Archimedes - numbers for counting, otherwise lengths, areas, etc. Kepler - geometry for planetary motion

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1 Topics i Aalysis 3460:589 Summer 007 Itroductio Ree descartes - aalysis (breaig dow) ad sythesis Sciece as models of ature : explaatory, parsimoious, predictive Most predictios require umerical values, hece fuctios. Historical Archimedes - umbers for coutig, otherwise legths, areas, etc. Kepler - geometry for plaetary motio Newto - plaetary motio ad optics via geometry, Calculus to study brachistochroe, etc. Leibitz - Calculus Taylor, Maclauri - represet fuctios by power series Fourier - represet periodic fuctios by trigoometric series Weierstrass, Riema, Abel, Lebesgue et al - clea up mess Basic Cotiuity Questio : Why do we care about cotiuity? Defiitio : f(x) is cotiuous at x=a Defiitio : f(x) is cotiuous o [a,b] Questio : Why cotiuity? The spaces C([a, b]) Theorem : The Itermediate Value Theorem Proof : Note that, if f(a) > 0 ad f(x) is cotiuous at x = a, the f(x) > 0 i a eighborhood of x = a. Cosider the simple case where f(x) is cotiuous o [a, b], ad f(a) < 0, f(b) > 0. Let [a, c) be the largest iterval for which f(x) < 0. The, c < b ad f(c) = 0. 1

2 Theorem : If f exists ad f (x) M, the δ(f, c, ɛ) ɛ M Metric ad Normed Spaces Need more geeralized cocepts of distace ad legth Defiitio : Metric Space Defiitio : Cotiuity Defiitio : Norm Defiitio : Covergece of a sequece Itegral orm ad sup orm o C([a, b]) what is the differece (completeess)? Defiitio : Cauchy sequece Defiitio : Complete metric space Theorem : C([a, b]) is complete i the sup orm. Proof : Start with {f } Cauchy Fix x [a, b], let ɛ > 0 be give. f (x) is Cauchy for each x, hece piecewise limit. Choose N 1 so that f f m < ɛ/4 for m, N 1. Choose N so that f (x) f(x) < ɛ/4 for N. Choose N = max{n 1, N }. There exists δ > 0 such that f N (x) f N (y) < ɛ/4 wheever x y < δ. Suppose that y [a, b] ad x y < δ. Choose M N 1 such that f M (y) f(y) < ɛ/4. f(x) f(y) f(x) f N (x) + f N (x) f N (y) + f N (y) f M (y) + f M (y) f(y) < ɛ

3 a Theorem : C([a, b]) is ot complete i the itegral orm. Proof : f (x) = x o [0, 1]. Theorem : If f f(x)dx f i C([a, b]) with sup orm, the a f (x)dx Compactess Cotiuity of fuctios as predictability. Defiitio : Ope set Defiitio : Closed set Defiitio : Bouded set Defiitio : Ope cover Defiitio : Compact set Theorem : A set C is closed iff, give ay coverget sequece {x } C, we have lim x C Theorem : A subset of RI is compact if ad oly if it is closed ad bouded (Heie-Borel) Proof : a) K compact implies K bouded : cover by (, ) b) K compact implies K closed : Pic p / K, cover K by balls of ceter x K, radius ρ(x, p)/ = x p /. Fiite subcover does t cover p, so it is ot a limit poit. c) Usual proof for closed iterval. Theorem : Bolzao-Weierstrass Defiitio : C is sequetially compact Theorem : C is sequetially compact iff C is compact Proof : 1. C compact implies C sequetially compact. C ot compact implies C ot sequetially compact, by costructio. Uiform Cotiuity 3

4 Defiitio : Uiform cotiuity o S. Theorem : If f(x) is cotiuous o (S, ρ) ad S is compact, the f(x) is uiformly cotiuous o S. Proof : Let ɛ > 0 be give. For each x [a, b], fid the δ x for the defiitio. Cover [a, b] with itervals of radius δ x /. Get a fiite sub-cover. Size meas both are i distace for x j Riema Itegratio Defiitio : partitio of [a, b], orm of partitio Defiitio : lower, upper Riema sum Defiitio : Riema sum Defiitio : lower, upper Riema itegral, Riema itegral, Riema itegrable Theorem : The Riema itegral of f C([a, b]) exists Defiitio : set of measure 0 Defiitio : Property X holds almost everywhere (a.e.) Theorem : f is Riema-itegrable o [a, b] iff f is bouded ad cotiuous a.e. { 0 if x ratioal Example : f(x) = 1 if x irratioal The f(x) dx is udefied for b > a. a { 0 if x irratioal Example : g(x) = 1 if x = p i lowest terms q q The a g(x) dx = 0. 4

