在特定步长间隔内近似连续函数的导数
2022-09-04 23:25:48
我希望用Java编写一个方法,为连续函数找到一个导数。这些是为该方法所做的一些假设 -
- 该函数是从 x = 0 到 x = 无穷大的连续函数。
- 导数在每个区间都存在。
- 需要将步长定义为参数。
- 该方法将找到给定区间内连续函数的最大值/最小值 [a:b]。
例如,函数cos(x)可以显示为在0,pi,2pi,3pi,...新产品。
我希望编写一个方法,该方法将找到所有这些最大值或最小值,前提是给定一个函数,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)。以下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(这绝对有效)。我只把它包含在我的测试程序中,因为我认为我可以以某种方式在查找导数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?下面是示例输出。
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