衍生塔以及如何使用矢量空间包（haskell） [英] Derivative Towers and how to use the vector-space package (haskell)
问题描述
到目前为止，我正在研究一个计算一些严重问题的项目，比如从一个给出结构。
我跟着博客写的 Conal Elliott （这里有更多线性地图），它是很不错，根本。
不幸的是，我缺乏一个简单的例子：）
更准确地说，我有一条曲线
f：[0,1] in R - > R³
t - > a * e_y + 2 * t * e_z
这是（0，a， 2 * T）。
当我想计算f的导数时，例如对于曲线的长度，我知道数学结果，这很简单（0,0,2），但是我怎么在Haskell中完成这项工作，特别是使用 vector-space 包？
我真的想用这个库，因为它的功能正是我所要采取的方法（但我在Haskell的道路上并不遥远）
到目前为止，我所得到的是：
{ - ＃LANGUAGE Rank2Types，TypeOperators，FlexibleContexts，TypeFamilies＃ - }
{ - ＃OPTIONS_GHC -Wall＃ - }
导入Numeric.GSL.Integration
导入Data.VectorSpace
导入Data.Basis
导入Data.Cross
导入Data.Derivative
导入数据。 LinearMap
类型Vec3 s =三个s
prec ::双
prec = 1E-9
f1 ::（浮动s ，VectorSpace s，Scalar s〜s）=> s - > s
f1 = id
c1 :: Double - > Vec3 Double
c1 = \ t - > linearCombo [（（v 0 0 1），f1 t），（（v 0 1 0），2）]
derivC :: Double - > Vec3（Double：> Double）
derivC t = c1（pureD t）
是pureD函数的实际实现，到目前为止我没有尝试使用这个代码片段进行编译。我得到以下错误：
tests.hs：26：12：
无法匹配预期的类型`Double产品：> Double'
，实际类型为'Double'
预期类型：Vec3（Double：> Double）
实际类型：Vec3 Double
在`c1调用的返回类型'
在表达式中：c1（pureD t）
失败，模块加载：无。
还有一个使用向量空间的图形库，甚至在圆环上也有一个例子， pureD被使用。我试图推断出这个例子，但我没有看到如何将它映射到我的问题。
任何帮助将不胜感激。
预先致谢
PS：我无法发布所有我想要的链接，但愿意提供
这是一个有趣的图书馆。感谢分享。
虽然我不明白库的概念，但是
这段代码如何：
{ - ＃LANGUAGE Rank2Types，TypeOperators，FlexibleContexts，TypeFamilies＃ - }
模块Main其中
导入Data.LinearMap
导入Data.Maclaurin
diff ::（Double：〜>（Double，Double，Double）） - > （Double：〜>（Double，Double，Double））
diff f = \ x - > （atBasis（derivative（f x））（））
eval ::（Double：〜>（Double，Double，Double）） - >双 - > （Double，Double，Double）
eval f x = powVal（f x）
f :: Double：〜> （double，double，double）
f x = tripleD（pureD 0，pureD 1，（2 * idD）x）
* Main> map（eval f）[0,0.2..1]
[（0.0,1.0,0.0），（0.0,1.0,0.4），（0.0,1.0,0.8），（0.0,1.0,1.2000000000000002），
（0.0,1.0,1.6000000000000003），（0.0,1.0,2.0000000000000004）]
* Main> map（eval（diff f））[0,0.2..1]
[（0.0,0.0,2.0），（0.0,0.0,2.0），（0.0,0.0,2.0），（0.0,0.0， 2.0），（0.0,0.0,2.0），
（0.0,0.0,2.0）]
* Main> map（eval（diff $ diff f））[0,0.2..1]
[（0.0,0.0,0.0），（0.0,0.0,0.0），（0.0,0.0,0.0），（0.0， 0.0,0.0），（0.0,0.0,0.0），（0.0,0.0,0.0）]
试试还有gx = tripleD（pureD 0，idD x，（idD * idD）x）（它似乎代表曲线（0，x，x ^ 2））。
I am working with Haskell for quite a while now, but I am far from being an expert. But I see that the functional approach to programming suits me the best.
