February 2012 - mathfun@mailman.xmission.com

[math-fun] Diophantine equation a^4+b^4=d*c^2 and the specialness of 17, 68, 82...
by Warren Smith 05 Mar '12

05 Mar '12

Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive integer solutions. Also a^4-b^4=2*c^2 has no positive integer solutions. These results are shown in T.Nagell: Intro to number theory, Wiley 1951, pp.227-230. However, a^4+b^4=2*c^2 trivially has a ton of solutions a=b, c=a^2. If we exclude the trivial by demanding b>a>0 though, I don't know of any solutions. Let us generalize. Ask about a^4 + b^4 = d*c^2 with a,b,c,d integers and b>a>0. My computer looked for solutions. Search I: a^4+b^4=d*c^2 with 0<a<b<65536 and 0<d<65. Result: a ton of solutions for d=17, but zero solutions for any other d. a=1, b=2, d=17, c=1 a=2, b=4, d=17, c=4 a=2, b=13, d=17, c=41 a=3, b=6, d=17, c=9 a=4, b=8, d=17, c=16 ... Well, that's interesting: 17 is special. Search II: a^4+b^4=d*c^2 with 0<a<b<=1024 and 0<d<=1024. Result: The following d are "special": 17, 68, 82, 97, 113, 153, 193, 257, 272, 274, 328, 337, 388, 425, 433, 452, 514, 577, 593, 612, 626, 641, 673, 706, 738, 772, 833, 873, 881, 914, 1017. [This sequence is not in OEIS.] What is going on here? I do not really know, but an initial speculation is this. When d=17, there is the easy solution a=1, b=2, d=17, c=1. But here is the thing: if there is one solution, then it can usually be used to generate an infinite set of solutions. I got that from Tito Piezas's online book http://sites.google.com/site/tpiezas/018 where he attributes it to A.Desboves (probably about 1880?). So there are "special" d with an infinity of solutions, and there are "usual" d with zero solutions (and conceivably there are a few "crazy" d with a finite nonzero number of solutions because the below Desboves method for generating more solutions happens merely to regenerate old ones, which would be an amazing miracle). How to generate more solutions from one: Piezas says Desboves said (as edited by me): if p^4+q^4 = d*r^2, then x^4+y^4 = d*z^2 where x = p*(4*m^2-3*(m+n)^2), y = q*(4*n^2-3*(m+n)^2), z = r*(4*(m+n)^4-3*(m-n)^4) where m=p^4 and n=q^4. So for example starting from p=1, q=2, d=17, r=1 we get x=863, y=314, z=182209 and sure enough 863^4 + 314^4 = 17*182209^2 is another solution. Questions: Q1. Can you find a parameterized family of d=17 solutions? Q2. Can you understand the set of special d? Q3. Can you prove there are no "crazy" d (or find them all)?

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[math-fun] Tiling a square
by Kerry Mitchell 29 Feb '12

29 Feb '12

Hi all, I know that 7 * 5^2 + 9 * 3^2 = 16^2, which suggests that 7 5x5 squares plus 9 3x3 squares can possibly tile a 16x16 square. Is there a relatively simple way to figure out if either: a) no, they won't, or b) yes, and here's how? Ideally, I'd like to learn the process, not just the answer, as I will probably have other combinations to consider. Thanks, Kerry Mitchell -- lkmitch(a)gmail.com www.kerrymitchellart.com

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[math-fun] Provide & Require
by Robert Smith 28 Feb '12

28 Feb '12

Hey all. Computing history question. What's the earliest instance of PROVIDE and REQUIRE functionalities provided in languages today? Common Lisp uses *FEATURES* with PROVIDE and REQUIRE. Emacs Lisp the same. I am wondering if it comes any earlier, especially with Lisp implementations. Thanks, Robert P.S., if you can make this math related, you get my good graces.

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[math-fun] Equilateral Triangulations with a sense of humour!
by Stuart Anderson 28 Feb '12

28 Feb '12

http://www.interactivetriangulation.com/ click away! (thanks to Geoffrey Morley for the link!)

