是否有用于集成复杂函数的Haskell库? [英] Are there any Haskell libraries for integrating complex functions?

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问题描述


  1. 如何在Haskell中以数值方式集成复杂的复值函数?

  2. 是否有任何现有的库? 数字工具仅适用于reals。

我知道在复杂的平面上只有线积分,所以我感兴趣的界面如下所示:

  i =整合fxab精度

计算沿直线的积分从 a b 函数 f on point X
i x a b 都是复合双倍或更好 Num a =>复杂的a 类型。



请......:)

你可以自己做这样的事情。假设你有一个函数 realIntegrate 类型为(Double - > Double) - > (Double,Double) - > Double ,接受一个函数和一个包含下限和上限的元组,并将结果返回一定的精度。例如,您可以使用数字工具定义 realIntegrate f(lo,hi)= quadRomberg defQuad(lo,hi)f

然后,我们可以按照如下方式创建你想要的函数 - 我现在忽略了精度(而且我不明白你的 x 参数是用于什么的! ):

  integrate ::(Complex Double  - > Complex Double) - >复合双 - >复合双 - > Complex Double 
integrate fab = r:+ i其中
r = realIntegrate realF(0,1)
i = realIntegrate imagF(0,1)
realF t = realPart(f t)) - 或realF = realPart。 F 。插值
imagF t = imagPart(f(interpolate t))
interpolate t = a +(t:+ 0)*(b - a)
pre>

因此,我们表示从 a b 作为通过线性插值从0到1的实际区间的函数,沿着该路径取值 f ,分别对实部和虚部进行积分(我不虽然),并将它们重新组合成最终答案。

我没有测试过这段代码,因为我没有numeric-工具安装,但至少它typechecks: - )


  1. How to numerically integrate complex, complex-valued functions in Haskell?
  2. Are there any existing libraries for it? numeric-tools operates only on reals.

I am aware that on complex plane there's only line integrals, so the interface I am interested in is something like this:

i = integrate f x a b precision

to calculate integral along straight line from a to b of function f on point x. i, x, a, b are all of Complex Double or better Num a => Complex a type.

Please... :)

解决方案

You can make something like this yourself. Suppose you have a function realIntegrate of type (Double -> Double) -> (Double,Double) -> Double, taking a function and a tuple containing the lower and upper bounds, returning the result to some fixed precision. You could define realIntegrate f (lo,hi) = quadRomberg defQuad (lo,hi) f using numeric-tools, for example.

Then we can make your desired function as follows - I'm ignoring the precision for now (and I don't understand what your x parameter is for!):

integrate :: (Complex Double -> Complex Double) -> Complex Double -> Complex Double -> Complex Double
integrate f a b = r :+ i where
    r = realIntegrate realF (0,1)
    i = realIntegrate imagF (0,1)
    realF t = realPart (f (interpolate t)) -- or realF = realPart . f . interpolate
    imagF t = imagPart (f (interpolate t))
    interpolate t = a + (t :+ 0) * (b - a)

So we express the path from a to b as a function on the real interval from 0 to 1 by linear interpolation, take the value of f along that path, integrate the real and imaginary parts separately (I don't know if this can give numerically badly behaving results, though) and reassemble them into the final answer.

I haven't tested this code as I don't have numeric-tools installed, but at least it typechecks :-)

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