回答編集履歴
4
あ
test
CHANGED
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@@ -142,7 +142,7 @@
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142
142
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143
143
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for (i = 1; i <= K_MAX; i++) {
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144
144
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145
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-
if (df(x1) < EPS)
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145
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+
if (fabs(df(x1)) < EPS)
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146
146
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147
147
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return false; // ニュートン法が失敗する場合
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148
148
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3
a
test
CHANGED
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@@ -142,7 +142,7 @@
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142
142
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143
143
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for (i = 1; i <= K_MAX; i++) {
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144
144
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145
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-
if (df(x1)
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145
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+
if (df(x1) < EPS)
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146
146
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147
147
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return false; // ニュートン法が失敗する場合
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148
148
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2
あ
test
CHANGED
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@@ -52,7 +52,7 @@
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52
52
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53
53
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double df(double x);
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54
54
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55
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-
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55
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+
bool newton(double init_x, double *x, int *num_iters);
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56
56
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57
57
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58
58
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@@ -74,9 +74,13 @@
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74
74
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75
75
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76
76
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77
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-
newton(init_x, &x, &num_iters)
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77
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+
if (newton(init_x, &x, &num_iters))
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78
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-
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78
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+
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79
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-
printf("number of iteration: %d, x=%.15f\n", num_iters, x);
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79
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+
printf("number of iteration: %d, x=%.15f\n", num_iters, x);
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80
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+
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81
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+
else
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82
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+
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83
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+
printf("newton method failed.");
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80
84
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81
85
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}
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82
86
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@@ -126,7 +130,7 @@
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126
130
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127
131
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*/
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128
132
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129
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-
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133
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+
bool newton(double init_x, double *x, int *num_iters)
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130
134
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131
135
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{
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132
136
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@@ -138,15 +142,29 @@
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138
142
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139
143
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for (i = 1; i <= K_MAX; i++) {
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140
144
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145
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+
if (df(x1) == 0)
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146
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+
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147
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+
return false; // ニュートン法が失敗する場合
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148
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+
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149
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+
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150
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+
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141
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-
x2 = x1 - f(x1) /
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151
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+
x2 = x1 - f(x1) / df(x1);
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142
152
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143
153
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printf("%d: x=%.15f\n", i, x2);
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144
154
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145
155
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146
156
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|
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147
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-
if (fabs(x2 - x1) < EPS)
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157
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+
if (fabs(x2 - x1) < EPS) {
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148
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-
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158
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+
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149
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-
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159
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+
*x = x2;
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160
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+
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161
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+
*num_iters = i;
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162
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+
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163
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+
return true;
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164
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+
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165
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+
}
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166
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+
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167
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+
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150
168
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151
169
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x1 = x2;
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152
170
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@@ -154,15 +172,9 @@
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154
172
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155
173
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156
174
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157
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-
//
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175
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+
return false; // 収束しなかった場合
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158
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-
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159
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-
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176
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+
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160
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-
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161
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-
*num_iters = i;
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162
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-
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163
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-
}
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177
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+
}
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164
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-
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165
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-
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166
178
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167
179
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```
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168
180
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1
a
test
CHANGED
|
@@ -138,7 +138,7 @@
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|
138
138
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|
139
139
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for (i = 1; i <= K_MAX; i++) {
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140
140
|
|
|
141
|
-
x2 = x1 - f(x1) / df(x1);
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141
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+
x2 = x1 - f(x1) / (df(x1) + EPS);
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142
142
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143
143
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printf("%d: x=%.15f\n", i, x2);
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144
144
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