos(6000 * pi * x + pi / 6); double c = exp(-x / 2); cout << x << "\t" << a << "\t" << b << "\t" << c << "\t" << a * b * c << endl; system("pause"); } int main(){ double x; cout << "Enter x : "; cin >> x; cout << endl; f(x); return 0; }
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