On Some Estimates of the Remainder in Taylor s Formula

Transcription

1 Journal of Mathematical Analysis and Applications 263, (2) doi:.6/jmaa , available online at on On Some Estimates of the Remainder in Taylor s Formula G. A. Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, Tennessee ganastss@memphis.edu and S. S. Dragomir School of Communications and Informatics, Victoria University of Technology, Melbourne City MC, Victoria 8, Australia sever@matilda.vu.edu.au Submitted by Jerome Goldstein Received April 3, 2 New estimates of the remainder in Taylor s formula are given. Key Words: Taylor s formula. 2 Academic Press. INTRODUCTION The following theorem is well known in the literature as Taylor s formula or Taylor s theorem with the integral remainder. Theorem. Let f a b and let n be a positive integer. If f is such that f is absolutely continuous on a bx a b, then for all x a b we have (.) f x T n f x x+r n f x x where T n f x is Taylor s polynomial of degree n, i.e., n f k x T n f x x x x k (.2) k! X/ $35. Copyright 2 by Academic Press All rights of reproduction in any form reserved. k 246

2 the remainder in taylor s formula 247 (note that f f and! ), and the remainder can be given by (.3) R n f x x x x t n f n+ t dt For a mapping g a b and two arbitrary points x x a b, define and g x xp gt p dt /p x g x x ess sup gt t x x t x x p Using Hölder s inequality, we may state the following corollary. Corollary. (.4) With the above assumptions, we have x x n f n+ x x x x n+/q R n f x x nq+ f n+ /q x xp x x n+ f n+ n+! x x if f n+ L ab; if f n+ L q ab, p> p + q ; if f n+ L ab. For some applications of (.4) for particular functions, see[]. 2. SOME NEW BOUNDS FOR THE REMAINDER The following simple result was considered by G. A. Anastassiou in [2]. Lemma. Assume that the mapping f a b is such that f n is absolutely continuous on a b and x a b. Then for all x a b the remainder R n f x x in (.) can be represented by (2.) R n f x x [ f t f x n! ] x x t n dt n

3 248 anastassiou and dragomir Proof. We apply Taylor s formula with the integral remainder for n, obtaining f x T n f x x+ x t n f t dt n! x T n f x x x x n f x + x t n f t dt n! x T n f x x+ n! which produces the representation (2.). The following theorem holds. x [ f t f x ] x t n dt Theorem 2. the bounds Assume that f x, and x are as in Lemma. Then we have (2.2) R n f x x x x n f f x x x if f L ab; x x n +/q n!n q+ f f if f L /q q ab, x x xp p> p + q ; x x n f f x x x if f L ab. Proof. We have R n f x x f t f x n! x t n dt x f t f x n! x t n dt Mx x x If f L a b, then Mx x n! sup x t n x t x x t x x x x n f f x x x and the first inequality in (2.2) is proved. f t f x dt x

4 the remainder in taylor s formula 249 Using Hölder s integral inequality, we have, for f L p a b, that Mx x f t f x n! p dt /p x x t n q dt x x f f x n! [ x x n q+ ] /q x xp n q + n!n q + /q x x n +/q f f x x xp and the second inequality in (2.2) is proved. Finally, we have for f L a b that Mx x ess sup f t f x x t n dt t x x n! x t xx /q and the theorem is proved. x x n f f x x x The following result for Hölder type mappings also holds. Theorem 3. Assume that the mapping f a b is such that f is of H r-hölder type. That is, (2.3) f t f s Ht s r for all t s a b and H> is given. Then we have the inequality HBr + (2.4) R n f x x x x n! r+n where B is Euler s beta function. Proof. As f is of H r-hölder type, we may write R n f x x f t f x n! x t n dt x H t x n! r x t n (2.5) dt Nx x x Assume that x x. Then Nx x t x r x t n dt x x r+n + t r t n dt x x x r+n Br +

