How to speed up the n-th Laplacian or derivative of function?

Question

We wish to solve for the multiple Laplacian of a function, and here we use sum of E-exponential functions as an example：

Clear[g]
t[x_, y_, z_, i_] := Random[] Exp[I (Random[] x + Random[] y + Random[] z)]
g[x_, y_, z_] := {Sum[t[x, y, z, i], {i, 1, i0}], Sum[t[x, y, z, i], {i, 1, i0}], Sum[t[x, y, z, i], {i, 1, i0}]}
g1[x, y, z] = g[x, y, z]

Here's the questions：

1. It takes a long time to compute multiple Laplacians using Nest, how to speed it up? (Functions in real-problems are more complex than E-exponential functions, and thus take more time to the point of being unacceptable.)

n0 = 200;
i0 = 10;
lap1[n_, f_] := Nest[Laplacian[#, {x, y, z}] &, f, n]
AbsoluteTiming[lap1[n0, g1[x, y, z]]][[1]]

2.By writing the Laplace in the form of second-order derivative, the computation time is sharply reduced; can the computation be speeded up further on this basis?

lap2[n_, f_] := Nest[(D[#, {x, 2}] + D[#, {y, 2}] + D[#, {z, 2}]) &, f, n]
AbsoluteTiming[lap2[n0, g1[x, y, z]]][[1]]

3.It can be noticed that the Precision of the second-order derivative and Laplacian is not the same, what is the reason?

s1 = lap1[n0, g1[x, y, z]] /. {x -> 0.3, y -> 7, z -> 0.5}
s2 = lap2[n0, g1[x, y, z]] /. {x -> 0.3, y -> 7, z -> 0.5}
s1 - s2

4. Is it possible to use Compile to speed up multi-order derivative? If it is possible how to do.

Answer 1

At 3.) You mix up the notions of (absolute) accuracy and (relative) precision. The relative error is computed by Abs[s1 - s2]/Abs[s1] and it s{1.94629*10^-15, 1.09421*10^-14, 1.82465*10^-15}. So we have roughly 14 digits of precision here. That's basically as good as double precision floating point number can get.

At 4.) No, Compile cannot handle symbolic manipulations.

Answer 2

Consider the simple polynomial expansion of higher derivatives like powers of squares

Function@@{f,((ExpandAll[(x^2+y^2+z^2)^200]  /.
  { Plus->Inactive[Plus],Times->List,x_Symbol^n_:>{x,n}})/.
    {{x_Symbol,n_Integer}:> Inactive[D][f,{x,n}],
     {n_Integer,{ a_,m_Integer},b___List}:> n Inactive[D][f,a,b]})/.
  {Inactive[Plus][a___,b_Symbol,c___]:>Inactive[Plus][a,Inactive[D][f,b] ,c]}}