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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