Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006
Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer kernel s, nd eventully use these to prove form of Fejer s theorem generlized to functions of rbitrry period. We lso discuss one specific exmple of Fourier Series, pply Fejer s theorem to see tht it converges to the correct function.
. Introduction We ssume the reder is fmilir with Fourier Series. So, we merely stte, without justifiction, the form of the Fourier series ssocited with function f(x which is periodic. f(x 0 + n cos nπx where the constnts n nd b n re + b nsin nπx, n = f(tcos nπt dt, b n = f(xsin nπt dt. We will notte the Nth prtil sum of Fourier series s S N (x = N 0 + n cos nπx + b n sin nπx. We will lso use σ N (x, the rithmetic men of the prtil sums up to N, defined s follows. σ N (x = S n (x Additionlly, in this pper, the followingtrigonometric identities will be ssumed without proof. ( cos(xcos(y = (cos(x y+cos(x + y, ( sin(xsin(y = (cos(x y cos(x + y, (3 sin(xcos(y = (sin(x + y sin(y x, (4 sin (x = ( cos(x.. The Dirichlet Kernel We substitute the definitions of n nd b n into S N (x togive + ( f(tcos nπt S N (x = f(t dt nπx cos dt + f(tsin nπt nπx sin dt
We then turn the sum of integrls into the integrl of sum, nd pply ( nd ( in the Introduction. After cncelingterms, we hve S N (x = ( N f(t + nπ(t x cos dt. Then, we mke the substitution u = t x. Since f(x hsperiod, this substitution doesn t require us to chnge the limits of integrtion. We then hve S N (x = ( N f(x + u + cos nπu dt. sin πu Now, note tht we my tke ( + N cos nπu =sin πu N + sin πu nπu cos. Applyingthe third trigonometric identity in the Introduction, nd cnceling terms, we rrive t + N cos nπu (N+/πu sin = sin πu. Substituting this result into the previous integrl eqution gives wht is commonly referred to s Dirichlet s Integrl. Dirichlet s Integrl S N (x = f(x + ud N (udu. Here, D N is Dirichlet s Kernel. Dirichlet s Kernel D N (u = (N+/πu sin sin. πu We will now prove one importnt property of the Dirichlet Kernel, to be used lter. Usingthe summtion form of D N, we write ( D N (xdx = + cos nπx dx. Agin, we chnge the integrl of sum to sum of integrls nd evlute. D N (xdx = + N πn (sin(nπ+sin( nπ The right hnd term evlutes to zero, nd we hve (5 D N (xdx =
3. The Fejer Kernel We define Fejer s Kernel K N (x in the followingmnner. Fejer s Kernel K N (x = N N+ D n(x However, we desire closed form expression for Fejer s Kernel. Towrd this end, we multiply the kernel by sin πx, which, fter pplyingthe definition of D n,gives sin πx K N(x = sin Agin pplying( nd cncelingterms, we get ( cos sin πx K N (x = (n +/πx Finlly, pplying(4 nd solvingfor K N (x yields K N (x = (πx cos (N+πx cos πx. sin πx. We wish to prove four properties of Fejer s Kernel. Firstly, tht (6 K N (x 0. This fct is obvious if we pply (4 to both cosine terms, giving K N (x = ( sin (N+πx sin πx.. For the second property, we use the definition of K N to write K N (x = D N (xdx. Using (5 to evlute the right hnd integrl to for ech n immeditely shows tht (7 K N (xdx =. For the third property, we let 0 <δ x. This implies tht cos πδ. This, in turn, implies tht cos πx K N (x = cos (N+πx cos πx 3 cos (N+π x cos πδ.
Additionlly, it is cler tht cos (N+πx. So, we rrive t (8 K N (x cos πδ whenever 0 <δ x. The lst property is wht is clled Fejer s Integrl. Reclling the definition of σ N (x nd Dirichlet s Integrl, we my write σ N (x = S n (x = f(x + ud n (udu. Agin tking the sum inside the integrl, we find σ N (x = ( f(x + u D n (u du. Rememberingthe definition of K N (x t the strt of this section, we rrive t Fejer s Integrl. (9 σ N (x = f(x + uk N (udu. 4. Convergence by Arithmetic Mens Before provingfejer s Theorem, we will give brief proof of the following fct: Convergence Theorem If the limit of {S n } exists, then lim n S n = lim n σ n, where S n nd σ n hve the sme definitions s in the introduction. Hving proof of this theorem will mke Fejer s Theorem much more meningful. Proof: Suppose the sequence in question converges to S. Tht is, lim n S n = S. Then, there exists n integer N such tht for every n N, S S n < ɛ. Then, we my write S σ n = (n +S n + n + n S k. We let n N. We then tke S inside the sum nd divide the sum into two prts, giving S σ n = n + (S S k + n + k=0 4 k=0 n k=n+ (S S k.
