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Copy pathIntegration.cpp
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260 lines (235 loc) · 5.49 KB
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#include "Integration.h"
#include <iostream>
#include <iomanip>
using namespace std;
double CompoundSimpson(double a, double b, int n,double(*function)(double)){
double h = (b - a) / n;
double XI0 = function(a) + function(b);
//f(x2i-1)的和
double XI1 = 0;
//f(x2i)的和
double XI2 = 0;
for (int i = 1; i <n; i++){
double X = a + i*h;
if (i % 2 == 0){
XI2 = XI2 + function(X);
cout << "X2:" << XI2 << endl;
}
else{
XI1 = XI1 + function(X);
cout << "XI1:" << XI1 << endl;
}
}
double XI = h*(XI0 + 2 * XI2 + 4 * XI1) / 3;
return XI;
}
double functionCompoundSimpson(double d){
return exp(d);
}
void testCompoundSimpson(){
double result = CompoundSimpson(0, 4, 10,functionCompoundSimpson);
cout << "积分值为:" << result << endl;
}
double** Romberg(double a, double b, int n,double(*function)(double)){
double** R = new double *[n];
for (int i = 0; i < n; i++){
R[i] = new double[n];
}
double h = b - a;
R[0][0] = (h / 2)*(function(a) + function(b));
for (int i = 1; i < n; i++){
double sum = 0;
for (int k = 1; k <=pow(2, i - 1); k++){
sum = sum + function(a + (k - 0.5)*h);
}
//梯形方法近似
R[i][0] = 0.5*(R[i-1][0] + h*sum);
for (int j = 1; j <=i; j++){
//外推
R[i][j] = R[i][j-1] + (R[i][j-1] - R[i-1][j-1]) / (pow(4, j) - 1);
}
h = h / 2;
}
return R;
}
void testRomberg(){
double **p=Romberg(0, 3.1415926, 6,functionRomberg);
for (int i = 0; i < 6; i++){
for (int j = 0; j < i + 1; j++){
cout << "R[" << i << "][" << j << "]:" << setprecision(8)<<p[i][j] << " ";
}
cout << endl;
}
}
double functionRomberg(double x){
return sin(x);
}
double AdaptiveIntegration(double a, double b, double tol, int N, double(*function)(double)){
double app = 0;
int i = 1;
double toli = 10 * tol;
double ai = a;
double hi = (b - a) / 2;
double FAI = function(a);
double FCI = function(a + hi);
double FBI = function(b);
//对整个区间的Simpson方法近似
double SI = hi*(FAI + 4 * FCI + FBI) / 3;
double LI = 1;
while (i > 0){
double FD = function(ai + hi / 2);
double FE = function(ai + 3 * hi / 2);
//对子区间一半的Simpson方法近似
double S1 = hi*(FAI + 4 * FD + FBI) / 6;
double S2 = hi*(FCI + 4 * FE + FBI) / 6;
double v1 = ai;
double v2 = FAI;
double v3 = FCI;
double v4 = FBI;
double v5 = hi;
double v6 = toli;
double v7 = SI;
double v8 = LI;
//删除层次
i = i - 1;
if (fabs(S1 + S2 - v7) < v6){
cout << "达到精度" << endl;
app = app + (S1 + S2);
}
else{
if (v8 >= N){
cout << "算法失败！" << endl;
return -1;
}
else{
cout << "继续细分" << endl;
//增加一个层次
//右半子区间的数据
i = i + 1;
ai = v1 + v5;
FAI = v3;
FCI = FE;
FBI = v4;
hi = v5 / 2;
toli = v6 / 2;
SI = S2;
LI = v8 + 1;
//左子区间的数据
i = i + 1;
ai = v1;
FAI = v2;
FCI = FD;
FBI = v3;
hi = hi/2;
toli = toli/2;
SI = S1;
LI = LI;
}
}
}
return app;
}
void testAdaptiveIntegration(){
cout << "积分值：" << AdaptiveIntegration(0, 3.1415926, 0.0001, 5, functionRomberg);
}
double DoubleIntegration(double a, double b, double c, double d, int m, int n,
double(*function)(double, double)){
double h = (b - a) / n;
double J1 = 0; //首选项
double J2 = 0;
double J3 = 0;
for (int i = 0; i < n+1; i++){
double x = a + i*h;
double HX = (d - c) / m;
double K1 = function(x, c) + function(x, d);
double K2 = 0;
double K3 = 0;
for (int j = 1; j < m; j++){
double y = c + j*HX;
double Q = function(x, y);
if (j % 2 == 0){
K2 = K2 + Q;
}
else{
K3 = K3 + Q;
}
}
double L = (K1 + 2 * K2 + 4 * K3)*HX / 3;
if (i == 0 || i == n){
J1 = J1 + L;
}
else if (i % 2 == 0){
J2 = J2 + L;
}
else{
J3 = J3 + L;
}
}
double J = h*(J1 + 2 * J2 + 4 * J3) / 3;
return J;
}
double functionDoubleIntegration(double x, double y){
return log(x + 2 * y);
}
void testDoubleIntegration(){
double result = DoubleIntegration(1.4, 2.0, 1.0, 1.5, 4, 4, functionDoubleIntegration);
cout << "二重积分值：" << setprecision(16)<<result << endl;
}
//m为4,n为4
double GaussDoubleIntegration(double a, double b, double c, double d,int m,int n,
double(*function)(double,double)){
double h1 = (b - a) / 2;
double h2 = (b + a) / 2;
double J = 0;
for (int i = 0; i < m ; i++){
double JX = 0;
double x = h1 * r4[i] + h2;
double d1 = d;
double c1 = c;
double k1 = (d1 - c1) / 2;
double k2 = (d1 + c1) / 2;
for (int j = 0; j < n; j++){
double y = k1 * r4[j] + k2;
double Q = function(x, y);
JX = JX + c4[j] * Q;
}
J = J + c4[i] * k1*JX;
}
J = h1*J;
return J;
}
void testGaussDoubleIntegration(){
double result = GaussDoubleIntegration(1.4, 2.0, 1.0, 1.5, 4,4,functionDoubleIntegration);
cout << "高斯二重积分求值：" << setprecision(16)<<result << endl;
}
double GaussTrebleIntegration(double a, double b, double c, double d, double e, double f,
int m, int n, int p, double(*function)(double, double, double)){
double h1 = (b - a) / 2;
double h2 = (b + a) / 2;
double J = 0;
for (int i = 0; i < m ; i++){
double JX = 0;
double x = h1 * r4[i] + h2;
double d1 = d;
double c1 = c;
double k1 = (d1 - c1) / 2;
double k2 = (d1 + c1) / 2;
for (int j = 0; j < n ; j++){
double JY = 0;
double y = k1 * r4[j] + k2;
double e1 = e;
double f1 = f;
double l1 = (f1 - e1) / 2;
double l2 = (f1 + e1) / 2;
for (int k = 0; k < p; k++){
double z = l1 * r4[k] + l2;
double Q = function(x, y, z);
JY = JY + c4[k] * Q;
}
JX = JX + c4[j] * JY;
}
J = J + c4[i] * k1*JX;
}
J = h1*J;
return J;
}