Fourier series ad the δ process Eyuel Abebe* Jameso Graber* Charles N. Moore Worcester Polytechic Istitute Washigto ad Lee Uiversity Kasas State Uiversity Worcester, MA 0609 Lexigto, VA 4450 Mahatta, KS 66506 eyueld@wpi.edu graberp@wlu.edu cmoore@math.su.edu Itroductio Because of the widespread utility of Fourier series, it is of iterest, especially i applicatios, to aalyze their speed of covergece. Particularly eticig is the prospect of applyig methods to accelerate this covergece. Various methods of acceleratio of covergece of sequeces have bee applied with some success to the partial sums of Fourier series. Throughout, we will cosider a fuctio defied o [ π, π] which is itegrable. For such a fuctio f we defie the Fourier coefficiets by ˆf() := π f(x)e ix dx π π for each iteger, ad we defie the th partial sum of the Fourier series as S f(x) := = ˆf()e ix, where is a positive iteger ad x is i [ π, π]. A fudametal cetral questio of the theory is: whe ad i what sese does S f(x) f(x) as? Whe f is square-itegrable, this holds i the mea-square sese, that is, π π S f(x) f(x) dx 0 as. The very deep result of Carleso [] asserts that for square-itegrable f, S f(x) f(x) at every poit except for a set of zero Lebesgue measure. Other classical results, due to Dii-Lipschitz, Lebesgue ad Dirichlet-Jorda give coditios for poitwise covergece (see e.g., Zygmud [0] for these). Typical of these results is the Dirichlet-Jorda theorem: If f is of bouded variatio over [ π, π], the S f(x) coverges to f(x) at each poit of cotiuity of f. However, eve for fairly smooth fuctios, this covergece ca be quite slow ad it is desirable to attempt to fid a way to speed this covergece. For a umerical sequece {s } which has limit s, we say a trasformatio t of s accelerates covergece if there exists a such that each t depeds oly o s 0,..., s + ad t coverges to s faster tha s. May sequece trasformatios have bee developed to speed covergece of umerical sequeces which arise i may cotexts (see e.g. Delahaye [3] or Wimp [9]). I this paper, we discuss a particular o-liear trasformatio, ow as the δ process, usually attributed to Aite [], although the idea had appeared earlier i the wors of other authors. Geeralizatios of this trasformatio, which trasform a sequece {s } ito a family of sequeces {e (s )} were studied extesively by Shas [7] ad for this reaso these trasformatios are sometimes called Shas trasformatios. I this otatio, e (s ) is idetical to the δ process. This process taes the sequece s ad trasforms it to e (s ) = t := where we set t = s if the deomiator of the fractio is zero. s +s s s + + s s = s (s + s )(s s ) (s + s ) (s s ), () The authors were partially supported by the Kasas State Uiversity REU ad NSF grat GOMT53075; AMS subject classificatio 65B0, 4A0
I the case whe the s are the partial sums of a geometric series, the trasform () produces a costat sequece, each term the sum of the series. Shas [7] geeralizes this to show a improved rate of covergece uder trasform () for partial sums of series which he calls early geometric. Cosider the fuctio f o the iterval [ π, π] give by f(x) = if 0 x π, f(x) = if π x < 0. After computatio ad simplificatio, its Fourier series is: At x = π f 4 π = we obtai the slowly coverget Leibiz series: si( )x. () = 4 π ( 3 + 5 7 +... ) Shas [7] applies the trasformatio () ad iteratios of this trasform to the sequece of partial sums ad the results are dramatic: t 8 is accurate to three sigificat figures, ad iteratig the trasform four times gives a sequece whose fifth term is correct to eight sigificat figures. By cotrast, Shas otes it would tae over 40,000 summads