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[math-fun] 1, 2, 3.
by Bill Gosper 09 Jun '19

09 Jun '19

Is apparently the complete sequence of solutions to PrimeQ[1+Hyperfactorial@x]. (Checked through a couple of hundred.) (Unconvincing) probabilistic arguments "predict" finiteness of prime subsequences of (sufficiently) rapidly growing sequences, but {1, 2, 3} seems a little too pat. Maybe RCS can conjure one of his magic case arguments. The complete list of prime 1+BarnesG seems to be 1, 2, 3, 4, 5, 9, 16. (Checked through 278.) —rwg

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Re: [math-fun] Jacobian and related
by Dan Asimov 09 Jun '19

09 Jun '19

Wait. I don't see why an onto and locally one-to-one continuous function f : C —> C needs to be a covering map (even assuming it's analytic). In fact it doesn't have to be. An example by Lawrence Zalcman defined by f(z) = Integral exp(z^2) dz (integrating over any curve from 0 to z), is such an example. —Dan ————— * Proof: f'(z) = exp(z^2) is never 0, so f is locally one-to-one. Also f(0) = 0 and f(-z) = f(z), so if any nonzero value c were omitted, then -c must also be omitted — contradiction since an entire function cannot omit two values (Picard's theorem). Hence f is onto. But it's not a covering map C —> C since the analytic covering maps C —> C are only the functions of form az + b as Andy mentions. Andy Latto wrote: ----- On Sun, Jun 9, 2019, 14:08 Eugene Salamin via math-fun <math-fun(a)mailman.xmission.com> wrote: The only analytic bijections of the Riemann sphere are the homographic functions w=(az+b)/(cz+d). If the function is analytic on the finite complex plane, it cannot have a pole, so c=0. Thus the only such functions are linear, w=az+b. This shows impossibility under the added assumption that the function can be extended from the complex plane to the Riemann sphere, but that additional assumption is not needed. F is a covering map on the plane. Since the plane is simply connected, it can't have more than one sheet. Andy ----Gene On Sunday, June 9, 2019, 10:52:44 AM PDT, Dan Asimov <dasimov(a)earthlink.net> wrote: Now assume further that the entire function f : C —> C is onto. (Unlike exp(z), which omits 0.) Fun question: ----- Does there exist an analytic function f : C —> C that is a) locally one-to-one (equivalently, f'(z) is nowhere zero), and b) onto ??? ----- -----

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Re: [math-fun] Jacobian and related
by Dan Asimov 09 Jun '19

09 Jun '19

Now assume further that the entire function f : C —> C is onto. (Unlike exp(z), which omits 0.) Fun question: ----- Does there exist an analytic function f : C —> C that is a) locally one-to-one (equivalently, f'(z) is nowhere zero), and b) onto ??? ----- —Dan Gene Salamin wrote: ----- On Saturday, June 8, 2019, 11:12:30 PM PDT, Dan Asimov <dasimov(a)earthlink.net> wrote: ----- Obviously a *necessary* condition is that P be locally one-to-one Q x Q —> Q. Which is just saying that for all (x,y) in Q^2, we have dP/dx * dP/dY ≠ 0 (i.e., not both derivatives are zero). ------------------------------------------------------------ The relation dP/dx * dP/dY ≠ 0 does not imply that P is locally one-to-one.. Counterexample: P(x,y) = xy. ---------------------------------------------------------------- Puzzle (tricky!): Find an entire function f(z) — i.e., f(z) is analytic on all of C — that is locally one-to-one everywhere without being globally one-to-one ... or else prove that such cannot exist. —Dan -------------------------------------------------————— f(z) = exp(z). --------------------------------------------------------------------- -----

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[math-fun] Jacobian and related
by Dan Asimov 09 Jun '19

09 Jun '19

The Jacobian conjecture is one of my favorite unsolved conjectures, because it's so simple to state (C = the complexes): Suppose F : C^2 —> C^2 is a polynomial map F(z, w) = (P(z,w), Q(z,w)) meaning: P and Q are each a polynomial over C. Then: If* we assume that F is locally one-to-one, then F is a) globally one-to-one and b) onto. I believe the truth status of neither a) nor b) is known at present. Amazing! Here's another related question I just saw in Terry Tao's math blog: ----- Does there exist a *polynomial bijection* P : Q x Q —> Q where P is a polynomial in two variables with rational coefficients i.e., P is a member of Q[X, Y]. ----- He says he thinks this very unlikely, but says that a proof (if true) is likely to be very hard to come by. So, it's evidently an open problem! Obviously a *necessary* condition is that P be locally one-to-one Q x Q —> Q. Which is just saying that for all (x,y) in Q^2, we have dP/dx * dP/dY ≠ 0 (i.e., not both derivatives are zero). So the obvious more general questions: Given *any* field K (or given any subfield K of Q^ = the algebraic closure of the rationals): ----- Does there exist a bijection F : K x K —> K, where F(X,Y) is a polynomial with coefficients in K ??? ----- [[[[[ Puzzle (tricky!): Find an entire function f(z) — i.e., f(z) is analytic on all of C — that is locally one-to-one everywhere without being globally one-to-one ... or else prove that such cannot exist. ]]]]] —Dan ————— * "Locally one-to-one" means, for each point pt = (z,w) of C^2, there is a neighborhood U of pt = (z,w) in C^2 (C^2 is topologically R^4) such that if for any pt_1, pt_2 both in U we have F(pt_1) = F(pt_2), then pt_1 = pt_1.

