Series of functions. Chapter The Dirichlet kernel

Transcription

1 Chapter 3 Series of fuctios 3.9 The Dirichlet kerel Our argumets so far have bee etirely abstract we have ot really used ay properties of the fuctios e (x) = e ix except that they are orthoormal. To get better results, we eed to take a closer look at these fuctios. I some of our argumets, we shall eed to chage variables i itegrals, ad such chages may take us outside our basic iterval [, π], ad hece outside the regio where our fuctios are defied. To avoid these problems, we exted our fuctios periodically outside the basic iterval such that f(x + 2π) = f(x) for all x R. The figure shows the extesio graphically; i part a) we have the origial fuctio, ad i part b) (a part of ) the periodic extesio. As there is o dager of cofusio, we shall deote the origial fuctio ad the extesio by the same symbol f. a) b) π 3π π 3π Figure Let us begi by lookig at the partial sums N s N (x) = f, e e (x) = N

2 2 CHAPTER 3. SERIES OF FUNCTIONS of the Fourier series. Sice we have s N (x) = 2π N = N α = f, e = 2π ( f(t)e it dt ) f(t)e it dt e ix = 2π = 2π f(x u) N = N e iu du f(t) N = N e i(x t) dt = where we i the last step has substituted u = x t ad used the periodicity of the fuctios to remai i the iterval [, π]. If we itroduce the Dirichlet kerel N D N (u) = e iu we may write this as s N (x) = 2π = N f(x u)d N (u) du Note that the sum N = N eiu = N = N (eiu ) is a geomtric series. For u = 0, all the terms are ad the sum is 2N +. For u 0, we use the sum formula for a fiite geometric series to get: D (u) = e inu e i(n+)u e iu = e i(n+ 2 )u e i(n+ 2 )u e i u 2 e i u 2 = si((n + 2 )u) si u 2 where we have used the formula si x = eix e ix 2i twice i the last step. This formula gives us a ice, compact expressio for D N (u). If we substitute it ito the formula above, we get s N (x) = f(x u) si((n + 2 )u) 2π si u du 2 If we wat to prove that partial sums s N (x) coverge to f(x) (i.e. that the Fourier series coverges poitwise to f), the obvious strategy is to prove that itegral above coverges to f. I 829, Dirichlet used this approach to prove: Theorem 3.9. (Dirichlet s Theorem) If f D has oly a fiite umber of local miima ad maxima, the the Fourier series of f coverges poitwise to f.

3 3.9. THE DIRICHLET KERNEL 3 Dirichlet s result must have come as somethig of a surprise; it probably seemed ulikely that a theorem should hold for fuctios with jumps, but ot for cotiuous fuctios with a ifiite umber of extreme poits. Through the years that followed, a umber of mathematicias tried ad failed to prove that the Fourier series of a periodic, cotiuous fuctio always coverges to the fuctio. I 873, the Germa mathematicia Paul Du Bois-Reymod explaied why they failed by costructig a periodic, cotiuous fuctio whose Fourier series diverges at a dese set of poits. It turs out that the theory for poitwise covergece of Fourier series is quite complicated, ad we shall ot prove Dirichlet s theorem here. Istead we shall prove a result kow as Dii s test which allows us to prove covergece for may of the fuctios that appear i practice. But before we do that, we shall take a look at a differet otio of covergece which is easier to hadle, ad which will also give us some tools that are useful i the proof of Dii s test. This alterative otio of covergece is called Cesaro covergece or covergece i Cesaro mea. But first of all we shall collect some properties of the Dirichlet kerels that will be useful later. Let us first see what they look like. The figure below shows Dirichlet s kerel D for = 5, 0, 5, 20. Note the chagig scale o the y-axis; as we have alresdy observed, the maximum value of D is 2 +. As grows, the graph becomes more ad more domiated by a sharp peak at the origi. The smaller peaks ad valleys shrik i size relative to the big peak, but the problem with the Dirichlet kerel is that they do ot shrik i absolute terms as goes to ifiity, the area betwee the curve ad the x-axis (measured i absolute value) goes to ifiity. This makes the Dirichlet kerel very difficult to work with. Whe we tur to Cesaro covergece i the ext sectio, we get aother set of kerels the Fejér kerels ad they tur out ot to have this problem. This is the mai reaso why Cesaro covergece works much better tha ordiary covergece for Fourier series.