5 25 anastassiou and dragomir A similar equality can be obtained if x<x. Consequently, in general and then, by (2.5), we deduce (2.4). Nx xx x r+n Br + Corollary 2. Assume that the mapping f a b is such that f is L-Lipschitian on a b, i.e., (2.6) f t f s Lt s for all t s a b where L> is given. Then we have the inequality R n f x x Lx x n+ (2.7) n +! Proof. For r we have B2 t n t dt nn + Using (2.4), we deduce (2.7). We can now state the following result as well. Theorem 4. Let f x, and x be as in Theorem. Then the remainder R n f x x satisfies the bound x t n t x n! x f n+ dt x t if f n+ L ab; x t n t x n! /q x R n f x x f n+ dt x tp if f n+ L q ab, (2.8) Proof. p> p + q ; x t n n! x f n+ dt x t if f n+ L ab. As f is absolutely continuous on a b we may write that f t f x t x f n+ u du and then, by (2.), we have the representation ( t ) (2.9) R n f x x f n+ u du x t n dt n! x x

6 the remainder in taylor s formula 25 By (2.9) we may write t (2.) R n f x x f n+ u du x t n dt n! x x Now, if f n+ L a b, then t f n+ u du t x f n+ x t x Also, by Hölder s integral inequality we have (for p>, p + ) that q t f n+ u du t x /q t f n+ u p du /p x x and t x /q f n+ x tp t f n+ t u du f n+ u du f n+ x t x x Consequently, we have t x f n+ x t t f n+ u du t x /q f n+ x tp x (2.) f n+ x t Using (2.) and (2.), we easily deduce (2.8). 3. SOME FURTHER BOUNDS OF THE REMAINDER Let us consider the Chebychev functional defined by (3.) T g h b a b a b hxgx dx hx dx b a 2 a b a gx dx where h g a b are measurable on a b and the involved integrals exist on a b. The following identity which can be proved by direct computation is well known in the literature as Korkine s identity: b b (3.2) T g h hx hygx gy dx dy 2b a 2 a a

7 252 anastassiou and dragomir The following lemma holds. Lemma 2. Assume that the mapping f a b and x x are as in Lemma. Then we have the representation R n f x x [[ f n x x ] f x ] (3.3) Proof. x x x x n + 2n!x x ( f t f s ) x x ( x t n x s n ) dt ds Applying Korkine s identity, we may write x f t f x x t n dt x x 2 x ( f t f x ) dt 2x x 2 x which is clearly equivalent to x f t f x x t n dt x x x x t n dt x ( f t f s ) x t n x s n dt ds [ f n x f x x x f x ] x x n n ( + f t f s )( x t n x s n ) dt ds 2x x x x from which we get (3.3). The following theorem holds. Theorem 5. Assume that the mapping f a b has the property that f L 2 a b and x x a b. Then we have the inequality R n f x x f n x x f x x x n + n x x n 2n [ f ( [f 2 x x n x x x2 x ]) ] 2 /2 (3.4) where f x x f t 2 dt x2 /2 x

8 the remainder in taylor s formula 253 Proof. We have, by (3.3), that R n f x x [ f n x x ] f x x x n + 2n!x x ( f t f s ) x x [ x t n x s n ] (3.5) dtds Using the Cauchy Buniakowski Schwart inequality, we have ( f t f s )[ x t n x s n ] dt ds x x ( f t f s ) /2 2 dt ds x x [ x t n x s n ] /2 2 dt ds x x [ [ 2 x x f t ] ( ) 2 2 ] /2 dt f t dt x x [ ( ) 2 ] /2 x x x t 2n dt x t n dt x x 2 [x x f 2xx ( f n x f n x 2 ) ] 2 /2 [x x x x 2n + ( x x n ) 2 ] /2 2n + n [ 2x x f ( [f 2 x x n x x x 2 x ]) ] 2 /2 [ x x 2n 2n [ 2x x x x 2n n 2 ] /2 x x f 2 x x 2 ( [f n x x ]) 2 ] /2 ] /2 [ n x x n 2 2n + n 2 2n [ 2x x n+ f ( [f 2 x x n x x x 2 x ]) ] 2 /2 n n 2n 2n x x n+ [ n f 2 2n x x ([ f n x x x 2 x ]) ] /2 2

9 254 anastassiou and dragomir and then 2n! x n x x n 2n Using (3.5), we deduce (3.4). ( f t f s )[ x t n x s n ] dt ds x [ f ( [f 2 x x n x x x 2 x ]) ] 2 /2 4. SOME INEQUALITIES FOR SPECIAL CASES In this section we assume that x x a b and x x. The following theorem holds. Theorem 6. Let f a b be such that f is monotonic nondecreasing (nonincreasing) on x x. Then we have the inequality (4.) f x T n f x x+ [[ f n x x ] f x ] x x n or, equivalently, (4.2) f x T n f x x+ x x n [ f n x x ] Proof. We use the Chebychev inequality (4.3) T g h provided that g h are synchronous (asynchronous), i.e., we recall that the mappings g h are synchronous (asynchronous) if (4.4) gx gyhx hy for all x y a b As the mapping ht x t n is monotonic nonincreasing on x x, then we have, for f nondecreasing, and then, by (3.3) we deduce that T ( f x n ) R n f x x [[ f n x x ] f x ] x x n (4.5) The case where f is monotonic nonincreasing goes likewise and we omit the details.

