Approximation Properties of G-Bernoulli Polynomials

Home , Bernoulli number

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n−1 j N (a; q)0 =1, (a; q)n = 1 − q a , n ∈ , j=0 Y ∞ j C (a; q)∞ = 1 − q a , |q| < 1, a ∈ . j=0 Y n (q; q) = n , k (q; q) (q; q) q n−k k q-standard terminology and notation can be found at [1] and [6]. We call βn,q Bernoulli number and n βn,q = βn,q(0). In addition, q-analogue of (x − a) is deﬁned as;

1, n =0, (x − a)n = , q (x − a)(x − aq)...(x − aqn−1), n 6=0. In the standard approach to the q−calculus, two exponential function are used.These q−exponentials are deﬁned by;

∞ ∞ zn 1 1 e (z)= = , 0 < |q| < 1, |z| < , q [n] ! (1 − (1 − q) qkz) |1 − q| n=0 q X kY=0 ∞ 1 n(n−1) n ∞ q 2 z E (z)= e (z)= = 1+(1 − q) qkz , 0 < |q| < 1, z ∈ C, q 1/q [n] ! n=0 q k=0 X Y q−shifted factorial can be expressed by Heine’s binomial formula as follow;

n − n k(k 1) k k (a; q) = q 2 (−1) a n k q Xk=0 Let for some 0 ≤ α < 1, the function |f(x)xα| is bounded on the interval (0, A], then Jakson integral deﬁnes as [1];

∞ i i f (x)dqx = (1 − q)x q f(q x) i=0 Z X Above expression converges to a function F (x) on (0, A] , which is a q−antiderivative of f(x). Suppose 0

b ∞ i i f (x)dqx = (1 − q)b q f(q b) i=0 Z0 X b b a

f (x)dqx = f (x)dqx − f (x)dq x Za Z0 Z0

2 We need to apply some properties of the q-Bernoulli polynomials to prepare the approximation conditions. These properties are listed below as a lemma;

Lemma 1 Following statements holds true:

a) Dq (βn,q(t)) = [n]q βn−1,q(t), (3) n n b)β (t)= β tn−k, (4) n,q k k,q k=0 q n X n βk,q c) = δ0,n where δ is kronecker delta (5) k q [n − k + 1]q Xk=0 1

d) βn,q(t)dq t =0 n 6=0. (6) Z0

Proof. taking q−derivative from both side of (2) to prove part (a). If we apply Cauchy product for generating formula (4), it leads to (b). Put x = 0 at part (b) and change boundary of summation to reach (c). If we take Jackson integral directly from βn,q(t) with the aid of part (c), we can reach (d).

A few numbers of these polynomials and related q−Bernoulli numbers can be expressed as follows β0,q =1 β0,q(t)=1 β = − 1 β (t)= t − 1 1,q [2]q 1,q [2]q 2 2 β = q β (t)= t2 − t + q 2,q [3]q! 2,q [3]q ! 3 2 3 β = q (1−q) β (t)= t3 − [3]q t2 + q t + q (1−q) 3,q [5]q +[4]q−1 3,q [2]q [2]q [5]q +[4]q−1

Deﬁnition 2 q−Bernoulli polynomials of two variables are deﬁned as a generating function as follow

te (xt)e (yt) ∞ tn q q = β (x, y) , |t| < 2π. (7) e (t) − 1 n,q [n] ! q n=0 q X A simple calculation of generating function leads us to

t te (xt) ∞ tn te (xt)+ e (tx)= q e (t)= β (x, 1) , |t| < 2π. (8) q e (t) − 1 q e (t) − 1 q n,q [n] ! q q n=0 q X The LHS can be written as

∞ tn+1xn ∞ tn + β (x) (9) [n] ! n,q [n] ! n=0 q n=0 q X X By comparing the nth coeﬃcients, we drive to diﬀerence equations as follow

n−1 βn,q(x, 1) − βn,q(x) = [n]q!x n ≥ 1. (10)

If we put x =0, then βn,q(0, 1) − βn,q(0) = δn,1.

