Chapter 8. Uniform Convergence and Differentiation.

Chapter 8 Uiform Covergece ad Differetiatio This chapter cotiues the study of the cosequece of uiform covergece of a series of fuctio I Chapter 7 we have observed that the uiform limit of a sequece of cotiuous fuctio is cotiuous (Theorem 4 Chapter 7) We shall ow ivestigate whether the uiform limit of a differetiable fuctio is differetiable Covergece is most effectively treated i the settig of metric spaces which allow for geeralizatio to the space of bouded fuctios, whose codomai is a complete metric space But we shall ot itroduce this settig here, preferrig to use the equivalet techique ot phrased i that settig Some observatio as etesio to "complete metric space" is apparet We shall cofie strictly to real valued fuctios o subset of the real umbers 8 The Weierstrass M Test The first test for uiform covergece of a series of fuctio is a form of compariso test Theorem (Weierstrass M Test) Suppose ( f k : E R, k =, 2, ) is a sequece of fuctios Suppose ( M k ) is a sequece of o-egative real umbers for which is coverget ad that for each iteger k, the fuctio f k is M k k= bouded by M k, ie, f k () [ M k for all i E The the series k= f k () coverges uiformly o E Proof Sice for each i E, f k () [ M k ad sice M k is coverget, k= k= f k () is coverget for each i E, by the Compariso Test (Propositio 2 Chapter 2 Series) It follows by Propositio 4 of Chapter 3 that f k () is coverget for each i E Hece f k () is poitwise coverget Uiform covergece of k= f k () is a cosequece of M k is uiformly coverget (sice it is idepedet k= k= of ) Here is how we deduce this is a Cauchy series, sice it is coverget Hece give ay ε > 0, there eists M k k= a positive iteger N such that for all N ad for ay p i P, +p M k < k=+ 2 Thus, it follows that for all N, for ay p i P ad for ay i E, +p +p +p f k () [ f k () [ ------------------- () k=+ k=+ k=+ M k < 2 Therefore, for all N ad for all i E, f k () [ 2 < k=+ Hece, for ay iteger N ad for all i E, k= Ng Tze Beg 2007

Chapter 8 Uiform Covergece ad Differetiatio k= f k () k= f k () = k=+ f k () < Therefore, this says that f k () coverges uiformly to f ()= f k () k= k= Remark Coditio () above defies a otio which we shall call "uiformly Cauchy" We may formulate a criterio for uiform covergece i terms of iequality () or beig uiformly Cauchy, but it is the M-test that is more readily applicable, easy to apply 82 A criterio for Uiform Covergece: Uiformly Cauchy Defiitio 2 A sequece of fuctios ( f k : E R ) is said to be uiformly Cauchy if give ay ε > 0, there eists a iteger N such that for all > m N ad for all E f () f m () < Theorem 3 The sequece of fuctios ( f k : E R ) coverges uiformly if ad oly if ( f k : E R ) is uiformly Cauchy Proof If the sequece ( f k ) coverges uiformly to f, the give ay ε > 0, there eists a iteger N such that for all N ad for all E, f () f () < 2 Therefore, for all itegers, m such that > m N ad for all E, f () f m () = f () f ()+ f () f m () [ f () f () + f m () f () <, 2 + 2 = Thus, by Defiitio 2, ( f k ) is uiformly Cauchy Coversely ow suppose ( f k ) is uiformly Cauchy The give ay ε > 0, there eists a iteger N such that for all m > N ad for all E f () f m () < ----------------------- () 2 Hece for each, ( f k () ) is a Cauchy sequece ad so (by Cauchy Priciple of Covergece), ( f k () ) coverges to a fuctio f () poitwise Thus for ay i E, f () f () = f () lim f k () = lim f () f k () ------------- (2) kd kd Now by (), for ay k > N ad for all E f () f k () < 2 Therefore, for all i E, lim f () f k () [ kd 2 < It follows, the from (2), that for all N ad for all E, is to say, f f uiformly o E f () f () < That Eample 4 The followig three statemets are cosequece of the Weierstrass M Test () is uiformly coverget o the closed ad bouded iterval [,] = 2 Sice for all i [,] ad for all positive itegers, ad is 2 [ 2 = 2 coverget, by Weierstrass M Test (Theorem ), the series is uiformly coverget Ng Tze Beg 2007 2

Chapter 8 Uiform Covergece ad Differetiatio (2) is uiformly coverget o R by Weierstrass M Test sice for each = 2 + 2 positive iteger ad for all i R, ad is coverget 2 + 2 [ 2 = 2 (3) is uiformly coverget o the closed ad bouded iterval [ a, a], = 2 + 2 where a > 0 Sice for each positive iteger ad for all i [ a, a], ad is coverget, by the Weierstrass M Test, the series 2 + 2 [ a a 2 = 2 is uiformly coverget o the [ a, a] 2 + 2 = Eample 5 We ca have a sequece of fuctios covergig o-uiformly to a costat fuctio For eample the sequece of fuctios ( f ) where for each positive iteger, f :R R is defied by f ()=, is such a sequece + 2 2 For each i R, f () 0 We deduce this as follows For each 0 i R, f ()= as For = 0, for each + d 0 2 0 + 2 = 0 2 positive iteger, f (0) = 0 Hece f (0) 0 Thus the poitwise limit of the sequece ( f ) is the zero costat fuctio, ie, f f poitwise, where f () = 0 for all To see that the covergece is ot uiform, we eamie what it meas for a covergece ot to be uiform We start with the egatio of the defiitio of uiform covergece i Defiitio Chapter 7 The sequece ( f ) does ot coverge uiformly to f meas there eists a ε > 0 such that for ay positive iteger N, there eists a iteger N ad a elemet i the domai of f such that f ( ) f ( ) m So we shall proceed to fid this ε by eamiig the values that f () f () ca take Recall that f () = 0 for all Hece for all positive iteger ad for all i R, f () f () = f () = + 2 2 = + 2 2 [ Thus the set { f () f () : c R} is bouded above by for all positive iteger Therefore, sup{ f () f () : R} eists for each positive iteger Now by quick ispectio of the fuctio rule for f, we see that f ( )= for each positive 2 iteger Cosequetly, sup{ f () f () : R} f ( )= for all positive 2 iteger We ca thus take ε = /2 Thus for each positive iteger N i P, choose = N ad N = the we have N f ( ) f ( ) = f N ( N ) = 2 m This meas the covergece is ot uiform We may also show that the sequece ( f ) is ot uiformly Cauchy so that by Theorem 3 the covergece is ot uiform Observe that for ay positive itegers ad m, Ng Tze Beg 2007 3

