[math-fun] Compassless tangent construction
To www.tweedledum.com/rwg/tan.htm I added code for a Mathematica animation: (* Intersect line segments *) Interseg[{x1_, y1_}, {x2_, y2_}, {x3_, y3_}, {x4_, y4_}] := {((x2 - x1)*(x3*y4 - x4*y3) - (x4 - x3)*(x1*y2 - x2*y1)), -((y2 - y1)*(y3*x4 - y4*x3) - (y4 - y3)*(y1*x2 - y2*x1))}/((x2 - x1)*(y4 - y3) - (x4 - x3)*(y2 - y1)) Animate[Block[{pts = {Cos[#], Sin[#]} & /@ {Pi/2, -8*Pi/12 + Sin[u]/2, 3*Pi/10 + Sin[2*u]/3, 5*Pi/5 - Sin[2*u]/3 + Sin[3*u]/4, -Pi/4 + Sin[4*u]/1.5}, p, r}, p = Interseg[pts[[1]], pts[[2]], pts[[4]], pts[[5]]]; r = Interseg[p, Interseg[pts[[2]], pts[[3]], pts[[5]], pts[[1]]], pts[[3]], pts[[4]]]; Graphics[{Line[{pts[[1]], r}], Red, Line[{p, r, pts[[4]]}], Opacity[.5], Polygon[pts], Yellow, Disk[]}]], {u, 0, 2*Pi}] along with a frame image. Also mentioned on that page: Modulo some simplification, one can neatly *compute* the definition of the interseg function as the vanishing of two triangular areas. In Macsyma define(interseg(x1,y1,x2,y2,x3,y3,x4,y4), ([x,y],subst(linsolve([area_polygon([x1,y1],%%,[x2,y2]), area_polygon([x3,y3],%%,[x4,y4])],%%),%%))) The %% symbol just abbreviates [x,y]. Actually, since DEFINE evaluates its 2nd arg, better form is to quote the point pairs, define(interseg(x1,y1,x2,y2,x3,y3,x4,y4), ('[x,y],subst(linsolve([area_polygon('[x1,y1],%%,'[x2,y2]), area_polygon('[x3,y3],%%,'[x4,y4])],%%),%%))) to guard against any of the ten x,y,... symbols having an unintended preexisting value. --rwg
Student Daisuke Minematsu and his classmates have noticed that the Josephus problem has fractal behavior. http://demonstrations.wolfram.com/TheJosephusProblemInBothDirections/ Here's a specific image: http://demonstrations.wolfram.com/TheJosephusProblemInBothDirections/HTMLIma... Seems like a great result for a classic problem. It's new to me. Is it actually new? --Ed Pegg Jr
If you try to approximate e^-x on [0,oo) with a ratio of polynomials of degree n, then instead of vanishing, the error for large x will approach the ratio of the leading coefficients, which will also be the ripple height in the best (minimax) approximation. For successively larger n, this ripple height goes down by a factor which was erroneously conjectured to be 1/9, but is more like .107653919, and is somewhat wryly called the One Ninth constant. But it is sometimes also called Halphen's constant, because its square root coincidentally solves the equation 0=Theta_2''(0,i q), posed by Halphen. This square root (.32810656687497825302548) also has nice geometric properties w.r.t. the family of curves (or surface) whose parametric equation is (x,y) = (Theta_2(t), Theta_1(t)), 0<q<1. For small q this is a tiny "circle" which enlarges and grows "squarer" as q -> .328, in the manner of |x|^n + |y|^n = r^n, but rotated by pi/4, so that the incircle is tangent at +-x = +-y (t= (2k+1) pi/4). At q=.328, the incircle is maximal (r = 1.34883916748787) and the curvature vanishes at the four tangencies. The "CRT screen" exponent is n=5.419, a very close fit, but x^n+y^n has nonvanishing curvature at the axes. For q>.328, the curve develops concavities, eventually approaching an infinite "plus" sign. The equation stating the vanishing curvature at t=pi/4 is Theta_2'(pi/4) Theta_1''(pi/4) = Theta_1'(pi/4) Theta_2''(pi/4). But Theta_2'(pi/4) = Theta_1'(pi/4), and obviously doesn't vanish, and since Theta_2''(pi/4) = Theta_1''(pi/4) for all q, the Theta'' must vanish on both sides of the eqn. (I.e., the parametric curves must have their inflections at pi/4.) Maximality of the inradius is d Theta_2(pi/4)/dq = 0, but d Theta_s(z)/dq = -Theta_s''(z)/4/q, which for z=pi/4 is equivalent to the vanishing curvature. Alternative expressions for d Theta_2(pi/4)/dq are d Theta_2(0,i q)/dq /sqrt(2)/i^(1/4) = - Theta_2''(0,i q)/4/sqrt(2)/i^(1/4)/q = - Theta_1''(pi/4,q)/4/q . The area is maximal (7.40051953747301) at q = 0.43939962899098. Adding a z coordinate of q produces a surface resembling a squat, blunt-nosed bomb with four infinitely spreading tailfins, volume = 4.231765651557. Anyway, howsabout we honor Halphen with .328 and relieve "One ninth" of its extra, though more decorous synonym? Plouffe's Inverter might be the most influential current reference using "Halphen's" to mean "One ninth", which is (fortunately) discouraged in the current edition of Finch's (more influential) Constants book. The Weissopedia sez "See One Ninth constant", but doesn't come right out and say they're the same. --rwg (This exercise exposed many deficiencies in my Macsyma Theta function package (http://www.tweedledum.com/rwg/thet.gif, not with I.E.), but surprisingly more in Mma 6.0.) Great, politically incorrect Mythbusters question: Is it actually possible to break a (healthy) camel's back with straw?
