Re: Question on finding eta expansions

["LNU, Swati" <S10@email.sc.edu>]

Dear Professor,

Thank you for your reply. I have a follow-up question:

The first few terms of series expansion of (eta(2z)^3 *  eta(3z)^2 *  eta(12z)^2) / (eta(6z)^2) = q - 3q^3 - q^4 + 6q^6 + 6q^7 - 3q^9 + ...

On writing the code:

D(n) = {q * eta(q^(24 * 2) + O(q^(24 * ( n + 1))))^(3) * eta(q^(24 * 3) + O(q^(24 * (n + 1))))^(2) * eta(q^(24 * 12) + O(q^(24 * (n + 1))))^(2) * eta(q^(24 * 6) + O(q^(24 * (n + 1))))^(-2);} 

The output is as follows:

q - 3*q^49 - 2*q^73 + 6*q^121 + 6*q^145 - 3*q^193 - 12*q^217 + 6*q^265 - 6*q^337 + 4*q^361 + 6*q^433 + 12*q^505

The powers are not comparable on taking q^(1/24). Am I missing something?

Thanks,

Swati

"The pursuit of science is at its best when it is a part of a way of life" - Alladi Ramakrishnan.

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From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
Sent: Friday, September 13, 2024 3:37 PM
To: LNU, Swati <S10@email.sc.edu>
Cc: pari-users@pari.math.u-bordeaux.fr <pari-users@pari.math.u-bordeaux.fr>
Subject: Re: Question on finding eta expansions

 

On Fri, Sep 13, 2024 at 07:22:49PM +0000, LNU, Swati wrote:
> Hello,
> I am trying to find the expansion of the eta quotient f(z) = \eta(7z)
> \eta(z)^2 in fractional powers of q^(1/24) and I tried the following code:
> D(n) = {q^(9/24) * eta(q^7 + O(q^(n+1))) * eta(q + O(q^(n+1)))^(2) ;}

PARI just cannot define power series with fractional exponents.

You have two options:
- Do the multiplication by q^(9/24) 'in you head'
D(n) = { eta(q^7 + O(q^(n+1))) * eta(q + O(q^(n+1)))^(2) ;}
? D(10)
%6 = 1-2*q-q^2+2*q^3+q^4+2*q^5-2*q^6-q^7-q^9-q^10+O(q^11)

Set Q = q^(1/24) (in your head) and then do
D(n) = {Q^9 * eta(Q^(24*7) + O(Q^(24*(n+1)))) * eta(Q^24 + O(Q^(24*(n+1))))^(2) ;}

? D(10)
%4 = Q^9-2*Q^33-Q^57+2*Q^81+Q^105+2*Q^129-2*Q^153-Q^177-Q^225-Q^249+O(Q^273)

For exampke the second exponent is 9/24 + 1 = 33/24

Cheers,
Bill.