4 4 CHAPTER 3. SERIES OF FUNCTIONS Let us ow prove some of the crucial properties of the Dirichlet kerel. Recall that a fuctio g is eve if g(t) = g( t) for all t i the domai: Lemma The Dirichlet kerel D (t) is a eve, realvalued fuctio such that D (t) D (0) = 2 + for all t. For all, π D (t) dt = 2π but lim D (t) dt Proof: That D N is realvalued ad eve, follows immediately from the formula D (t) = si((+ 2 )t) To prove that D si t (t) D (0) = 2 +, we just 2 observe that D (t) = k= Similarly for the itegral e ikt π D (t) dt = 2π k= k= e ikt = 2 + = D (0) π e ikt dt = 2π

5 3.0. THE FEJÉR KERNEL 5 as all itegrals except the oe for k = 0 is zero. To prove the last part, we observe that sice si u u for all u, we have D (t) = si(( + 2 )t) si t 2 2 si(( + 2 )t) t Usig the symmetry ad the substitutio z = ( + 2 )t, we see that D (t) dt = 0 2 D (t) dt 0 4 si(( + 2 )t) t dt = = (+ 2 )π 0 4 si z z dz k= kπ (k )π 4 si z kπ dz = 8 π The expressio o the right goes to ifiity sice the series diverges. k= k The last part of the lemma is bad ews. It tells us that whe we are doig calculatios with the Dirichlet kerel, we have to be very careful i puttig i absolute values as the itegrals are likely to diverge. For this reaso we shall ow itroduce aother kerel the Fejér kerel where this problem does ot occur. Exercises for Sectio 3.9. Let f : [, π] C be the fuctio f(x) = x. Draw the periodic extesio of f. Do the same with the fuctio g(x) = x 2. si((+ 2. Check that D (0) = 2 + by computig lim 2 )t) t 0. si t 2 3. Work out the details of the substitutio u = x t i the derivatio of the formula D N = si((n+ 2 )t). si t 2 4. Explai the details i the last part of the proof of Lemma (the part that proves that lim D (t) dt ). 3.0 The Fejér kerel Before studyig the Fejér kerel, we shall take a look at a geeralized otio of covergece for sequeces. Certai sequeces such at 0,, 0,, 0,, 0,,... do ot coverge i the ordiary sese, but they do coverge i average i the sese that the average of the first elemets approaches a limit as goes to ifiity. I this sese, the sequece above obviously coverges to 2. Let us make this otio precise:

6 6 CHAPTER 3. SERIES OF FUNCTIONS Defiitio 3.0. Let {a k } be a sequece of complex umbers, ad let S = a k. We say that the sequece coverges to a C i Cesaro mea if a = lim S a 0 + a + + a = lim We shall write a = C- lim a. The sequece at the begiig of the sectio coverges to 2 i Cesaro mea, but diverges i the ordiary sese. Let us prove that the opposite ca ot happe: Lemma If lim a = a, the C-lim a = a. Proof: Give a ɛ > 0, we must fid a N such that S a < ɛ whe N. Sice {a } coverges to a, there is a K N such that a a < ɛ 2 whe K. If we let M = max{ a k a : k = 0,, 2,...}, we have for ay K: S a = (a 0 a) + (a a) + + (a K a) + (a K a) + (a a) (a 0 a) + (a a) + + (a K a) + (a K a) + (a a) MK + ɛ 2 Choosig large eough, we get MK < ɛ 2, ad the lemma follows, The idea behid the Fejér kerel is to show that the partial sums s (x) coverge to f(x) i Cesaro middel; i.e. that the sums coverge to f(x). Sice S (x) = s 0(x) + s (x) + + s (x) s k (x) = 2π where D k is the Dirichlet kerel, we get S (x) = ( ) π f(x u) D k (u) 2π f(x u)d k (u) du du = f(x u)f (u) du 2π where F (u) = D k(u) is the Fejér kerel. We ca fid a closed expressio for the Fejér kerel as we could for the Dirichlet kerel, but the argumets are a little loger:

