Geeral Mathematics Vol. 4, No. 2 (26, 43 54 O a class of coverget sequeces defied by itegrals Dori Adrica ad Mihai Piticari Abstract The mai result shows that if g : [, ] R is a cotiuous fuctio such that eists ad it is fiite, the for ay cotiuous > fuctio f : [, ] R f(g( d = f( d. The order of covergece i the above relatio, cosequeces ad some applicatios are give. 2 Mathematics Subject Classificatio: 26D5, 65D3 Key words ad phrases: quadrature rule, umerical itegratio, error bouds, optimal quadrature Received March 6, 26 Accepted for publicatio (i revised form April, 26 43
44 Dori Adrica ad Mihai Piticari Itroductio There are may importat classes of sequeces defied by usig Riema itegrals. We metio here oly two. The first oe is called the Riema- Lebesgue Lemma ad it asserts that if g : [, + R is a cotiuous ad T -periodic fuctio, the for ay cotiuous fuctio f : [a, b] R, a < b, the followig relatio holds: ( b a f(g(d = T T d b a f(d For the proof we refer to [3] (i special case a =, b = T ad [8]. I the paper [] we have proved that a similar relatio as ( holds for all cotiuous ad bouded fuctios g : [, + R havig fiite Cesaro mea. The secod oe was give i our paper [2] ad shows that if f : [, + R is a cotiuous fuctio such that f( eists ad it is fiite, the (2 for ay real umber a >. a f( d = f( d, I this paper we ivestigate the class of sequeces defied by f(g( d, where f, g : [, ] R are cotiuous fuctios. The mai results i [6] are obtaied as cosequeces ad some applicatios are give. 2 The mai results We begi with two preiary results.
O a class of coverget sequeces defied by itegrals 45 > Lemma. Let g : [, ] R be a cotiuous fuctio such that eists ad it is fiite. The (3 g( d = g(u u du. Proof. Defie the fuctio h : [, ] R, if (, ] (4 h( = if = > It is clear that h is cotiuous ad deote We have g( d = = H( H( = h(tdt. h( d = H( H( d = If < a <, the we ca write H( d H( d = d a H( d + H( d H( d. a H( d (5 a H(α + ( am, where α [, a] ad M ma H(t. t [,]
46 Dori Adrica ad Mihai Piticari Cosider ε > such that a > ε. Because 2M H(α =, it follows that a H(α < ε for all positive itegers N(ε. From (5 we 2 get i.e. > Lemma 2. class C, H( d ε 2 + ( am < ε ( 2 + + ε 2M H( d = ad the coclusio follows. M = ε, Let g : [, ] R be a cotiuous fuctio such that eists ad it is fiite. The for ay fuctio f : [, ] R of (6 f(g( d = f( d Proof. Deote G( = f(g( d = = G( f( = G(f( g(t dt, [, ], ad ote that t (7 = f( d We will prove that f( g( d (f ( + f(g( d (f ( + f(g( d (f ( + f(g( d. (f ( + f(g( d =.
O a class of coverget sequeces defied by itegrals 47 Ideed, by cosiderig M = ma [,] f ( + f( we have Usig that (f ( + f(g( d f ( + f( G( d relatio (6 follows from (7. > M Our mai results are the followig. Theorem. G( d. G( d = (see the proof of Lemma the desired Let g : [, ] R be a cotiuous fuctio such that eists ad it is fiite. f : [, ] R the relatio (6 holds. The for ay cotiuous fuctio Proof. Accordig to the well-kow Weierstrass approimatio theorem, cosider (f m m a sequece of polyomials uiformly coverget to f o the iterval [, ]. Let ε > be a fied real umber. We will show that we ca fid a positive iteger N(ε such that for ay N(ε ad for ay [, ], we have (8 f(g( d f( d < ε ( From techical reasos, take ε = ε/ 2 d + ad cosider the positive iteger N(ε with the property that f m ( f( < ε for ay [, ]. Because f ad g are bouded it follows that we ca assume that f ad g. For m N(ε we have f m (g( ε g( f(g( f m (g( + ε g(,
48 Dori Adrica ad Mihai Piticari hece f m (g( d ε g( d f(g( d (9 ad From Lemma 2 we have f m (g( d + ε g( d f m (g( d = f m ( d ε g( d = ε d ad it follows that for ay positive iteger N (ε ad f m (g( d ε ε f m (g( d + ε +ε g( d f m ( d d ε g( d f m ( d d + ε But f( ε < f m ( < f( + ε imply for all N (ε (f( ε ( d ε d + (f( + ε ( d + ε d + f(g( d.
