I don't see the symmetry argument. Consider the 1-dimensional case; then R[1,0] = 1, not 1/2. Additionally, I would be wary of injecting current into one node without also extracting an equal current. Current must flow out to infinity. In a 2 or more dimensional lattice, how can you be sure you have a unique solution without imposing boundary conditions at infinity? -- Gene ________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 10:29:04 AM Subject: Re: [math-fun] The people you know... R[m_, n_] := Integrate[(1 - Cos[m p + n q])/(2 - Cos[p] - Cos[q]), {p, 0, 2 Pi}, {q, 0, 2 Pi}]/(2 Pi)^2 R[5,0] = (-3760 + 1203 pi)/(6 pi]) = 1.02580 R[4,3] = (48 - 5 pi)/(10 pi) = 1.02789 You can superimpose two symmetric circuit solutions, one where unit current is injected at [0,0] and the other where unit current is extracted at [1,0], to argue R[1,0] = 1/2. Veit On Jan 28, 2011, at 12:53 PM, Cordwell, William R wrote:
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points.
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