April 2012 - mathfun@mailman.xmission.com

[math-fun] sqrt(pi)
by Bill Gosper 24 Mar '13

24 Mar '13

Pairwise combining the terms of A&S 6.5.31 (upper Gamma CF), then adding last week's lower gamma CF: In[185]:= Equal[Gamma[ z], ((y^z*((1/(cfk[n*(z - n), -z + y + 2*n + 1, List[n, Infinity]] - z + y + 1)) + (1/(z + cfk[n*y, z - y + n, List[n, Infinity]] - y))))/(E^y))] z 1 Gamma[z] == (y (------------------------------------------- + -y + z + cfk[n y, n - y + z, {n, Infinity}] 1 y -----------------------------------------------------------)) / E 1 + y - z + cfk[n (-n + z), 1 + 2 n + y - z, {n, Infinity}] where cfk:=ContinuedFractionK and y>0 can be chosen to optimize convergence. Specializing, In[186]:= % /. z -> 1/2 /. y -> 1 Out[186]= 1 Sqrt[Pi] == (-------------------------------------- + 1 1 -(-) + cfk[n, -(-) + n, {n, Infinity}] 2 2 1 ------------------------------------------) / E 3 1 3 - + cfk[(- - n) n, - + 2 n, {n, Infinity}] 2 2 2 In[192]:= N[List @@ %186 /. cfk[num_, den_, _] :> ContinuedFractionK[num, den, {n, 9}]] Out[192]= {1.77245, 1.77244} This is the first non-asymptotic sum or CF I've seen for sqrt(pi). But the sum behind the lowergamma cf has been there all along, so I must not have been paying attention. --rwg

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[math-fun] my biggest lie yet
by Bill Gosper 28 May '12

28 May '12

http://www.tweedledum.com/rwg/stanfordn3.pdf (p 14) (Also: Gosper, Strip mining in the abandoned orefields of Nineteenth Century mathematics, in Computers in Mathematics (D. Chudnovsky & R. Jenks, eds.) pp 261-283, m. dekker, N. Y.) "8. Continued fractions This is three N - K planes, j = −1, 0, 1, joined by two J matrices. In the top (j = 1) plane, the k direction computes a continued fraction, and on the bottom (j = −1), it’s the n direction. [...] Thus, to get a relation between two continued fractions, we must j-hop between planes before switching directions. If k runs from c to ∞ and n runs from d to ∞, and the corresponding continued fractions converge, then they are insensitive to whatever subsequent matrices accrue on the right by way of path closure. I.e., the two paths can be (j,k,n) = (0,c,d) → (1,c,d) → (1,∞,d) and (0,c,d) → (−1,c,d) → (−1,c,∞). This yields the identity [an undersimplified version of that symmetric CF identity I sent the other day]" All of a sudden, I'm deriving false identities with this! E.g., with K and N matrices (c85) [km(k,n),nm(k,n)] [ 4 n 4 k 11 1 ] [ n k 13 1 ] [ --- + --- + -- - ] [ - + - + -- - - ] [ 5 5 30 5 ] [ 5 5 60 5 ] (d85) [[ ], [ ]] [ k 1 1 ] [ k 1 1 ] [ - - - - k + - ] [ - + - n + - ] [ 2 6 3 ] [ 2 6 4 ] Check path invariance: (c81) CHECKM(); [ 0 0 ] (d81) [ ] [ 0 0 ] Confirm convergence of a dozen N terms: (c82) (PRUD(NM(0,N),N,0,11),DFLOAT(%%[1]/%%[2])); (d82) [- 1.12262566430759d0, - 1.12264219198252d0] This says that as a function of the tail t, the PRUD (matrix product) computes (A t + B)/(a t +b), with A/a ~ B/b ~ -1.12262. This is the "insensitivity" claimed above. Continuing to wander in the +n and +k directions shouldn't change this value much. But tacking on 69 k steps: (c88) (PRUD(NM(0,N),N,0,11) . PRUD(KM(K,12),K,0,69),DFLOAT(%%[1]/%%[2])); (d88) [0.30872240339868d0, 1.27956902429756d0] This is batbleep! Not only was the value bashed, the convergence was destroyed. Note that by path invariance, it doesn't matter how we got here: (c89) (PRUD(KM(K,0),K,0,69) . PRUD(NM(70,N),N,0,11),DFLOAT(%%[1]/%%[2])); (d89) [0.30872240339868d0, 1.27956902429756d0] There's nothing obviously wrong with the K matrix except for being a bit slow to converge: (c86) (PRUD(KM(K,0),K,0,11),DFLOAT(%%[1]/%%[2])); (d86) [1.10980568563031d0, 1.48744247508643d0] (c87) (PRUD(KM(K,0),K,0,69),DFLOAT(%%[1]/%%[2])); (d87) [1.62173336398064d0, 1.62175009332196d0] Except it's not even the same sign as the N product! There's nothing magic about 69 terms of K. Using 11 instead still fails bigtime; (c83) (PRUD(NM(0,N),N,0,11) . PRUD(KM(K,12),K,0,11),DFLOAT(%%[1]/%%[2])); (d83) [- 0.77240969070225d0, - 1.11033801208252d0] (c84) (PRUD(KM(K,0),K,0,11) . PRUD(NM(12,N),N,0,11),DFLOAT(%%[1]/%%[2])); (d84) [- 0.77240969070225d0, - 1.11033801208252d0] Maybe the kids can help get to the bottom of this. --rwg (j-hopping to force continued fraction form is just cosmetic, and only adds to the confusion above.)

