Absolutely hilarious. Reminds me of Alan Mendelsohn's "missile whistle" from Alan Mendelsohn, the Boy from Mars. He shoots an incredibly loud whistle right at someone's head, and as they turn in astonishment, they adjust their course just slightly and run smack into a trash can.
I'm a physicist and I had that problem on my Math Methods final... never could figure out what it had to do with anything that we covered in that class (complex analysis, crazy PDEs and other continuous stuff).
Last edited by DragonMudd on Wed Dec 12, 2007 5:35 am UTC, edited 1 time in total.
theyellowhobbit wrote:In this case it's a good thing. We're smart enough not to get hit by trucks...
(Though I'm more archaeology/religion nerd, but it's the same principle. Though I suppose a shiny artifact would have the same effect on me...)
Agreed. But I guess this comic just made me think in general, all of the comics. I guess the description of the comic excludes literature though. Only language... (A webcomic of romance, sarcasm, math, and language.)
Wouldn't the resistance be 0? I actually came up with a number that would satisfy this in my Calculus class: 1-.99999... It comes out to .000...0001 (infinite zeros). It can be represented as "the limit of y such that y=1/x as x goes to infinity". Since it could never really reach zero it would be as close to zero as was possible. Just a thought.
chardish wrote:Gosh, I want to say that it's infinite. That'd be too easy though, right?
I would say it'd have to be less than three ohms... since there's a 3 ohm path between the two nodes, and adding extra, parallel paths, can only make the resistance less...
Also... is it a bad sign that, despite the entire setup of the strip warning me about it, I still almost stopped reading the strip in the middle and started trying to solve the puzzle, before I realised what I was doing, and finished reading the strip?
vwlou89: Please don't bring up the 0.999... = 1 thing again... I'm still recovering from last time.
While no one overhear you quickly tell me not cow cow. but how about watch phone?
Oh-oh. This could be dangerous. I suspect this is doing to be one busy thread.
(under breath: Let's see, resistance is cumulative in series, but an inverse ratio in parallel... the grid is infinite, so we've got an unlimited number of both types... though the parallel paths will always also have some series paths, but there are series paths without any parallel elements... I wonder if that could lead to a canceling effect, making for a "neater" answer... but maybe not, and looking for that is red herring... hmmmm.... looks at clock Yikes. Okay, punt.)
I have discovered a truly marvelous answer to this, which this post is too small to contain...
Last edited by DragonHawk on Wed Dec 12, 2007 5:43 am UTC, edited 1 time in total.
I didn't have time to read the derivation, as I need to get studying for a non-ece final I have in 8 hours, but... if the solution I did find is only for a single square diagonal, can't we assume that since this is an infinite symetrical grid, with each indidivual box having equivalent resistance, the equivalent resistance of the problem in question should be related by a simple triangle, ie, taking the triangle formed in the single square to have signs equal to an arbitrary unit of 1, the hyptoneuse then has a length of sqrt(2). For the rectangle depicted here, using the same units, the hyptoneneuse will have a length of sqrt(5). Assuming symmetry, etc, etc, we therefore have the answer here is (2/pi)*sqrt(5)/sqrt(2)...?
edit: this comes out to sqrt(10)/pi ohms
Last edited by ep103 on Wed Dec 12, 2007 5:53 am UTC, edited 1 time in total.
This problem is discussed in J. Cserti's 1999 arXiv preprint. It is also discussed in The Mathematica GuideBook for Symbolics, the forthcoming fourth volume in Michael Trott's GuideBook series, the first two of which were published just last week by Springer-Verlag. The contents for all four GuideBooks, including the two not yet published, are available on the DVD distributed with the first two GuideBooks.
You know if xkcd has this kind of motivating power, Randall ought to put it to good use. For example slip in a subtle problem that is equivalent to a proof of NP-completeness or some other intractable task... within hours of being posted someone on the fora might have a solution!