[math-fun] Hirschhorn's modular identity
rwg> http://reference.wolfram.com/mathematica/ref/QPochhammer.html (Neat examples) gives "Hirschhorn's modular identity ": [I just pasted the above from my Sent Mail. It ends with ((q;q)_oo)^5=(q^5,q^5)_oo, pasted from the Mma doc, but apparently vaporized on transmission.] rwg> QPochhammer[q, q]^3 = QPochhammer[q^3, q^3] *except* at q^binomial(n,2), n>1. It seems that QPochhammer[q, q]^3 - QPochhammer[q^3, q^3] == -3*Sum[(-1)^n*q^Binomial[n, 2]*Floor[2*n/3], {n, 0, Infinity}] which ought to be some kind of theta derivative, but I can't place it. --rwg
What one needs here is the Jacobi identity for QPochhammer[q, q]^3 which is Sum[(-1)^n*q^Binomial[n, 2]*(2*n+1), {n, 0, Infinity} together with Euler's Pentagonal number theorem and the fact that 3*(3*n+1)*n/2 = binomial(3*n+1,2) George On Mon, 20 Feb 2012, Bill Gosper wrote:
rwg> http://reference.wolfram.com/mathematica/ref/QPochhammer.html (Neat examples) gives "Hirschhorn's modular identity [18.gif] ":
[I just pasted the above from my Sent Mail. It ends with ((q;q)_oo)^5=(q^5,q^5)_oo, pasted from the Mma doc, but apparently vaporized on transmission.]
rwg> QPochhammer[q, q]^3 = QPochhammer[q^3, q^3] *except* at q^binomial(n,2), n>1.
It seems that QPochhammer[q, q]^3 - QPochhammer[q^3, q^3] == -3*Sum[(-1)^n*q^Binomial[n, 2]*Floor[2*n/3], {n, 0, Infinity}]
which ought to be some kind of theta derivative, but I can't place it. --rwg
On Mon, Feb 20, 2012 at 2:32 AM, Bill Gosper <billgosper@gmail.com> wrote:
rwg> http://reference.wolfram.com/mathematica/ref/QPochhammer.html (Neat examples) gives "Hirschhorn's modular identity ":
[I just pasted the above from my Sent Mail. It ends with ((q;q)_oo)^5=(q^5,q^5)_oo,
mod 5
pasted from the Mma doc, but apparently vaporized on transmission.]
And yes, it's still there in the Sent Mail record of this reply, too.
rwg> QPochhammer[q, q]^3 = QPochhammer[q^3, q^3] *except* at q^binomial(n,2), n>1.
It seems that QPochhammer[q, q]^3 - QPochhammer[q^3, q^3] == -3*Sum[(-1)^n*q^Binomial[n, 2]*Floor[2*n/3], {n, 0, Infinity}]
A welcome note from George Andrews proves this by expanding
QPochhammer[q, q]^3 ==Sum[(-1)^n*q^Binomial[n+1, 2]*(2*n+1), {n, 0, Infinity} ] (Jacobi. And in my old q-trig paper<http://books.google.com/books?id=eVtRHYcNMmYC&pg=PA79&lpg=PA79&dq=Experiments+and+discoveries+in+q-trigonometry&source=bl&ots=9iAHSTt6vo&sig=E-X7JkNDxu2fCMZiW1P3LXDVtDw&hl=en&sa=X&ei=oq5CT7SxFomMiAL75dGMAQ&sqi=2&ved=0CFkQ6AEwBg#v=onepage&q=Experiments%20and%20discoveries%20in%20q-trigonometry&f=false>!, p88.) Then George expands QPochhammer[q^3, q^3] by Euler's pentagonal number thm, getting a result equivalent to 6*Sum[(n + 1)*(-1)^n*q^((6*n + 5)^2/8), {n, 0, Infinity}] + 6*Sum[n*(-1)^n*q^((6*n + 1)^2/8), {n, 0, Infinity}] - 3*Sum[(2*n + 1)*(-1)^n*q^((9*(2*n + 1)^2)/8), {n, 0, Infinity}] == \[Eta][q]^3 - \[Eta][q^3] where, to test this, you need % /. \[Eta][q_] -> DedekindEta[Log[q]/2/I/Pi] The lhs/q^(1/8) is now equivalent to the trisection of -3*Sum[(-1)^n*q^Binomial[n, 2]*Floor[2*n/3], {n, 0, Infinity}] rwg>which ought to be some kind of theta derivative, but I can't place it.
--rwg
Despite all this, I still can't. --rwg
participants (2)
-
Bill Gosper -
George E Andrews