[math-fun] AGM for Gamma(n/12)
In case anybody besides me missed it, Gamma(1/3) = 2^(11/18)*(sqrt(3)+1)^(1/3)*%pi^(2/3)/(3^(1/4)*agm(sqrt(2)*(sqrt(3)+1),1)^(1/3)) 11/18 1/3 2/3 1 2 (sqrt(3) + 1) %pi gamma(-) = ------------------------------------- 3 1/4 1/3 3 AGM (sqrt(2) (sqrt(3) + 1), 1) follows from Borwein&Borwein, pp 15(a) and 28(d), so Gamma(1/3) is nearly as easy as pi. Similarly, Gamma(1/4) = 2^(3/4)*%pi^(3/4)/sqrt(agm(sqrt(2),1)) 3/4 3/4 1 2 %pi Gamma(-) = --------------------- 4 sqrt(AGM(sqrt(2), 1)) finishes off Gamma(n/12). Interesting that, unlike pi, there are no hypergeometric series for these, although the sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) is a degenerate bibasic series. ((1 = AGM(eta(q^2)^10/(eta(q)^4*eta(q^4)^4),eta(q)^4/eta(q^2)^2)) = AGM((16*eta(q^4)^8+eta(q)^8)^(5/12)/eta(q)^(2/3),eta(q)^(10/3)/(16*eta(q^4)^8+eta(q)^8)^(1/12))/eta(q^4)^(2/3)) = AGM(sqrt(64*eta(q^4)^24+eta(q^2)^24)-8*eta(q^4)^12,eta(q^2)^12)/(eta(q^2)^2*eta(q^4)^2*sqrt(sqrt(64*eta(q^4)^24+eta(q^2)^24)-8*eta(q^4)^12)) 10 2 4 eta (q ) eta (q) 1 = AGM(----------------, --------) 4 4 4 2 2 eta (q) eta (q ) eta (q ) 8 4 8 5/12 10/3 (16 eta (q ) + eta (q)) eta (q) AGM(---------------------------, ---------------------------) 2/3 8 4 8 1/12 eta (q) (16 eta (q ) + eta (q)) = ------------------------------------------------------------- 2/3 4 eta (q ) 24 4 24 2 12 4 12 2 AGM(sqrt(64 eta (q ) + eta (q )) - 8 eta (q ), eta (q )) = --------------------------------------------------------------------. 2 2 2 4 24 4 24 2 12 4 eta (q ) eta (q ) sqrt(sqrt(64 eta (q ) + eta (q )) - 8 eta (q )) --rwg Does anyone besides Salamin remember the meaning of the agm error in Peter Samson's (1960s) PDP-1 music compiler?
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Aug 28. 2008 13:29]:
In case anybody besides me missed it,
Gamma(1/3) = 2^(11/18)*(sqrt(3)+1)^(1/3)*%pi^(2/3)/(3^(1/4)*agm(sqrt(2)*(sqrt(3)+1),1)^(1/3))
11/18 1/3 2/3 1 2 (sqrt(3) + 1) %pi gamma(-) = ------------------------------------- 3 1/4 1/3 3 AGM (sqrt(2) (sqrt(3) + 1), 1)
follows from Borwein&Borwein, pp 15(a) and 28(d), so Gamma(1/3) is nearly as easy as pi. Similarly,
Gamma(1/4) = 2^(3/4)*%pi^(3/4)/sqrt(agm(sqrt(2),1))
3/4 3/4 1 2 %pi Gamma(-) = --------------------- 4 sqrt(AGM(sqrt(2), 1))
finishes off Gamma(n/12). Interesting that, unlike pi, there are no hypergeometric series for these, although the sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) is a degenerate bibasic series.
cf. http://arxiv.org/abs/math/0403510 (section 5) gives similar expressions also for gamma(1/8) and gamma(1/24)
[... AGM and eta ...]
lovely!
