On a class of convergent sequences defined by integrals 1

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1 Geeral Mathematics Vol. 4, No. 2 (26, 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

2 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.

3 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 [,]

4 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 ( 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 =.

5 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(,

6 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.

7 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

8 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

9 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.

10 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,

11 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( 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 = 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, [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, [3] Dumitrel, F., Problems i Mathematical Aalysis (Romaia, Editura Scribul, 22.

12 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, [6] Muşuroia, N., O the its of some sequeces of itegrals, Creative Math., 4(25, [7] Păltăea, E., O reprezetare a uui şir de itegrale, G.M. No. (23, [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 address: dadrica@math.ubbcluj.ro Dragoş-Vodă Natioal College Câmpulug Moldoveesc, Romaia