THÔNG TIN TÀI LIỆU
Mathematical and Computer Modelling 52 (2010) 522–528 Contents lists available at ScienceDirect Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm New inequalities of Simpson-like type involving n knots and the mth derivative ´ Vu Nhat Huy a , Quôc-Anh Ngô a,b,∗ a Department of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viet Nam b Department of Mathematics, National University of Singapore, Block S17 (SOC1), 10 Lower Kent Ridge Road, Singapore 119076, Singapore article abstract info Article history: Received May 2009 Received in revised form March 2010 Accepted 22 March 2010 Based on recent results due to Nenad Ujević, we obtain some new inequalities of Simpsonlike type involving n knots and the mth derivative where n, m are arbitrary numbers Our method is also elementary © 2010 Elsevier Ltd All rights reserved Keywords: Inequality Error Integral Taylor Simpson Introduction In recent years, a number of authors have considered error inequalities for some known and some new quadrature formulas Sometimes they have considered generalizations of these formulas, for example, the Simpson inequality (which gives an error bound for the well-known Simpson rule) is considered in [1–10] In [5], we can find the inequality, b f (t ) dt − b−a a f (a) + 4f a+b + f ( b) Γ −γ 12 ( b − a) , (1) where Γ , γ are real numbers, such that γ < f (t ) < Γ , t ∈ [a, b] We define the Chebyshev functional, T (f , g ) = b b−a f (t ) g (t ) dt − a b ( b − a) b g (t ) dt f (t ) dt a a Then T (f , f ) = b−a f L2 − (b − a)2 b f (t ) dt a We also define σ (f ) = (b − a)T (f , f ) In [10], the author proved the following result ∗ Corresponding author at: Department of Mathematics, College of Science, Viêt Nam National University, Hà Nôi, Viet Nam E-mail addresses: nhat_huy85@yahoo.com (V.N Huy), bookworm_vn@yahoo.com (Q.-A Ngơ) 0895-7177/$ – see front matter © 2010 Elsevier Ltd All rights reserved doi:10.1016/j.mcm.2010.03.049 (2) V.N Huy, Q.-A Ngô / Mathematical and Computer Modelling 52 (2010) 522–528 523 Theorem (See [10], Theorem 1) Let f : [a, b] → R be an absolutely continuous function, whose derivative f ∈ L2 [(a, b)] Then b f (t ) dt − b−a a a+b f (a) + 4f ( b − a) + f (b) σ (f ), where σ (·) is defined by (2) Inequality (3) is sharp in the sense that the constant (3) cannot be replaced by a smaller one Since the constant 16 in (3) is sharp, in order to strengthen (3) we have to replace the exponent 32 on the right-hand side of (3) This leads us to strengthen (3) by enlarging the number of knots (6 knots in (3)) and replacing f in (3) (see [11] for more details) Before stating our main result, let us introduce the following notation b f (x) dx I (f ) = a Let m, n < ∞ and p ∞ For each i = 1, n, we assume < xi < such that n x1 + x2 + · · · + xn = , ··· n xj1 + xj2 + · · · + xjn = , j+1 ··· n m−1 m−1 m−1 x1 + x2 + · · · + xn = m , xm + xm + · · · + xm = n n m+1 Put Q (f , n, m, x1 , , xn ) = b−a n n f (a + xi (b − a)) i =1 Remark With the above notations, inequality (3) reads as follows I (f ) − Q f , 6, 1, 0, 1 1 (b − a) 2 2 , , , ,1 σ (f ) (4) We are now in a position to state our main result Precisely, we shall apply the Fundamental