i the origial series to obtai this accuracy. Smith ad Ford [8] used umerical tests to compare differet methods of covergece acceleratio to the partial sums of Fourier series. Usig a set of five poits they tested a slowly ad a rapidly covergig Fourier series ad foud that, o the average, the trasformatio () improves covergece of the slowly covergig series but slightly degrades covergece i the rapidly coverget case. Drummod [4] discusses may methods of covergece acceleratio ad icludes discussio of their applicatio to Fourier series. Although with specific series it is possible to get improved, eve dramatic results at some poits, i this ote we show that with the trasformatio δ the results are, i a geeral sese, very bad. We will show that for a fairly geeral set of fuctios, for a large set of values x [ π, π], the δ process turs the sequece {S f(x) : =,,... } ito a set with multiple limit poits. Theorem. Suppose that f C ([ π, π]) ad that f( π) f(π). Cosider the sequece t (x) formed by applyig the trasformatio () to the sequece S f(x). The t (x) diverges at every x of the form x = πa, where a [.5,.5] is irratioal. I fact, the same difficulty also occurs o a dese set of umbers of the form πq, where q is ratioal. At these poits, we ca be eve more precise. Theorem. Suppose f is as i Theorem. Suppose x := πj where j is i lowest terms ad is odd. α The t (x) has three limit poits, f(x) ad f(x) ± si (x/) α si(x/)+β cos(x/), where α = [f(π) f( π)]/π ad β = [f (π) f ( π)]/π. Remars. Cosider the π periodic extesio of f to the whole real lie. The hypotheses of the theorems imply that this fuctio is i C, except for a sigle jump discotiuity every period. By periodicity, the Theorems remai valid if this jump occurs aywhere i the iterval [ π, π]. At a jump discotiuity of f the partial sums of the Fourier series exhibit the Gibbs pheomeo i a eighborhood of the discotiuity ad may authors report difficulties with acceleratio methods ear this discotiuity. I fact, the Theorems imply that, at least for the δ process, a jump discotiuity causes difficulties just about everywhere. It is well ow that the δ process ca tur a coverget sequece ito oe with multiple coverget subsequeces. (See Wimp [9] for a example due to Lubi.) Thus, our theorems provide may more examples of that pheomeo. Prelimiaries, bacgroud ad techical lemmas If S f(x) deotes the sequece of partial sums of the Fourier series of f, applyig the δ trasform () results i the sequece of fuctios:
( ˆf( ( + ))e i(+)x + ˆf( + )e i(+)x ) ( ˆf( )e ix + ˆf()e ix ) t (x) = S f(x) ( ˆf( ( + ))e i(+)x + ˆf( ) ( + )e i(+)x ˆf( )e ix + ˆf()e ). (3) ix Our theorems will be established by demostratig the bad behavior of the fractio o the right. I this sectio we prove some lemmas which will be used to aalyze this term. Lemma. Let f C ([ π, π]). The for every iteger, where α ad β are as i the Theorem. Proof. Itegratio by parts. Remar. [0].) Notice that f () ˆf() = ( )+ i α + ( ) β f (), = o( ) as by the Riema-Lebesgue lemma. (See, e.g. Zygmud Let a be a irratioal umber ad let a 0, a, a,... be its uique expressio as a cotiued fractio. Defie sequeces {h } ad { } as follows: h = 0, h =, h i = a i h i + h i for i 0, =, = 0, i = a i i + i for i 0. (4) Note that = 0 < < 3 <. The value r := h is called the th coverget to a, ad r = a 0, a,..., a. (See Nive ad Zucerma [6], Chapter 7 for these results.) Lemma. Let a be a irratioal umber. The there exist ifiitely may ratioal umbers h with odd such that a h <. { } h Proof. Let be the sequece of covergets to a as defied above. By Theorem 7. i Nive ad Zucerma [6], we have a h < <, + where the secod iequality