MEAN VALUE THEOREMS IN q-calculus. Predrag M. Rajkovic, Miomir S. Stankovic, Sladana D. Marinkovic

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1 MATEMATHQKHBECHHK 54 (2002), UDK oprrr-anaznra HaY'-IHVl pazr research paper MEAN VALUE THEOREMS IN q-calculus Predrag M. Rajkovic, Miomir S. Stankovic, Sladana D. Marinkovic Abstract. In this paper, some properties of continuous functions in q-analysis are investigated. The behavior of q-derivative in a neighborhood of a local extreme point is described. Two theorems are proved which are q-analogons of the fundamental theorems of the differential calculus. Also, two q-integral mean value theorems are proved and applied to estimating remainder term in q-taylor formula. Finally, the previous results are used in considering some new iterative methods for equation solving. 1. Introduction At the last quarter of the XX century, q-ca1culus appeared as a connection between mathematics and physics ([5], [6]). It has a lot of applications in different mathematical areas, such as number theory, combinatorics, orthogonal polynomials, basic hyper-geometric functions and other sciences-quantum theory, mechanics and theory of relativity. Let q E ffi.+ \ {I}. A q-natural number [n] q is defined by [n]q := 1 + q «:'. n E N. 1 _ qn Generally, a q-complex number [a]q is [a]q := -1--' -q number [n]q is [O]q! := 1, [n]q!:= [n]q[n - l]q'" [1]q, ti E N. Let q-derivative of a function j(z) be a E c.. The factorial of a (Dqj)(z) := j(z) j(qz), z #- 0, (Dqj)(O):= lim(dqj)(z), z - qz z--+o and high q-derivatives are D~j:= j, D~ j := Dq(D~-l1), n = 1,2,3,.... Notice, that a continuous function on an interval, which does not include 0, IS continuous q-differentiahle. AMS Subject Classification: 26 A 99 Keywords and phrases: q-calculus. Communicated at the 5th International Symposium on Mathematical Analysis and its Applications, Niska banja, Yugoslavia, October, 2-6, Supported by Ministry of Sci. & Techn. Rep. Serbia, the projects No. 1409/

2 172 P. M. Rajkovic, M. S. Stankovic, S. D. Mariukovic 2. Extreme values and q-derivative We will consider relations between, on one side, extreme value of a continuous function and, on the other side, derivatives and q-derivatives. THEOREM 2.1. Let f(x) be a continuous function on a segment [a, b] c E (a, b) bea point of its local maximum. (i) If 0 < a < b, then there exists q E (0,1) such that and let D { 2: 0, v«e (q,l) ( qf)(c) < 0, v«e (1,q-1). (ii) If a < b < 0, then there exists q E (0,1) such D )(){:::; 0, v«e (q,l) ( qf c 2: 0, v«e (1, q-1). Furthermore, ('iq E (q, 1) U (1, q-1) )(3~ E (a, b)) (Dqj)(~) = o. Proof. Since the proofs of (i) and (ii) are very similar, we will expose only the first one. Since c is a point of local maximum of the function f (x), there exists e > 0, such that f(x) :::; f(c), for all x E (c - c, c + c) C (a, b). Let qo E (0,1) such that c - e < qoc < c. Now, for all q E (qo, 1), it is valid qc < c and f(qc) :::; f(c), wherefrom (Dqf)(c) 2: O. In a similar way, there exists qi E (0,1) such that c < C/q1 < c+c and for all q E (1,ql 1) it holds (Dqf)(c):::; o. At last, denote by q = max{qo,qd Let q E (q, 1) be an arbitrary real number. Then 1] = c/q E (c, c+c), wherefrom f(c) 2: f(1]), i.e., f(q1]) 2: f(1]) From q1] < 1] we conclude (Dqf)(1]) :::; o. As f(x) is a continuous function, (Dqf)(x) is continuous in (a, b), too. Since (Dqf)(c) 2: 0, (Dqf)(1]) :::; 0, where C,1] E (a, b), there exists ~ E (c,1]) C (a, b), such that (Dqf)(~) = o. Analogously, for an arbitrary q E (1, q-1), the number 1] = cl q E (c - c, c), wherefrom (Dqf)(1]) 2: O. Since (Dqf)(c) :::; 0, we have proved the existence of a zero of (Dqf)(x) for q E (1, q-1). EXAMPLE 2.1. Let us consider f(x) = (x - 1)(3 - x) + 2. Its maximum is at c = 2, but q-derivative is (Dqf)(x) = -[2]x + 4 and it vanishes at the point ~ = 4/(1 + q). So, here is q = 1/3. For q = 3/4, we have (D 3 / 4 f )(2) = 1/2, 1] = 2~ and ~ = 2~. In a similar way, we can prove the next theorem. THEOREM 2.2. Let f(x) be a continuous function on a segment [a, b] c E (a, b) be a point of its local minimum. (i) If 0 < a < b, then there exists q E (0,1) such that and let D { :::; 0, v«e (q,l) ( qf)(c) 2: 0, v«e (1,q-1).