Theorem of Calculus, Taylor’s formula and the Hölder inequality to establish the following result Theorem Let I ⊂ R be an open interval such that [a, b] ⊂ I and let f : I → R be an m-times differentiable function such that f (m) ∈ L2 (a, b) Then we have |I (f ) − Q (f , n, m, x1 , , xn ) | √ 2m + +√ 2m − 1 (b − a)m+ m! σ f (m) (5) This work can be considered as a continued and complementary part to our recent papers [11–13] Remark It is worth noticing that the right-hand side of (5) does not involve xi , i = 1, n and that m can be chosen arbitrarily This means that our inequality (5) is better in some sense, especially when b − a However, the constant √ 2m + +√ 2m − 1 m! in the inequality (5) is not sharp This is because of the restriction of the technique that we use It is better if we leave these to be solved by the interested reader Proofs Before proving our main theorem, we need an essential lemma below It is well known in the literature as Taylor’s formula or Taylor’s theorem with the integral remainder 524 V.N Huy, Q.-A Ngô / Mathematical and Computer Modelling 52 (2010) 522–528 Lemma (See [14]) Let f : [a, b] → R and let r be a positive integer If f is such that f (r −1) is absolutely continuous on [a, b], x0 ∈ (a, b) then for all x ∈ (a, b) we have f (x) = Tr −1 (f , x0 , x) + Rr −1 (f , x0 , x) where Tr −1 (f , x0 , ·) is Taylor’s polynomial of degree r − 1, that is, r −1 Tr −1 (f , x0 , x) = f (k) (x0 ) (x − x0 )k k! k=0 and the remainder can be given by x Rr −1 (f , x0 , x) = x0 (x − t )r −1 f (r ) (t ) dt (r − 1)! (6) By a simple calculation, the remainder in (6) can be rewritten as x−x0 Rr −1 (f , x0 , x) = (x − x0 − t )r −1 f (r ) (x0 + t ) dt (r − 1)! which helps us to deduce a similar representation of f as follows r −1 f (x + u) = uk (k) f (x) + k! k=0 u (u − t )r −1 (r ) f (x + t ) dt (r − 1)! (7) Proof of Theorem Define x f (t ) dt F (x) = a By Fundamental Theorem of Calculus I ( f ) = F ( b ) − F ( a) Applying Lemma to F (x) with x = a and u = b − a, we get m F (b) = F (a) + k=1 (b − a)k (k) F (a) + k! b −a (b − a − t )m (m+1) F (a + t ) dt m! which yields m I (f ) = k=1 (b − a)k (k) F (a) + k! b −a (b − a − t )m (m+1) F (a + t ) dt m! Equivalently, m−1 I (f ) = k=0 For each i (b − a)k+1 (k) f (a) + (k + 1)! b −a (b − a − t )m (m) f (a + t ) dt m! n, applying Lemma to f (x) with x = a and u = xi (b − a), we get xki (b − a)k (k) f (a) + k! k=0 xi (b−a) m−1 f (a + xi (b − a)) = xki (b − a)k (k) f (a) + k! k=0 m−1 = b −a (xi (b − a) − t )m−1 (m) f (a + t ) dt (m − 1)! m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! (8) V.N Huy, Q.-A Ngô / Mathematical and Computer Modelling 52 (2010) 522–528 525 By applying (8) to i = 1, n and then summing, we deduce that xki (b − a)k (k) f ( a) + k! i=1 k=0 i=1 m−1 n n f (a + xi (b − a)) = i =1 n n xki (b − a)k m−1 i =1 = k! k=0 m−1 = k=0 f (k) (a) + n ( b − a) (k + 1)! n f (k) (a) + n b−a i=1 m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! b−a i=1 k b −a m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! Thus, m −1 Q (f , n, m, x1 , , xn ) = k =0 b−a (b − a)k+1 (k) f (a) + n (k + 1)! n b−a i=1 m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! Therefore, b −a |I (f ) − Q (f , n, m, x1 , , xn )| = b−a (b − a − t )m (m) f (a + t ) dt − m! n n b−a × i=1 m−1 xm i (b − a − u) f (m) (a + xi u) du (m − 1)! b−a (b − x)m (m) f (x) dx − m! n b = a n m−1 xm i ( b − x) f (m) ((1 − xi ) a + xi x) dx , (m − 1)! b × i=1 a which yields b |I (f ) − Q (f , n, m, x1 , , xn )| = a − b−a n b−a b a − n b m−1 xm i (b − x) (m − 1)! a i =1 b − + (b − x)m (m) f ( x) − m! b−a n f (m) ((1 − xi ) a + xi x) a n b a i =1 b b−a f (m) (t ) dt dx a f (m) (t ) dt dx (b − x)m (m) f ( x) − m! b−a b−a b m−1 xm i ( b − x) (m − 1)! b f (m) (t ) dt dx a f (m) ((1 − xi ) a + xi x) f (m) (t ) dt dx a We note by the Hölder inequality that b a (b − x)m (m) f (x) − m! b−a b a (b − x)m m! b 2 f (m) (t ) dt dx a b f dx a (m) ( x) − b−a f a 2 b (m) (t ) dt dx (9) 526 V.N Huy, Q.-A Ngô / Mathematical and Computer Modelling 52 (2010) 522–528 We now compare the last integral on the right-hand side of the above inequality with b (m) f (x) − a b f (m) (x) − = f b−a (m) = f (m) (x) (t ) dt dx a − 2f f (m−1) (b) − f (m−1) (a) (x) = f (m) (x) dx − f (m−1) (b) − f (m−1) (a) = f f (m−1) (b) − f (m−1) (a) (m) (x) dx − b f (m) (x) = dx − f (m−1) (b) − f (m−1) (a) = b−a b−a √ = b b−a f (m) f (m) b b−a (x) dx + f (m−1) (b) − f (m−1) (a) b−a f (m−1) (b) − f (m−1) (a) b−a 2 dx 2 (b − a) 2 f (m−1) (b) − f (m−1) (a) (x) dx − 2 2 b−a a dx b−a a √ b a + 2 b−a a b−a a (m) f b−a a b b f (m−1) (b) − f (m−1) (a) + b−a a b dx b−a (m) 2 f (m−1) (b) − f (m−1) (a) a b 2 b σ (f (m) ) More precisely, one has b (x) dx − f (b − a)2 a (m) (x) dx a σ f (m) = Hence, (b − x)m (m) f (x) − m! b−a b a b (b − a)m+ σ f (m) √ m! 2m + f (m) (t ) dt dx a (10) Again by the Hölder inequality, one obtains b m−1 xm i (b − x) (m − 1)! a f (m) ((1 − xi ) a + xi x) − m−1 xm i ( b − x) b b dx (m − 1)! a 2 f (m) b b−a f (m) (t ) dt dx a ((1 − xi ) a + xi x) − a b−a 2 b f (m) (t ) dt dx a Clearly, b m−1 xm i (b − x) 2 (m − 1)! a = dx b xm i (b − x) 2m−2 (m − 1)! dx a and b f (m) ((1 − xi ) a + xi x) − a = (1−xi )a+xi b xi f (y) − a b xi (m) b−a f a (m) (y) − b−a f a (t ) dt (m) dx 2 b f (m) (t ) dt a 2 b f (m) a b−a 2 b (t ) dt dy dy = (b − a)m− , √ (m − 1)! 2m − xm i V.N Huy, Q.-A Ngô / Mathematical and Computer Modelling 52 (2010) 522–528 b f (m) (y) − = √ xi a 1 σ f (m) = √ xi xi b b−a f (m) (t ) dt 527 dy a σ f (m) Therefore, b m−1 xm i (b − x) f (m) ((1 − xi ) a + xi x) − (m − 1)! a b b−a f (m) (t ) dt dx a −1 xm i (b − a)m− σ f (m) √ (m − 1)! 2m − Thus, n b i =1 m−1 xm i ( b − x) (m − 1)! a f (m) ((1 − xi ) a + xi x) − b b−a f (m) (t ) dt dx a n (b − a)m− σ f (m) √ m! 2m − (11) Combining (10) and (11) gives |I (f ) − Q (f , n, m, x1 , , xn ) | b − a n (b − a)m− (b − a)m+ + √ √ n m! 2m − m! 2m + σ f (m) , or equivalently, |I (f ) − Q (f , n, m, x1 , , xn ) | √ 2m + +√ 2m − 1 (b − a)m+ m! σ f (m) which completes the proof Examples In this section, by applying our main theorem, we will obtain some new inequalities which cannot be easy obtained from [10] Example Assume n = 6, m = 1, 2, or Clearly x1 = 0, x2 = x3 = x4 = x5 = system , and x6 = satisfy the following linear x1 + x2 + · · · + x6 = , ··· j j j x1 + x2 + · · · + x6 = , j + ··· xm + xm + · · · + xm = m+1 Therefore, we obtain the following inequalities b f (t ) dt − a b−a f (a) + 4f a+b + f (b) √ 2m + Example Assume n = 3, m = By solving the following linear system x1 + x2 + x3 = , 2 x1 + x2 + x3 = , x3 + x3 + x3 = , +√ 2m − 1 (b − a)m+ m! σ f (m) 528 V.N Huy, Q.