The presentation of a new type of quantum calculus
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1 DOI.55/tmj The presenttion of new type of quntum clculus Abdolli Nemty nd Mehdi Tourni b Deprtment of Mthemtics, University of Mzndrn, Bbolsr, Irn E-mil: nmty@umz.c.ir, mehdi.tourni@gmil.com b Abstrct In this pper we introduce new type of quntum clculus, the p-clculus involving two concepts of p-derivtive nd p-integrl. After fmilirity with them some results re given. 2 Mthemtics Subject Clssifiction. 5A3. 34A25 Keywords. p-derivtive, p-ntiderivtive, p-integrl. Introduction Simply put, quntum clculus is ordinry clculus without tking limit. In ordinry clculus, the derivtive of function f(x) is defined s f (x) = lim. However, if we void tking the f(y) f(x) y x y x limit nd lso tke y = x p, where p is fixed number different from, i.e., by considering the following expression: f(x p ) f(x) x p, (.) x then, we crete new type of quntum clculus, the p-clculus, nd the corresponding express is the definition of the p-derivtive. The formul (.) nd severl of the results derived from it which will be mentioned in the next sections, pper to be new. In [8] the uthors developed two types of quntum clculus, the q-clculus nd the h-clculus. If in the definition of f (x), s hs been stted bove, we do not tke limit nd lso tke y = qx or y = x + h, where q is fixed number different from, nd h fixed number different from, the q-derivtive nd the h-derivtive of f(x) re defined. For more detils, we refer the reders to [, 2, 4, 7]. Generlly, in the lst decdes the q-clculus hs developed into n interdisciplinry subject, which is briefly discussed in chpters 3 nd 7 of [3] nd lso hs interesting pplictions in vrious sciences such s physics, chemistry, etc [5, 6]. A history of the q-clculus ws given by T.Ernst [3]. The purpose of this pper is to introduce nother type of quntum clculus, the p-clculus, lso we re going to give some results by it. The pper hs been orgnized s follows. In section 2, we define the p-derivtive, lso some of its properties will be expressed. In section 3, we introduce the p-integrl, including sufficient condition for its convergence is given. In section 4, we will define the definite p-integrl, followed by the definition of the improper p-integrl. Finlly, we will conclude our discussion by fundmentl theorem of p-clculus. 2 p-derivtive Throughout this section, we ssume tht p is fixed number different from nd domin of function f(x) is [, + ). Corresponding uthor Tbilisi Mthemticl Journl (2) (27), pp Tbilisi Centre for Mthemticl Sciences. Received by the editors: 4 My 26. Accepted for publiction: 25 December 26. Unuthenticted Downlod Dte 6/7/8 8:5 AM
2 6 A. Nemty, M. Tourni Definition 2.. Let f(x) be n rbitrry function. We define its p-differentil to be d p f(x) = f(x p ) f(x). In prticulr, d p x =. By the p-differentil we cn define p-derivtive of function. Definition 2.2. Let f(x) be n rbitrry function. We define its p-derivtive to be nd D p f(x) = f(xp ) f(x) x p, ifx, x D p f() = lim x + D pf(x), D p f() = lim x D p f(x). Remrk 2.3. If f(x) is differentible, then lim p D p f(x) = f (x), nd lso if f (x) exists in neighborhood of x =, x = nd is continuous t x = nd x =, then we hve D p f() = f +(), D p f() = f (). Definition 