3 2 q-analogue of Euler-Maclaurin formula

This section introduces q−operator to ﬁnd q−analogue of Euler-Maclaurin formula. The q−analogue of Euler-Maclaurin formula has been studied in [29]. They applied q-integral by parts to reach their formula. We can not apply that approach to approximation, because it was written in term of p(x) = βn,q(x − [x]). Moreover, the errors has not been studied and our q−operator which is totally new, leads us to the more applicable function. That is why, this study situate q−Taylor theorem ﬁrst[6]

Theorem 3 If the function f(x) is capable of expansion as a convergent power series and if q 6= root of unity, then

∞ (x − a)n f(x)= q Dnf (a) (11) [n] ! q n=0 q X f(xq)−f(x) where Dq(f(x)) = x(q−1) is q−derivative of f(x), for x 6=0, we can deﬁne it at x =0 as a normal derivative.

Deﬁnition 4 We deﬁne Hq and Dh operators as follow

h h h h Hn(x)= + x + [2] x ... + [n − 1] x (1 − q)n (12) q 1 − q 1 − q 1 − q q 1 − q q n n x n k(k−1) n−k k = (x + h) ; n = q 2 (x + h) (−x) (13) k x + h q q Xk=0 F (x + h) − F (x) D (F (x)) = = f(x) (14) h h In this definition we assume that n ∈ N and the functions and values are well-defined (h 6=0, q 6= 1) . The fist equality is hold because of Heine’s binomial formula.

For a long time, mathematician have worked on the area of operators and they solved several types of diﬀerential equations by using shifted-operators. The H−operators rules like a bridge between the ordinary expansions and q-analogue of these expansions. We may rewrite several formulae of these area to the form of q−calculus such as [23, 24, 25, 26, 27]. Actually we may write q−expansion of these functions. In a letter, Bernoulli concerned the importence of this expansion for Leibnitz by these words” Nothing is more elegent than the agreement, which you have observed between the numerical power of the binomial and diﬀeretial expansions, there is no doubt that something is hidden there” [26]

Theorem 5 (Fundemental Theorem of h-Calculus) If F (x) is an h-derivative of f(x) and b − a ∈ hZ, we have[1]

f(x)dhx = F (b) − F (a) (15) Z where we deﬁne h-integral as follows

h (f(a)+ f(a + h)+ ... + f(b − h)) if ab  4 Now in the aid of q−Taylor expansion, we may write F (x + h) as follows;

∞ ((x + h) − x)j ∞ Dj Hj F (x + h)= q Dj F (x) = q q (F (x)) = e (D H ) (F (x)) (17) [j] ! q [j] ! q q q j=0 q j=0 q X X Here, eq is q−shifted operator and can be expressed as expansion of Dq and Hq. We may assume that Dh (F (x)) = f(x) and then

F (x + h) − F (x) e (D H ) − 1 f(x)= = q q q F (x) (18) h Hq

Since the Jackson integral is q−antiderivative, we have

H D q q f(x)d x = F (x) (19) e (D H ) − 1 q q q q Z And in the aid of (2), the left hand side of the equation can be expressed as a q-Bernoulli numbers, therefore

∞ (H D )n F (x)= β q q f(x)d x (20) n,q [n] ! q n=0 q X Z h ∞ β = f(x)d x − f(x)+ n,q Dn−1f(x) Hn(x) (21) q [2] [n] ! q q q n=1 q Z X ∞ n n−1 h n k(k−1) n−k k Dq f(x) βn,q = f(x)d x − f(x)+ q 2 (h + x) (−x) (22) q [2] k [n] ! q n=1 q ! q Z X Xk=0 Thus we can state the following q−analogue of Euler-Maclaurian formula Theorem 6 If the function f(x) is capable of expansion as a convergent power series and f(x) decrease so rapidly with x such that all normal derivatives approach zero as x → ∞, then we can express the series of function f(x) as follow