5 Covergece of Sequeces of Fuctios Defiitio : {f (x)} coverges poitwise to f(x) Defiitio : {f (x)} coverges uiformly to f(x) o [a, b] Defiitio : f (x) coverges poitwise/uiformly o [a, b] Example : =0 0 if 0 x < ( ) x 1 1 if 4 f (x) = x ( ) x 3 4 if 4 x if < x The lim f (x) = 0 poitwise, but f (x) dx = 1 0 Theorem : If f ad f are both cotiuous o [a, b] [c, d], the x d d d f f(x, y) dy = (x, y) dy dx c c x Proof : Sice f (x, y) ad f(x, y) are cotiuous, if (x, y) is iterior to x the rectagle, we may write (for sufficietly small h) f(x + h, y) = f(x, y) + h f (x, y) + hg(x + h, y) x where g(x + h, y) 0 uiformly as h 0. Hece, ( 1 d ) d d f d f(x + h, y) dy f(x, y) dy = h c c c x (x, y) dy + c g(x + h, y) dy 5

6 Theorem : (Summatio by Parts) Cosider sequeces {a } ad {b }, ad let S = partial sum of a. The, =1 a deote the th =1 1 a b = [S b S m 1 b m ] + S (b b +1 ). =m =m Theorem : (Lagrage) For ay positive iteger, cos x = 1 =1 + si ( ) + 1 x si x, ad si x = cos x cos ( ) + 1 x =1 si x Proof : e ix = eix 1 + e ix e i(+1)x =1 (1 cos x) Separatig real ad imagiary parts ad applyig some trigoometric idetities geerates the desired results. Corollary If x is ot a iteger multiple of π, the cos x 1 =1 ( 1 + csc x ), ad si x csc x =1. 6

7 ad Theorem : If x is ot a iteger multiple of π ad p > 0, the =1 cos x p both coverge. =1 si x p Proof : Use the estimates above, with summatio by parts, otig that 1 1 p ( + 1) p p p+1 si x I particular, f(x) = is defied for all x, but the term-by-term =1 series for f (x) does ot coverge aywhere, except at iteger multiples of π. Lebesgue Itegratio Defiitio : For c RI ad a < b, defie φ {a,b,c} (x) = { c if a < x < b 0 otherwise Defiitio : Defie S 0 = {f(x) = φ {aj,b j,c j }(x)} j=1 The, S 0 is a liear space. Defiitio : For each f(x) = φ {aj,b j,c j }(x) S 0, defie j=1 f(x) dx = c l((a, b) (a j, b j )), a j=1 where l(i) deotes the legth of the iterval I, that is, l( ) = 0 ad l((a, b)) = b a for a < b. For ay closed iterval [a, b], let S 1 ([a, b]) deote the set of fuctios f such that ( ){f } S 0 which are icreasig (a.e.) ad for which lim f (x) = f(x) (a.e.) 7

8 For ay f S 1 ([a, b]) with lim f (x) = f(x) (a.e.), defie the Lebesgue itegral of f : Note : a f(x) dx = lim f (x) dx. a 1. C([a, b]) S 1 ([a, b]), ad the itegrals agree o this set. { } b. This itegral is well-defied as f (x) dx is o-decreasig. 3. Sice C([a, b]) S 1 ([a, b]), there are cases where the itegral is ot the Riema itegral. a Theorem : (Lebesgue Domiated Covergece Theorem) If {f } S 1 ([a, b]) is poitwise coverget ad uiformly bouded, the lim f (x) dx = lim f (x) dx a a Note : If the limit fuctio is cotiuous, this gives the iterchage result that we would lie to have, but the result is more geeral tha that. Approximatio Theory Exactess ad approximatio Theorem :! = 0 t e t dt Proof : Theorem : (Stirlig s Approximatio)! π e for large Proof : Via u = t, we have! = e e u / ( 1) exp ( ) u du =3 8

9 Theorem : (Cetral Limit Theorem) Give a Biomial trial of size, P r(success) = p, q = 1 p, we have ( lim P r α p ) β = 1 β e x / dx pq π ( ) Proof : Defie f(t) = p t q t. For itegers r s, we have t ( ) s s+1 P r(r s) = p q = f(t) dt r =r Settig x = p pq ad cosiderig t < + 1, we have, usig Stirlig s approximatio (through logs), that f(t) = α 1 πpq e x /+o(1), as. Thus, with the trasformatio x = t p pq, we have P r ( α p ) β pq = β α e x /+o(1) dx + o(1), hece the result usig uiform boudedess, poitwise limits ad the Lebesgue Domiated Covergece Theorem. Theorem : (The Weierstrass Approximatio Theorem) Let f(x) be cotiuous o [a, b]. The, give ɛ > 0, there exists a positive iteger (ɛ) ad a polyomial P (x) of degree such that P (x) f(x) < ɛ for all x [a, b] Proof : (Berstei) Assume that [a, b] = [0, 1]. If ot, we ca mae the trasformatio t = x a b a. Defie the Berstei polyomial of degree to f(x) via B (f; x) = =0 ( ) x (1 x) f 9 ( )