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[math-fun] Three piece Golden Rectangle to square dissection
by Bill Gosper 09 Jun '19

09 Jun '19

Golden rectangle↔︎square <http://gosper.org/gold2square.gif> I forget where I stole this, but can you find the superfluous step I just noticed? (Not the retrace, which was to tell you a dimension.) —rwg (Rebuild in progress)

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[math-fun] a bracelet sum
by Wouter Meeussen 08 Jun '19

08 Jun '19

is it easy (or trivial) to prove that Sum[Exp[2 \[Pi] I Mod[(6k+1)^2, p]/p], {k,1,p}] equals Sqrt[p] for p prime and Mod[p,4]==1 and I*Sqrt[p] for p prime and Mod[p,4]==3 ? It also has a combinatorial significance cfr. https://math.stackexchange.com/questions/2899878 Wouter.

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[math-fun] Stirlingoid figure of merit
by Bill Gosper 07 Jun '19

07 Jun '19

n!/√(2πn)e^n/n^n ~ 1 + O(1/n), whose infinite product over n blows up, giving an infinite figure of demerit. For convergence, we can improve Stirling's approximation: n! ~ √(2π(n+1/6)) n^n/e^n ~ n! (1+1/(144 n^2)+. . .). Then the infinite product of all the relative errors is In[86]:= Product[n!/√(2 π (n + 1/6)) n^n/E^n), {n, ∞}] Out[87]= E^( 2 Zeta'[-1] - 1/12) √(⅙! √(2π))) Out[88]= 1.00781097654253 (BtW, Stirling's contribution was "only" the √(2π); de Moivre had already found the rest. This must have been nearly as stunning as Euler's 𝜁(2) = π²/6.) Stirling's approximation can be improved no end. Perhaps the next better is E^-z √(2 π z) (1/(12 z) + z)^z ~ z! (1-1/(1440 z^3)+O[1/z]^4) whose "figure of merit" is Product[z!/(E^-z √(2 π z) (1/(12 z) + z)^z),{z,∞}] == E^(2 Zeta'[-1]) √√(2π)/(BarnesG[1 - I/(2√3)] BarnesG[1 + I/(2√3)]) ~ 1.001178221812, which seems to cry out for the BarnesG reflection formula. Which, unfortunately, only applies to G(z)/G(-z), not G(z)G(-z): E.g. BarnesG[1 - I/(2 √3)]/BarnesG[1 + I/(2 √3)] == (-1)^(1/24) E^(-((I PolyLog[2, E^(-(π/√3))])/(2 π))) (π Csch[π/(2 √3)])^(-(I/(2 √3))) —rwg

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[math-fun] Stupid me
by Bill Gosper 07 Jun '19

07 Jun '19

PG&E pulled the plug as I stepped into the bathroom to brush my teeth. (60 years ago, in an era of utility civility and iron lungs, power companies tried very hard to avoid interruptions.) [Artifically intelligent GMail will *not* allow me to delete the period it inserted at the beginning of two lines above!! But it hides it while I add recipients!] Curious to know when the internet might resume, I switched on the bathroom light. But then I remembered that you sometimes have to flick that switch multiple of times, thanks to gypsy electricianship, and figured I couldn't be sure 'til I got back to my laptop. Why was that stupid? SPOILER: StringReverse@".eldarc gnigrahc sti ni ti decalper I nehw ffo tuhs dluow hsurbhtoot cirtcele ym ,demuser rewop fI" —rwg

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[math-fun] I've only now come to understand that 𝜁[s,a] ≠ HurwitzZeta[s,a],
by Bill Gosper 07 Jun '19

07 Jun '19

and is instead completely batbleep: \[Zeta](s,a)==\!\(\* UnderoverscriptBox["\[Sum]", RowBox[{"k", "=", "0"}], "\[Infinity]", LimitsPositioning->True] \*SuperscriptBox[\((k + a)\), \(-s\)]\), where any term with k+a==0 is excluded. (Decipher Box Notation by pasting into Mma.) Can someone provide motivation or precedent for this definition? E.g., why not arbitrarily *define* (-n)!, for positive integer n, as Limit[k! + (-1)^n/(k + n)/Gamma@n, k -> -n] ? (See A060746 <http://oeis.org/A060746>.) Or some more useful sequence, say, Bernoulli numbers? —rwg >From the Mma doc: "HurwitzZeta[s,a] gives the Hurwitz zeta function 𝜁(s,a)." Please feel welcome to share my confusion.

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[math-fun] Summarizing and clarifying my recent Stirlingoid noises
by Bill Gosper 04 Jun '19

04 Jun '19

Stirling stuff <http://gosper.org/3stirlings.pdf> —rwg

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