10 the remainder in taylor s formula 255 The following refinement of Chebychev s inequality is known (see, for example, [3]): (4.6) T g h maxt g h T gh T g h Using (2.5), we may improve (4.) as follows. Theorem 7. Let f a b be such that f is monotonic nonincreasing on x x. Then we have the inequality (4.7) f x T n f x x [ f n x x ] x x n f t x t n dt n! x n x x n f t dt x Proof. (4.8) Apply inequality (4.6) for g f hx n to obtain T ( f x n ) maxa B C where A T ( f x n ) T ( f x n ) B T ( f x n ) f t x t n dt x x x and f t dt x t n dt x x 2 x x x x x f t x t n dt f t dt x x n x x 2 x n C T f x n T f x n B Now, using the fact that (see Lemma 2) (4.9) f x T n f x x [[ f n x x ] f x ] x x x x n n! T ( f x n ) then by the inequality (4.8), we may deduce (4.7).

11 256 anastassiou and dragomir The following theorem also holds. Theorem 8. Let f a b be such that f is convex (concave) on x x. Then we have the inequality f x T n f x x (4.) n +! f n x x x n+ Proof. As f is convex (concave) on x x, we may write that f t f x f n x t x t x x which implies that [ f t f x ] x t n f n x t x x t n t x x Integrating over t on x x and using the representation (.), we may obtain R n f x x f n x n! t x x t n dt x n! f n x t x x t n dt x n! f n x x x n+ B2 n +! f n x x x n+ and the inequality (4.) is proved. 5. TAYLOR-MULTIVARIATE CASE ESTIMATES Let Q be a compact convex subset of k k 2; k x x x k Q. Let f Q be such that all partial derivatives of order n are coordinatewise absolutely continuous functions, n. Also, f C n Q. Each nth order partial derivative is denoted by f α α f/ x α, where α α α k, α i + i k, and α k i α i n. Consider g t fx + t x t. Then g j t [( k i x i ) j ] f x i i (5.) x + t x x k + t k x k for all j 2n. Note that g t is given in a similar way.

12 and Example. the remainder in taylor s formula 257 Let n k 2. Then g t f x + t x x 2 + t 2 x 2 t g t x f x x + t x + 2 x 2 f x 2 x + t x In addition, ( ) f g t x x x + t x ( ) f + 2 x 2 x x + t x 2 x { x f 2 f 2 } + x 2 2 x 2 x 2 x { f x 2 x + x x 2 x 2 f 2 } 2 Thus, g x 2 2 t x 2 f 2 f 2 + x 2 x 2 x 2 x x 2 f 2 + x 2 x 2 + x x 2 x 2 2 f 2 2 Similarly, we obtain the case for n k for g t. Notice that if f α x exists for all α such that α n, then g also exists; is any type of p-norm p. Therefore, we obtain the multivariate Taylor Theorem: Theorem 9. With the above assumptions, we have (5.2) where (5.3) or (5.4) R n f k g R n ( t n! n j g j + R j! n x 2 2 ( tn ) ) g t n g dt n dt θ n ( g θ g ) dθ

13 258 anastassiou and dragomir Asimpler form is (5.5) f k g n j g j + j! R n where (5.6) R n ( t or (5.7) R n n! Notice that g fx. ( tn ) ) g t n dt n dt θ n g θ dθ For a mapping f Q, x Q, Q k compact and convex, we define f x p f y p p (5.8) dy p x Here x x x x is a kth multiple integral. Also, (5.9) f x ess sup f y y x y x where x x are line segments in Q. We first find estimates for R n as in (5.7). Remark. Let be any norm on the functions from Q to. Let fα x max αn f α x. Then g t [( k i x i ) n ] f x i i x + t x x k + t k x k that is, (5.) ( k ) n i x i fα x g i t x l n f α x Here, x may be any kind of p-norm p.

