QEtaQSeriesTools(C, xiord, CX, xi)ΒΆ
qetaqseriestool.spad line 118 [edit on github]
- xiord: PositiveInteger 
- CX: Algebra C 
- xi: CX 
QEtaQSeriesTools provides functions for the transformation of $q$-series.
- laurent: (QEtaPuiseuxSeries CX, Fraction Integer) -> QEtaLaurentSeries CX
- Necessary input condition for laurent( - x,- w) is that- w>0. If- p(- q)- =s(- z) is a Puiseux series- pexpressed as a Laurent series- sin the variable z=q^r, then laurent(- x,- w) returns the laurent series- lsuch that- l(- x)- =s(- z)- =p(- q) where x=q^(1/w) in case- r*wis an integer. It might happen that- r*wis not an integer. That is even to be expected since non-modular eta-quotients involve a factor in terms of- q^(1/24). If r=s/t, then we take only every- t-th term and check that the intermediate terms come indeed with a zero coefficient.
- laurent: (QEtaPuiseuxSeries CX, PositiveInteger) -> QEtaLaurentSeries CX
- laurent(p, w)returns laurent(- p,- w/1)
- substitute: (QEtaLaurentSeries CX, Fraction Integer, Fraction Integer) -> QEtaPuiseuxSeries CX
- If - s(- q) is a series in- q, then substitute(- s,- u,- v) returns a series- t(- q) such that- t(- q)- =s(q^u*xi^v). Note that we assume that the series- shas only- qpowers that are divisible by denom(- v). Otherwise the function will abort with an error message if during the expansion of- tthe unity root- xiwould have to be raised to a fractional power.
- substitute: (QEtaLaurentSeries CX, Fraction Integer, NonNegativeInteger) -> QEtaPuiseuxSeries CX
- If - s(- q) is a series in- q, then substitute(- s,- u,- v) returns a series- t(- q) such that- t(- q)- =s(q^u*xi^v).