EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 1005-1023 ISSN 1307-5543 – ejpam.com Published by New York Business Global Construction of Fourier Series Expansion of Apostol-Frobenius-Type Tangent and Genocchi Polynomials of Higher-order Roberto B. Corcino1,2,∗, Cristina B. Corcino1,2, Karl Patrick Casas1,3, Allan Roy Elnar1,3, Gibson Maglasang1,3 1 Research Institute for Computational Mathematics and Physics, Cebu Normal University, 6000 Cebu City, Philippines 2 Mathematics Department, Cebu Normal University, 6000 Cebu City, Philippines 2 Physics Department, Cebu Normal University, 6000 Cebu City, Philippines Abstract. In this study, the Fourier series expansions of the Apostol-Frobenius type of Tangent and Genocchi polynomials of higher order are derived using the Cauchy residue theorem. Some novel and intriguing results are obtained by applying the Fourier series expansion of these types of polynomials. 2020 Mathematics Subject Classifications: 11B68, 42A16 Key Words and Phrases: Cauchy residue theorem; Fourier series, Tangent polynomials, Bernoulli polynomials, Genocchi polynomials 1. Introduction There are numerous well-known special functions, numbers, and polynomials, such the Bernoulli, Tangent, and Genocchi numbers and polynomials, and derivative polynomials, that are well studied in the current literature due to their broad applications ranging from number theory and combinatorics to other fields of applied mathematics[5, 6, 13]. In the literature, other variants and extensions of these functions, numbers, and polyno- mials have appeared. Some versions have been created by combining two or three special functions, integers, or polynomials. Poly-Bernoulli numbers and polynomials, for exam- ple, were created by combining the notions of polylogarithm and Bernoulli numbers and polynomials[9, 14]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4700 Email addresses: rcorcino@yahoo.com (R. Corcino), corcinoc@cnu.edu.ph (C. Corcino), casask@cnu.edu.ph (K. Casas), elnara@cnu.edu.ph (A.R. Elnar), maglasangg@cnu.edu.ph (G. Maglasang) https://www.ejpam.com 1005 © 2023 EJPAM All rights reserved. R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1006 The principles of Apostol, Frobenius, Genocchi, and Euler polynomials were combined to create the Apostol-Genocchi polynomials, Frobenius-Euler polynomials, Frobenius- Genocchi polynomials, and Apostol-Frobenius-type poly-Genocchi polynomials in the pa- per of Ryoo et.al.[10–12]. Another intriguing combination of special polynomials may be created by combining the principles of Apostol and Frobenius polynomials with Tangent, Bernoulli, and Genocchi polynomials, which will be the topic direction of this paper. This will be carried out using the Fourier series expansion method[7]. Currently, there was no literature or related works that mentioned the Fourier series expansion of Apostol- Frobenius: Tangent, Bernoulli, and Genocchi polynomials that were accessible at the time of the research. Fourier series is