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Re: [math-fun] Limit puzzle
by Bill Gosper 28 Feb '12

28 Feb '12

* Bill Gosper <billgosper(a)gmail.com> [Feb 28. 2012 08:14]: > [...] > > Dawk! And better still as one Lambert series! > Joerg>(first:) > Sum[((-1)^k*q^((1/2)*(-1 + k)*k)*x^k*QPochhammer[q, q, -1 + k])/ > (QPochhammer[-1, q, 1 + k]*QPochhammer[-x, q, k]), {k, 1, Infinity}] > == (second:) > Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] + > (1/2)*x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}] > == (third:) > (-(1/2))*x*Sum[(-q)^n/(1 + q^n*x), {n, 0, Infinity}] > Prettier still (except for taking x -> +or-1) (fourth:) (-(1/2))*x*Sum[(-q)^n/(1 + q^n*x), {n, 0, Infinity}] == (1/2)*Sum[(-x)^j/(1 + q^j), {j, 1, Infinity}] These Lambert transformations are all the same easy trick, which I'm not stating because I've already asked Funster Neil (who owes us his sliding block report) to figure it out for tomorrow. Joerg> The equality (first) == (third) appears to be a specialization of my relation (7): \begin{equation}\label{rel:lambert-qxt-alt} % sum(n=0,N, t^n/(1-x*q^n) ); \sum_{n\geq{}0}{ \frac{t^n}{1-x\,q^{n}} } = % % sum(n=0,N, (qbin(q,q,n)) / (qbin(x,q,n+1)*qbin(t,q,n+1)) * (-x*t)^n * q^((n^2-n)/2) ); \sum_{n\geq{}0}{ \frac{ (q;q)_n }{ (x;q)_{n+1} \, (t;q)_{n+1} } \, (x\,t)^n\, q^{(n^2-n)/2} } \end{equation} Joerg>The equality (second) == (third) resembles the "well known" sum(n>=1, x*q^n/(1-x*q^n) ) == -x/(1-x) + sum(n>=1, x^n/(1-q^n) ) (but I cannot morph this one into your's right now, replacing x by -x may be a good start). > [...] This whole limit puzzle arose from the false hope that a->-1 would nicely simplify the lhs of (QPochhammer[a, q]*QPochhammer[(a*q)/x, q]*QPochhammer[a*x, q]* (Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}] + (a*(x + q)* Sum[(QPochhammer[-a, q, n]*q^n*QPochhammer[a^2*q, q^2, n])/ (QPochhammer[a, q, n + 1]*QPochhammer[(a*q)/x, q, n + 1]* QPochhammer[a*x, q, n]), {n, 0, Infinity}])/x))/ (QPochhammer[-a, q]*QPochhammer[a^2*q, q^2]) == Sum[(-1)^n*Sqrt[q]^((n - 1)*n)*x^n, {n, 0, Infinity}] giving a "unilateral triple product identity". --rwg Re Provide and Require Macsyma was written in Lisp, but its SETUP_AUTOLOAD may predate Lisp's equivalent functionality. However, the autoloading occurs at runtime, not compile time.

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Re: [math-fun] rwg's matrix product for Clausen's Lambert identity
by Bill Gosper 28 Feb '12

