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  1. // x = sqrt(2) の x を求める
  2. // x^2 = 2
  3. // f(x) = x^2 - 2 = 0
  4. #include <stdio.h>
  5.  
  6. int main()
  7. {
  8. 	double	x, fx, dfdx;
  9. 	int	i;
  10.  
  11. 	x = 1;
  12. 	for (i = 0; i < 10; i++) {
  13. 		fx = (x * x) - 2;
  14. 		dfdx = 2 * x;
  15. 		x -= fx / dfdx;
  16. 		printf("%d %.20f\n", i, x);
  17. 	}
  18. 	return 0;
  19. }
  20.  
Success #stdin #stdout 0.02s 1676KB
comments (?)
stdin 
Standard input is empty
stdout 
0 1.50000000000000000000
1 1.41666666666666674068
2 1.41421568627450988664
3 1.41421356237468986983
4 1.41421356237309514547
5 1.41421356237309514547
6 1.41421356237309514547
7 1.41421356237309514547
8 1.41421356237309514547
9 1.41421356237309514547
https://ideone.com/CI65D
language:
C (gcc 8.3)
created:
12 years ago
visibility:
public

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