AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE

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1 Bulleti o Mathematical Aalysis ad Applicatios ISSN: 8-9, URL: Volume 3 Issue 3), Paes AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE COMMUNICATED BY HAJRUDIN FEJZIC) DAN ŞTEFAN MARINESCU, MIHAI MONEA Abstract. I this ote, we aalize the cotext i which the calculatio o its o sequeces is perormed usi the iteratio o product o Riema iterable uctios. We will also study the deree o coverece o such sequeces.. INTRODUCTION The iteratio is oe o the most useul cocept i itroductory calculus. Its developmet was started with Newto which cosidered this process like a reverse o dieretiatio. Later, Leibiz ad Riema have see the iteratio as summatio o may quatities. This approach led to the curret deiitio o Riema iteral see e.. [5], pa. 6). To evaluate Riema iteral without derivative is ot easy. Thereore, i this article, we will aalize the Riema iteral as a it. First, we recall a classic resultsee e.. [8], pa. 49). Lemma.. Let : [a, b] R be Riema iterable. The b a a + k ) b a) = b a x)dx. I [8], p. 49, Polya ad Szeo have ormulated the problem o the deree o approximatio see problem ) ad ave a aswer p. 3) which is preseted i the ext lemma. Mathematics Subject Classiicatio. 6A4, 6A4. Key words ad phrases. Riema iterable, it o sequece, deree o coverece. c 8 Uiversiteti i Prishtiës, Prishtië, Kosovë. Submitted April,. Accepted May 5,. 5

2 6 D.Ş. MARINESCU, M. MONEA Lemma.. Let : [a, b] R be cotiuously dieretiable. The b a a + k ) b b a) x) dx b) a) = b a). a I this paper we wat to exted these results or product o two uctios. I order to do this, we recall the ext results about the product o two iterable uctios see e.. [9], theorem 7.9). Theorem.3. I, : [a, b] R are two Riema iterable uctios, the is Riema iterable. Sice the trasormatios x x a b a chae the iterval [a, b] ito [, ], we will preset all our results or the uctios deied o the iterval [, ]. I this cotext, by usi theorem.3, lemmas. ad. admit the ollowi extesios: Lemma.4. I, : [, ] R are two Riema iterable uctios, the ) k ) k = x) x)dx. Lemma.5. I, : [, ] R are two cotiuously dieretiable uctios, the ) ) k k x) x) dx ) ) ) ) =. We will eeralize these lemmas i the ext sectio. The mai result is the propositio 3., which will be preseted i sectio 3. This result will ive more iormatio about the coverece s order o a sequece, more eeral the the oe rom lemma.5.. SEQUENCES WHOSE LIMITS ARE CALCULATED USING RIEMANN INTEGRALS First, we ormulate the ext lemma to establish a eeral ramework or the mai results o this ote. Lemma.. Let, : [, ] R be two Riema iterable uctios. For ay N ad k {,,..., }, we deote I k = [ k, ] k. We deie Ak = sup x), x I k a k = i x), B k = sup x) ad b k = i x). We cosider the sequeces x I k x I k x I k x = a k b k

3 AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE 7 ad The: a) y x ) = ; b) x = y = y = A k B k. x) x)dx. Proo. As the uctios are Riema iterable, we obtai rom Darboux s criterio, that A k a k ) = ad B k b k ) =. a) First, we assume that the uctios ad are oeative. The y x = A k B k a k b k ) = A k a k ) B k + sup x) x [,] B k b k ) a k A k a k ) + sup x) x [,] B k b k ). But, are iterable, so they are bouded. The, we obtai ) sup x) A k a k ) + sup x) B k b k ) = x [,] x [,] ad y x ) =. I the absece o the oeative assumptio, we ote a = respectively b = i x [, ] i x) ad, x [, ] x). Cosider the uctios F : [, ] R, F x) = x) a ad, G : [, ] R, G x) = x) b, which are oeative uctios. The : y x ) = A k B k a k b k ) = because A k a) B k b) a k a) b k b)) + a B k b k ) + A k a) B k b) a k a) b k b)) =, i accordace with previous calculatios or the uctios F, respectively G. b) I, are oeative, the or ay iterval I [, ], the ollowi iequalities are satisied: ad sup x I x) x)) sup x I x) sup x) x I i x) x) i x) i x). x I x I x I ) b A k a k ) =,

