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Re: How to evaluate the Laplacian of a function as a


to understand whats happening and why this works please check carefully
the difference between Set (=) and SetDelayed (:=)

<< VectorAnalysis`
func[x_, y_] := 3/Pi*(1 - x^2 + y^2)^2
funcLapl[Xx_, Yy_] = Laplacian[func[Xx, Yy]]

funcLapl[a, b]


On Wed, 2010-12-01 at 02:15 -0500, Iliyan Georgiev wrote:
> Hi,
> I have a stupid problem and cannot find a solution anywhere. I have
> the function
> func[x_,y_] := 3/Pi*(1 - x^2 + y^2))^2
> which I need the Laplacian of:
> funcLapl := Laplacian[func[Xx,Yy]]
> The Laplacian is computed correctly, but how can I treat the result as
> a function? I want to be able to evaluate/plot the Laplacian. I tried
> many things and I'm frustrated. My current solution is to manually
> copy the derived Laplacian and create a function from that. It's
> obviously not a good solution.
> Any suggestions?

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