3 Mean value theorems in q-calculus 173 (ii) If a < b < 0, then there exists q E (0,1) such that ( D )(){;:::: 0, v«e (q,l) qf c :s 0, v«e (1, q-i). Moreover, ('iq E (q, q-i)) (::J~ E (a, b)) (Dq.f) (~) = 0. REMARK. If f(x) is differentiable for all x E (a, b), then limqtl Dqf(x) = j'(x). So, if c E (a,b) is a point oflocal extreme of f(x), we have j'(c) = DIf(c) = Some q-mean value theorems By using the previous results, we can establish and prove analogons of wellknown mean value theorems in q-calculus. THEOREM 3.1. (q-rolle) Let f(x) be a continuous function on [a, b] f(a) = f(b). Then there exists q E (0,1) such that ('iq E (q, 1) U (1,q-I)) (::J~ E (a, b)): (Dqf)(~) = 0. satisfying Proof. If f(x) is not a constant function on [a, b], then it attains its extreme value in some point in (a,b). But, according to Theorems 2.1-2, (Dq)f(x) vanishes at a point ~ E (a, b). - THEOREM 3.2 (q-lagrange) Let f(x) be a continuous function on [a, b]. there exists q E (0, 1) such that Then ('iq E (q, 1) U (1,q-I)) (::J~ E (a, b)): f(b) - f(a) = (Dqf)(~)(b - a). Proof. The statement follows by applying the previous theorem to the Iunction f(x) - x(f(b) - f(a))/(b - a) Mean value theorems for q-integrals In q-analysis, we define q-integral by Notice that I(f) = r f(t) dt = lim Iq(f). io qtl THEOREM 4.1 Let f(x) be a continuous function on a segment [O,a] (a> 0). Then ('iq E (0, 1))(::J~ E [0, ad: Iq(f) = l a f(t) dq(t) = a f(~).

4 174 P. M. Rajkovic, M. S. Stankovic, S. D. Mariukovic Proof. Since f(x) is a continuous function on the segment [O,a], it attains its minimum m and maximum M and takes all values between. According to assumption 0 < q < 1, we have 0 < aq" < a and m :S f(aqn) :S M. Now, = = = a(1- q) L mq" :S a(1- q) L f(aqn)qn :S a(1- q) L Mqn, n-o n-o n-o wherefrom m :S ~ 1q(j) :S M. So, there exists ~ E [O,a] such that a-i 1q(j) = f(~) - Moreover, if we define then the next theorem is valid. THEOREM 4.2. Let f(x) be a continuous function on a segment [a,b]. there exists q E (0, 1) such that (Vq E (q, 1))(:3~ E (a, b)): 1q(j) =lb f(t) dq(t) = f(~)(b - a). Then Proof. It is easy to prove that lim q t1 1q(j) = 1(j), i.e. (VE> O)(:3qO E (0, l))(vq E (qo, 1)): 1(j) - E < 1q(j) < 1(j) + E. According to the well known mean value theorem for integrals, we have (:3c E (a, b)): 1(j) = f(c)(b - a). Let E:S (b - a) min{m - f(c), f(c) - m}, where m and M are the minimum and maximum of f(x) on [a, b]. Now, (:3q E (0, l))(vq E (q, 1)): f(c) - b ~ a < b ~ a 1q(j) < f(c) + b ~ a' hence m < 1q(j)j(b- a) < M. Since f(x) is a continuous function on the segment [a, b], it takes all values between m and JM, i.e. what we wanted to prove. _ THEOREM 4.3 Let f(x) and g(x) be some continuous functions on a segment [a, b]. Then there exists q E (0,1) such that (Vq E (q, 1))(:3~ E (a, b)): 1q(jg) = g(~)1q(j).

