Re: [math-fun] Theta_3(0,q) near the unit circle
----- Original Message ---- From: "rwg@sdf.lonestar.org" <rwg@sdf.lonestar.org> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, October 7, 2008 4:02:59 AM Subject: [math-fun] Theta_3(0,q) near the unit circle ... PS, how can a natural boundary grow only like 1/sqrt(-log q)? The other thetas are *much* wilder. _______________________________________________ Einar Hille gives the following example (Analytic Function Theory, vol. 1, p.133). sum( z^(2^n) / n!, n=0..infinity) This function and its derivatives of all orders are continuous and bounded in the closed unit disk. Yet the unit circle is a natural boundary. Gene
----- Original Message ----
From: "rwg@sdf.lonestar.org" <rwg@sdf.lonestar.org> To: math-fun <math-fun@mailman.xmission.com> Sent: Tuesday, October 7, 2008 4:02:59 AM Subject: [math-fun] Theta_3(0,q) near the unit circle
... PS, how can a natural boundary grow only like 1/sqrt(-log q)? The other thetas are *much* wilder. _______________________________________________
Einar Hille gives the following example (Analytic Function Theory, vol. 1, p.133).
sum( z^(2^n) / n!, n=0..infinity)
This function and its derivatives of all orders are continuous and bounded in the closed unit disk. Yet the unit circle is a natural boundary.
Gene ?? Surely it blows up at z=+-1. Interestingly, it has an infinitude of roots (but trickily, no accumulation point) inside the unit circle: 0, -.65862675430016392241347283058, .120314841052762693451935272875 +- .934605942791339747870826388582 i, -.685206279747129651080553110888 +- .670534105899025904030180437365 i,... .
ParametricPlot3D[{r*Cos[t], r*Sin[t], Abs[Sum[(r*E^(I*t))^2^n, {n, 0, Infinity}]]}, {r, 0, 1}, {t, 0, 2*Pi}] reveals the first few and suggests a few more, but they're so abrupt Mma would need to rootfind to look convincing. If it found too many, the inner ones would be hidden behind a curtain of needles at the boundary, although in a super-accurate plot, those ought to be pretty transparent. --rwg
(Oops, sorry, Gene, you're right--I completely missed the 1/n!. Unbelievable that something with that smooth an edge could have a curtain of poles there. Even less believable with the edge hidden by your challenge continuation. Nice plot. only one nontrivial root.) eta(%i*%e^-(%pi/(2*sqrt(3)))) = 3^(3/8)*gamma(1/3)^(3/2)/(2^(5/6)*%i^(1/6)*%pi) 3/8 3/2 1 3 gamma (-) i 3 eta(----------) = ---------------- . pi 5/6 1/6 --------- 2 i pi 2 sqrt(3) e eta(-%e^-(%pi/sqrt(3))) = 3^(3/8)*%i^(1/12)*gamma(1/3)^(3/2)/(2*%pi) 3/8 3/2 1 1/12 3 gamma (-) i 1 3 eta(- --------) = ---------------------- . pi 2 pi ------- sqrt(3) e theta[4](0,%i*%e^-(%pi/(2*sqrt(3)))) = 3^(3/8)*gamma(1/3)^(3/2)/(2^(2/3)*%i^(5/12)*%pi) 3/8 3/2 1 3 gamma (-) i 3 theta (0, ----------) = ---------------- . 4 pi 2/3 5/12 --------- 2 i pi 2 sqrt(3) e eta(%e^(2*%i*%pi/3-%pi/(3*sqrt(3)))) = (sqrt(3)+1)^(1/4)*(2^(5/6)*((2^(2/3)+2*2^(1/3)+1)*sqrt(3)-3*2^(1/3)+3)+%i*(2*2^(1/6)*(2^(1/3)-1)*sqrt(3)+2^(5/6)*(sqrt(3)-3)))*gamma(1/3)^(3/2)/(8*2^(3/8)*3^(13/24)*%pi) 2 i pi pi ------ - --------- 3 3 sqrt(3) 1/4 3/2 1 eta(e ) = (sqrt(3) + 1) gamma (-) 3 1/6 1/3 5/6 ((2 2 (2 - 1) sqrt(3) + 2 (sqrt(3) - 3)) i 5/6 2/3 1/3 1/3 3/8 13/24 + 2 ((2 + 2 2 + 1) sqrt(3) - 3 2 + 3))/(8 2 3 pi) . --rwg PS, to compute the volume of a tetraroller, I need to run the prismatoid formula on a wedge cut off the base of a cone. This requires finding two saggitae. Since osage orange is most highly prized for making bows, it is therefore completely germane to math-fun. PRISMATOID DIATROPISM, CONSIDERATE DESECRATION
(Oops, sorry, Gene, you're right--I completely missed the 1/n!. Unbelievable that something with that smooth an edge could have a curtain of poles there. Even less believable with the edge hidden by your challenge continuation. Nice plot. only one nontrivial root.)
Foo, there are no poles. You can write in closed form the Taylor expansion coefficients for a point on the circle. The only problem is the radius of convergence is 0. So natural boundaries have nothing to do with curtains of poles--they're just places you can't continue past because of vanishing convergence circles. I had no idea of the inequivalence. But then why do the radii vanish? Does this mean there are infinitely many solutions to Gene's challenge? Looks like there's also a nice Fourier series on the circle. --rwg
rwg>(http://gosper.org/tetraroller1.gif, http://gosper.org/tetraroller2.gif)
PS, to compute the volume of a tetraroller, I need to run the prismatoid formula on a wedge cut off the base of a cone. This requires finding two saggitae. Since osage orange is most highly prized for making bows, it is therefore
Can't work. The saggita and radius are linear functions of the vertical coordinate. If the segment area were a polynomial function of height, we could solve the transcendental Kepler's Pizza equation (www.tweedledum.com/rwg/pizza.htm). Did anyone else try for the volume formula? I got a couple of "challenge integrals" that defeated Mma and Macsyma. But this is fairly nice: dissect the tetraroller into twelve congruent conical wedges. These are "tetrahedroids", with one edge joining a tetrahedron vertex to the centroid, and the opposite "edge" being an elliptical arc, the intersection of two cones. For a tetrahedron centered at the origin with edge length sqrt3, apex on the z axis, one of the three tetrahedroids sharing the apex has parametric edge formula E(t) = [cos(t)*(sqrt(cos(t)^2+3)-cos(t))/2,sqrt(3)*(sqrt(cos(t)^2+3)-cos(t))*sin(t)/4, cos(t)*(sqrt(cos(t)^2+3)-cos(t))/(2*sqrt(2))], |t|<acos 1/5. Now a volume element(t) is vol_tetrahedron(E(t),E(t+dt),apex,origin), a big mess that to first order simplifes down to sqrt(2)*sqrt(3)*(2*cos(t)*sqrt(cos(t)^2+3)-2*cos(t)^2-3)*dt/64 . Integrating -acos(1/5)<t<acos(1/5) (times 12) gave 3*(sqrt(6)*asin(sqrt(6)*(2*sqrt(19)-1)/25)/2-3*sqrt(19)/25+3/50) ~1.76132660170417, nearly thrice the tetrahedron's sqrt(6)/4 ~0.6123724356958. It would be nice to get an independent check of this, and then use it as the rhs of the challenge integrals. --rwg PS, did anybody look at intersecting four cylinders instead of cones? I'm not even sure how many faces this makes (http://gosper.org/face.jpg). Likewise the intersections of 6 and 10 cylinders corresponding to the higherhedra. Nice workouts for somebody's solid-intersector software. True or False Quickie: The volume of a tetrahedron is determined by the areas of its faces.
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