On Monday 18 August 2008, rwg@sdf.lonestar.org wrote:
For successively larger n, this ripple height goes down by a factor which was erroneously conjectured to be 1/9, but is more like .107653919, and is somewhat wryly called the One Ninth constant. But it is sometimes also called Halphen's constant,
...
Anyway, howsabout we honor Halphen with .328 and relieve "One ninth" of its extra, though more decorous synonym?
"But I would rather not call it the Caratheodory constant C, since Mr Caratheodory determined that it already has another name, namely 1/16." -- Edmund Landau -- g
Adding a z coordinate of q produces a surface resembling a squat, blunt-nosed bomb with four infinitely spreading tailfins, volume = 4.231765651557.
Actually, it looks rather hypersonic if you crop the infinite tail and stretch it a bunch: http://gosper.org/sst.png .
volume = 4.231765651557. Per Macsyma. Mma is still working on the twelve digits I asked for yesterday.
This exercise exposed many deficiencies in my Macsyma Theta function package
Mostly in the area of simplifying special values and derivatives. Analogous to trig (under control of the same (%piargs) switch), it now translates out quarterperiods and quarterquasiperiods (which can get messy): (c430) (block([%piargs : false],thetaderiv[1](z+%pi/2+%i*log(q)/2,3,q,1)), %% = radcan(subst([q = 1/q,q = 1/q],%%))) %i log(q) %pi (d430) thetaderiv (z + --------- + ---, 3, q, 1) = 1 2 2 %i z %i z - (%i %e theta (z, q) + %e thetaderiv (z, 5, q) 3 3 %i z %i z + 5 %i %e thetaderiv (z, 4, q) - 10 %e thetaderiv (z, 3, q) 3 3 %i z %i z 5/4 - 10 %i %e thetaderiv (z, 2, q) + 5 %e thetaderiv (z, 1, q))/(4 q ) 3 3 Note that the rhs is free of d/dq. (c431) taylor(exponentialize(makeseries(%)),q,0,6) (d431)/T/ %i z %i z 4 %i z 8 11/4 %e %i (243 (%e ) - 1) %i ((3125 (%e ) - 243) %i) q - --------- - ---------------------- - --------------------------------- +... 5/4 %i z 1/4 %i z 3 4 q (4 %e ) q 4 (%e ) %i z %i z 4 %i z 8 11/4 %e %i (243 (%e ) - 1) %i ((3125 (%e ) - 243) %i) q = - --------- - ---------------------- - --------------------------------- +... 5/4 %i z 1/4 %i z 3 4 q (4 %e ) q 4 (%e ) The peculiar subst([q=1/q,q=1/q],...) is a sneaky way of temporarily asserting assume(not equal(q,0)). The special values of the eighth-periods of theta': (c407) for s thru 4 do block([%piargs : false,fancy_display : false], thetaderiv[s](%pi/4,1,q),print(%% = (if %piargs : true then resimplify(%%))))$ 3 2 %pi sqrt(2) eta (- q ) thetaderiv (---, 1, q) = - ------------------ 1 4 1/4 %i 3 2 %pi sqrt(2) eta (- q ) thetaderiv (---, 1, q) = ------------------ 2 4 1/4 %i %pi 3 8 thetaderiv (---, 1, q) = - 4 eta (q ) 3 4 %pi 3 8 thetaderiv (---, 1, q) = 4 eta (q ) 4 4 The 9th derivative wrt q at q=0 of the nth derivative wrt z of theta[3](z,q): (c436) thetaderiv[3](z,n,0,9) n %pi n (d436) 725760 6 cos(6 z + -----) 2 I wonder if WRI would like one of these.-) --rwg> --rwg