2.4. The p-derivtive of higher order of function f is defined by (D pf)(x) = f(x), (D n p f)(x) = D p (D n p f)(x), n N. Exmple 2.5. Let f(x) = c, g(x) = x n nd h(x) = ln(x) where c is constnt nd n N. Then we hve (i) D p f(x) =, (ii) D p g(x) = g(xp ) g(x) x p = xpn x n x = x(p )n x p xn, (iii) D p h(x) = h(xp ) h(x) = (p ) ln(x) = (p ) ln(x) x p Notice tht the p-derivtive is liner opertor, i.e., for ny constnts nd b, nd rbitrry functions f(x) nd g(x), we hve x. D p (f(x) + bg(x)) = D p f(x) + bd p g(x). We wnt now to compute the p-derivtive of the product nd the quotient of f(x) nd g(x). Thus D p (f(x)g(x)) = f(xp )g(x p ) f(x)g(x) = f(xp )g(x p ) f(x)g(x p ) + f(x)g(x p ) f(x)g(x) = (f(xp ) f(x))g(x p ) + f(x)(g(x p ) g(x)) x p. x D p (f(x)g(x)) = g(x p )D p f(x) + f(x)d p g(x). (2.) Unuthenticted Downlod Dte 6/7/8 8:5 AM
3 The presenttion of new type of quntum clculus 7 Similrly, we cn interchnge f nd g, nd obtin D p (f(x)g(x)) = g(x)d p f(x) + f(x p )D p g(x), (2.2) which both of (2.) nd (2.2) re vlid nd equivlent. Here let us prove quotient rule. By chnging f(x) to f(x) in (2.), we hve g(x) nd thus Using (2.2) with functions f(x) g(x) Both of (2.3) nd (2.4) re vlid. D p f(x) = D p ( f(x) g(x) g(x)) = g(xp )D p ( f(x) g(x) ) + f(x) g(x) D pg(x), D p ( f(x) g(x) ) = g(x)d pf(x) f(x)d p g(x) g(x)g(x p. (2.3) ) nd g(x), we obtin D p ( f(x) g(x) ) = g(xp )D p f(x) f(x p )D p g(x) g(x)g(x p. (2.4) ) Note 2.6. We do not hve generl chin rule for p-derivtives, but in most cses we cn hve the following rule: D p [f(u(x))] = D p u(x)d h(x) f(u(x)), ln(u(x)) where h(x) is depended on u(x). Exmple 2.7. If α > nd u(x) = αx β, then becuse, u(x) D p [f(u(x))] = f(αxpβ ) f(αx β ) = f(αxpβ ) f(αx β ) αx pβ αx β αxpβ αx β = D ln(α) + pβ ln(x) f(u(x))d p u(x), ln(u(x)) ln(α) + pβ ln(x) ln(u(x)) = αx pβ. Exmple 2.8. If α > nd u(x) = αe x, then ) f(αe x ) D p [f(u(x))] = f(αexp ) f(αe x ) = f(αexp αe xp αe x αex p αe x = D ln(α) + x p f(u(x))d p u(x), ln(u(x)) Unuthenticted Downlod Dte 6/7/8 8:5 AM
4 8 A. Nemty, M. Tourni becuse, u(x) 3 p-integrl ln(α) + x p ln(u(x)) = αe xp. The first thing tht comes to our mind fter studying the derivtive of function is its integrl topic. Before investigting it, let us define p-ntiderivtive of function. Definition 3.. A function F (x) is p-ntiderivtive of f(x) if D p F (x) = f(x). It is denoted by F (x) = f(x)d p x. Notice tht s in ordinry clculus, the p-ntiderivtive of function might not be unique. We cn prove the uniqueness by some restrictions on the p-ntiderivtive nd on p. Theorem 3.2. Suppose < p <. Then, up to dding constnt, ny function f(x) hs t most one p-ntiderivtive tht is continuous t x =. Proof. Suppose F nd F 2 re two p-ntiderivtive of f(x) tht re continuous t x =. Let Φ(x) = F (x) F 2 (x). Since F nd F 2 re continuous t x = nd lso by the definition of p-derivtive tht led to D p Φ(x) =, we hve Φ is continuous t x = nd Φ(x p ) = Φ(x) for ny x. Since for some sufficiently lrge N >, Φ(x pn ) = Φ(x pn+ ) =... = Φ(x) nd lso by the continuity Φ t x =, it follows tht Φ(x) = Φ(). As ws mentioned we denote the p-ntiderivtive of f(x) by function F (x) such tht D p F (x) = f(x). Here we re going to construct the p-ntiderivtive. For this purpose, we use of n opertor. We define n opertor ˆM p, by ˆM p (F (x)) = F (x p ). Then we hve: ( ˆM p )F (x) = F (xp ) F (x) x p = D p F (x) = f(x). x Since ˆM p(f j (x)) = F (x pj ) for j {,, 2, 3,...