10 We show that lim B (f; x) = f(x) uiformly o [0, 1] We use the followig idetities : =0 ( =0 =0 ) Combiig these, we obtai ( ) x (1 x) = 1 ( ) x (1 x) = x x (1 x) = ( ) ( ) =0 x x (1 x) = ( 1 1 ) x + 1 x x(1 x) ad so [ f(x) B (f; x) = f(x) f =0 ( )] ( ) x (1 x) f(x) is cotiuous o [0, 1], hece uiformly cotiuous ad bouded, so there exist δ, M > 0 such that f(x) < M for all x [0, 1] ad f(x 1 ) f(x ) < ɛ if x 1, x [0, 1] ad x 1 x < δ. Choose ay x [0, 1] Let A be the set of = 0, 1,, such that x < δ, B the set of the remaiig values of. The, [ ( )] ( ) f(x) f x (1 x) < ɛ ( ) x (1 x) ɛ A ad, usig the fact that 0 x(1 x) 1/4 o [0, 1] : [ ( )] ( B f(x) f ) x (1 x) M = M B 10 A B ( M δ x(1 x) ) x (1 x) (/ x) (/ x) ( M δ ) x (1 x)

11 Choosig so that completes the proof. M δ < ɛ Theorem : For ay o-egative itegers m,, we ca write cos m t si t m+ as a trigoometric sum a 0 + (a cos t + b si t). =1 Proof : Euler s formula ad de Moivre s Theorem. Theorem : (Weierstrass) Let F (t) be cotiuous ad periodic of period π. The, there exists a positive iteger (ɛ) ad a trigoometric sum S (t) = a 0 + (a cos t + b si t) =1 which uiformly approximates F (t) for all t. F (t) + F ( t) F (t) F ( t) Proof : Set φ(t) =, ψ(t) = si t. The, φ(t) ad ψ(t) are eve, so we ca assume that 0 t π. By settig cos t = x, there exist polyomials P, Q such that φ(t) P (cos t) ɛ 4, ψ(t) Q(cos t) ɛ 4 The, settig U(t) = Q(cos t) + P (cos t) si t, we have F (t) si t U(t) = φ(t) si t + ψ(t) U(t) ɛ Similarly, we ca fid V (t) such that ( ) π F t si t V (t) ɛ ad so F (t) cos t V ( π t ) ɛ 11

12 Thus, F (t) cos t V ( π t ) cos t ɛ, F (t) si t U(t) si t ɛ from which F (t) V ( π t ) cos t U(t) si t ɛ This latter quatity ca the be writte as a trigoometric sum. Example : Discuss the covergece of s (x) = x + x x x o the real lie. Clearly, we do have poitwise covergece to s(x) = { x if x 1 x if x 1 Sice the limit fuctio is cotiuous o RI, it is plausible that the covergece is uiform o some sets, but the quatities x mea that there is liely to be a slower rate of covergece close to x = 1. Sice s (x) ad s(x) are eve, it suffices to cosider x 0. Cosider first the case 0 x 1. We have s (x) s(x) = f (x) = x (1 x ) 1 + x 0 which satisfies f (0) = 0 = f (1), ad must have a maximum. Now, f (x) = 4x 1 [ ( + 1)x x + ] (1 + x ) Applyig Descartes rule of sigs, we have a uique extremum for x 0, which must occur i [0, 1]. Let this maximum be a, which satisfies ( + 1)a a + = 0, 1

13 or equivaletly a + = ( + 1)a Substitutig to fid the maximum value of f (x), ad usig the above relatioship o a yields ( ) [ ] + 1 f (a ) = + 1 a Cosider the defiig equatio for a, amely (+1)x x + = 0. Settig x = ( ) i the polyomial yields < 0, ad + 1 settig x = = ( ) yields 1 > 0, from + 1 which we deduce ad hece f (x) <. Now cosider the case 1 x. I this case, so that g (1) = 0, maximum. We obtai < a < + 1 s (x) s(x) = g (x) = (x 1) 1 + x 0 lim g (x) = 0 for > 1, ad so g (x) also has a x + g (x) = 4x [1 + x ( 1)x ] (1 + x ) Agai applyig Descartes rule of sigs shows that g (x) has a uique extremum o [1, + ), which is a maximum. Call this value b, which satisfies which ca be writte as 1 + b b = ( 1)b = 0 b ( 1)b 13

