FUNCTION : qseries[findnonhomcombo] - tries to express a q-series as
a polynomial of degree n of a
given list of q-series.
CALLING SEQUENCE : findnonhomcombo(L,q,n,topshift,etaoption)
findnonhomcombo(L,q,n,topshift)
findnonhomcombo(L,q,n,etaoption)
findnonhomcombo(L,q,n)
PARAMETERS : L - list of q-series
q - variable
n - positive integer
topshift - integer greater than -20
etaoption - yes, no
GLOBAL VARIABLE : X
SYNOPSIS :
findnonhomcombo(f,L,q,n,topshift,etatoption) returns a set of potential
set of polynomials in elements of L for the given q-series f.
The value of topshift is usually taken to be zero. However if
it appears that spurious relations are being generated then a higher
value of topshift should be taken.
If etaoption=yes then each monomial in the polynomial is %converted%
into an eta-product.
NOTE: There is a global variable X that is reassigned each time the
function is called. This variable is used to display the polynomials.
If the list L is linearly dependent more than one linear combination
may be returned.
EXAMPLES :
> with(qseries):
> c:=q*etaq(q,3,100)^9/etaq(q,1,100)^3:
> a:=radsimp(theta3(q,100)*theta3(q^3,40)+theta2(q,100)*theta2(q^3,40)):
> c:=3*q^(1/3)*etaq(q,3,100)^3/etaq(q,1,100):
> x:=radsimp(c^3/a^3):
> z:=a:
> N:=q-> 1 - 504*sum(n^5*q^n/(1-q^n),n=1..100):
> findnonhomcombo(N(q)/z^6,[x],q,2);
# of terms , 24
matrix is , 4, x, 24
-----possible linear combinations of degree------, 2
2
{- 8 X[1] - 20 X[1] + 1}
> xi:=q^2*etaq(q,49,100)/etaq(q,1,100):
> T:=q*(etaq(q,7,100)/etaq(q,1,100))^4:
> findnonhomcombo(T^2,[T,xi],q,7,-15,no);
# of terms , 42
matrix is , 37, x, 42
-----possible linear combinations of degree------, 7
7 6 5 2 3
{343 X[2] + 343 X[2] + 147 X[2] + 35 X[1] X[2] + 49 X[1] X[2] + X[2]
4 3 2
+ 49 X[2] + 21 X[2] + 7 X[1] X[2] + 7 X[2] }
DISCUSSION :
If we define
3 3
a = theta (q) theta (q ) + theta (q) theta (q )
3 3 2 2
3
eta(3 tau)
c := 3 -----------
eta(tau)
3
c
x := ----
3
a
/infinity \
| ----- 5 n|
| \ n q |
and N := 1 - 504 | ) ------|
| / n|
| ----- 1 - q |
\ n = 1 /
then it seems that
6 2
N = z (1 - 20 x - 8 x ).
If we define
eta(49 tau)
xi := -----------
eta(tau)
4
eta(7 tau)
T := -----------
4
eta(tau)
then it seems that
2 2 3 7 6 5 4 3
T = (35 xi + 49 xi + 7 xi) T + 343 xi + 343 xi + 147 xi + 49 xi + 21 xi
2
+ 7 xi + xi
SEE ALSO : findhom, findnonhom, findhomcombo