4 8 D.Ş. MARINESCU, M. MONEA This implies that x x) x) dx y, or all N, as ay Riema sum attached to the uctio lies betwee x ad y. So, we obtai which meas that Similarly ad we obtai y y = x) x) dx y x x) x)dx. x) x) dx x y x x = x) x)dx. I, are ot ecessarily oeative, we obtai the same result by applyi the same reasoi or uctios F ad G, deied at the previous poit. The result rom. help us to prove the ext lemma. We should metio that this result was preseted by Béyi ad Niţu i [3] but without a solutio. Lemma.. Let, : [, ] R be two Riema iterable uctios. For ay k {,, 3,..., } ad or ay α k, β k [ k, k ], the sequece z = α k ) β k ) coveres to x) x) dx. Proo. We have two cases. First, we assume that the uctios ad are oeative. We choose the sequeces x ad y rom lemma.. Obviously we have x z y ad the coclusio ollows. I the absece o the oeative assumptio, we ote a = respectively b = i x [, ] i x) ad, x [, ] x). Similar with the proo o lemma.., we cosider the

5 AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE 9 uctios F : [, ] R, F x) = x) a ad G : [, ] R, G x) = x) b, which are oeative uctios. The α k ) β k ) = F α k ) + a) G β k ) + b) = = = F α k ) G β k ) + b F x) G x) dx + b which coclude the proo. F α k ) + a F x) dx + a F x) + a) G x) + b) dx = ) G β k ) + ab G x) dx + ab x) x) dx A similar result could be oud i [], where Spivak has posed the same problem see problem, p.63) but or cotiuous uctios. Evidetly, our result is more eeral. A cosequece o lemma. is the ollowi result rom [3]. Corollary.3.Béyi, Niţu) Let : [, ] R be a Riema iterable uctios.the ) ) k k = x) dx. Proo. We apply. or = ad the sequeces α = k ad β = k. 3. THE ORDER OF CONVERGENCE Lemma. ives us a aswer about the its o sequeces o the type a k ) b k ), but we wat to kow more details about these its. We would like to study their order o coverece ad we ca deliver a ood estimatio i the case o the sequece ) ) k α k β, with α, β [, ]. The results are icluded i ext propositio.

6 3 D.Ş. MARINESCU, M. MONEA Propositio 3.. Let, : [, ] R be two cotiuously dieretiable uctios. Let deie α, β [, ] ad the sequece a = or all N. The: x) x) dx ) k α a = ) ) ) )) + α k β x) x) dx + β Proo. Ater some alebra we have ) ) k α k β a = x) x) dx = But x) x) dx + k β x) x) dx ) k ) ) k + ) k ) k ) k k α ) k = )). ), x) x) dx. ) k )) k β ) ) ) ). [ ] However, or ay k {,, 3.., } ad ay iterval k β, k, we apply mea value theorem or the uctio ad we id c k k, ) k which veriies the relatio ) ) k k β = c k ) β. So From., we obtai ) k β ) k )) k β = β ) ) k c k ) = β ) k c k ). x) x) dx. Similarly, or the uctio o the iterval [ k α, ] k, applyi mea value theorem, we id b k k, ) k so that ) ) k k α = b k ) α. The k β ) ) k )) k α = α ) k β b k ).

7 AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE 3 Applyi Lemma.. we id that α ) ) k β b k ) = α Fially, we obtai a = ) ) ) )) + α x) x) dx. x) x) dx + β x) x) dx. I we choose the particular values or the umbers α ad β rom propositio 3., we obtai some results as: For α = ad β =, we obtai x) x) dx = ) ) ) )) + ) k ) k x) x) dx; For α = ad β =, we obtai x) x) dx ) ) k k = )) ))) + x) x) dx + Usi the ormula o iteratio by parts, we obtai x) x) dx ) ) k k = x) x) dx. x) x) dx. I we choose x) = we obtai the ext result, which represet the theorem rom [4]. Corollary 3.. Chiţescu) Let : [, ] R be a cotiuously dieretiable uctios. Let deie α [, ] ad the sequece or all N. The: a = x)dx ) k α, a = α ) [) )].

