January 2008 - mathfun@mailman.xmission.com

Re: [math-fun] church of the sub-genus?
by James Propp 04 Jan '08

04 Jan '08

I wrote: >the set >of functions from [0,1] to {0,1} such that the pre-images of both 0 and 1 >consist of finitely many points and intervals is something like an infinite >dimensional polytope, and it seems to be trying to have Euler characteristic >1/2. Replace [0,1] here by (0,1). (The mumbo-jumbo way to compute the Euler characteristic of the set of polyhedral maps from (0,1) to {0,1} is to take the Euler characteristic of {0,1} to the power of the Euler characteristic of (0,1) --- that's 2 to the power of -1, or 1/2.) And before you object that the Euler characteristic of the interval (0,1) is +1 rather than -1, let me hasten to add that the kind of Euler characteristic I'm using is not the usual homotopy-invariant kind, but a more naive sort that has different good properties (such as the property of being additive). To compute this kind of "combinatorial" Euler characteristic, think "V-E+F": for {0,1}, V-E+F = 2-0+0 = 2, while for (0,1), V-E+F = 0-1+0 = -1. Jim Propp

2 1

Re: [math-fun] church of the sub-genus?
by James Propp 03 Jan '08

03 Jan '08

Back on December 11, Marc LeBrun asked: >If we can have "fractional dimension", can we somehow have "fractional >genii"? Dan Asimov replied: >There are uses for fractional Euler characteristics in describing orbifolds, >but as far as I know ordinary surfaces have integer genera. There's also a sense in which certain sorts of infinite-dimensional objects can be sensibly said to have fractional Euler characteristic; e.g., the set of functions from [0,1] to {0,1} such that the pre-images of both 0 and 1 consist of finitely many points and intervals is something like an infinite dimensional polytope, and it seems to be trying to have Euler characteristic 1/2. See "Exponentiation and Euler Characteristic" (math.CO/0204009). I also recall that there's a sense in which the classifying space of a finite group G has 1/|G| elements. See http://www.math.ucr.edu/home/baez/counting/ for relevant references. Jim Propp

1 0

Re: [math-fun] xkcd points out dangers of math fun
by Fred lunnon 03 Jan '08

03 Jan '08

On 12/15/07, Steve Witham <sw(a)tiac.net> wrote: > Here's a comic with an interesting random-walk problem: > > http://www.xkcd.com/356/ > Apparently quoted from Google Labs aptitude test --- On an infinite square lattice with equal unit resistances along each edge, find the resistance between two nodes at a knight's move from one other. I got an extraordinary answer for this --- has anybody else had a go at it? WFL

1 2

Re: [math-fun] Proofs without words
by Dan Asimov 02 Jan '08

02 Jan '08

Michael Reid wrote: << ____ /\ \ ____/ \___\ /\ \ / / ____/ \ \/___/ /\ \ \ \ ____/ \ \ \_______\ /\ \ \ \ / / / \ \ \ \/_______/ \ \ \ \ \ \ \ \ \___________\ \ \ \ / / \ \ \/___________/ \ \ \ \ \_______________\ \ / / \/_______________/ . . . . . . >> Lovely proof . . . and top-notch ASCII art as well! --Dan

1 0

Re: [math-fun] Proofs without words
by Michael Reid 02 Jan '08

02 Jan '08

____ /\ \ ____/ \___\ /\ \ / / ____/ \ \/___/ /\ \ \ \ ____/ \ \ \_______\ /\ \ \ \ / / / \ \ \ \/_______/ \ \ \ \ \ \ \ \ \___________\ \ \ \ / / \ \ \/___________/ \ \ \ \ \_______________\ \ / / \/_______________/ x 3 = ________________ /\ \ ____/ \____ \ /\ \ / /\ \ / \ \/___/ \____ \ / \ \ / /\ \ / \ \/___/ \____ \ / \ \ / /\ \ / \ \/___/ \___\ / \ \ / / / \_______________\/___/ \ / / / \ / ____/ / \ / / / \ / ____/ / \ / / / \ / ____/ / \ / / / \/___/_______________/ > b) sum of the first n squares = n(n+1)(2n+1)/6 ok, i guess i'll need some words now. six copies of the "square pyramid" pack an n x (n + 1) x (2n + 1) box, whence the formula. there are six such blocks in the math common area at smith college, where i learned of this construction. i later saw a similar, or perhaps the same construction, i believe as a "proof without words" in math intelligencer. (i'm not sure which came first.) i don't have the reference at my fingertips; perhaps someone can find it. (it's also possible that the intelligencer proof only used 3 pyramids, then sliced and re-assembled to get an n x (n + 1) x (n + 1/2) block.) mike

