On the fundamental theorem of (p, q)-calculus and some (p, q)-Taylor formulas P. Njionou Sadjang∗ September 17, 2013 Abstract In this paper, the (p, q)-derivative and the (p, q)-integration are investigated. Two suitable polynomials bases for the (p, q)-derivative are provided and various properties of these bases are given. As application, two (p, q)-Taylor formulas for polynomials are given, the fundamental theorem of (p, q)-calculus is included and the formula of (p, q)-integration by part is proved. Keywords: (p, q)-derivative, (p, q)-integration, (p, q)-Taylor formula, fundamental theorem, (p, q)-integration by part. AMS Subject Classification (2010): 33D15, 33D25, 33D35. 1 Introduction The Taylor formula for polynomials f (x) evaluates the coefficients fk in the expansion ∞ (k) k f (c) f (x)= ∑ fk(x − c) , fk = . (1) k=0 k! It is possible to generalize (1) by considering other polynomial bases and suitable operators. The fundamental theorem of calculus can be stated as follows. Theorem 1. If f is a continuous function on an interval (a; b), then f has an antiderivative on (a; b). Moreover, if F is any antiderivative of f on (a; b), then b f (x)dx = F(b) − F(a). (2) arXiv:1309.3934v1 [math.QA] 22 Aug 2013 Za The q version of this theorem was stated in [5] as follows. Theorem 2. If F(x) is an antiderivative of f (x) and if F(x) is continuous at x = 0, then b f (x)dq x = F(b) − F(a), 0 ≤ a ≤ b ≤ ∞. (3) Za ∗Email adress: [email protected] 1 Here the q-integral is defined by a ∞ k k f (x)dq x =(1 − q)a ∑ q f (aq ). (4) Z0 k=0 In this paper, two generalizations of (1) are given and a generalization of (3) is stated. The paper is organised as follows. • In Section 2, we introduce and give relevant properties of the (p, q)-derivative. The (p, q)- power basis is given and main of its properties are provided. The properties of the (p, q)- derivative combined with those of the (p, q)-power basis enable to state two (p, q)-Taylors for polynomials. It then follows connection formulas between the canonical basis and the (p, q)-power basis. • In Section 3, the (p, q)-antiderivative, the (p, q)-integral are introduced and sufficient con- dition for their convergence are investigated. Finally the fundamental theorem of (p, q)- calculus is proved and the formula of (p, q)-integration by part is derived. 2 The (p, q)-derivative and the (p, q)-power basis In this section, we introduce the (p, q)-derivative, the (p, q)-power and provide some of their rel- evant properties. Two (p, q)-Taylor formulas for polynomials are stated and some consequences are investigated. 2.1 The (p, q)-derivative Let f be a function defined on the set of the complex numbers. Definition 1. The (p, q)-derivative of the function f is defined as (see e.g. [4, 1]) f (px) − f (qx) D f (x)= , x 6= 0, (5) p,q (p − q)x ′ and (Dp,q f )(0)= f (0), provided that f is differentiable at 0. The so-called (p, q)-bracket or twin-basic number is defined as pn − qn [n] = . (6) p,q p − q n n−1 It happens clearly that Dp,qx = [n]p,qx . Notealso that for p = 1, the (p, q)-derivative reduces to the Hahn derivative given by f (x) − f (qx) D f (x)= , x 6= 0. q (1 − q)x As with ordinary derivative, the action of the (p, q)-derivative of a function is a linear operator. More precisely, for any constants a and b, Dp,q(a f (x)+ bg(x)) = aDp,q f (x)+ bDp,qg(x). The twin-basic number is a natural generalization of the q-number, that is 1 − qn lim[n]p,q = [n]q = , q 6= 1. (7) p→1 1 − q 2 The (p, q)-factorial is defined by n [n]p,q! = ∏[k]p,q!, n ≥ 1, [0]p,q! = 1. (8) k=1 Let us introduce also the so-called (p, q)-binomial coefficient