follows from the fact that < + as oted above. Thus, it suffices to show that there exist ifiitely may values such that is odd. We offer a proof by cotradictio. Suppose m is the greatest value for which m is odd. The m+ ad m+ must be eve. But equatio (4) gives us m+ = a m+ m+ + m, which is odd. Hece m+ is both eve ad odd, a cotradictio. Propositio. Let a be irratioal ad let x = πa. The there exists a strictly icreasig sequece {m } of positive itegers such that lim if si m x si(m + )x (m + ) si m x + m si(m + )x > 0. Proof. By the previous lemma, we ca defie sequeces {m } ad {l } of itegers such that {m } is strictly icreasig, each m is positive, ad, for every, a l m + < (m + ) < 4m. + m 3
Thus, for each, si(m + /)x = si((m + /)x l π) (m + /)x l π = (m + )πa l π = (m + )π a l m + < (m + )π 4m = π. + m m Now observe that It follows that, for each N, si m x = si(m + /)x cos x si x cos(m + /)x si(m + )x = si(m + /)x cos x + si x cos(m + /)x. si m x si(m + )x (m + ) si m x + m si(m + )x = si (m + /)x cos x si x cos (m + /)x si m x + m cos x si(m. + /)x Notice that the deomiator of this expressio is bouded: si m x + m cos x si(m + /)x + m si(m + /)x < + π. Moreover, the limit of the umerator is clearly si x, sice si (m + /)x 0 as ad hece cos (m + /)x as. It follows that this ratio is bouded away from 0 as. Lemma 3. Suppose x = πj, where j ad are o-zero itegers with o commo factors ad is odd. The, for itegers, si x = si( + )x if ad oly if = + m, where m is some iteger. Proof. Suppose si x = si( + )x. Note the followig idetities: (a) si x = si( + /)x cos x si x cos( + /)x ad (b) si( + )x = si( + /)x cos x + si x cos( + /)x. We quicly see that si( + /)x = 0, which implies that ( + /)x = lπ for some iteger l. Multiplyig both sides by /jπ ad substitutig for x we obtai + = l/j. Sice l/j must be a odd iteger ad j ad have o commo factors, l/j must be a odd iteger. Hece we write l/j = m + for some iteger m ad obtai + = m +, which yields = + m, as desired. To prove the coverse, ote that wheever = + m, we have ( + /)x = (m + )jπ, which implies si( + /)x = 0 ad cos( + /)x = ( ) j. By idetities (a) ad (b) we have si x = ( ) j+ si x = si( + )x, as desired. Remar: Note the followig idetities: (c) cos x = cos( + /)x cos x + si x si( + /)x ad (d) cos( + )x = cos( + /)x cos x si x si( + /)x. Now observe that si x = si( + )x implies si( + /)x = 0, which by (c) ad (d) implies cos x = cos( + )x = ( ) j cos x. This relatio will be ecessary i the proof of Theorem. 4
3 Proofs of theorems ad We first give a proof of Theorem. For x as described i the statemet of the theorem, we compute the fractio o the right of (3). From Lemma we get ˆf( )e ix + ˆf()e ix = ( )+ α si x + ( ) β cos x + ε(), ( ) where we have defied ε() := f ( ) e ix + f () e ix. Usig this otatio, we may write the fractio o the right had side of (3) as: ( ) ( ) ( ) + α si( + )x + ( )+ β cos(+)x (+) + ε( + ) ( ) (+) α si x + ( ) β cos x + ε() ( ) ( ) = ( ) + α si( + )x + ( )(+) β cos(+)x (+) + ε( + ) ( ) + α si x + ( ) β cos x + ε() ( ) ( ) ( ) α si( + )x β cos(+)x + + ( ) β cos x ( + )ε( + ) α si x + + ( ) ε() ( ). (+) α ( si( + )x + ( + ) si x) β + cos( + )x + cos x + ( + )( ) (ε( + ) ε()) (5) Let m be the sequece of icreasig itegers give by Propositio. We substitute this sequece for i the last expressio. Notice that m ε(m ), (m + )ε(m + ), β cos(m+)x β cos mx m +, m all go to 0 as. Furthermore, m (m + )(ε(m + ) ε(m )) 0 as by the Riema-Lebesgue lemma. m Both of the terms m cos(m + + ) ad (m+) m cos m x remai bouded as. As i the proof of Propositio, the umerator is asymptotically ( ) + α si ( x ) ad the deomiator remais bouded. As, this