14 which geerates the maximum value ( 1 g (b ) = ) [ b ] 1 As above, we may examie the defiig polyomial for b ad obtai ad hece g (b ) <. 1 < b < I coclusio, combiig these results, we have that max 0 x s (x) s(x) = max{f (a ), g (b )} < ad so the covergece is i fact uiform o RI. 14

15 Examples (some of which were homewor) ( ) 1 1. Fid δ x, c, ɛ for c > 0.. Fid δ (e x, c, ɛ). 3. Prove that δ(f + g, c, ɛ) mi { δ ( f, c, ɛ 4. For f, g C([a, b]), defie d 1 (f, g) = a ) ( )}, δ g, c, ɛ. f(x) g(x) dx Show that d 1 defies a metric o C([a, b]). 5. For f, g C([a, b]), defie d (f, g) = max f(x) g(x) a x b Show that d defies a metric o C([a, b]). 6. Let {x } be a bouded sequece i RI. Cosider the rage S = {x : NI }. If S is compact, what ca be said about {x }? 7. Let K be compact, ad S = {x } be a sequece i K. Cover K with the ope itervals {(y 1/, y + 1/)}, ad extract a fiite sub-cover. A ifiite umber of x must lie i oe iterval, say (y 1 1/, y 1 + 1/). Let x 1 be the first term of S that lies i this iterval, ad let {x (1) } be those x for which > 1, ad x is i this iterval. Cover K with the ope itervals (y 1/4, y + 1/4), ad extract a fiite subcover. A ifiite umber of x (1) must lie i oe iterval, say (y 1/4, y + 1/4). Let x be the first term of {x (1) } that lies i this iterval, ad {x () } be those x (1) for which > ad x is i this iterval. 15

16 Repeat this process idefiitely, so that, at step, we cover K with the ope itervals (y 1/, y + 1/ ), ad extract a fiite sub-cover. Oe iterval, say (y 1/, y + 1/ ), must cotai a ifiite umber of {x ( 1) ad {x () }. Let x be the first term of {x ( 1) } those terms for which > ad x ( 1) Show directly that lim x = L exists, ad that L K. } which lies i this iterval, is i this iterval. 8. Use summatio by parts to fid a simplified closed form for x. =1 Hit: Use a = x. 9. Fid the largest set for which the series x coverges ui- 1 + x formly. 10. For each positive iteger, defie s = = Show that lim s exists, ad evaluate this limit. 11. For each positive iteger, defie f (x) = x + 1 x + x + 1 (a) Prove that lim f (x) exists for all real x, ad evaluate this limit. (b) Where is the covergece uiform? 1. Defie the sequece {x } via x 1 = 1 ad x +1 = x+3 x +1 for 1. (a) Show that lim x exists, ad fid this limit. (b) Discuss the rate of covergece of this sequece. 13. Defie f(x) = =1 1 cos x 1 + x + x 4 16

17 (a) For which real x is f(x) defied? (b) Where does this series coverge uiformly? (c) Is f(x) cotiuous at x = 0? 14. Defie Q 1 (x) = 1 ad Q +1 (x) = 1 ( Q (x) + 1 x ) for 1. Show that lim Q (x) = 1 x for every x [ 1, 1]. 15. Let A = RI { : NI }, ad defie f(x) = =1 ( 1) x + (a) Show that f(x) is defied for every x A. (b) Where is the covergece uiform? (c) Show that f(x) is cotiuous o A. 16. Let {s } be a sequece of real umbers, ad s be a real umber such that s s ad s +1 1 (s + s) for all 1. Prove that lim s exists. 17. Defie the sequece {a } via a 1 = 1 ad a +1 = 1 (a + 3 ) a for 1. (a) Show that lim a exists, ad fid this limit. (b) Discuss the rate of covergece. 18. Cosider the sequece of fuctios i C([0, )), defied by C 1 (x) = 1 ad C +1 (x) = 3 C x (x) + for 1. 3C(x) (a) Show that C +1 (x) x 1/3 for 1 ad x 0. Hit: The algebra is easier if you write x = t 3. 17

18 (b) Show that {C +1 (x)} is decreasig for each x. (c) Discuss the rate of covergece to the limit fuctio. 19. Estimate the rate of covergece of 0. Give g = t (1 t) dt. t e t dt, what is the best way to compute g 0? 18

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges