Converting Laplacian to Polar

Someone asked me to show them how to get Mathematica to do the polar computation. Here is the easy way.

Laplacian[ f[r,θ],{r,θ},"Polar"]

Here is a more hands on way.

Thing1=Simplify[
D[ f[(Sqrt[(x^2)+(y^2)]), ArcTan[y/x]],{x,2}] + D[ f[(Sqrt[(x^2)+(y^2)]), ArcTan[y/x]],{y,2}]
]
Thing2=Thing1/.{x->r Cos[θ], y ->r Sin[θ]}
Thing3=Simplify[Thing2,Assumptions->r>0]
Thing4 = Thing3/.ArcTan[Tan[θ]]->θ

The second way is essentially what you did manually.

HW Questions

n=123;
A= SparseArray[{
Band[{2,1}]->1.0,
Band[{1,1}]->-2,
Band[{1,2}]->1
}, {n,n}];
MatrixForm[A];
ListPlot[- Eigenvalues[A]]

ListPlot[- Eigenvalues[A,12]]

λ=Eigenvalues[A];
i=-1;
λ[[i]]
V=Eigenvectors[A];
ListPlot[ 
Table[
((V[[-i]])/(Max[V[[-i]]])),
{i,1,5}],
Joined->True]

n=123;
A= Table[
Which[
i==j,-2,
i==j+1,1,
True, 0],
 {i,n},{j,n}];
MatrixForm[A]

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