[math-fun] On to 67 and 163 !
Gack! Apologies Fred, I meant The INsignificance of your irreverence cannot be underestimated! Bit of good fortune: The Jacobi symbol collapses those gruesome Gamma products in the eta identities. See Out[24] in http://gosper.org/nuetas.html . --rwg
eta(%e^-(2*%pi*sqrt(n))) = (prod(gamma(k/n)^((jacobi(k,n)+1)/4),k,1,n-1))/(sqrt(2)*(a*2^n*%pi^(n+1))^(1/8)) n - 1 jacobi(k, n) + 1 /===\ ---------------- | | 4 k | | gamma (-) | | n - 2 %pi sqrt(n) k = 1 eta(%e ) = ------------------------------ n n + 1 1/8 sqrt(2) (a 2 %pi ) where a is the real root of 3*a^2*\j((%i*sqrt(n)+1)/2)^(1/3)*n+2*a^3 = n^3 2 1/3 %i sqrt(n) + 1 3 3 3 a J (--------------) n + 2 a = n 2 and J is Klein's absolute invariant, IFF sqrt(-n) is a UFD! Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients). There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums. --rwg
Oops! Forgot to stipulate: n > 7 . (And don't know why.)
eta(%e^-(2*%pi*sqrt(n))) = (prod(gamma(k/n)^((jacobi(k,n)+1)/4),k,1,n-1))/(sqrt(2)*(a*2^n*%pi^(n+1))^(1/8))
n - 1 jacobi(k, n) + 1 /===\ ---------------- | | 4 k | | gamma (-) | | n - 2 %pi sqrt(n) k = 1 eta(%e ) = ------------------------------ n n + 1 1/8 sqrt(2) (a 2 %pi )
where a is the real root of
3*a^2*\j((%i*sqrt(n)+1)/2)^(1/3)*n+2*a^3 = n^3
2 1/3 %i sqrt(n) + 1 3 3 3 a J (--------------) n + 2 a = n 2
and J is Klein's absolute invariant, IFF sqrt(-n) is a UFD! Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients).
There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums. --rwg
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eta(%e^-(2*%pi*sqrt(n))) = (prod(gamma(k/n)^((jacobi(k,n)+1)/4),k,1,n-1))/(sqrt(2)*(a*2^n*%pi^(n+1))^(1/8))
n - 1 jacobi(k, n) + 1 /===\ ---------------- | | 4 k | | gamma (-) | | n - 2 %pi sqrt(n) k = 1 eta(%e ) = ------------------------------ n n + 1 1/8 sqrt(2) (a 2 %pi )
where a is the real root of
3*a^2*\j((%i*sqrt(n)+1)/2)^(1/3)*n+2*a^3 = n^3
2 1/3 %i sqrt(n) + 1 3 3 3 a J (--------------) n + 2 a = n 2
and J is Klein's absolute invariant, IFF sqrt(-n) is a UFD! Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients).
Actually, this works for n down to 5. 7 gives a = 14, which satisfies a cubic!-)
There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums.
Joerg> RWG: congrats on that 163 thing! Thanks! Oddity: DedekindEta[I*Sqrt[13]]==(2^(5/104)*Gamma[7/52]*Gamma[15/52]* (Cos[Pi/26]*Cos[(5*Pi)/26]*Gamma[11/52]*Gamma[19/52])^(3/4)* Sqrt[(Gamma[23/52]*Gamma[31/52]*Gamma[35/52])/ (13*Pi*Gamma[5/52]*Gamma[17/52]*Gamma[25/52]*Gamma[29/52])]* ((Gamma[27/52]*Sin[(2*Pi)/13])/(Gamma[37/52]*Gamma[10/13]* Gamma[45/52]*Gamma[53/52]))^(1/4))/(Gamma[2/13]^(3/4)*Gamma[21/52]) I.e., no algebraic factor other than the trigs, which are all intances of Gamma reflection. Pretty clearly not in simplest form. By contrast, DedekindEta[I*Sqrt[17]] == (Gamma[9/34]^(3/8)*(Gamma[15/34]/ ((5 + Sqrt[17] + Sqrt[26 + 10*Sqrt[17]])*Gamma[2/17]* Gamma[4/17]*Gamma[8/17]))^(1/8)*(17*Gamma[3/68]*Gamma[7/68]* Gamma[11/68]*Gamma[23/68]*Gamma[13/34]*Gamma[27/68]*Gamma[31/68]* Gamma[39/68]*Gamma[63/68])^(1/4)*Sqrt[Gamma[35/34]])/ (2*2^(15/68)*Pi^(11/8)*Gamma[1/17]^(5/8)), and the algebraic factor appears not to be expressible in Gammas. I.e., I couldn't eliminate it with LatticeReduce. ---rwg
eta(%e^-(2*%pi*sqrt(n))) = (prod(gamma(k/n)^((jacobi(k,n)+1)/4),k,1,n-1))/(sqrt(2)*(a*2^n*%pi^(n+1))^(1/8))
n - 1 jacobi(k, n) + 1 /===\ ---------------- | | 4 k | | gamma (-) | | n - 2 %pi sqrt(n) k = 1 eta(%e ) = ------------------------------ n n + 1 1/8 sqrt(2) (a 2 %pi )
[...]