} nd lso by the geometric series expnsion, we formlly hve F (x) = ˆM p ((x x p )f(x)) = ˆM p((x j x p )f(x)) = (x pj x pj+ )f(x pj ). (3.) It is worth mentioning tht we sy tht (3.) is forml becuse the series does not lwys converge. Definition 3.3. The p-integrl of f(x) is defined to be the series (x pj x pj+ )f(x pj ). (3.2) Remrk 3.4. Generlly, the p-integrl does not lwys converge to p-ntiderivtive. Here we wnt to give sufficient condition for convergence the p-integrl to p-ntiderivtive. Unuthenticted Downlod Dte 6/7/8 8:5 AM
5 The presenttion of new type of quntum clculus 9 Theorem 3.5. Suppose < p <. If f(x)x α is bounded on the intervl (, A] for some α <, then the p-integrl defined by (3.2) converges to function H(x) on (, A], which is p-ntiderivtive of f(x). Moreover, H(x) is continuous t x = with H() =. Proof. We consider the following two cses. Cse. x (, A]. Suppose f(x)x α < M on (, A]. For ny < x A, j Thus, for ny < x A, we hve Since f(x pj ) < M(x pj ) α < M. (x pj x pj+ )f(x pj ) (x pj x pj+ )M. (x pj x pj+ )M = M(x ), thus, it follows from the comprison test tht the p-integrl converges to function F (x). It follows directly from (3.) tht F () =. We wnt now to prove tht F (x) is p-ntiderivtive of f(x), but before of it let us show F is right continuous t x =. For < x A, F (x) = (x pj x pj+ )f(x pj ) M(x ), which pproches s x +. Since F () =, thus F is right continuous t x =. To prove tht F (x) is p-ntiderivtive, it is sufficient to p-differentite it: D p F (x) = F (xp ) F (x) x = (xpj+ pj+2 )f(x pj+ ) x (xpj pj+ )f(x pj ) = f(x). Cse 2. x (, ). Suppose f(x)x α < M on (, ). For ny < x <, j f(x pj ) < M(x pj ) α Mx α. Thus, for ny < x <, we hve (x pj x pj+ )f(x pj ) (x pj+ x pj )Mx α, nd since (x pj+ x pj )Mx α = Mx α ( x), Unuthenticted Downlod Dte 6/7/8 8:5 AM
6 2 A. Nemty, M. Tourni hence, it follows from the comprison test tht the p-integrl converges to function G(x) nd by (3.) we hve G() =. Similr to proof of cse, it is esy to verify tht G is left continuous t x = nd is lso p-ntiderivtive of f(x). We now define H(x) = G(x)χ (,) (x) + F (x)χ (,A] (x). It is esy to see p-integrl converges to H(x) on (, A] nd lso H(x) is p-ntiderivtive of f(x) on (, ) (, A] nd is continuous t x = with H() =. If f(x) is continuous in x =, then D p H() = lim D p H(x) = f() nd it concludes tht H(x) is p-ntiderivtive of f(x) on (, A], x hence the proof is complete. Corollry 3.6. If the ssumption of Theorem 3.5 is stisfied, then by Theorem 3.2, the p-integrl gives the unique p-ntiderivtive tht is continuous t x =, up to dding constnt. Exmple 3.7. Let < p < nd f(x) = c, i.e., f(x) is constnt. Since for α <, f(x)x α is bounded on intervl (, A], hence by Theorem 3.5, p-integrl of f(x) converges whose it is vlid, becuse (x pj x pj+ )f(x pj ) = c (x pj x pj+ ) = c(x )χ (,A] (x). Exmple 3.8. Let < p < nd f(x) = x x p. The p-integrl gives (x pj x pj+ )f(x pj ) = (x pj x pj+ ) =, x pj x pj+ nd lso f(x)x α is not bounded on (, ) (, A] nd α <. 