Ed Pegg had me snooping in http://functions.wolfram.com/EllipticFunctions/DedekindEta/ where I was royally shocked to find an (unattributed) evaluation of eta(e^(-2 pi)). Which Mma 6.0 knows! This opens the floodgates, permitting closed forms for eta(e^(-r pi)) for *all* positive rational r. E.g., 3/4 1/4 1/6 1 (5 3 - 3 sqrt(2) sqrt(3) - 3 3 ) gamma(-) - 12 %pi 4 [eta(%e ) = -------------------------------------------------, 1/24 3/4 2 2 sqrt(6) %pi 1/6 1 (sqrt(3) - 1) gamma(-) - 6 %pi 4 eta(%e ) = -------------------------, 1/12 3/8 3/4 2 2 3 %pi 1 1 gamma(-) gamma(-) - 4 %pi 4 - 2 %pi 4 eta(%e ) = -------------, eta(%e ) = --------, 3/8 3/4 3/4 2 2 %pi 2 %pi 1 2 %pi gamma(-) - ----- - %pi 4 3 eta(%e ) = -----------, eta(%e ) = 7/8 3/4 2 %pi 1/8 1/6 1 3 (sqrt(3) - 1) gamma(-) 4 - %pi/3 ------------------------------, eta(%e ) = 1/12 3/4 2 2 %pi 3/4 1/4 1/6 1 (5 3 - 3 sqrt(2) sqrt(3) - 3 3 ) gamma(-) 4 -------------------------------------------------] 1/24 3/4 2 2 %pi which in turn provide the corresponding theta[1]'(0,q), and theta[s](0,q): - 3 %pi [theta (0, %e ) = 2 1/6 3/4 1/4 2/3 1 (sqrt(3) - 1) (- 3 - 3 + 3 sqrt(2)) gamma(-) 4 --------------------------------------------------------, 2/3 13/24 3/4 2 2 3 %pi 1 gamma(-) - 3 %pi 4 theta (0, %e ) = ----------------------------------, 3 1/4 3/8 3/4 2 3 sqrt(sqrt(3) - 1) %pi - 3 %pi theta (0, %e ) = 4 5/6 1 (sqrt(3) - 1) gamma(-) 4 ------------------------------------------------] 1/3 5/24 3/4 1/4 2/3 3/4 2 3 (- 3 - 3 + 3 sqrt(2)) %pi There was also an evaluation of eta', giving such goodies as - %pi theta''' (0, %e ) = 1 2 1 %pi - %pi/3 3 1 gamma (-) theta'' (---, %e ) 3 gamma (-) 4 1 3 4 ----------------------------------- = - -----------. 3/2 13/4 2 sqrt(3) %pi 4 %pi (I just noticed %pi 1/3 2 2 2 theta'' (---, q ) eta (q ) 1 3 theta''' (0, q) = ------------------------------.) 1 sqrt(3) Do we agree with Rich that the simplifier should prefer theta''(pi/3) to theta'''(0)? So far, no luck finding eta(e^-(pi sqrt(2))) etc, and eta' of anything besides e^-(2 pi). (One more of the latter would open another floodgate.) Likewise, LatticeReduce found no polynomial for q^(1/24)*(i q;q)_oo for q=e^-pi. --rwg
More specifically, I pointed him to the formula search. http://functions.wolfram.com/formulasearch/ --Ed Pegg Jr --- On Fri, 8/22/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote: Ed Pegg had me snooping in http://functions.wolfram.com/EllipticFunctions/DedekindEta/ where I was royally shocked to find an (unattributed) evaluation of eta(e^(-2 pi)).
2^(1/4)*3^(3/8)*Gamma(1/4)*Gamma(1/3)=sqrt(sqrt(3)-1)*sqrt(%pi)*Gamma(1/12) 1/4 3/8 1 1 1 2 3 Gamma(-) Gamma(-) = sqrt(sqrt(3) - 1) sqrt(%pi) Gamma(--) 4 3 12 Can any CAS turn this into 0=0 and 1=1, or merely True? It must be easy--you can even carelessly switch factorial with Gamma(-: 1/4 3/8 1 1 1 2 3 (-)! (-)! = sqrt(sqrt(3) - 1) sqrt(%pi) (--)! 4 3 12 A general approach would be to reduce the factorialand denominator, but how short of making this one of the "axioms"? --rwg
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Aug 23. 2008 18:17]:
Ed Pegg had me snooping in http://functions.wolfram.com/EllipticFunctions/DedekindEta/ [...]
