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;
}