数学函数微分与C＃？ [英] Mathematical function differentiation with C#?

问题描述

我知道我可以声明一个函数（说）

I see that I can declare a function with (say)

public double Function(double parameter)

但是如果我想采取这一函数的导数是什么？

but what if I do want to take the derivative of that function?

推荐答案

您无法计算使用计算机程序（除非你做符号数学函数的导数确切...但这是另一个，方式更加复杂，主题）。

You can't calculate the exact derivative of a function using a computer program (unless you're doing symbolic math... but that's another, way more complicated, topic).

有几种方法计算数值函数的导数。最简单的是为本三点法：

There are several approaches to computing a numerical derivative of a function. The simplest is the centered three-point method:

• 取一数H


• 评估 [F（X + H） - F（X-H）] /小时


• 瞧，F'（X）的近似值，只有两个功能评价

另一种方法是在中心五点法：

Another approach is the centered five-point method:

• 取一数H


• 评估 [F（X-2H） - 8F（X-H）+ 8F（X + H） - F（X + 2H）/ 12小时


• 瞧，F'（x）的更好的近似，但它需要更多的功能评价

另外一个话题是如何实现这一点使用C＃。首先，你需要一个委托，它重新presents的实数的一个子集映射到一个实数的另一个子集的功能：

Another topic is how to implement this using C#. First, you need a delegate that represents a function that maps a subset of the real numbers onto a another subset of the real numbers:

delegate double RealFunction(double arg);

然后，你需要评估衍生物的路由：

Then, you need a routing that evaluates the derivative:

public double h = 10e-6; // I'm not sure if this is valid C#, I'm used to C++

static double Derivative(RealFunction f, double arg)
{
    double h2 = h*2;
    return (f(x-h2) - 8*f(x-h) + 8*f(x+h) - f(x+h2)) / (h2*6);
}

如果你想要一个面向对象的实现，你应该创建以下类：

If you want an object-oriented implementation, you should create the following classes:

interface IFunction
{
    // Since operator () can't be overloaded, we'll use this trick.
    double this[double arg] { get; }
}

class Function : IFunction
{
    RealFunction func;

    public Function(RealFunction func)
    { this.func = func; }

    public double this[double arg]
    { get { return func(arg); } }
}

class Derivative : IFunction
{
    IFunction func;
    public static double h = 10e-6;

    public Derivative(IFunction func)
    { this.func = func; }

    public double this[double arg]
    {
        get
        {
            double h2 = h*2;
            return (
                func[arg - h2] - func[arg + h2] +
                ( func[arg + h]  - func[arg - h] ) * 8
                ) / (h2 * 6);
        }
    }
}

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