# Some Properties of  theq-Extension of  the p-Adic  Gamma Function

Authors: Hamza Menken; Adviye Körükçü
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We study theq-extension of the p-adic gamma function Γp,q. We give a new identity for the q-extension of the p-adic gamma Γp,q in the case p=2. Also, we derive some properties and new representations of the q-extension of the p-adic gamma Γp,q in general case.


---

## Body

## 1. Introduction

Letp be a prime number and let ℤp,ℚp, and ℂp denote the ring of p-adic integers, the field of p-adic numbers, and the completion of the algebraic closure of ℚp, respectively. It is well known that the analogous of the classical gamma function Γ in p-adic context depends on modifying the factorial function n! [1]. The factorial function (n!)p in ℚp is defined as
(1)(n!)p=∏j<n(p,j)=1‍j.
The p-adic gamma function Γp is defined by Morita [2] as the continuous extension to ℤp of the function n→(-1)n(n!)p. That is, Γp(x) is defined by the formula
(2)Γp(x)=limn→x(-1)n∏j<n(p,j)=1‍j
for x∈ℤp, where n approaches x through positive integers. The p-adic gamma function Γp(x) had been studied by Diamond [3], Barsky [4], and others. The relationship between some special functions and the p-adic gamma function Γp(x) were investigated by Gross and Koblitz [5], Cohen and Friedman [6]. and Shapiro [7].Theq-extension of the p-adic gamma function Γp,q(x) is defined by Koblitz as follows.Definition 1 (see [8]).
Letq∈ℂp,|q-1|p<1,q≠1. The q-extension of the p-adic gamma function Γp,q(x) is defined by formula
(3)Γp,q(x)=limn→x(-1)n∏j<n(p,j)=1‍1-qj1-q
for x∈ℤp, where n approaches x through positive integers. We recall that limq→1Γp,q=Γp.Theq-extension of the p-adic gamma function Γp,q(x) was studied by Koblitz [8, 9], Nakazato [10], Kim et al. [11], and Kim [12].