5 Mean value theorems in q-calculus 175 Proof. According to the second mean value theorem for integrals, we have (:=Jc E (a, b)): l(fg) = g(c)l(f). Hence lim lq(fg) = g(c)l(f) = g(c) lim lq(f), i.e., lim lq((fg)) = g(c). Now, qt1 qt1 qt1 l q f I (fg) (:=JqO E (0, l))(\fq E (qo' 1)): g(c) - e < ;q(f) < g(c) + c. Since g(x) is a continuous function on the segment [a, b]' it attains its minimum m g and maximum Mg. Let c:s min{mg - g(c),g(c) - m g}. Hence (:=Jq E (0, l))(\fq E (q, 1)) : lq(fg) m g < lq(f) < Mg. Since f(x) takes all values between m g and M g, we conclude that lq(fg) (:=J~ E (a, b)): lq(f) = g(~). 5. Estimation of remainder term in q-taylor formula Let f(x) be a continuous function on some interval (a,b) and c E [a,b]. Jackson's q-taylor formula (see [3], [4] and [2]) is given by where = ~ (D~f)(c) (~_ )(k) (b) f( z) LJ [k]' c c, z E a,, k=o q' k-l (z - c)(o) = 1, (z - c)(k) = IT (z - cqi) (k EN). i=o T. Ernst [2] have found the next q-taylor formula _ n-l (D~f)(c) _ (k) f(z) - k'f [k]q! (z c) + Rn(f, z, c, q), o where Rn(f, z, c, q) r is the remainder term determined by (z - t)(n) (Dn j)(t) Rn(f, z, c, q) = [ q ]' dq(t). t=c z - t n - 1 q. (5.1) (5.2) THEOREM 5.1. Let f(x) be a continuous function on [a,b] and Rn(f,z,c,q), z, c E (a, b) be the remainder term in q-taylor formula. Then there exists q E (0,1) such that for all q E (q, 1), ~ E (a, b) can be found between c and z, which satisfies _ (D~ j)(~)t: (z - t)(n) Rn(f, z, c, q) - [ ], dq(t). n - 1 q' t=c z - t (5.3)

6 176 P. M. Rajkovic, M. S. Stankovic, S. D. Mariukovic Proof. Since f(x) is a continuous function on [a, b], it can be expanded by q-taylor formula (5.1) with the remainder term (5.2). Notice that the functions (z _ t)(n) n-1. Z _ t = II(z - tq') ;=1 and (D~f) (t)/ [n - 1]q! are continuous on the segment between c and z which is contained in (a, b). According to Theorem 4.3., there exists ij E (0,1), such that for all q E (ii, 1) can be found ~ between c and z such that (5.3) is valid. - THEOREM 5.2. Let f(x) be a continuous function on [a,b] and z,c E (a, b). Then there exists (j E (0,1) such that for all q E (ii, 1), ~ E (a, b) can be found between c and z, which satisfies f(z) = nf:1 (D; f),(c) (z _ C)(k) + (D~ f),(~) (z _ c)(n). k=o [k]q. [n]q. (z - t)(n) ((z - t)(n)) Proof. Applying z _ t = -Dq,t [n]q, to the integral in (5.3) we have r:(z _ t)(n) _ r: ((z _ t).(n)) It=c z - t dq(t) - - It=c Dq,t [n]q dq(t) = _ (z - t)(n) It=z (z - c)(n) [n]q t=c S (f ) (D~ f)(~) ( _ )(n) 0, R n,z, c, q - []' z c._ n q' THEOREM 5.3 Let f(x) be a continuous function on [a,b] and Rn(j,z,c,q), z, c E (a, b) be the remainder term in q-taylor formula. Then there exists ij E (0,1) such that for all q E (ii, 1), ~ E (a, b) can be found between c and z, which satisfies [n]q _ (D~f)(~)( _ )(n) Rn ([, z, c, q) - []' z c. PTOOJ. Since f(x) is a continuous function on [a, b]' it can be expanded by q-taylor formula (5.1) with the remainder term (5.2). Notice that the functions n q' (z _ t)(n) n-1. Z _ t = II(z - tq') ;=1 and (D~f) (t)/ [n - 1]q! are continuous on the segment between c and z which is contained in (a, b). According to Theorem 4.3., there exists ij E (0,1), such that for all q E (ii, 1), ~ between c and z can be found such that _ (D~ f)(~)t: (z - t)(n) Rn(j, z, c, q) - [ ], dq(t). n - 1 q' t=c z - t

