April 2011 - mathfun@mailman.xmission.com

[math-fun] Simple problem
by David Wilson 09 Apr '11

09 Apr '11

A box holds 37 cubic inches. Add an inch to each side and it holds 83 cubic inches. Add another inch to each side and it holds 157 cubic inches. Add yet another inch to each side, and how much does it hold?

5 9

[math-fun] A blog post about the Minsky Circle Algorithm.
by Neil Bickford 08 Apr '11

08 Apr '11

I've made a blog post about the (non)recent discoveries about the Minsky Circle algorithm: http://nbickford.wordpress.com/2011/04/03/the-minsky-circle-algorithm/ Comments welcome! --Neil Bickford

2 1

[math-fun] High Precision Computing ...
by Guy Haworth 08 Apr '11

08 Apr '11

Maybe a reference to http://people.maths.ox.ac.uk/trefethen/tools.html, esp to the section on 'Symbolic and High Precision Computing', is useful. Guy

1 0

[math-fun] Algorithms for high precision computation of things
by quad 08 Apr '11

08 Apr '11

Hello all! Many months ago, maybe even a couple years ago (I lost count), I wrote a high precision floating point arithmetic library in C. Here's a sample demo of it: WINDOWS: http://dl.dropbox.com/u/734346/rpn.exe [open a command prompt, cd to the right directory, then type "rpn" to see the options] LINUX (32-bit): http://dl.dropbox.com/u/734346/rpn [should work on 64-bit] The LINUX executable might run on OS X. In fact, I'd like to know if it does, if any one could test it! I can assure you it's safe. It was written in ANSI C by me in a very portable fashion. No file writing, no system calls, etc. If you think it's dodgy, then feel free to disassemble it or something. It's an RPN calculator. Basic usage is along the lines of $ ./rpn -p1000 1 tta 4 x which does 4*atan(1) (tta = trig tan arc; x = multiply) to 1000 digits. Anyway, you don't need to run it, and it's not the point of this email. The algorithms I use are here: http://www.symbo1ics.com/files/algs.pdf [Section 3] Any corrections would be helpful. Anyway, I'm looking: 1. to include more algorithms that are known to work well for computing special functions and whatever to high precision. 2. to include improvements to what I have at the moment, algorithmically. Is the way I compute sine slow? Are there faster ways? 3. for criticism on the way I represent big floats. I don't know if there is a better way to express arbitrary precision floats than what I have. I prefer *not* to go the "do everything with rationals then divide" because that's memory intensive and has a good amount of overhead. Of course some algorithms work best by using rationals, but I don't want to use rationals entirely across all algorithms. Mostly I'd like to improve what I have, but also document things I don't yet have, like Gosper's theta function products, and whatever. Thanks all! Robert

3 3

[math-fun] computing Penrose tilings
by James Propp 07 Apr '11

07 Apr '11

I'm having trouble finding explicit recipes on the web for having a computer generate a finite excerpt of a Penrose tiling (in any of its equivalent forms; personally I like the rhombus-tilings). One sort of solution would be a prescription of the form "To generate the set of vertices of the rhombus-tiling that lie within distance d of the origin in C, let zeta = exp(2 Pi i / 5) and take all sums of the form a_1 zeta + a_2 zeta^2 + a_3 zeta^3 + a_4 zeta^4 with a_1, a_2, a_3, a_4 in Z satisfying the following inequalities: ..." Another sort of recipe would be an algebraic version of the inflation rule, where one starts with a finite set of colored points in R^2 and then replaces it by a larger set of colored points, applying some linear maps to get new points from old (where the choice of map depends on the color of the point, and new points get colored according to some rule). Surely something of this sort is on the web, but I haven't been able to find it. For that matter, I recall that there's a nice way to characterize one of the versions of Penrose tilings by means of the algebraic conjugacy of Q(zeta) (with zeta as above) that sends zeta^k to zeta^2k: we take those points in Q(zeta) such that the image of the point under the conjugacy lies in some finite window (a pentagon of some size centered on 0, as I recall). I learned about this from Veit Elser at an Oberwolfach meeting many years ago, and I believe it may be due independently to Veit and Rick Kenyon; was a description of this point of view ever published? Jim Propp