4 Definite p-integrl We now re in position to define the definite p-integrl. Generlly, one of the principle tools to define the definite p-integrl of function is use of prtition on set. we will use of it to chieve our gol. As proof of the Theorem 3.5, we consider the following three cses. Then, the definite p-integrl relted to ech cse is given. Cse. Let < < b where, b R +, p (, ) nd function f is defined on (, b]. Notice tht for ny j {,, 2, 3,...}, b pj (, b]. We now define the definite p-integrl of f(x) on intervl (, b]. Definition 4.. The definite p-integrl of f(x) on the intervl (, b] is defined s lim (b pj b pj+ )f(b pj ) = (b pj b pj+ )f(b pj ), (4.) nd f(x)d p x f(x)d p x. (4.2) Unuthenticted Downlod Dte 6/7/8 8:5 AM
7 The presenttion of new type of quntum clculus 2 Note 4.2. Geometriclly, the integrl in (4.) corresponds to the re of the union of n infinite number of rectngles. On [ + ε, b], where ε is smll positive number, the sum consists of finitely mny terms, nd is Riemnn sum. Therefore, s p, the norm of prtition pproches zero, nd the sum tends to the Riemnn integrl on [ + ε, b]. Since ε is rbitrry, provided tht f(x) is continuous in the intervl [, b], thus we hve lim p f(x)dx. Exmple 4.3. Let b = 3 nd f(x) = c where c is constnt. 3 cd p x = lim (3 pj 3 pj+ )f(3 pj ) = c lim [(3 3p ) + (3 p 3 p2 ) + (3 p2 3 p3 ) (3 pn 3 pn+ )] = c lim [3 3pN+ ] = c(3 ) = 2c, nd if = 2, 3 2 cd p x = 3 cd p x Exmple 4.4. Let b = 2 nd f(x) = ln(x) x x p. 2 2 cd p x = 2c c = c. ln(2 pj ) (2 pj 2 pj+ ) = p j ln(2) = ln(2) 2 pj 2 pj+ p. Cse 2. Let < < b < nd p (, ). It should be noted tht for ny j {,, 2, 3,...}, b pj [b, ) nd b pj < b pj+. We will define the definite p-integrl of f(x) on intervl [b, ) s follows. Definition 4.5. The definite p-integrl of f(x) on the intervl [b, ) is defined s b lim (b pj+ b pj )f(b pj ) = (b pj+ b pj )f(b pj ). Exmple 4.6. Let b = 2 nd f(x) = c 2 cd p x = lim (( ( )c 2 )pj+ 2 )pj = c lim [(( 2 )p ( 2 )) + (( ( 2 )p2 2 )p ) + (( ( ) (( )pn+ ( )] 2 )p3 2 )p2 2 2 )pn = c lim [ 2 + ()pn+ ] = c( ) = 2 c. Unuthenticted Downlod Dte 6/7/8 8:5 AM
8 22 A. Nemty, M. Tourni Note 4.7. The bove two definite p-integrls re lso denoted by b I p +f(b), I p f(b). Cse 3. Let < < b < nd p (, ). Then for ny j {,, 2, 3,...}, b p j b p j < b p j. Let us stte the definite p-integrl of f(x) on intervl (, b]. (, b] nd Definition 4.8. The definite p-integrl of f(x) on the intervl (, b] is defined s I p f(b) = lim (b p j b p j )f(b p j ) = (b p j b p j )f(b p j ), (4.3) nd Exmple 4.9. Let = 4, b = 2 2 cd p x = lim nd f(x) = c. (( ( )c 2 )p j 2 )p j f(x)d p x f(x)d p x. (4.4) = c lim [(( 2 ) ( ) + (( ( ) (( ( )] 2 )p 2 )p 2 )p 2 2 )p N 2 )p N = c lim [( 2 ) ()p N ] = 2 2 c. Similrly, thus we hve 4 cd p x = 4 c, 2 4 cd p x = 2 4 cd p x cd p x = 4 c. Note 4.. We cn lso pply Note 4.2 for the p-integrls defined in the cses 2 nd 3 on the intervls [b, ε] nd [ε, b] respectively, nd by it define the Riemnn integrl. Definition 4.. Suppose < < b. Then by Note 4.2 nd Note 4., we hve f(x)d p x + f(x)d p x. Corollry 4.2. By the definitions of p-integrls, we derive more generl formul: Unuthenticted Downlod Dte 6/7/8 8:5 AM
