使用Math.NET C#计算导数

Question

我正在寻找一个简单的函数,它将接受一个double值数组并返回数学导数.

Math.NET似乎有这样的功能,但它要求以下语法:

double FirstDerivative(Func<double, double> f, double x)

我不确定为什么我需要指定一个函数.我只想要一个可以传递数据的预先存在的函数.

Answer 1

获取数据点并创建Math.NET Numerics Cubic Spline对象.然后使用该.Differentiate()方法获得所需的每个点的斜率.

例

请尝试以下代码:

static class Program
{
    const int column_width = 12;
    /// <summary>
    /// The main entry point for the application.
    /// </summary>
    [STAThread]
    static void Main(string[] args)
    {
        var xvec = new DenseVector(new double[] { 0.0, 1.0, 2.0, 3.0, 4.0 });
        var yvec = new DenseVector(new double[] { 3.0, 2.7, 2.3, 1.6, 0.2 });
        Debug.WriteLine("Input Data Table");
        Debug.WriteLine($"{"x",column_width} {"y",column_width}");
        for(int i = 0; i < xvec.Count; i++)
        {
            Debug.WriteLine($"{xvec[i],column_width:G5} {yvec[i],column_width:G5}");
        }
        Debug.WriteLine(" ");
        var cs = CubicSpline.InterpolateNatural(xvec, yvec);

        var x = new DenseVector(15);
        var y = new DenseVector(x.Count);
        var dydx = new DenseVector(x.Count);
        Debug.WriteLine("Interpoaltion Results Table");
        Debug.WriteLine($"{"x",column_width} {"y",column_width} {"dy/dx",column_width}");
        for(int i = 0; i < x.Count; i++)
        {
            x[i] = (4.0*i)/(x.Count-1);
            y[i] = cs.Interpolate(x[i]);
            dydx[i] = cs.Differentiate(x[i]);
            Debug.WriteLine($"{x[i],column_width:G5} {y[i],column_width:G5} {dydx[i],column_width:G5}");
        }


    }
}

看看调试输出:

Input Data Table
           x            y
           0            3
           1          2.7
           2          2.3
           3          1.6
           4          0.2

Interpoaltion Results Table
           x            y        dy/dx
           0            3     -0.28214
     0.28571        2.919     -0.28652
     0.57143       2.8354     -0.29964
     0.85714       2.7469      -0.3215
      1.1429       2.6509     -0.35168
      1.4286       2.5454     -0.38754
      1.7143        2.429     -0.42864
           2          2.3       -0.475
      2.2857        2.154     -0.55809
      2.5714       1.9746      -0.7094
      2.8571       1.7422     -0.92894
      3.1429       1.4382      -1.1979
      3.4286       1.0646      -1.4034
      3.7143      0.64404      -1.5267
           4          0.2      -1.5679

Answer 2

如果您不反对 Math.Net 以外的库，您可以尝试AlgLib及其spline1ddiff函数

构建样条线很容易，而且 Akima 样条线在各点上看起来非常漂亮且平滑。如果您想要一个接受一组数据并返回导数的方法，这里有一个使用 AlgLib 数学库的示例：

public static void CalculateDerivatives(this Dictionary<double, double> inputPoints, out Dictionary<double, double> firstDerivatives, out Dictionary<double, double> secondDerivatives)
{
        var inputPointsXArray = inputPoints.Keys.ToArray();
        var inputPointsYArray = inputPoints.Values.ToArray();

        spline1dinterpolant akimaSplineToDifferentiate;
        alglib.spline1dbuildakima(inputPointsXArray, inputPointsYArray, out akimaSplineToDifferentiate);

        firstDerivatives = new Dictionary<double, double>();
        secondDerivatives = new Dictionary<double, double>();
        foreach (var pair in inputPoints)
        {
            var xPoint = pair.Key;
            double functionVal, firstDeriv, secondDeriv;
            alglib.spline1ddiff(akimaSplineToDifferentiate, xPoint, out functionVal, out firstDeriv, out secondDeriv);

            firstDerivatives.Add(point, firstDeriv);
            secondDerivatives.Add(point, secondDeriv);
        }
}

请注意：Akima 样条线在数据集范围之外具有不可预测的行为。