4 3

Re: [math-fun] FBI Needs Cryptanalysis Help (possible answer)
by Michael Beeler 06 Apr '11

06 Apr '11

Unfortunately, it looks like not much math in this puzzle. While no one has figured it out for certain yet, a commenter on Yahoo News (oddly) seems to have the best suggestion for what the notes actually are. Apparently, it looks a lot like a medication log written in a sort of pseudo-code. In a later comment he mentions that people who are bipolar are sometimes told to keep their medication and episode notes secret in order to prevent them being used in a court if they were to commit a violent act (which sounds a bit fishy, but the rest of it makes a fair amount of sense). Here is part of the comment from Yahoo. It is a shorthand log of historic episodes in the mid seventies on (page 1, actually written second, but numbered one to keep events in chrono order) and medications taken with the effects listed. The key at the end is day week month year morning day latenight. It was started on page 2 and then page 1 was added as a log of the earlier childhood which is the basis for diagnosis and the "page 2" is indepth records of changes in meds. The 3 month periods are normal with bipolar episodes in the 4th QTR (September through December in the seventies. These seasons suggest seasonal disorder. ALPNTE GLSE-SE ERTE A: Latenight, Phenergan, taken in evening G: Latenight Serenace/Seroquel or Seroquel/Serenace Extended Release Taken Evening link (the commenter's name is John): http://news.yahoo.com/s/yblog_thelookout/20110329/ts_yblog_thelookout/fbi-a…

1 0

[math-fun] formula for the distance between two disjoint affine subspaces
by Jean Gallier 05 Apr '11

05 Apr '11

There is a formula giving the distance between two disjoint affine subspaces involving Gram determinants. Recall that if e_1, ..., e_m are vectors in R^n (m <= n) then Gram(e_1, .., e_m) = det(<e_i, e_j>), where <e_i, e_j> is the inner product of e_i and e_j. If U_1 and U_2 are disjoint affine subspaces, let V_1 and V_2 be their linear directions, that is, unique vector spaces such that U_1 = a_1 + V_1 and U_2 = a_2 + V_2, for any a_1\in U_1 and any a_2\in U_2. Pick any basis e_1, ..., e_m of V_1 + V_2 , then the square of the distance d(U_1, U_2) between U_1 and U_2 is given by d(U_1, U_2)^2 = Gram(a_1 - a_2, e_1, ..., e_m)/Gram(e_1, ..., e_m). In the special case where U_1 and U_2 are skew lines (i.e., not parallel), we also have the formula d(U_1, U_2)^2 = Gram(a - a', a - b, a' - b')/Gram(a - b, a' - b'), where a, b are any two distinct points on U_1 and a', b' any two distinct points on U_2. The above formula can be found in Berger, Geometry I, Chapter 9, Section 2. I found the more general formula in "Methodes Modernes en Geometrie", by Jean Fresnel, Part C, Section 1.4.2. The proof is not entirely trivial. See also Problem 7.13 of my "Geometric Methods and Applications", TAM 38, Springer-Verlag. Best, -- Jean Gallier