8 3 D.Ş. MARINESCU, M. MONEA Proo. We apply 3. or x) =. 4. APPLICATIONS At the ed o this ote, we preset three applicatios whose solutios are obtaied usi the results rom the previous pararaph. The irst problem reads Problem. Let the sequece a = or ay N. Show that a = l k k 3 + k, ad evaluate l a ). Solutios: By usi Lema. or the uctios, : [, ] R, x) = x) = x, we have so But x + x dx = k k 3 + k = k k ) + k) = a = k + k x + ) dx = x + x k k + k = x From Prop. 3., we obtai l ) a = 4 + = x) x) dx = ) ) ) )) + x + x dx = 4 + k, x ) k x + x dx, + l + x) ) k x) x) dx ) dx = 4 + x +x l x + ) x +x, = l. = 3 l. 4 Problem. Let N, ad the uctio : [, ] R, x) = xe [x]. We ote V, the volume o the solid obtaied by rotati the raph o about x axis. Fid V.

9 AN EXTENSION OF A RESULT ABOUT THE ORDER OF CONVERGENCE 33 Solutio: As V = π x) dx = π We have The But so xe [x] dx = k k xe [x] dx = V = π xe x dx = xe x + xe [x] dx, we evaluate irst e k ) k k k k ) e = e x dx = xe x V = π 3e. 4 xdx = xe x dx. e x 4 xe [x] dx. k e k ). = 3e, 4 Problem 3. Pr. 535 rom AMM 9/) Let be a cotiuously dieretiable uctio o [, ]. Let A = ) ad let B = x / x) dx. Evaluate i terms o A ad B. x) dx ) ) k k ) k ) Solutio: We perorm the chae o variables x = y, so we have x) dx = y y ) dt ad x / x) dx = y ) dt. Usi the otatio y) = y ), the hypothesis becomes ) = A ad The, we compute = x) dx y) dy = B. ) ) k k ) k ) y y) dy = y y) dy k ) k

10 34 D.Ş. MARINESCU, M. MONEA rom propositio 3.. But y y) dy = y y) y) dy = ) Fially we obtai that the it value is A B. y) dy = A B. Reereces [] V. Arsite, Probleme de Calcul Iteral, Ed. Uiv. Bucureşti, 995. [] A. Béyi, M. Bobeş, Asupra uei probleme de opiadă, Gazeta Matematică, CXIII8), o.. pa [3] A. Béyi, C. Niţu, Asupra uei probleme deschise, Gazeta Matematică, CXV), o.7-8-9, pa [4] M. Chiriţă, I. Chiţescu, A. Costatiescu, Aplicaţii ale iteralei Riema, Gazeta Matematică, XCII988), o.4. pa [5] J. B. Dece, Th. P. Dece, Advaced Calculus, Elsevier, Lodo,, [6] D. Ş. Mariescu, V. Corea, Asupra uor probleme cu şiruri, Gazeta Matematică, CV), o.5-6. pa. -. [7] W.J. Kaczor, M.T. Nowak, Problems i Mathematical Aalysis III, AMS, 3. [8] G. Pólya, G. Szeö, Problems ad Theorems i Aalysis I, Sprier, Berli, 998. [9] J.A. Fridy, Itroductory Aalysis, Academic Press,. [] M. Spivak, Calculus, Cambride Uiversitary Press, 994. [] *** - America Mathematical Mothly, 7), o. 9, problem 535. Da Ştea Mariescu Natioal Collee Iacu de Huedoara, Romaia, Huedoara, 3378, Victoriei o. address: mariescuds@yahoo.com Mihai Moea Natioal Collee Decebal, Romaia, Deva, 338, Decembrie o. address: mihaimoea@yahoo.com