我正在写一个小型的微积分库供个人使用。它是标准的微积分工具 - 取一阶导数，n阶导数，Riemann和等。我遇到的一个问题是n阶导数函数对某些h值（有限差分）显然返回虚假结果。
Code here及以下：
typedef double(*math_func)(double x);
inline double max ( double a, double b ) { return a > b ? a : b; }
//Uses the five-point-stencil algorithm.
double derivative(math_func f,double x){
double h=max(sqrt(DBL_EPSILON)*x,1e-8);
return ((-f(x+2*h)+8*f(x+h)-8*f(x-h)+f(x-2*h))/(12*h));
}
#define NDEPS (value)
double nthDerivative(math_func f, double x, int N){
if(N<0) return NAN; //bogus value of N
if(N==0) return f(x);
if(N==1) return derivative(f,x);
double* vals=calloc(2*N+9,sizeof(double)); //buffer region around the real values
if(vals==NULL){ //oops! no memory
return NAN;
}
int i,j;
//don't take too small a finite difference
double h=max(sqrt(DBL_EPSILON)*x,NDEPS);
for(i=-(N+4);i<=N+4;i++){
vals[i+N+4]=derivative(f,x+h*i);
}
//for(i=0;i<2*N+9;i++){printf("%.1e ",vals[i]);}putchar('\n');
for(j=1;j<N;j++){
double *vals2=calloc(2*N+9,sizeof(double));
for(i=2;i<2*N+7;i++){
vals2[i]=(-vals[i+2]+8*vals[i+1]-8*vals[i-1]+vals[i-2])/(12*h);
}
free(vals);
vals=vals2;
//for(i=0;i<2*N+9;i++){printf("%.1e ",vals[i]);}putchar('\n');
}
double result=vals[N+4];
free(vals);
return result;
}
我给测试此函数的一个示例问题是当x = pi时sin（x）的五阶导数。众所周知，答案是-1。当我试图改变NDEPS（“Nth derivative epsilon”）的值时会出现问题。
为什么会这样？它与sin（）函数的性质有关吗？还是因为浮点精度问题？
答案 0 :(得分：6)
数值微分是一个难题。关键问题是有限差分近似
f'(x) =(approx) (f(x+h)-f(x-h)) / (2*h)
是cancellation的配方。
这意味着您必须在两个错误之间选择谨慎的权衡：如果h太大，您将在有限差分中产生大的近似误差。如果h太小，则会产生大的（可能是灾难性的）数值误差。这篇关于numerical differentiation的维基百科文章很好地说明了这一点。随着衍生物的顺序增加，这些问题会加剧。
这就是为什么automatic differentiation是如此重要的原因 - 实质上，当所有给出的算法都是计算基函数的算法时，它允许你计算精确的导数。如果函数足够简单，你可以象征性地确定导数，那么你应该这样做。
如果你确实需要进行数值微分，那么你可以应用很多数字技巧。 Numerical Recipes有一个很好的处理方法 - 看一下第186页开始的第5.7节。
答案 1 :(得分：1)
以下是您的代码的改编。除了使用更多空格之外，主要修改是将NDEPS作为nthDerivative()函数的参数，以便可以使用不同的值调用它并添加大量打印。我还必须在普通的derivative()函数上发挥创造力;代码编译正确（但我真的不是试图用断言assert(sin == fun);制作哲学或神学语句，但它确实意味着代码编译时没有警告，并且它识别出这个派生函数的局限性）。
#include <assert.h>
#include <float.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define PRIe_double "%21.15e"
typedef double(*math_func)(double x);
static double derivative(math_func fun, double x) { assert(sin == fun); return cos(x); }
static double nthDerivative(math_func f, double x, int N, double NDEPS)
{
if (N < 0) return NAN; //bogus value of N
if (N == 0) return f(x);
if (N == 1) return derivative(f, x);
double* vals = calloc(2*N+9, sizeof(double)); //buffer region around the real values
if (vals == NULL) //oops! no memory
return NAN;
int i, j;
//don't take too small a finite difference
double h = max(sqrt(DBL_EPSILON)*x, NDEPS);
printf("h = " PRIe_double "\n", h);
for (i = -(N+4); i <= N+4; i++)
{
vals[i+N+4] = derivative(f, x+h*i);
printf("%2d: deriv(" PRIe_double ") = " PRIe_double "\n", i, x+h*i, vals[i+N+4]);
}
for (j = 1; j < N; j++)
{
printf("Iteration %d\n", j);
double *vals2 = calloc(2*N+9, sizeof(double));
for (i = 2; i < 2*N+7; i++){
vals2[i] = (-vals[i+2] + 8*vals[i+1] - 8*vals[i-1] + vals[i-2]) / (12*h);
}
free(vals);
vals = vals2;
for (i = 0; i < 2*N+9; i++)
printf("%2d: " PRIe_double "\n", i, vals[i]);
}
double result = vals[N+4];
free(vals);
return result;
}