Chapter 8 Uiform Covergece ad Differetiatio f () f m () = + 2 2 m ( m) + m(m )3 + m 2 2 = ( + 2 2 )( + m 2 2 ) ( m) + m(m )3 = ( + 2 2 )( + m 2 2 ) Thus, f ( ) f m( ) = ( m )+ m ( m ) --------------------------- () 2 + m2 2 For each positive iteger N, choose ay N, choose m = 3 ad take N = The we have usig (), f ( ) f m( ( 3)+3(2) ) = 2(0) = 5 So takig =, we 5 have show that, for each positive iteger N, we ca fid itegers ad m N ad a elemet N i the domai R, such that f ( N ) f m ( N ) m = 5 Thus, by Defiitio 2,( f ) is ot uiformly Cauchy 83 Uiform Covergece ad Differetiatio Theorem 4 of Chapter 7 says that cotiuity behaves well uder uiform covergece, ie, the uiform limit of a sequece of cotiuous fuctios is cotiuous But differetiability is less well behaved ad eve less well behaved tha itegrability The uiform limit of differetiable fuctios eed ot be differetiable There are various possibilities Each f of the sequece ( f ) may be differetiable but the sequece of the derivatives ( f ' ) may ot be coverget ad whe ( f ' ) is coverget, the covergece eed ot be uiform Thus, if we were to formulate a result usig the uiform covergece of derivatives, the uiform covergece of the sequece of derivatives will have to be assumed I this way by usig the good behaviour of itegratio uder uiform covergece, we use the Fudametal Theorem of Calculus to deduce result about the derivatives of the limitig fuctio of the sequece ( f ) ad the uiform covergece of ( f ) if the uiform covergece of the derivatives ( f ' ) is assumed ad that the derivatives f ' are all cotiuous Eample 6 A sequece ( f ) covergig uiformly to a fuctio f but ( f ') does ot coverge to f ' Let ( f ) be a sequece of fuctio defied o R by f ()= for i R + 2 The f f poitwise, where f is the zero costat fuctio The covergece is uiform We deduce this as follows f () f () = for 0 ad Now ote that + 2 = f (0) f (0) = 0 + + achieves its miimum i (0, ) at = Hece the maimum of the reciprocal is sup{ f () f () : c R} =f ( As give ay )= 2 2 d 0, ε > 0, there eists a positive iteger N such that m N u Hece for all 2 < i R, m N u f () f () [ sup{ f (y) f (y) : y c R} = 2 < This meas by Defiitio 3 Chapter 7, f f uiformly Ng Tze Beg 2007 4

Chapter 8 Uiform Covergece ad Differetiatio Now for each positive iteger, f is differetiable ad f ()= 2 ( + 2 ) 2 2 2 For 0, f ()= as Plaily f ' (0) = as ( + 2 ) 2 d 0 4 = 0 Therefore, f ' g poitwise, where g()= 0,! 0 Plaily g f ' = 0, = 0 We ote that each f is cotiuously differetiable, ie, f ' is cotiuos Therefore, sice g is ot cotiuous at 0, ( f ' ) does ot coverge uiformly 84 Uiform Covergece ad Itegratio It is ot ureasoable i the light of Eample 6, to make the requiremet that ( f ' ) be uiformly coverget ad that each f ' be cotiuous Perhaps the we may be able to deduce f ' = lim f With the coditio that each f ' is cotiuous ad the d sequece ( f ' ) is uiformly coverget, by Theorem 4 Chapter 7, g = lim d f is cotiuous ad hece Riema itegrable o ay bouded iterval If we have Riema itegrability how ca we the show that g = f '? The et questio is the whe does the followig equatio a g(t)dt =d lim a f (t)dt -------------------------- (*) hold? That is, dose itegratig each f ' ad fidig its limit the same as itegratig g? The right had side of (*) by the Fudametal Theorem of Calculus (Theorem 43 Chapter 5) is just lim d a f (t)dt = lim d (f () f (a)) = f () f (a) assumig f f poitwise So if we assume (*), the we have a g(t)dt = f () f (a) It will the follow by the Fudametal Theorem of Calculus (Theorem 45 Chapter 5) that g() = f ' () for each sice g is cotiuous Hece g = f ' Thus f ' f ' uiformly I this fashio, iformatio about itegratio ca tell us iformatio about differetiatio What we require is a simple result regardig the covergece of Riema itegrals So we state the result below Theorem 7 Suppose ( g :[a, b] R ) is a sequece of cotiuous fuctio covergig uiformly to g:[a, b] R The g is cotiuous o [a, b], b b b lim g ad d a (t) g(t) dt = 0 a g(t)dt =d lim g a (t)dt Proof For each positive iteger, g is cotiuous o [a, b] ad so g is itegrable o [a, b] (see eg, Theorem 23 Chapter 5) Sice g g uiformly, by Theorem 9 Chapter 7, g is cotiuous o [a, b] ad hece itegrable o [a, b] Therefore, g g is Riema itegrable o [a, b] Now for each positive iteger, --- () a b g () a b g()d = a b ( g () g())d [ a b g () g() d by Theorem 53 Chapter 5 Itegratio Sice g g uiformly, give ay ε > 0 there eists a positive iteger N such that for all iteger, Ng Tze Beg 2007 5

Chapter 8 Uiform Covergece ad Differetiatio Thus, m N u g () g() < m N u a b g () g() d [ a b 2(b a) for all i [a, b] 2(b a) d = 2 < Therefore, lim g d a (t) g(t) dt = 0 Now, b b b m N u g a () a g()d [ g a () g() d by (2) This meas < ε b b g a () d a g()d This completes the proof 85 Differetiatig A Sequece Now we formulate our theorem about differetiatio -------------- (2) by () Theorem 8 Let I be a o-empty iterval (bouded or ubouded) Suppose we have a sequece of cotiuously differetiable fuctios ( f : I R ) That is, for each positive iteger, f is differetiable ad the derived fuctio f ' : I R is cotiuous Suppose the followig two coditios are satisfied: () ( f : I R) coverges poitwise to a fuctio f : I R ; (2) ( f ' : I R) coverges uiformly to a fuctio g : I R The f : I R is differetiable, g : I R is cotiuous f ' = g ad f f uiformly o ay closed ad bouded iterval [a, b] I Proof Fi a poit a i I We shall proceed to use Theorem 7 For each positive iteger, sice f : I R is cotiuous, by the Fudametal Theorem of Calculus (Theorem 43 Chapter 5) a f (t)dt = f () f (a) -------------------- () By Theorem 7, sice f ' g uiformly as give by coditio (2), a f (t)dt d a g(t)dt for each i I Therefore, for each i I by (), a g(t)dt =d lim a f (t)dt = lim d f () lim d f (a)=f () f (a) -------- (2) by coditio () Note that g is cotiuous o I by Theorem 4 Chapter 7 Thus by the Fudametal theorem of Calculus (Theorem 45 Chapter 5), the fuctio G: I R defied by G()= a g(t)dt is differetiable ad G'() = g() for each i I Therefore, it follows from (2) that for each i I, g() = f ' () sice G() = f () f (a) Hece we have proved the first two assertios Observe that sice f ' g uiformly, the sequece of fuctios ( F :I R ), where for each positive iteger, F is defied by F ()= a f (t)dt for i I, also Ng Tze Beg 2007 6