9 The presenttion of new type of quntum clculus 23 i) If b >, ii) If < b <, Becuse, f(x)d p g(x) = f(x)d p g(x) = f(b pj )(g(b pj ) g(b pj+ )). f(b p j )(g(b p j ) g(b p j )). f(x)d p g(x)d p x = = = (b pj b pj+ )(f(b pj )D p g(b pj )) g(bpj+ ) g(b pj ) (b pj b pj+ )f(b pj )( ) b pj+ b pj f(b pj )(g(b pj ) g(b pj+ )). Since D p g(x) = d pg(x), hence we hve d p x Similrly, it is esy to prove (b). f(x)d p g(x) = f(b pj )(g(b pj ) g(b pj+ )). Definition 4.3. The p-integrl of higher order of function f is given by 5 Improper p-integrl (I pf)(x) = f(x), (I n p f)(x) = I p (I n p f)(x), n N. In this section we wnt to define the improper p-integrl of f(x) nd lso give sufficient condition for its convergence. We strt this section by computing the following p-integrl. Let p (, ), thus p > nd consider p = b. For ny j {, ±, ±2,...}, we hve b pj >, b pj+ < b pj nd thus ccording to (4.2), we obtin p j b pj+ = = p j p j+ f(x)d p x f(x)d p x ((b pj ) pk (b pj ) pk+ )f((b pj ) pk ) ((b pj+ ) pk (b pj+ ) pk+ )f((b pj+ ) pk ) k= k= k= (b pk+j b pk+j+ )f(b pk+j ) (b pk+j+ b pk+j+2 )f(b pk+j+ ), k= Unuthenticted Downlod Dte 6/7/8 8:5 AM
10 24 A. Nemty, M. Tourni nd thus, p j b pj+ (b p We now define the improper p-integrl s follows. j b pj+ )f(b pj ). Definition 5.. Let p (, ) nd p = b. The improper p-integrl of f(x) on [, + ) is defined to be j= p j b pj+ = (b pj b pj+ )f(b pj ) j= (b pj b pj+ )f(b pj ) + (b p j b p j+ )f(b p j ). j= Definition 5.2. If p (, ), then for ny j {, ±, ±2,...}, we hve p pj (, ), p pj < p pj+ nd Becuse, ccording to (4.4) p p j+ p pj p p j+ j= (p pj+ p pj )f(p pj ). p p j f(x)d p x (p pj+ p pj )f(p pj ). Hence, p p j+ j= j= p pj (p pj+ p pj )f(p pj ). Definition 5.3. Let p (, ). Then the improper p-integrl of f(x) on [, ] is defined to be f(x)d p x + f(x)d p x. Definition 5.4. Let p (, ). Then the improper p-integrl of f(x) on [, ] is defined s follows: i) If < <, then ii) If >, then f(x)d p x + lim j= p j f(x)d p x. p j+ f(x)d p x. Unuthenticted Downlod Dte 6/7/8 8:5 AM
11 The presenttion of new type of quntum clculus 25 Here we give sufficient condition for convergence the improper p-integrl. Proposition 5.5. Let p (, ), < r < nd ε is smll positive number. inequlity f(x) < min{rx α, x x p (ln x) 2α } Assume tht holds in neighborhood of x = with some α < nd for sufficiently lrge x with some ε α <. Then, the improper p-integrl of f(x) converges on [, ). Proof. Consider b = p. According to Definition 5., we hve (b pj b pj+ )f(b pj ) + (b p j b p j+ )f(b p j ). By the ssumptions nd lso Theorem 3.5, the convergence of the first sum is proved. For lrge x, we hve f(x) < x x p (ln x) 2α where ε α <. Then, we hve for sufficiently lrge j, j= Hence f(b p j ) < (b p j b p j+ ) (ln b p j ) 2α. (b p j b p j+ )f(b p j ) (b p j b p j+ )(b p j b p j+ ) (ln b p j ) 2α = (ln b p j ) 2α = (p j ln b) 2α = (ln b) 2α (p 2α ) j. Therefore, by the comprison test, the second sum lso converges. 6 Fundmentl Theorem of p-clculus Since we re fmilir with the concepts of p-derivtive nd p-integrl, so we re going to study the reltion between them s follows. We begin this section with the following lemm. Lemm 6.. If x > nd p (, ), then D p I p +f(x) = f(x), nd lso if function f is continuous t x =, then we hve I p +D p f(x) = f(x) f(). Proof. According to definitions of p-derivtive nd p-integrl, we hve Hence I p +f(x) = x f(s)d p s = (x pj x pj+ )f(x pj ). D p I p +f(x) = I p +f(xp ) I p +f(x) x = (xpj+ pj+2 )f(x pj+ ) x (xpj pj+ )f(x pj ) = [(xp x p2 )f(x p ) + (x p2 x p3 )f(x p2 ) +...] [(x x p )f(x) + ( p2 )f(x p ) +...] = (xp x)f(x) x p = f(x). x Unuthenticted Downlod Dte 6/7/8 8:5 AM