We my of course pply the tringle inequlity over both sums to give bsolute vlues. Since, the right hnd sum is less thn (n + ɛ.sowehve S σ n < n + S S k + ɛ. As N is constnt, we my let n grow lrge until the left hnd sum is less thn ɛ. So we then hve S σ n <ɛ for ll n M N, where M is some constnt integer. Therefore, σ n converges to S. k=0 5. Fejer s Theorem We now come to Fejer s Theorem, which is stted below. Fejer s Theorem If f is rel vlued, continuous function with period, then σ n (x converges uniformly to f(x. Proof: Remembering(7 nd (9, we write the difference between σ N (x nd f(x s σ N (x f(x = f(x + uk N (udu f(x K N (udu. We my combine these integrls, nd pply the tringle inequlity to get bsolute vlues on both sides. We write K N (u outside the bsolute vlue since we hve lredy shown tht it is lwys greter thn or equl to zero. σ N (x f(x f(x + u f(x K N (udu. Note tht [, ] is compct, nd f is continuous on tht intervl. Therefore, f is uniformly continuous on tht intervl. Since f is uniformly continuous, there exists δ>0 (which does not depend on x such tht x y <δimplies f(y f(x < ɛ. Without loss of generlity, we tke δ<. Additionlly, since f is continuous nd periodic, it must be bounded bove nd below. We let M = sup f(x. We then hve tht f(x + u f(x M. Keepingll of this in mind, we divide the bove integrl into three prts: to δ, δ to δ, nd δ to. Tkingdvntge of the continuity of f in the middle term gives ( σ N f(x < δ MK N (udu + ɛ δ δ K N (udu + MK N (udu. δ 5
The combintion of both (6 nd (7 requires tht the middle term be less thn or equl to ɛ. Using this nd (8 on the outer integrls gives ( M δ σ N (x f(x < ( cos πδ du + δ cos πδ du + ɛ. Evlutingboth integrls, we now rrive t 4M( δ σ N (x f(x < ( ( cos πδ + ɛ. Everythingon the right hnd side is constnt except for N. Thus, for some sufficiently lrge A, σ N (x f(x <ɛ whenever N A. Note tht nowhere is there dependence on x in the right hnd side, s M depends only on δ, which doesn t depend on x. This detil is of gret importnce, becuse it plces bounds on the difference between σ N nd f everywhere. One cn imgine the nightmre in engineering pplictions if the bound on the error of Fourier pproximtion vried from point to point. Such scenrio would significntly decrese the usefulness of Fourier Series. This concludes our proof of Fejer s Theorem. 6. The Tringle Wve To conclude, we discuss n explicit exmple of function to which the methods of Fourier Series re pplicble. The function is the so clled tringle wve, defined s follows: T (x = x 3 for x 5, T(x +=T (x. This is n odd function of period. Strightforwrd ppliction of the definitions given in the Introduction to this pper shows tht the Nth prtil sum of the Fourier Series of T (x is S N = 8 π N sin nπ n sin nπx. Notingtht for even n, sin nπ nπ = 0, nd tht for odd n, sin chieve the followingform for the prtil sums: =( n,we S N = 8 π N,3,5,.. ( n n sin nπx. Included below re three grphs of S 3 (x, S 5 (x, nd S 9 (x, respectively. 6
Note tht the sequence of prtil sums clerly converges by the Weierstrss M-test. We simply note tht ( n sin nπx. Then, we simply need tht,3,.. n converges, which it most certinly does. Note tht this not only proves tht the prtil sums converge, but tht they converge uniformly. Applyingthe formul for σ N (x from the introduction to this cse, we get σ N (x = 8 π n k=,3,.. ( k k sin kπx. Notingtht the nth prtil sum ppers n times in σ N (x, we see tht σ N (x = 8 π (,3,.. n n ( n sin nπx. It is not so immeditely cler tht this sequence converges s N goes to infinity. However, the theorem proved in Section 4 gurntess tht it does. Since {S N (x} converges, {σ N (x} must s well. Furthermore, they converge to the sme thing. Fejer s Theorem, s proved in Section 5, gurntees tht lim σ N(x =T (x. n In fct, Fejer s Theorem implies even more: tht this convergence is uniform. If we gin mke use of the theorem from Section 4, we see tht lim n S N (x =T (x s well. In this cse, this convergence is lso uniform, s shown by the Weierstrss M-test bove. Included below, for those who desire visul clrifiction, re three grphs: S 3 (x, S 5 (x, nd S 9 (x, respectively. Note tht ech successive grph looks more similr to T (x thn the previous one. This concludes our discussion of the tringle wve, nd the pper. 7
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