Chapt.12: Orthogonal Functions and Fourier series

1 : Orthogonal Functions and Fourier of Mathematics & StatisticsMcMaster UniversityHamilton, ON, CanadaLecture: January 10, 2011 p. 1 : Orthogonal Functions Recall thatRn={(x1, .. , xn), xi R, i= 1, .. n}. Ifu= (u1.. , un)andv= (v1.. , vn)belong toRn, theirdotproductis the number(u,v) =u1v1+ +unvn=n i=1uivi. The dot product has the following properties: (u,v) = (v,u) ( u,v) = (u,v) = (u, v), R (u+v,w) = (u,w) + (v,w) (u,u) 0and(u,u) = 0iffu= 0 Lecture: January 10, 2011 p. 2/30 Orthogonal collections Thenormof a vector:kuk= u21+ +u2n= (u,u)1/2 Orthogonality of two vectors:u viff(u,v) = 0. Orthogonality of a collection of vectors:{u1, .. ,um}is anorthogonal collection of vectors iff(ui,uj) = 0ifi6=j. Orthogonal basis: Ifm=n, the dimension of the space, then anorthogonal collection{u1, .. ,un}whereui6= 0for alli, forms anorthogonal basis. In that case, any vectorv Rncan be expandedin terms of the Orthogonal basis via the formulav=n i=1(v,ui)uikuik2.

2 Orthonormal basis: Orthogonal basis{u1, .. ,un}withkuik= : January 10, 2011 p. 3/30 Orthogonal Functions In what follows, we will alwaysassume that the functionsconsidered arepiecewise continuouson some interval[a, b]. Inner product: Iff1, f2are two Functions defined on[a, b], we definetheir inner product as(f1, f2) = baf1(x)f2(x)dx Orthogonality: Two functionsf1, f2are Orthogonal on[a, b]if(f1, f2) = 0. Example:f(x) = sin(3x),g(x) = cos(3x). We have sin(3x) cos(3x)dx= 0sincesin(3x) cos(3x)is odd and the interval[ , ]is symmetricabout0. Thusf(x) = sin(3x)andg(x) = cos(3x)are Orthogonal on[ , ].Lecture: January 10, 2011 p. 4/30 Orthogonal Functions contd. Example:f(x) = sin(3x),g(x) = cos(3x). We have sin(3x) cos(3x)dx= 0sincesin(3x) cos(3x)is odd and the interval[ , ]is symmetricabout0. Thusf(x) = sin(3x)andg(x) = cos(3x)are Orthogonal on[ , ]. Orthogonal collections: A collection of Functions { 0(x), 1(x).}

3 , m(x), ..}defined on[a, b]is called orthogonalon[a, b]if( i, j) = ba i(x) j(x)dx= 0,wheni6= : January 10, 2011 p. 5/30An exampleThe collection{1,cos(x),cos(2x),cos(3x), ..}={cos(k x), k 0}isorthogonal on[ , ]. To show this, we use the identitycosAcosB=cos(A+B) + cos(A B)2. We have, ifm, n 0are integers withm6=n, cos(mx) cos(nx)dx= cos((m+n)x) + cos((m n)x)2dx=12[sin((m+n)x)m+n+sin((m n)x)m n] = 0 Lecture: January 10, 2011 p. 6/30 Orthonormality Iff(x)is a function defined on[a, b], we define thenormoffto bekfk= (f, f)1/2=( baf(x)2dx)1/2 A collection of Functions { 0(x), 1(x), .. , m(x), ..}defined on[a, b]is calledorthonormalon[a, b]if( i, j) = ba i(x) j(x)dx={0, i6=j1, i=j. Note that if the collection{ 0(x), 1(x), .. , m(x), ..}isorthogonal on[a, b]andk ik6= 0, the collection{ 0(x)k 0k, 1(x)k 1k, .. , m(x)k mk, ..}is orthonormal on[a, b].Lecture: January 10, 2011 p. 7/30An example Consider the collection{1,cos(x),cos(2x),cos(3x).}