## 2. Main Results

In the present work, we give a new identity for theq-extension of the p-adic gamma function Γp,q(x) in special case p=2. Also, we derive some properties and representations for the q-extension of the p-adic gamma function Γp,q(x).Theorem 2.
Ifp=2, then for all x∈ℤ2(4)Γ2,q(x)Γ2,q(1-x)=(-1)1+σ1(x)limn→x∏j<n(2,j)=1qj,
where σ1 is defined by the formula
(5)σ1(∑j=0∞‍aj2j)=a1.Proof.
Letp=2 and n∈ℕ. From Proposition  3 in [12] we known that
(6)Γ2,q(n+1)Γ2,q(-n)=(-1)n+1-[n/2]∏j<n+1(2,j)=1‍qj.
Here, [·] is the greatest integer function. Taking n-1 in place of n, the relation becomes
(7)Γ2,q(n)Γ2,q(1-n)=(-1)n-[(n-1)/2]∏j<n(2,j)=1‍qj.
Now, let n=a0+a12+a222+⋯ in base 2. If a0≠0, then a1=1 in base 2 and
(8)[n-12]=[(a0-1+a12+a222+⋯)2]=[a1+a22+⋯]≡a1(mod2).
Thus, we get
(9)(-1)n-[(n-1)/2]=(-1)n(-1)-[(n-1)/2]=(-1)1(-1)-a1=(-1)1-a1=(-1)1+a1=(-1)1+σ1.
If a0=0, then
(10)[n-12]=[(-1+a12+a222+⋯)2]=[(1+(a1-1)2+a222+⋯)2]≡a1-1(mod2).
Hence,
(11)(-1)n-[(n-1)/2]=(-1)n(-1)-[(n-1)/2]=(-1)2(-1)-(a1-1)=(-1)2+a1-1=(-1)1+a1=(-1)1+σ1(n).
Thus, we have
(12)Γ2,q(n)Γ2,q(1-n)=(-1)1+σ1(n)∏j<n(2,j)=1‍qj
and thus, we obtain
(13)Γ2,q(x)Γ2,q(1-x)=(-1)1+σ1(x)limn→x∏j<n(2,j)=1‍qj.We recall that theq-factorial [n;q]! is defined in [13] by the formula
(14)[n;q]!=[n;q][n-1;q]⋯[2;q][1;q]
for n≥1, where
(15)[x;q]=1-qx1-q.
Note that for n=0, we can define [0;q]!=1.We use the following theorem to prove our results.Theorem 3 (see [12]).
Letn∈ℕ. Then,
(16)Γp,q(n+1)=(-1)n+1[n;q]![p;q][n/p][[n/p];qp]!,
where [·] is the greatest integer function. In particular,
(17)[pn-1;q]!=(-1)p[p;q]pn-1-1[pn-1-1;qp]!Γp,q(pn).Theorem 4.
Letn∈ℕ and let sn be the sum of the digits of n=∑j=0sajpj(as≠0) in base p. Then (a)
[[n/ps];q]!=(-1)n+1-s(-[p;q])(n-sn)/(p-1)∏j=0s-1[[n/pj+1];qp]!/[[n/pj];q]!∏j=0sΓp,q([n/pj]+1)(b)
[n;q]!=(-1)n+1-s(-[p;q])(n-sn)/(p-1)[[n/p];qp]!∏j=1s[[n/pj+1];qp]!/[[n/pj];q]! ∏j=0sΓp,q([n/pj]+1).Proof.
From the Theorem3 we know that
(18)[n;q]!=(-1)n+1[p;q][n/p][[np];qp]!Γp,q(n+1).
By taking [n/p0],[n/p1],…,[n/ps] instead of n, respectively, we get the relations(19)[[np0];q]!=(-1)[n/p0]+1[p;q][n/p1]×[[np1];qp]!Γp,q([np0]+1),[[np1];q]!=(-1)[n/p1]+1[p;q][n/p2]×[[np2];qp]!Γp,q([np1]+1),⋮[[nps];q]!=(-1)[n/ps]+1[p;q][n/ps+1]×[[nps+1];qp]!Γp,q([nps]+1).
By multiplying of the equalities above, we can easily obtain
