**comp.graphics.algorithms**

## Subject: **Re: rotationally symmetry of laplacian**

"news.cse.ohio-state.edu"

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> In the context of edge-detection using lapacian operators, I came across

> the statement that the lapacian is rotationally invariant. I am unable to

> see why this is true. Any pointers or ideas?

Consider a C2-function F(x,y); then Laplacian(F) = Fxx + Fyy

where Fx = dF/dx, Fy = dF/dy, Fxx = d^2 F/dx^2,

Fxy = d^2 F/dx dy, and Fyy = d^2 F/dy^2 (all partial derivatives).

Define u = c*x - s*y and v = s*x + c*y, where c = cos(A) and

s = sin(A). This is a rotational change of variables. Define

G(u,v) = F(x,y). Use the chain rule

Fx = Gu*du/dx + Gv*dv/dx = c*Gu +s*Gv

Fy = Gu*du/dy + Gv*dv/dy = -s*Gu + c*Gv

Differentiate again,

Fxx = c^2*Guu + 2*s*c*Guv + s^2*Gvv

Fyy = s^2*Guu - 2*s*c*Guv + c^2*Gvv

Add these together,

Fxx + Fyy = Guu + Gvv

so Laplacian(F) = Laplacian(G).

--

Dave Eberly

http://www.geometrictools.com

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