7 3.0. THE FEJÉR KERNEL 7 Lemma The Fejér kerel is give by for u 0 ad F (0) =. Proof: Sice F (u) = si2 ( u 2 ) si 2 ( u 2 ) F (u) = D k (u) = si( u 2 ) si((k + 2 )u) we have to fid si((k + 2 )u) = 2i ( ) e i(k+ 2 )u e i(k+ 2 )u The series are geometric ad ca easily be summed: ad e i(k+ 2 )u = e i u 2 e iku = e i u 2 e i(k+ 2 )u = e i u 2 e iku = e i u 2 Hece e iu e iu = e iu e iu = e iu e i u 2 e i u 2 e iu e i u 2 e i u 2 si((k + 2 )u) = ( e iu + e iu ) 2i e i u 2 e i = ( e iu 2 + e iu ) u 2 2i e i u 2 e i = u 2 ad thus = 2i (ei u 2 e u 2 ) 2 e i u 2 e i u 2 = ( ) e i u 2 e u 2 2 ) 2i e i u 2 e i u 2 2i = si2 ( u 2 ) si u 2 F (u) = si( u 2 ) si((k + 2 )u) = si2 ( u 2 ) si 2 u 2 To prove that F (0) =, we just have to sum a arithmetic series F (0) = D k (0) = (2k + ) =

8 8 CHAPTER 3. SERIES OF FUNCTIONS The figure below shows the Fejer kerels F for = 5, 0, 5, 20. At first glace they look very much like the Dirichlet kerels i the previous sectio. The peak i the middle is growig slower tha before i absolute terms (the maximum value is compared to 2+ for the Dirichlet kerel), but relative to the smaller peaks ad values, it is much more domiat. The fuctios are ow positive, ad the areas betwee their graphs ad the x-axis is always equal to oe. As gets big, almost all this area belogs to the domiat peak i the middle. The positivity ad the cocetratio of all the area i the ceter peak make the Fejér kerels much easier to hadle tha their Dirichlet couterparts. Let us ow prove some of the properties of the Fejér kerels. Propositio For all, the Fejér kerel F fuctio such that π F (x) dx = 2π is a eve, positive For all ozero x [, π] 0 F (x) π2 x 2

9 3.0. THE FEJÉR KERNEL 9 Proof: That F is eve ad positive follows directly from the formula i the lemma. By Propositio 3.9.2, we have π F (x) dx = 2π 2π D k dx = π D k dx = 2π = For the last formula, observe that for u [ π 2, π 2 ], we have 2 π u si u (make a drawig). Thus F (x) = si2 ( x 2 ) si 2 x 2 ( 2 π π2 x 2 )2 x 2 We shall ow show that S (x) coverges to f(x), i.e. that the Fourier series coverges to f i Cesaro mea. We have already observed that S (x) = 2π f(x u)f (u) du If we itroduce a ew variable t = u ad use that F is eve, we get S (x) = f(x + t)f ( t) ( dt) = 2π π = f(x + t)f (t) dt = f(x + u)f (u) du 2π 2π If we combie the two expressios we ow have for S (x), we get S (x) = (f(x + u) + f(x u)) F (u) du Sice 2π F (u) du =, we also have f(x) = f(x)f (u) du 2π Hece S (x) f(x) = (f(x + u) + f(x u) 2f(x)) F (u) du To prove that S (x) coverges to f(x), we oly eed to prove that the itegral goes to 0 as goes to ifiity. The ituitive reaso for this is that for large, the kerel F (u) is extremely small except whe u is close to 0, but whe u is close to 0, the other factor i the itegral, f(x+u)+f(x u) 2f(x), is very small. Here are the techical details.