widely known as an expansion of a periodic function f(x) in terms of an infinite sum of sine and cosine functions. It makes use of the orthogonality relationships of the sine and cosine functions. Fourier series is expressed as[1] s(x) = a0 2 + ∞∑ n=1 [ an cos ( 2π T nx ) + bn sin ( 2π T nx )] (1) where T is the function’s period. The above expression can also be cast into its exponential form as follows: s(x) = ∞∑ n=−∞ cn · ei2πnx/T (2) where the coefficients cn are computed as cn = 1 T ∫ T 0 e−i2πnx/T · s(t)dt (3) The Fourier expansion of several well-known polynomials has recently piqued the curiosity of mathematicians. The Fourier expansions for the Apostol-Bernoulli and Apostol-Euler polynomials are given by Lou (2009). Using the Lipschitz summation formula, Lou derives the Fourier expansions and integral representation for Genocchi polynomials the same year. Araci-Acikgoz (2018) made a significant finding about the Fourier expansion of the Apostol Frobenius-Euler and Genocchi polynomials. With this motivation, we are interested in determining the Fourier series expansions of higher order Apostol-Frobenius- type Tangent and Genocchi polynomials using the Cauchy residue theorem and a complex integral over a contour[2], which they found to be particularly useful. 2. Main Results In this section, we use the Cauchy Residue theorem and Bayad’s method[2] in evalu- ating the complex integral over a circle C to obtain the Fourier series expansion for the Frobenius type of Apostol-Tangent and Apostol-Genocchi polynomials. R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1007 2.1. Fourier expansion and Integral representation of Apostol-Frobenius- Tangent Polynomials: The Tangent polynomials with complex argument x are defined as coefficients of the following generating function[8] ∞∑ n=0 Tn(x) tn n! = ( 2 e2t + 1 ) ext (4) where Tn(0) = Tn. The Apostol-Frobenius-type Tangent polynomials, which are another variation of the Tangent polynomials, are defined as follows: ∞∑ n=0 Tn(x;u, λ) tn n! = 1− u λe2t − u ext (5) where u, λ ∈ C with u ̸= 1, λ ̸= 1 and u ̸= λ. By Cauchy Integral formula, we have Tn(x;u, λ) n! = 1 2πi ∫ C 1− u λe2t − u ext dt tn+1 (6) Consider now the function inside the integral f(t) = 1− u λe2t − u ext tn+1 (7) which has a pole at t = 0 of order n + 1. We then find the other poles of the function fn(t) as follows: λe2t − u = 0 λe2t = u λ 2t = log u λ + 2kπi tk := t = log (u λ )1/2 + kπi, for k ∈ Z By Cauchy Residue Theorem, we have 1 2πi ∫ CN fn(t)dt = Res(f(t), t = 0) + ∑ k∈Z (f(t), t = tk) (8) In Eq. (8), we integrate fn(t) around the circle with radius (N + ϵ)π where ϵ ∈ R. That is, ϵπi ± log ( u λ )1/2 ̸= 0 (mod 2πi). This radius guarantees that the circle CN does not pass through any of the poles tk. The following lemma contains the limit of the integral in Eq. (8) as N → ∞ R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1008 Lemma 1. Let u, λ ∈ C{0, 1} with |λ| ≠ |u|. For 0 < x ≤ 1 lim N→∞ ∫ CN fn(t)dt = ∫ CN 1− u λe2t − u ext dt tn+1 = 0 where CN = {t : |t| = (N + ϵ)π and ϵ ∈ R, ( ϵπi± log(u/λ)1/2 ) ̸= 0} Proof. We first take the modulus of the integral given as∣∣∣∣∫ CN 1− u λe2t − u ext dt tn+1 ∣∣∣∣ ≤ ∫ CN |1− u||ext| |λe2t − u||tn+1||dt| ≤ ∫ CN |ext| |λe2t − u| |dt| Consider the function in the last integral |ext| |λe2t − u| = |ext| | − u| ∣∣−u λe 2t + 1 ∣∣ ≤ eRe(t) |α||u||e2t + 1 α | , where α = −λ u ≤ 1 | − λ| So that, ∣∣∣∣∫ CN 1− u λe2t − u ext dt tn+1 ∣∣∣∣ ≤ 1 | − λ| ∫ CN |dt| |tn+1| = 2n+1 | − λ|((2N + ϵ)π)n+1 As N → ∞ for n ≥ 1 ∫ CN 1− u λe2t − u ext dt tn+1 → 0 Using Lemma 1, Eq. (8) becomes Res(fn(t), t = 0) = − ∑ k∈Z Res(fn(t), t = tk) We then compute the Res(fn(t), t = 0,) and ∑ k∈ZRes(fn(t), t = tk) to obtain the Fourier series expansion of Apostol-Frobenius-Tangent polynomials. The following theorem ex- plicitly shows the Fourier series representation of the said polynomials Theorem 1. Let u, λ ∈ C{0, 1} with |λ| ≠ |u|. For 0 < x ≤ 1 Tn(x;u, λ) = n! 2 u− 1 u (u λ ) 1 2 x∑ k∈Z eiπkx 2λ [ log ( u λ )1/2 + kπi ]n+1 (9) R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1009 Proof. We compute Res(fn(t), t = 0) and ∑ k∈ZRes(fn(t), t = tk) as follows: Res(fn(t), t = 0) = lim t→0 1 n! dn dtn (t− 0)n+1 1 tn+1 ∞∑ m=0 Tm(x; , u, λ) tm m! = lim t→0 1 n! dn dtn ∞∑ m=0 Tm(x;u, λ) = lim t→0 1 n! ∞∑ m=0 Tm(x;u, λ) tm−n (m− n)! = Tm(x;u, λ) n! (10) and Res(fn(t), t = tk) = lim t→tk (t− tk) 1− u λe2t − u ext 1 tn+1 = (1− u) tn+1 k extk lim t→tk t− tk λe2t − u = 1− u tn+1 k extk lim t→tk 1 2λe2t = (1− u)e(x−2)tk 2λtn+1 k , where tk = log (u λ )1/2 + kπi = (1− u) exp { (x− 2) [ log(u/λ)1/2 + kπi ]} 2λ [ log ( u λ )1/2 + kπi ]n+1 = (1− u)ex log(u/λ)1/2ekπixe−2 log(u/λ)1/2e−2kπi 2λ [ log ( u λ )1/2 + kπi ]n+1 = (1− u)(u/λ) 1 2 x−1ekπix 2λ [ log ( u λ )1/2 + kπi ]n+1 (11) Combining the results of the residues equations (10) and (11) and substitute it to Eq. (8), we get Tn(x;u, λ) n! = − ∑ k∈Z (1− u) ( u λ ) 1 2 x−1 ekπix 2λ [ log ( u λ )1/2 + kπi ]n+1 Simplifying the above expression we then have Tn(x;u, λ) = n! 2 u− 1 u (u λ ) 1 2 x∑ k∈Z eiπkx[ log ( u λ )1/2 + kπi ]n+1 R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1010 When u = −1, the above expression will then be Tn(x;−1, λ) = T (x;λ). That is Tn(x;u = −1, λ) = n! 2 ( −1 λ ) 1 2 x∑ k∈Z eikπx · 2n+1[ log (−1 λ ) + 2kπi ]n+1 = n! 2 e−iπk x 2 λ 1 2 x ∑ k∈Z eiπkx · 2n+1 [(2k − 1)π − log λ]n+1 = n! 2 1 (λ) 1 2 x ∑ k∈Z eiπkxe−iπ x 2 · 2n+1[ log (−1 λ ) + 2kπi ]n+1 T (x;λ) = n! 2 1 (λ) 1 2 x ∑ k∈Z e x 2 (2k−1)πi · 2n+1 [(2k − 1)πi− log λ]n+1 Note that, the above expression is the known Apostol-Tangent polynomial as shown in the paper of Corcino et al.[3] Now, consider an integral formulation of the polynomials Apostol-Frobenius-Tangent. Theorem 2. For n ∈ N (set of natural numbers), 0 < x < 1, ξ < 1 2 , ξ ∈ R, we have Tn ( x;u;−ue2ξπi ) = ( u− 1 u ) 2n−1e−ξπix × [∫ ∞ 0 M(n;x; v) cosh(2ξπv) + iN(n;x; v) sinh(2ξπv) cosh(2πvs.) + cos(πx) vndv ] (12) where M(n;x; v) = eπv cos ( −π 2 x+ (n+ 1)π 2 ) − e−πv cos ( π 2 x+ (n+ 1)π 2 ) (13) N(nn;x; v) = eπv sin ( −π 2 x+ (n+ 1)π 2 ) − e−πv sin ( π 2 x+ (n+ 1)π 2 ) (14) Proof. From Eq.