FUNCTION : findmaxind - find a maximal independent subset of q-series
CALLING SEQUENCE : findmaxind()
findmaxind(XFL,T)
PARAMETERS : XFL - list of q-series (q-polynomials)
T - nonnegative integer (used in findhom)
SYNOPSIS :
Find a maximal independent subset of q-series
[P,NXFL] is returned where P is maximal independent subset of XFL, and
and NXFL is a list of indices.
EXAMPLES :
> with(qseries):
> findmaxind();
-------------------------------------------------------------
findmaxind(XFL,T)
XFL is list of q-polynomials.
T is an integer (ussually 0).
Returns [P,NXFL],
where P is maximal independent subset of XFL, and
and NXFL is a list of indices.
-------------------------------------------------------------
> gp2qs:=L->q^((add(L[2*n]*L[2*n-1],n=1..nops(L)/2)/24))*mul(etaq(q,L[2*n-1],1000)^L[2*n],n=1..nops(L)/2):
> GP1:= [[2, -1, 10, 5], [2, 2, 10, 2], [1, 1, 2, -1, 5, 3, 10, 1],
> [1, 1, 2, 2, 5, 3, 10, -2], [1, 2, 2, -2, 5, -2, 10, 6],
> [1, 2, 2, 1, 5, -2, 10, 3], [1, 3, 2, -2, 5, 1, 10, 2],
> [1, 3, 2, 1, 5, 1, 10, -1]];
GP1 := [[2, -1, 10, 5], [2, 2, 10, 2], [1, 1, 2, -1, 5, 3, 10, 1],
[1, 1, 2, 2, 5, 3, 10, -2], [1, 2, 2, -2, 5, -2, 10, 6],
[1, 2, 2, 1, 5, -2, 10, 3], [1, 3, 2, -2, 5, 1, 10, 2],
[1, 3, 2, 1, 5, 1, 10, -1]]
> GL1:=map(gp2qs,GP1):
> EB1:=map(x->etamake(x,q,100),GL1);
5
eta(10 tau) 2 2
EB1 := [------------, eta(10 tau) eta(2 tau) ,
eta(2 tau)
3 3 2
eta(10 tau) eta(5 tau) eta(tau) eta(5 tau) eta(2 tau) eta(tau)
--------------------------------, --------------------------------,
eta(2 tau) 2
eta(10 tau)
6 2 3 2
eta(10 tau) eta(tau) eta(10 tau) eta(2 tau) eta(tau)
-----------------------, ---------------------------------,
2 2 2
eta(5 tau) eta(2 tau) eta(5 tau)
2 3 3
eta(10 tau) eta(5 tau) eta(tau) eta(5 tau) eta(2 tau) eta(tau)
---------------------------------, -------------------------------]
2 eta(10 tau)
eta(2 tau)
> nops(EB1);
8
> Y:=findmaxind(GL1,0):
> Y[2];
[1, 2, 3, 4, 6, 7, 8]
> findhom(GL1,q,1,0);
{-X[2] + X[3] + X[5]}
DISCUSSION :
EB1 is a list of 8 eta-products.
EB1[1], EB1[2], EB1[3], EB1[4], EB1[6], EB1[7], EB1[8],
form a linearly independent subset apparently.
This is confirmed by findhom.
SEE ALSO :
findhom