4 }or{cos(k x), k 0}which we have shown to be is Orthogonal on[ , ]. We havek1k2= 12dx= 2 ,sok1k= 2 . Form 1, we havekcos(mx)k2= cos2(mx)dx= 1 + cos(2mx)2dx=[x2+sin(2mx)4m] = . Thuskcos(mx)k= . The collection{1 2 ,cosx ,cos(2x) ,cos(3x) , ..}is thus orthonormalon[ , ].Lecture: January 10, 2011 p. 8/30 Section continued Suppose that the collection{ n(x)}n 0is an Orthogonal collection(or system ) on[a, b]and that the functionf(x)defined on[a, b]can be expanded as a seriesf(x) =c0 0(x) +c1 1(x) + +cn n(x) +.. ,(1)how can we compute the coefficientsc0, c1, c2, ..? Note that if(1)holds, we have, for eachn 0,(f, n) = baf(x) n(x)dx= ba{ k=0ck k(x)} n(x)dx= k=0ck ba k(x) n(x)dx=cn ba 2n(x)dx=cnk : January 10, 2011 p. 9/30 Orthogonal systems It follows thus that, if(1)holds, thencn=(f, n)k nk2, n (x) = n=0(f, n)k nk2 n(x).(2) However, the expansion formula(2)does not hold in general for anarbitrary Orthogonal system on[a, b].

5 For example, it could happenthatf6= 0butf(x)is Orthogonal to each function n(x)in thesystem and thus the RHS of(2)would be0in that case whilef(x)6= 0. In order for(2)to hold for an arbitrary functionf(x)defined on[a, b], there must be enough Functions nin our : January 10, 2011 p. 10/30 Completeness Definition: An Orthogonal system{ n(x)}n 0on[a, b]iscompleteifthe fact that a functionf(x)on[a, b]satisfies(f, n) = 0for alln 0implies thatf 0on[a, b], or, more precisely, thatkfk2= baf2(x)dx= 0. If{ n(x)}n 0on[a, b]is a complete Orthogonal system on[a, b],then every (piecewise continuous) functionf(x)on[a, b]has theexpansionf(x) n=0(f, n)k nk2 n(x).(3)on[a, b]in theL2-sensewhich means thatlimN ba f(x) N n=0(f, n)k nk2 n(x) 2dx= : January 10, 2011 p. 11/30 Some remarks If{ n(x)}n 0on[a, b]is a complete Orthogonal system on[a, b],the expansion formula(3)holds for every (pwc) functionf(x)on[a, b]in theL2-sense, but not necessarily pointwise , for a fixedx [a, b]the series on the RHS of(3)might not necessarilyconverge and, even if it does, it might not converge tof(x).

6 The system{1,cos(x),cos(2x),cos(3x), ..}={cos(k x), k 0}isorthogonal on[ , ]but it isnot completeon[ , ]. Indeed, iff(x)any odd function on[ , ](f( x) = f(x)) withkfk6= 0, such asf(x) =xorf(x) = sinx, we have f(x) cos(n x)dx= 0, n 0,sincef(x) cos(n x)is odd and[ , ]is symmetric : January 10, 2011 p. 12/30 Section : Fourier series Theorem:The systemT:={1,cos(x),sin(x),cos(2x),sin(2x ),cos(3x),sin(3x), ..}is acomplete Orthogonal systemon[ , ]. To show the orthogonality of this system, one needs to show that cos(mx) cos(nx)dx= 0, m, n 0, m6=n,(a) sin(mx) sin(nx)dx= 0, m, n 1, m6=n,(b) cos(mx) sin(nx)dx= 0, m 0, n 1.(c)Lecture: January 10, 2011 p. 13/30 Fourier series contd. For example, to show(b), we use the formulasinAsinB=cos(A B) cos(A+B)2. We have then, ifm, n 1andm6=n, sin(mx) sin(nx)dx= cos((m n)x) cos((m+n)x)2dx=[sin((m n)x)m n sin((m+n)x)m+n] = 0. We have also, form, n 1,k1k2= 2 ,kcos(mx)k2= ,ksin(nx)k2=.

7 Lecture: January 10, 2011 p. 14/30 Fourier series expansions Note that the completeness of the systemTis much more difficultto prove. Using the previous theorem, it follows that every (pwc) functionf(x)on[ , ]admits the expansionf(x) a02+ n=1{ancos(nx) +bnsin(nx)}(4),wherea02=(f,1)k1k2=12 f(x)dx=average offon[ , ],an=(f,cos(nx))kcos(nx)k2=1 f(x) cos(nx)dx, n 1,bn=(f,sin(nx))ksin(nx)k2=1 f(x) sin(nx)dx n : January 10, 2011 p. 15/30 Fourier series on general intervals The series expansion(4)in terms of the trigonometric systemTiscalled theFourier series expansionoff(x)on[ , ]. More generally, ifp >0andf(x)is pwc on[ p, p], then it will havea Fourier series expansion on[ p, p]given byf(x) a02+ n=1{ancos(n xp)+bnsin(n xp)}(4),where theFourier coefficientsan,bnare defined byan=1p p pf(x) cos(n xp)dx, n 0,bn=1p p pf(x) sin(n xp)dx n : January 10, 2011 p. 16/30An example The functionf(x) ={0, < x 0x,0< x < has a Fourier series expansion on[ , ]given byf(x) 4+ n 1nodd( 2 n2)cos(nx) + n 1( 1)n+1nsin(nx) 4 2 cos(x) 29 cos(3x) 225 cos(5x) +.}