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Re: [math-fun] geometry puzzle
by Michael Reid 18 May '12

18 May '12

> > What a great problem! I was still working through the first few minutes > > of "but... how... isn't..." when Jessica said "Okay, how about this?..." > > and solved it. > > > > But is there more than one solution? I know about the answer depicted in > > http://i.imgur.com/iOfRI.png, but is that unique? > > > > Ah: it's not unique, in that there are solutions like > http://i.imgur.com/bxtbB.png as well. > > Are there any solutions that aren't a refinement of the basic dissection > you get by gluing together pairs of pieces from either of the images above, > though? That seems like it might be provable. There are also dissections into 6n pieces, of which 6 border on the center, and the others do not. By the way, this has been around for a while. I first(?) saw it in Croft, Falconer and Guy's "Unsolved Problems in Geometry". It's on page 89 of my 1991 edition, well, the first solution Michael gave. The generalization (to his second solution) I mention above is more or less "obvious". Your mileage may vary. Michael Reid > > --Michael > > > > > > On Wed, Apr 18, 2012 at 10:50 PM, Bill Gosper <billgosper(a)gmail.com>wrote: > > > >> Colin Wright (e.g., > >> *Colin Wright*: The Mathematics of *Juggling* on > >> Vimeo<http://vimeo.com/27998521> > >> ) > >> > >> posed this at G4GX (and G9?): Dissect a disk into congruent parts > >> > >> at least one of which avoids the center by a positive distance. Neil > >> > >> just beat me to a solution. Grr. > >> > >> --rwg

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[math-fun] The Online RILYBOT Inverse Equation Solver
by Robert Munafo 30 Apr '12

30 Apr '12

Hello math-funsters, After seeing some of the desires expressed by xkcd forum members, I decided to implement an online server version of my RIES program [1] at the address: http://www.mrob.com/pub/ries/index.html (or simply mrob.com/ries ) It's a pretty basic service, mainly due to the constraints of the server. But it's still kind of fun and certainly adequate for a lot of the recreational maths that the program was intended for, like the approximations in last Wednesday's xkcd [2]. For example, 1.234321 ~ Sqrt[Pi-Phi]. I've bashed on it over the weekend, and gotten a fair amount of traffic from around the world. I think I have it working pretty well now but it would be nice to see what some more math-oriented folks think. So go ahead and throw a few numbers at ORIES. - Robert [1] (RIES is a semi-silly acronym for "RILYBOT Inverse Equation Solver"; the program source code has been online since 2000) [2] xkcd.com/1047/ "Approximations", is full of examples of just the sort of thing my program is "meant for". Although I and RIES are mentioned, Randall Munroe and I did not communicate about it beforehand and it came as a complete surprise to me. Munroe clearly likes his numbers almost as much as I do. (See for example xkcd's 221, 798, 759, 718, and 715) -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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[math-fun] geometry puzzle
by Bill Gosper 30 Apr '12

30 Apr '12

Colin Wright (e.g., *Colin Wright*: The Mathematics of *Juggling* on Vimeo<http://vimeo.com/27998521> ) posed this at G4GX (and G9?): Dissect a disk into congruent parts at least one of which avoids the center by a positive distance. Neil just beat me to a solution. Grr. --rwg

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[math-fun] IEG puzzle
by Veit Elser 30 Apr '12

30 Apr '12

A puzzle I gave my mechanics class: The average orbital periods of the Jovian moons Io, Europa and Ganymede are: TI=1.769137786 TE=3.551181041 TG=7.15455296 These are taken from Wikipedia; the time unit is days. 1. (math) On the basis of just these numbers, infer the existence of a much longer period (on the order of hundreds of days). 2. (physics) What is the physical origin of the long period? -Veit

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[math-fun] Fwd: IEG puzzle
by Bill Gosper 29 Apr '12