(9), we let λ = −ue2ξπi and k 7→ −k T (x;u;−ue2ξπi) = u− 1 u 2n ( −e−2ξπi ) 1 2 x n! ∑ k∈Z e−iπkx [−2kπi+ log(e−πi · e−2ξπi)] n+1 = u− 1 u 2n ( e−πi−2ξπi ) 1 2 x n! ∑ k∈Z e−iπkx [−2kπi− πi− 2ξπi]n+1 = u− 1 u 2n ( e−( 1 2 +ξ)πix ) n! ∑ k∈Z e−iπkx (−πi)n+1(2k + 2ξ + 1)n+1 = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 n! ∑ k∈Z e−iπkx (2k + 2ξ + 1)n+1 R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1011 = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 n! [ ∞∑ k=0 e−πkx (2k + 2ξ + 1)n+1 + ∞∑ k=1 eiπkx (−2k + 2ξ + 1)n+1 ] = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 n! [ ∞∑ k=0 e−πkx (2k + 2ξ + 1)n+1 + (−1)n+1 ∞∑ k=1 eiπkx (2k − 2ξ − 1)n+1 ] = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 [ ∞∑ k=0 e−iπkx n! (2k + 2ξ + 1)n+1 + (−1)n+1 ∞∑ k=1 eiπkx n! (2k − 2ξ − 1)n+1 ] where (2k + 2ξ + 1) > 0 if |ξ| < 1 2 , ξ ∈ R and k ≥ 0, (2k − 2ξ − 1) > 0 if |ξ| < 1 2 , ξ ∈ R and k ≥ 0. We then apply the integral formula given as∫ ∞ 0 tne−atdt = n! an+1 (n = 0, 1, ...;R(a) > 0) So that, T (x;u;−ue2ξπi) = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 [ ∞∑ k=0 e−iπkx ∫ ∞ 0 tne−(2k+2ξ+1)tdt + (−1)n+1 ∞∑ k=1 eiπkx ∫ ∞ 0 tne−(2k−2ξ−1)tdt ] = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 [∫ ∞ 0 tne−(2ξ+1)t ∞∑ k=0 e−(iπx+2t)kdt + (−1)n+1 ∫ ∞ 0 e(2ξ+1)t ∞∑ k=1 e(iπx−2t)ktndt ] = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 [∫ ∞ 0 tne−(2ξ+1)t 1 1− e−(iπx+2t) dt + (−1)n+1 ∫ ∞ 0 e(2ξ+1)t eiπx−2t 1− eiπx−2t tndt = u− 1 u 2n e−( 1 2 +ξ)πix (−πi)n+1 [∫ ∞ 0 tne−(2ξ+1)t eiπx eiπx − e−2t dt + (−1)n+1 ∫ ∞ 0 e(2ξ+1)t eiπx e2t−eiπx t ndt R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1012 = u− 1 u 2n e−ξπix (−πi)n+1 [∫ ∞ 0 e 1 2 πix eπix − e−2t e−(2ξ+1)ttndt + (−1)n+1 ∫ ∞ 0 e 1 2 πix e2t − eiπx e(2ξ+1)ttndt ] = u− 1 u 2n e−ξπix (−πi)n+1 [∫ ∞ 0 e−(2ξ+1)t e 1 2 πix − e−2t− 1 2 πix tndt + (−1)n+1 ∫ ∞ 0 e(2ξ+1)t e2t− 1 2 πix − e 1 2 πix tndt ] = u− 1 u 2n e−ξπix (−πi)n+1 [∫ ∞ 0 e−(2ξ+1)t e 1 2 πix − e−2t− 1 2 πix · ( e2t − eπix ) e− 1 2 πix e2t− 1 2 πix − e 1 2 πix tndt + (−1)n+1 ∫ ∞ 0 e(2ξ+1)t e2t− 1 2 πix − e 1 2 πix · ( eπix − e−2t ) e− 1 2 πix e 1 2 πix − e−2t− 1 2 πix tndt ] = u− 1 u 2n−1 e−ξπix (−πi)n+1 [∫ ∞ 0 e−(2ξ+1)t(e2t − eπix)e− 1 2 πix cosh(2t)− cos(πx) tndt + (−1)n+1 ∫ ∞ 0 e(2ξ+1)t(eπix − e2t)e− 1 2 πix cosh(2t)− cos(πx) tndt ] We then make the substitution t = πv, (−1/i)n+1 = e(n+1)πi/2 and (−1)n+1 = e−(n+1)πi, we find that T (x;u;−ue2ξπi) = u− 1 u 2n−1 e−ξπix (−πi)n+1 [∫ ∞ 0 e−(2ξ+1)πv(e2πv − eπix)e− 1 2 πix cosh(2πvs.)− cos(πx) (πv)nπdv + (−1)n+1 ∫ ∞ 0 e(2ξ+1)πv(eπix − e2πv)e− 1 2 πix cosh(2πvs)− cos(πx) (πv)nπdv ] = u− 1 u 2n−1 e −ξπix πn+1 [∫ ∞ 0 e(n+1)πi/2(e2πv − eπix)e−(2ξ+1)πve− 1 2 πix cosh(2πvs)− cos(πx) πn+1vndv + (−1)n+1 ∫ ∞ 0 e−(n+1)πi/2(eπx − e2πv)e(2ξ+1)πve− 1 2 πix cosh(2πvs.)− cos(πx) πn+1vndv ] = u− 1 u 2n−1e−ξπix [∫ ∞ 0 eπve i [ −π 2 x+ (n+1)π 2 ] e−2ξπv − e−πve i [ π 2 x+ (n+1)π 2 ] e−2ξπv cosh(2πvs.)