回答編集履歴
4
あ
answer
CHANGED
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@@ -70,7 +70,7 @@
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70
70
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int i;
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71
71
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72
72
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for (i = 1; i <= K_MAX; i++) {
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73
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-
if (df(x1) < EPS)
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73
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+
if (fabs(df(x1)) < EPS)
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74
74
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return false; // ニュートン法が失敗する場合
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75
75
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76
76
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x2 = x1 - f(x1) / df(x1);
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3
a
answer
CHANGED
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@@ -70,7 +70,7 @@
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70
70
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int i;
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71
71
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72
72
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for (i = 1; i <= K_MAX; i++) {
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73
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-
if (df(x1)
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73
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+
if (df(x1) < EPS)
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74
74
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return false; // ニュートン法が失敗する場合
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75
75
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76
76
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x2 = x1 - f(x1) / df(x1);
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2
あ
answer
CHANGED
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@@ -25,7 +25,7 @@
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25
25
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void input(double *init_x);
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26
26
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double f(double x);
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27
27
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double df(double x);
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28
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-
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28
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+
bool newton(double init_x, double *x, int *num_iters);
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29
29
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30
30
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int main()
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31
31
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{
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@@ -36,8 +36,10 @@
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36
36
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printf("input x: ");
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37
37
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scanf("%lf", &init_x);
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38
38
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39
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-
newton(init_x, &x, &num_iters)
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39
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+
if (newton(init_x, &x, &num_iters))
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40
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-
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40
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+
printf("number of iteration: %d, x=%.15f\n", num_iters, x);
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41
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+
else
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42
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+
printf("newton method failed.");
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41
43
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}
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42
44
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43
45
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/**
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@@ -62,25 +64,29 @@
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62
64
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@param x 解
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63
65
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@param num_iters 反復回数
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64
66
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*/
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65
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-
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67
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+
bool newton(double init_x, double *x, int *num_iters)
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66
68
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{
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67
69
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double x1 = init_x, x2;
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68
70
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int i;
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69
71
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70
72
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for (i = 1; i <= K_MAX; i++) {
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73
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+
if (df(x1) == 0)
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74
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+
return false; // ニュートン法が失敗する場合
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75
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+
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71
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-
x2 = x1 - f(x1) /
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76
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+
x2 = x1 - f(x1) / df(x1);
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72
77
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printf("%d: x=%.15f\n", i, x2);
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73
78
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74
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-
if (fabs(x2 - x1) < EPS)
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79
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+
if (fabs(x2 - x1) < EPS) {
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75
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-
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80
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+
*x = x2;
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81
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+
*num_iters = i;
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82
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+
return true;
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83
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+
}
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84
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+
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76
85
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x1 = x2;
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77
86
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}
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78
87
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79
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-
//
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88
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+
return false; // 収束しなかった場合
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80
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-
*x = x2;
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81
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-
*num_iters = i;
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82
89
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}
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83
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-
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84
90
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```
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85
91
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86
92
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```
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1
a
answer
CHANGED
|
@@ -68,7 +68,7 @@
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68
68
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int i;
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69
69
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70
70
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for (i = 1; i <= K_MAX; i++) {
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71
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-
x2 = x1 - f(x1) / df(x1);
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71
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+
x2 = x1 - f(x1) / (df(x1) + EPS);
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72
72
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printf("%d: x=%.15f\n", i, x2);
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73
73
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74
74
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if (fabs(x2 - x1) < EPS)
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