(20)[[nps];q]!=(-1)[n/p0]+⋯+[n/ps]+s+1[p;q][n/p1]+⋯+[n/ps+1]×[[nps+1];qp]!∏j=0s-1‍[[n/pj+1];qp]![[n/pj];q]!×∏j=0s‍Γp,q([npj]+1)=(-1)(n-sn)/(p-1)(-1)n+1-s[p;q](n-sn)/(p-1)×[[nps+1];qp]!∏j=0s-1‍[[n/pj+1];qp]![[n/pj];q]!×∏j=0s‍Γp,q([npj]+1).
Therefore, we get the relation (a)
(21)[[nps];q]!=(-1)n+1-s(-[p;q](n-sn)/(p-1))×∏j=0s-1‍[[n/pj+1];qp]![[n/pj];q]!∏j=0s‍Γp,q([npj]+1),[n;q]!=(-1)[n/p0]+⋯+[n/ps]+s+1[p;q][n/p1]+⋯+[n/ps+1]×[[np];qp]!∏j=1s‍[[n/pj+1];qp]![[n/pj];q]!×∏j=0s‍Γp,q([npj]+1)=(-1)(n-sn)/(p-1)(-1)n+1-s[p;q](n-sn)/(p-1)×[[np];qp]!∏j=1s‍[[n/pj+1];qp]![[n/pj];q]!×∏j=0s‍Γp,q([npj]+1).
Therefore, we get the relation (b)
(22)[n;q]!=(-1)n+1-s(-[p;q])(n-sn)/(p-1)[[np];qp]!×∏j=1s‍[[n/pj+1];qp]![[n/pj];q]!∏j=0s‍Γp,q([npj]+1).Theorem 5.
Letn∈ℕ and let n=∑j=0sajpj(as≠0). Then
(23)[pn-1;q]!=(-1)p(-[p;q])(pn-1)/(p-1)×[p;q]-n[pn-1-1;qp]!×∏j=0n-2‍[pj-1;qp]![pj+1-1;q]!∏j=0n‍Γp,q(pj).Proof.
From Theorem3 it follows that
(24)[pj-1;q]!=(-1)p[p;q]pj-1-1[pj-1-1;qp]!Γp,q(pj).
Taking of 0,1,…,n instead of j, respectively, we have the equalities
(25)[p0-1;q]!=1=(-1)Γp,q(p0),[p1-1;q]!=(-1)p[p;q]p0-1[p0-1;qp]!Γp,q(p1),[p2-1;q]!=(-1)p[p;q]p1-1[p1-1;qp]!Γp,q(p2),⋮[pn-1;q]!=(-1)p[p;q]pn-1-1[pn-1-1;qp]!Γp,q(pn).
By multiplying of the equalities above, we can easily obtain
(26)[pn-1;q]!=(-1)np+1[p;q]p0+p1+⋯+pn-1-n[pn-1-1;qp]!×∏j=0n-2‍[pj-1;qp]![pj+1-1;q]!∏j=0n‍Γp,q(pj).
Thus,
(27)[pn-1;q]!=(-1)p(-[p;q])(pn-1)/(p-1)×[p;q]-n[pn-1-1;qp]!×∏j=0n-2‍[pj-1;qp]![pj+1-1;q]!∏j=0n‍Γp,q(pj).Lemma 6.
Letn∈ℤ+, n=∑j=0sajpj(as≠0), and let p be a prime number. Then, for j=0,1,…,s(28)[[n/pj];q]![p;q][n/pj][[n/pj];qp]!=∏k=1[n/pj]‍1-qk1-qkp(0≤k≤s).Proof.
Forj=0(29)[n;q]![p;q]n[n;qp]!=[1;q][2;q]⋯[n,q][p;q]n[1;qp][2;qp]⋯[n;qp]=(1-q1-q1-q21-q⋯1-qn1-q)×((1-qp1-q)n1-qp1-qp⋯1-qnp1-qp)-1=(1-q1-q1-q21-q⋯1-qn1-q)×(1-qp1-q1-q2p1-q⋯1-qnp1-q)-1=(1-q)(1-q2)⋯(1-qn)(1-qp)(1-q2p)⋯(1-qnp).
For 1≤j≤s it follows that
(30)[[n/pj];q]![p;q][n/pj][[n/pj];qp]!=[1;q][2;q]⋯[[n/pj],q][p;q][n/pj][1;qp][2;qp]⋯[[n/pj];qp]=(1-q1-q1-q21-q⋯1-q[n/pj]1-q)×((1-qp1-q)[n/pj]1-qp1-qp⋯1-q[n/pj]p1-qp)-1=(1-q)(1-q2)⋯(1-q[n/pj])(1-qp)(1-q2p)⋯(1-q[n/pj]p).
Then, we obtain(31)[[n/pj];q]![p;q][n/pj][[n/pj];qp]!=∏k=1[n/pj]‍1-qk1-qkp.Theorem 7.
Letn∈ℕ and let sn be the sum of the digits of n=∑j=0sajpj(as≠0) in base p. Then
(32)[n;q]!=(-1)((n-sn)/(p-1))+n+1-s∏k=1[n/p1]‍(1-qkp)(1-qk)⋯∏k=1[n/ps]‍(1-qkp)(1-qk)∏j=0s‍Γp,q([npj]+1).Proof.
This theorem can be proved by using Theorem4 and Lemma 6.



---

*Source: 176470-2013-03-14.xml*