28 Feb '12

Your perseverance rewards us all. Glitch: $K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$ would be clearer as $N(k,\infty)$ for $\lim_{n\to\infty}{N(k,n)}$ --rwg Joerg>Finally. Typset document at http://www.jjj.de/lambert-paper/ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Gosper's matrix form of Clausen's identity} % \jjfile{gosper-path-invariant.gp} % Gosper defines \cite{Gosper-mathfun} matrices $K(k,n)$ and $N(k,n)$ as % \[ %% rwg: % {k, {{q^(1 + 2*k + n), -((q*(-1 + q^(2*k + n)))/((-1 + q^k)*(-1 + q^(k +n))))}, {0, 1}}}, %% jj: % Km(k,n)={ [ q^(n+2*k+1), q*(1-q^(2*k+n))/( (1-q^(k)) * (1-q^(k+n)) ) ; 0, 1 ]; } K(k,n)= \begin{bmatrix} q^{n+2k+1} & {q\,(1-q^{2k+n})}/{\left((1-q^k) \, (1-q^{k+n})\right)} \\ 0 & 1 \end{bmatrix} \] % \[ %% rwg: % {n, {{q^k, q/(1 - q^(k + n))}, {0, 1}}} %% jj: % Nm(k,n)={ [ q^(k), q/(1-q^(k+n)); 0, 1 ]; } N(k,m)= \begin{bmatrix} q^{k} & {q}/{(1-q^{k+n})} \\ 0 & 1 \end{bmatrix} \] % These matrices satisfy \[ %% rwg: % handy invariance checker % Nm(k,n) Km(k,n+1) = Km(k,n) Nm(k+1,n), %% jj: % Nm(k,n)*Km(k,n+1) - Km(k,n)*Nm(k+1,n) \\ == zero, OK N(k,n) \cdot K(k,n+1) = K(k,n) \cdot N(k+1,n) \] % Writing $K(\infty,n)$ for $\lim_{k\to\infty}{K(k,n)}$ and $K(k,\infty)$ for $\lim_{n\to\infty}{K(k,n)}$ we obtain \eqref{rel:clausen-lambert} as upper right entries on both sides of \[ %% rwg: % MProd[[N] /. k -> 1, {n, 0, oo}] . MProd[Limit[[K], n -> oo], {k, 1, oo}] %% jj: % Ln = prod(n=0,N, Nm(1,n) ) \\ == [0, Lam; 0, 1] % Lk = prod(k=1,N, Km(k,oo) ) \\ == [0, 1; 0, 1]] % L = Ln * Lk \\ == [0, Lam; 0, 1] \\ OK \prod_{n\geq{}0}{N(1,n)} \cdot \prod_{k\geq{}1}{K(k,\infty)} = % == %% rwg: % MProd[[K] /. n -> 0, {k, 1, oo}] . MProd[Limit[[N], k -> oo], {n, 0, oo}]] %% jj: % Rk = prod(k=1,N, Km(k,0) ) \\ == [0, Lam; 0, 1] % Rn = prod(n=0,N, Nm(oo,n) ) \\ == [0, q; 0, 1] % R = Rk * Rn \\ [0, Lam; 0, 1] \\ OK \prod_{k\geq{}1}{K(k,0)} \cdot \prod_{n\geq{}0}{N(\infty,n)} % % d = L - R \\ == zero \] %%%%%%%%%%%%%%%%%%%%%

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Re: [math-fun] Limit puzzle
by Bill Gosper 28 Feb '12

28 Feb '12

* Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> [Feb 27. 2012 14:27]:> On Sun, Feb 26, 2012 at 12:32 AM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:> > > (-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0,> > Infinity}])/> > (1 + a)> >> > as a -> -1+ . (In terms of q-hypergeometric(s) in x and q).> > > Here's a much better answer as two Lambert series:> > Limit[(-1/2 +> Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n],> {n, 0, Infinity}])/(1 + a), a -> -1, Direction -> -1] ==> 1/4 + 2*Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] +> x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}] Dawk! And better still as one Lambert series! Sum[((-1)^k*q^((1/2)*(-1 + k)*k)*x^k*QPochhammer[q, q, -1 + k])/ (QPochhammer[-1, q, 1 + k]*QPochhammer[-x, q, k]), {k, 1, Infinity}] == Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] + (1/2)*x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}] == (-(1/2))*x*Sum[(-q)^n/(1 + q^n*x), {n, 0, Infinity}] rwg> This came from combining Heine's first and second> transformations http://dlmf.nist.gov/17.6 (17.6.6, 17.6.7) to get> QHypergeometricPFQ[{a, b}, {c}, q, z]==> QHypergeometricPFQ[{c/b, z}, {a z}, q, b] *> QPochhammer[b, q] QPochhammer[a z, q])/> (QPochhammer[c, q] QPochhammer[z, q])> > This transforms the sum in the limitand into one with numerator> parameter a^2, which means all its terms after the 0th will contain> a removable factor of 1-a^2 = (1-a)(1+a), and the 0th minus 1/2> and then divided by a+1 contributes one of the Lambert sums> as a -> -1.> > [...] Joerg>Not sure this is pertinent here, but see relation (3.1) on p.604 of %\bibitem{Chan}{Song Heng Chan: % {Generalized Lambert Series Identities}, % Proceedings of the London Mathematical Society, vol.91, no.3, pp.598-622, (2005). %}%% rel.(3.1): expression for bilateral Lambert series "Subscription required." Joerg>Can send the pdf in case the paper isn't open access. That would be sweet. --rwg Joerg> IIRC there is something similar in Fine, but can only point to his relation (18.4) on p.20 right now.