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Aug 28. 2008 13:29]:
In case anybody besides me missed it,
Gamma(1/3) = 2^(11/18)*(sqrt(3)+1)^(1/3)*%pi^(2/3)/(3^(1/4)*agm(sqrt(2)*(sqrt(3)+1),1)^(1/3))
11/18 1/3 2/3 1 2 (sqrt(3) + 1) %pi gamma(-) = ------------------------------------- 3 1/4 1/3 3 AGM (sqrt(2) (sqrt(3) + 1), 1)
follows from Borwein&Borwein, pp 15(a) and 28(d), so Gamma(1/3) is nearly as easy as pi. Similarly,
Gamma(1/4) = 2^(3/4)*%pi^(3/4)/sqrt(agm(sqrt(2),1))
3/4 3/4 1 2 %pi Gamma(-) = --------------------- 4 sqrt(AGM(sqrt(2), 1))
finishes off Gamma(n/12). Interesting that, unlike pi, there are no hypergeometric series for these, although the sum(n/(%e^(2*sqrt(3)*%pi*n)-1),n,1,inf) is a degenerate bibasic series.
Joerg Arndt> cf. http://arxiv.org/abs/math/0403510 (section 5)
gives similar expressions also for gamma(1/8) and gamma(1/24)
That's the sort of thing I'd much prefer be done by an active CAS. Same goes for most of http://functions.wolfram.com . But it's probably too hard to automate the singular value stuff, so we're stuck with tables. (Of course, we could hide tables inside the CAS. Macsyma once had an error: "Factor ran out of primes!") Likewise for the singular eta' values I'm finding, and their consequent eta'/eta sums: 'Sum(n/(%e^(2*sqrt(7)*%pi*n)-1),n,1,inf) = -9*Gamma(1/7)^2*Gamma(2/7)^2*Gamma(4/7)^2/(3584*%pi^4)-1/(8*sqrt(7)*%pi)+1/24 inf 2 1 2 2 2 4 ==== 9 Gamma (-) Gamma (-) Gamma (-) \ n 7 7 7 > --------------------- = - ------------------------------- / 2 sqrt(7) %pi n 4 ==== %e - 1 3584 %pi n = 1 1 1 - ------------- + -- 8 sqrt(7) %pi 24 'Sum(n/(%e^(2*%pi*n/sqrt(7))-1),n,1,inf) = 9*Gamma(1/7)^2*Gamma(2/7)^2*Gamma(4/7)^2/(512*%pi^4)-sqrt(7)/(8*%pi)+1/24 inf 2 1 2 2 2 4 ==== 9 Gamma (-) Gamma (-) Gamma (-) \ n 7 7 7 sqrt(7) 1 > ------------- = ------------------------------- - ------- + -- / 2 %pi n 4 8 %pi 24 ==== ------- 512 %pi n = 1 sqrt(7) %e - 1 Unfortunately, tables have bugs. E.g., for K(k_11) in terms of Betas, Eric defines a constant R as "the real root of x^3-4x=4=0". The obvious guesses x^3-4x=4 and x^3-4*x+4=0 don't seem to work. Earlier, they use the undenested radicals sqrt(2+-sqrt(3)), which also make me nervous. --rwg rwg>Does anyone besides Salamin remember the meaning of the agm error in
Peter Samson's (1960s) PDP-1 music compiler?
TK> That argument greedily masticates. Would you believe Samson himself had forgotten this? Seemed incredulous, even. It suggests that they don't have that compiler at the restoration project.
[... AGM and eta ...]
lovely!
Tnx!
I haven't read Borwein & Borwein, and I don't know much about the AGM beyond its definition.
(Or one def. at least: for a,b > 0 define F(a,b) := ((a+b)/2, sqrt(ab)) and iterate; the limit is then (AGM(a,b), AGM(a,b)) if memory serves.)*
Is much known about when AGM(a,b) is rational, algebraic, or transcendental?
