Question

如何计算在无穷远处涉及指数和奇点的函数的数值二阶导数。不幸的是，Ridder方法中提供的数值导数“C中的数值配方”只能计算出一阶导数（它需要预先对函数进行解析表达式。）此外，我尝试了Chebyshev近似并对函数进行了区分，但给出的值是方式关闭实际值。我也尝试过在数学论文中提供的一些有限差分算法，但它们也容易出错。函数是e ^（x / 2）/ x ^ 2。对此事我感激不尽。

提前致谢

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获得

编辑：

// The compilation command used is given below
// gcc Q3.c nrutil.c DFRIDR.c -lm -o Q3

#include <stdio.h>
#include <math.h>
#include "nr.h"

#define LIM1 20.0
#define a -5.0
#define b 5.0
#define pre 100.0 // This defines the pre
/* This file calculates the func at given points, makes a 
 * plot. It also calculates the maximum and minimum of the func
 * at given points and its first and second numerical derivative.
*/
float func(float x)
{
    return exp(x / 2) / pow(x, 2);
}

int main(void)
{
    FILE *fp = fopen("Q3data.dat", "w+"), *fp2 = fopen("Q3results.dat", "w+");
    int i; // Declaring our loop variable
    float x, y, min, max, err, nd1, nd2;
    // Define the initial value of the func to be the minimum
    min = func(0); 
    for(i = 0; x < LIM1 ; i++)
    {   
        x = i / pre; // There is a singularity at x = 0 
        y = func(x);
        if(y < min)
            min = y;
        fprintf(fp, "%f \t %f \n", x, y);
    }
    fprintf(fp, "\n\n");
    max = 0;
    for(i = 0, x = a; x < b; i++)
    {   
        x = a + i / pre;
        y = func(x);
        nd1 = dfridr(func, x, 0.1, &err); 
        //nd2 = dfridr((*func), x, 0.1, &err);
        fprintf(fp, "%f \t %f \t %f \t %f \n", x, y, nd1);
        if(y > max)
            max = y;
    }

    fprintf(fp2, "The minimum value of f(x) is %f when x is between 0 and 20. \n", min);
    fprintf(fp2, "The maximum value of f(x) is %f when x is between -5 and 5. \n", max);
    fclose(fp);
    fclose(fp2);
    return 0;
}

编辑：切比雪夫

// The compilation command used is given below
//gcc Q3.c nrutil.c CHEBEV.c CHEBFT.c CHDER.c -lm -o Q3


#include <stdio.h>
#include <math.h>
#include "nr.h"
#define NVAL 150 // Degree of Chebyshev polynomial
#define LIM1 20.0
#define a -5.0
#define b 5.0
#define pre 100.0 // This defines the pre
/* This file calculates the func at given points, makes a 
 * plot. It also calculates the maximum and minimum of the func
 * at given points and its first and second numerical derivative.
*/
float func(float x)
{
    return exp(x / 2) / pow(x, 2);
}

int main(void)
{
    FILE *fp = fopen("Q3data.dat", "w+"), *fp2 = fopen("Q3results.dat", "w+");
    int i; // Declaring our loop variable
    float x, y, min, max;
    float nd1, nd2, c[NVAL], cder[NVAL], cder2[NVAL];
    // Define the initial value of the func to be the minimum
    min = func(0); 

    for(i = 0; x < LIM1 ; i++)
    {   
        x = i / pre; // There is a singularity at x = 0 
        y = func(x);
        if(y < min)
            min = y;
        fprintf(fp, "%f \t %f \n", x, y);
    }
    fprintf(fp, "\n\n");
    max = 0;
    // We make a Chebyshev approximation to our function our interval of interest 
    // The purpose is to calculate the derivatives easily
    chebft(a,b,c,NVAL,func);
    //Evaluate the derivatives
    chder(a,b,c,cder,NVAL); // First order derivative
    chder(a,b,cder,cder2,NVAL); // Second order derivative
    for(i = 0, x = a; x < b; i++)
    {   
        x = a + i / pre;
        y = func(x);
        nd1 = chebev(a,b,cder,NVAL,x);
        nd2 = chebev(a,b,cder2,NVAL,x);

        fprintf(fp, "%f \t %f \t %f \t %f  \n", x, y, nd1, nd2);
        if(y > max)
            max = y;
    }

    fprintf(fp2, "The minimum value of f(x) is %f when x is between 0 and 20. \n", min);
    fprintf(fp2, "The maximum value of f(x) is %f when x is between -5 and 5. \n", max);
    fclose(fp);
    fclose(fp2);
    return 0;
}

Answer 1

该功能是可区分的，因此使用数值方法可能不是最好的。二阶导数是：

6 * exp（x / 2）/（x ^ 4）-2 * exp（x / 2）/ x ^ 3 + exp（x / 2）/（4 * x ^ 2）

以上可以简化，以加快计算速度。编辑：第一次有原始公式错误。

Answer 2

如果您想要100％数字方法，请查看numerical recipes以进行公共样条插值（Charter 3.3）。它会在任何位置为您提供第二个衍生物。

使用spline()和x值调用y以返回y2中的第二个导数。二阶导数在每个区间内线性变化。所以，例如，如果你有

x      y      y2
0     10     -30
2      5     -15
4     -5     -10

然后x=1的二阶导数为y2=-22.5，介于-30和-15之间。

您还可以创建一个新的splint()函数来返回二阶导数a*y2a[i]+b*y2a[i+1]

数值微分

2 个答案: