实施C / C衍生物++ [英] implementing the derivative in C/C++

问题描述

如何衍生物的函数f（x）通常计算的编程，以确保最高的准确度？我实施牛顿 - 拉夫逊的方法，它需要采取的函数的导数。

感谢

解决方案

我同意erikkallen的（F（X + H） - F（X - h）条）/ 2H是数值逼近衍生品通常的做法。然而，获得正确的步长h是有点微妙。

在近似误差（函数f（x + H） - F（X - h））进行/小时降低为h变小，这表示，你应该采取H作为越小越好。但为h变小，从浮点减法增加了错误，因为分子需要减去几乎相等。如果h过小，可以在减法宽松了很多precision。因此，在实践中你必须选择的h的不太小的值最小化相结合的逼近的错误和数字的错误。

作为一个经验法则，你可以尝试H = SQRT（DBL_EPSILON），其中DBL_EPSILON是最小的双precision查到，使得1 + E！= 1机precision。 DBL_EPSILON约为10 ^ -15所以你可以使用H = 10 ^ -7或10 ^ -8。

有关详细信息，请参阅以下说明上采摘的步长微分方程。

How is derivative of a f(x) is typically calculated programmatically to ensure maximum accuracy? I am implementing the Newton-Raphson method and it requires taking of the derivative of a function.

thanks

解决方案

I agree with erikkallen that (f(x + h) - f(x - h))/2h is the usual approach for numerically approximating derivatives. However, getting the right step size h is a little subtle.

The approximation error in (f(x + h) - f(x - h))/2h decreases as h gets smaller, which says you should take h as small as possible. But as h gets smaller, the error from floating point subtraction increases since the numerator requires subtracting nearly equal numbers. If h is too small, you can loose a lot of precision in the subtraction. So in practice you have to pick a not-too-small value of h that minimizes the combination of approximation error and numerical error.

As a rule of thumb, you can try h = SQRT(DBL_EPSILON) where DBL_EPSILON is the smallest double precision number e such that 1 + e != 1 in machine precision. DBL_EPSILON is about 10^-15 so you could use h = 10^-7 or 10^-8.

For more details, see these notes on picking the step size for differential equations.

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