09/07/03 23:07:12
>>540
// ニュートン法で「f(x) = 0」を解く
#include <iostream>
// f(x) および f'(x)
double f1(double x) { return x*x - 57; }
double f1d(double x) { return 2*x; }
double f2(double x) { return x*x*x - 7; }
double f2d(double x) { return 3*x*x;}
double f3(double x) { return 3*x*x*x + 2*x*x + 5*x - 15; }
double f3d(double x) { return 9*x*x + 4*x + 5; }
int main() {
int n; double x; // 繰り返し回数、初期値
n = 10; x = 1.0;
while (n--) { x -= f1(x) / f1d(x); }
std::cout << "(1) " << x << std::endl;
n = 10; x = 1.0;
while (n--) { x -= f2(x) / f2d(x); }
std::cout << "(2) " << x << std::endl;
n = 10; x = 1.0;
while (n--) { x -= f3(x) / f3d(x); }
std::cout << "(3) " << x << std::endl;
return 0;
}