在某些步骤间隔内近似连续函数的导数

Question

我希望用Java编写一个方法,找到连续函数的导数.这些是对该方法做出的一些假设 -

1. 该函数从x = 0到x =无穷大是连续的.
2. 衍生物存在于每个区间.
3. 步长需要定义为参数.
4. 该方法将在给定间隔[a:b]中找到连续函数的最大值/最小值.

例如,函数cos(x)可以显示为0,pi,2pi,3pi,... npi的最大值或最小值.

我正在寻找一个方法,它将找到所有这些最大值或最小值,只要给出函数,lowerBound,upperBound和步长.

为了简化我的测试代码,我为cos(x)编写了一个程序.我使用的函数与cos(x)非常相似(至少在图形上).这是我写的一些测试代码 -

public class Test {
    public static void main(String[] args){
        Function cos = new Function () 
        {
        public double f(double x) {
        return Math.cos(x);
        }
    };

        findDerivative(cos, 1, 100, 0.01);      
    }

    // Needed as a reference for the interpolation function.
    public static interface Function {
    public double f(double x);
    }

     private static int sign(double x) {
    if (x < 0.0)
            return -1;
        else if (x > 0.0)
            return 1;
        else
            return 0;
    }

     // Finds the roots of the specified function passed in with a lower bound,
    // upper bound, and step size.
    public static void findRoots(Function f, double lowerBound,
                  double upperBound, double step) {
    double x = lowerBound, next_x = x;
    double y = f.f(x), next_y = y;
    int s = sign(y), next_s = s;

    for (x = lowerBound; x <= upperBound ; x += step) {
        s = sign(y = f.f(x));
        if (s == 0) {
        System.out.println(x);
        } else if (s != next_s) {
        double dx = x - next_x;
        double dy = y - next_y;
        double cx = x - dx * (y / dy);
        System.out.println(cx);
        }
        next_x = x; next_y = y; next_s = s;
    }
    }

    public static void findDerivative(Function f, double lowerBound, double 
            upperBound, double step) {
    double x = lowerBound, next_x = x;
    double dy = (f.f(x+step) - f.f(x)) / step;

    for (x = lowerBound; x <= upperBound; x += step) {
        double dx = x - next_x;
        dy = (f.f(x+step) - f.f(x)) / step;
        if (dy < 0.01 && dy > -0.01) {
            System.out.println("The x value is " + x + ". The value of the "
                    + "derivative is "+ dy);
            }
        next_x = x;
        }
    }   
}

查找根的方法用于查找零(这肯定有效).我只将它包含在我的测试程序中,因为我认为我可以在找到衍生物的方法中使用类似的逻辑.

方法

public static void findDerivative(Function f, double lowerBound, double 
            upperBound, double step) {
    double x = lowerBound, next_x = x;
    double dy = (f.f(x+step) - f.f(x)) / step;

    for (x = lowerBound; x <= upperBound; x += step) {
        double dx = x - next_x;
        dy = (f.f(x+step) - f.f(x)) / step;
        if (dy < 0.01 && dy > -0.01) {
            System.out.println("The x value is " + x + ". The value of the "
                    + "derivative is "+ dy);
            }
        next_x = x;
        }
    }   

绝对可以改进.我怎么能以不同的方式写这个？这是样本输出.

The x value is 3.129999999999977. The value of the derivative is -0.006592578364594814
The x value is 3.1399999999999766. The value of the derivative is 0.0034073256197308943
The x value is 6.26999999999991. The value of the derivative is 0.008185181673381337
The x value is 6.27999999999991. The value of the derivative is -0.0018146842631128202
The x value is 9.409999999999844. The value of the derivative is -0.009777764220086915
The x value is 9.419999999999844. The value of the derivative is 2.2203830347677922E-4
The x value is 12.559999999999777. The value of the derivative is 0.0013706082193754021
The x value is 12.569999999999776. The value of the derivative is -0.00862924258597797
The x value is 15.69999999999971. The value of the derivative is -0.002963251265619693
The x value is 15.70999999999971. The value of the derivative is 0.007036644660118885
The x value is 18.840000000000146. The value of the derivative is 0.004555886794943564
The x value is 18.850000000000147. The value of the derivative is -0.005444028885981389
The x value is 21.980000000000636. The value of the derivative is -0.006148510767989279
The x value is 21.990000000000638. The value of the derivative is 0.0038513993028788107
The x value is 25.120000000001127. The value of the derivative is 0.0077411191450771355
The x value is 25.13000000000113. The value of the derivative is -0.0022587599505241585

Answer 1

在 f 计算成本昂贵的情况下，我可以看到提高性能的主要方法是，您可以保存 f(x) 的先前值，而不是每次迭代计算两次。而且 dx 从未被使用过，并且无论如何总是等于步长。next_x也从未使用过。可以在循环内声明一些变量。将变量声明移到内部可以提高可读性，但不能提高性能。

public static void findDerivative(Function f, double lowerBound, double upperBound, double step) {
    double fxstep = f.f(x);

    for (double x = lowerBound; x <= upperBound; x += step) {
        double fx = fxstep;
        fxstep = f.f(x+step);
        double dy = (fxstep - fx) / step;
        if (dy < 0.01 && dy > -0.01) {
            System.out.println("The x value is " + x + ". The value of the "
                    + "derivative is " + dy);
        }
    }
}