我以前使用Simpsons复合规则创建了此代码来评估积分,现在需要对其进行修改以评估双积分。有谁知道如何去做这件事?我将不胜感激任何帮助..
#include <iostream>
#include <cmath>
using namespace std;
float f(float x); //returns the function we are integrating
void simpsons_rule(float a, float b, int n); // Carries out Simpsons composite rule
int main()
{
cout << "This program finds an approximation of the integral of a function under two limits using Simpsons composite rule." << endl;
// User is introduced to program and given explanation of what it does
float a,b;
int n;
cout << "Enter the lower limit " << endl;
cin >> a;
cout << "Enter the upper limit " << endl;
cin >> b; // User has freedom to specify the upper and lower limits under which the function will be integrated
cout << "Enter the number of intervals (must be an even number) " << endl;
do
{
cin >> n; // User can also choose the number of intervals
if(n%2 == 0 && n!=0) // If an even number of intervals was entered, the program continues and simpsons rule is carried out
{
simpsons_rule(a,b,n);
}
else
{
cout << "Invalid number of intervals, please re-enter a value" << endl; //If an odd number of intervals was entered, the user is prompted to enter a valid number
}
}while(cin); // do-while loop used to ensure even number of intervals was entered
return 0;
}
float f(float x)
{
return(pow(x,4) - 3*pow(x,3) + pow(x,2) + x + 1); // function to be integrated
}
void simpsons_rule(float a, float b, int n)
{
float s2 = 0;
float s3 = 0;
int i;
float h = (b-a)/n; // step length is calculated
for(i=1; i <= (n/2)-1; i++)
{
if((i%2) == 0) // checks if i is even
{
s2 = s2 + f(a+(i*h)); // Finds the sum of all the values of f(x) where x is even
}
}
for(i=1; i<=(n/2); i++)
{
if((i%2) == 1) // checks if i is odd
{
s3 = s3 + f(a+2*(i*h)); // Finds the sum of all the values of f(x) where x is odd
}
}
float s = (h/3)*(f(a)+ f(b) + (2*s2) + (4*s3)); // This variable contains the final approximaton of the integral
cout << "The value of the integral under the specified limits is: " << s << endl;
答案 0 :(得分:3)
不幸的是,辛普森的规则无法直接应用于多个积分。您需要做的是分别为双重或三重积分导出插值曲面或超曲面。对于双积分,您最终会在九个点的网格上评估函数,而不是在单个积分情况下使用的三个点。对于三重积分,您使用27点的3-D点阵。不用说,它变得非常复杂。
更简单的方法是某种Monte Carlo integration,其中您可以多次随机抽样函数,取所有函数样本的平均值,然后乘以积分区域。这里的缺点是误差与样本数的平方根成反比,因此4倍的样本只会使预期误差减半。如果你有足够的时间并且你的准确度要求不高,这可能是你应该尝试的方法。