10 0 CHAPTER 3. SERIES OF FUNCTIONS Theorem If f D, the S coverges to f o [, π], i.e. the Fourier series coverges i Cesaro mea. The covergece is uiform o each subiterval [a, b] [, π] where f is cotiuous. Proof: Give ɛ > 0, we must fid a N N such that S (x) f(x) < ɛ whe N. Sice f is i D, there is a δ > 0 such that f(x + u) f(x u) 2f(x) < ɛ whe u < δ (keep i mid that sice f D, f(x) = 2 lim u 0(f(x + u) f(x u))). We have S (x) f(x) = = + + δ δ δ δ For the first itegral we have δ δ f(x + u) + f(x u) 2f(x) F (u) du = f(x + u) + f(x u) 2f(x) F (u) du+ f(x + u) + f(x u) 2f(x) F (u) du+ f(x + u) + f(x u) 2f(x) F (u) du f(x + u) + f(x u) 2f(x) F (u) du δ ɛf (u) du δ For the secod itegral we get δ ɛf (u) du = ɛ 2 f(x + u) + f(x u) 2f(x) F (u) du δ π 2 4 f δ 2 du = π2 f δ 2 Exactly the same estimate holds for the third itegral, ad by choosig N > 2 f, we get the sum of the last two itegrals less tha ɛ ɛδ 2 2. But the S (x) f(x) < ɛ, ad the covergece is proved. So what about the uiform covergece? We eed to check that we ca choose the same N for all x [a, b]. Note that N oly depeds o x through the choice of δ, ad hece it suffices to show that we ca use the same δ for all x [a, b]. Oe might thik that this follows immediately from the fact that a cotiuous fuctio o a compact iterval [a, b] is uiformly cotiuous, but we eed to be a little careful as x + u or x u may be outside the

11 3.. THE RIEMANN-LEBESGUE LEMMA iterval [a, b] eve if x is iside. The quickest way to fix this, is to observe that sice f is i D, it must be cotiuous ad hece uiformly cotiuous o a slightly larger iterval [a η, b + η]. This meas that we ca use the same δ < η for all x ad x ± u i [a η, d + η], ad this cliches the argumet. We have ow fially proved Theorem which we restate here: Corollary The trigoometric polyomials are dese i C p i - orm, i.e. for ay f C P there is a sequece of trigoometric polyomials covergig uiformly to f. Proof: Accordig to the theorem, the sums S N (x) = N N =0 s (x) coverge uiformly to f. Sice each s is a trigoometric polyomial, so are the S N s. Exercises to Sectio 3.0. Let {a } be the sequece, 0,, 0,, 0,, 0,.... Prove that C-lim a = Show that C-lim (a + b ) = C lim a + C lim b 3. Check that F (0) = by computig lim u 0 si 2 ( u 2 ) si 2 u 2 4. Show that S N (x) = N = (N ) α ( N )e (x), where α = f, e is the Fourier coefficiet. 5. Assume that f C P. Work through the details of the proof of Theorem ad check that S coverges uiformly to f.. 3. The Riema-Lebesgue lemma The Riema-Lebesgue lemma is a seemigly simple observatio about the size of the Fourier coefficiets, but it turs out to be a very efficiet tool i the study of poitwise covergece. Theorem 3.. (Riema-Lebesgue Lemma) If f D ad α = 2π f(x)e ix dx, Z, are the Fourier coefficiets of f, the lim α 0. Proof: Accordig to Bessel s iequality 3.6.9, = α 2 f 2 2 <, ad hece α 0 as. Remark: We are cheatig a little here as we oly prove the Riema- Lebesgue lemma for fuctio which are i D ad hece square itegrable.