So far I am working on a project to calculate some serious stuff, like currents and potentials radiated from a given structure.
I followed the blog written by Conal Elliott (here is some more Linear Maps) which is very nice and fundamental.
Unfortunately, I am lacking a simple example :)
To be more precise, I have a curve
f:[0,1] in R -> R³
t -> a*e_y + 2*t*e_z
which is a simple straight line at (0,a,2*t). When I want to calculate the derivative of f, e.g. for the length of the curve, I know the mathematical result, which is quite simple (0,0,2), but how do I accomplish this in Haskell, especially with the vector-space package?
I really want to use this library because of its functionality, it is exactly the approach I would have take too (but I am not that far ahead on the Haskell road)
What I have so far is this:
{-# LANGUAGE Rank2Types, TypeOperators, FlexibleContexts, TypeFamilies #-}
{-# OPTIONS_GHC -Wall #-}
import Numeric.GSL.Integration
import Data.VectorSpace
import Data.Basis
import Data.Cross
import Data.Derivative
import Data.LinearMap
type Vec3 s = Three s
prec :: Double
prec = 1E-9
f1 :: (Floating s, VectorSpace s, Scalar s ~ s) => s -> s
f1 = id
c1 :: Double -> Vec3 Double
c1 = \t -> linearCombo [((v 0 0 1),f1 t),(( v 0 1 0),2)]
derivC :: Double -> Vec3 (Double :> Double)
derivC t = c1 (pureD t)
It is the the actual implementation of the pureD function, so far nothing that I have tried works to get this snippet to compile. I get the following error:
tests.hs:26:12:
Couldn't match expected type `Double :> Double'
with actual type `Double'
Expected type: Vec3 (Double :> Double)
Actual type: Vec3 Double
In the return type of a call of `c1'
In the expression: c1 (pureD t)
Failed, modules loaded: none.
There is also a graphics library which uses vector-space and there is even an example on a torus, where pureD is used. I tried to deduce the example but I don't see how I can map it to my problem.
Any help would be greatly appreciated.
Thanks in advance
PS: I cannot post all the links I'd like to, but am willing to provide
That's an interesting library.. Thanks for sharing. Although I don't understand the concept of the library yet, how about this code:
{-# LANGUAGE Rank2Types, TypeOperators, FlexibleContexts, TypeFamilies #-}
module Main where
import Data.LinearMap
import Data.Maclaurin
diff :: (Double :~> (Double,Double,Double) ) -> (Double :~> (Double,Double,Double))
diff f = \x -> (atBasis (derivative (f x)) ())
eval :: (Double :~> (Double,Double,Double)) -> Double -> (Double,Double,Double)
eval f x = powVal (f x)
f :: Double :~> (Double,Double,Double)
f x = tripleD (pureD 0,pureD 1,(2*idD) x)
*Main> map (eval f) [0,0.2 .. 1]
[(0.0,1.0,0.0),(0.0,1.0,0.4),(0.0,1.0,0.8),(0.0,1.0,1.2000000000000002),
(0.0,1.0,1.6000000000000003),(0.0,1.0,2.0000000000000004)]
*Main> map (eval (diff f)) [0,0.2 .. 1]
[(0.0,0.0,2.0),(0.0,0.0,2.0),(0.0,0.0,2.0),(0.0,0.0,2.0),(0.0,0.0,2.0),
(0.0,0.0,2.0)]
*Main> map (eval (diff $ diff f)) [0,0.2 .. 1]
[(0.0,0.0,0.0),(0.0,0.0,0.0),(0.0,0.0,0.0),(0.0,0.0,0.0),(0.0,0.0,0.0),(0.0,0.0,0.0)]
Try also g x = tripleD (pureD 0,idD x,(idD*idD) x) (which seem to represent the curve (0,x,x^2)).
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