May 2012 - mathfun@mailman.xmission.com

[math-fun] Sum csc(π k coth 1)/(k log k) converges? [Was: Teacher trolling]
by Bill Gosper 18 May '12

18 May '12

rwg>try sum (csc π φ n)/n. Julian showed the largest terms have n = fib(k), and Csc[π*GoldenRatio*Fibonacci[k]]/Fibonacci[k] -> (-1)^Ceiling[k/3]*Sqrt[5]/π) so the partial sums of just the "spike" terms oscillate around 0 with period 6. So, what about csc(π k coth 1)/(k log k)? Mathematica startled me with a closed form ("obvious" in retrospect) for the convergents of coth 1: In[612]:= MatrixForm[Out[585]] Out[612]//MatrixForm= a[n] a[-1 + n] b[n] b[-1 + n] n 1 1 (-1) Pi BesselI[- + n, 1] + 2 E BesselK[- + n, 1] Cosh[1] 2 2 {a[n_] :> ----------------------------------------------------------, E Sqrt[2 Pi] n 1 1 -(-1) Pi BesselI[- + n, 1] + 2 E BesselK[- + n, 1] Sinh[1] 2 2 b[n_] :> -----------------------------------------------------------} E Sqrt[2 Pi] In[589]:= Table[MatrixForm[FullSimplify[FunctionExpand[%585 /. %588]]], {n, 0, 6}] Out[589]= {1 0, 1 1, 4 1, 21 4 , 151 21 , 1380 151 , 15331 1380 } 0 1 1 0 3 1 16 3 115 16 1051 115 11676 1051 In[594]:= Dot @@ Table[MatrixForm[{{2*k - 1, 1}, {1, 0}}], {k, 6}] Out[594]= 1 1 . 3 1 . 5 1 . 7 1 . 9 1 . 11 1 1 0 1 0 1 0 1 0 1 0 1 0 In[595]:= Thread[%, MatrixForm] Out[595]//MatrixForm= 15331 1380 11676 1051 In[571]:= Convergents[Coth[1], 6] Out[571]= {1, 4/3, 21/16, 151/115, 1380/1051, 15331/11676} whereupon Julian determined that the sequence {Csc[π*k*Coth[1]]/k} spikes when k=b[n], approaching (-1)^Floor[(n+2)/3]*(2*n+1)/π: Out[610]= Pi Csc[Pi b[n] Coth[1]] ------------------------------------- Floor[(2 + n)/3] (-1) ((1 + 2 n) b[n]) In[611]:= N[Table[N[% /. %588, 69], {n, 6, 9}]] Out[611]= {1.01203, 1.009, 1.00699, 1.00559} We have not yet examined the intervening partial sums. --rwg In a stunning breakthrough, Google has automated translation into Engrish, providing a bountiful new source of examples of this mellifluous language: 一方、3月31日も事態の安定に向けた努力が続く原発では、世界の原子力大国の注目が高まっている。 YouTubeに掲載された映像に映っていたのは、水平線から迫り来る大津波で、まさに発電所に牙をむこうとしていた。 On the other hand, in the primary followed by efforts to stabilize the situation also March 31, the attention of the world's nuclear power has been increasing. It was reflected in the video are posted to YouTube, looming large in the tsunami from the horizontal, son-in-law was trying to power plant just fangs .

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[math-fun] Another puzzle
by James Aaronson 16 May '12

16 May '12

Dear all, Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp). It turns out that, even if the two ends of the hole are not antipodal, the particle will oscillate with the same half-period, and proving this is not much harder than in the previous case. I talk about the half-period because that is the time taken by the particle to cross the planet. In the case in which the points are not antipodal, this might not be the shortest time our particle could take to travel between the two points. For example, if the path travels towards the centre and then away, in a "curvy" shape, then the particle will be travelling faster for the journey, and this might compensate for the fact that the particle has a longer distance to travel. So the problem is to decide what the path is which will lead to the shortest time of travel. We'll assume air resistance and friction don't exist, and that no time is wasted turning (if that were possible). Regarding my own efforts on this problem, I've managed to prove some ostensibly wrong results, so am interested to see what you think. -- James

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[math-fun] A puzzle
by James Aaronson 15 May '12

15 May '12

Dear all, For anyone who wants a little puzzle to start the week: if I set z_0 = i, and z_(n+1) = 0.5 (z_n + |z_n|), what is the limit of z_n as n tends to infinity? -- James

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Re: [math-fun] A puzzle
by Michael Reid 15 May '12

15 May '12

> For anyone who wants a little puzzle to start the week: if I set z_0 = i, > and z_(n+1) = 0.5 (z_n + |z_n|), what is the limit of z_n as n tends to > infinity? (Spoiler below) . . . . . . . . . . . . . . . . . . . . . . . . . . Cute. The limit in question is 2 / pi . You can do this for any initial value of z_0 . For real values of z_0 , it's easy, so suppose instead that z_0 is non-real. Write z_0 = r (cotan(alpha) + i) , where r = Im(z_0) and alpha = Arg(z_0) is in the interval (-pi , pi) . With this choice of argument, r and alpha have the same sign. We claim that z_n = 2^(-n) r (cotan(alpha / 2^n) + i) , which is immediate for n = 0 . Then we have |z_n| = 2^(-n) |r| |cosec(alpha / 2^n)| = 2^(-n) r cosec(alpha / 2^n) because r and alpha / 2^n and cosec(alpha / 2^n) all have the same sign. Now we have z_(n+1) = 2^(-(n+1)) r (cotan(alpha / 2^n) + cosec(alpha / 2^n) + i) = 2^(-(n+1)) r (cotan(alpha / 2^(n+1)) + i) , by using the "half angle" formula cotan(theta) + cosec(theta) = cotan(theta / 2) . This establishes the claim, and now the limit is easily evaluated as r / alpha = Im(z_0) / Arg(z_0) . Michael Reid

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[math-fun] Droste effect in Life, but smoothly
by Mike Stay 13 May '12

13 May '12

Last year I made a video of Life in Life using the OTCA metapixel http://www.youtube.com/watch?v=QtJ77qsLrpw and stitched frames of it together into some "Print Gallery"-like images http://reperiendi.wordpress.com/2011/03/10/print-gallery-metapixel/ http://reperiendi.wordpress.com/2011/03/10/more-views-of-the-metapixel/ A user named phillipping, who knows Golly much better than I do, took the idea and rendered it using a smoothly increasing zoom and speed: http://www.youtube.com/watch?v=xP5-iIeKXE8 It's beautiful. -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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Re: [math-fun] Droste effect in Life, but smoothly
by Dan Asimov 13 May '12

13 May '12

Mike, your images are very nice. Print Gallery is my favorite of all Escher drawings, and the only one I have a print of (since 1973, even). I've been known to call it a "reality manifold" since locally (though not globally) it's realistic, just as a manifold can be locally but not globally Euclidean. (Or like integrating f(z) = 1/z on C, which is possible in an open set not surrounding the origin, but not in one that does.) But I've never believed that it's related to complex variables in the way claimed in < http://www.ams.org/notices/200304/fea-escher.pdf >. --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] 1000s of rogue planets per star?
by Warren Smith 11 May '12

11 May '12

http://www.sciencedaily.com/releases/2012/05/120510100217.htm is a press release claiming our galaxy harbors 1000s of free-floating "rogue" planets for every star. I have not read the original paper (paywall) but this sounds very hard to believe. (Their theory is these planets formed soon after the big bang and constitute a lot of the "dark matter.") "The scientists calculate that such a planetary body would cross the inner solar system every 25 million years on the average." I'd like it if somebody were to tell me whether there is any actual evidence or is this pure theoretical speculation.

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[math-fun] Oscillation and the perils of numerical pessimism
by Bill Gosper 10 May '12