12 2 CHAPTER 3. SERIES OF FUNCTIONS The lemma holds for itegrable fuctios i geeral, but eve i that case the proof is quite easy. The Riema-Lebesgue lemma is quite deceptive. It seems to be a result about the coefficiets of certai series, ad it is proved by very geeral ad abstract methods, but it is really a theorem about oscillatig itegrals as the followig corollary makes clear. Corollary 3..2 If f D ad [a, b] [, π], the Also b lim a b lim a f(x) cos(x) dx = f(x)e ix dx = 0 b lim a f(x) si(x) dx = 0 Proof: Let g be the fuctio (this looks more horrible tha it is!) 0 if x / [a, b] the g is i D, ad b a f(x) if x (a, b) g(x) = 2 lim x a f(x) if x = a 2 lim x b f(x) if x = b f(x)e ix dx = g(x)e ix dx = 2πα where α is the Fourier coefficiet of g. By the Riema-Lebesgue lemma, α 0. The last two parts follows from what we have just proved ad the idetities si(x) = eix e ix 2i ad cos(x) = eix +e ix 2 Let us pause for a momet to discuss why these results hold. The reaso is simply that for large values of, the fuctios si x, cos x, ad e ix (if we cosider the real ad imagiary parts separately) oscillate betwee positive ad egative values. If the fuctio f is relatively smooth, the positive ad egative cotributios cacel more ad more as icreases, ad i the limit there is othig left. This argumet also idicates why rapidly oscillatig, cotiuous fuctios are a bigger challege for Fourier aalysis tha jump discotiuities fuctios with jumps average out o each side of the jump, while for wildly oscillatig fuctios the averagig procedure may ot work. Sice the Dirichlet kerel cotais the factor si(( + 2 )x), the followig result will be useful i the ext sectio:

13 3.2. DINI S TEST 3 Corollary 3..3 If f D ad [a, b] [, π], the b lim a f(x) si ( ( + 2 )x) dx = 0 Proof: Follows from the corollary above ad the idetity si ( ( + 2 )x) = si(x) cos x 2 + cos(x) si x 2 Exercises to Sectio 3.. Work out the details of the si(x)- ad cos(x)-part of Corollary Work out the details of the proof of Corollary a) Show that if p is a trigoometric polyomial, the the Fourier coefficiets β = p, e are zero whe is sufficietly large. b) Let f be a itegrable fuctio, ad assume that for each ɛ > 0 there is a trigoometric polyomial such that π 2π f(t) p(t) dt < ɛ. Show that if α = π 2π f(t)e it dt are the Fourier coefficiets of f, the lim α = Dii s test We shall fially take a serious look at poitwise covergece of Fourier series. As aready idicated, this is a rather tricky busiess, ad there is o ultimate theorem, just a collectio of scattered results useful i differet settigs. We shall cocetrate o a criterio called Dii s test which is relatively easy to prove ad sufficietly geeral to cover a lot of differet situatios. Recall from Sectio 3.9 that if s N (x) = N = N is the partial sum of a Fourier series, the s N (x) = 2π f, e e (x) f(x u)d N (u) du If we chage variable i the itergral ad use the symmetry of D N, we see that we also get s N (x) = f(x + u)d N (u) du 2π

14 4 CHAPTER 3. SERIES OF FUNCTIONS Combiig these two expressios, we get s N (x) = ( f(x + u) + f(x u) ) DN (u) du Sice 2π D N(u) du, we also have f(x) = f(x)d N (u) du 2π ad hece s N (x) f(x) = (f(x + u) + f(x u) 2f(x)) D N (u) du (ote that the we are ow doig exactly the same to the Dirichlet kerel as we did to the Fejér kerel i Sectio 3.0). To prove that the Fourier series coverges poitwise to f, we just have to prove that the itegral coverges to 0. The ext lemma simplifies the problem by tellig us that we ca cocetrate o what happes close to the origi: Lemma 3.2. Let f D ad assume that there is a η > 0 such that lim N η η (f(x + u) + f(x u) 2f(x)) D N (u) du = 0 The the Fourier series {s N (x)} coverges to f(x). Proof: Note that sice si x 2 tells us that lim N is a bouded fuctio o [η, π], Corollary 3..3 π (f(x + u) + f(x u) 2f(x)) D N (u) du = η π [ = lim (f(x + u) + f(x u) 2f(x)) N η si u 2 ] si ( (N + 2 )u) du = 0 The same obviously holds for the itegral from to η, ad hece s N (x) f(x) = = + η η η (f(x + u) + f(x u) 2f(x)) D N (u) du = (f(x + u) + f(x u) 2f(x)) D N (u) du+ (f(x + u) + f(x u) 2f(x)) D N (u) du+