∞ ∞ 1 f(n)= f(x)d x + (f(a)) − (23) q [2] n=a a q X Z ∞ n n n−1 n k(k−1) n−k k n−k k h Dq f(a) βn,q lim q 2 (1 + b) (−b) − (1 + a) (−a) k b→∞ q ! [n]q! n=1 k=0 X X (24) Proof. In the aid of theorem (5) and (18), If we suppose that h = 1 and b − a ∈ N, we have

b−1 b f(n)= f(x)d1x = F (b) − F (a) (25) n=a X Za b 1 = f(x)d x − (f(b) − f(a)) (26) q [2] Za q ∞ n n−1 n−1 n k(k−1) n−k k n−k k Dq f(b) − Dq f(a) βn,q + q 2 (1 + b) (−b) − (1 + a) (−a) k q ! [n]q! n=1 k=0 X X (27)

5 −f(x) Let f(x) decrease so rapidly with x then Dqf(x) can be estimated by x(q−1) , tend x to inﬁnity and use ′ −f (x) L’Hopital to reach (q−1) .Now If normal derivatives of f(x) is tend to zero, then q−derivatives is also tending n−1 zero. The same discussion make Dq f(b) = 0 for big enough b. Therefore

∞ ∞ 1 f(n)= f(x)d x + (f(a)) − (28) q [2] n=a a q X Z ∞ n n n−1 n k(k−1) n−k k n−k k h Dq f(a) βn,q lim q 2 (1 + b) (−b) − (1 + a) (−a) k b→∞ q ! [n]q! n=1 k=0 X X (29)

Example 7 Let f(x)= xs where s is a positive integer, then we can apply this function at (22) where a =0 and b − 1 ∈ N

b−1 bs+1 bs ns = − (30) [s + 1] [2] n=0 q q ! X s+1 n s−n+1 n k(k−1) n−k k s b βn,q + q 2 (1+ b) (−b) (31) k n − 1 [n] n=2 q ! q q X Xk=0 This relation shows the sum of power in the combination of q−Bernoulli numbers. when q → 1 from the right side, we have an ordinary form of this relation. For the another forms of sum of power, see [22]

−x 1 Example 8 Let f(x) = e = x then this function decreases so rapidly with x such that all normal q Eq x derivatives approach zero as x → ∞. For comﬁrming this, let us mention that Eq , for some ﬁxed 0 < |q| < 1 ∞ ∞ 1 d x j k and |− x| < |1−q| is increasing rapidly, since dx Eq = (1 − q)q 1+(1 − q)q x > 0. j=1 k=0 P kQ6=j ∞ ∞ n 1 n k(k−1) n−k k βn,q −n 2 eq =1+ − lim q (1 + b) (−b) (32) [2] b→∞ k [n] ! n=0 q n=1 q ! q X X Xk=0 ∞ −n −1 3 ∞ β2n Moreover, this is q-analogue of n=0 e = e−1−1 = 2 + n=1 (2n)! , where β2n is a normal Bernoulli −1 numbers that is generated by − . e 1−1 P P 3 approximation by q-Bernoulli polynomial

We know that the class of q-Bernoulli polynomials are not in a form of orthogonal polynomials. In addition, we may apply Gram-Schmit algorithm to make these polynomials orthonormal then, according to Theorem 8.11 [19] we have the best approximation in the form of Fourier series. Now, Instead of us- 2 ing that algorithm, we apply properties of lemma (1) to achieve an approximation. Let H = Lq [0, 1] = 1 2 fq : [0, 1] → [0, 1] | fq (t) dqt< ∞ be the Hilbert Space[18], then βn,q(x) ∈ H for n = 0, ..., N and 0 Y = Span {β (x),βR (x), ..., β (x)} is ﬁnite dimensional vector subspace of H. Unique best approxima- 0,q 1,q N,q tion for any arbitrary elements of H like h, is h ∈ Y such that for any y ∈ Y the inequality h − h ≤ 2,q 1 1 2 b 2 b kh − yk2,q , where the norm is deﬁned by kfk2,q := fq (t) dqt . Following proposition determine the 0 R

