Hi Sean,

AD relies on overloading to instrument functions so they can be differentiated.

f :: Num a => m1 (m2 a) -> m3 a -> a

By looking at the type of "grad", one can see that f will be specialized  at type "Reverse s Double":

f :: m1 (m2 (Reverse s Double)) -> m3 (Reverse s Double) -> Reverse s Double

So if "x :: m1 (m2 Double)", you will need to apply "auto :: Double -> Reverse s Double" to lift x to the right type.

df x theta = grad (f x') theta
  where x' = (fmap . fmap) auto x

Cheers,
Li-yao

On 11/20/2017 05:05 PM, Sean Matthews wrote:

I'm having some problems with Numeric.AD (translation, things are not
working for reason that I don't understand).  Note I don't have much
experience with this package, so these are newbie questions, thus
appropriate answers may involve pointing me to a document somewhere out
there on the net.

Anyway,  here is my problem:

I have a function (call it f x theta) which I have defined purely in terms
of basic arithmetic functions  (+/-/(/)/*/**) glued together using standard
applicative functor operations has type

f  :: m1 (m2 Double) -> m3 Double -> Double
f x theta = ...

m1 m2 and m3 are all Traversable.
f is defined purely in terms of basic arithmetic operations, (+/-/(/)/*/**)
glued together using standard applicative functor operations, and m1 m2 and
m3 are all pretty trivial record types (no recursion, even).

I would like to write

df x theta = grad (f x) theta

But it refuses to type, even though (admittedly quite a lot) simpler
versions do.

So what am I missing? Does AD not go through Applicative?  That seems
unlikely to me.

Any advice / suggestions, etc. gratefully received.

Sean Matthews





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