Re: [math-fun] xkcd points out dangers of math fun
----- Original Message ---- From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, January 6, 2008 12:50:21 PM Subject: Re: [math-fun] xkcd points out dangers of math fun On 1/6/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
... On a square grid of 1 ohm resistors, the effective resistance between the node at (0,0) and the node at (p,q) is given by the following expression.
R[p,q] = (1/(4 pi^2)) int (F(x,y), x=-pi..pi, y=-pi..pi),
F(x,y) = [1 - cos(p x + q y)] / [2 - (cos x + cos y)].
This is pretty much identical to formula (18) in Cserti, for the 2-D case. No doubt one could be convinced of its correctness simply by checking that it satisfied the Kirchhoff-inspired equations I quoted earlier [ my u^{kl} being your R[p,q] ] : [Lat1] u^{00} = 0; [Lat2] u^{01} = 1/2; [Lat3] \del^2 u^{kl} = 0 for (k,l) <> (0,0); where \del^2 u^{kl} == u^{k,l-1} + u^{k,l+1} + u^{k-1,l} + u^{k+1,l} - 4 u_{kl} . But how did you arrive at it; and exactly what assumptions did you make in order to do so? Fred Lunnon _____________________________________________ We have the Kirchhoff equations (1/4)(v[j+1,k] + v[j-1,k] + v[j,k+1] + v[j,k-1]) - v[j,k] = 0, except at the nodes where current enters or exits. If 1 A of current exits at a node, the current is divided equally among the 4 resistors, and the 4 adjacent nodes will be (1/4) V higher. So at that node, the RHS should be (1/4). If 1 A exits at (p,q) and enters at (r,s), the RHS is (1/4)(delta(j,p) delta(k,q) - delta(j,r) delta(k,s)). Now define the Fourier series V(x,y) = sum(v[j,k] exp(i j x + i k y)) summed over all integers j and k. Applying this to the Kirchhoff equations, we can solve for V(x,y). It is a fraction whose numerator is exp(i r x + i s y) - exp(i p x + i q y), and whose denominator is 4 - 2(cos x + cos y). To calculate v[j,k], multiply V(x,y) by exp(- i j x - i k y)/(2 pi)^2 and integrate over (x,y) in [-pi, pi]^2. The effective resistance between (r,s) and (p,q) is equal to the voltage difference v[r,s] - v[p,q]. You can set (p,q) = (0,0), and get my solution, where I replaced (r,s) by (p,q). Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ
On 1/6/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
... We have the Kirchhoff equations
(1/4)(v[j+1,k] + v[j-1,k] + v[j,k+1] + v[j,k-1]) - v[j,k] = 0,
except at the nodes where current enters or exits. If 1 A of current exits at a node, the current is divided equally among the 4 resistors, and the 4 adjacent nodes will be (1/4) V higher. So at that node, the RHS should be (1/4). If 1 A exits at (p,q) and enters at (r,s), the RHS is
(1/4)(delta(j,p) delta(k,q) - delta(j,r) delta(k,s)).
Now define the Fourier series
V(x,y) = sum(v[j,k] exp(i j x + i k y))
summed over all integers j and k. Applying this to the Kirchhoff equations, we can solve for V(x,y). It is a fraction whose numerator is
exp(i r x + i s y) - exp(i p x + i q y),
and whose denominator is
4 - 2(cos x + cos y).
To calculate v[j,k], multiply V(x,y) by exp(- i j x - i k y)/(2 pi)^2 and integrate over (x,y) in [-pi, pi]^2. The effective resistance between (r,s) and (p,q) is equal to the voltage difference v[r,s] - v[p,q]. You can set (p,q) = (0,0), and get my solution, where I replaced (r,s) by (p,q).
OK --- I can now see that this was essentially what Cserti was trying to say, apart from being three times shorter and ten times clearer! So the cryptic "periodic boundary condition" assumption Cserti mentions boils down to assuming that you can substitute in the Fourier series. Mind you, Atkinson et al does something similar as well, with no such assumption stated: instead, they talk about "vanishing current at infinity" [whatever that might mean, or how impact on the computation]. By the way, according to Cserti, Atkinson et al, Mathematica manages to evaluate these integrals explicitly. Sadly my Maple 9 can't hack 'em at all --- not even u^{01}, which ought to be essentially easier than the general node. Fred Lunnon
participants (2)
-
Eugene Salamin -
Fred lunnon