为了整合形式的二维函数
cmdkey
我一直在尝试使用以下代码(用C ++编写),这些代码主要来自于Numerical Recipes一书,该书调用了高斯求积法来进行积分:
cmdkey /list |
Select-String -Pattern $ComputerName -Context 2 |
Select-Object -Property @(
@{N='ComputerName';E={$ComputerName}}
@{N='User';E={$PSItem.Context.PostContext[-1] -replace '\s*User:\s*'}}
)
此代码适用于具有有限限制的二维积分,但由于外部限制被视为无穷大而失败。为了解释这一点,我试图使用变量的变化:
$$\int_{1}^\infty \int_{-\sqrt{x^2-1}}^{\sqrt{x^2-1}} e^{-x} \rm{d}y \rm{d}x,$$
但这仍然没有给出正确的答案。它应该是
static float xsav;
static float (*nrfunc)(float,float);
float quad2d(float (*func)(float, float), float x1, float x2)
{
float qgaus(float (*func)(float), float a, float b);
float f1(float x);
nrfunc=func;
return qgaus(f1,x1,x2);
}
float f1(float x)
{
float qgaus(float (*func)(float), float a, float b);
float f2(float y);
float yy1(float),yy2(float);
xsav=x;
return qgaus(f2,yy1(x),yy2(x));
}
float f2(float y)
{
return (*nrfunc)(xsav,y);
}
但是给出了
#define FUNC(x) ((*funk)(-log(x))/(x))
float qgaus(float (*funk)(float), float aa, float bb)
{
int j;
float xr,xm,dx,s,a,b;
b=exp(-aa);
a=0.0;
static float x[]={0.0,0.1488743389,0.4333953941,
0.6794095682,0.8650633666,0.9739065285};
static float w[]={0.0,0.2955242247,0.2692667193,
0.2190863625,0.1494513491,0.0666713443};
xm=0.5*(b+a);
xr=0.5*(b-a);
s=0;
for (j=1;j<=5;j++)
{
dx=xr*x[j];
s += w[j]*(FUNC(xm+dx)+FUNC(xm-dx));
}
return s *= xr;
}
float f(float x, float y)
{
float a = exp(-x);
return a;
}
float yy1(float x)
{
float y = -sqrt(x*x-1);
return y;
}
float yy2(float x)
{
float y = sqrt(x*x-1);
return y;
}
static float xsav;
static float (*nrfunc)(float, float);
float quad2d(float (*func)(float, float), float x1, float x2)
{
float qgaus(float (*func)(float), float aa, float bb);
float f1(float x);
nrfunc=func;
float t = qgaus(f1,x1,x2);
return t;
}
float f1(float x)
{
float qgaus(float (*func)(float), float aa, float bb);
float f2(float y);
float yy1(float);
float yy2(float);
xsav=x;
float r = qgaus(f2,yy1(x),yy2(x));
return r;
}
float f2(float y)
{
float k = (*nrfunc)(xsav,y);
return k;
}
int main ()
{
float z;
z = quad2d(f, 1.0, 20.0);
cout << z << endl;
}
有关如何修改此代码以解释无限限制的任何建议将非常感谢!