Newton法パネェ

#include <stdio.h>

int main()
{
	double a, x;
	int count, i;
	printf("aの3乗根を求めます。\na:");
	scanf("%lf", &a);
	printf("x0:");
	scanf("%lf", &x);
	printf("反復回数:");
	scanf("%d", &count);
	
	for(i = 1; i <= count; i++)
	{
		x = (2 * x + a / (x * x)) / 3;
		printf("x%d = %.20f (x%d^3 = %.20f)\n", i, x, i, x*x*x);
	}
	return 0;
}

aの3乗根を求めます。
a:10
x0:10
反復回数:10
x1 = 6.70000000000000017800 (x1^3 = 300.76300000000003360000)
x2 = 4.54092225439964369800 (x2^3 = 93.63370300166980087000)
x3 = 3.18893705034208307900 (x3^3 = 32.42931976553633433000)
x4 = 2.45374135571122797800 (x4^3 = 14.77360039874827358000)
x5 = 2.18945956481615100500 (x5^3 = 10.49568497518751542000)
x6 = 2.15499199671155583500 (x6^3 = 10.00776237299924886800)
x7 = 2.15443483414562608800 (x7^3 = 10.00000200675034633900)
x8 = 2.15443469003189314000 (x8^3 = 10.00000000000013145000)
x9 = 2.15443469003188381400 (x9^3 = 10.00000000000000177600)
x10 = 2.15443469003188381400 (x10^3 = 10.00000000000000177600)

適当な初期値でも9回で収束した、だと…!? まさかこれほどとは… 大した奴だ。