--Dan
Sort of. That formula (a) on p15 of B&B effectively says (AGM(a,b) = AGM(b,a)) = %pi*b/(2*\k(sqrt(1-a^2/b^2))) %pi b agm(a, b) = agm(b, a) = ----------------- 2 a 2 K(sqrt(1 - --)) 2 b where K is the complete elliptic integral. Unfortunately, if you try this with Macsyma (elliptic_kc) or Mma (EllipticK, ArithmeticGeometricMean) (but not Maple (Elliptick)), you'll lose until you lose the sqrt, because the first two expect the "parameter" (:=k^2) instead of the modulus (:=k). K then takes on special ("singular") values, of the form algebraic*powers(Gammas), which are algebraic multiples (of sqrt) of those etas I'm flaming about. These are fractional Gammas over sqrt pi, so they're all transcendental, and nobody can prove it. Btw, A great virtue of AGM is that it converges quadratically (a'-b' ~ (a-b)^2), providing the fastest numerical methods for Gamma of n/12 and n/8, except that for Gamma(1/2) the advantage over series methods may be purely theoretical. --rwg ___________________________________
* Btw, is there a natural continuum of "means" with the AM & GM at the extremes, and the AGM halfway between them?
Beaucoup. See if you can locate a paper by DH Lehmer on iterated means.
rwg>>Still open: eta'', eta(e^-(pi phi)), e.g.. rwg> Well, not the latter, which is a singular value. NO! i phi is *not* a modular transformation of i sqrt 5. I have no idea what eta(e^-(pi phi)) is. But here is some phi and sqrt 5 merriment: 'sum(n/(%e^(2*sqrt(5)*%pi*n)-1),n,1,inf) = -%phi*Gamma(1/20)^2*Gamma(9/20)^2/(960*5^(3/4)*%pi^3)-1/(8*sqrt(5)*%pi)+1/24 inf 2 1 2 9 ==== %phi Gamma (--) Gamma (--) \ n 20 20 > --------------------- = - -------------------------- / 2 sqrt(5) %pi n 3/4 3 ==== %e - 1 960 5 %pi n = 1 1 1 - ------------- + -- 8 sqrt(5) %pi 24 'sum(n/(%e^(2*%pi*n/sqrt(5))-1),n,1,inf) = %phi*Gamma(1/20)^2*Gamma(9/20)^2/(192*5^(3/4)*%pi^3)-sqrt(5)/(8*%pi)+1/24 inf 2 1 2 9 ==== %phi Gamma (--) Gamma (--) \ n 20 20 sqrt(5) 1 > ------------- = -------------------------- - ------- + -- / 2 %pi n 3/4 3 8 %pi 24 ==== ------- 192 5 %pi n = 1 sqrt(5) %e - 1 Joerg Arndt> [arxiv table of rational Gamma reductions] rwg>That's the sort of thing I'd much prefer be done by an active CAS. Same goes for
most of http://functions.wolfram.com .
Notice that the above results are considerably simpler than the 5th singular values tabulated in MathWorld and Borwein&Borwein (p 298), due partly to using the 20th singular value instead, but more to a Gamma simplification that was not in http://arxiv.org/abs/math/0403510 . If we can't find a rigorous algorithm to simplify them, there's always LatticeReduce! That's how I'm finding the (eta')s. --rwg PS, these log derivate sums can also be written 'sum(n/(1/q^n-1),n,1,inf) = 'sum(q^n/(1-q^n)^2,n,1,inf) inf inf ==== ==== n \ n \ q > ------- = > --------- / -n / n 2 ==== q - 1 ==== (1 - q ) n = 1 n = 1
Notice that the above results are considerably simpler than the 5th singular values tabulated in MathWorld and Borwein&Borwein (p 298), due partly to using the 20th singular value instead, but more to a Gamma simplification that was not in http://arxiv.org/abs/math/0403510 .
