Questions about Numeric.AD / Automatic differentiation
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 -- Sean Matthews seanmatthews1@gmail.com / +49 1515 800 1901
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
_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.
After working through the other problems, that fills the last remaining
hole (at least for the moment).
Many thanks,
Sean
On Tue, Nov 21, 2017 at 7:48 AM, Li-yao Xia
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
_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.
-- Sean Matthews seanmatthews1@gmail.com / +49 1515 800 1901
participants (2)
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Li-yao Xia -
Sean Matthews