10 May '12

The Minsky recurrence handily generates long sequences of a*sin(n*ω+φ). One can initialize the general case minsksin[ω_, φ_: 0, ε_: 2 Sin[ω/2], a_: 1] := {a*Sin[φ], -(a/2) ε Cos[φ + ω/2] Csc[ω/2], (4 Sin[ω/2]^2)/ε, ε} where ε is arbitrary. E.g., In[138]:= minsksin[ω, φ, ε, a] Out[138]= {a Sin[φ], -(1/2) a ε Cos[φ + ω/2] Csc[ω/2], (4 Sin[ω/2]^2)/ε, ε} The general Minsky step is stepminsk[{x_, y_, δ_, ε_}] := {#, y + ε*#, δ, ε} &[x - δ*y] E.g., running three steps, In[139]:= Simplify[Nest[stepminsk, %, 3]] Out[139]= {a Sin[φ + 3 ω], -(1/2) a ε Cos[φ + (7 ω)/2] Csc[ω/2], 4 Sin[ω/2]^2/ε, ε} (The 3rd and 4th positions are just along for the ride.) Nice choices of ε are 2 Sin[ω/2] and 1, where the latter saves a multiply in the stepper: stepm1nsk[{x_, y_, δ_}] := {#, y + #, δ}&[x - δ*y] E.g., for amplitude 1, phase 0, In[140]:= Drop[minsksin[w, 0, 1], -1] Out[140]= {0, (-(1/2))*Cot[w/2], 4*Sin[w/2]^2} Run four steps: In[144]:= FullSimplify[Nest[stepm1nsk, Out[140], 4]] Out[144]= {Sin[4*w], (-(1/2))*Cos[(9*w)/2]*Csc[w/2], 2 - 2*Cos[w]} But instead of Minsky, we could use good old Chebychyov*: In[37]:= chebsin[ω_, φ_: 0] := {Sin[φ], Sin[φ - ω], 2*Cos[ω]} whose stepper is stepcheb[{x_, y_, δ_}] := {δ*x - y, x, δ} E.g., In[146]:= FullSimplify[Nest[stepcheb, chebsin[a], 3]] Out[146]= {Sin[3 a], Sin[2 a], 2 Cos[a]} So let's run cheb vs minsky a few million and see who's better. Using ω := π φ (which will figure in a later email): In[148]:= chebsin[N[π*GoldenRatio]] Out[148]= {0, 0.932032, 0.72475} In[149]:= Timing[Nest[stepcheb, %, 9999999]] Out[149]= {22.9417, {0.748998, 0.888957, 0.72475}} In[150]:= minsksin[N[π*GoldenRatio], 0, 1] Out[150]= {0, 0.730862, 1.27525, 1} In[151]:= Timing[Nest[stepm1nsk, Drop[%, -1], 9999999]] Out[151]= {40.471, {0.748998, 0.858748, 1.27525}} Well, cheb is faster, but which is better? In[152]:= Sin[9999999*π*GoldenRatio] - %149[[2, 1]] Out[152]= 3.01301*10^-9 In[153]:= Sin[9999999*π*GoldenRatio] - %151[[2, 1]] Out[153]= 3.01292*10^-9 They're the same! But wait, this is a trap. You can't accurately compute sin(huge) with single precision! sin(huge) is like asking which way your valve stem will point after you drive across the country. In[154]:= Sin[9999999*π*GoldenRatio] - SetPrecision[%149[[2, 1]], 22] Out[154]= 1.9186385802598*10^-9 Significantly different, but In[155]:= Sin[9999999*π*GoldenRatio] - SetPrecision[%151[[2, 1]], 22] Out[155]= 1.9185395483660*10^-9 They're *still* the same! Something else is fishy. MachinePrecision is ~16, so these values are ~ 10^6 LSBs, not 10^(6/2). Let's drive ten times further. In[156]:= Timing[Nest[stepcheb, %149[[2]], 90000000]] Out[156]= {214.079, {-0.722388, -0.906264, 0.72475}} and now we're off by In[159]:= Sin[99999999*π*GoldenRatio] - SetPrecision[%156[[2, 1]], 22] Out[159]= -2.00240499632334*10^-8 Ten times as far! Whereas with Minsky In[157]:= Timing[Nest[stepm1nsk, %151[[2]], 90000000]] Out[157]= {382.448, {-0.722388, -0.866577, 1.27525}} In[160]:= Sin[99999999*π*GoldenRatio] - SetPrecision[%157[[2, 1]], 22] Out[160]= -2.00242976539902*10^-8 The same! The explanation: These recurrences don't compute Sin[n*π*GoldenRatio], but rather Sin[n*N[π*GoldenRatio]], which is off by In[161]:= SetPrecision[N[π*GoldenRatio], 22] - π*GoldenRatio Out[161]= -3.34860*10^-16 We are doomed to linearly creeping error. But hey, we can tack on a dozen extra digits and promise not to drive past a trillion. But first, let's drive just 999 to see how long it will take. (This is where it hits the fan.) In[166]:= Nest[stepcheb, chebsin[N[π*GoldenRatio, 33]], 999] Out[166]= {0.*10^121, 0.*10^121, 0.72474978016096023991729327494997} WAH? In[167]:= Nest[stepm1nsk, Drop[minsksin[N[π*GoldenRatio, 33], 0, 1], -1], 999] Out[167]= {0.*10^434, 0.*10^434, 1.27525021983903976008270672505003} WHA?? This is too gross: In[168]:= Nest[stepm1nsk, Drop[minsksin[N[π*GoldenRatio, 288], 0, 1], -1], 999] Out[168]= {0.*10^179, 0.*10^179, 1.27525021983903976008270672505002620127826891198880290709968642251975\ 2987505102052895607541471314178683631575021864167072493028631467693604\ 7459492180699315909580232204640529914388966962377507409450195194836649\ 3221557106208837998002035152821718325478473977678977255669689503145798\ 805151318} "Precision Tracking" (precision trashing, interval arithmetic) *totally* destroys the computation. Whereas the actual loss during the fixed machine precision iteration was negligible. --rwg *(The standard counterexample to the Continuum Hypothesis: The cardinal between ℵ0 and C is the number of ways to spell Chebychyov.)