Chapter 8 Uiform Covergece ad Differetiatio coverges uiformly to G()= a g(t)dt o ay closed iterval [a, b] i I This is easily deduced as follows Sice f ' g uiformly, for ay ε > 0, there eists a positive iteger N such that for all iteger, m N u f () g() < for all i [a, b] 2(b a) Hece m N u F () G() = a f (t)dt a g(t)dt [ a f (t) g(t) dt [ ( a) a = 2(b a) 2(b a) < for all i [a, b] Thus F G uiformly o [a, b] Sice for each positive iteger, F () = f () f (a) for all i I (by ()) ad sice F G uiformly o [a, b], f () f (a) coverges uiformly to f () f (a) uiformly o [a, b] ad so sice f (a) f (a) uiformly, f f uiformly o [a, b] This proves the last assertio ad thus completes the proof Remark If I is a closed ad bouded iterval, say [a, b], the the coclusio of Theorem 8 will give uiform covergece of ( f ) 2 Sice by Theorem 3, ( f ' ) coverges uiformly is equivalet to ( f ' ) beig uiformly Cauchy, we may replace coditio (2) of Theorem 8 by requirig that ( f ' ) be uiformly Cauchy 3 Coditio () of Theorem (8) may be replaced by a simpler lookig coditio ()' : "There eists a elemet a i I such that the sequece ( f (a) ) coverges" The coditio (2) would imply poitwise covergece for ( f ) o I We deduce this as follows By () i the proof of Theorem 8, f ()= a f (t)dt + f (a) By Theorem 7, sice f ' g uiformly by coditio (2), a f (t)dt coverges poitwise to a g(t)dt Therefore, if ( f (a) ) is coverget ad coverges to, say f (a), the f coverges poitwise to a g(t)dt + f (a) 4 A stroger versio of Theorem 8 is also true Uder the hypothesis () that each f is differetiable o I (ot ecessarily cotiuously differetiable), (2) there eists a elemet a i I such that the sequece ( f (a) ) coverges ad (3) ( f ' : I R) coverges uiformly to a fuctio g : I R, we ca coclude that the sequece ( f ) coverges to a fuctio f such that f ' = g The proof is more delicate sice f ' may ot be itegrable ad we shall eed to use oly the cosequece of differetiability We shall prove this below Theorem 8' Let I be a o-empty iterval (bouded or ubouded) Suppose we have a sequece of differetiable fuctios ( f : I R ) Suppose the followig two coditios are satisfied: () There eists a poit 0 such that the sequece ( f ( 0) ) is coverget (2) ( f ' : I R) coverges uiformly to a fuctio g : I R The ( f : I R ) coverges o I to a differetiable fuctio f : I R such that f ' = g ad f f uiformly o ay closed ad bouded iterval [a, b] I Ng Tze Beg 2007 7

Chapter 8 Uiform Covergece ad Differetiatio Proof Take a closed ad bouded iterval [a, b] cotaiig 0 i I We shall show that ( f : I R ) is uiformly coverget o [a, b] Sice ( f ' : I R) coverges uiformly to a fuctio g : I R, ( f ' : I R) is uiformly Cauchy o [a, b] This meas give ε > 0, there eists a iteger N such that for all i I ad, m N f ' () f m' () < ε/ (2L), () where L = b a is the legth of a closed ad bouded [a, b] i I Now for ay iteger, m > 0, f () f m() = ( f () f m() ) ( f ( 0 ) f m ( 0 )) + ( f ( 0 ) f m ( 0 )) = ( f '(c) f m' (c) )( 0 ) + ( f ( 0 ) f m ( 0 )) for some c betwee ad 0 by the Mea Value Theorem ( f '(c) f m' (c) ) 0 + f ( 0 ) f m ( 0 )) ---------- (2) by the triagle iequality Sice ( f ( 0) ) is coverget, it is Cauchy Hece there eists a iteger M, such that,, m M f ( 0 ) f m ( 0 ) < ε / 2 ----------------------------- (3) Therefore, it follows from (2) ad (3) that for all i [a, b],, m ma (N, M) f () f m () < ε 0 /(2L)+ ε / 2 < ε This proves that ( f ) is uiformly Cauchy o [a, b] Therefore, by Theorem 3, ( f ) coverges uiformly to a fuctio, say f o [a, b] For ay i I, there eists a closed ad bouded iterval D cotaiig both ad 0 Thus, by we have just proved, f coverges uiformly to a fuctio, f o D By uiqueess of limit, the limitig fuctio f is uique Hece f coverges poitwise to a fuctio, f o the iterval I I particular, by the above proceedig we ca coclude that f coverges uiformly to a fuctio o ay closed ad bouded iterval D i I We shall ow show that the limitig fuctio f is differetiable ad that f ' = g Take ay c i I We shall show that f ' (c) = g(c) f () f (c) Defie g ()= c,! c The g is cotiuous o I sice f is f (c), = c differetiable at c ad g (c) = f '(c) Observe that the sequece ( g ) is poitwise covergece o I {c}, sice ( f ) is Because the sequece ( ( f '(c) ) is coverget ad coverges to g(c), ( g ) is poitwise coverget o I We shall show that ( g ) is uiformly coverget o I For ay c, g () g m () = f () f (c) c f m () f m (c) c = f (d) f m (d) for some d betwee c ad by the Mea Value Theorem Sice ( f ' : I R) is uiformly Cauchy o I, for all i I, there eists a iteger N 0 such that, m N 0 f ' () f m' () < ε It follows that for ay c,, m m N 0 u g () g m () = f (d) f m (d) < Also,, m m N 0 u g (c) g m (c) = f (c) f m (c) < Hece ( g ) is uiformly coverget o I Ng Tze Beg 2007 8

Chapter 8 Uiform Covergece ad Differetiatio f () f (c) Note that for c, g () d c ad g (c)= f (c) d g(c) Thus f () f (c) g () d G()= c,! c Sice each g is cotiuous, the uiform limit g(c), = c G is cotiuous o I Therefore, f () f (c) Lim d c c = Lim d c G()=G(c)=g(c) This shows that f is differetiable at c ad f ' (c) = g(c) This completes the proof 86 Differetiatig Power Series We shall apply Theorem 8 to power series First a eample Eample 9 For each positive iteger, let f ()= + + 2 for 2! + + ( )! i R (This is the familiar trucated epoetial epasio) f () is the -th partial sum of the series =0! Defie f 0 () = 0 for all i R By the Ratio Test (see eg Theorem 8 Chapter 7) the series coverges for all, as! ( )! = d 0 < Thus, for each i R, the sequece ( f ()) coverges to a value which we deote by f ( ) I this way we defie a fuctio f : R R This is the well kow epoetial fuctio Note that f f poitwise o R Now fi a positive umber K ad cosider the closed iterval [ K, K] The we claim that f f uiformly o [ K, K] We ow proceed to prove just this fact Note that for each o-egative iteger, ad for all i [ K, K], K! [ K! Therefore, sice is coverget, by the Weierstrass M Test (Theorem ), =0! coverges uiformly o [ K, K] That is to say f f uiformly o [ K, K]! =0 Now for each positive iteger, f is a polyomial fuctio ad so is cotiuous o R ad hece o [ K, K] Therefore, by Theorem 4 Chapter 7, f is cotiuous o [ K, K] For each positive iteger, the derived fuctio is give by f ()= + + 2 2! + + ( 2 2)! = f () Thus the sequece ( f ') = ( f - ) coverges uiformly o [ K, K] to f Hece, by Theorem 8, f is differetiable ad f ()= lim d f ()= lim d f ()=f() for all i [ K, K] Sice this is true for all K > 0, f ' () = f () for all i R Ng Tze Beg 2007 9

