思考题1

改进cosx？优化算法

关键点在于cos计算过于麻烦，而每次都要求sinx的值
故直接简化为cosx的导数 -sinx
即：
原：//double daoshu(double x) {
// return 18 * x - cos(x);
//}
改：double daoshu(double x) {return 18 * x + sin(x);
}

证明：

思考题2：

c++代码如下：
// 二分法
double bin(double l, double r) {double eps = 1e-3; while (r - l > eps) {double mid = (l + r) / 2;double f = exp(mid) + 10 * mid - 2;if (f > 0) {r = mid;} else {l = mid;}}return (l + r) / 2;
}
// 迭代法
double ite() {double x = 0;double eps = 1e-3; while (1) {double x1 = (2 - exp(x)) / 10;if (x - x1 > eps) {break;}x = x1;}return x;
}
// 牛顿法
double new() {double x = 0;double eps = 1e-3; while (1) {double f = exp(x) + 10 * x - 2;double df = exp(x) + 10;double x1 = f / df;if (x1 > eps) {break;}x = x - x1;}return x;
}

思考题3：

求谱半径？

思考题4：

代码实现：

double Max(vectorx) {double max = x[0];int n = x.size();for (int i = 0; i < n; i++)if (x[i] > max) max = x[i];return max;
}
void Jacobi(vector<vector > A, vector B, int n) {vector X(n, 0); vector Y(n, 0); vector D(n, 0); int k = 0; do {X = Y;for (int i = 0; i < n; i++) {double tem = 0;for (int j = 0; j < n; j++) {if (i != j) tem += A[i][j] * X[j];}Y[i] = (B[i] - tem) / A[i][i];cout << left << setw(8) << Y[i] << " ";}cout << endl;k++;if (k > 100) { return;}
for (int a = 0; a < n; a++) {
D[a] = X[a] - Y[a];
}
} while (fabs(Max(D)) > 1e-4);
return;
}
int main() {int n;cout << "未知数的个数n：";cin >> n;cout << endl;vector<vector>A(n, vector(n, 0));vectorB(n, 0);cout << "系数矩阵：" << endl;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {cin >> A[i][j];}}cout << endl;cout << "值矩阵：" << endl;for (int k = 0; k < n; k++) {cin >> B[k]; }cout << endl;cout << "您输入的方程组为：" << endl;for (int a = 0; a < n; a++) {for (int b = 0; b < n; b++) {cout << A[a][b] << " "; }cout << " " << B[a] << endl;}cout << endl;cout << "方程组的解：" << endl;Jacobi(A, B, n);return 0;
}

结果：

思考题5：

int main()
{
//甲乙丙丁:ABCD
int a,b,c,d;
for(a=1;a<=4;++a)
{
for(b=1;b<=4;++b)
{
if(a!=b)
for(c=1;c<=1;++c)
{
d=10-a-b-c;
if(((a==1)+(b==3)==1)&&( (c==1)+(d==4 )==1)&&((a==3)+(d==2)==1 ) )
{
cout<<"甲："<<a<<"乙："<<b<<"丙："<<c<<"丁："<<d<<endl;break;
}
}
}
}
return 0;
}