This can be simplified to eta(%e^-(2*%pi*sqrt(n)))^8 = prod(gamma(k/n)^(2*jacobi(k,n)),k,1,n-1)/(32*%pi^2*a*n) n - 1 /===\ | | 2 jacobi(k, n) k | | Gamma (-) | | n 8 - 2 pi sqrt(n) k = 1 eta (e ) = ---------------------------- . 2 32 pi a n where a is the real root of 3*a^2*\j((%i*sqrt(n)+1)/2)^(1/3)*n+2*a^3 = n^3 2 1/3 %i sqrt(n) + 1 3 3 3 a J (--------------) n + 2 a = n 2 and J is Klein's absolute invariant, IFF n in {7, 11, 19, 43, 67, 163} (= -Heegners > 3). But this prod(gamma(k/n)^jacobi(k,n),k,1,n-1) construct appears in the 60 year old(!) Chowla-Selberg formula for certain products of etas, and the n = 11 case of the above is (hairily!) *derived* (as opposed to guessed from numeric evidence) in http://www.warwick.ac.uk/~masfaw/BillRobin2.pdf , wherein it is suggested that Robin Chapman and Alf van der Poorten have been doing these for >= 9 years. What I haven't seen them claim is the the cubicsurd(J^(1/3)) business, but I wouldn't be surprised. At this point I think I could automate the guessing process for eta(rational+sqrt(-rational)). Also for J(such), e.g., In[158]:= KleinInvariantJ[((Sqrt[13]*I + 1)/2)]== -((125*(9*Sqrt[13] - 31)^3)/8) Out[158]= 1 125 3 KleinInvariantJ[- (1 + I Sqrt[13])] == -(---) (-31 + 9 Sqrt[13]) 2 8 --rwg
Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients).
There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums.
eta(%e^-(2*%pi*sqrt(n))) = (prod(gamma(k/n)^((jacobi(k,n)+1)/4),k,1,n-1))/(sqrt(2)*(a*2^n*%pi^(n+1))^(1/8))
n - 1 jacobi(k, n) + 1 /===\ ---------------- | | 4 k | | gamma (-) | | n - 2 %pi sqrt(n) k = 1 eta(%e ) = ------------------------------ n n + 1 1/8 sqrt(2) (a 2 %pi )
[...]
This can be simplified to
eta(%e^-(2*%pi*sqrt(n)))^8 = prod(gamma(k/n)^(2*jacobi(k,n)),k,1,n-1)/(32*%pi^2*a*n)
n - 1 /===\ | | 2 jacobi(k, n) k | | Gamma (-) | | n 8 - 2 pi sqrt(n) k = 1 eta (e ) = ---------------------------- . 2 32 pi a n
where a is the real root of
3*a^2*\j((%i*sqrt(n)+1)/2)^(1/3)*n+2*a^3 = n^3
2 1/3 %i sqrt(n) + 1 3 3 3 a J (--------------) n + 2 a = n 2
and J is Klein's absolute invariant,
IFF n in {7, 11, 19, 43, 67, 163} (= -Heegners > 3).
But this prod(gamma(k/n)^jacobi(k,n),k,1,n-1) construct appears in the 60 year old(!) Chowla-Selberg formula for certain products of etas, and the n = 11 case of the above is (hairily!) *derived* (as opposed to guessed from numeric evidence) in http://www.warwick.ac.uk/~masfaw/BillRobin2.pdf ,
I should have cautioned that there is a bogus reciprocation at eqn (17) near the end of this derivation.
wherein it is suggested that Robin Chapman and Alf van der Poorten have been doing these for >= 9 years. What I haven't seen them claim is the the cubicsurd(J^(1/3)) business, but I wouldn't be surprised.