Chapter 8 Uiform Covergece ad Differetiatio Remark We could have proved the uiform covergece of f o [ K, K] i Eample 9 by a Compariso Test just as i the proof of the Weierstrass M Test It is really also a test for absolute covergece Hece the test is restricted i this way for applicatio Let us follow the argumet of the proof For each o-egative iteger k ad for all i [ K, K], +p k=+ k k! Therefore, for ay positive iteger, k k! K k [ Kk k! +p [ k=+ k k! +p [ k=+ K k K k k! ------------------ () Sice we kow is coverget, the series is a Cauchy series Hece for k=0 k! k=0 k! ay ε > 0, there eists a positive iteger N such that for all N ad for all p i P, +p K k k=+ k! < It the follows from () that for all N ad for all p i P, +p k +p [ K k=+ k! k=+ k k! < for all i [ K, K] Therefore, the series =0! so by Theorem 3 it coverges uiformly o [ K, K] Our et result is about the disk of covergece of the power series is uiformly Cauchy o [ K, K] ad Lemma 0 The power series a ad =0, = a, =2 ( )a 2 =0 a + + all have the same radius of covergece Proof It is sufficiet to show that =0 a ad a = covergece Let r be the radius of covergece of =0 have the same radius of a ad r ' the radius of covergece of a Let be such that < r ' The is = = a coverget Sice for each iteger, a [ a, by the Compariso Test for Series (Propositio 2 Chapter 6), a is coverget (for < r ' ) Therefore, a = is coverget for < r ' Thus, r Hece = = a r ' r (This is because if r ' > r, the we ca choose a 0 such that r < 0 < r ' The by the above argumet we ca show that a 0 is coverget ad cosequetly cotradictig that a 0 is diverget sice 0 > r ) =0 Now we shall show that r r ' Suppose < r Choose a real umber c such that < c < r The both series a coverge It follows that a c 0 =0 ad =0 a c (see Propositio 0 Chapter 6) Therefore, give ay ε > 0, there eists a positive iteger N such that for all iteger N, a c < ε Now take ε = c > 0 It the follows that for all iteger, = =0 Ng Tze Beg 2007 0

Chapter 8 Uiform Covergece ad Differetiatio N a c < Therefore, for all iteger N, ka k ------------------ () k=n k = k=n a k c k k c k < k=n k c k Now otice that k c k is coverget by the Ratio Test (Theorem 2 Chapter 6) k=n ( + ) because for 0, c ad it is plaily coverget c =( + ) c d c < for = 0 Therefore, usig (), by the Compariso Test (Propositio 2 Chapter 6), ka k is coverget It follows that, is coverget Therefore, k=n k k= ka k k r ' It the follows that r r ' ( This is because if r > r' the choose such that r > > r ' But we have show that r ' ad this cotradicts > r ' ) Therefore, r = r ' So a have the same radius of covergece For each =0 ad a = positive iteger, let b = (+)a + The =2 = a = b ad =0 ( )a 2 = ( )b 2 = =2 = b Therefore, by what we have just show, =0 b ad b have the same radius = of covergece It follows that a have the same radius = ad =2 ( )a 2 of covergece Now if we let c 0 = 0 ad for each iteger, let c = a The a =0 + + = =0 c + + = =0 c ad a =0 = =0 ( + )c + = = c Thus agai by what we have proved covergece ad so =0 =0 a + + ad =0 a c ad c have the same radius of = have the same radius of covergece We deduce from Lemma 0 that the power series obtaied from oe by differetiatig term by term have the same radius of covergece We shall ow show that we ca ideed obtai the derivative of the fuctio represeted by the power series by term by term differetiatio withi the radius of covergece Theorem If f ()= a is a real power series with radius of covergece r =0 ad D r = {: < r} is the disc of covergece, the the fuctio f : D r R is differetiable ad f ()= a for each i D r Moreover, both f () ad f ' () = as power series coverge uiformly (ad absolutely) o ay closed iterval [ c, c] D r Proof We show that for ay power series with disc of covergece D r, the power series coverges uiformly o ay closed ad bouded iterval [ c, c] i D r = ( r, r) Ng Tze Beg 2007

Chapter 8 Uiform Covergece ad Differetiatio Let 0 < c < r Take a fied real umber K such that c < K < r The =0 a K coverges absolutely (Theorem 4 Chapter 7) Now sice 0 < c < K, for all i [ c, c], < K Therefore, for ay iteger 0, ad for all i [ c, c], a a K Hece, by the Weierstrass M Test (Theorem ) the iterval [ c, c] Thus, if we write f () for =0 =0 a is uiformly coverget o a for each i [ c, c], the the -th partial sum s ()= a k k d f () uiformly o [ c, c] Similarly, sice by Lemma 0 k=0 =0 a ad = a have the same radius of covergece ad hece the same disc of covergece, s ()= ka k k coverges uiformly o [ c, c] Therefore, by k= Theorem 8, f is differetiable o [ c, c], s ' coverges uiformly to f ' o [ c, c] That is f ()= a = o [ c, c] Sice this is true for ay c with 0 < c < r, f is differetiable o D r = ( r, r) ad f ()= a for each i D r This completes the proof = Remark Theorem says that we ca differetiate a power series term by term i its disc of covergece This is a very importat property of power series fuctio 2 It is Taylor's Theorem that liks power series with other theory of fuctios 3 Thus a real power series represets a ifiitely differetiable fuctio f o its iterval of covergece ad all the derivatives ca be obtaied by term wise differetiatio (Theorem ) We ca thus epress the coefficiet a i terms of the derivatives of f 4 We may prove Theorem directly without usig Theorem 8 as follows: Take with < r ad T, S such that < T < S < r For real, ad h such that 0 < h T, we have h ( ( + h) ) =( + h) +( + h) 2 + + Hece, h ( f ( + h) f ()) = = g (h) where g (h)=a (( + h) +( + h) 2 + + ) for h 0 Defie g (0)=a The g is cotiuous for all h i R I particular g (h) [ a T for h T ------------------- () Now ( T S ) d 0 because 0 < T/S < Therefore, for all sufficietly large we have ( T S ) [ T or T [ S Ng Tze Beg 2007 2