− cos(πx) vndv + ∫ ∞ 0 eπve i [ π 2 x− (n+1)π 2 ] e2ξπv − e−πve i [ −π 2 x− (n+1)π 2 ] e2ξπv cosh(2πvs)− cos(πx) vndv ] R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1013 Let A = −π 2x+ (n+1)π 2 and B = π 2x+ (n+1)π 2 . So that, T (x;u;−ue2ξπi) = u− 1 u 2n−1e−ξπix [∫ ∞ 0 eπveiAe−2ξπv − e−πveiBe−2ξπv cosh(2πvs.)− cos(πx) vndv + ∫ ∞ 0 eπve−iAe2ξπv − e−πve−iBe2ξπv cosh(2πvs.)− cos(πx) vndv ] = u− 1 u 2n−1e−ξπix [∫ ∞ 0 (eπv cosA− e−πv cosB)(e−2ξπv + e2ξπv) cosh(2πvs.)− cos(πx) vndv + i ∫ ∞ 0 (eπv sinA− e−πv sinB)(e2ξπv − e2ξπv) cosh(2πvs.)− cos(πx) vndv ] = u− 1 u 2n−1e−ξπix [∫ ∞ 0 (eπv cosA− e−πv cosB)(cosh(2ξπv)) cosh(2πvs.)− cos(πx) vndv + i ∫ ∞ 0 (eπv sinA− e−πv sinB)(sinh(2ξπv)) cosh(2πvs.)− cos(πx) vndv ] Simplifying the above equation gives us the integral representation of the Apostol-Frobenius- Tangent polynomials T (x;u;−ue2ξπi) = u− 1 u 2n−1e−ξπix [∫ ∞ 0 M(n;x; v) cosh(2ξπv) + iN(n;x; v) sinh(2ξπv) cosh(2πvs.)− cos(πx) vndv ] (15) where M(n;x; v) = eπv cos ( −π 2 x+ (n+ 1)π 2 ) − e−πv cos ( π 2 x+ (n+ 1)π 2 ) N(nn;x; v) = eπv sin ( −π 2 x+ (n+ 1)π 2 ) − e−πv sin ( π 2 x+ (n+ 1)π 2 ) 2.2. Fourier expansion of Apostol-Frobenius-Tangent Polynomials of Higher- order The Apostol-Frobenius-Tangent polynomials of higher order, denoted by T (r) n (x;u, λ), are defined as coefficients of the following generating function ∞∑ n=0 T (r) n (x;u, λ) tn n! = ( 1− u λe2t − u )r ext (16) where r ≥ 1, u, λ ∈ C with u ̸= 1, λ ̸= 1 and u ̸= λ. In this section, we derive the Fourier expansion for Apostol-Frobenius-Tangent poly- nomials of higher order as shown in the following theorem. R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1014 Theorem 3. For 0 ≤ x ≤ 1, T (r) n (x;u, λ) = −n! ∑ k∈Z ( 1− u 2u )r r−1∑ j=0 (−1)j−12j ( n+ r − 1− j r − 1− j ) B (r) l ( x 2 ) j! ( u λ )x 2 exkπi[ log ( u λ ) 1 2 + kπi ]r+n−j (17) where B (r) n ( x 2 ) is the Bernoulli polynomials of order r defined by( t et − 1 )r ext = ∞∑ n=0 B(r) n (x) tn n! Proof. By Cauchy Residue Theorem, 1 2πi ∫ CN f(t)dt = Res(f(t), t = 0) + ∑ k∈Z Res(f(t), t = tk) where tk = log (u λ ) 1 2 + kπi, k ∈ Z f(t) = ( 1− u λe2t − u )r ext tn+1 Consider the left-hand side of the equation in the Cauchy Residue Theorem as N → ∞, we have lim N→∞ ∫ CN f(t)dt = lim N→∞ ∫ CN ( 1− u λe2t − u )r ext dt tn+1 = 0 We use the same proof as in the case when r = 1 as shown in Lemma 1. We shall evaluate the first term of the right-hand side of the equation in the Cauchy Residue Theorem as t = 0, we have Res(f(t), t = 0) = lim t→0 1 n! dn dtn (t− 0)n+1 1 tn+1 ∞∑ m=0 T (r) m (x;u, λ) tm m! = 1 n! T (r) n (x;u, λ). Now we will evaluate the second term of the right-hand side of the equation of the Cauchy Residue theorem as t = tk. That is, for r ≥ 2 Res(f(t), t = tk) = 1 (r − 1)! lim t→tk dr−1 dtr−1 (t− tk) r ( 1− u λe2t − u )r ext tn+1 Consider the function (t− tk) r ( 1− u λe2t − u )r ext tn+1 = (t− tk) r (1− u)r( λ ue 2t − 1 )r 1 ur ext tn+1 R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1015 = (t− tk) r (1− u)ru−r( λ ue 2t − 1 )r extt−(n+1) = (t− tk) r (1− u)ru−r( λ ue 2t − 1 )r extt−(n+1) = (t− tk) r (1− u)ru−r( λ ue 2(t−tk) u λ − 1 )r extt−(n+1) since u λ e−2tk = 1 = (1− u)r (2u)r [2(t− tk)] r (e2(t−tk) − 1)r ext tn+1 = (1− u)r (2u)r ∞∑ n=0 B(r) n [2(t− tk)] r n! t−(n+1)ext where ( w ew − 1 )r = ∞∑ n=0 B(r) n wn n! To get the derivative, we employ the Leibniz Rule which gives us dr−1 dtr−1 { (t− tk) r ( 1− u λe2t − u )r ext tn+1 } = ( 1− u 2u )r dr−1 dtr−1 { (t− tk) r [ ∞∑ n=0 B(r) n [2(t− tk)] n n! ] extt−(n+1) } = ( 1− u 2u )r dr−1 dtr−1 {([ B(r) n [2(t− tk)] n n! ] ext ) t−(n+1) } = ( 1− u 2u )r r−1∑ j=0 ( r − 1 j ) dr−1−j dtr−1−j t−(n+1) d j dtj ext ∞∑ n=0 B(r) n 2n (t− tk) n n!︸ ︷︷ ︸ H(t)  Consider now, dj dtj (H(t)) = j∑ l=0 ( j l ) xj−lext ∞∑ n=0 B(r) n 2n(n)l n! (t− tk) n−l = ext j∑ l=0 ( j l ) xj−l ∞∑ n=0 B(r) n 2n (t− tk) n−l (n− l)! So that, the derivative becomes dr−1 dtr−1 { (t− tk) r ( 1− u λe2t − u )r ext tn+1 } = ( 1− u 2u )r r−1∑ j=0 ( r − 1 j ) dr−1−j dtr−1−j t−(n+1) R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1016 × ext j∑ l=0 ( j l ) xj−l ∞∑ n=0 B(r) n 2n (t− tk) n−l (n− l)! Thus, Res(f(t), t = tk) = 1 (r − 1)! lim t→tk dr−1 dtr−1 (t− tk) r ( 1− u λe2t − u )r ext tn+1 = 1 (r − 1)! lim t→tk ( 1− u 2u )r r−1∑ j=0 ( r − 1 j ) dr−1−j dtr−1−j t−(n+1) × ext j∑ l=0 ( j l ) xj−l ∞∑ n=0 B(r) n 2n (t− tk) n−l (n− l)! Note that B (r) n (t−tk) n−l (n−l)! → 0 as t → tk except when n = l. So that Res(f(t), t = tk) = 1 (r − 1)! ( 1− u 2u )r r−1∑ j=0 ( r − 1 j ) (−1)r−1−j(n+ r − 1− j)r−1−jt −(n+r−j) k × extk j∑ l=0 ( j l ) xj−l2lB (r) l = 1 (r − 1)! (1− u)r (2u)r r−1∑ j=0 (r − 1)! j!(r − 1− j)! (−1)r−1−j(n+ r − 1− j)r−1−jt −(n+r−j) k × extk j∑ l=0 ( j l ) xj−l2lB (r) l = ( u− 1 2u )r r−1∑ j=0 ( n+ r − 1− j r − 1− j ) (−1)−j−1 t j−n−r j! extk2j j∑ l=0 ( j l ) xj−l 2j−l B(r) n . We use the identity that B (r) l = ∑j j=0 ( j l ) B (r) l ( x 2 )j−l . Thus, Res(f(t), t = tk) = ( u− 1 2u )r r−1∑ j=0 ( n+ r − 1− j r − 1− j ) (−1)j−1B (r) l ( x 2 ) j! extk tr+n−j k Substituting tk = log ( u λ ) 1 2 + kπi, we get Res(f(t), t = tk) = ( u− 1 2u )r r−1∑ j=0 ( n+ r − 1− j r − 1− j ) (−1)j−1B (r) l ( x 2 ) j! e x ( log(u λ) 1 2+kπi ) 1 2 [ log ( u λ ) 1 2 + kπi ]r+n−j R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1017 This gives T (r) n (x;u, λ) = −n! ∑ k∈Z ( u− 1 2u )r r−1∑ j=0 (−1)j−12j ( n+ r − 1− j r − 1− j ) B (r) l ( x 2 ) j! ( u λ )x/2 exkπi[ log ( u λ ) 1 2 + kπi ]r+n−j where ( t et − 1 )r ext = ∞∑ n=0 B(r) n (x) tn n! 2.3. Fourier expansion of Apostol-Frobenius-Genocchi Polynomials of Higher Order The Genocchi polynomials can be defined as [4] ∞∑ n=0 Gn(x) tn n! = ( 2t et + 1 ) ext (18) The Apostol-Frobenius-Genocchi polynomials which are certain variation of the Genocchi polynomials are defined by Araci and Acikgoz [1] as coefficients of the following generating function: ∞∑ n=0 Gn(x;u, λ) tn n! = (1− u)t λet − u ext (19) where u, λ ∈ C with u ̸= 1, λ ̸= 1 and u ̸= λ. By Cauchy Integral formula, we observe that Gn(x;u, λ) n! = 1 2πi ∫ C (1− u)t λet − u ext dt tn+1 If we consider the