# Evaluate pure function through round brackets

In have some matrices of pure functions or numbers which I multiply by other matrices or vectors. I would like the functions to be evaluated in the result, but all that I get is something like this:

(I*(D[#1, x] & ))[f[{x, y}]]


Instead I would like something like this:

I*Derivative[{1, 0}][f][{x, y}]


All that I need is that the derivative could pass through the most external round brackets, so that the function could be applied to the argument.

Since I work with a large amount of matrices, I cannot delete manually all the parenthesis after getting the result; is there a way to let the function know it can pass through the brackets?

Thanks.

EDIT:

The solution which best fits my needs is a modified version of @kglr

f1=#/.head_[a___][d___]:>If[StringMatchQ[ToString[{a}//FullForm],"*Function*"], head[##&@@({a}/.w_Function:(1*#&):>w[d])], head[a][d]]&;]

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• see Here for the underlying problem. – andre314 Apr 30 '16 at 11:32
• Thank you, now I understand the reason for that behaviour. Nevertheless, the brackets are put automatically by mathematica, even if they are not present when the list which they are part of is defined. – PPeg Apr 30 '16 at 12:24

Restructure the FullForm using ReplaceAll:

f1 =  # /. head_[a___][d___] :> head[## & @@ ({a} /. w_Function :> w[d])] &


Examples:

{#, f1 @ #} &[(((Log@(D[#1, x] &))))[h[{x, y}]] ] {#, f1 @ #} &[(((I (D[#1, x] &) ((D[#1, y] &)))))[h[{x, y}]] ] Alternatively, define a function that processes the box forms of the expressions to remove the parentheses:

f2 = ToExpression[Replace[ToBoxes[#], {RowBox[{"(", RowBox[{a__}], ")"}] :>
RowBox[{a}], RowBox[{RowBox[{a___}], b___, RowBox[{c___}], d___}] :>
RowBox[{a, b, c, d}]}, {0, Infinity}]] &;


Examples:

{I*D[#1, x] &[3 x y z], (I*(D[#1, x] &))[3 x y  z], (I*(D[#1, x] &))[3 x y  z] // f2} I*D[#1, x] &[h[{x, y}]] (I*(D[#1, x] &))[h[{x, y}]] (I*(D[#1, x] &))[h[{x, y}]] // f2 • ... not sure how robust this is for general expressions. – kglr Apr 30 '16 at 13:15
• Thank you, it seems a good starting point to solve my problem. – PPeg Apr 30 '16 at 13:53
• I don't understand very well what the f1 function does, could you explain it to me, please? – PPeg May 2 '16 at 17:11
• @PPeg, if you look at the FullForm of the expression in your post using FullForm[(I*(D[#1, x] &))[3 x y z]] you see that it is seen by mma as Times[Complex[0,1], Function[D[Slot,x]]][Times[3,x,y,z]] which has the pattern head[arg1][arg2]. The function f1 transforms such patterns by replacing any Function object, i.e., Function[D[Slot, x]] within args1 with Function[D[Slot, x]][args2]; that is, it pushes args2 inside the parantheses into the appropriate place. Hope this helps. – kglr May 2 '16 at 17:26
• Thank you very much. This could be the answer to my question. I will work a bit on it and eventually close the question. – PPeg May 3 '16 at 6:55