Chapter 8 Uiform Covergece ad Differetiatio Sice =0 =0 a S is coverget, it follows by the Compariso Test that a T is coverget The by the Weierstrass M Test (Theorem ), it follows from () that g (h) coverges uiformly o {h: h T } = Therefore, its sum, ie, its limitig fuctio is cotiuous at 0 by Theorem 3 Chapter 7 This meas as h 0 h ( f ( + h) f ()) = = g (h) d = g (0)= a = Hece f ()= a = Thus by Lemma 0, both f () ad f ' () have the same radius of covergece ad so coverge uiformly (ad absolutely) o ay closed iterval [ c, c] D r 5 Puttig = 0 we ca deduce from Theorem that f '(0) = a ad f () (0) =!a This shows that oe ca deduce the coefficiet a from the sum fuctio, that is the limitig fuctio f 87 Usig Taylor's Theorem Eample 2 Use of Taylor's Theorem Suppose we have the followig differetial equatio: f ' = f with iitial coditio f (0) = Suppose we have proved that sufficietly well behaved differetial equatios have uique solutios The suppose this equatio has a solutio f o the iterval [ K, K] The f is differetiable ad so it is cotiuous ad hece bouded o [ K, K] Thus f () M for all [ K, K] Sice f ' = f, f ' () M for all [ K, K] Now Apply Taylor's Theorem (Theorem 44 Chapter 4) with epasio aroud 0 = 0 The we have for each 2, f ()=f (0)+f (0)+ + k! k f (k) (0) +! f () (0)+ + f (+) (, ) ( + )! where θ, is some poit betwee 0 ad Now by the iitial coditio f '(0) = f (0) = It follows that f () (0) = f (0) = for ay positive iteger Thus the Taylor epasio becomes f ()= + + + k! k +! + + f (,) ( + )! Now we kow that the series ( see Eample 9) + + + k! k +! + coverges uiformly o [ K, K] I particular, the modulus of the Lagrage form of the remaider f (,) + [ K Sice as, ( + )! + M K ( + )! + M ( + )! d 0 f (,) + d 0 as by the Compariso Test for sequeces Hece for each ( + )! i [ K, K], f () as Therefore, by the k=0 k! k = f (,) + d 0 ( + )! Ng Tze Beg 2007 3,

Chapter 8 Uiform Covergece ad Differetiatio Compariso Test for sequeces (Propositio 8 Chapter 2), f ()= for ay =0! i [ K, K] Sice this is true for ay K > 0, f ()= for all i R Thus, =0! the vaishig of the Lagrage remaider R ()= f (+) ( +, ) plays a critical role ( + )! i showig that the solutio of the differetial equatio is give by the ifiite Taylor series k=0 k! k Eample 3 A fuctio that does ot admit a ifiite Taylor series epasio Let f : R R be defied by f ()= e 2,! 0 0, = 0 The sice the epoetial fuctio is differetiable, f is thus a compositio of two differetiable fuctios o 0 ad so is differetiable at 0 Now f () f (0) lim = lim e 2 d0 + 0 d0 + = lim / = lim /2 d0 + e /2 d0 + e /2 ( 2 ) = lim = 0 d0 + 2e /2 3 by L'Hôpital's Rule ad that lim = 0 We ca show i eactly the same maer d0 + e /2 that f () f (0) f () f (0) lim = 0 Hece, lim = 0 ad so f d0 0 d0 0 f () f (0) (0)= lim = 0 d0 0 ( We ca also use the fact that for 0, e 2 m so that Thus for 0, 2 [ e 2 /2 0 < / [ ad so by the Compariso Test, lim / = 0 It follows that e /2 d0 e /2 f f () f (0) (0)= lim = lim ) d0 0 / = 0 d0 e /2 For 0, f ()=e 2 2, where is a polyomial i 3 = 2 = p 3 e ( )e 2 p ( ) /2, p (y) = 2y 3 We ow eamie the limit of the first derivative lim f ()= lim p d0 + d0 + ( )e 2 = lim td p (t)e t2 = lim p (t) td = lim 6t2 e t2 td = lim 3t 2te t2 td = lim 3 e t2 td = 0 2te t2 by L'Hôpital's Rule ad that lim = 0 because lim te I eactly the same td te t2 = t2 td way we show that lim f ()=0 Hece d0 lim f ()=0 d0 Note that f is cotiuous at = 0, sice lim f ()=0 = f (0) Therefore, the eistece of lim f () implies that f is differetiable at = 0 ad that d0 f (0)= lim f ()=0 This meas that f ' is cotiuous at = 0 d0 We shall show that for each positive iteger, lim f () ()=0 ad cosequetly, d0 f () ()=0 First we claim that for each positive iteger, f () ()=p ( )e 2 ------------------------------ () where p ( ) is a polyomial i d0 Ng Tze Beg 2007 4

Chapter 8 Uiform Covergece ad Differetiatio We shall prove this statemet by iductio Note that () is true for =, as we have observed Now assume that () is true for, ie, f () ()=p ( )e 2 Differetiatig we have, f (+) ()=p ( )( 2 )e 2 + p ( )( 2 )e 3 2 = 2 p ( )+ 2 p 3 ( ) e 2 But is a polyomial i Therefore, lettig 2 p ( )+ 2 p 3 ( ) p + ( )=, we see that Thus () is 2 p ( )+ 2 p 3 ( ) f (+) ()=p + ( )e 2 true for + ad so by mathematical iductio, () is true for all positive itegers We et eamie the limit lim f () () We shall ow show that for all positive iteger, d0 lim f () ()=0 d0 Now ote lim f () ()= lim p by a repeated use d0 + d0 + ( )e 2 = lim td p (t)e t2 = lim p (t) td = 0 e t2 of the L'Hôpital's Rule [We ca first compute the limit lim t2k+ (2k + ) = lim t 2k (2k + )!! td e t2 td = lim t (2k + )!! 2 e t2 td 2 k = lim e t2 td 2 k+ = 0 te t2 by a repeated use of the L'Hôpital's Rule ] Similarly, lim f () ()= lim p d0 d0 ( )e 2 = td lim p (t)e t2 = td lim p (t) = 0 e t2 Therefore, lim f () ()=0 Note that, lim f () ()=0 ad f (-) is cotiuous at = 0 implies that d0 d0 f () (0)= lim f () ()=0ad cosequetly f () is cotiuous at 0 sice it is d0 differetiable there [We ca use L'Hôpital's Rule, for this deductio f () (0)= lim f ( ) () f ( ) (0) d0 0 lim f () () Rule, sice eists ad equals 0] d0 We thus have for iteger, = lim f ( ) () d0 = lim f () () d0 = 0 by L'Hôpital's f () ()= p ( )e 2,! 0 0, = 0 Therefore, the -th degree Taylor epasio of f about = 0 gives, f ()=f(0)+f (0)+ + k! k f (k) (0) +! f () (0)+ f (+) ( +, ) ( + )! = 0 + $ 0 + 2! 0 $ 2 + + k! k $ 0 +! $ 0 + f (+) ( +, ) ( + )! = f (+) ( +, ) ( + )! for some θ, betwee 0 ad Hece the remaider, f (+) ( +, ) caot coverge to 0 as teds to for ( + )! otherwise f would be idetically the zero costat fuctio ad thus givig a cotradictio as f is ot a costat zero fuctio Therefore, we caot write f () as a ifiite Taylor series I particular the sequece (f (+) (, )) caot be bouded 88 Covergece of Taylor Polyomials We ow state Taylor's Theorem with Lagrage form of the remaider without proof Ng Tze Beg 2007 5

