June 2008 - mathfun@mailman.xmission.com

Re: [math-fun] becoming a 21st century mathematician
by James Propp 09 Jun '08

09 Jun '08

>In any case, I presume you want to output to some file format (e.g. PDF), >rather than interactive display? Though one could always screen grab, in >small quantities anyway ... PDF would be fine. (If I wanted an animation, I'd probably take the trouble of learning Java or Pytthon, which I may actually do, one of these days.) >> OS X puts all the files and directories in one's Desktop into RAM at >> startup. Aside from the fact that it's undocumented, it's just not >> the sort of thing a UNIX system would do. > > That's gross! How did you find out about it? I took my Mac to the nearest Apple Genius Bar because I was having power issues (I couldn't figure out why my battery had lasted only a year), and the guy who worked there said "Well, one problem is the you've got so many files on your desktop." >The Mac Desktop has other problems as well. There's an associated reference >file which apparently goes on getting larger, more convoluted, and slower >indefinitely; and can't be deleted without super-user / root privilege >[tricky if, >like mine, your SU password mysteriously fails to work]. Can someone provide the name of the reference file, or better yet, information on how to unclutter it? I still remember my root password so I'm in a position to do something about the problem. Jim

1 0

Re: [math-fun] Analytic continuation software?
by Dan Asimov 08 Jun '08

08 Jun '08

Andy wrote: << On Sun, Jun 8, 2008 at 5:05 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: <<< Speaking of math software, can anyone tell me if there exists analytic continuation software? . . . . . . The software would also output a routine that will evaluate the function f_k inside the disk D_k. (Or a statement that the function cannot be continued along a(t), 0 <= t <= L.) . . . . . . and all this to a pre-specified level of numerical accuracy. >>> I don't think this is possible. In finite time the software can only evaluate your original function f at a finite number of points. And there are analytic functions that have exactly the given value at that finite number of points, but have any desired value at any other point. >> Right you are! So please take my request for analytic continuation software as an ideal. Any approximation thereto would be of interest. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

Re: [math-fun] Superfluous Multinomials ( and conjecture to Lyndon words)
by Dan Asimov 08 Jun '08

08 Jun '08

Wouter wrote: << Let q[n] represent the n-th root of -1, or exp(2 i pi/n) given that Sum(j=1 to n, q[n]^j) equals zero by symmetry, it is evident that (q[1]+q[2]+..+q[n])^k also equals zero for any k. . . . >> Did you intend to omit the 2: q[n] := exp(i pi/n) (so that q[n]^n = -1) ? or perhaps q[n] represents the n-th root of +1 ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

1 0

Re: [math-fun] Superfluous Multinomials ( and conjecture to Lyndon words)
by wouter meeussen 08 Jun '08