Could anyone make the pdf available? WRI managed to break the download mechanism: after entering email, name, etc, enabling javascript and cookies, trying two different browsers, I get: nothing. These people need a new brain implant.
From: "Joerg Arndt" <arndt@jjj.de>
WRI managed to break the download mechanism: after entering email, name, etc, enabling javascript and cookies, trying two different browsers, I get: nothing. These people need a new brain implant.
I was also surprised at first, but then found out that after returning back (after submitting the guestbook entry) to the DedekindEta page, and trying to download the pdf, it works. Alec
Ed Pegg had me snooping in http://functions.wolfram.com/EllipticFunctions/DedekindEta/ where I was royally shocked to find an (unattributed) evaluation of eta(e^(-2 pi)). Which Mma 6.0 knows! This opens the floodgates, permitting closed forms for eta(e^(-r pi)) for *all* positive rational r. [...] (I just noticed %pi 1/3 2 2 2 theta'' (---, q ) eta (q ) 1 3 theta''' (0, q) = ------------------------------.) 1 sqrt(3)
Do we agree with Rich that the simplifier should prefer theta''(pi/3) to theta'''(0)?
I'm leaning the other way, because of block([%piargs:all,fancy_display:false], print(intosum(niceindices(subst([q=1/q,q=1/q],thetaderiv[s](z,n,q,k))))),0)$ k ==== Stirling_s1(k, i) thetaderiv (z, n + 2 i, q) \ s
--------------------------------------------, / i k k ==== 4 (- 1) q i = 0
i.e., d/dq ~ d^2/dz^2 . So you can completely eliminate k "dq"s at a cost of 2k "dz"s, but not vice versa, where you wind up with Floor(n/2) "dq"s, and a possible leftover dz. The additional k terms are a bit unflavorful, but are a cost typical of kth derivatives, e.g. of f(q)*g(q).
So far, no luck finding eta(e^-(pi sqrt(2))) etc,
LatticeReduce just churned up 1 3 sqrt(gamma(-) gamma(-)) - 2 sqrt(2) %pi 8 8 eta(%e ) = -----------------------, 1/8 3/4 2 8 %pi 2 %pi 1/8 1 2 4 - ------- 7 sqrt(gamma(-) gamma(-) gamma(-)) sqrt(7) 7 7 7 eta(%e ) = -------------------------------------, 2 sqrt(2) %pi 4 %pi - ------- sqrt(7) eta(%e ) = 1/8 1/4 1 2 4 7 (sqrt(7) + 3) sqrt(gamma(-) gamma(-) gamma(-)) 7 7 7 ------------------------------------------------------, 1/8 4 2 %pi 1/8 3/2 1 3 gamma (-) - 2 sqrt(3) %pi 3 eta(%e ) = ----------------, 1/3 2 2 %pi 2 %pi 3/8 3/2 1 - ------- 3 gamma (-) sqrt(3) 3 eta(%e ) = ----------------, 1/3 2 2 %pi 4 %pi 3/8 1/4 3/2 1 - ------- 3 (sqrt(3) + 1) gamma (-) sqrt(3) 3 eta(%e ) = ---------------------------------, 1/8 1/4 2 2 8 %pi and might be coaxed into e^-(pi sqrt(n)) for any given positive integer, from which we can get pi <rational> sqrt(n) with the trivariate polynomials, at least for <rational> made of small primes. Automating this for eta|theta(0) simplification would make a fairly weird piece of nonrigorous (due to the LatticeReduce) computer algebra. Actually, the trivariate polynomials (for general eta(q)) are nonrigorous too, having been found with a Taylor series analog of LatticeReduce. But I dimly recall there being a theorem bounding the number of terms you need to check to guarantee equality. This pushes the luck frontier back to, e.g., e^-(pi sqrt(3/2)),
and eta' of anything besides e^-(2 pi). Sadly. (One more of the latter would open another floodgate.)
