Question

When using the vector-space package for derivative towers (see derivative towers) I come across the need to differentiate integrals.
From math it is quite clear how to achieve this:

f(x) = int g(y) dy from 0 to x

with a function

g : R -> R

for example.

The derivative with respect to x would be:

f'(x) = g(x)

I tried to get this behaviour by first defining a class “Integration”

class Integration a b where
--standard integration function
integrate :: (a -> b) -> a -> a -> b

a basic instance is

instance  Integration Double Double where
  integrate f a b = fst $ integrateQAGS prec 1000 f a b

with integrateQAGS from hmatrix

the problem comes with values b which represent towers of derivatives:

instance Integration Double (Double :> (NC.T Double)) where
  integrate = integrateD

NC.T is from Numeric.Complex (numeric-prelude).
The function integrateD is defined as follows (but wrong):

integrateD ::(Integration a b, HasTrie (Basis a), HasBasis a, AdditiveGroup b) =>  (a -> a :> b) -> a -> a -> (a :> b)
integrateD f l u = D (integrate (powVal . f) l u) (derivative $ f u)

The function does not return what I want, it derives the integrand, but not the integral. The problem is, that I need a linear map which returns f u. The a :> b is defined as follows:

data a :> b = D { powVal :: b, derivative :: a :-* (a :> b) }

I don’t know how to define derivative. Any help will be appreciated, thanks

edit:

I forgot to provide the instance for Integration Double (NC.T Double):

instance  Integration Double (NC.T Double) where
  integrate f a b = bc $ (\g -> integrate g a b) <$> [NC.real . f, NC.imag . f]
      where bc (x:y:[]) = x NC.+: y

and I can give an example of what I mean:
Let’s say I have a function

f(x) = exp(2*x)*sin(x)

>let f = \x -> (Prelude.exp ((pureD 2.0) AR.* (idD x))) * (sin (idD x)) :: Double :> Double 

(AR.*) means multiplication from Algebra.Ring (numeric-prelude)

I can easily integrate this function with the above function integrateD:

>integrateD f 0 1 :: Double :> Double
D 1.888605715258933 ...

When I take a look at the derivative of f:

f'(x) = 2*exp(2*x)*sin(x)+exp(2*x)*cos(x)

and evaluate this at 0 and pi/2 I get 1 and some value:

> derivAtBasis (f 0.0) ()
D 1.0 ...

> derivAtBasis (f (pi AF./ 2)) ()
D 46.281385265558534 ...

Now, when deriving the integral, I get the derivation of the function f not its value at the upper bound

> derivAtBasis (integrate f 0 (pi AF./ 2)) ()
D 46.281385265558534 ...

But I expect:

> f (pi AF./ 2)
D 23.140692632779267 ...

Answer 1

I finally found a solution to my question.
The key to the solution is the >-< function from the vector-space package, it stands for the chain rule.

So, I define a function integrateD' like this:

integrateD' :: (Integration a b, HasTrie (Basis a), HasBasis a, AdditiveGroup b , b ~ Scalar b, VectorSpace b) => (a -> a :> b) -> a -> a -> (a:>b) -> (a :> b)
integrateD' f l u d_one =  ((\_ -> integrate (powVal . f) l  (u)) >-< (\_ ->  f u)) (d_one)

the d_one means a derivation variable and its derivative must be 1. With this function I can now create some instances like

instance Integration Double (Double :> Double) where
integrate f l u = integrateD' f l u (idD 1)

and

instance Integration ( Double) (Double :> (NC.T Double)) where
integrate f l u = liftD2 (NC.+:) (integrateD' (\x -> NC.real <$>> f x) l u (idD 1.0 :: Double :> Double)) (integrateD' (\x -> NC.imag <$>> f x) l u (idD 1.0 :: Double :> Double))

unfortunately I can’t use integrateD on complex values out of the box, I have to use liftD2. The reason for this seems to be the idD function, I don’t know if there is a more elegant solution.

When I look at the example in the question, I now get my desired solution:

*Main> derivAtBasis (integrateD' f 0 (pi AF./ 2) (idD 1.0 :: Double :> Double )) ()
D 23.140692632779267 ...

or by using the instance:

*Main> derivAtBasis (integrate f 0 (pi AF./ 2)) ()
D 23.140692632779267 ...