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Infinite resistors (nerd sniping)

Started by Ertxiem, April 25, 2012, 02:33:31 PM

Ertxiem

Quote from: Duckling on April 24, 2012, 10:49:17 PM
Huh? My quality of life hasn't gone down any. In fact, I've been working on this problem:
http://xkcd.com/356/

I can't quite figure out how to get past infinite resistors in one dimension and move on to two... sorry, that's a bit off-topic. Anyway, the point is that I still appreciate me some maths.

I was caught and the truck ran over me. I may count as 5 points to you Duckling because I'm a physicist and a mathematician.

My first approach at solving it is:

Compute for each path j the electric intensity,
Ij = V / nj,
where nj is the number of one ohm resistors in the path j.

Then sum the intensity for all paths (an infinite sum, of course)
Itot = sumj=1 to ininity Ij.

And finally compute Req = V / Itot.

Of course, the hardest part is to organize the sums in a way that allows us to compute somehow that sum, for instance, a way to compute how many paths of size n exist, for all n

I have a feeling that there may exist near sqrt(n) paths of length n, thereby making Req = 0. I'll think about it a bit more if you, Duckling, or anyone else is interested in this thread. Or if Craig shuts down the forums in pure horror when he looks at this thread. Whatever happens first.
With 2 small Mandelbrot sets as the spots.

ArtDrake

Hold on. Exactly what are you defining as a path?

I assume it would have to be nonintersecting, but since circuits can be expressed as directed graphs, are we only considering the paths which follow these?

Ertxiem

I was calling "path" to of any subset of the circuit that starts in the 1st resistor, ends in the 2nd resistor and all the resistors in between are only connected to other 2 resistors. So, the path does not intercepts itself.
With 2 small Mandelbrot sets as the spots.

ArtDrake

#3
I see. So we're also considering the paths which have negative or zero current along them?

Because if we weren't, I think the terms of the series would be easier to find, if only because self-intersection as a limiting factor does not become an issue -- that is, you don't have to rule them out, because they're not there.

Ertxiem

I've been thinking about my approach and it isn't correct.
It needs some sort of weight in the sum, to account for the parts of a path that go "backwards" and how many times a resistor is being used. Something like the effect of the phase in quantum mechanics (I'm thinking about Feynman diagrams).

So, a second approach would be to compute the intensity at each resistor k as being the sum of the intensities of all paths that go through that resistor, having in consideration the direction (positive if we're going from left to right or bottom to up, otherwise negative)
Ik = sumj=1 to infinity Ij,
with Ij = V / nj, where n is the number of resistors in path j.
This is also near from what we do when we apply Kirchhoff's circuit laws.
While I'm writing this I'm thinking that this approach is also wrong! :(
I'll have to think more about it.
With 2 small Mandelbrot sets as the spots.

ArtDrake

Well, if it helps, I count 3 paths of length 3, 11 of length 5, and I think either 35 or 36 of length 7.

Also, the way I solved a version with infinite resistors in a single direction was to work from the nonexistent ends, and deduce the ratios of parts of the current. Maybe working from the corners could help?

ArtDrake

#6
So... if an electric field can be expressed as a vector field, which I'm assuming it can,
and you have poles terminals at (-1, 0, 0) and (1, 0, 0),
is there any concise way of expressing the magnitude of the vector in each direction with respect to its position?

Ertxiem

The electric field is a vector field. But I'm failing to see an easy way of computing its value as a function of the position.