--rwg
rwg>This pushes the luck frontier back to, e.g., e^-(pi sqrt(3/2)). eta(%e^-(2*sqrt(2)*%pi/sqrt(3))) = Gamma(1/24)*sqrt(sin(%pi/24))*csc(%pi/8)^(1/6) /(2^(23/24)*3^(1/8)*(sqrt(3)-1)^(1/4)*sqrt(%pi)*sqrt(Gamma(1/12))) 2 sqrt(2) %pi - ------------- sqrt(3) eta(%e ) = 1 %pi 1/6 %pi Gamma(--) sqrt(sin(---)) csc (---) 24 24 8 ------------------------------------------------------, 23/24 1/8 1/4 1 2 3 (sqrt(3) - 1) sqrt(%pi) sqrt(Gamma(--)) 12 eta(%e^-(4*sqrt(2)*%pi/sqrt(3))) = (sqrt(3)-1)^(1/8)*Gamma(1/24)*sin(%pi/24)^(1/4) /(2*2^(17/24)*3^(1/8)*sqrt(%pi)*sqrt(Gamma(1/12))*SIN(%PI/8)^(1/12)) 4 sqrt(2) %pi - ------------- sqrt(3) eta(%e ) = 1/8 1 1/4 %pi (sqrt(3) - 1) Gamma(--) sin (---) 24 24 ----------------------------------------------------. 17/24 1/8 1 1/12 %pi 2 2 3 sqrt(%pi) sqrt(Gamma(--)) sin (---) 12 8 (Jacobi's imaginary transformation gives sqrt(6) pi and sqrt(3/2) pi.) All unproven, of course. rwg>So far, no luck finding eta(e^-(pi sqrt(2))) etc, and eta' of anything
besides e^-(2 pi). (One more of the latter would open another floodgate.)
(c665) 'at('diff(eta(q),q,1),q = %e^-(3*%pi)) = -%e^(3*%pi)*thetaderiv[1](%pi/3,2,%e^-(%pi/6))/(24*sqrt(3))) = -2^(1/8)*%e^(3*%pi)*Gamma(1/4)*(Gamma(1/4)^4/(16*%pi^(15/4))-1/%pi^(7/4))/8 | d | (d665) -- (eta(q))| dq | - 3 %pi |q = %e 3 %pi %pi - %pi/6 %e thetaderiv (---, 2, %e ) 1 3 = - -------------------------------------- 24 sqrt(3) 4 1 gamma (-) 1/8 3 %pi 1 4 1 2 %e gamma(-) (---------- - ------) 4 15/4 7/4 16 %pi %pi = - ------------------------------------------- 8 (c666) dfloat(%) | d | (d666) -- (eta(q))| dq | |q = 8.06995175703047d-5 = - 77.8356998979772d0 = - 77.8356998979791d0 So now we can do those log derivative sums, e.g., 'sum(n/(%e^(3*%pi*n)-1),n,1,inf) = 1/24-(2-sqrt(2)*3^(1/4))^(1/4)*((2-sqrt(2)*3^(1/4))^(3/4)*(5*sqrt(2)*3^(3/4)+6*sqrt(3)+9*sqrt(2)*3^(1/4)+6)*Gamma(1/4)^4/(288*(sqrt(3)-1)^(5/12)*%pi^2)+(sqrt(3)-1)^(7/12)/(2-sqrt(2)*3^(1/4))^(1/4))/(12*(sqrt(3)-1)^(7/12)*%pi) inf ==== \ n 1 1/4 1/4
------------- = -- - (2 - sqrt(2) 3 ) / 3 %pi n 24 ==== %e - 1 n = 1
1/4 3/4 3/4 1/4 ((2 - sqrt(2) 3 ) (5 sqrt(2) 3 + 6 sqrt(3) + 9 sqrt(2) 3 7/12 4 1 5/12 2 (sqrt(3) - 1) + 6) gamma (-)/(288 (sqrt(3) - 1) %pi ) + ---------------------) 4 1/4 1/4 (2 - sqrt(2) 3 ) 7/12 /(12 (sqrt(3) - 1) %pi) (I didn't say they'd be nice. The only opportunity here for further simplification is that quadrinomial.) (c658) bfloat(apply_nouns(subst(22,inf,%)),33) (d658) 8.07190569092038104656260759325017b-5 = 8.07190569092038104656260759325078b-5 The e^(2 pi n) case of this is very simple, but I don't see a way of searching for nonhypergeometric series identities on the Wolfram site, nor am I clear whether there are identities there that the latest version of Mma doesn't know. I repeat my plea: Does anybody know where WRI got those valuations of DedekindEta[I] and DedekindEta'[I] that made this all possible? --rwg PS, that Abel-Plana Theta integral further simplifies: 'integrate(sin(y)/((%e^-(2*%pi*y/log(q))-1)*(cos(2*y)+(q^4+1)/(2*q^2))),y,0,inf) = (((q+1)^2/(q^2+1)-theta[3](0,q)^2)*log(q)-4*atan(q))/(4*(1/q-q)) inf / [ sin(y) I ------------------------------------- dy ] 2 %pi y / - ------- 4 0 log(q) q + 1 (%e - 1) (cos(2 y) + ------) 2 2 q 2 (q + 1) 2 (-------- - theta (0, q)) log(q) - 4 atan(q) 2 3 q + 1 = --------------------------------------------. 1 4 (- - q) q (Which we can now give in Gammas for q=e^-(sqrt(rational) pi).)