Chapter 8 Uiform Covergece ad Differetiatio Theorem 4 Taylor's Theorem (with Lagrage form of the remaider) Let I be a ope iterval cotaiig the poit 0 ad be a o-egative iteger Suppose f : I R has + derivatives The for ay i I, f ()=f( 0 )+! ( 0) f ( 0 )+ + k! ( 0) k f (k) ( 0 ) +! ( 0) f () ( 0 )+R (), where the term R () is the Lagrage form of the remaider ad is give by R ()= ( + )! ( 0) + f (+) () for some η betwee ad 0 (Referece: Theorem 44 Chapter 4) As we have see i Eample 2 ad 3, i order to write f as a Taylor series we eed to show that the remaider R () coverges to 0 as for all Oe advatage of havig the series represetatio of a fuctio is to cosider differetiatig the fuctio by simply differetiatig the terms of the series withi the disk of covergece or to cosider itegratig the fuctio term by term withi the disk of covergece Now if f is a fuctio defied o a ope iterval I havig derivatives of all order, ie, f is a smooth fuctio, the Theorem 4 says that for all iteger, f has a Taylor polyomial p ()= k=0 k! ( 0) k f (k) ( 0 ) about the poit 0 i I ad f () = p () + R () If p () f () for i I, the a Taylor series epasio of the fuctio f : I R about the poit 0 is the series ( k=0 k! 0 ) k f (k) ( 0 ) I particular, at each poit i I, lim p d () f () = 0 which is equivalet to lim R Thus f admits a Taylor series epasio if d () = 0 ad oly if it has derivatives of all order ad lim R d ()=0 We shall ow ivestigate the covergece of p () to f () We do this via the Lagrage form of the remaider R () The et result makes use of a criterio of the covergece of R () to 0 Theorem 5 Suppose f : I R is a fuctio defied o the ope iterval I, havig derivatives of all order Let 0 be a poit i I Suppose there eists a closed iterval [ 0 r, 0 +r] i I such that for every iteger ad for all i [ 0 r, 0 +r], there eists M 0 such that f () () [ M The f ()= ( if 0 r k=0 k! 0 ) k f (k) ( 0 ) Proof By Theorem 4, for all i [ 0 r, 0 +r], Ng Tze Beg 2007 6

Chapter 8 Uiform Covergece ad Differetiatio p () f () [ ( + )! ( 0) + f (+) () [ ( + )! r+ M + Sice [ ( Mr ) + ( + )! (Mr ) + ( + )! d 0 for some η betwee ad 0 sice f (+) () [ M + for all such that 0 r, as, by the Compariso Test, p f uiformly o [ 0 r, 0 +r] Hece, f ()= ( if 0 r k=0 k! 0 ) k f (k) ( 0 ) Theorem 5 ca be applied to fuctios with easily observed bouded derivatives of all order Thus sie ad cosie are such fuctios Eample 6 ( ) si()= =0 (2 + )! 2+ = 3 3! + 5 5! + 2 +( )+ (2 )! + for all i R Let f () = si() We oly eed to kow that si' () = cos() ad cos' () = si() Hece f ' () = cos(), f (2) () = si(), f (3) () = cos(), f (4) () = si(), f (5) () = cos(), ad i geeral, f (2+) () = ( ) cos(), f (2) () = ( ) si() for iteger 0 Therefore, f (2+) (0) = ( ) ad f (2) (0) = 0 Thus the Taylor polyomial about = 0 has oly odd powers of f For iteger 0, p 2+ ()= (2k+) (0) ( ) ad p 2+2 () = k=0 (2k + )! 2k+ == k=0 k (2k + )! 2k+ p 2+ () For all iteger 0, f () (), assumig that si(), cos() We deote here f () by f (0) () Therefore, by Theorem 5, p f uiformly o [ K, K], for ay K > 0 Hece p () f () = si() for all i R The followig is a applicatio of Taylor's series Propositio 7 The Euler costat e is irratioal Proof By Taylor's Theorem (Theorem 4), for each iteger 0, e = + + + k! k +! + ( + + )! e for some θ betwee 0 ad Hece for i [0, ], e + + + k! k +! = ( + + )! e Hece takig =, 0 < e + + 2! + k! +! = ( + )! e [ ( + )! e This meas for ay iteger 0, 0 < e + + 2! + k! +! [ ( + )! e -------------- () We have show i Chapter 6 i the sectio o Euler costat γ, that k= k + < t dt = l()< k= k Ng Tze Beg 2007 7

Chapter 8 Uiform Covergece ad Differetiatio Hece we have l(4)> ad cosequetly takig epoetiatio we 2 + 3 + 4 > get e < 4 Thus from () we get 0 < e + + -------------- (2) 2! + k! +! [ 4 ( + )! for ay iteger 0 Thus if e is ratioal, say e = p q i its lowest terms, the from (2) we get for ay iter 0 0 < p q + + -------------- (3) 2! + k! +! [ 4 ( + )! Let = ma(4, q) Multiply (3) by! we get 0 <!p q 2! +! --------- (4) 2! + +! k! + [ + 4 [ 4 5!p But the term q 2! +! is a iteger sice every term i the 2! + +! k! + epressio is a iteger (4) the says it is a iteger i (0, 4/5], cotradictig that there is o iteger i (0, 4/5] Hece e is irratioal 89 Cotiuity of Power Series, Abel's Theorem Now we go back to the questio of cotiuity of a power series fuctio at the boudary of the disc of covergece, if the power series is coverget there For real power series, if the series is coverget at the boudary of the disc of covergece, the it is also cotiuous there, a result attributed to Abel Eve if we do ot have covergece at the boudary, for istace if R is the radius of covergece ad if lim a eists, though is diverget, the oe has a defiitio of "sum" dr =0 =0 a R for the diverget series to take o this limit This meas that it is possible to defie the sum of a series i etirely ew ways that give fiite sum to series that are diverget i Cauchy's sese For series that are coverget i Cauchy's sese ad if it is also coverget i these ew ways of summig the series, the we call this a regularity or cosistecy result The otios of Abel summability ad Cesaro summability are regular oes The results called Tauberia theorems that give coditio so that give the summability i whatever ew way of a series, it will also be coverget i Cauchy's sese For eample, Alfred Tauber (886-942) proved that if a is Abel summable to the value A ad if a 0 as, the a =0 =0 coverges to A i the sese of Cauchy We shall prove Abel's regularity theorem Theorem 8 (Abel's Theorem, Abel, Niels Herik, 802-29) Suppose the real power series a has radius of covergece R > 0 If it =0 coverges at = R, the it coverges uiformly i [0, R] Similarly, if it coverges at R, the it coverges uiformly i [ R, 0] Proof We may assume that the radius of covergece is This makes the proof easier ad more elegat We may use the chage of variable = Ry to chage the power series if eed be to oe with radius of covergece With this chage of variable, a, =0 = =0 a R y = =0 b y Ng Tze Beg 2007 8

