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I think it is best to approach these kinds of problems with pure functions, which minimizes nasty surprises from variable scoping issues. Let's see how well this idea works out when applied to your problem. Generate test data. With[{u = Range[-180., 180., 6.] Degree}, data = With[{x = #[[1]], y = #[[2]]}, {x, y, Sin[x] Cos[ y]}] & /@ ...


3

If I understand what you want: SetAttributes[fun, HoldAll]; fun[expr_] := Identity @@ Identity @@ MapAll[Unevaluated, Hold[expr], Heads -> True] The two Identity operations fix the fact that MapAll applies Unevaluated to the entire expression as well - MapAll[g, f[x,y]] is g[g[f][g[x],g[y]] when we want g[f][g[x],g[y] - and remove the Hold we ...


0

It seems that Residue (and Series on which it is based) is unable to analyze all the cases in which the formula for the residue might vary. Reduce can do it, within its limitations, if we reduce the system 1/function == 0 && nu == pole. Here's the OP's example. fn = (w^(n/2 + I nu) ws^(-(n/2) + I nu))/(n/2 + I nu)^2; We need to add some ...



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