n [n]p,q! = , 0 ≤ k ≤ n. (9) k [k] ![n − k] ! p,q p,q p,q are called (p, q)-binomial coefficients. Note that as p → 1, the (p, q)-binomial coefficients reduce to the q-binomial coefficients. Proposition 1. The (p, q)-derivative fulfils the following product rules Dp,q( f (x)g(x)) = f (px)Dp,qg(x)+ g(qx)Dp,q f (x), (10) Dp,q( f (x)g(x)) = g(px)Dp,q f (x)+ f (qx)Dp,qg(x) (11) Proof. From the definition of the (p, q)-derivative, we have f (px)g(px) − f (qx)g(qx) D ( f (x)g(x)) = p,q (p − q)x f (px)[g(px) − g(qx)] + g(qx)[ f (px) − f (qx)] = (p − q)x = f (px)Dp,qg(x)+ g(qx)Dp,q f (x). This proves (10). (11) is obtained by symmetry. Proposition 2. The (p, q)-derivative fulfils the following product rules f (x) g(qx)Dp,q f (x) − f (qx)Dp,qg(x) D = (12) p,q g(x) g(px)g(qx) f (x) g(px)Dp,q f (x) − f (px)Dp,qg(x) D = (13) p,q g(x) g(px)g(qx) Proof. The proof of this statements can be deduced using (10). 2.2 The (p, q)-power basis Here, we introduce the so-called (p, q)-power and investigate some of its relevant properties. The expression n n−1 n−1 (x ⊖ a)p,q =(x − a)(px − aq) · · · (px − aq ) (14) is called the (p, q)-power. These polynomials will be useful to state our Taylor formulas. Proposition 3. The following assertion is valid. n n−1 Dp,q(x ⊖ a)p,q = [n]p,q(px ⊖ a)p,q , n ≥ 1, (15) 0 and Dp,q(x ⊖ a)p,q = 0. 3 Proof. The proof follows by a direct computation. Proposition 4. Let γ be a complex number and n ≥ 1 be an integer, then n n−1 Dp,q(γx ⊖ a)p,q = γ[n]p,q(γpx ⊖ a)p,q . (16) Proof. The proof is done exactly as the proof of (15). We now generalize (15) in the following proposition. Proposition 5. Let n ≥ 1 be an integer, and 0 ≤ k ≤ n, the following rule applies k [n]p,q! k n (2) k n−k Dp,q(x ⊖ a)p,q = p (p x ⊖ a)p,q . (17) [n − k]p,q! Proof. The prove is done by induction with respect to k. Remark 1. For the classical derivative, it is known that for any complex number α, one has d xα = αxα−1. dx In what follows, we would like to state similar result for the Dp,q derivative as done for the Dq derivative in [5]. Proposition 6. Let m and n be two non negative integers. Then the following assertion is valid. m+n m m m n (x ⊖ a)p,q =(x ⊖ a)p,q(p x ⊖ q a)p,q. (18) In Proposition 6, if we take m = −n, then we get the following extension of the (p, q)-power basis. Definition 2. Let n be a non negative integer, then we set the following definition. −n 1 (x ⊖ a)p,q = −n −n n . (19) (p x ⊖ q a)p,q Proposition 7. For any two integers m and n, (18) holds. Proof. The case m > 0 and n > 0 has already been proved, and the case where one of m and n is zero is easy. Let us first consider the case m = −m′ < 0 and n > 0. Then, m m m n −m′ −m′ −m′ n (x ⊖ a)p,q(p x ⊖ q a)p,q = (x ⊖ a)p,q (p x ⊖ q a)p,q −m′ −m′ n (p x ⊖ q a)p,q by (19) = −m′ −m′ m′ (p x ⊖ q a)p,q ′ m −m m −m n−m ′ (p (p x) ⊖ q (q a))p,q if n ≥ m by (18) = 1 < ′ m′−n if n m qn(q−m′ x)⊖qn(q−m′ a) ( )p,q n−m′ n+m by (19) = (x ⊖ a)p,q =(x ⊖ a)p,q . 4 If m ≥ 0 and n = −n′ < 0, then m m m n m m m −n′ (x ⊖ a)p,q(p x ⊖ q a)p,q = (x ⊖ a)p,q(p x ⊖ q a)p,q m (x ⊖ a)p,q = m−n′ m−n′ n′ (p x ⊖ q a)p,q m−n′ m−n′ m−n′ n′ (x⊖a)p,q (p x⊖aq )p,q > ′ m−n′ m−n′ n′ if m n (p x⊖q a)p,q = (x⊖a)m p,q < ′ m−n′ m−n′ n′−m n′−m m−n′ n′−m m−n′ m if m n (p x⊖q a)p,q (p (p )x⊖q (q a))p,q m−n′ > ′ (x ⊖ a)p,q if m n = 1 < ′ m−n′ m−n′ n′−m if m n ( (p x⊖q a)p,q m−n′ m+n = (x ⊖ a)p,q =(x ⊖ a)p,q . Lastly, if m = −m′ < 0 and n = −n′ < 0, m m m n −m′ −m′ −m′ −n′ (x ⊖ a)p,q(p x ⊖ q a)p,q = (x ⊖ a)p,q (p x ⊖ q a)p,q 1 = −m′ −m′ m′ −n′−m′ −n′−m′ n′ (p x ⊖ q a)p,q(p x ⊖ q a)p,q 1 = −n′−m′ −n′−m′ n′ +m′ (p x ⊖ q a)p,q −m′−n′ m+n = (x ⊖ a)p,q =(x ⊖ a)p,q .