Actually, it sort of was, giving Gamma of n/20 in terms of 1/20, 1/5, and 2/5. But that's three species vs two. They also give n/24 in terms of 1/24, 1/3, and 1/4, which with
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 gives us AGMs for Gamma(n/24). The table also indicates that eta(e^-(pi sqrt 15)) needs only Gamma of 1/15 an 4/15. Rashly assuming I am the only current source for these results, if anyone wants a particular eta, eta', or logderiv series(e^(-pi sqrt(rational)), send it along. (Meanwhile, I'll resume jobseeking.) Finally, those log derivative sums came in pairs by virtue of the imaginary transformation: x*('sum(n/(%e^(2*%pi*n*x)-1),n,1,inf)-1/24)+('sum(n/(%e^(2*%pi*n/x)-1),n,1,inf)-1/24)/x = -1/(4*%pi) inf inf ==== ==== \ n 1 1 \ n 1 1 x ( > --------------- - --) + - ( > ------------- - --) = - -----. / 2 %pi n x 24 x / 2 %pi n 24 4 %pi ==== %e - 1 ==== ------- n = 1 n = 1 x %e - 1 --rwg
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Aug 29. 2008 10:17]:
[...]
Unfortunately, tables have bugs. E.g., for K(k_11) in terms of Betas, Eric defines a constant R as "the real root of x^3-4x=4=0". The obvious guesses x^3-4x=4 and x^3-4*x+4=0 don't seem to work. Earlier, they use the undenested radicals sqrt(2+-sqrt(3)), which also make me nervous.
k_11 = 0.02184972261085453699635444... minpoly(k_11^2) == 4096*x^6 - 12288*x^5 + 13056*x^4 - 5632*x^3 + 2864*x^2 - 2096*x + 1 (?)
[...]
On Aug 28, 2008, at 4:07 AM, rwg@sdf.lonestar.org wrote:
rwg>Does anyone besides Salamin remember the meaning of the agm error in Peter Samson's (1960s) PDP-1 music compiler?
TK> That argument greedily masticates.
Would you believe Samson himself had forgotten this? Seemed incredulous, even. It suggests that they don't have that compiler at the restoration project.
[... AGM and eta ...]
lovely!
Tnx!
To clarify.... The music compiler we are running at the PDP-1 restoration project was assembled by me from source code in my possession dated 5/21/63. It does not contain the agm error; in fact it contains neither the string ag nor the string gm. Perhaps that error was added later by someone else. Regards, --prs.
On Aug 28, 2008, at 4:07 AM, rwg@sdf.lonestar.org wrote:
rwg>Does anyone besides Salamin remember the meaning of the agm error in Peter Samson's (1960s) PDP-1 music compiler?
TK> That argument greedily masticates.
Would you believe Samson himself had forgotten this? Seemed incredulous, even. It suggests that they don't have that compiler at the restoration project.
To clarify.... The music compiler we are running at the PDP-1 restoration project was assembled by me from source code in my possession dated 5/21/63. It does not contain the agm error; in fact it contains neither the string ag nor the string gm. Perhaps that error was added later by someone else.
Regards, --prs.
Wow, thanks for that, mystifying though it is. What about wlk and dnt wlk ? I don't suppose there was a compiler doc in that binder I donated. This all recalls another seemingly counterfactual memory of that era: There was on the front panel an indicator lamp labeled HALT STORE, right up there with RUN, CYCLE, and DEFER. No one had ever seen it so much as glimmer until a very short tape appeared on the console labeled Turn on HALT STORE, which did just that, while extinguishing RUN. But at the rollout, I could find no one else who shared this memory, not even Kotok (R.I.P.). So I ran down to examine the machine, and to my horror, there was no HALT STORE! Finally, Jack Holloway remembered that the very earliest PDP-1s (was RLE's Serial # 3?) had different sequence break logic, and used that bit to remember interrupting out of the IO wait state. I'll be dzmed. --rwg
You might be able to track down Stu Nelson for his take on some of these questions. I contacted him about 5 years ago through the Systems Concepts answering machine number in Las Vegas (or Reno?). He said that he and Mike Levitt now lived in France (Paris?) most of the time but kept an answering machine in Nevada for old customers to contact them through. Samson may have better contact info. -----Original Message----- From: rwg@sdf.lonestar.org [mailto:rwg@sdf.lonestar.org] Sent: Sunday, August 31, 2008 7:50 AM To: Peter Samson Cc: rwg@sdf.lonestar.org; math-fun Subject: PDP-1 lore
On Aug 28, 2008, at 4:07 AM, rwg@sdf.lonestar.org wrote:
rwg>Does anyone besides Salamin remember the meaning of the agm error in Peter Samson's (1960s) PDP-1 music compiler?
