Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction


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1 Czechoslovk Mthemticl Journl, 55 (130) (2005), ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued function of one vrible is pproximted by its nth degree Tylor polynomil, the reminder is estimted using the Alexiewicz nd Lebesgue pnorms in cses where f (n) or f (n+1) re HenstockKurzweil integrble. When the only ssumption is tht f (n) is HenstockKurzweil integrble then modified form of the nth degree Tylor polynomil is used. When the only ssumption is tht f (n) C 0 then the reminder is estimted by pplying the Alexiewicz norm to Schwrtz distributions of order 1. Keywords: Tylor s theorem, HenstockKurzweil integrl, Alexiewicz norm MSC 2000 : 26A24, 26A39 1. Introduction In this pper the HenstockKurzweil integrl is used to give vrious estimtes of the reminder in Tylor s theorem in terms of Alexiewicz nd Lebesgue pnorms. Let [, b] be compct intervl in nd let f : [, b]. Let n be positive integer. When f is pproximted by its nth degree Tylor polynomil bout, the reminder is estimted using the Alexiewicz norm of f (n+1) nd pnorms of f (n) in the cse when f (n+1) is HenstockKurzweil integrble (Theorem 4). When the only ssumption is tht f (n) is HenstockKurzweil integrble then f (n) need not exist t. In this cse we use modified form of the Tylor polynomil where f (n) is evluted t point x 0 [, b] t which f (n) exists. The resulting modified reminder is then estimted in the Alexiewicz nd pnorms (Theorem 3). And, when the only ssumption is tht f (n) C 0, the Alexiewicz norm is used in the spce of Schwrtz distributions of Reserch prtilly supported by the Nturl Sciences nd Engineering Reserch Council of Cnd. An djunct ppointment in the Deprtment of Mthemticl nd Sttisticl Sciences, University of Albert, mde vluble librry nd computer resources vilble. 933
2 order 1 to estimte the reminder (Theorem 6). The results extend those in [1]. See lso [7] for nother form of the reminder. For the HenstockKurzweil integrl we hve the following version of Tylor s theorem. Theorem 1. Let f : [, b] nd let n be positive integer. If f (n) ACG then for ll x [, b] we hve f(x) = P n (x) + R n (x) where (1) P n (x) = n f (k) ()(x ) k k=0 k! nd (2) R n (x) = 1 x f (n+1) (t)(x t) n dt. A proof under the ssumption tht f (n) is continuous on [, b] nd f (n+1) exists nerly everywhere on (, b) is given in [6]. The generl cse follows from the Fundmentl Theorem: If F : [, b] nd F ACG then F exists lmost everywhere, F is HenstockKurzweil integrble nd x F = F (x) F () for ll x [, b]. For the wide Denjoy integrl the corresponding function spce is ACG. If F ACG then x F p = F (x) F () for ll x [, b]. Here the integrl is the wide Denjoy integrl nd the pproximte derivtive is used. All the results proved in the pper hve suitble extension to the wide Denjoy integrl. The function spces ACG nd ACG re defined in [2] nd [4]. Note tht AC ACG ACG C 0. The set of continuous functions tht re differentible nerly everywhere is proper subset of ACG, which is itself proper subset of the set of continuous functions tht re differentible lmost everywhere. The other hlf of the Fundmentl Theorem sys tht