12 26 A. Nemty, M. Tourni Also I p +D p f(x) = lim = lim = lim (x pj x pj+ )D p f(x pj ) f(xpj+ ) f(x pj ) (x pj x pj+ )( ) x pj+ x pj (f(x pj ) f(x pj+ )) = lim (f(x) f(xpn+ )) = f(x) f(). The lst equlity is true becuse f is continuous t x =. following lemms. Similrly, it is esy to obtin the Lemm 6.2. If x, p (, ), then D p I p f(x) = f(x), nd lso if function f is continuous t x =, then we hve I p D p f(x) = f() f(x). Lemm 6.3. If x, p (, ) nd I p f(x) = x f(s)d ps, then D p I p f(x) = f(x) nd lso if function f is continuous t x =, then we hve I p D p f(x) = f(x) f(). We re now in position to express fundmentl theorem for p-clculus. Theorem 6.4. (Fundmentl theorem of p-clculus) Let p (, ). If F (x) is n ntiderivtive of f(x) nd F (x) is continuous t x = nd x =, then for every < b, we hve Proof. We consider the following cses. F (b) F (). (6.) Cse. Let < < b nd,b re finite. Since F (x) is n ntiderivtive of f(x), hence D p F (x) = f(x). By Lemm 6., we hve which implies, Using (4.2), thus we hve F (x) F () = I p +f(x) = f(s)d p s = F () F (), x F (b) F (). f(s)d p s, f(s)d p s = F (b) F (). Unuthenticted Downlod Dte 6/7/8 8:5 AM
13 The presenttion of new type of quntum clculus 27 Cse 2. Let < b <. Since D p F (x) = f(x), by Lemm 6.3, we hve which implies, Using (4.4), thus we hve F (x) F () = I p f(x) = f(s)d p s = F () F (), x F (b) F (). f(s)d p s, f(s)d p s = F (b) F (). Cse 3. Let < < < b nd b is finite. According to Note 4. nd lso by Lemm 6.2, we hve Similrly, Thus I p f() = I p D p F () = F () F (). I p +f(b) = I p +D p F (b) = F (b) F (). F (b) F (). For b = +, without loss of generlity, we consider > nd by the Definition 5.4, we hve + lim j= = lim j= p j p j+ f(x)d p x (F ( p j ) F ( p j+ )) = lim (F (p N ) F ()), nd if lim x F (x) exists, so (6.) is true for b =. Corollry 6.5. If f(x) is continuous t x = nd x =, then we hve D p f(b) f(). Corollry 6.6. If f(x) nd g(x) re continuous t x = nd x =, then we hve f(x)d p g(x) = f(b)g(b) f()g() g(x p )d p f(x). Unuthenticted Downlod Dte 6/7/8 8:5 AM
14 28 A. Nemty, M. Tourni Proof. Using the product rule (2.), we hve D p (fg)(x)d p x = By Corollry 6.5, we hve f(b)g(b) f()g() = Since, D p g(x)d p x = d p g(x), thus (f(x)d p g(x) + g(x p )D p f(x))d p x. f(x)d p g(x)d p x + f(x)d p g(x) = f(b)g(b) f()g() g(x p )D p f(x)d p x. g(x p )d p f(x). References [] M.H. Annby, Z.S. Mnsour, q-frctionl Clculus nd Equtions, Springer-Verlg, Berlin Heidelberg, 22. [2] A. Arl, V. Gupt, R.P. Agrwl, Applictions of q-clculus in Opertor Theory, New York, Springer, 23. [3] T. Ernst, The history of q-clculus nd new method, Thesis, Uppsl University, 2. [4] T. Ernst, A comprehensive tretment of q-clculus, Springer Science, Business Medi, 22. [5] R.J. Finkelstein, Symmetry group of the hydrogen tom, J. Mth. Phys. 8 (967), no. 3, [6] R.J. Finkelstein, The q-coulomb problem, J. Mth. Phys. 37 (996), no. 6, [7] K.R. Prthsrthy, An introduction to quntum stochstic clculus, Springer Science, Business Medi, 22. [8] V. Kc, P. Cheung, Quntum clculus, Springer Science, Business Medi, 22. Unuthenticted Downlod Dte 6/7/8 8:5 AM
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