TK> That argument greedily masticates.
Would you believe Samson himself had forgotten this? Seemed incredulous, even. It suggests that they don't have that compiler at the restoration project.
To clarify.... The music compiler we are running at the PDP-1 restoration project was assembled by me from source code in my possession dated 5/21/63. It does not contain the agm error; in fact it contains neither the string ag nor the string gm. Perhaps that error was added later by someone else.
Regards, --prs.
Wow, thanks for that, mystifying though it is. What about wlk and dnt wlk ? I don't suppose there was a compiler doc in that binder I donated. This all recalls another seemingly counterfactual memory of that era: There was on the front panel an indicator lamp labeled HALT STORE, right up there with RUN, CYCLE, and DEFER. No one had ever seen it so much as glimmer until a very short tape appeared on the console labeled Turn on HALT STORE, which did just that, while extinguishing RUN. But at the rollout, I could find no one else who shared this memory, not even Kotok (R.I.P.). So I ran down to examine the machine, and to my horror, there was no HALT STORE! Finally, Jack Holloway remembered that the very earliest PDP-1s (was RLE's Serial # 3?) had different sequence break logic, and used that bit to remember interrupting out of the IO wait state. I'll be dzmed. --rwg
Although eta goes berserk on the unit circle, it really is 0 at e^(i pi rational) by virtue of infinitely many terms of the product formula vanishing identically. Just how emphatically 0 is clear from what you have to multiply by to keep it finite as you (nontangentially) approach the boundary. E.g., for rational = 2n/11, 0<n<11: 2 %pi 2 %i %pi n - ---------- ---------- 726 log(q) 11 limit %e sqrt(- log(q)) eta(%e q) q -> 1 %i %pi k(n) ---------- 2 %pi 132 = sqrt(-----) %e , 11 where k(1),k(2),... = -89, 28, -15, -14, 35, -24, 25, 26, 39, 100 (mod 264). There is probably a nifty formula for these if anyone wants to decrypt it. For more data, you should have no trouble guessing how to generalize to denominators besides 11. Mma 6.0 is surprisingly good at the numerics (but with the tau = log(q)/2/i/pi argument convention. tau = I/999 is plenty small enough.) Or I can crank out a few more k(n) sets if you tell me which you want. --rwg PS, this limit stuff applies also to (q;q)_oo = eta(q)/q^24.