Chapter 8 Uiform Covergece ad Differetiatio where b = a R Thus =0 b y coverges absolutely for y < ad diverges whe y > Now we assume the radius of covergece is Suppose at the boudary =, a is coverget We may assume that =0 = =0 a =0 a = lim d k=0 a k = 0 (If eed be, we may redefie the ew a 0 to be the old a 0 (old)-lim d a k Suppose k=0 lim The we let c k = a k for iteger k > 0, c 0 =a 0 L The ad d k=0 a k = L =0 c = 0 a is coverget if ad oly if c is coverget Plaily is uiformly =0 =0 =0 a coverget o [0, ] if ad oly if =0 c is uiformly coverget o [0, ] because the costat term a 0 ad c 0 do ot affect the Cauchy coditio) Thus we may assume that (i) the radius of covergece of a is ; (ii) a =0 (iii) a = 0 =0 is coverget ad =0 For each iteger 0, let s = a k The (iii) says s 0 Plaily, a = s s - k=0 for iteger ad a 0 = s 0 We shall rewrite the partial sums of =0 a i a more useful form For each iteger 0, a k -------------------- () k=0 k = a 0 + k= (s k s k ) k = s 0 + k= (s k s k ) k We shall show that a is uiformly Cauchy o [0, ] =0 (We ca actually use () to deduce that the power series is cotiuous at = We shall pursue this later) For ay iteger N ad for ay iteger p, k=n+ a k k = k=n+ = k=n+ (s k s k ) k = s k k k=n =( ) k=n+ k=n+ s k k s k k+ = k=n+ k=n+ s k k + s s N N+ s k k s k k k=n s k k s k k + s s N N ------------------------ (2) =( ) k=n Therefore, it follows from (2) ad triagle iequality that for ay iteger N, ay iteger p ad for [0, ], a k k k=n+ [ ( ) k=n s k k + s + s N N [ ( ) k=n s k k + s + s N ------------------ (3) Ng Tze Beg 2007 9

Chapter 8 Uiform Covergece ad Differetiatio Now sice s 0, s 0 For each iteger 0, let M = sup { s, s +, }= sup { s j : j is a iteger ad j } The for each iteger 0, M 0 ad lim M sice d = lim sup s = 0 lim s d d = 0 From (3) we have that for ay iteger N, ay iteger p ad for [0, ], a k k k=n+ [ ( ) k=n [ M N k=n M N k + 2M N = M N k k=n k+ + 2M N k=n k ( )+2M N [ M N ( N ) + 2M N = M N N ( p )+2M N [ M N + 2M N = 3M N ------------------------------------- (4) Now sice M 0 as, give ay ε > 0, there eists a positive iteger N 0 such that for ay iteger, N 0 M < ε /3 Thus it follows from (4) that for ay iteger N N 0, ay iteger p ad for ay [0, ], k=n+ a k k [ 3M N < Therefore, a k is uiformly Cauchy o [0, ] Thus, by Theorem 3, k=0 k k=0 a k k coverges uiformly o [0,] The case that a k is coverget at the other ed poit - ad is k=0 k k=0 a k ( ) k = 0 similar Just ote that for ay iteger N, for ay iteger p ad for i [,0], a k k = ( ) k a k k, k=n+ k=n+ k=n+ (s k s k ) k = k=n+ s k k k=n+ s k k where s = ( ) k a k We ca the deduce (3) ad (4) with the same otatio but k=0 with i place of ad deduce i like maer the uiform covergece of k=0 a k k o [,0] Corollary 9 Suppose the real power series a has radius of covergece R > 0 If it coverges at = R to a value L, the lim a = L That is to say the d R =0 power series fuctio a is cotiuous at = R If the series coverges at = =0 R to a value L', the lim a = L, hece the power series fuctio is d R + =0 =0 a cotiuous at = R Proof By Theorem 8, if a is coverget at R, the is uiformly =0 =0 a coverget o [0, R] Therefore, by Theorem 4 Chapter 7, =0 =0 a is cotiuous o [0, R] because for each iteger, the -th partial sum s ()= a k k is a k=0 cotiuous polyomial fuctio ad (s ) coverges uiformly o [0, R] Thus lim a = Similarly if is coverget at R, the is d R =0 =0 a R = L =0 a =0 a Ng Tze Beg 2007 20

Chapter 8 Uiform Covergece ad Differetiatio uiformly coverget o [ R, 0] by Theorem 8 Agai by Theorem 4 of Chapter 7, the cosequece of uiform covergece is cotiuity at = R The coclusio about the right limit at R the follows Eample 20 We shall illustrate the techique of usig Abel's formula i the proof of Theorem 8 to deduce cotiuity at a ed poit of the iterval of covergece Suppose the radius of covergece of a is ad The =0 =0 a = 0 lim a = 0 d =0 Proof For each iteger 0, a k, ------------ () k=0 k = a 0 + k= (s k s k ) k = s 0 + k= (s k s k ) k where s = a k k=0 The followig () for all i [0, ], k=0 a k k = s 0 + k= s k k k= s k k = k= s k k k=0 s k k+ + s 0 = k= s k k k=0 s k k + s 0 =( ) k= s k k + s s 0 + s 0 =( ) s k k + s k = 0 Suppose N is a positive iteger The for > N+, for all i [0, ], a k k=0 k =( ) k = 0 Thus, for > N+, for all i [0, ], iequality k=0 a k k N [ ( ) k = 0 N [ ( ) k = 0 N s k k +( ) s k k +( ) k =N+ s k k + s k =N+ s k k + s s k k +( ) k =N+ s k k + s by triagle sice 0 Note that s 0, s 0 For each iteger 0, let M = sup { s, s +, }= sup { s j : j is a iteger ad j }, ---------------- (2) The M 0 for all positive iteger ad lim M sice d = lim sup s = 0 lim s d d = 0 So if N+, s M N It the follows from (2) that for > N+ ad for all i [0, ], k=0 a k k ( ) k =N+ k [ N [ ( ) k = 0 N [ ( ) k = 0 s k k +( ) k =N+ M N k + M N s k k + M N + M N sice Ng Tze Beg 2007 2