2 3

[math-fun] q-exponentials
by Bill Gosper 05 Apr '11

05 Apr '11

"Shew that" Limit (z*(q-1);q)_oo q->1 = 1/Limit (z*(1-q);q)_oo q->1 = e^z . I thought I once did by directly manipulating the infinite product(s), but can't recall how. The usual proof uses identities for equivalent summations whose termwise limits -> (+-z)^n/n! . But recall 26 Jan 11, Re Plouffe's 2 surprising approximations: rwg>Yes, it's immediate. Rewrite eta as its infinite product; log(prod) -> sum(logs); log -> powerseries; sumswap; closed form inner sum. (c1) [%E^(%PI*K*N) = Q^-N,%E^-(%PI*K) = Q,%PI*K = -LOG(Q)]; %pi k n 1 - %pi k (d1) [%e = --, %e = q, %pi k = - log(q)] n q (c2) SUBST(%,SUM(1/(N*(%E^(%PI*K*N)-1)),N,1,INF) = -LOG(ETA(%E^-(%PI*K)))-%PI*K/24); inf ==== \ 1 log(q) (d2) > ---------- = ------ - log(eta(q)) / 1 24 ==== n (-- - 1) n = 1 n q [...]<rwg This easily generalizes to (c1) POQIFY(LOG(%E^-SUM(Z^I/(I*(1-Q^I)),I,1,INF) = PRODUCT(1-Q^I*Z,I,0,INF))); inf ==== i \ z (d1) - > ---------- = log(qpoch(z, q, inf)) / i ==== i (1 - q ) i = 1 which Macsyma can test directly: (c3) BLOCK([SIMPSUM : TRUE],TAYLOR(SUBST((Q-1)*Z,Z,MAKEPROD(D1)),Z,0,3)); 2 2 3 (q - 1) z (q - 2 q + 1) z (d3)/T/ z + ---------- + ----------------- + . . . = 2 q + 2 2 3 q + 3 q + 3 2 2 3 (q - 1) z (q - 2 q + 1) z z + ---------- + ----------------- + . . . 2 q + 2 2 3 q + 3 q + 3 But we only need the first term for the q->1 limit. The two usual q-exponential(z) functions are the cases p=1 and -1 of the continuum ((p*z*(q-1);q)_oo)^p . Based on q-trig, Julian proposes yet another candidate, qexp(pi z) := (theta[4](z ln q)-theta[1](z ln q))/theta[4](0)/q^z^2 = 1 - 2 qpi z ln q/(1-q^2) + z^2 ln q (theta''[4](0) ln q/2/theta[4] - 1) ... (where qpi := (1+q)* "q Wallis's product) but I don't know what nice properties it might have. Presumably one can recast q-trig double angle formulas etc in terms of it. --rwg Mike Hirschhorn privately remarked that the QPochhammer reflection formula can't be new. I agree, and conjecture that BHS omitted it to skirt the gigantic subject of theta functions.

1 1

[math-fun] Lambert series
by Bill Gosper 04 Apr '11

04 Apr '11

I didn't realize we could do these. Sum[Csch[n*x]^2, {n, 1, Infinity}] == 1/6 + Derivative[0, 1, 0][EllipticThetaPrime][1, Pi/3, E^(-x/3)]/ (6*Sqrt[3]*DedekindEta[(I*x)/Pi]) Mma doesn't realize it either: In[517]:= % /. x -> 1/2 Out[517]= False Yet In[518]:= List @@ %% /. x -> 1/2 Out[518]= {4 (-1 + QPolyGamma[1, 1, Sqrt[E]] + QPolyGamma[1, 1 - 2 I Pi, Sqrt[E]]), (0,1,0) Pi 1 EllipticThetaPrime [1, --, ----] 3 1/6 1 E - + --------------------------------------} 6 I 6 Sqrt[3] DedekindEta[----] 2 Pi In[519]:= N[%, 33] Out[519]= {4.74640293405957128234200051279399 + 0.*10^-35 I, 4.74640293405957128234200051279836} --rwg

1 1

Re: [math-fun] Gould's sequence and 'mean spectra'
by N. J. A. Sloane 03 Apr '11

03 Apr '11

Adam, those statistics about A001316 are very interesting. (Especially since I was recently looking at similar properties for a closely related sequence.) Could you add them to the entry for A001316? Or, since this might enlarge the entry too much, put them in a text file (or pdf file) similar to your math-fun posting, and upload it to the OEIS web site as a file to be attached to A001316? Thanks! Neil PS You would need to register first, but this is easy. Announcement, Nov 17 2010: New Version of OEIS! =============================================== The OEIS is now running on the web. There are two parts: the main OEIS page is oeis.org, which has all the sequence data, and a wiki, oeis.org/wiki, which has discussion pages. To contribute a new sequence or edit a sequence, you must first register at oeis.org/wiki - click Register, enter your real name (use John M. Smith, NOT john smith, NOT johnny33), enter your email address, and a brief bio. You will be sent an automatic email (from admin(a)oeis.org, I think), which you must answer to get your password. When you get the password, go to the main page, oeis.org, and login with the same user name and password. Neil Sloane President, OEIS Foundation Inc (see http://oeisf.org/)

1 0