29 Apr '12

Veit> Isn't this obsession -- with looking for relationships where there almost certainly aren't any -- a waste of time? [...] rwg> That same day, I sent this to the kids: ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Thu, Apr 12, 2012 at 6:13 AM Subject: Re: [math-fun] IEG puzzle To: Neil Bickford <techie314(a)gmail.com>, Julian Ziegler Hunts , Corey Ziegler Hunts Strange: This is two Venusian "days". --Bill >A puzzle I gave my mechanics class:>> The average orbital periods of the Jovian moons Io, Europa and Ganymede > are:>> TI=1.769137786> TE=3.551181041> TG=7.15455296>> These are taken from Wikipedia; the time unit is days.>> 1. (math) On the basis of just these numbers, infer the existence of a > much> longer period (on the order of hundreds of days). That's a nice puzzle. The first thing I did was to enter TE/TI, TG/TE and TG/TI into a continued fraction calculator, to produce best rational approximations. The first few convergents for TE/TI are: 2/1 275/137 * 3027/1058 3302/1645 And those for TG/TE are: 2/1 137/68 * 3153/1565 3920/1633 Finally, the convergents for TG/TI are: 4/1 89/22 93/23 275/68 * The asterisked convergents suggest a ratio of: TI : TE : TG = (1/275) : (1/137) : (1/68) This means there is a large period of 275 TI = 137 TE = 68 TG. For each of the orbital periods you have provided, this results in the following approximations to the large period: 275 TI = 486.51289115 days 137 TE = 486.511802617 days 68 TG = 486.50960128 days = approx 486.51 days. (Technically, a better method than using continued fractions is the LLL lattice reduction algorithm. However, since the data are so precise, the continued fraction method was adequate.) Sincerely, Adam P. Goucher

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[math-fun] Figure 33
by Bill Gosper 29 Apr '12

29 Apr '12

http://gosper.org/x=1, y=1, dmin=emin=3o2, dmax=emax=8o3, dg=eg=1o2574, de<=49o12.png , corrects and refines the "δ-ℇ plot" on p23 of http://www.blurb.com/books/2172660 (also Neil's blog, http://nbickford.files.wordpress.com/2011/03/x1-y1-dminemin3o2-dmaxemax8o3-… ). It's a map of the periods of the Minsky circle algorithm, starting with x0=1, y0=1, with multipliers 3/2 ≤ δ,ℇ ≤ 8/3. With a zoomable, nonblurring viewer, I find it fascinating and surprising. The (δ,ℇ) which blow up exponentially (because δ ℇ>4) are shown in white. But note that some colored rectangles transgress δ ℇ = 4 ! The floor function in the Minsky iteration can actually tame an infinite sequence. Contrariwise, imposing the floor operation on x0=1, y0=1/2, δ 3^(n+1), ℇ = 3^-n changes a mere period 3 into infinite linear growth! Any such growers in the graphic appear black, along with any orbits longer than 10000, of which there are >57000 (in this plot). Prior to this hi-res plot, I had conjectured that no rectangle had two corners transgressing the hyperbola. Julian completed my humiliation by finding the period 159 region {x=1, y=1, 83/49 ≤ δ < 61/36, 85/36 ≤ ℇ < 111/47} with *three* transgressing corners. I still conjecture the impossibility of four. Zooming at a crack between large rectangles often shows a single layer of variegated pixels, due to the sampler landing exactly on the crack, which, remarkably, is often composed of numerous truly 1-dimensional line segment subregions. Now zoom at the top of the upper largish pink rectangle. Euclid's orchard? Other high-frequency patches resemble peering into factory windows or out of skyscrapers. Also note the apparent, possibly provable accumulation point at (2,2). See book, p 23, for Julian's proof why this would guarantee unbounded periods, despite the lack of thin black rectangles at this sampling density (1/2574)). Also, note the optically illusory reverse curvature when you zoom at (2,2). The plot was computed by Corey's old program, and took 14 hrs, and produced a blank screen due to a bug in Mathematica. Exporting it took several more minutes, returning the filename, but writing no file! I spent a day fighting back creeping insanity, until Julian found an ingenious workaround. --rwg I have proofread this message, and disavow it. --Julian

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Re: [math-fun] pi^4 + pi^5 ~ e^6
by Robert Munafo 29 Apr '12

