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(Funcf, 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.
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.
请尝试以下代码:
static class Program { const int column_width = 12; ////// The main entry point for the application. /// [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