At this point I think I could automate the guessing process for eta(rational+sqrt(-rational)). Also for J(such),
Actually, this remark is redundant because (\j(-%i*log(q)/%pi)^(1/3) = (16*eta(q^2)^24+eta(q)^24)/(12*eta(q)^8*eta(q^2)^16)) = (256*eta(q^4)^24+eta(q^2)^24)/(12*eta(q^2)^16*eta(q^4)^8) 24 2 24 24 4 24 2 1/3 i log(q) 16 eta (q ) + eta (q) 256 eta (q ) + eta (q ) J (- --------) = ----------------------- = ------------------------- pi 8 16 2 16 2 8 4 12 eta (q) eta (q ) 12 eta (q ) eta (q )
e.g., In[158]:= KleinInvariantJ[((Sqrt[13]*I + 1)/2)]== -((125*(9*Sqrt[13] - 31)^3)/8)
Out[158]= 1 125 3 KleinInvariantJ[- (1 + I Sqrt[13])] == -(---) (-31 + 9 Sqrt[13]) 2 8 --rwg
Apparently you can UFD-test n by solving the eta identity for a and seeing if it satisfies a cubic (with integer coefficients).
There are similar identities for eta(%e^-(%pi*sqrt(n))), eta(%e^-(4*%pi*sqrt(n))), and those log-derivative Lambert type sums.
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Reminder|apology: Math-fun doesn't do attachments, so this is Macsyma's 40+ year old way of displaying Sums and Products on a teletype. You'll need to switch your mail reader to a fixed pitch font, e.g., Courier. Actually, this is a way to evaluate any finite or infinite sum of a rational function a(n), provided you can express the poles and roots of 1-x*a(n) (as a fn(n)), and therefore the product(1-x*a(n)) in terms of Gammas|factorials. For any f, this is a telescoping identity: sum(f(n)*prod(1-f(k)*z,k,0,n-1),n,0,m) = (1-prod(1-f(k)*z,k,0,m))/z m /===\ | | m n - 1 1 - | | (1 - f(k) z) ==== /===\ | | \ | | k = 0 (d154) > f(n) | | (1 - f(k) z) = ---------------------- / | | z ==== k = 0 n = 0 It's 0=0 for m=-1, and differencing both sides wrt m finishes the induction. So now let m=oo and f(x) = (-)^x/(2x+1): (c155) subst([m = inf,f = lambda([x],(-1)^x/(2*x+1))],%) n - 1 inf /===\ k /===\ k n | | (- 1) z | | (- 1) z inf (- 1) | | (1 - --------) 1 - | | (1 - --------) ==== | | 2 k + 1 | | 2 k + 1 \ k = 0 k = 0 (d155) > --------------------------- = ------------------------ / 2 n + 1 z ==== n = 0 Bisect the rhs product to ditch the (-)^k, because the CLOSEDFORM command is stupid: (c156) resimplify(substpart(linchange(piece,2*k),%,2)) n - 1 /===\ k n | | (- 1) z inf (- 1) | | (1 - --------) ==== | | 2 k + 1 \ k = 0 (d156) > --------------------------- = / 2 n + 1 ==== n = 0 inf /===\ | | z z 1 - | | (1 - -------) (------- + 1) | | 4 k + 1 4 k + 3 k = 0 ------------------------------------- z CLOSEDFORM the rhs (it's actually smart enough to do both): (c157) substpart(closedform(piece),%,2) n - 1 /===\ k n | | (- 1) z inf (- 1) | | (1 - --------) ==== | | 2 k + 1 \ k = 0 (d157) > --------------------------- = / 2 n + 1 ==== n = 0 %pi z %pi 1 - sqrt(2) cos(----- + ---) 4 4 ---------------------------- z Now let z -> 0, again sidestepping cleverness on the left: (c158) subst(0,z,lhs(%)) = limit(rhs(%),z,0) inf ==== n \ (- 1) %pi (d158) > ------- = --- / 2 n + 1 4 ==== n = 0 --rwg Another use for (d154): f(k) := q^k gets a sum for q-Pochhammer which is easy to accelerate.
For a long time now I have been trying to figure out how to get your (rgw's) formulas to display correctly on my screen. So far I have failed. I use gmail, in text mode, under windows XP. My browser is the latest firefox. If I cut and paste messages to a text window set to courier font they display perfectly. Is there an easier way, under my setup, which must be a common one?. --JB The problem seems to be in how the mailer handles blanks and maybe tabs. becaise On Wed, Jul 22, 2009 at 6:11 AM, <rwg@sdf.lonestar.org> wrote:
Reminder|apology: Math-fun doesn't do attachments, so this is Macsyma's 40+ year old way of displaying Sums and Products on a teletype. You'll need to switch your mail reader to a fixed pitch font, e.g., Courier.