8 Sin(x) 12sin(2x) +13sin(3x) 14sin(4x) +..( )Lecture: January 10, 2011 p. 17/30 Periodic extension If a functionf(x)defined on the interval[ p, p]is expanded as theFourier seriesf(x) a02+ n=1{ancos(n xp)+bnsin(n xp)}(5),we can view the RHS of(5)as a function defined om all ofR. Sincecos(n (x+ 2p)p)= cos(n xp+ 2n )= cos(n xp),sin(n (x+ 2p)p)= sin(n xp+ 2n )= sin(n xp),the RHS of(5)is2p-periodic and thus equal to the2p-periodicextensionoff(x)to the real : January 10, 2011 p. 18/30 Piecewise continuity Recall that a functionf(x)defined on the interval[a, b]ispiecewisecontinuous(pwc) on[a, b]if[a, b]can be divided intoNsubintervals[ai, ai+1],i= 0, .. , N 1witha=a0< a1< a2< < aN 1< aN=band such thatf(x)iscontinuous on each open interval(ai, ai+1),i= 0, .. , N 1andlimx a+if(x) =f(a+i),limx a i+1f(x) =f(a i+1)both exist (and are finite) for eachi= 0, .. , N 1. A functionf(x)defined onRis pwc if it is pwc on every interval[a, b] : January 10, 2011 p.

9 19/30 Pointwise convergence Note that, in the theory of Fourier series , iff(x)is pwc, the value ofthe functionf(x)at the end pointsaiwheref(x)is discontinuousis unimportant (as they do not affect the integral to computetheFourier coefficients off(x)). Definition:If a functionf(x)defined onRis2p-periodic(f(x+ 2p) =f(x)), its Fourier series is the Fourier series of itsrestriction to the interval[ p, p]. Theorem:Letf(x)be a2p-periodic function defined onRsuch thatbothf(x)andf (x)are pwc onR. Then, the Fourier series off(x)converges for allxto a Functions (x)whereS(x) ={f(x),iff(x)is continuous atx,f(x+)+f(x )2,iff(x)is not continuous : January 10, 2011 p. 20/30 Section : Fourier cosine and sine series Definition:Letf(x)be a function defined on[ p, p] f(x)iseveniff( x) =f(x). f(x)isoddiff( x) = f(x). Note that iff(x)is even, then p pf(x)dx= 2 p0f(x)dx. On the other hand, iff(x)is odd, p pf(x)dx= 0. Note that f(x)even andg(x)even= f(x)g(x)even f(x)even andg(x)odd= f(x)g(x)odd f(x)odd andg(x)odd= f(x)g(x)evenLecture: January 10, 2011 p.}

10 21/30 Fourier cosine and sine series Iff(x)is even on[ p, p], we havean=1p p pf(x) cos(n xp)dx=2p p0f(x) cos(n xp)dxforn 0, andbn=1p p pf(x) sin(n xp)dx= 0, n 1. Similarly, iff(x)is odd on[ p, p], we havean=1p p pf(x) cos(n xp)dx= 0, n 0and, forn 1,bn=1p p pf(x) sin(n xp)dx=2p p0f(x) sin(n xp)dxLecture: January 10, 2011 p. 22/30 Fourier cosine and sine series contd. The Fourier series of anevenfunctionf(x)on[ p, p]is thus aFourier cosine seriesf(x) a02+ n=1ancos(n xp)x [ p, p],(6)wherean=2p p0f(x) cos(n xp)dx, n 0. Similarly, the Fourier series of anoddfunctionf(x)on[ p, p]is aFourier sine seriesf(x) n=1bnsin(n xp)x [ p, p],(7)wherebn=2p p0f(x) sin(n xp)dx, n : January 10, 2011 p. 23/30 Fourier sine series :an example The functionf(x) = sin(x/2), < x < , is odd. Its Fourier series on[ , ]is thus a sine Fourier series . It is given explicitly byf(x) 2 n=1( 1)n+1nn2 1/4sin(nx), x [ , ]. : January 10, 2011 p.