shows that (5) stays bouded away from 0, ad hece by (3), t (x) has a subsequece which fails to coverge to f(x). To show that t (x) diverges, we show that there are subsequeces which coverge to f(x). Cosider the term ( + ) si x + si( + )x i the deomiator of (5). Rewrite this as: si x + cos( x ) si( + )x. By Weyl s equidistributio theorem (see Körer [5]), there are a ifiite umber of for which (+ )x [ 3π 4, π 4 ] [ π 4, 3π 4 ]. For these, si x+ cos( x ) si(+ )x is bouded below by a positive costat times, which forces (5) 0. To see Theorem, we examie (5) i this case. Cosider a x := πj, where is odd ad the fractio j is i lowest terms. Cosider of the form = + m for positive itegers m (as i the statemet of Lemma 3) i the expressio (5). As before, ε(), (+)ε(+), β cos(+)x β cos x +, ad all go to 0 as ad ( + )(ε( + ) ε()) 0 as by the Riema-Lebesgue lemma. Furthermore, for this x ad such, si x = ( ) j+ si(x/), si( + )x = ( ) j si(x/), ad cos x = cos( + )x = ( ) j cos(x/); thus expressio (5) acts asymptotically as ( ) +j+ α si (x/), which has oe sig for odd α si(x/) + β cos(x/) m ad the opposite sig for eve m. If + m for every iteger m, si x si( + )x ad thus i (5) the term ( + ) si x + si( + )x is bouded below by a costat times, which causes (5) 0 as. (We are also usig the fact that as varies {x (mod π) : =,,... } is oly a fite set.) Remar. I coclusio, i the case of Theorem, t (x) has a subsequece, {t (x) : = + m ad m = 0,,,... } which produces two limit poits differet from f(x) ad alog the subsequece of all other idices t (x) f(x). 4 Iterated ad higher-order trasforms I his extesive study, Shas [7] also cosiders the effect of iteratig the trasform () as well as higher order trasforms. We will mae oly a few commets o these. With regards to the iterated trasform, e, we ca assert that agai, o a dese set of x, this trasform seds S f(x) to a sequece with diverget subsequeces. Ideed, we oly eed cosider x of the form x = πj, odd, ad where j ad have o commo factors. The remar at the ed of the previous sectio describes the behavior of the sequece t (x) = e (S f(x)) ad the fact that e (S f(x)) has subsequeces which still do ot coverge to f(x) is a immediate cosequece of the followig: 5
Lemma 4. If a is a sequece ad m, l, are positive itegers with m variable ad l, fixed,, such that a a if l + m ad a a + b for = l + m the e (a ) a + b for = + ml as m. Proof. Notice that for = l + m, e (a ) = a (a + a )(a a ) [a (a + b)][(a + b) a] (a + b) (a + a ) (a a ) [a (a + b)] [(a + b) a] = a + b. Remar. Aalyzig e (S f(x)) seems to be much more difficult. (See e.g. Shas [7] for the defiitio of e.) As i Theorems ad, let S f(x) be the sequece of partial sums of a Fourier series of at a x [ π, π], ad t (x) be the sequece obtaied by applyig the trasformatio (). We ca obtai e (S f(x)) via the ε algorithm (see [3]): ε () := 0 for every, ε 0 () := S f(x) ad ε + () = ε ( + ) + ε (+) ε (). The e (S f(x)) = t (x) = ε () ad e (S f(x)) = ε 4 (). For coveiece, write s for S f(x), ad t for t (x). The computig ε 4 () gives: ε 4 () = t + + [ t + s + ] + [ t +3 t + ] [ t + t + ] Cosider x of the form x := πj where j ad have o commo factors ad is odd. By Theorem, ad the remar after its proof, t (x) has a subsequece, = + m, m = 0,,,... alog which α t (x) f(x) + si (x/) α si(x/)+β cos(x/). For coveiece write this limit as f(x) + γ. For = + m ±, as, t (x) f(x). With i place of i the above formula, ad lettig alog the subsequece = + m, m = 0,,,..., yields: ε 4 ( ) = t + [ t s ] + [ t + t ] [ t t ] f(x) + γ + γ + γ = f(x). γ So curiously, the stray subsequece of t (x) which diverged from f(x) has bee trasformed bac to the correct limit. However, the aalysis is too delicate to tell whether there are ow ew subsequeces which go astray, ad certaily i the case whe the limit is the actual value f(x) we caot determie whether covergece has bee accelerated. 