In this section, we give some applications related to eigenvalues for the total Hecke-type operators of and . We derive many new identities which are related not only to the total Hecke-type operators, but also to the Riemann zeta function, the Hurwitz zeta function, Bernoulli and Euler numbers, Euler identities and the convolution of Bernoulli and Euler numbers and polynomials.
Throughout this section, we use the following notation:
The partial zeta function is defined by
where , and () (cf. [4, 8, 10, 12, 14]).
Theorem 4.1 The polynomials are eigenfunctions for the operators with eigenvalues , that is,
where is a partial zeta function.
Proof
Therefore,
Substituting , and into the above equation, after using Theorem 1.3, we arrive at the desired result. □
Theorem 4.2 Let with . Then we have
(10)
Proof Putting in Theorem 1.3 and using
we have
(11)
We recall from the definition of and that we have
(12)
(cf. [[10], p.96]). Combining (11) and (12), we get
(13)
If we replace n by 2n in the above equation, we obtain
(14)
From the work of Srivastava and Choi [[4], p.98], we recall that
(15)
where with and
(16)
By substituting (15) and (16) into (14), after some elementary calculations, we arrive at the desired result. □
Theorem 4.3 Let . Then
(17)
Proof Combining (14) and (16), we easily complete the proof of the theorem, that is,
□
By using (10) and (17), we obtain a convolution formula (Euler identity) for Bernoulli numbers.
Theorem 4.4 Let . Then
(18)
Proof Since the left-hand sides of (10) and (17) are equal, the right-hand sides of (10) and (17) must be equal. Thus, we obtain
After some elementary calculation in the above equation, we get the desired result. □
Observe that the proof of (18) is also given in [4].
Theorem 4.5 Let . Then
Proof For all , we have
(19)
(cf. [[4], p.131]). By using (14) and (19), we obtain
Thus, the proof is completed. □
Theorem 4.6 Let . Then we have
(20)
Proof Consider that n is replaced by in (13), we have
(21)
For all , one can easily get
(22)
(cf. [[4], p.99, Eq. (21)]). Hence, we have
Thus, the proof is completed. □
Theorem 4.7 Let . Then we have
(23)
Proof Note that, for all , we have
(24)
(cf. [[4], p.99, Eq. (22)]). By using (21) and (24), we have
Thus, the proof is completed. □
Theorem 4.8 Let . Then
Proof Substituting into Theorem 1.3 and by , we have
(25)
If n is replaced by 2n in the above equation, we get
(26)
By using (16), we have
(27)
Thus, the proof is completed. □
Theorem 4.9 Let with . Then we have
Proof By using (26), (15) and (16), we have
Thus, the proof is completed. □
Theorem 4.10 Let . Then
Proof By using (26) and (19), we have
Thus, the proof is completed. □
Theorem 4.11 Let . Then
(28)
Proof By replacing n by in (25), we have
(29)
By substituting (29) into (22), we get
Thus, the proof is completed. □
Theorem 4.12 Let . Then we have
(30)
Proof By using (29) and (24), we have
Thus, the proof is completed. □
By comparing (20) and (23) or (28) and (30), we arrive at the following result.
Corollary 4.13