14 the remainder in taylor s formula 259 We may now state the first result in estimating the remainder R n. Theorem. With the above assumptions, we have g n! L if g R n n!pn p + /p Lq if Proof. That is, (5.) We have g Rn n! g g L ; L q, p + p>; q if g L. n! θ n g g Rn θ dθ θ dθ n! g L n! g t L given that g L, the last is implied by all f α L x. Again we see that Rn θ n g θ dθ n! ( ) /p ( θ n p dθ n! ( ) /p θ pn p dθ g n! That is, (5.2) where p q >, L q x. g n!pn p + /p L q Rn g Lq ) /q g θ q dθ Lq /p n!pn p + p + q g L q ; the last is implied by all f α

15 26 anastassiou and dragomir Also it holds that Rn θ n g θ dθ n! ( ) θ n dθ g n! g that is, Rn g (5.3) if g L the last is implied by all f α L x. Remark 2. Observe that [( g k i x i ) n ] f x x i i where x x x k. Similarly, we get g t g L (5.4) R n n! given that g L, which holds when all f α L x. Also, g t g (5.5) Lq R n /p n!pn p + where p q >, p + ; g q L q, when all f α L q x. Furthermore, we find g t g (5.6) R n if g L when all f α L x. Assume now that (for all α such that α n) (5.7) f α x f α y L x y β l <β for all x y Q, L> where l is the l norm in k. Here f α is any partial derivative of order n. Then clearly (for all α such that α n) (5.8) f α x + t x f α x L t β x β l where x l k i i x i.

16 the remainder in taylor s formula 26 Thus, if x, then for at least one i k, we have i x i, i.e., x l. So, without loss of generality assume that x, which implies that x l. Hence by (5.3) ( ( ( t tn k R n i i x i α i α αn! α k! ) ) )dt n L x β l t β n k i i x i α i L x α αn! α k! β l ( t ( tn ) ) tn β dt n dt ( ) n k L i x i x β l n j β + j L x β+n l β + j i Consequently, we may state the following result. Theorem. (5.9) Let f α satisfy (5.7). Then R n L x β+n l n jβ + j dt Another matter to discuss is the following. We have (5.2) where (5.2) R n f k n 2! n 2! n j n j for all Qα n g j + R j! n θ n 2( g n θ n 2 ( θ and ( θ (5.22) R n θ n 2 n 2! Now, if all f α L x, then g θ g u du θ g θ g n ) g udu dθ L, and thus L θ ) dθ ) g u du dθ

17 262 anastassiou and dragomir Let p q > such that p +. Then q θ g u ( θ ) /q ( θ du q du θ /q g where g Also, Lp θ L p, when all f α L p x. θ g g u du g L θ where g L, when all f α L x. Thus we may state the following result. Theorem 2. With the above assumptions, we have (5.23) R n n 2! n 2! n 2! θ n 2 u p du if g L θ g L θdθ all f α L x ; θ n 2 θ /q ) /p if g L p p + q g Lp θdθ pq > all f α L p x θ n 2 if g L g L θdθ all f α L x. Remark 3. (a) Using Lemma 2, we have the representation [( ) ] R n g n g n g ( + g t g s ) 2n! (5.24) ( t n s n ) dt ds Next, assume that g L 2 2. By Theorem 5 we get 2 ( g n g n ) g R n + n [ g (g 2 2 2n n g ) n 2 ] /2 (5.25)

18 the remainder in taylor s formula 263 where ( g 2 g t /2 dt) 2 is nondecreasing (nonincreas- (b) By Theorem 6, assuming that g ing) over, we have (5.26) That is, (5.27) f k n j g j + j! n f k j [ g n g n ( g j + j! g n ] g g n ) (c), we find that (5.28) By Theorem 7, assuming that g n f k j n! ( g j j! g g n is monotonic increasing on g n t t n dt n ) g t dt (5.29) (d) Last, by Theorem 8, for g f k n j convex (concave) on, weget g j j! g n n +! REFERENCES. S. S. Dragomir, New estimation of the remainder in Taylor s formula using Grüss type inequalities and applications, Math. Ineq. Appl. 2, No. 2 (999), G. A. Anastassiou, Ostrowski type inequalities, Proc. Amer. Math. Soc. 23 (995), S. S. Dragomir and J. Pečarić, Refinements of some inequalities for isotonic functionals, Anal. Num. Theor. Approx. (989), 6 65.