Moron> (c665) 'at('diff(eta(q),q,1),q = %e^-(3*%pi)) [...] No! Just e^-pi: (c675) ('at('diff(eta(q),q,1),q = %e^-%pi) = -%e^(3*%pi)*thetaderiv[1](%pi/3,2,%e^-(%pi/6))/(24*sqrt(3))) = %e^(3*%pi)*(16*gamma(1/4)-gamma(1/4)^5/%pi^2)/(64*2^(7/8)*%pi^(7/4)) | d | (d675) -- (eta(q))| dq | - %pi |q = %e 3 %pi %pi - %pi/6 %e thetaderiv (---, 2, %e ) 1 3 = - -------------------------------------- = 24 sqrt(3) 5 1 gamma (-) 3 %pi 1 4 %e (16 gamma(-) - ---------) 4 2 %pi = --------------------------------- 7/8 7/4 64 2 %pi (c676) dfloat(%); | d | (d676) -- (eta(q))| dq | |q = 0.04321391826377d0 = - 77.8356998979772d0 = - 77.8356998979791d0 --rwg
Idiot>Moron> (c665) 'at('diff(eta(q),q,1),q = %e^-(3*%pi)) [...] Sorry, shoulda slept first. ('at('diff(eta(q),q,1),q = %e^-%pi) = -%e^%pi*thetaderiv[1](%pi/3,2,%e^-(%pi/6))/(24*sqrt(3))) = %e^%pi*(16*Gamma(1/4)-Gamma(1/4)^5/%pi^2)/(64*2^(7/8)*%pi^(7/4)) | d | (d682) -- eta(q)| dq | - %pi |q = %e %pi %pi - %pi/6 %e theta'' (---, %e ) 1 3 = - ------------------------------ 24 sqrt(3) 5 1 gamma (-) %pi 1 4 %e (16 gamma(-) - ---------) 4 2 %pi = ------------------------------- 7/8 7/4 64 2 %pi (c683) dfloat(%) | d | (d683) -- eta(q)| dq | |q = 0.04321391826377d0 = - 0.14535371204188d0 = - 0.14535371204189d0 --rwg
rwg>I repeat my plea: Does anybody know where WRI got those valuations of
DedekindEta[I] and DedekindEta'[I] that made this all possible?
Well, just the former, which Jacobi imaginary-transforms to itself. But if you differentiate the transformation before plugging in tau = I, you immediately get | 2 %pi - 2 %pi d | %e eta(%e ) -- (eta(q))| = ---------------------- dq | - 2 %pi 4 %pi |q = %e | d | - -- (eta(q))| dq | - 2 %pi |q = %e
(I didn't say they'd be nice. [...] But some of them are: 'sum(n/(%e^(6*%pi*n)-1),n,1,inf) = -sqrt(2)*3^(3/4)*Gamma(1/4)^4/(864*(sqrt(3)-1)*%pi^3)-1/(24*%pi)+1/24
inf 3/4 4 1 ==== sqrt(2) 3 Gamma (-) \ n 4 1 1
------------- = - ---------------------- - ------ + -- / 6 %pi n 3 24 %pi 24 ==== %e - 1 864 (sqrt(3) - 1) %pi n = 1
Taking only four terms of the sum, (c848) bfloat(apply_nouns(subst(4,inf,%)),32) (d848) 6.5124122633144373234994796373366b-9 = 6.512412263314437323499479556926b-9 Notice how the rhs approximates 0. Can its baby brother be new? 'Sum(n/(%e^(%pi*n)-1),n,1,inf) = Gamma(1/4)^4/(64*%pi^3)-1/(4*%pi)+1/24 inf 4 1 ==== gamma (-) \ n 4 1 1 > ----------- = --------- - ----- + -- / %pi n 3 4 %pi 24 ==== %e - 1 64 %pi n = 1 Still open: eta'(e^-(pi sqrt r)), and thus the corresponding sums like the above. --rwg