5 A fuctio with two jumps Recall the example of equatio (). This case is ot covered by Theorem or Theorem because it has two jumps whe cosidered as a π periodic fuctio. Nevertheless, it exhibits problems uder the Shas trasformatio aalogous to problems we have already see. I some sese, we cheat by trasformig the series cosistig of oly the odd terms i the Fourier series of f. But all the eve Fourier coefficiets i this example are 0, so if we were to tae the series cosistig of all the terms i the Fourier series of f, the () would give us t = s. This would give us o isight. Hece our cheat is justified. Lemma 5. Let x := πj, where j, are itegers with o commo factors ad j is odd. The for itegers, 4 si( )x = si( + )x if ad oly if = m, m odd, i which case this commo value is ( ) r cos x, where r = (mj )/. Proof. First, suppose si( )x = si( + )x. From the idetities (a) si( )x = si x cos x si x cos x ad (b) si( + )x = si x cos x + si x cos x, we quicly see that cos x = 0. It follows that x = lπ + π/ for some iteger l. Multiply by /πj ad substitute for x to get = (l + )/j. Sice is a iteger, ad j ad have o commo factors, we must have that m := (l + )/j is a iteger. Sice mj is odd, it follows that m is odd; this proves the first implicatio of the lemma. We prove the coverse by otig that wheever = m, m odd, we have x = mjπ/, i which case cos x = 0 ad si x = ( ) r where r := (mj )/. Thus, by idetities (a) ad (b) we have as desired. si( )x = si( + )x = si x cos x = ( ) r cos x, 6
Propositio. Let f be defied as i equatio (). Suppose x [ π, π] satisfies the hypotheses of the previous lemma. The t (x) has three limit poits: f(x) ad f(x) ± π cos x. Proof. The th partial sum of the Fourier series of f at x is Substitutig this ito equatio (), we get t (x) = S f(x) 4 π S f(x) = 4 π = si( )x. si( )x si( + )x ( ) si( + )x ( + ) si( )x. We ow that S f(x) f(x). Now as o subsequeces for which si( )x si(+)x 0, the deomiator of the secod term i the above equatio is ubouded, forcig the ratio to go to zero. Therefore, t (x) f(x) as o those subsequeces. Now suppose is icreasig o a subsequece of odd multiples of. The by Lemma 5, si( )x = si( + )x = ± cos x for each. Thus the secod term i the above equatio is simply ± π cos x for each alog this subsequece, ad so as, t (x) f(x) ± π cos x. Refereces [] Aite, A.C., O Beroulli s umerical solutio of algebraic equatios, Proc. Roy. Soc. Ediburgh 46 (96) 89-305. [] Carleso, L., O covergece ad growth of partial sums of Fourier series, Acta. Math. 6 (966), 35-57. [3] Delahaye, Jea-Paul, Sequece Trasformatios, Spriger Series i Computatioal Mathematics, Spriger Verlag, Berli, 988. [4] Drummod, J.E., Covergece speedig, covergece ad summability, Joural of Computatioal ad Applied Mathematics (984), 45-59. [5] Körer, T. W., Fourier Aalysis, Cambridge Uiversity Press, Cambridge, 988. [6] Nive, Iva ad Zucerma, Herbert, A Itroductio to The Theory of Numbers, Third Editio, Joh Wiley ad Sos, New Yor, 97 [7] Shas, Daiel, No-liear trasformatios of diverget ad slowly coverget sequeces, J. Math. Phys. 34 (955), -4. [8] Smith, David A. ad Ford, William F., Numerical Comparisos of Noliear Covergece Accelerators, Mathematics of Computatio 38, o. 58 (98), 48-499. [9] Wimp, Jet, Sequece Trasformatios ad Their Applicatios, Academic Press, New Yor. 98. [0] Zygmud, A., Trigoometric Series, Secod Editio, Cambridge Uiversity Press, Cambridge, 959. 7