-A Ngô / Mathematical and Computer Modelling 52 (2010) 522–528 we obtain {x1 , x2 , x3 } is a permutation of √ ,1 − √ 1± , 2 1± Therefore, we obtain the following inequalities √ b f (x) dx − b−a f a a+ 1− 1± 2 ( b − a) √ + f a+ (b − a) + f a+ 1− 1± 2 √ √ 7+ ( b − a) √ 35 (b − a) σ (f ) Example If n = 2, m = 2, then by solving the following system x1 + x2 = , x2 + x2 = , we obtain √ (x1 , x2 ) = √ ± , ∓ We then obtain √ b f (x) dx − a √ b−a f a+ − √ (b − a) + f a+ + ( b − a) √ 3+ √ 15 (b − a) σ (f ) Acknowledgements The authors wish to express their gratitude to the anonymous referees for a number of valuable comments and suggestions References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] P Cerone, Three points rules in numerical integration, Nonlinear Anal 47 (2001) 2341–2352 S.S Dragomir, On Simpson’s quadrature formula and applications, Mathematica 43 (66) (2001) 185–194 S.S Dragomir, R.P Agarwal, P Cerone, On Simpson’s inequality and applications, J Inequal Appl (2000) 533–579 W.J Liu, Several error inequalities for a quadrature formula with a parameter and applications, Comput Math Appl 56 (2008) 1766–1772 C.E.M Pearce, J Pečarić, N Ujević, S Varošanec, Generalizations of some inequalities of Ostrowski–Grüss type, Math Inequal Appl (2000) 25–34 N Ujević, Error inequalities for a quadrature formula of open type, Rev Colombiana Mat 37 (2003) 93–105 N Ujević, Error inequalities for a quadrature formula and applications, Comput Math Appl 48 (2004) 1531–1540 N Ujević, Error inequalities for a quadrature formula of open type, Rev Colombiana Mat 37 (2003) 93–105 N Ujević, New error bounds for the Simpson’s quadrature rule and applications, Comput Math Appl 53 (2007) 64–72 N Ujević, Sharp inequalities of Simpson type and Ostrowski type, Comput Math Appl 48 (2004) 145–151 V.N Huy, Q.A Ngô, New inequalities of Ostrowski-like type involving n knots and the Lp -norm of the m-th derivative, Appl Math Lett 22 (2009) 1345–1350 [12] V.N Huy, Q.A Ngô, New bounds for the Ostrowski-like type inequalities, Bull Korean Math Soc (2010) (in press) [13] V.N Huy, Q.A Ngô, A new way to think about Ostrowski-like type inequalities, Comput Math Appl 59 (2010) 3045–3052 [14] G.A Anastassiou, S.S Dragomir, On some estimates of the remainder in Taylor’s formula, J Math Anal Appl 263 (2001) 246–263 ... Simpson type and Ostrowski type, Comput Math Appl 48 (2004) 145–151 V .N Huy, Q.A Ngô, New inequalities of Ostrowski-like type involving n knots and the Lp -norm of the m-th derivative, Appl Math... Since the constant 16 in (3) is sharp, in order to strengthen (3) we have to replace the exponent 32 on the right-hand side of (3) This leads us to strengthen (3) by enlarging the number of knots. .. can be considered as a continued and complementary part to our recent papers [11–13] Remark It is worth noticing that the right-hand side of (5) does not involve xi , i = 1, n and that m can
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