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牛頓法範例
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// 比較可以重複利用牛頓法的寫法(使用 function pointer) | |
#include <iostream> | |
using namespace std; | |
typedef double (*Equation)(double x); // 把 function pointer 命名為 Equation | |
double diff(Equation f, double x, double h=1e-6) { | |
return (f(x + h) - f(x)) / h; | |
} | |
// The format should be the left part of the equation f(x) = 0 | |
double f1(double x) { | |
// You can change the function by yourself | |
return x * x - 2; | |
} | |
// Another example of the equation | |
double f2(double x) { | |
// to solve the equation x^2 -5x + 6 = 0, you will get different answer for different initial values | |
// You will get x = 2 if the initial value is for example, 0 | |
// You will get x = 3 if the initial value is for example, 5 | |
// You can try different initial values! | |
return x*x - 5*x + 6; | |
} | |
// Another example of the equation | |
double f3(double x) { | |
// to solve the equation x^3 - 6x^2 + 11x - 6 = 0 | |
return x*x*x - 6*x*x + 11*x - 6; | |
} | |
double newton(Equation f, double x_init, int N) { | |
double x = x_init; | |
for (int i = 0; i < N; i++) { | |
x = x - (f(x) / diff(f, x)); | |
} | |
return x; | |
} | |
int main() { | |
double x; | |
int N; | |
cout << "Please enter a initial point: "; | |
cin >> x; | |
cout << "Please enter the times to apply the Newton method (such as 100): "; | |
cin >> N; | |
cout << "We have the root: " << newton(f3, x, N) << endl; | |
return 0; | |
} |
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// 用牛頓法計算根號2的範例 | |
#include <iostream> | |
using namespace std; | |
double f(double x) { | |
return x * x - 2; // f(x) = x^2 - 2 = 0, to find x^2=2, x = ? | |
} | |
double f_diff(double x, double h=1e-6) { | |
return (f(x + h) - f(x)) / h; | |
} | |
double newton(double x_init, int N) { | |
double x = x_init; | |
for (int i = 0; i < N; i++) { | |
double x_next = x - (f(x) / f_diff(x)); | |
x = x_next; | |
} | |
return x; | |
} | |
int main() { | |
double x; | |
int N; | |
cout << "Please enter a initial point: "; | |
cin >> x; | |
cout << "Please enter the times to apply the Newton method: "; | |
cin >> N; | |
cout << "We have the root: " << newton(x, N) << endl; | |
return 0; | |
} |
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