I wonder if I got to the muffler shop I'm sitting in soon enough to save my brain from monoxide. My yahoo page is excitedly telling me that Obama may soon announce his running mate, and there is a large, six colored Rand McNally US map on the waiting room wall, with Virginia, Kentucky, and Illinois forming a single-colored blob. Copyright 2001. If anyone replied to my eta(e^(i pi n/11)) item, please cc me here, until Yahoo breaks out of its stasis field. --rwg
"Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0. --rwg
The b(k)^n in 'sum(b(k),k,1,n) = 'sum(b(k)^n/'prod(b(k)-b(j),j,1,k-1)/'prod(b(k)-b(j),j,k+1,n),k,1,n) n n ==== ==== n \ \ b (k) > b(k) = > ------------------- / / n ==== ==== /===\' k = 1 k = 1 | | | | (b(k) - b(j)) | | j = 1 looks improbable but works. For what I'm unsure. qpoch(q,q,inf)*(1-qpoch(z,q,inf)) = z*'sum((-1)^k*q^(k*(k+3)/2)*qpoch(-z/q^k,q^2,inf)/(qpoch(q,q,k)*(q^k*z+1)),k,0,inf) k (k + 3) --------- k z 2 2 oo (- 1) (- --; q ) q ==== k oo \ q (q; q) (1 - (z; q) ) = z > ------------------------------ oo oo / k ==== (q; q) (q z + 1) k = 0 k has a somewhat oddball lhs, a naked z, and a rather stubborn infinite product in its summand. Computationally (e.g., for evaluating eta(q)/q^(1/24)), it's generally inferior to the q-Exponential qpoch(z,q,inf) = 'sum((-1)^n*q^((n-1)*n/2)*z^n/qpoch(q,q,n),n,0,inf) (n - 1) n oo --------- ==== n 2 n \ (- 1) q z (z; q) = > -------------------- oo / (q; q) ==== n n = 0 (as opposed to the q-exponential). Both sums (and the equivalent infinite product) bog down as |q|->1, which doesn't matter for eta due to the imaginary transformation, but thwarts Pochhammer numerics in general. A partial remedy it to use (z;q)_oo = (z;q)_k (z q^k;q)_oo, qpoch(z,q,inf) = qpoch(z,q,k)*'sum((-1)^n*q^(n*(n+2*k-1)/2)*z^n/qpoch(q,q,n),n,0,inf) n (n + 2 k - 1) oo --------------- ==== n 2 n \ (- 1) q z (z; q) = (z; q) > -------------------------- oo k / (q; q) ==== n n = 0 (of which the previous sum is the k=0 case), which improves the term ratio to -q^(n+k)*z/(1-q^(n+1)) n + k q z - ----------, n + 1 1 - q for however large k that optimizes the sum vs product tradeoff. --rwg
That reminds me of an old puzzle which I posted to the internet in 1993, and it is still there: http://groups.google.com/group/rec.puzzles/browse_thread/thread/281a9299b200... Not sure if it can be Brahmaguptaed, though :) On Wed, Sep 3, 2008 at 7:52 AM, <rwg@sdf.lonestar.org> wrote:
"Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0. --rwg
I think this fully characterizes eta (and (q;q)_oo) near the cyclotomic points: Empirically, 'limit(sqrt(d)*%e^-(%pi^2/(6*d^2*log(q)))*sqrt(-log(q))*eta(%e^(2*%i*%pi*n/d)*q)/(sqrt(2)*sqrt(%pi)),q,1) = exp(%i*%pi*g[n,d]/12) 2 pi - ----------- 2 i pi n 2 ---------- 6 d log(q) d sqrt(d) e sqrt(- log(q)) eta(e q) limit -------------------------------------------------------- q -> 1 sqrt(2 pi) g n, d i pi ------, 12 = e where g[1,d] := 3-d-1/d, g[n,1] := n, and g[n,d] is a sum over the continued fraction expansion of n/d: g[n,d] := if d = 1 then n else (if n = 1 then -d-1/d+3 else (if ?numgcd(n,d) = 1 then block([k : nummod(n,d),f : nummod(d,n)],n/d+g[k,f]-k/d+f/k+1/(f*k)-d/k-1/(d*k)) else "")) 1 (d226) g := if d = 1 then n else (if n = 1 then - d - - + 3 n, d d else (if gcd(n, d) = 1 then block([k : mod(n, d), f : mod(d, n)], n k f 1 d 1 - + g - - + - + --- - - - ---) else "")) d k, f d k f k k d k
2 %pi 2 %i %pi n - ---------- ---------- 726 log(q) 11 limit %e sqrt(- log(q)) eta(%e q) q -> 1
%i %pi k(n) ---------- 2 %pi 132 = sqrt(-----) %e , 11
where k(1),k(2),... = -89, 28, -15, -14, 35, -24, 25, 26, 39, 100 (mod 264). There is probably a nifty formula for these if anyone wants to decrypt it.
(c227) makelist(g[n,11]*11,n,1,10) (d227) [- 89, - 28, - 15, - 14, 35, - 24, 25, 26, 39, 100] I'm no longer shocked Rich didn't see it immediately. And if some Cretaceous Old Legend managed all this by hand, I'd sure like to know how. I think. --rwg
I think this fully characterizes eta (and (q;q)_oo) near the cyclotomic points: Empirically, Here it is again, tested and slightly simplified,
'limit(%e^-(%pi^2/(6*d^2*log(q)))*sqrt(-log(q))*eta(%e^(2*%i*%pi*n/d)*q),q,1) = sqrt(2*%pi/d)*%e^(%i*%pi*g(n/d)/12) 2 i n pi -------- i g(n/d) pi d ----------- sqrt(- log(q)) eta(e q) 12 2 pi limit ------------------------------- = e sqrt(----), q -> 1 2 d pi ----------- 2 6 d log(q) e with a greatly simplified phase fudger g, now with one (rational) argument, more clearly showing the continued fraction recursion, r -> 1/mod(r,1): g(r) := if integerp(r) then r else 3+floor(r)-1/(denom(r)^2*nummod(r,1))-g(1/nummod(r,1)) 1 1 g(r) := if integerp(r) then r else - g(---------) - ------------------- mod(r, 1) 2 denom (r) mod(r, 1) + floor(r) + 3 . --rwg PS, I was startled to see the Merriam Webster dictionary program offer adjective and adverb definitions of "nation". Unable to imagine such usages, I gave up and peeked, expecting to kick myself. I didn't.
I think this fully characterizes eta (and (q;q)_oo) near the cyclotomic points: Empirically, Here it is again, tested and slightly simplified,
Actually, what I tested were the Mma definitions: HoldForm[Limit][(DedekindEta[n/d - I*Log[q]]*Sqrt[-Log[q]])/ E^(Pi/(12*d^2*Log[q])), q -> 1] == E^((1/12)*I*Pi*g[n/d])/Sqrt[d] n DedekindEta[- - I Log[q]] Sqrt[-Log[q]] 1/12 I Pi g[n/d] d E Limit[---------------------------------------, q -> 1] == ----------------- 2 Sqrt[d] Pi/(12 d Log[q]) E g[r_?ExactNumberQ] := If[IntegerQ[r], r, 3 + Floor[r] - 1/(Denominator[r]^2*Mod[r, 1]) - g[1/Mod[r, 1]]] g[r_?ExactNumberQ] := 1 1 If[IntegerQ[r], r, 3 + Floor[r] - ------------------------- - g[---------]] 2 Mod[r, 1] Denominator[r] Mod[r, 1] where DedekindEta takes argument tau instead of q=e^(2 i pi tau), and has period 24, just like the rhs of the limit, which is good for all relatively prime n and d, d>0. But like Perkin-Elmer, I failed to all-up test after converting back to eta(q) := q^(1/24) (q;q)_oo, which has branch problems. The formula
'limit(%e^-(%pi^2/(6*d^2*log(q)))*sqrt(-log(q))*eta(%e^(2*%i*%pi*n/d)*q),q,1) = sqrt(2*%pi/d)*%e^(%i*%pi*g(n/d)/12)
2 i n pi -------- i g(n/d) pi d ----------- sqrt(- log(q)) eta(e q) 12 2 pi limit ------------------------------- = e sqrt(----), q -> 1 2 d pi ----------- 2 6 d log(q) e
is good only for 2 |n| < d, n/=0, which is barely enough, with 2 %pi - -------- 6 log(q) eta(q) %e sqrt(- log(q)) -> sqrt(2 %pi), to describe eta(e^(i t pi)) for rational t. A simple example of the eta(q) branch problem: Suppose you try to roll your own in Mma with the handy identity eta[q_] := EllipticTheta[1, Pi/3, q^(1/6)]/Sqrt[3] Now Plot[Arg[{DedekindEta[tau],eta[E^(2*I*Pi*tau)]}],{tau,I/2-3,I/2+3}] (It refuses because Plot needs educating, so mut-mut) and you'll get a wavy ramp and a sawtooth instead of superposed ramps. This is because eta[E^(2*I*Pi*tau)] gives EllipticTheta[1, Pi/3, (E^(2*I*Pi*tau))^(1/6)]/Sqrt[3] instead of EllipticTheta[1, Pi/3, E^((I*Pi*tau)/3)]/Sqrt[3], whose period is 6 times greater. One of those cases where PowerExpand fixes things instead of breaking them: eta[q_] := EllipticTheta[1, Pi/3, PowerExpand[q^(1/6)]]/Sqrt[3] --rwg
Here's the limit in terms of a (not obviously) integer "triangle" h(n,d): 'limit(%e^(%pi/(12*d^2*y))*sqrt(y)*e\ta(%i*y+n/d),y,0) = %e^(%i*%pi*h(n,d)/(12*d))/sqrt(d) pi i h(n, d) pi ------- ------------ 2 12 d 12 d y n e limit e sqrt(y) eTa(i y + -) = ------------- y -> 0 d sqrt(d) where maybe we could spell eta with a tau to indicate it's not the q flavor. h(n,d) is just d*g(n/d), but the definition of g(r) I gave had the peculiarity of recursing exclusively with r>1, even though we "always" call it with r<1, thus wasting a recursion and memoizing space. This definition h(n,d) := 3*d-(if n = 1 then 1+d^2 else (1+d*h(nummod(d,n),n))/n+d*floor(d/n)) 2 d h(mod(d, n), n) + 1 d h(n, d) := 3 d - (if n = 1 then d + 1 else --------------------- + d floor(-)) n n while even more peculiar, maintains n<d throughout the recursion, but I put "triangular" in quotes because it also works for n>d: \n 1 2 3 4 5 6 7 8 9 10 11 12 d ------------------------------------------------------------------------- 1| 1 2 3 4 5 6 7 8 9 10 11 12 | 2| 1 3 5 7 9 11 | 3| - 1 4 2 7 5 10 8 13 | 4| - 5 9 - 1 13 3 17 | 5| - 11 2 3 16 - 6 7 8 21 - 1 12 | 6| - 19 25 - 13 31 | 7| - 29 - 4 9 - 2 11 36 - 22 3 16 5 18 | 8| - 41 - 3 11 49 - 33 5 | 9| - 55 - 14 20 - 11 23 64 - 46 - 5 | 10| - 71 3 7 81 - 61 | 11| - 89 - 28 - 15 - 14 35 - 24 25 26 39 100 - 78 | 12| - 109 7 5 121 | 13| - 131 - 46 - 9 16 5 54 - 41 8 - 3 22 59 144 | 14| - 155 - 33 - 31 45 47 | 15| - 181 - 68 - 34 77 - 62 49 | 16| - 209 - 27 35 25 - 9 - 19 | 17| - 239 - 94 - 57 4 - 7 - 54 - 5 104 - 87 22 71 24 | 18| - 271 - 11 23 - 5 | 19| - 305 - 124 - 51 - 62 - 61 60 - 11 26 135 - 116 - 7 30 | 20| - 341 - 87 - 83 51 - 31 h(n,d) requires only integer arithmetic because the .../n always comes out even. Presumably. Does anyone see why mod(d h(mod(d, n), n), n) = - 1 ? Neil, how do you handle 2D arrays with undefined entries? If this is already in EIS, I've wasted a lot of work. --rwg Magic moment: The mass of the lady being sawn times the separation of her respective halves.
participants (6)
-
James Buddenhagen -
Joerg Arndt -
Peter Samson -
Russell Noftsker -
rwg＠sdf.lonestar.org -
Thomas Knight