EXAMPLE (5) =========== The maple calculations below are used in the proof of Ramanujan's identity for p(5n+4). ============================================================================== > read "FUNCS.txt": > x:=expand( (J0-q+q^2*J2)^3 ); 3 2 2 2 2 3 4 2 3 x := J0 - 3 J0 q - 3 J0 q J2 + 3 J0 q + 6 J0 q J2 + 3 J0 q J2 - q 4 5 2 6 3 - 3 q J2 - 3 q J2 - q J2 > dissectq(x,q,5); 3 5 2 2 5 3 2 3 J0 - 3 q J2 + q (-3 J0 + q J2 ) + 3 J0 (J0 J2 + 1) q - (6 J0 J2 + 1) q 4 + 3 J2 (J0 J2 + 1) q > zeta:=RootOf(1+z+z^2+z^3+z^4=0): > simplify(zeta^5); 1 > v:=q->J0 - q - q^2/J0: > y:=mul(v(zeta^j*q),j=1..4); / 2 2\ / 4 2\ / 6 2\ | %1 q | | 2 %1 q | | 3 %1 q | y := |J0 - %1 q - ------| |J0 - %1 q - ------| |J0 - %1 q - ------| \ J0 / \ J0 / \ J0 / / 8 2\ | 4 %1 q | |J0 - %1 q - ------| \ J0 / 2 3 4 %1 := RootOf(1 + _Z + _Z + _Z + _Z ) > expand(y): > simplify(%); 8 7 3 5 2 6 4 4 5 3 6 2 7 (J0 + q J0 + 3 q J0 + 2 q J0 + 5 q J0 - 3 q J0 + 2 q J0 - q J0 8 / 4 + q ) / J0 / > a:=expand(%); 5 6 7 8 4 3 3 2 2 4 3 q 2 q q q a := J0 + q J0 + 3 J0 q + 2 q J0 + 5 q - ---- + ---- - --- + --- J0 2 3 4 J0 J0 J0 > dissectq(a,q,5); 5 5 5 5 2 5 5 3 5 5 -J0 + 3 q q (J0 + 2 q ) q (-2 J0 + q ) q (3 J0 + q ) 4 - ----------- + -------------- - ---------------- + --------------- + 5 q J0 2 3 4 J0 J0 J0 > ==============================================================================