Page 28 - Textos de Matemática Vol. 34

18
LU´IS DANIEL DE ABREU
3. Special cases: The q−trigonometric functions (3)
As special cases of the functions Jν (z; q) we have, following [5],
z ∞ 1−q n=0
∞ cos(z;q)= (−1)
n=0 (2)
q n ( n + 1 ) 1 − q n22n 12
sin(z;q)=
(−1) z = (q2;q3;q2)n
G1(zq4;q ) z 2
and
q n ( n − 1 )
n 2 2n −1 2
z =G−1(zq 4;q) (q; q2; q2)n 2
and as special cases of Jν (z; q) [6]:
z∞qn2+1 z1
and
It is of interest to see what the formulas in the preceding section mean in terms of this special cases. A big simplification occurs: the double sums in Theorem 2.1 and Theorem 2.2 become single sums.
and
∞ 41111∞q2
S (z) = (−1)n z2n = F1 (zq2 ;q2) q221222
(−qz ;q )∞ n=0 (q2 ;q;q)n (−qz ;q )∞ 1∞ qn2−1 1 1
C(z)= (−1)n z2n= F 1(zq−2;q2). q22122−2
(−qz ;q )∞ n=0 (q2 ;q;q)n (−qz ;q )∞
Corollary 3.1. For every z ∈ C,
sin(q−1 z;q) = 1 q 4 −qk−1z2;q2 S (qk−1 z)
∞
(−q;q2 )∞ k=0 (q2 ;q2 )k
k(k−1) 4111∞q2
k(k+1)
cos(q1 z;q) = 1 q 4 −qk+1z2;q2 C (qk+1 z).
(−q2 ;q2 )∞ k=0 (q2 ;q2 )k Proof. Setting ν = 1 and ν = −1 in (2.7) gives
22
n+3n+1 n+11kq4nk+k
n+1n+1 n+11kq4nk+k (−q 2 ; −q ; q)∞ = (−q 2 ; q 2 )∞ = (−1) 1 1 q 2 .
∞
k(k−1) (−1) 1 1 q
(3.1)
and (3.2)
(−q 2;−q
;q)∞ =(−q
;q2)∞ =
(q2 ;q2 )k k(k−1)
k=0
∞
k=0
(q2 ;q2 )k