數值計算:一重積分計算的C++實現
阿新 • • 發佈:2019-01-06
#include<iostream> #include<iomanip> #include<cmath> using namespace std; //求積:梯形公式 double Func_Integral_Trapezoid (double lo, double hi, double(*Func)(double), int n = 1000) {//lo:下限;hi:上限;Func:函式;n:等分數 if (n <= 0) n = 100; double x; double step = (hi - lo) / n; double result = 0.0; x = lo; for (int i = 1; i < n; i++) { x += step; result += Func(x); } result += (Func(lo) + Func(hi)) / 2; result *= step; return result; } //求積:Simpson公式 double Func_Integral_Simpson (double lo, double hi, double(*Func)(double), int n = 1000) {//lo:下限;hi:上限;Func:函式;n:等分數 if (n <= 0) n = 100; double x; double step = (hi - lo) / n; double result1 = 0.0; x = lo; for (int i = 1; i < n; i++) { x += step; result1 += Func(x); } result1 *= 2; double result2 = 0.0; x = lo + step / 2; for (int i = 0; i < n; i++) { result2 += Func(x); x += step; } result2 *= 4; double result = result1 + result2 + Func(lo) + Func(hi); result *= step / 6; return result; } //求積:Cotes公式 double Func_Integral_Cotes (double lo, double hi, double(*Func)(double), int n = 1000) {//lo:下限;hi:上限;Func:函式;n:等分數 if (n <= 0) n = 100; double x; double step = (hi - lo) / n; double result1 = 0.0; x = lo; for (int i = 1; i < n; i++) { x += step; result1 += Func(x); } result1 *= 14; double result2 = 0.0; x = lo + step / 2; double result3 = 0.0; double x1 = lo + step / 4; double x2 = lo + step / 4 * 3; for (int i = 0; i < n; i++) { result2 += Func(x); result3 += Func(x1) + Func(x2); x += step; x1 += step; x2 += step; } result2 *= 12; result3 *= 32; double result4 = (Func(lo) + Func(hi)) * 7; double result = result1 + result2 + result3 + result4; result *= step / 90; return result; } //求積:Romberg公式 double Func_Integral_Romberg (double lo, double hi, double(*Func)(double), int k = 4) {//lo:下限;hi:上限;Func:函式;k:等分指數(區間等分成2^k份) int size = k + 1; double *matrix = new double[size*size]; for (int i = 0; i < size*size; i++) matrix[i] = 0.0; double step = hi - lo; matrix[0] = Func_Integral_Trapezoid(lo, hi, Func, 1); for (int i = 1; i < size; i++) { int n = 1 << (i - 1); for (int k = 0; k < n; k++) { matrix[i*size + 0] += Func(lo + (k + 0.5)*step); } matrix[i*size + 0] *= step; matrix[i*size + 0] += matrix[(i - 1)*size]; matrix[i*size + 0] /= 2.0; step /= 2.0; } double temp = 1.0; double factor1, factor2; for (int j = 1; j < size; j++) { temp *= 4.0; factor1 = temp / (temp - 1); factor2 = 1 / (temp - 1); for (int i = j; i < size; i++) { matrix[i*size + j] = factor1*matrix[i*size + j - 1] - factor2*matrix[(i - 1)*size + j - 1]; } } double result = matrix[k*size + k]; delete[] matrix; return result; } //測試用的被積函式,0到1積分為PI double Func_test1(double x) { return 4 / (1 + x*x); } int main() { cout << "Trapezoid Numerical Integration" << endl; cout << setprecision(15) << Func_Integral_Trapezoid(0, 1, Func_test1, 1024) << endl << endl; cout << "Simpson Numerical Integration" << endl; cout << setprecision(15) << Func_Integral_Simpson(0, 1, Func_test1, 1024) << endl << endl; cout << "Cotes Numerical Integration" << endl; cout << setprecision(15) << Func_Integral_Cotes(0, 1, Func_test1, 1024) << endl << endl; cout << "Romberg Numerical Integraion" << endl; cout << setprecision(15) << Func_Integral_Romberg(0, 1, Func_test1, 10) << endl << endl; getchar(); return 0; }