inf 4 1 ==== Gamma (-) \ n 4 1 1
----------- = --------- - ----- + --. / %pi n 3 4 %pi 24 ==== %e - 1 64 %pi n = 1
Still open: eta'(e^-(pi sqrt r)),
'at('diff(eta(q),q,1),q = %e^-(2*sqrt(3)*%pi)) = %e^(2*sqrt(3)*%pi)*Gamma(1/3)^(3/2)*(2^(1/3)*sqrt(3)*gamma(1/3)^6+32*%pi^3)/(512*2^(1/3)*3^(3/8)*%pi^5) | d | -- (eta(q))| dq | - 2 sqrt(3) %pi |q = %e 2 sqrt(3) %pi 3/2 1 1/3 6 1 3 %e gamma (-) (2 sqrt(3) gamma (-) + 32 %pi ) 3 3 = -------------------------------------------------------------- 1/3 3/8 5 512 2 3 %pi 'at('diff(eta(q),q,1),q = %e^-(2*%pi/sqrt(3))) = 3*%e^(2*%pi/sqrt(3))*Gamma(1/3)^(3/2)*(32*%pi^3-2^(1/3)*sqrt(3)*Gamma(1/3)^6)/(512*2^(1/3)*3^(1/8)*%pi^5) | d | 2 %pi -- (eta(q))| - ------- dq | sqrt(3) |q = %e 2 %pi ------- sqrt(3) 3/2 1 3 1/3 6 1 3 %e Gamma (-) (32 %pi - 2 sqrt(3) Gamma (-)) 3 3 = ---------------------------------------------------------- 1/3 1/8 5 512 2 3 %pi
and thus the corresponding sums like the above.
'sum(n/(%e^(2*%pi*n/sqrt(3))-1),n,1,inf) = 3*2^(1/3)*Gamma(1/3)^6/(256*%pi^4)-3/(8*sqrt(3)*%pi)+1/24 inf 1/3 6 1 ==== 3 2 Gamma (-) \ n 3 3 1
------------- = ---------------- - ------------- + -- / 2 %pi n 4 8 sqrt(3) %pi 24 ==== ------- 256 %pi n = 1 sqrt(3) %e - 1
'sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) = -2^(1/3)*Gamma(1/3)^6/(256*%pi^4)-1/(8*sqrt(3)*%pi)+1/24 inf 1/3 6 1 ==== 2 Gamma (-) \ n 3 1 1
--------------------- = - -------------- - ------------- + -- / 2 sqrt(3) %pi n 4 8 sqrt(3) %pi 24 ==== %e - 1 256 %pi n = 1
These were a lot of work, so they'd better be new! Still open: eta'', eta(e^-(pi phi)), e.g.. --rwg
inf 4 1 ==== Gamma (-) \ n 4 1 1
> ----------- = --------- - ----- + --.
/ %pi n 3 4 %pi 24 ==== %e - 1 64 %pi n = 1
Still open: eta'(e^-(pi sqrt r)),
'at('diff(eta(q),q,1),q = %e^-(2*sqrt(3)*%pi)) = %e^(2*sqrt(3)*%pi)*Gamma(1/3)^(3/2)*(2^(1/3)*sqrt(3)*gamma(1/3)^6+32*%pi^3)/(512*2^(1/3)*3^(3/8)*%pi^5)
| d | -- (eta(q))| dq | - 2 sqrt(3) %pi |q = %e
2 sqrt(3) %pi 3/2 1 1/3 6 1 3 %e gamma (-) (2 sqrt(3) gamma (-) + 32 %pi ) 3 3 = -------------------------------------------------------------- 1/3 3/8 5 512 2 3 %pi
'at('diff(eta(q),q,1),q = %e^-(2*%pi/sqrt(3))) = 3*%e^(2*%pi/sqrt(3))*Gamma(1/3)^(3/2)*(32*%pi^3-2^(1/3)*sqrt(3)*Gamma(1/3)^6)/(512*2^(1/3)*3^(1/8)*%pi^5)
| d | 2 %pi -- (eta(q))| - ------- dq | sqrt(3) |q = %e
2 %pi ------- sqrt(3) 3/2 1 3 1/3 6 1 3 %e Gamma (-) (32 %pi - 2 sqrt(3) Gamma (-)) 3 3 = ---------------------------------------------------------- 1/3 1/8 5 512 2 3 %pi
and thus the corresponding sums like the above.