O a class of coverget sequeces defied by itegrals 49 The last relatio is equivalet to f(g( d f( d (2 < ε ad the coclusio follows. = ε, for all N (ε, Remarks. Cosider the fuctio h : [, ] R, if h( = > if =. d + Because eists ad it is fiite, it follows that fuctio h is > cotiuous o [, ]. Applyig the result i Theorem we obtai that for ay cotiuous fuctios f, h : [, ] R the followig relatio holds: ( f(h( d = f( h(d Relatio ( was proved i [6] i the case whe f is differetiable ad f is cotiuous o [, ]. 2 If u : [, ] R is a cotiuous fuctio such that its right derivative at eists ad it is fiite, the the fuctio = u( u( satisfies the hypotheses i Theorem. From (6 it follows i.e. f(g( d =, ( f(u( d = u( f(d
5 Dori Adrica ad Mihai Piticari I the paper [6] (see also [3] is proved that the above relatio holds eve f, u are oly cotiuous o [, ]. 3 If f =, the costat fuctio o [, ], from ( we get the result i paper [7]. The order of covergece i ( is give i the followig result. Theorem 2. Let f : [, ] R be a fuctio of class C ad let h : [, ] R be a cotiuous fuctio. The [ f( h(d ] f(h( d (2 = (f( + f ( where H( = h(tdt. Proof. We ca write Therefore [ f( f(h( d = = f(h( h(d H( d, f((h( d (f( H( d. ] f(h( d = (f( H( d. Fuctios f(, H( satisfy the hypothesis i Theorem, hece we have (f( H( d = (f( + f ( H( d
O a class of coverget sequeces defied by itegrals 5 ad the desired relatio follows. Remarks. Writig h( = if ad h( =, where > g : [, ] R is a cotiuous fuctio such that > fiite, from ( we derive the followig relatio [ f( = (f( + f ( d ( ] f(g( d g(t dt d. t This is the order of covergece i (6 whe f is of class C. eists ad it is 2 If h =, the costat fuctio o [, ], from ( we derive Problem 2.83.b i [3]. 3 Some applicatios Applicatio. If f : [, ] R is a cotiuous fuctio, the f( + d = π 2 4 f(. 2 If f : [, ] R is a fuctio of class C, the [ π ] 4 f( f( d = (f( + f ( + 2 These results follows from (6 ad (3, where =, [, ]. + 2 arctg d. If f( = for all [, ], the we get Problem 2 of the 2th Form i fial Roud of Romaia Natioal Olympiad 26.
52 Dori Adrica ad Mihai Piticari Applicatio 2. (Romaia Natioal Olympiad, Couty roud 2, partial statemet If a >, the 2 The followig relatio holds a + d = l a + a. ( (4 l a + a a + d = = ( + a 2. Ideed, takig i (6 f = ad = we easily derive the first a + relatio. For the secod oe we use (3 for the same choosig of fuctios. The right had side i (3 becomes = ( dt d = a + t ( l + d = a l( + a l a d ( + ( d = a = If a =, from (4 we get the iterestig relatio ( (5 l 2 + d = π2 2 = ( + a 2. Applicatio 3. If f : [, ] R is a cotiuous fuctio, the (6 f( l( + d = π2 2 f( 2 If f : [, ] R is a fuctio of class C, the [ π 2 (7 2 f( where ζ is the well-kow Riema, s fuctio. ] f( l( + d = 3 4 (f( + f (ζ(3,
O a class of coverget sequeces defied by itegrals 53 To prove (6 we take i (6, = l( +. We have d = = ( + = 2 l( + d = ( + d = = ζ(2 2 2 2 ζ(2 = 2 ζ(2π2 2. I order to prove (7 we use relatio (3 ad observe that i the right had side we obtai = ( ( l( + t dt d = t ( + d = = Refereces 2 = ( 3 ( + t dt d = ( + = ζ(3 2 2 ζ(3 = 3 3 4 ζ(3. [] Adrica, D., Piticari, M., A etesio of the Riema-Lebesgue lemma ad some applicatios, Proc. Iteratioal Cof. o Theory ad Applicatios of Mathematics ad Iformatics (ICTAMI 24, Thessaloiki, Acta Uiversitatis Apulesis, No.8(24, 26-39. [2] Adrica, D., Piticari, M., O a class of sequeces defied by usig Riema itegral, Proc. Iteratioal Cof. o Theory ad Applicatios of Mathematics ad Iformatics (ICTAMI 25, Albac, Acta Uiversitatis Apulesis, No. (25, 38-385. [3] Dumitrel, F., Problems i Mathematical Aalysis (Romaia, Editura Scribul, 22.
54 Dori Adrica ad Mihai Piticari [4] Lag, S., Aalysis I, Addiso-Wesley, 968. [5] Mocau, M., Sadovici, A., O the covergece of a sequece geerated by a itegral, Creative Math., 4(25, 3-42. [6] Muşuroia, N., O the its of some sequeces of itegrals, Creative Math., 4(25, 43-48. [7] Păltăea, E., O reprezetare a uui şir de itegrale, G.M. No. (23, 49-42. [8] Sireţchi, Gh., Mathematical Aalysis II. Advaced Problems i Differetial ad Itegral Calculus, (Romaia, Uiversity of Bucharest, 982. Babeş-Bolyai Uiversity Faculty of Mathematics ad Computer Sciece Cluj-Napoca, Romaia E-mail address: dadrica@math.ubbcluj.ro Dragoş-Vodă Natioal College Câmpulug Moldoveesc, Romaia