if f HK then d x dx f = f(x) lmost everywhere (nd certinly t points of continuity of f). The Alexiewicz norm of n integrble function f is defined x (3) f = sup f. x b We will write f I when it needs to be mde cler the norm is over intervl I. See [5] for discussion of the Alexiewicz norm nd the HenstockKurzweil integrl. An equivlent norm is d c f. It leds to similr estimtes to those obtined sup (c,d) (,b) for the Alexiewicz norm in Theorems 3, 4 nd 6. Denote the spce of HenstockKurzweil integrble functions on [, b] by HK. Note tht if f (n) ACG then f (n+1) HK. 934
3 2. Estimtes when f (n 1) ACG If f (n 1) ACG (i.e., f (n) HK) then f (n) need only exist lmost everywhere on (, b). If f (n) exists t x 0 [, b] then we cn modify the Tylor polynomil (1) so tht the nth derivtive is evluted t x 0 nd then obtin estimtes on the resulting reminder term. The following lemm is used. Lemm 2. If f : [, b] nd f (n 1) ACG then let x 0 [, b] such tht f (n) exists t x 0. Define the modified Tylor polynomil by (4) P n,x0 (x) = P n 1 (x) + f (n) (x 0 )(x ) n nd define the modified reminder by R n,x0 (x) = f(x) P n,x0 (x). x [, b] Then for ll (5) R n,x0 (x) = 1 x [f (n) (t) f (n) (x 0 )](x t) n 1 dt. The proof is the sme s for Lemm 1 in [1] (which is flse without the proviso tht f (n) exists t x 0 ). Of course we cn tke x 0 s close to s we like. See [3] for reminder estimtes bsed on estimtes of f (n) (t) f (n) (x 0 ). Theorem 2 in [1] gives pointwise estimtes of R n in terms of pnorms of f (n) when f (n 1) AC nd f (n) L p (1 p ). We hve the following nlogue when f (n 1) ACG. Note tht if f (n 1) ACG \ AC (i.e., f (n) HK \ L 1 ) then for ech 1 p we hve f (n) L p. However, we cn use the Alexiewicz norm to estimte R n,x0. Theorem 3. With the nottion nd ssumptions of Lemm 2, (6) R n,x0 For ll x [, b], (b )n f (n) ( ) f (n) (x 0 ). (7) R n,x0 (x) And, (8) R n,x0 p (x )n 1 f (n) ( ) f (n) (x 0 ) [,x]. (b ) n 1+ [(n 1)p + 1] f (n) ( ) f (n) (x 0 ), 1 p <, (b ) n 1 f (n) ( ) f (n) (x 0 ), p =. 935
4 . Let c b. Using Lemm 2 nd the reversl of integrls criterion in [2, Theorem 57, p. 58], (9) c R n,x0 = = 1 (c )n c ξ [f (n) (t) f (n) (x 0 )] c t (x t) n 1 dx dt [f (n) (t) f (n) (x 0 )] dt for some ξ [, c]. Eqution (9) comes from the second men vlue theorem for integrls [2]. Tking the supremum over c [, b] now gives (6). Similrly, (10) R n,x0 (x) = (x )n 1 ξ [f (n) (t) f (n) (x 0 )] dt for some ξ [, x] (x )n 1 f (n) ( ) f (n) (x 0 ). This gives (7). The other estimtes in (8) follow from this. An lterntive pproch in Theorem 3 is to ssume f (n 1) is ACG on [, b] nd f (n) exists t. Then eqution (6) is replced with R n (b ) n f (n) ( ) f (n) () / with similr chnges in (7) nd (8). This is Young s expnsion theorem [8, p. 16] but with more precise error estimte. 3. Estimtes when f (n) ACG Corollry 1 in [1] gives pointwise estimte of R n in terms of pnorms of f (n+1) when f (n) AC nd f (n+1) L p. When f (n) ACG \ AC (i.e., f (n+1) HK \ L 1 ) then for no 1 p do we hve f (n+1) L p. But, we cn estimte R n using the Alexiewicz norm of f (n+1) nd pnorms of f (n). Theorem 4. If f : [, b] such tht f (n) ACG then (11) R n For ll x [, b], (b )n+1 f (n+1). (n + 1)! (12) R n (x) 936 (x )n f (n+1) [,x].