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Re: [math-fun] Visualizing f(z), mark II
by Dan Asimov 28 Feb '12

28 Feb '12

James, what does f(z) stand for here? Any function f: C -> C ? Any entire function f: C -> C ? Something else? Thanks, Dan James Cloos wrote: << My previous code to attempt to visualize f(z) failed to usefully illustrate the map's behaviour. Perhaps this attempt will succeed. . . . >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Visualizing f(z), mark II
by James Cloos 27 Feb '12

27 Feb '12

My previous code to attempt to visualize f(z) failed to usefully illustrate the map's behaviour. Perhaps this attempt will succeed. My intent, this time, is to do an (inverse?) steregraphic projection of the CC plane onto the unit sphere, put the sphere in the CIELAB colour space and then scale that sphere to fit nicely within (human-)visable (or real) colours. Given the typical CIELAB scaling, L* \in [0,100] and a*, b* \in [-128,128], this maps the origin to the black point (0,0,0), the point at infinity to the white point (100,0,0) and the unit circle to the circle with radius=128 on the a*b* plane where L* = 50. (For those not in the know, CIELAB attempts to be a perceptually uniform colour space where L* represents lightness, a* is green vs magenta and b* is blue vs yellow.) Longer term I intend to write out PDF files in CIELAB space, and then convert it to a library which would allow one to use higher level languages to define f(z). (Notably a python api would allow sage to use it.) The mapping still may be imperfect; there is, eg, more white than I’d expect. It might (??) be better to scale the sphere a bit smaller in the a*b* plane. An example of Warren’s e·(Γ(z+½)/√(2π))^(1/z) is at: http://jhcloos.com/wsmith-gamma/Lab/ Any critique is welcome. -JimC -- James Cloos <cloos(a)jhcloos.com> OpenPGP: 1024D/ED7DAEA6

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[math-fun] Limit puzzle
by Bill Gosper 27 Feb '12

27 Feb '12

(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/ (1 + a) as a -> -1+ . (In terms of q-hypergeometric(s) in x and q). This gave me quite a tussle. If you crack it, please say how. After all else failed, I StringReverse["]}ytinifnI ,1 ,k{ ,)]k ,q ,x-[remmahhcoPQ*]k + 1 ,q ,1-[remmahhcoPQ( /)]k + 1- ,q ,q[remmahhcoPQ *k^x*)k*)k + 1-(*)2/1((^q*k^)1-(([muS*2 + 4/1 si ,1-=a rof ,hcihw ]}ytinifnI ,1 ,k{ ,)]k ,q ,x*a[remmahhcoPQ*]k + 1 ,q ,a[remmahhcoPQ( \ /)]k + 1- ,q ,q*2^a[remmahhcoPQ*k^x*)k*)k + 1-(*)2/1((^q*k^)1-(([muS *)a + 1-(- )))a + 1-(*2(/1(- :1+a eht delecnac hcihw morf eno gnidnif yllanif ,dnatimil eht demrofsnart hcihw sruotnoc xirtam dehcraes ylmodnar tsuj"] --rwg The spellchecker likes demrofsnart? Oh, it's on the same line as (").

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