08 Jun '08

Let q[n] represent the n-th root of -1, or exp(2 i pi/n) given that Sum(j=1 to n, q[n]^j) equals zero by symmetry, it is evident that (q[1]+q[2]+..+q[n])^k also equals zero for any k. Now, looking at the expansion of this power, and dropping (setting =1) the multinomial coefficients, it is (was) mildly surprising this expression is still zero, except when k=n, in which case it is 1. reason: use q[n]^z = q[n]^mod(z,n) and find that Sum(j=1 to n, q[n]^j)^k without the usual multinomials, and with the use of q[n]^z = q[n]^mod(z,n) gives oh, darn, now that we've done the mod thing, we can drop the [n] bit, and just write {{1}, {1 + q, 2 + q}, {1 + q + q^2, 2 + 2 q + 2 q^2, 4 + 3 q + 3 q^2}, {1 + q + q^2 + q^3, 3 + 2 q + 3 q^2 + 2 q^3, 5 + 5 q + 5 q^2 + 5 q^3, 10 + 8 q + 9 q^2 + 8 q^3}, {1 + q + q^2 + q^3 + q^4, 3 + 3 q + 3 q^2 + 3 q^3 + 3 q^4, 7 + 7 q + 7 q^2 + 7 q^3 + 7 q^4, 14 + 14 q + 14 q^2 + 14 q^3 + 14 q^4, 26 + 25 q + 25 q^2 + 25 q^3 + 25 q^4}} etc.. so, for each pair (n,k), we got ourselves a polynomial in q, but, a surprisingly simple one, at first sight. I expected a straight 1+q+q^2+..+q^n everywhere, but no! There seems to be a bit more to it. The least we can do, is to look at the coefficients of q^0 and q^n, no? And that, dear reader, if you've not run off by now, severely contorted with abdominal cramps with crap overload, is where two unseemly conjectures get pushed into the open (excuse my french) given that W*q+W*q^2+...+W*q^n equals W*(ziltch)=0 as we very well could and should expect, where does the W factor come from? conjecture 1: the coefficient of q^n is the constant term, q^n being 1, and equals Sum(d | gcd(n+k,k) ; phi(d) C(n/d+k/d,n/d) ) Cf. A004526, A007997, A008610, A008646, A032191, A032192, A032193 (Number of necklaces with n black beads and k white beads check item 1 below. conjecture 2: the coefficient of q^(n-1) is the factor of the highest power of q, and equals Sum(d | gcd(n+k,k) ; mu(d) C(n/d+k/d,k/d) ) A130513: Subtriangle of triangle in A051168 : remove central column of A051168 and all columns to the right; now read by upwards diagonals. check item 2 below. please take a sec to notice their similarities, phi versus mu, C(..,n/d) versus C(..,k/d) the other coefficients? I dun'no, phu versus mi? Wouter. --------------------------------- item 1 : conjecture 1. z=Table[Expand[ Plus@@ Table[q[n,j],{j,n}]^k /k! ],{n,5},{k,n}]/. q[n_,j_]^(e_:1) -> e! q[n]^(e j) a=Table[ Plus@@ Apply[Times,(q[n]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] i assume z==a 'by nature', .. handwaving...then, b=a/. q[n_]^(e_:1) :> q^Mod[e,n] (* just take 'a', replace q[n]^e with q^mod(e,n) as we should for roots of 1 *) b /. q->0 (* and set q equal to 0 to get the constant term.. *) {{1}, {1, 2}, {1, 2, 4}, {1, 3, 5, 10}, {1, 3, 7, 14, 26}, {1, 4, 10, 22, 42, 80}, {1, 4, 12, 30, 66, 132, 246}, {1, 5, 15, 43, 99, 217, 429, 810}, {1, 5, 19, 55, 143, 335, 715, 1430, 2704}, {1, 6, 22, 73, 201, 504, 1144, 2438, 4862, 9252}} and that looks like (=conjecture) Table[1/(n+k) Plus@@ (EulerPhi[#]Binomial[n/#+k/#,n/#]&/@ Divisors[GCD[n+k, k]]),{n,10},{k,n}] --------------------------------- item2 : conjecture 2. Map[Coefficient[q #,q^Exponent[q #,q]]&,b] (* get the highest exponent's factor *) {{1}, {1, 1}, {1, 2, 3}, {1, 2, 5, 8}, {1, 3, 7, 14, 25}, {1, 3, 9, 20, 42, 75}, {1, 4, 12, 30, 66, 132, 245}, {1, 4, 15, 40, 99, 212, 429, 800}, {1, 5, 18, 55, 143, 333, 715, 1430, 2700}, {1, 5, 22, 70, 200, 497, 1144, 2424, 4862, 9225}} and that looks like (=conjecture) Table[1/(n+k) Plus@@ (MoebiusMu[#]Binomial[(n+k)/#,k/#] &/@ Divisors[GCD[n+k,k]]),{n,12},{k,n}] ---------------------------------- minor OEIS bonus (to maybe submit after the summer recess) A051168: T(h,k)=number of Lyndon words of k 1's and h-k 0's. See http://www.theory.cs.uvic.ca/~cos/inf/neck/NecklaceInfo.html Sum(d | gcd(n,k) ; mu(d)/n C(n/d,k/d) ) Table[1/n Plus@@ (MoebiusMu[#]Binomial[n/#,k/#] &/@ Divisors[GCD[n,k]]),{n,12},{k,0,n}] A130513: Subtriangle of triangle in A051168 : remove central column of A051168 and all columns to the right; now read by upwards diagonals. Column order reversed: Sum( d | gcd(n+k,k) ; mu(d) C(n/d+k/d,k/d) /(n+k) ) or, in Mma dialect: Table[1/(n+k) Plus@@ (MoebiusMu[#]Binomial[n/#+k/#,k/#] &/@ Divisors[GCD[n+k,k]]),{n,12},{k,n}] ------------------------------------ end-of-tale

1 0

[math-fun] Morpion Solitaire
by Christian Boyer 06 Jun '08

06 Jun '08

An easy game: Only paper and pencil needed. A game for one player. And an open game: Who will improve the current world records??? Here is my new website dedicated to "Morpion Solitaire": www.morpionsolitaire.com Fully bilingual, English/French. Feel free to report any error, or any missing information. As you can read in the "Origin" page, any new information on the origin of this game will be welcomed. Some other questions. Is it possible to do best than the two grids given in "Other 5"? Or worst than my very bad grids at the end of the "Limits" page? Christian.

1 2

Re: [math-fun] becoming a 21st century mathematician
by N. J. A. Sloane 04 Jun '08

04 Jun '08

For any work involving group theory, number theory, algebraic geometry, combinatorics, algebra, coding theory, ring theory, etc., etc., - mathematics generally - MAGMA is indispensable Neil

4 3

[math-fun] Kepler's pizza
by rwg＠sdf.lonestar.org 04 Jun '08

04 Jun '08

(my favorite math teacher embarrasser) is now a click at http://www.tweedledum.com/rwg/, per prodding by MLB. The .mfe is http://gosper.org/pizza.mfe (not https, as I mistakenly sent someones last nite). Caution: earlier versions contain a bogus definition of "mean anomaly". --rwg

2 2