15 3.2. DINI S TEST 5 + (f(x + u) + f(x u) 2f(x)) D N (u) du η = 0 Theorem (Dii s test) Let x [, π], ad assume that there is a δ > 0 such that δ f(x + u) + f(x u) 2f(x) u du < δ The the Fourier series coverges to the fuctio f at the poit x, i.e. s N (x) f(x). Proof: Accordig to the lemma, it suffices to prove that lim N δ δ (f(x + u) + f(x u) 2f(x)) D N (u) du = 0 Give a ɛ > 0, we have to show that if N N is large eough, the δ δ (f(x + u) + f(x u) 2f(x)) D N (u) du < ɛ Sice the itegral i the theorem coverges, there is a η > 0 such that η f(x + u) + f(x u) 2f(x) u du < ɛ η Sice si v 2 v π for v [ π 2, π 2 ] (make a drawig), we have D N(u) = si((n+ 2 )u) si u π 2 u for u [, π]. Hece η (f(x + u) + f(x u) 2f(x)) D N (u) du η η η By Corollary 3..3 we ca get δ η f(x + u) + f(x u) 2f(x) π u du < ɛ 2 (f(x + u) + f(x u) 2f(x)) D N (u) du as small as we wat by choosig N large eough ad similarly for the itegral from δ to η. I particular, we ca get δ δ (f(x + u) + f(x u) 2f(x)) D N (u) du =

16 6 CHAPTER 3. SERIES OF FUNCTIONS = η + + δ η η δ η (f(x + u) + f(x u) 2f(x)) D N (u) du+ (f(x + u) + f(x u) 2f(x)) D N (u) du+ (f(x + u) + f(x u) 2f(x)) D N (u) du less tha ɛ, ad hece the theorem is proved. Dii s test have some immediate cosequeces that we leave to the reader to prove. Corollary If f D is differetiable at a poit x, the the Fourier series coverges to f(x) at this poit. We may eve exted this result to oe-sided derivatives: Corollary Assume f D ad that the limits ad f(x + u) f(x + ) lim u 0 u f(x + u) f(x ) lim u 0 u exist at a poit x. The the Fourier series s N (x) coverges to f(x) at this poit. Exercises to Sectio 3.2 2( ) +. Show that the Fourier series = si(x) i Example 3.7. coverges to f(x) = x for x (, π). What happes i the edpoits? 2. Prove Corollary Prove Corollary Termwise operatios I Sectio 3.3 we saw that power series ca be itegrated ad differetiated term by term, ad we ow wat to take a quick look at the correspodig questios for Fourier series. Let us begi by itegratio which is by far the easiest operatio to deal with. The first thig we should observe, is that whe we itegrate a Fourier series α e ix term by term, we do ot get a ew Fourier series sice the costat term α 0 itegrates to α 0 x, which is ot a term i a Fourier