6 2 coeﬃcient of q−Bernoulli polynomials, when we estimate any f ∈ Lq [0, 1] by truncated q−Bernoulli series. In fact, in order to see how well a certain partial sum approximates the actual value, we would like to drive a formula similar to (19), but with the inﬁnite sum on the right hand side replaced by the Nth partial sum SN, plus an additional term RN. Sippose a ∈ Z and b = a +1.Consider the Nth partial sum

N β S = k,q HkDk (f) (a + 1) − HkDk (f) (a) N [k] ! q q q q k=0 q X Let g(x, y)= βN,q(x,y) , then in the aid of lemma (1) we have Dn−kg(x, y)= βk,q (x,y) and we have [N]q ! q [k]q !

N βk,q(0, 0) k k βk,q(0, 1) k k SN = Hq Dq (f) (a + 1) − Hq Dq (f) (a) − Dq (f) (a) [k]q! [k]q! ! kX=0 N N−k k k N−k k k = Dq g(0, 0)Hq Dq (f) (a + 1) − Dq g(0, 1)Hq Dq (f) (a) − Dq (f) (a) kX=0 N N−k k k x=1 = Dq (f) (a) − Dq g(0, x)Hq Dq (f) (a +1 − x)|x=0 kX=0 By taking q−derivative of the summation, we may rewrite this partial sum by integral presentation, Moreover this relation give a boundary for RN (x).

2 N Proposition 9 Let f ∈ Lq [0, 1] and be estimated by truncated q−Bernoulli series n=0 Cnβn,q(x). Then 1 1 (n) Cn coeﬃcients for n =0, 1, ..., N can be calculated as an integral form Cn = [n] ! DqPf(x)dqx. In addition q 0 we can write f(x) in the following form R

N 1 1 h 1 f(x)= f(x)d x − f(x)+ D(n)f(x)d x β (x)+ R (x) q [2] [n] ! q q  n,q q,n 0 q n=1 q Z X Z0  2n  (n) Moreover Rq,n(x) as a reminder part is bounded by Supx∈[0,1]|βn,q(x)|Supx∈[0,1]|Dq f(x)| [N]q ! Proof. In fact we approximate f(x) as a linear combinations of q−Bernoulli polynomials and we assume N that f(x) ≃ n=0 Cnβn,q(x), so take the Jackson integral from both sides lead us to P 1 N 1 1 1 1 f(x)dq x ≃ Cn βn,q(x)dq x = C0 β0,q(x)dq x + C1 β1,q(x)dq x + ... + CN βN,q(x)dq x 0 n=0 0 0 0 0 Z X Z Z Z Z In the aid of Lemma 1 part (d) all the terms at the right side except the ﬁrst one has to be zero. we 1 1 calculate β0,q(x)=1, so C0 = [0] ! f(x)dqx.Using part (a) of that lemma for q−derivative of q−Bernoulli q 0 polynomial gather by taking q-derivativeR of f(x) leads to

1 N 1 1

Dq (f(x)) dqx ≃ Cn [n]q! βn−1,q(x)dq x = C1 β0,q(x)dq x = C1 0 n=0 0 0 Z X Z Z Repeating this procedure n-times, yields the form of Cn as we mention it at theorem. We mention that n n if x ∈ [0, 1] and 0 < |q| < 1, then Hq (x) for h = 1 is bounded by 2 .In addition, the reminder part can be presented by integral forms and this boundary can be found easily.

7 3.1 Further works In this paper, we introduced a proper tool to approximate a given capable function by combinations of q- Bernoulli polynomials. In spite of a lot of investigations on q-Bernoulli polynomials, study of approximation properties of the q-Bernoulli polynomials are not studied. We may apply these results to solve q-diﬀerence equation, q-analogue of the things that is done in [20] or [21]. The techniques of H-operator can be applied in operator theory.

4 conﬂict of interest

The authors declare that there is no conﬂict of interest regarding the publication of this paper

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