Fast Hessian size increase

[lferrant]

@Sunny

I mean that I’d like to have the result with three dimensions (2,2,2) for 2x2 array and
the hessian with shape (2,2,10) for (2,10) array

This makes me think you actually have a \mathbb{R}^2\rightarrow\mathbb{R} function and you want to evaluate the hessian at n distinct points (e.g if your array is (2,10) at 10 points) is this what you want to do?

A possible way I can think of is the following (there might be better ones)

julia> f(x) = sum(x.^3)
f (generic function with 1 methods)

julia> pnts = [[1,2], [3,4], [5,6], [7,8]]
4-element Vector{Vector{Int64}}:
 [1, 2]
 [3, 4]
 [5, 6]
 [7, 8]

julia> ForwardDiff.hessian.(f, pnts)
4-element Vector{Matrix{Int64}}:
 [6 0; 0 12]
 [18 0; 0 24]
 [30 0; 0 36]
 [42 0; 0 48]