17 3.3. TERMWISE OPERATIONS 7 series whe α 0 0. However, we may, of course, still itegrate term by term to get the series α 0 x + ( iα ) e ix Z, 0 The questio is if this series coverges to the itegral of f. Propositio 3.3. Let f D, ad defie g(x) = x 0 f(t) dt. If s is the partial sums of the Fourier series of f, the the fuctios t (x) = x 0 s (t) dt coverge uiformly to g o [, π]. Hece x g(x) = f(t) dt = α 0 x + ( iα ) e ix 0 Z, 0 where the covergece of the series is uiform. Proof: By Cauchy-Schwarz s iequality we have g(x) t (x) = 0 (f(t) s (t)) dt f(t) s (t) dt ( π ) 2π f(s) s (s) ds = 2π f s, 2π 2π f s 2 2 = 2π f s 2 By Theorem 3.8.6, we see that f s 2 0, ad hece t coverges uiformly to g(x). If we move the term α 0 x to the other side i the formula above, we get g(x) α 0 x = ( iα ) e ix Z, 0 where the series o the right is the Fourier series of g(x) α 0 x. As always, termwise differetiatio is a must trickier subject. I Example of Sectio 3.7, we showed that the Fourier series of x is 2( ) + si(x), = ad by what we ow kow, it is clear that the series coverges poitwise to x o (, π). However, if we differetiate term by term, we get the hopelessly diverget series 2( ) + cos(x) = Fortuately, there is more hope whe f C p, i.e. whe f is cotiuous ad f() = f(π):

18 8 CHAPTER 3. SERIES OF FUNCTIONS Propositio Assume that f C P ad that f is cotiuous o [, π]. If =0 α e ix is the Fourier series of f, the the differetiated series =0 iα e ix is the Fourier series of f, ad it coverges poitwise to f at ay poit x where f (x) exists. Proof: Let β be the Fourier coefficiet of f. By itegratio by parts β = 2π f (t)e it dt = [ f(t)e it ] π 2π f(t)( ie it ) dt = 2π = 0 + i f(t)e it dt = iα 2π which shows that =0 iα e ix is the Fourier series of f. The covergece follows from Corollary Fial remark: I this chapter we have developed Fourier aalysis over the iterval [, π]. If we wat to study Fourier series over aother iterval [a r, a + r], all we have to do is to move ad rescale the fuctios: The basis ow cosists of the fuctios the ier product is defied by f, g = 2r ad the Fourier series becomes e (x) = e iπ r (x a), = a+r a r f(x)g(x) dx α e iπ r (x a) Note that whe the legth r of the iterval icreases, the frequecies iπ r of the basis fuctios e iπ r (x a) get closer ad closer. I the limit, oe might imagie that the sum = α e iπ r (x a) turs ito a itegral (thik of the case a = 0): α(t)e ixt dt This leads to the theory of Fourier itegrals ad Fourier trasforms, but we shall ot look ito these topics here. Exercises for Sectio 3.3. Use itegratio by parts to check that Z, 0 ( iα ) e ix is the Fourier series of g(x) α 0 x (see the passage after the proof of Propositio 3.3.).

19 3.3. TERMWISE OPERATIONS 9 2. Show that si 2x k= cos((2k )x) = 2 si x. 3. I this problem we shall study a feature of Fourier series kow as Gibbs pheomeo. Let f : [, π] R be give by for x < 0 f(x) = 0 for x = 0 for x > The figure shows the partial sums s (x) of order = 5,, 7, 23. We see that although the approximatio i geeral seems to get better ad better, the maximal distace betwee f ad s remais more or less costat it seems that the partial sums have bumps of more or less costat height ear the jump i fuctio values. We shall take a closer look at this pheomeo. Alog the way you will eed the solutio of problem 2. a) Show that the partial sums ca be expressed as s 2 (x) = 4 si((2k )x) π 2k k= b) Use problem 2 to fid a short expressio for s 2 (x). c) Show that the local miimum ad maxima of s 2 closest to 0 are x = π 2 ad x + = π 2. d) Show that s 2 (± π 2 ) = ± 4 si (2k )π 2 π 2k k=

20 20 CHAPTER 3. SERIES OF FUNCTIONS e) Show that s 2 (± π 2 ) ± 2 π as a Riema sum. 0 si x x dx by recogizig the sum above f) Use a calculator or a computer or whatever you wat to show that dx.8 2 π 0 si x x These calculatios show that the size of the bumps is 9% of the size of the jump i the fuctio value. Gibbs showed that this umber holds i geeral for fuctios i D.