5 And, (b ) n+ (np + 1) f (n+1), 1 p <, (13) R n p (b ) n f (n+1), p =. Also, (14) R n p (b ) n f (n) () + where 1/α + 1/β = 1 nd (b ) n+ (np + 1) f (n) () + A i, 1 p <, A 1 = (b )n 1+ f (n), for n 1, [(n 1)p + 1] A 1 = f( ) f() p, for n = 1, A 2 = (b )n f (n), p =, (b ) n 1++1/β f (n) α, for 1 < α, [(n 1)β + 1] 1/β [(n 1 + 1/β)p + 1] A 2 = (b )n 1+ f (n) 1, for α = 1, [(n 1)p + 1] (b ) 1 ( A 3 = f (n) (t) p (b t) dt) (n 1)p+1, [(n 1)p + 1] A 4 = (b )n(1 ) ( f (n) (t) p (b t) dt) n.. The proof of (11) is very similr to the proof of (6), except tht we begin with the reminder in the form of (2). Using the second men vlue theorem, we hve (15) R n (x) = (x )n ξ f (n+1) (t) dt for some ξ [, x]. The estimtes in (12) nd (13) now follow directly. Integrte (2) by prts to get (16) R n (x) = f (n) ()(x ) n 1 + x f (n) (t)(x t) n 1 dt. 937
6 Then (17) R n p f (n) () where I I 2 (18) nd (19) I 1 = I 2 = (x ) np dx = x (b )np+1 np + 1 p f (n) (t)(x t) n 1 dt dx. A 1 is obtined from I 2 using the second men vlue theorem nd A 2 using Hölder s inequlity. Writing (20) I 2 ( x Jensen s inequlity nd Fubini s theorem give ) p f (n) (t) (x t) n 1 dt (x ) p dx, x (21) (22) I 2 f (n) (t) p (x t) (n 1)p (x ) p 1 dx dt t (b b )p 1 f (n) (t) p (b t) (n 1)p+1 dt. (n 1)p + 1 From this we obtin A 3. Using Jensen s inequlity in the form (23) (24) I 2 1 n p 1 ( x f (n) (t) (x ) p t)n 1 n dt (x ) np (x ) n n p dx x f (n) (t) p (x t) n 1 (x ) n(p 1) dt dx, we cn pply Fubini s theorem to get (25) I 2 (b )n(p 1) n p f (n) (t) p (b t) n dt, which gives A 4. The cse p = follows directly from (16). 938
7 Note tht A 2, A 3 nd A 4 ll led to estimtes of form A k C n,p (b ) n f (n) p (k = 2, 3, 4) where C n,p is independent of, b nd f. The integrl over x in (21) cn be evluted using hypergeometric functions. However, this does not mrkedly improve the estimte for A 3. Similrly with A 4. In (12) we hve R n (x) = o[(x ) n ] s x. Note tht if f (n) ACG then f (n+1) = mx x b f (n) (x) f (n) (). This ffects how the reminder is written in Theorems 3 nd 4. Exmple 5. Let 0 < n 1 be sequence tht decreses to 0. Let b n be positive sequence tht decreses to 0 such tht the intervls ( n b n, n + b n ) re disjoint. For this it suffices to tke b n min([ n 1 n ]/2, [ n n+1 ]/2). Let f n (x) = (x n + b n ) 2 (x n b n ) 2 for x n b n nd 0, otherwise. Let α > 0 nd define f(x) = n α f n (x). For x n < b n we hve f n(x) = 4(x n + b n )(x n b n )(x n ) nd f n (x) = 4[3(x n) 2 b 2 n ]. Suppose b n = o( n ) s n nd n α b 3 n 0. Then mx f n(x) = O(b 3 n) so f C[0,1] 1 x n b n but f C[0,1] 2. If n α b 3 n = then f HK \ L 1. Let = 0 in Theorems 3 nd 4. (i) If n α b 3 n / n 0 then f (0) does not exist. An exmple of this cse is n = 1/n, b n = c n β for β 2 nd smll enough c, nd 3β 1 α < 3β. Let x 0 (0, 1)\{ n ±b n } n. Then f (x 0 ) exists nd we hve the modified second degree Tylor polynomil P 2,x0 (x) = f(0) + f (0)x + f (x 0 )x 2 /2 = f (x 0 )x 2 /2. The modified reminder is R 2,x0 (x) = x 