29 Apr '12

Yes! USENET archives! The "new Google Groups" entirely fails to work, because its search engine has been dumbed down too much. But with the old Google Groups, I found the following, from 1989. That pushes it back 9 years earlier than Weisstein/CRC. "Doug Ingram" was closest to the source, but "it warn't him neether..." ---8<--snip-here---- Newsgroups: sci.math From: do...(a)utastro.UUCP (Doug Ingram) Date: 5 Jul 89 15:59:59 GMT Local: Wed, Jul 5 1989 11:59 am Subject: Relation between pi and e? A couple of days ago, someone posted this in their .sig file: 4 5 6 Pi + Pi = e Strange enough to be true????? Well, it comes close on my calculator, but it may be off due to roundoff error. Is this a coincidence, or can someone offer proof of this relation, maybe using e^(i*phi) = cos(phi) + i*sin(phi)? -------------------- Doug Ingram (dougi) do...(a)astro.as.utexas.edu do...(a)utastro.uucp {charm,ut-sally,emx,noao}!utastro!dougi dougi%astro.as.utexas.edu(a){CUNYVM.CUNY.EDU,WISCVM.WISC.EDU} 'Airline food is of the highest quality, and critics who think otherwise are not objective at all...pardon me while I go make lasagne from used bus seats.' - Airline Food Employee -- Dave Barry Newsgroups: sci.math From: dkei...(a)Gang-of-Four.Stanford.EDU (Dave Eisen) Date: 6 Jul 89 10:31:08 GMT Local: Thurs, Jul 6 1989 6:31 am Subject: Re: Relation between pi and e? >> 4 5 6 >> Pi + Pi = e Strange enough to be true????? >pi**4 + pi**5 doesn't equal e**6, and if we are given the values of e and >pi to about 13 or more digits we can prove it. This is sci.math? Then how come I've seen a half dozen computations that show $$ \pi^4 + \pi^5 \ne e^6 $$ by numerical coincidence. Don't all mathematicians know from birth that $\pi$ and $e$ are algebraically independent? I'm not a number theorist so I don't know the reasoning behind it and I don't know the references that well. Still, this is the kind of fact that smells like it has to be discussed in Hardy and Wright. Experts?? Dave Eisen dkei...(a)Gang-of-Four.Stanford.EDU (415)723-2963 "It is poignant but fundamental that the flag protects those who hold it in contempt." --- Associate Justice Anthony Kennedy Newsgroups: sci.math From: mo...(a)hpcupt1.HP.COM (John Mount) Date: 6 Jul 89 00:24:05 GMT Local: Wed, Jul 5 1989 8:24 pm Subject: Re: Relation between pi and e? / hpcupt1:sci.math / do...(a)utastro.UUCP (Doug Ingram) / 8:59 am Jul 5, 1989 / >A couple of days ago, someone posted this in their .sig file: > 4 5 6 > Pi + Pi = e Strange enough to be true????? pi**4 + pi**5 doesn't equal e**6, and if we are given the values of e and pi to about 13 or more digits we can prove it. we put some rational bounds on e and pi and then do infinite precision arithmetic (actually we do 99 digit precsion using bc calculator but since we have only about 12 digits in the numbers and deal with only 6th powers we need only 6x12 = 72 digits to acheive calculations that have *no* roundoff error) on these numbers to get bounds on the results (left one rounded down right one rounded up). If pi to about 13 digits after the decimal is: 3.14159265358979323 and e to a lot of figures is: 2.7182818284590452 (from math.h so if wrong not my fault) we know 3.14159265358 <= p <= 3.14159265359 therefore 403.428775 <= pi**4 + pi**5 <= 403.428776 and 2.7182818284 <= e <= 2.7182818285 therfore 403.428793 <= e**6 <= 403.428794 and we see that these two intervals do not overlap so the two quantities are not equal. John Mount ----8<--snippy-be-done---- - Robert Munafo On 2012-04-28, David Wilson <davidwwilson(a)comcast.net> wrote: > I first started working on the OEIS back in 1997 (My earliest > contributions were in the A02xxxx range). I'm pretty sure I knew about > the identity at that time, and I guess this clinches it. > > But still, it came from an email signature in a Usenet group, so perhaps > we could find the originator in the Usenet archives, if there is such a > thing. > > On 4/28/2012 11:24 AM, Robert Munafo wrote: >> David, you must have told Eric Weisstein about it at least as early as >> 1998. >> >> I just checked my copy of "The CRC concise encyclopedia of >> mathematics", dated 1998 (CRC Press, ix + 1969 pages, LC number >> 98-22385). It includes the near-identity and credits "D. Wilson". It >> is on page 33, right column, equation (7) near the bottom. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] pi^4 + pi^5 ~ e^6
by David Wilson 29 Apr '12

29 Apr '12

Apparently I lied earlier. More likely my math was off. Rhode Island, perhaps? On 4/29/2012 11:22 AM, George Hart wrote: > David, > > Can you double check that? I get 1/(355/113-Pi) inches as roughly a > 59 mile diameter, which gives a circle of area 2749 square miles, > while CT has area 5543 square miles. > > George > http://georgehart.com/ > > > On 4/29/2012 10:34 AM, David Wilson wrote: >> >> If you computed the circumference of a circle using 355/113 for pi, and >> were 1 inch off from the true circumference, the circle would enclose >> the state of Connecticut. >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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