7 Mean value theorems in q-calculus 177 Applying (z - t)e n) = -D ((z - t)e n)) z - t q.t: [n]q, to the previous integral we have I t =z(z - t)e n) _ I t =z ((z - t)e n)) t=c Z - t dq(t) - - t=c Dq,l [n]q dq(t) = _ (z - t)en)it=z (z - c)e n ) [n]q t=c S (f ) (D~ f)(~) ( )en ) o,rn,z,c,q - []' z-c. n q' [n]q 6. Application Here we will apply the previous theorems in analyzing an iterative method for solving equations. Suppose that an equation f (x) = 0 has a unique isolated solution x = T. If X n is an approximation for the exact solution T, using Jackson's q-taylor formula, we have 0= f(t) :::::: f(xn) + (Dqf)(xn)(T - x n), f(x n ) hence T :::::: Xn - (Dqf)(x n) So, we can construct q-neuitori method (6.1) More simply, it looks like Xn+l = Xn{ 1-1 ~~ }. This method written in the form fex n ) Xn+l = X n - f(x:) =~~;Xn) f(x n) reminds to the method of chords (secants). THEOREM 6.1. Suppose that a function f(x) is continuous on a segment [a, b] and that the equation f (x) = 0 has a unique isolated solution T E (a, b). Let the conditions I(Dqf)(x)1 2: M 1 > 0, I(D~f)(x)1 :::; M 2 are satisfied for all x E (a,b). Then there exists I} E (0,1), such that for all q E (I}, 1), the iterations obtained by q-neuitoti method satisfy

8 178 P. M. Rajkovic, M. S. Stankovic, S. D. Mariukovic Proof. From the formulation of q-newton method (6.1), we have f(xk) Xk+l - T = Xk - T - (Dqf)(Xk) ' hence f(xk)+(dqf)(xk)(t-xk) = (Dqf)(Xk)(T-Xk+d. By using q-taylor formula at the point x k of order n = 2 for f (T) we have f(t) = f(xk) + (Dqf)(Xk)(T - Xk) + R2(f,T, Xk, q). Since f(t) = 0, we obtain (Dqf)(Xk)(T - xk+d = I - 1- IR2(f,T,Xk,q)1 T Xk+l - I(Dqf)(Xk)1. According to Theorem 5.l., there exists q E ~ E (a, b) can be found such that Now, -R 2(f,T,xk,q), i.e. _ (D~f)(~) (_ (2) R 2 ( f,t,xk,q) - [2]q T Xk). I(D~f)(~)1 I(T - Xk)(2)1 IT - Xk+ll = I(Dqf)(Xk)1 1 + q (0,1) such that for all q E (1i,1), Using the conditions which function f(x) and its q-derivatives satisfy we obtain the statement of the theorem. _ REMARK. In our papers [7] and [8] we have discussed q-iterative methods in details. REFERENCES [1: N. S. Bakhvalov, Numerical Methods, Moscow State University, Moscow, [2: T. Ernst, A new notation for q-calculus and a new q- Taylor formula, Uppsala University Report Depart. Math., (1999), [3: F.H. Jackson, A q-form of Taylor's theorem, Mess. Math. Math. 38, (1909), [4: S. C. Jing and H. Y. Fan, q- Taylor's formula with its q-remainder, Comm. Theoret. Phys. 23, 1 (1995), [5: E. Koelink, Eight lectures on quantum groups and q-special functions, Revista Colombiana de Matematicas 30, (1996), [6: T. H. Koornwinder and R. F. Swarttow, On q-analogues of the Fourier and Hankel transforms, Trans. Amer. Math. Soc. 333, (1992), [7: P. M. Rajkovic, M. S. St.aukovic and S. D. Marinkovic, On q-iterative methods for equation 8ulving, Conf. Applied Math. PRIM, Zlatibor, Yugoslavia, [8: M.S. Stankovic, P.M. Rajkovic and S.D. Marinkovic, Solving equations by q-iterative methods and q-sendov conjecture, Constructive Function Theory (Ed. B. Bojanov), Varna Bulgaria (received ) Predrag M. Rajkovic, Faculty of Mechanical Engineering, University of Nis, Nis, Yugoslavia pecar@masfak.ni.ac.yu Miomir S. Stankovic, Faculty of Occupational Safety, University of Nis, Nis, Yugoslavia mstan@znrfak.znrfak.ni.ac.yu Sladana D. Marinkovic, Faculty of Electrical Engineering, University of Nis, Nis, Yugoslavia sladjana@elfak.ni.ac.yu