int main(void)
{
double val = M_PI;
double eps;
double r;
eps = 1.0 / 64.0;
r = nthDerivative(sin, val, 5, eps);
printf("5th Derivative of sin(x) at x = " PRIe_double " = " PRIe_double " (eps = %f)\n", val, r, eps);
eps = 1.0 / 100000.0;
r = nthDerivative(sin, val, 5, eps);
printf("5th Derivative of sin(x) at x = " PRIe_double " = " PRIe_double " (eps = %f)\n", val, r, eps);
return(0);
}
在Mac OS X 10.7.5上使用GCC 4.7.1，输出为：
h = 1.562500000000000e-02
-9: deriv(3.000967653589793e+00) = -9.901285883701071e-01
-8: deriv(3.016592653589793e+00) = -9.921976672293290e-01
-7: deriv(3.032217653589793e+00) = -9.940245152582091e-01
-6: deriv(3.047842653589793e+00) = -9.956086864580017e-01
-5: deriv(3.063467653589793e+00) = -9.969497940760287e-01
-4: deriv(3.079092653589793e+00) = -9.980475107000991e-01
-3: deriv(3.094717653589793e+00) = -9.989015683384429e-01
-2: deriv(3.110342653589793e+00) = -9.995117584851364e-01
-1: deriv(3.125967653589793e+00) = -9.998779321710066e-01
0: deriv(3.141592653589793e+00) = -1.000000000000000e+00
1: deriv(3.157217653589793e+00) = -9.998779321710066e-01
2: deriv(3.172842653589793e+00) = -9.995117584851364e-01
3: deriv(3.188467653589793e+00) = -9.989015683384429e-01
4: deriv(3.204092653589793e+00) = -9.980475107000991e-01
5: deriv(3.219717653589793e+00) = -9.969497940760287e-01
6: deriv(3.235342653589793e+00) = -9.956086864580017e-01
7: deriv(3.250967653589793e+00) = -9.940245152582091e-01
8: deriv(3.266592653589793e+00) = -9.921976672293291e-01
9: deriv(3.282217653589793e+00) = -9.901285883701071e-01
Iteration 1
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: -1.091570566584531e-01
3: -9.361273104952932e-02
4: -7.804555123490846e-02
5: -6.245931771829009e-02
6: -4.685783565504131e-02
7: -3.124491392324735e-02
8: -1.562436419383969e-02
9: -1.776356839400250e-15
10: 1.562436419384087e-02
11: 3.124491392324558e-02
12: 4.685783565504250e-02
13: 6.245931771828653e-02
14: 7.804555123490846e-02
15: 9.361273104953050e-02
16: 1.091570566584471e-01
17: 0.000000000000000e+00
18: 0.000000000000000e+00
Iteration 2
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: -3.577900251527073e+00
3: 1.660540592568783e+00
4: 9.969497901146779e-01
5: 9.980475067341610e-01
6: 9.989015643694564e-01
7: 9.995117545137316e-01
8: 9.998779281977730e-01
9: 9.999999960264082e-01
10: 9.998779281978486e-01
11: 9.995117545137319e-01
12: 9.989015643693869e-01
13: 9.980475067340947e-01
14: 9.969497901149178e-01
15: 1.660540592568512e+00
16: -3.577900251527123e+00
17: 0.000000000000000e+00
18: 0.000000000000000e+00
Iteration 3
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: 6.553266640232313e+01
3: 1.898706817407991e+02
4: -5.267598134705870e+01
5: 3.608762837830827e+00
6: 4.685783548517186e-02
7: 3.124491378285654e-02
8: 1.562436412277357e-02
9: 3.226456139297321e-12
10: -1.562436412279785e-02
11: -3.124491378869069e-02
12: -4.685783548888563e-02
13: -3.608762837816180e+00
14: 5.267598134704988e+01
15: -1.898706817408119e+02
16: -6.553266640231030e+01
17: 0.000000000000000e+00
18: 0.000000000000000e+00
Iteration 4
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: 8.382087654791743e+03
