Question

我目前正在找到原始数据的三次样条：

    double y[] = { 0.0, 1.0, 2.0,3.0,4.0,5.0 };
    double x[] = { 0.0, 1.0, 8.0, 27.0, 64, 125};
    //these arrays are just an example

    UnivariateInterpolator interpolator = new SplineInterpolator();
    UnivariateFunction spline = interpolator.interpolate(y, x);
    double interpolatedY = spline.value(5.0);

我在程序的另一部分中使用了interpolatedY变量。我还需要找到UnivariateFuction样条的导数，以便可以计算y'（x），其中x可以是任何值。我尝试使用以下方法：

  // function to be differentiated
    UnivariateFunction basicF = new UnivariateFunction() {
                                    public double value(double x) {
                                        return spline.value(x);
                                    }
                                };

// create a differentiator using 5 points and 0.01 step
 FiniteDifferencesDifferentiator differentiator =
         new FiniteDifferencesDifferentiator(51, 0.01);

 UnivariateDifferentiableFunction completeF = differentiator.differentiate(basicF);

System.out.println("y'(x=2) = " + completeF.value(new DerivativeStructure(1,1,0,2).getPartialDerivative(1)));

但是，这看起来有些冗长，但却无法返回正确的答案。有什么建议吗？

谢谢

Answer 1

interpolator.interpolate(y, x)返回扩展了以下接口的PolynomialSplineFunction：

DifferentiableUnivariateFunction (deprecated)

UnivariateDifferentiableFunction

UnivariateFunction

您可以做的一件事是，可以像下面这样使用UnivariateFunction和UnivariateInterpolator来代替UnivariateDifferentiableFunction和SplineInterpolator：

SplineInterpolator interpolator = new SplineInterpolator();
UnivariateDifferentiableFunction spline = interpolator.interpolate(y, x);
double interpolatedY = spline.value(5.0);

DerivativeStructure ds = spline.value(new DerivativeStructure(1, 1, 0, 2));
System.out.println(ds.getPartialDerivative(1));

输出

// Function to interpolate f(x) = x^3
// First order derivative  f'(x) = 3x^2
// f'(2) = 12.0
Approximation: 12.167464114832537
Theoretical:   12.0

注释

在内部，样条插值器计算Natural Cubic Spline。 Natural Cubic Spline和x的{{1}}是
由于您尝试评估y的一阶导数，因此感兴趣的函数为：

2.0

取一阶导数：

f(x) = 1.1196 x^3 - 0.43062 x^2 + 0.45455 x - 0.14354

现在将2插入f'(x) = 3.3588 x^2 - 0.8612 x + 0.4546

f'(x)

证实了我们先前获得的结果。

查找Java中样条函数的派生词

1 个答案: