[math-fun] theta[s](pi/6,q), or eta as theta again
A couple of years ago I restated an almost forgotten identity equivalent to %pi %pi 6 theta (---, q) = theta (---, q) = sqrt(3) eta(q ) 2 6 1 3 3 2 6 theta (0, q ) theta (0, q ) 2 4 1/3 = sqrt(3) (---------------------------) , 2 in this form as special values of Theta(pi/n) in terms of eta or equivalently the "theta constants" theta[s](0,q^k). What is the analogous value with the lhs 1 and 2 switched? 2 2 12 %pi %pi eta (q ) eta(q ) theta (---, q) = theta (---, q) = ----------------- 2 3 1 6 4 6 eta(q ) eta(q ) 3 9 theta (0, q ) theta (0, q) - 3 theta (0, q ) 2 2 1/3 2 2 = theta (0, q ) (---------------) = ------------------------------. 4 6 2 2 theta (0, q ) 4 This is a rich area of nonobvious(?) identities, e.g., 2 6 6 theta (0, q) theta (0, q ) theta (0, q ) 2 2 3 2 2 2 3 = theta (0, q ) theta (0, q ) theta (0, q ), 2 3 2 or equivalently 2 6 theta (0, q) theta (0, q) theta (0, q ) 3 4 4 2 2 3 3 = theta (0, q ) theta (0, q ) theta (0, q ) 4 3 4 --rwg
A couple of years ago I restated an almost forgotten identity equivalent to
%pi %pi 6 theta (---, q) = theta (---, q) = sqrt(3) eta(q ) 2 6 1 3
3 2 6 theta (0, q ) theta (0, q ) 2 4 1/3 = sqrt(3) (---------------------------) , 2
in this form as special values of Theta(pi/n) in terms of eta or equivalently the "theta constants" theta[s](0,q^k). What is the analogous value with the lhs 1 and 2 switched? 2 2 12 %pi %pi eta (q ) eta(q ) theta (---, q) = theta (---, q) = ----------------- 2 3 1 6 4 6 eta(q ) eta(q )
3 9 theta (0, q ) theta (0, q) - 3 theta (0, q ) 2 2 1/3 2 2 = theta (0, q ) (---------------) = ------------------------------. 4 6 2 2 theta (0, q ) 4
This is a rich area of nonobvious(?) identities, e.g.,
2 6 6 theta (0, q) theta (0, q ) theta (0, q ) 2 2 3 2 2 2 3 = theta (0, q ) theta (0, q ) theta (0, q ), 2 3 2 or equivalently 2 6 theta (0, q) theta (0, q) theta (0, q ) 3 4 4
2 2 3 3 = theta (0, q ) theta (0, q ) theta (0, q ) 4 3 4
The other two thetas at pi/3|6 are 3 3 2 3 theta (0, q) theta (0, q) theta (0, q ) pi 2 4 3 1/6 theta (--, q) = (---------------------------------------) 3 3 3 3 4 theta (0, q ) theta (0, q ) 2 4 9 4 2 6 3 theta (0, q ) - theta (0, q) pi eta(q) eta(q ) eta (q ) 3 3 = theta (--, - q) = ------------------------ = ------------------------------ 4 3 2 3 12 2 eta(q ) eta(q ) eta(q ) pi = theta (--, q), 4 6 and 3 theta (0, q ) 2 2 3 pi 4 1/3 eta (q ) eta(q ) theta (--, q) = theta (0, sqrt(q)) (-----------------) = ---------------- = 4 3 2 3/2 6 4 theta (0, q ) eta(q) eta(q ) 2 9 3 theta (0, q ) - theta (0, q) 4 4 pi ------------------------------ = theta (--, q). 2 3 6 I started with 2 2 pi 2 2 pi sqrt(theta (0, q) theta (--, q) + theta (0, q) theta (--, q)) pi 3 1 3 4 1 6 theta (--, q) = --------------------------------------------------------------- 3 6 theta (0, q) 2 from a determinant identity and got absolutely stuck trying for monomials. One thing that should've worked was that power series analog of LatticeReduce that I keep hyping. It's failure in both Macsyma and Mma nearly convinced me there weren't any monomials. I still don't know where I screwed up, although a streamlined version that only proposes constant coefficients, vs polynomials in q, would've settled things promptly. [Later] I expected expansion to q^24 or so would suffice, but to be "sure", chose q^99, w/o success. But q^199 just now succeeded in Mma, taking most of an hour (prior experiences would predict a few seconds), scaling by a 5.5 page rational number! (So a lot of time was wasted reducing huge fractions. You can turn that off in Macsyma with gcd:false.) Aha. I'm using the series for the log eta and trying to find the powers. The q^(k/24) multiplier of the eta contributes a k*log(q) term which both systems (mis)treat as constant. Knowing the logs will wash out anyway, I replaced them by 0, destroying the washout information. Replacing them instead by 1/q gets the time down to 58 secs(!) and needs q^(<=79). (But q^(>69).) --rwg
We have the following curious pair of telescoping products: PROD(THETA[2](%PI/3,Q^(3*2^J))/THETA[1](%PI/3,Q^2^J),J,0,N-1) = THETA[1](%PI/3,Q)*THETA[1](%PI/3,Q^(3*2^N))/(THETA[1](%PI/3,Q^3)*THETA[1](%PI/3,Q^2^N)*3^(N/2)) j n n - 1 pi 3 2 pi pi 3 2 /===\ theta (--, q ) theta (--, q) theta (--, q ) | | 2 3 1 3 1 3 | | ----------------- = ----------------------------------- | | j n j = 0 pi 2 n/2 pi 3 pi 2 theta (--, q ) 3 theta (--, q ) theta (--, q ) 1 3 1 3 1 3 PROD(THETA[2](%PI/3,Q^3^(J+1))/THETA[1](%PI/3,Q^3^J),J,0,N-1) = THETA[1](%PI/3,Q)*THETA[1](%PI/3,Q^(2*3^N))/(THETA[1](%PI/3,Q^2)*THETA[1](%PI/3,Q^3^N)*3^(N/2)) j n n - 1 pi 3 3 pi pi 2 3 /===\ theta (--, q ) theta (--, q) theta (--, q ) | | 2 3 1 3 1 3 | | ----------------- = ----------------------------------- | | j n j = 0 pi 3 n/2 pi 2 pi 3 theta (--, q ) 3 theta (--, q ) theta (--, q ) 1 3 1 3 1 3 where the latter isolated would be PROD(THETA[2](%PI/3,Q^3^J)/THETA[1](%PI/3,Q^3^J),J,1,N) = THETA[1](%PI/3,Q)^2*THETA[1](%PI/3,Q^(2*3^N))/(THETA[1](%PI/3,Q^2)*THETA[1](%PI/3,Q^3^N)^2*3^(N/2)) j n n pi 3 2 pi pi 2 3 /===\ theta (--, q ) theta (--, q) theta (--, q ) | | 2 3 1 3 1 3 | | --------------- = ----------------------------------- | | j n j = 1 pi 3 n/2 pi 2 2 pi 3 theta (--, q ) 3 theta (--, q ) theta (--, q ) 1 3 1 3 1 3 --rwg
Massaging the infinite products gives %i %pi ------ 24 3 2 %e eta (q ) eta(- q) = -----------------. 4 eta(q) eta(q ) Enormous screwing around gave 5 %i %pi - -------- 2 8 2 2 48 4 4 eta (q ) %i eta (q ) 1/4 eta(%i q) = %e eta(q ) (---------- + -----------) , 2 2 2 8 eta (q ) eta (q ) valid in the wedge -pi/8 < arg(q) <= pi/8. This is actually amazingly simple, given that the polynomial relating eta(q), eta(q^4), and eta(q^8) is degree 64, and is apparently too hard for Macsyma and Mma to factor over cis(pi/24), let alone 48. Using our earlier eta valuations, ETA(-%E^-(%PI/SQRT(3))) = 3^(3/8)*%I^(1/12)*GAMMA(1/3)^(3/2)/(2*%PI) pi 3/8 3/2 1 1/12 - ------- 3 gamma (-) i sqrt(3) 3 eta(- %e ) = ----------------------, 2 pi (which is actually simpler than ETA(%E^-(%PI/(2*SQRT(3)))) = 3^(3/8)*(5*SQRT(3)-4*SQRT(2)-3)^(1/8)*GAMMA(1/3)^(3/2)/(2*2^(13/48)*%PI) pi 3/8 1/8 3/2 1 - --------- 3 (5 sqrt(3) - 4 sqrt(2) - 3) gamma (-) 2 sqrt(3) 3 eta(%e ) = ----------------------------------------------- 13/48 2 2 pi ) and ETA(-%I*%E^-(%PI/(2*SQRT(3)))) = 3^(3/8)*%E^(%I*%PI/12)*GAMMA(1/3)^(3/2)/(2^(5/6)*%PI) i pi ---- pi 3/8 3/2 1 12 - --------- 3 gamma (-) %e 2 sqrt(3) 3 eta(- i %e ) = ----------------------- 5/6 2 pi This came up because q can get multiplied by a root of unity when expressing special values of thetas as theta constants: THETA[1](%PI/10,Q) = (%I*SQRT(SQRT(5)/8+5/8)*(2*SQRT(5)-2)-SQRT(5)-5)*THETA[2](0,%E^(4*%I*%PI/5)*Q)/8+(SQRT(SQRT(5)/8+5/8)*(SQRT(5)+1)+%I*SQRT(5))*THETA[2](0,%E^(2*%I*%PI/5)*Q)/4-(-SQRT(5)+4*%I*SQRT(SQRT(5)/8+5/8)-3)*THETA[2](0,Q)/8 %pi sqrt(5) 5 theta (---, q) = (%i sqrt(------- + -) (2 sqrt(5) - 2) - sqrt(5) - 5) 1 10 8 8 4 %i %pi -------- 5 sqrt(5) 5 theta (0, %e q)/8 + (sqrt(------- + -) (sqrt(5) + 1) 2 8 8 2 %i %pi -------- 5 + %i sqrt(5)) theta (0, %e q)/4 2 sqrt(5) 5 (- sqrt(5) + 4 %i sqrt(------- + -) - 3) theta (0, q) 8 8 2 - -----------------------------------------------------, 8 which I got only a few months ago but have totally forgotten how. With q=e^(-pi/10), EllipticTheta[1, Pi/10, (1/(E^(Pi/10)))]==((1/4)! * ((Sqrt[5] + 1) * ((1/(Sqrt[ - 5^(3/4) + 5^(1/4) + 2])) - ((Sqrt[5 * (Sqrt[5] + 2)])/2)) - 5^(3/4) * Sqrt[(2/(Sqrt[5] + 1))] + 5^(1/4) * (Sqrt[5] - 1))/(2^(5/8) * Pi^(3/4))) THETA[1](%PI/10,%E^-(%PI/10)) = (-SQRT(2)*5^(3/4)/SQRT(SQRT(5)+1)+(SQRT(5)+1)*(1/SQRT(2-5^(1/4)*(SQRT(5)-1))-SQRT(5)*SQRT(SQRT(5)+2)/2)+5^(1/4)*(SQRT(5)-1))*(1/4)!/(2^(5/8)*%PI^(3/4)) pi - pi/10 2 3/4 theta (--, e ) = (- sqrt(-----------) 5 1 10 sqrt(5) + 1 1 sqrt(5 (sqrt(5) + 2)) + (----------------------- - ---------------------) (sqrt(5) + 1) 3/4 1/4 2 sqrt(- 5 + 5 + 2) 1/4 1 5/8 3/4 + 5 (sqrt(5) - 1)) (-)!/(2 pi ) 4 = 0.02071028405516 This was a good day's work, including six new special values of eta. It took me a couple of hours to simplify, but that's faster than Mma's FullSimplify, which is still working on it from yesterday. If you guess the transcendental factors, you could in principle determine the algebraic factor numerically, but probably the only way to solve the (degree 8*128) polynomial in radicals would be to guess most of the extensions over which it factorizes. --rwg TORT-FEASING FORETASTING GRAFTONITES ICONODULIST DISLOCUTION NOSE TO NOSE ON ONE'S TOES
I want to ascribe the expression below to RWG. How should I cite a math-fun mail? Using "priv.comm." now, but that is not really true. I neither like "unpublished note". Any idea? * rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Feb 26. 2009 10:36]:
[...]
Enormous screwing around gave
5 %i %pi - -------- 2 8 2 2 48 4 4 eta (q ) %i eta (q ) 1/4 eta(%i q) = %e eta(q ) (---------- + -----------) , 2 2 2 8 eta (q ) eta (q )
[...]
I restate in \Eta(x) = \prod_{n=1}{\infty}{1-x^n} and give two more expressions: 1) RWG gives: # 4 [ 1 ]1/4 # \Eta(i x)= \Eta(x ) | - - 4 i x u | # [ u ] # # 8 2 # \Eta(x ) # \where u = --------- # 2 2 # \Eta(x ) # 2) Set P to # 4 8 # \Eta(x ) # \Eta(+i x) \Eta(-i x) = --------------------- # 2 3 8 3 # \Eta(x ) \Eta(x ) # and set R to # 2 2 8 7 # \Eta(+i x) \Eta(x ) \Eta(x ) # ---------- = ---------------------- # \Eta(-i x) 4 7 16 2 # \Eta(x ) \Eta(x ) # then # +---+ [ +---+ +---+ ] # \Eta(i x)=\|P/2 [ \|1+R - i \|1-R ] # 3) By solving # 16 4 8 8 4 16 2 24 # \Eta(x) \Eta(x ) + 16 x \Eta(x) \Eta(x ) - \Eta(x ) =0 # 8 for \Eta(x) we obtain # # [ +--------+ ] # | | 2 | # | -B +\|B -4 A C | # | - |1/8 # \Eta(i x)=| --------------- | \where # [ 2 A ] # # 4 8 # A = \Eta(x ) , # # 4 16 # B = 16 i x \Eta(x ) , # # # [ 4 3 ] # 2 24 | \Eta(x ) |24 # C = -\Eta(-x ) = -| ----------------- | # | 2 8 | # [ \Eta(x ) \Eta(x ) ] #
We write E_k for \Eta(x^k). One more expression for \Eta(i*x): # # [ 16 ] # | E | # 1 | 4 |1/4 4 2 2 # \Eta(x) = -- | --- - 4 x U | \where U = E E E # E [ U ] 8 4 2 # 4 # Expressions for \Eta(W*x) where W is the third root of unity are # # [ R 3 ]1/3 # \Eta(x) = | -- - 3 x E | \where # | E 9 | # [ 9 ] # # [ 3 12 12 ]1/3 # R = | 27 x E + E | # [ 9 3 ] # and # # [ 1 / 4 3 3 4 \ ]1/3 # \Eta(x) = | -- | E + 9 x E E - 3 x E | | # | E \ 3 3 27 9 / | # [ 9 ] # These follow from relations given in Somos' eta07.gp The following clean relations for fifth and seventh roots do not seem to be solvable (for E1): E1^5*E25 +5*x*E1^4*E25^2 +15*x^2*E1^3*E25^3 +25*x^3*E1^2*E25^4 \ +25*x^4*E1*E25^5 -1*E5^6 E1^7*E49 +147*x^8*E1^3*E49^5 +21*x^4*E1^5*E49^3 +343*x^10*E1^2*E49^6 \ +343*x^12*E1*E49^7 +35*x^3*E1^2*E7^4*E49^2 +49*x^5*E1*E7^4*E49^3 \ +49*x^6*E1^4*E49^4 +7*x*E1^3*E7^4*E49 +7*x^2*E1^6*E49^2 -1*E7^8
We write E_k for \Eta(x^k). One more expression for \Eta(i*x):
# # [ 16 ] # | E | # 1 | 4 |1/4 4 2 2 # \Eta(x) = -- | --- - 4 x U | \where U = E E E # E [ U ] 8 4 2 # 4 #
Expressions for \Eta(W*x) where W is the third root of unity are
# # [ R 3 ]1/3 # \Eta(x) = | -- - 3 x E | \where # | E 9 | # [ 9 ] # # [ 3 12 12 ]1/3 # R = | 27 x E + E | # [ 9 3 ] #
Using the conventional eta with q^(1/24) and tweaking your branches, ETA(%E^(2*%I*%PI/3)*Q) = (%E^(%I*%PI/12)*(27*ETA(Q^9)^12+ETA(Q^3)^12)^(1/3)/ETA(Q^9)+3*ETA(Q^9)^3/SQRT(%I))^(1/3) 2 i pi ------ 3 eta(%e q) = i pi ---- 12 12 9 12 3 1/3 3 9 %e (27 eta (q ) + eta (q )) 3 eta (q ) 1/3 (------------------------------------ + ----------) 9 sqrt(i) eta(q ) has a difference plot qualitatively similar to my eta(i q) formula, except that the wedge is |carg| < pi/9 and the fronds on the central sprout curl promptly out of the zero plane, leaving just a disk on a stem. Oops, except maybe the tiny 9th(?) pair, which seem(s) to be on the level! A super zoom might reveal others. I'm not even sure if the number of fronds is finite. --rwg
and
# # [ 1 / 4 3 3 4 \ ]1/3 # \Eta(x) = | -- | E + 9 x E E - 3 x E | | # | E \ 3 3 27 9 / | # [ 9 ] #
These follow from relations given in Somos' eta07.gp
The following clean relations for fifth and seventh roots do not seem to be solvable (for E1):
E1^5*E25 +5*x*E1^4*E25^2 +15*x^2*E1^3*E25^3 +25*x^3*E1^2*E25^4 \ +25*x^4*E1*E25^5 -1*E5^6
E1^7*E49 +147*x^8*E1^3*E49^5 +21*x^4*E1^5*E49^3 +343*x^10*E1^2*E49^6 \ +343*x^12*E1*E49^7 +35*x^3*E1^2*E7^4*E49^2 +49*x^5*E1*E7^4*E49^3 \ +49*x^6*E1^4*E49^4 +7*x*E1^3*E7^4*E49 +7*x^2*E1^6*E49^2 -1*E7^8
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We write E_k for \Eta(x^k). One more expression for \Eta(i*x):
# # [ 16 ] # | E | # 1 | 4 |1/4 4 2 2 # \Eta(x) = -- | --- - 4 x U | \where U = E E E # E [ U ] 8 4 2 # 4 #
Expressions for \Eta(W*x) where W is the third root of unity are
# # [ R 3 ]1/3 # \Eta(x) = | -- - 3 x E | \where # | E 9 | # [ 9 ] # # [ 3 12 12 ]1/3 # R = | 27 x E + E | # [ 9 3 ] #
and
# # [ 1 / 4 3 3 4 \ ]1/3 # \Eta(x) = | -- | E + 9 x E E - 3 x E | | # | E \ 3 3 27 9 / | # [ 9 ] #
These follow from relations given in Somos' eta07.gp
The following clean relations for fifth and seventh roots do not seem to be solvable (for E1):
E1^5*E25 +5*x*E1^4*E25^2 +15*x^2*E1^3*E25^3 +25*x^3*E1^2*E25^4 \ +25*x^4*E1*E25^5 -1*E5^6
It's clearly *possible*: P(1,5*a,5*b) mod P(1,5,25) will be at worst quartic in eta(q). The only trick is to find a remainder smaller than the entire math-fun archive. --rwg
E1^7*E49 +147*x^8*E1^3*E49^5 +21*x^4*E1^5*E49^3 +343*x^10*E1^2*E49^6 \ +343*x^12*E1*E49^7 +35*x^3*E1^2*E7^4*E49^2 +49*x^5*E1*E7^4*E49^3 \ +49*x^6*E1^4*E49^4 +7*x*E1^3*E7^4*E49 +7*x^2*E1^6*E49^2 -1*E7^8
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Enormous screwing around gave eta(%i*q)=(%e)^(-((5*%i*%pi)/48))*eta(q^4) *(((4*eta(q^8)^2)/(eta(q^2)^2))+((%i*eta(q^2)^2)/(eta(q^8)^2)))^(1/4)
5 %i %pi - -------- 2 8 2 2 48 4 4 eta (q ) %i eta (q ) 1/4 eta(%i q) = %e eta(q ) (---------- + -----------) , 2 2 2 8 eta (q ) eta (q )
valid in the wedge -pi/8 < arg(q) <= pi/8.
Ouch, no, a complicated subset of that. Something has gnawed holes at the outer edge of the pizza wedge extending in almost to |q|=.8 . Near |q|=1 there even appear to be holes in the holes where equality reappears. Wow, it looks like the cross section of a cabbage head (or clover bud). (http://gosper.org/cabbage.gif) At least it includes positive q. --rwg (Six or nine tries with firefox never loaded the entire image. Chrome loaded it all first try, but then clipped off the top!)
Enormous screwing around gave eta(%i*q)=(%e)^(-((5*%i*%pi)/48))*eta(q^4) *(((4*eta(q^8)^2)/(eta(q^2)^2))+((%i*eta(q^2)^2)/(eta(q^8)^2)))^(1/4)
5 %i %pi - -------- 2 8 2 2 48 4 4 eta (q ) %i eta (q ) 1/4 eta(%i q) = %e eta(q ) (---------- + -----------) , 2 2 2 8 eta (q ) eta (q )
valid in the wedge -pi/8 < arg(q) <= pi/8.
Ouch, no, a complicated subset of that. Something has gnawed holes at the outer edge of the pizza wedge extending in almost to |q|=.8 . Near |q|=1 there even appear to be holes in the holes where equality reappears. Wow, it looks like the cross section of a cabbage head (or clover bud). (http://gosper.org/cabbage.gif) At least it includes positive q. --rwg (Six or nine tries with firefox never loaded the entire image. Chrome loaded it all first try, but then clipped off the top!)
Not clover. The branches branch. http://gosper.org/cabbage.png . IWBN to know if that ganglion is > single-point connected to its stem and fronds. And the voids branch so that the level set isn't simply connected. (Wider view: http://gosper.org/cabbage2.png . Left edge is -pi/7, showing expected radial fault at -pi/8.) --rwg
participants (2)
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Joerg Arndt -
rwg＠sdf.lonestar.org