Chapter 8 Uiform Covergece ad Differetiatio N [ ( ) k s k = 0 + 2M N ------------------------ (3) Sice M 0, there eists a positive iteger L such that for all iteger, m L u M < 4 Thus, from (3), for all L+2 ad all i [0,], L a k k=0 k [ ( ) k s k + 2M L = 0 L L Therefore, a k k=0 k [ ( ) k s k + 2M L <( ) = 0 k s k + = 0 2 If we let = /2 L > 0, the we have + k s k = 0 L < < u a k k <( ) k=0 k s k + = 0 2 < 2 + 2 = This meas lim a = 0 = Hece is cotiuous at = d = =0 =0 a =0 a Eample 2 l (+ ) has the followig power series epasio for < l( + )=( ) + for < = We shall start with the geometric series for < + = + 2 3 + = =0 ( ) This is a power series epasio for The radius of covergece plaily is + Take ay real umber K such that 0< K < The ( ) coverges uiformly o [ K, K] by Theorem It follows from Theorem 7 that we ca itegrate the fuctio term by term i [ K, K] Thus 0 for all i [ K, K] + t dt = =0 ( ) + + But the left had side is l(+) Hece for ay real umber K such that 0 < K <, l( + )=( ) + for all i [ K, K] =0 + Therefore, l( + )=( ) + for all i (, ) Now for = =0 + = = ( ) +, the series ( ) is coverget by Leibitz's Alteratig series test = + Therefore, by Abel's Theorem (Corollary 9), lim ( ) + By the cotiuity of l(+) at =, d =0 + = lim l( + )=( ) + d = we the have l(2) = l(+) = ( ) + Thus for < = l( + )=( ) + = =0 Eercises 22 Ng Tze Beg 2007 22

Chapter 8 Uiform Covergece ad Differetiatio Determie whether each of the followig sequeces (of fuctios) coverge uiformly o the give domai si() (i) o [0, ] ; (ii) o [0, ] ; (iii) o [0, ] l 3 + 2 (iv) ( o [0, ] ; (v) - o [0, ] ; (vi) 2 + )2 o [a, b], a < b + 3 2 Use the Weierstrass M-Test to prove that each of the followig series is uiformly coverget o the give domai si() (i) o R ; (ii) o [ a, a], a > 0 ; = 2 2 + = 2 + 4 (iii) ( + ) o [ a, a], 0 < a < ; (iv) o [0, ] = = ( ) (Hit: Fid maimum value of ( ) i [0, ]) /2 3 Let f ()= Discuss how you might prove that f is cotiuous o [0, = (!) 2 ] 4 (Realizig fuctio as a power series) (i) Prove that for < + = + 2 + = =0 ( ) Discuss how you might prove that (ii) l( + )= 2 for <, 2 + 3 3 + = =0 ( ) + + ad (iii) for < ( + ) 2 = + 2 32 + = = ( ) 5 Use the power series epasio of for < to = + + 2 + = =0 prove that (a) if < ; = = ( ) 2 +( ) l( ) (b) if 0 < < ; =0 + = ( ) 0, if = 0 (c) = ( + ) = ( ) l( )+if0< < 0, if = 0 Give reasos for the steps you take (This questio is a eample of power series maipulatio) 6 (Optioal) Determie the radius of covergece of the Bessel fuctio of the first kid of order zero J 0 () give by J 0 ()= ( ) 2 Write out the first 4 =0 (!) 2 2 terms of J 0 () Show that J 0 () satisfies the differetial equatio d2 y d 2 + dy d + y = 0 (Bessel s differetial equatio of order zero) Ng Tze Beg 2007 23

Chapter 8 Uiform Covergece ad Differetiatio 7 Fid all those for which the followig series coverge 2 ( + 2) (i) ; (ii) ; (iii) = ( + 5)3 3 = 2 + 3 = 2 (2 + ) (Hit: Use ratio test) 8 Use trigoometric formula to prove that 4 si 3 () = 3 si() - si(3) Use this ad the power series epasio for si() to show that (i) si 3 ()= 3 for all real ; 4 = ( ) + 3 2 (2 + )! 2+ (ii) Use partial fractio ad obvious series epasio of the resultig ratioal fuctios, or otherwise, show that for < ½ + 2 2 = 3 = [ ( 2) ] 9 Assumig that y + y = 0, y(0) = 0, y (0) = has a solutio give by a power series Fid the power series ad determie its radius of covergece (Hit: Use the three coditios to obtai relatio amog the coefficiets of the power series ad solvig the relatio) 0 Fid the radius of covergece of y() =a 0 ( 2 ) a 2+, =0 (2 + )(2 ) where a 0 ad a are arbitrary real umbers Show that y() satisfies the differetial equatio ( 2 )y = 2 y o its iterval of covergece Show that f () = ( 2 ) coverges uiformly o [, ] ad fid its limitig fuctio g Hece coclude that f ()d d 0 0 2 Eplai what results you would use to show that is cotiuous o R = e 2 2 3 Show that ay power series is the Taylor's series of its sum 4 For the followig fuctios determie their Taylor series cetred at the poits idicated ad determie the radius of covergece i each case (a) si - (), 0 (b) cos(), π / 2 (c) ta - (), 0 (d) cosh(), 0 (e) l +, 0 (f) ta - (), 0 (g) l(+), 0 (h) a, (a > 0) (i), 0 (j) dt, 0 [Hit: cos(2) = 2si 2 () ] 0 e t 2 dt 0 si 2 () t 2 5 Prove that for < /2 9 ( + 2)( ) 2 = = 3 + 2 +( ) + 2 + 6 Prove that (i) l(( + ) (+) ) + l(( ) ( ) ) = 2 + 4 for <, 2 $ 3 + 3 6 $ 5 + 4 8 $ 7 + (ii) 2l() l( + ) l( )= for >, 2 + 2 4 + 3 6 + Ng Tze Beg 2007 24

Chapter 8 Uiform Covergece ad Differetiatio (iii) for > 0 2 l()= + + 3 + 3 5 + + 5 + 7 For each β >, prove that the series si coverges poitwise o the = iterval [0, ), to a cotiuous fuctio, but the covergece is ot uiform o [0, ), (Hit for o-uiform covergece: use the iequality for ay 0, si() 3 /6) 8 For each β such that 0 β, si diverges for all > 0 Show that wheever β is defied ad ot zero, si is diverget (Hit: use the hit for questio 7) = = 9 Prove that the series coverges uiformly to a differetiable = si fuctio o [ K, K] for ay costat K > 0 Hece deduce that = si coverges poitwise to a differetiable fuctio f o R such that However, prove that f ()= = the hit for questio 7) = si cos is ot uiformly coverget o R (Hit: see Ng Tze Beg 2007 25