3: -5.062815705775390e+03
4: -7.597917560836845e+03
5: 3.261984801532626e+03
6: -4.336626618856812e+02
7: 1.791410702361821e+01
8: -9.998779233550453e-01
9: -9.999999914294619e-01
10: -9.998779238596159e-01
11: 1.791410702341709e+01
12: -4.336626618847606e+02
13: 3.261984801532444e+03
14: -7.597917560838102e+03
15: -5.062815705774387e+03
16: 8.382087654792240e+03
17: 0.000000000000000e+00
18: 0.000000000000000e+00
5th Derivative of sin(x) at x = 3.141592653589793e+00 = -9.999999914294619e-01 (eps = 0.015625)
h = 1.000000000000000e-05
-9: deriv(3.141502653589793e+00) = -9.999999959500000e-01
-8: deriv(3.141512653589793e+00) = -9.999999968000000e-01
-7: deriv(3.141522653589793e+00) = -9.999999975500000e-01
-6: deriv(3.141532653589793e+00) = -9.999999982000000e-01
-5: deriv(3.141542653589793e+00) = -9.999999987500000e-01
-4: deriv(3.141552653589793e+00) = -9.999999992000000e-01
-3: deriv(3.141562653589793e+00) = -9.999999995500000e-01
-2: deriv(3.141572653589793e+00) = -9.999999998000000e-01
-1: deriv(3.141582653589793e+00) = -9.999999999500000e-01
0: deriv(3.141592653589793e+00) = -1.000000000000000e+00
1: deriv(3.141602653589793e+00) = -9.999999999500000e-01
2: deriv(3.141612653589793e+00) = -9.999999998000000e-01
3: deriv(3.141622653589793e+00) = -9.999999995500000e-01
4: deriv(3.141632653589793e+00) = -9.999999992000000e-01
5: deriv(3.141642653589793e+00) = -9.999999987500000e-01
6: deriv(3.141652653589793e+00) = -9.999999982000000e-01
7: deriv(3.141662653589793e+00) = -9.999999975500000e-01
8: deriv(3.141672653589793e+00) = -9.999999968000000e-01
9: deriv(3.141682653589793e+00) = -9.999999959500000e-01
Iteration 1
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: -7.000000116589669e-05
3: -5.999999941330713e-05
4: -5.000000598739025e-05
5: -3.999999683331386e-05
6: -2.999999600591015e-05
7: -2.000000257999327e-05
8: -1.000000082740371e-05
9: 0.000000000000000e+00
10: 1.000000082740371e-05
11: 2.000000165480742e-05
12: 2.999999508072429e-05
13: 3.999999590812801e-05
14: 5.000000413701854e-05
15: 5.999999663774957e-05
16: 6.999999746515327e-05
17: 0.000000000000000e+00
18: 0.000000000000000e+00
Iteration 2
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: -3.583333244325556e+00
3: 1.666666318844711e+00
4: 1.000000128999663e+00
5: 1.000000691821058e+00
6: 9.999995738881511e-01
7: 9.999997049561470e-01
8: 1.000000198388602e+00
9: 1.000000075030488e+00
10: 1.000000144419428e+00
11: 9.999996509869722e-01
12: 9.999995893079155e-01
13: 1.000000645561765e+00
14: 1.000000028771196e+00
15: 1.666666187776715e+00
16: -3.583333074708150e+00
17: 0.000000000000000e+00
18: 0.000000000000000e+00
Iteration 3
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: 1.027777535146502e+05
3: 2.972222191231725e+05
4: -8.263881528669108e+04
5: 5.555518108303881e+03
6: -6.636923522614542e-02
7: 4.677328483600658e-02
8: 1.991719546327412e-02
9: -3.148201866790915e-03
10: -2.319389535987426e-02
11: -4.176186143567406e-02
12: 6.726872146719150e-02
13: -5.555525175695851e+03
14: 8.263880834779718e+04
15: -2.972222015189417e+05
16: -1.027777456120210e+05
17: 0.000000000000000e+00
18: 0.000000000000000e+00
疯狂的迹象......