Actually, this is a way to evaluate any finite or infinite sum of a rational function a(n), provided you can express the poles and roots of 1-x*a(n) (as a fn(n)), and therefore the product(1-x*a(n)) in terms of Gammas|factorials.
For any f, this is a telescoping identity:
sum(f(n)*prod(1-f(k)*z,k,0,n-1),n,0,m) = (1-prod(1-f(k)*z,k,0,m))/z
m /===\ | | m n - 1 1 - | | (1 - f(k) z) ==== /===\ | | \ | | k = 0 (d154) > f(n) | | (1 - f(k) z) = ---------------------- / | | z ==== k = 0 n = 0
It's 0=0 for m=-1, and differencing both sides wrt m finishes the induction.
So now let m=oo and f(x) = (-)^x/(2x+1): (c155) subst([m = inf,f = lambda([x],(-1)^x/(2*x+1))],%)
n - 1 inf /===\ k /===\ k n | | (- 1) z | | (- 1) z inf (- 1) | | (1 - --------) 1 - | | (1 - --------) ==== | | 2 k + 1 | | 2 k + 1 \ k = 0 k = 0 (d155) > --------------------------- = ------------------------ / 2 n + 1 z ==== n = 0
Bisect the rhs product to ditch the (-)^k, because the CLOSEDFORM command is stupid:
(c156) resimplify(substpart(linchange(piece,2*k),%,2))
n - 1 /===\ k n | | (- 1) z inf (- 1) | | (1 - --------) ==== | | 2 k + 1 \ k = 0 (d156) > --------------------------- = / 2 n + 1 ==== n = 0
inf /===\ | | z z 1 - | | (1 - -------) (------- + 1) | | 4 k + 1 4 k + 3 k = 0 ------------------------------------- z
CLOSEDFORM the rhs (it's actually smart enough to do both): (c157) substpart(closedform(piece),%,2)
n - 1 /===\ k n | | (- 1) z inf (- 1) | | (1 - --------) ==== | | 2 k + 1 \ k = 0 (d157) > --------------------------- = / 2 n + 1 ==== n = 0
%pi z %pi 1 - sqrt(2) cos(----- + ---) 4 4 ---------------------------- z
Now let z -> 0, again sidestepping cleverness on the left: (c158) subst(0,z,lhs(%)) = limit(rhs(%),z,0)
inf ==== n \ (- 1) %pi (d158) > ------- = --- / 2 n + 1 4 ==== n = 0 --rwg Another use for (d154): f(k) := q^k gets a sum for q-Pochhammer which is easy to accelerate.
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hello, You need a client like thunderbird and force the fonts to be COURRIER everywhere, If you use gmail on the web then your browser need to be configure so that the web site can have their fonts. go to : Tools, content and advanced settings on fonts. Simon Plouffe
In the same little drop-down menu as the "Reply" button in the top right, you can click the down arrow and should then see a "show in fixed width font" option near the bottom. --Joshua Zucker On Wed, Jul 22, 2009 at 6:31 AM, James Buddenhagen<jbuddenh@gmail.com> wrote:
For a long time now I have been trying to figure out how to get your (rgw's) formulas to display correctly on my screen. So far I have failed. I use gmail, in text mode, under windows XP. My browser is the latest firefox. If I cut and paste messages to a text window set to courier font they display perfectly. Is there an easier way, under my setup, which must be a common one?.
--JB
Thanks, joshua, that did the trick! --JB On Wed, Jul 22, 2009 at 8:58 AM, Joshua Zucker<joshua.zucker@gmail.com> wrote:
In the same little drop-down menu as the "Reply" button in the top right, you can click the down arrow and should then see a "show in fixed width font" option near the bottom.
--Joshua Zucker
I use a greasemonkey script to show all my emails in fixed-font, both when composing and viewing. On Wed, Jul 22, 2009 at 7:42 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Thanks, joshua, that did the trick!
--JB
On Wed, Jul 22, 2009 at 8:58 AM, Joshua Zucker<joshua.zucker@gmail.com> wrote:
In the same little drop-down menu as the "Reply" button in the top right, you can click the down arrow and should then see a "show in fixed width font" option near the bottom.
--Joshua Zucker
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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
On Wed, Jul 22, 2009 at 9:58 AM, Joshua Zucker <joshua.zucker@gmail.com>wrote:
In the same little drop-down menu as the "Reply" button in the top right, you can click the down arrow and should then see a "show in fixed width font" option near the bottom.