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Re: [math-fun] [EXTERNAL] shapes with constant surface area
by Henry Baker 10 May '12

10 May '12

FYI -- I guess that this high school science project is no longer politically correct... http://en.wikipedia.org/wiki/Potassium_nitrate Into the 19th century, niter-beds were prepared by mixing manure with either mortar or wood ashes, common earth and organic materials such as straw to give porosity to a compost pile typically 1.5Ã2Ã5 meters in size.[9] The heap was usually under a cover from the rain, kept moist with urine, turned often to accelerate the decomposition, then finally leached with water after approximately one year, to remove the soluble calcium nitrate. Dung-heaps were a particularly common source: they contain ammonia from the decomposition of urea and other nitrogenous materials. It then undergoes bacterial oxidation (first by means of the Nitrosomonas bacteria) to produce (calcium) nitrite, and then (by means of the Spirobacter bacteria) to produce (calcium) nitrate. It is then converted to potassium nitrate by filtering through the potash of wood ashes. At 11:22 AM 5/9/2012, Charles Greathouse wrote: >> It is possible to unify these two subjects, as some fertilizers are also solid rocket fuel... > >I suspect the crossover appeal is limited! > >Charles Greathouse >Analyst/Programmer >Case Western Reserve University > >On Wed, May 9, 2012 at 12:12 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> It is possible to unify these two subjects, as some fertilizers are also solid rocket fuel... >> >> At 08:08 AM 5/9/2012, Cordwell, William R wrote: >>>In solid rocket motors, it can be desirable to have a roughly constant burning (surface) area. One solution is to start out with a star shaped hole down the middle; the star erodes as it burns, but the hole also gets larger. Of course, the outside surface is not burning in this case, so it doesn't directly apply to the fertilizer question, but it seems as though you could have a cylinder with an inside hole that gets larger at a similar rate as the outside surface gets smaller. >>> >>>-----Original Message----- >>>From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Warren Smith >>>Sent: Wednesday, May 09, 2012 8:58 AM >>>To: math-fun; David Langer; Monty McGovern; Holt, Robert >>>Subject: [EXTERNAL] [math-fun] shapes with constant surface area >>> >>>The surface of "slow release fertilizer" beads gradually erode, say at >>>1 centimeter per year. >>>With a spherical bead, the resulting curve of "fertilization rate" >>>would decline with time e.g. >>>it would be proportional to (1-t)^2 for 0<t<1 for a 1cm-radius bead. >>>It might be better to have beads with constant fertilization rate 1. >>> >>>QUESTION: is there a shape (or set of shapes) whose surface area stays constant as they erode? >>> >>>The answer in 1 and in 2 dimensions is "yes." >>>In 3D the answer is less obvious, but I have a (presently rather messy) argument the answer also is yes. >>> >>>More advanced question: >>>you might want the rate to increase with time (as your plants grow bigger they need more)... >>>or be some other curve... WHICH curves F(t) are achievable fertilization-rate curves? >>> >>>-- >>>Warren D. Smith

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[math-fun] Gosper's notes on higher factorials
by Bill Gosper 10 May '12

10 May '12

MRob> What is the date (year of authorship) of these ancient (mostly) POX notes? I want to add this as a reference for OEIS sequence A2109: https://oeis.org/A002109 ------------------------- There's a handwritten letter in the binder dated 1979. It must have coincided with the dawn of TeX, since the later pages are TeXed. MRob> It would also be nice to add a sequence for the (Gosper's) second factorial with integer arguments, which I take to be 1, 1, 16, 122009559759792, 16358..68032 (169 digits, approx. 1.63588 x 10 ^ 168) but I don't like to add sequences without a reference. - Robert [Munafo] I'm sure there are better references, but J.W.L. Glaisher uses the function on p81-87 of Products and series involving prime numbers only, part II, The Quarterly Journal of Pure and Applied Mathematics, Vol. XXVIII and on p82 refers to an apparently more relevant article in The Messenger of Mathematics, Vol XXIII, p155, presumably also by Glaisher. I haven't checked, but I suspect his ilg_2 function is the closely related double integral of log gamma. (A negapolygamma.) Glaisher also treats prod k^k^n, n<0. Barnes and Kinkelin and perhaps Alexiewsky(sp?) almost certainly treated this function. --rwg On 5/7/12, Bill Gosper <billgosper(a)gmail.com <http://www.gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:> By way of preservation, Neil has kindly scanned in my ancient,> (mostly) POX (pre-Tex), brief notes <http://gosper.org/facfun.pdf>> (3.7M) on higher factorials.> These stem from my astounding discovery that the dot at the> bottom of an exclamation point is actually hollow, representing> the number 0, and what happens when you change it to a 1.> --rwg On the subject of shapes with constant dissolution rates, there might be real monetary incentive from the drug companies, seeking to improve "timed release". To illustrate the lengths they've already reached, I take an expensive, insoluble plastic pill with a laser-punched pinhole, through which medication is pushed by a tiny expanding sponge. There must already be scads of patents on inhomogeneous pills, with solubility varying with internal depth, etc.

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