Iteration 4
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: 2.050347140226725e+10
3: -1.240740057099195e+10
4: -1.858796490195886e+10
5: 7.986101364079453e+09
6: -1.059021715700324e+09
7: 4.630176289959386e+07
8: -3.687893612405425e+03
9: -2.136279835945886e+03
10: -2.968840021291520e+03
11: 4.630204773690500e+07
12: -1.059022490436399e+09
13: 7.986100736580715e+09
14: -1.858796331554353e+10
15: -1.240739964045201e+10
16: 2.050347017082775e+10
17: 0.000000000000000e+00
18: 0.000000000000000e+00
5th Derivative of sin(x) at x = 3.141592653589793e+00 = -2.136279835945886e+03 (eps = 0.000010)
注意结果如何在最后一次迭代中使用数十亿的值。您至少有数值稳定性问题，或者您可能需要查看衍生公式。请注意，即使是较大的epsilon运行，也会在以后的迭代中出现较大的值。
derivative()功能将问题中现在存在的导数函数插入到上面的代码中会在第4次迭代中产生更不稳定的答案：
Iteration 4
0: 0.000000000000000e+00
1: 0.000000000000000e+00
2: -6.248925477935563e+09
3: 1.729549900845405e+11
4: -2.600544559219368e+11
5: -2.755286326619338e+11
6: 8.100546069731433e+11
7: -2.961111189495415e+09
8: -7.936480686806423e+11
9: 2.430177384467434e+11
10: 2.389084910067162e+11
11: -6.461168564124718e+10
12: -4.574822745530297e+10
13: -8.923883451146609e+10
14: 7.042030613792160e+10
15: 4.988386306820556e+10
16: -1.793262395787471e+10
17: 0.000000000000000e+00
18: 0.000000000000000e+00
5th Derivative of sin(x) at x = 3.141592653589793e+00 = 2.430177384467434e+11 (eps = 0.000010)
我想知道数组索引0,1,17和18处零的出现在多大程度上加剧了这个问题。
答案 2 :(得分：1)
这是一个棘手的主题。数值微分是病态的，并导致取消和舍入错误。在您重复区分的情况下，问题会明显放大。
通过使用m个点确定特定微分阶数N的系数，而不是重复区分，您将获得更好的高阶导数估计。
例如，正如您似乎使用一阶导数的系数：
{1, -8, 8, -1} / (12*h)
使用5个点逼近二阶导数的系数是：
{-1, 16, -30, 16, -1} / (12*h²)
您可以通过围绕采样点进行泰勒级数展开来求解一般问题的系数，找到这些扩展的线性组合，从而得到所需的导数。
对于m个点{x + n [1] * h，x + n [2] * h，x + n [3] * h，...，x + n [m] * h}，系数（ k）估计N阶导数可以通过以下方程组计算：
M*k=b
其中M是m * m矩阵：
M[i,j] = n[j]^(i-1), i,j = 1..m
和
b = [0, 0, 0, …, N!/h^N, …, 0, 0, 0],
其中N！表示N的阶乘。