The option is labeled "Message text garbled?" in my gmail, but it's in the same place and does the same thing. Andy -- Andy.Latto@pobox.com
The sum in Watson's identity can also be done with Jacobi's triple product, giving product = product - product. http://mathworld.wolfram.com/QuintupleProductIdentity.html gives this in terms of "Ramanujan theta functions", but with enough screwing around, it's equivalent to a strangish but sightly identity relating ordinary thetas: (EllipticTheta[1,x,q^6]*EllipticTheta[2,x,q^3])/EllipticTheta[1,Pi/3,q]== (1/3)*(-EllipticTheta[3,Pi/6+x,q^2]+EllipticTheta[4,Pi/3+x,q^2]) In:= FullSimplify[Series[List @@ %, {q, 0, 68}].{1, -1}] 69 Out= O[q ] theta[2](x,q^3)*theta[1](x,q^6)/theta[1](%pi/3,q) = (theta[4](x+%pi/3,q^2)-theta[3](x+%pi/6,q^2))/3 3 6 pi 2 pi 2 theta (x, q ) theta (x, q ) theta (x + --, q ) - theta (x + --, q ) 2 1 4 3 3 6 --------------------------- = --------------------------------------- . pi 3 theta (--, q) 1 3 Or, as a five card straight, theta[2](x,q^3)*theta[1](x,q^6)/eta(q^6) = (theta[4](x+%pi/3,q^2)-theta[3](x+%pi/6,q^2))/sqrt(3) 3 6 pi 2 pi 2 theta (x, q ) theta (x, q ) theta (x + --, q ) - theta (x + --, q ) 2 1 4 3 3 6 --------------------------- = ---------------------------------------, 6 sqrt(3) eta(q ) with one each theta_1, theta_2, theta_3, theta_4, and eta. --rwg
For n in {7, 11, 19, 43, 67, 163}, 2*%pi*sqrt(eta(%e^-(%pi*sqrt(n)))^8+16*eta(%e^-(4*%pi*sqrt(n)))^8)*n = product(gamma(k/n)^jacobi(k,n),k,1,n-1) n - 1 /===\ 8 1 8 1 | | jacobi(k, n) k 2 sqrt(eta (-----------) + 16 eta (-------------)) n pi = | | gamma (-) sqrt(n) pi 4 sqrt(n) pi | | n e e k = 1 (actually derived from the conjectural J^(1/3) result, using eta(%i*q)=eipi(-5/48)*eta(q^4)*(4*(eta(q^8)/eta(q^2))^2+%i*(eta(q^2)/eta(q^8))^2)^(1/4) 2 8 2 2 4 4 eta (q ) i eta (q ) 1/4 eta(q ) (---------- + ----------) 2 2 2 8 eta (q ) eta (q ) eta(i q) = ------------------------------------ 5 i pi ------ 48 e with its frizzy region of validity, http://gosper.org/cabbage.png , http://gosper.org/cabbage2.png.) Then we can go up and down by octaves via the easiest ("1,2,4", good for all n) eta relation eta(%e^-(2*%pi*sqrt(n)))^24 = eta(%e^-(4*%pi*sqrt(n)))^8*eta(%e^-(%pi*sqrt(n)))^16+16*eta(%e^-(4*% pi*sqrt(n)))^16*eta(%e^-(%pi*sqrt(n)))^8 24 1 8 1 16 1 eta (-------------) = eta (-------------) eta (-----------) 2 sqrt(n) pi 4 sqrt(n) pi sqrt(n) pi e e e 16 1 8 1 + 16 eta (-------------) eta (-----------) 4 sqrt(n) pi sqrt(n) pi e e (Jacobi's "aequatio identica satis abstrusa"). (If he only knew.) E.g., the double "angle": eta(%e^-(2*%pi*sqrt(n))) = ((prod(gamma(k/n)^jacobi(k,n),k,1,n-1))^2*eta(%e^-(%pi*sqrt(n)))^8*('prod(gamma(k/n)^jacobi(k,n),k,1,n-1))^2-4*%pi^2*eta(%e^-(%pi*sqrt(n)))^8*n^2)/(256*%pi^4*n^4))^(1/24) n - 1 /===\ 1 | | jacobi(k, n) k 2 8 1 eta(-------------) = expt(( | | gamma (-)) eta (-----------) 2 sqrt(n) pi | | n sqrt(n) pi e k = 1 e n - 1 /===\ | | jacobi(k, n) k 2 8 1 2 2 4 4 1 (( | | gamma (-)) - 4 eta (-----------) n pi )/(256 n pi ), --) | | n sqrt(n) pi 24 k = 1 (but n must be a Beegner.) --rwg
participants (6)
-
Andy Latto -
James Buddenhagen -
Joshua Zucker -
Mike Stay -
rwg＠sdf.lonestar.org -
Simon Plouffe