0 [f (t) f (x 0 )](x t) dt. If we tke x 0 = n ±b n / 3 for some n then f (x 0 ) = 0 nd we hve f ( ) f (x 0 ) [0,x] mx f (y) = 0 y 1/ 1/x sup n 1/x n α f n ( n ±b n / 3) = (8c 3 /3 3) sup n 1/x n α 3β = (8c 3 /3 3) 1/x α 3β. This llows us to obtin ll of the estimtes in Theorem 3. Note tht R 2,x0 (x) = O(x 1/x α 3β ) = O(x 1 α+3β ) s x 0 (nd 1 < 1 α + 3β 2). (ii) If n α b 3 n / n 0 then f (0) exists. An exmple is 0 < α < 3β 1 with n nd b n s in (i). This gives f L 1. Now P 2,0 (x) = 0 nd R 2,0 (x) = x 0 f (t)(x t) dt. The quntity f [0,x] cn be estimted s in (i). (iii) We cn use Theorem 4 with n = 1 to get P 1 (x) = 0 nd R 1 (x) = x 0 f (t)(x t) dt. This leds to the sme estimtes for f [0,x] s in (i) in the cse x 0 = n ± b n /
8 4. Estimtes when f (n) C 0 We now show tht (11) continues to hold when the only ssumption is tht f (n) C 0. Under the Alexiewicz norm, the spce of HenstockKurzweil integrble functions is not complete. It s completion is the subspce of distributions tht re the distributionl derivtive of continuous function, i.e., distributions of order 1 (see [5]). Thus, if f is in the completion of HK then f D (Schwrtz distributions) nd there is continuous function F, vnishing t, such tht F (ϕ) = F (ϕ ) = F ϕ = f(ϕ) for ll test functions ϕ D = {ϕ: [, b] : ϕ C nd supp(ϕ) (, b)}. And, we cn compute the Alexiewicz norm of f vi f = mx F (x) = F. x [,b] Theorem 6. If f : [, b] such tht f (n) C 0 then (26) R n (b )n+1 f (n+1) (b )n+1 = mx (n + 1)! (n + 1)! f (n) (x) f (n) (). x b. From Lemm 2, R n (x) = 1 x [f (n) (t) f (n) ()](x t) n 1 dt. Integrting x from to y nd reversing orders of integrtion gives y R n = 1 y [f (n) (t) f (n) ()](y t) n dt. Eqution (26) now follows. Exmple 7. Let g be continuous nd nowhere differentible on [0, 1]. Let n 1 nd define f(x) = (n 1)! 1 x 0 g(t)(x t)n 1 dt. By differentiting under the integrl nd then using the Fundmentl Theorem, f (n) = g C 0 but f (n) ACG. We hve R n (x) = 1 x 0 [g(t) g(0)](x t)n 1 dt nd R n (n + 1)! 1 mx g(x) 0 x 1 g(0). References [1] G. A. Anstssiou nd S. S. Drgomir: On some estimtes of the reminder in Tylor s formul. J. Mth. Anl. Appl. 263 (2001), [2] V. G. Čelidze nd A. G. Džvršeǐšvili: The Theory of the Denjoy Integrl nd Some Applictions. World Scientific, Singpore, [3] G. B. Follnd: Reminder estimtes in Tylor s theorem. Amer. Mth. Monthly 97 (1990), [4] S. Sks: Theory of the Integrl. Monogrfie Mtemtyczne, Wrsw, [5] C. Swrtz: Introduction to Guge Integrls. World Scientific, Singpore, [6] H. B. Thompson: Tylor s theorem using the generlized Riemnn integrl. Amer. Mth. Monthly 96 (1989), [7] R. Výborný: Some pplictions of KurzweilHenstock integrtion. Mth. Bohem. 118 (1993), [8] W. H. Young: The Fundmentl Theorems of the Differentil Clculus. Cmbridge University Press, Cmbridge, Author s ddress: Deprtment of Mthemtics nd Sttistics, University College of the Frser Vlley, Abbotsford, BC Cnd V2S 7M8, emil: 940
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