1 0

Re: Question Re: [math-fun] 2008 puzzles from the OEIS; answers next year
by Marc LeBrun 01 Jan '08

01 Jan '08

>=Dan Asimov >>=Marc LeBrun >> A104628: What 5 integer powers of the golden ratio sum to 2008? > May I inquire: Of course! (Please!<;-) > does "golden ratio" here mean (sqrt(5)+1)/2, or its reciprocal (sqrt(5)-1)/2 ? Heh: Yes! It doesn't matter much, right? Substituting the reciprocal will just negates all the powers. Isn't "golden ratio" (aka phi aka tau) generally understood to mean the larger (sqrt(5)+1)/2? Yet the "classic" defining relation a:b :: b:a+b would suggest a<b=1. What would Euclid say?

1 0

Question Re: [math-fun] 2008 puzzles from the OEIS; answers next year
by Dan Asimov 01 Jan '08

01 Jan '08

Marc LeBrun quoted: << A104628: What 5 integer powers of the golden ratio sum to 2008? May I inquire: does "golden ratio" here mean (sqrt(5)+1)/2, or its reciprocal (sqrt(5)-1)/2 ? --Dan

1 0

Correction Re: [math-fun] Proofs without words
by Dan Asimov 01 Jan '08

01 Jan '08

In error I wrote: << Sum_{k = 1 to n} 1/k^s = (1/Gamma(s))*Integral_{0 to oo} ((1-u^n)/(1-u))) (ln(1/u))^(s-1) du. >> but in fact the upper limit of the integral should be 1: Sum_{k = 1 to n} 1/k^s = (1/Gamma(s))*Integral_{0 to 1} ((1-u^n)/(1-u))) (ln(1/u))^(s-1) du.

1 0

Re: [math-fun] Proofs without words
by Dan Asimov 01 Jan '08

01 Jan '08

I asked: << Can someone please point me to proofs without words for a) nth tetrahedral number = n(n+1)(n+2)/6 b) sum of the first n squares = n(n+1)(2n+1)/6 c) sum of the first n cubes = (sum of first n numbers)^2 = (T_n)^2 (T_n = nth triangular #) ? >> Thanks for the nice answers. (Alas, Richard, my beloved copy of TBoN is still locked up in storage along with most of my math books.) * Fred's instantly clear induction step for a) immediately leads to a proof that generalizes the formulae for triangular and tetrahedral numbers to all "K-simplicial numbers" S_K(n): S_K(n) = (n+K-1)!/(K! (n-1)!). Which suggests the possibility of a proof directly based on this being the binomial coefficient n+K-1 choose K. (Or n-K+1 choose n-1: take your choice.) * Both a) and b), at least, look like discrete versions of the geometrical facts that a pyramid with base an equilateral right triangle (resp. unit square) and height 1 has volume = 1/6 of that of a unit cube (resp. 2x1x1 rectangular solid). (Perhaps this can be made to fit into an instant continuous-to-discrete theorem?) * I just read somewhere that for odd p, the (formula for the) sum of the first n pth powers is always a polynomial in the nth triangular number. QUESTION: For which (p,q) is the (formula for the) sum of the first n pth powers a polynomial in (the formula for) the sum of the first n qth powers? --Dan P.S. A formula I once found for the sum of the first n reciprocal sth powers (at least for real s >= 1) is: Sum_{k = 1 to n} 1/k^s = (1/Gamma(s))*Integral_{0 to oo} ((1-u^n)/(1-u))) (ln(1/u))^(s-1) du. Letting n -> oo gives an integral formula for the product of the Gamma and Zeta functions. I was crestfallen when in 1964 my freshman advisor (Henry McKean) showed me this limiting formula in Whittaker & Watson (1902).

2 1

[math-fun] Proofs without words
by Dan Asimov 01 Jan '08

01 Jan '08

Can someone please point me to proofs without words for a) nth tetrahedral number = n(n+1)(n+2)/6 b) sum of the first n squares = n(n+1)(2n+1)/6 c) sum of the first n cubes = (sum of first n numbers)^2 = (T_n)^2 (T_n = nth triangular #) ? Thanks, Dan

4 3