function f(t) = (1− u) λet − u ext tn , then it has a pole at t = 0 of order n. The other poles are found to be at λet − u = 0 λet = u et = u λ tk := t = log (u λ ) + 2kπi By Cauchy Residue Theorem, we have 1 2πi ∫ CN fn(t)dt = Res(fn(t), t = 0) + ∑ k∈Z Res(fn(t), t = tk) (20) In Eq. (20), we take the limit of the integral, ∫ CN fn(t)dt as N → ∞ as explicitly shown in the following lemma R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1018 Lemma 2. Let u, λ ∈ C{0, 1} with |λ| ≠ |u|. For 0 < x ≤ 1∫ CN (1− u)text (λet − u)tn+1 dt = 0 Proof. Consider∣∣∣∣∫ CN (1− u)text λet − u dt tn+1 ∣∣∣∣ ≤ ∫ CN |1− u||ext||dt| |λet − u||tn| = ∫ CN |1− u||ext||dt| |αu| ∣∣et + 1 α ∣∣ |tn| , where α = −λ u∫ CN |1− u||dt| | − λ||tn| So that ∣∣∣∣∫ CN (1− u)text λet − u dt tn+1 ∣∣∣∣ ≤ |1− u| | − λ| ∫ CN |dt| |tn| = 1 ((2N + ϵ)π)n As N → ∞ ∫ CN (1− u)text λet − u dt tn+1 dt → 0 Using Lemma 2, Eq. (20) becomes Res(fn(t), t = 0) = − ∑ k∈Z Res(fn(t), t = tk) With these, Araci and Acikgoz [1] obtained the following Fourier series expansion by calculating the residues of the function fn(t) at t = 0 and t = tk, respectively: Gn(x;u, λ) = n! 1− u u (u λ )x∑ k∈Z ei2πkx [log(u/λ) + 2kπi]−n (21) where u, λ ∈ C{0, 1} with |λ| ≠ |u| and 0 < x ≤ 1. Note that, when we take u = −1, we have G(x;u = −1, λ) = n! −1− 1 −1 ( −1 λ )x∑ k∈Z e2kπix [ log ( −1 λ ) + 2kπi ]−n = 2n! (−1)x λx ∑ k∈Z e2kπix[log(−1)− log λ+ 2kπi]−n = 2n! eπix λx ∑ k∈Z e2kπix[πi− log λ+ 2kπi]−n G(x;λ) = 2n! λx ∑ k∈Z e(2k+1)πix [− log λ+ (2k + 1)πi]n (22) R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1019 Note that Eq.(22) is the Apostol-Genocchi polynomials obtained in the paper of Corcino, et.al.[2]. That is, G(x;u = −1, λ) = G(x;λ). The Apostol-Frobenius-Genocchi polynomials of higher order, denoted by G (r) n (x;u, λ), are defined as coefficients of the following generating function: ∞∑ n=0 G(r) n (x;u, λ) tn n! = ( (1− u)t λet − u )r ext (23) where r ≥ 1, u, λ ∈ C with u ̸= 1, λ ̸= 1 and u ̸= λ. The following theorem contains the Fourier expansion of these polynomials. Theorem 4. For 0 ≤ x ≤ 1 G(r) n (x;u, λ) = −n! ( u− 1 u )r∑ k∈Z r−1∑ j=0 (−1)j−1 ( n− 1− j r − 1− j ) B (r) l (x) j! ( u λ )x e2xkπi[ log ( u λ ) + 2kπi ]n−j (24) where B (r) n (x) is the Bernoulli polynomials of order r. Proof. Using the Cauchy Residue Theorem (CRT), 1 2πi ∫ CN f(t)dt = Res(f(t), t = 0) + ∑ k∈Z Res(f(t), t = tk) where tk = log (u λ ) + 2kπi, k ∈ Z f(t) = ( (1− u)t λet − u )r ext tn+1 = ( 1− u λet − u )r ext tn+1−r Consider the left-hand side of the equation in the CRT as N → ∞, we have lim N→∞ ∫ CN f(t)dt = lim N→∞ ∫ CN ( (1− u) λet − u )r ext dt tn+1−r = 0, which can easily be shown using the same proof as in the case when r = 1 as shown in Lemma 2. We shall evaluate the first term of the right-hand side of the equation in the Cauchy Residue Theorem as t = 0, we have Res(f(t), t = 0) = lim t→0 1 n! dn dtn (t− 0)n+1 1 tn+1 ∞∑ m=0 G(r) m (x;u, λ) tm m! = lim t→0 1 n! dn dtn ∞∑ m=0 G(r) m (x;u, λ) tm m! R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1020 = lim t→0 1 n! ∞∑ m=0 G(r) m (x;u, λ) (m)n m! tm−n = lim t→0 1 n! ∞∑ m=0 G(r) m (x;u, λ) tm−n (m− n)! = 1 n! G(r) n (x;u, λ) Now we will evaluate the second term of the right-hand side of the equation of the Cauchy Residue theorem as t = tk. That is, for r ≥ 2 Res(f(t), t = tk) = 1 (r − 1)! lim t→tk dr−1 dtr−1 (t− tk) r ( 1− u λet − u )r extk tn−r+1 Consider now the function (t− tk) r ( 1− u λet − u )r extk tn−r+1 = (1− u)r ur ( t− tk λ ue t − 1 )r extk tn−r+1 = (1− u)r ur ( t− tk λ ue t−tk u λ − 1 )r extk tn−r+1 , since u λ e−tk = 1 = ( 1− u u )r ( t− tk et−tk − 1 )r extk tn−r+1 = ( 1− u u )r ∞∑ n=0 B(r) n (t− tk) n n! extk tn−r+1 where ( w ew − 1 )r = ∞∑ n=0 B(r) n wn n! Using the Leibniz Rule, the derivative part for our residue at t = tk, we get dr−1 dtr−1 ( t− tk et−tk − 1 )r extk tn−r+1 = dr−1 dtr−1 {[ ∞∑ n=0 B(r) n (t− tk) n n! ] extk tn−r+1 } = dr−1 dtr−1 {[ ∞∑ n=0 B(r) n (t− tk) n n! ] extk } t−(n−r+1) Performing the Leibniz derivative rule on the above equation, we get dr−1 dtr−1 {[ ∞∑ n=0 B(r) n (t− tk) n n! ] extk } t−(n−r+1) = r−1∑ j=0 ( r − 1 j ) dr−1−j dtr−1−j t−(n−r+1) R. B. Corcino et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1005-1023 1021 × dj dtj extk ∞∑ n=0 B(r) n (t− tk) n n!︸ ︷︷ ︸ H(t)  Now, consider the derivative dj dtj (extkH(t)) = j∑ l=0 ( j l ) xj−lextk ∞∑ n=0 B (r) n n! (n)l(t− tk) n−l = extk l∑ l=0 ( j l ) xj−l ∞∑ n=0 B(r) n (t− tk) n (n− l)! So that, Res(f(t), t = tk) = 1 (r − 1)! ( 1− u u )r lim t→tk r−1∑ j=0 ( r − 1 j ) dr−1−j dtr−1−j t−(n−r+1)extk × j∑ l=0 ( j l ) xj−l ∞∑ n=0 B(r) n (t− tk) n (n− l)! Note that B (r) n (t−tk) n−l (n−l)! → 0 as t → tk except when n = l Res(f(t), t = tk) = 1 (r − 1)! ( 1− u u )r r−1∑ j=0 ( r − 1 j ) (−1)r−1−j(n− 1− j)r−1−jt −(n−j) k × extk j∑ l=0 ( j l ) xj−lB (r) l = ( 1− u u )r r−1∑ j=0 (−1)r−1−j j!(r − 1− j)! (n− 1− j)r−1−jt −(n−j) k × extk j∑ l=0 ( j l ) xj−lB (r) l = ( 1− u u )r r−1∑ j=0 ( n− 1− j r − 1− j ) (−1)−j−1 t j−n k j! × extk j∑ l=0 ( j l ) xj−lB (r) l Recall that B (r) l (x) = ∑j l=0 ( j l ) B (r) l (x)j−l. Thus, Res(f(t), t = tk) = ( 1− u u )r r−1∑ j=0 ( n− 1− j r − 1− j ) (−1)−j−1B (r) l j! extk tn−j REFERENCES 1022 Substituting tk = log u λ + 2kπi, we get Res(f(t), t = tk) = ( 1− u u )r r−1∑ j=0 ( n− 1− j r − 1− j ) (−1)−j−1B (r) l j! ex(log u λ +2kπi)[ log u λ + 2kπi ]n−j This gives, G(r) n (x;u, λ) = −n! ∑ k∈Z ( 1− u u )r r−1∑ j=0 ( n− 1− j r − 1− j ) (−1)−j−1B (r) l j! ( u λ )x e2kxπi[ log u λ + 2kπi ]n−j . 3. Conclusion The researchers were able to obtain the Fourier series expansion of the Apostol- Frobenius type of: Tangent and Genocchi polynomials of higher order. Taking into con- sideration all of the generating function’s residues, together with the Cauchy Residue the- orem, proved to be a useful strategy for deriving the Fourier series of these polynomials of higher order. For future study, it will be interesting to derive the integral representations of these higher order polynomials. 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