This Week's Finds in Mathematical Physics (Week 297)
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From: Tom Roberts on 12 May 2010 03:20 John Baez wrote: > Next: electrical circuits! > [...] resistors only > [...] gluing terminals together > [...] permitting zero resistance How does gluing terminals together differ from permitting connections of zero resistance between terminals? If one prohibits infinite power dissipation, aren't they really equivalent (except for the labeling of terminals)? Tom Roberts
From: The Dougster on 24 May 2010 03:45 On May 10, 9:22�pm, b...(a)math.removethis.ucr.andthis.edu (John Baez) wrote: > Also available athttp://math.ucr.edu/home/baez/week297.html > > May 9, 2010 > This Week's Finds in Mathematical Physics (Week 297) > John Baez > > This week I'll talk about electrical circuits and Dirichlet forms. > But first: knot sculptures, special relativity in finance, lazulinos, > some peculiar infinite sums, and a marvelous fact about the number > 12. � (snip) ----- "Next, an observation from Robert Baille. Take this series: Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - ... Square each term, add them up... and you get the square of the previous sum: Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 + 1/11^2 + ... Don't tell undergraduates about this - they are already confused enough! " ----- The Freshman Exponentation Error is thinking that: (a + b)^n = a^n + b^n so we might call this the Freshman Series. :) For prime p: (a+b)^p == a^p + b^p mod p because all terms other than a^p and b^p in the expansion have a factor of p and so are zero mod p. Does (a+b+c)^p == a^p + b^p + c^p mod p? Note that (a+b+c)^2 = a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2 which is equal to a^2 + b^2 + c^2 + 2ab + 2ac + 2bc, which makes it look reasonable... Is this true always? Douglas (Dana) Goncz, CPS Replikon Research PO Box 4394 Seven Corners, VA 22044-0394
From: Gerry Myerson on 24 May 2010 18:58 In article <2707e274-ee35-4837-91d5-af2363ba35ef(a)s41g2000vba.googlegroups.com>, The Dougster <dgoncz.703.475.7456(a)gmail.com> wrote: > For prime p: > > (a+b)^p == a^p + b^p mod p > > because all terms other than a^p and b^p in the expansion have a > factor of p and so are zero mod p. > > Does > > (a+b+c)^p == a^p + b^p + c^p mod p? Of course. It follows from 2 applications of the previous formula. Now apply induction to get a general result. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Tony on 27 May 2010 10:11
[[Mod. note -- This seems to have drifted to the outer limits of *physics* research. Further discussion of number theory, while interesting, probably belongs over in one of the sci.math.* newsgroups. -- jt]] On May 24, 9:45 am, The Dougster <dgoncz.703.475.7...(a)gmail.com> wrote: [snip] > For prime p: > > (a+b)^p == a^p + b^p mod p > > because all terms other than a^p and b^p in the expansion have a > factor of p and so are zero mod p. > > Does > > (a+b+c)^p == a^p + b^p + c^p mod p? [snip] > Is this true always? Yes, of course. (a+b+c)^p = (a + (b+c))^p = a^p + (b+c)^p etc. But I wish you would change the title of this thread. |