'sum(n/(%e^(2*%pi*n/sqrt(3))-1),n,1,inf) = 3*2^(1/3)*Gamma(1/3)^6/(256*%pi^4)-3/(8*sqrt(3)*%pi)+1/24
inf 1/3 6 1 ==== 3 2 Gamma (-) \ n 3 3 1
------------- = ---------------- - ------------- + -- / 2 %pi n 4 8 sqrt(3) %pi 24 ==== ------- 256 %pi n = 1 sqrt(3) %e - 1
'sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) = -2^(1/3)*Gamma(1/3)^6/(256*%pi^4)-1/(8*sqrt(3)*%pi)+1/24
inf 1/3 6 1 ==== 2 Gamma (-) \ n 3 1 1
--------------------- = - -------------- - ------------- + -- / 2 sqrt(3) %pi n 4 8 sqrt(3) %pi 24 ==== %e - 1 256 %pi n = 1
These were a lot of work, so they'd better be new!
Joerg Arndt>You need formula 31.1-10a on p.628 of
http://www.jjj.de/fxt/#fxtbook (note my eta(q) is defined as prod(n=1,infty, 1-q^n))
YOW! I didn't know that!
and the evaluation of K(k_4) at http://mathworld.wolfram.com/EllipticIntegralSingularValue.html
And I was exposed to these within the last year, and completely failed to make the connection.
Attached a script, remains to show algebraically that 2^(-5/12) *k^(1/12)*kp^(1/3) * (sqrt(2)+1)^(1/2) == 1
The other evaluations K(k_n) on Eric's page will give you more eta-evaluations, but likely more complicated expressions. I suggest to try k_3 first. Well, yes, I've been tediously rediscovering these a few per day.
Is there a reason that all K(k_n) with n a square only involve gamma(1/4)?
Well, it's the same reason that eta(q^a), eta(q^b), and eta(q^c) are algebraically (and homogeneously) related when a, b, and c are positive integers. E.g., using h_n to mean (conventional) eta(q^n), 16*h[1]^8*h[4]^16+h[1]^16*h[4]^8-h[2]^24 8 16 16 8 24 16 h h + h h - h 1 4 1 4 2 387420489*h[1]^12*h[2]^24*h[3]^48+19131876*h[1]^24*h[2]^24*h[3]^36+19 6830*h[1]^36*h[2]^24*h[3]^24-16777216*h[2]^72*h[3]^12-196608*h[1]^24*h[2]^48*h[3]^12-12*h[1]^48*h[2]^24*h[3]^12-h[1]^72*h[3]^12+h[1]^60*h[2]^24 12 24 48 24 24 36 36 24 24 387420489 h h h + 19131876 h h h + 196830 h h h 1 2 3 1 2 3 1 2 3 72 12 24 48 12 48 24 12 72 12 - 16777216 h h - 196608 h h h - 12 h h h - h h 2 3 1 2 3 1 2 3 1 3 60 24 + h h 1 2 etc. These would be messier with your definition lacking the q^(1/24). Another formula that would be messier is eta(q) = theta_1(pi/3,q^(1/6))/sqrt(3). And I'm afraid your simplified definition will lead to needless confusion, especially since we already have the concise notation (q;q)_oo.
Still open: eta'', eta(e^-(pi phi)), e.g..
Well, certainly not the latter, which is a singular value. Can you damn my entire week with news that the eta' valuations are also old hat? --rwg
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Aug 25. 2008 13:37]:
[...]
I repeat my plea: Does anybody know where WRI got those valuations of DedekindEta[I] and DedekindEta'[I] that made this all possible? --rwg
[...]
My suggestion in the other mail seems to work (indeed without the eta-plus detour): You need formula 31.1-10a on p.628 of http://www.jjj.de/fxt/#fxtbook (note my eta(q) is defined as prod(n=1,infty, 1-q^n)) and the evaluation of K(k_4) at http://mathworld.wolfram.com/EllipticIntegralSingularValue.html Attached a script, remains to show algebraically that 2^(-5/12) *k^(1/12)*kp^(1/3) * (sqrt(2)+1)^(1/2) == 1 The other evaluations K(k_n) on Eric's page will give you more eta-evaluations, but likely more complicated expressions. I suggest to try k_3 first. Is there a reason that all K(k_n) with n a square only involve gamma(1/4)?
participants (6)
-
Alec Mihailovs -
Ed Pegg Jr -
Gareth McCaughan -
Joerg Arndt -
rwg＠sdf.lonestar.org -
Xeipon