Fourier Series Expansion - Government Engineering …

1 Fourier Series ExpansionDeepesh K PThere are many types of Series expansions for functions. The Maclaurin Series ,Taylor Series , Laurent Series are some such expansions. But these expansionsbecome valid under certain strong assumptions on the functions (those assump-tions ensure convergence of the Series ). Fourier Series also express a function asa Series and the conditions required are fairly good and suitable when we dealwith a real valued function fromRtoR. In this note, we deal withthe following three questions: When doesfhas a Fourier Series Expansion ?

2 How we find the Expansion ? What are the main properties of this Expansion ?1 Existance of a Fourier Series Expansion :There are three conditions which guarantees the existance of a valid Fourierseries Expansion for a given function. These conditions are collectively calledtheDirichlet a periodic function onR. This means that there exists a periodT 0such thatf(x) =f(x+T) for allx only a finite number of maxima and minima in a atmost a finite number of discontinuous points inside a integrable over the period of the should be noted that the second and third conditions are satisfied by many realvalued functions that we deal with, inside any finite interval.

3 But periodicity is acondition that is satisfied by very few functions, for example,constant function,sine, cos, tanand their combinations. But we can consider any function definedon a finite interval [a,b] (or (a,b)) as a periodic function onRby thinking thatthe function is extended toRby repeating the values in [a,b] to the remainingpart ofR. ThusMost of the functions, that we commonly use, definedon finite intervals can be expanded as Fourier seriesFigure12 Derivation of Fourier Series Expansion of afunction defined in[ , ]:In Fourier Series Expansion , we would like to write the function as a Series insine and cosine terms in the form:f(x) =a02+ n=1ancosnx+bnsinnxFor finding the above unknown co-efficientsa0,anandbnin the Fourier seriesexpansion of a function, one need to recall the value of certain integrals:1.

4 Sinmxdx= 0 for any cosmxdx= 0 for any sinmxcosnxdx= 0 sinmxsinnxdx= 0 for integersm6= cosmxcosnxdx= 0 for integersm6= sinmxsinnxdx= when the integersm= cosmxcosnxdx= when the integersm=n.[All the above integrals easily follow by evaluating using integration by parts]Now supposef(x) =a02+ j=1ajcosjx+ finda0:Observe that f(x)dx=a02 dx+ j=1(aj cosjxdx+bj sinjxdx)=a022 + j=1(0 + 0)This implies thata0=1 f(x)dx2To findan:Observe that f(x)cosnxdx=a02 cosnxdx+ j=1(aj cosnxcosjxdx+bj cosnxsinjxdx)=a020 +an + j=1bj0 This implies thatan=1 f(x)cosnxdxTo findbn:Observe that f(x)sinnxdx=a02 sinnxdx+ j=1(aj sinnxcosjxdx+bj sinnxsinjxdx)=a020 + j=1aj0 +bn.

5 This implies thatbn=1 f(x)sinnxdxThusf(x) =a02+ n=1ancosnx+bnsinnx,wherea0=1 f(x)dxan=1 f(x)cosnxdxbn=1 f(x)sinnxdx[This Expansion is valid at all those pointsx, wheref(x) is continuous.]Note:Note that the above mentioned results hold when we take any 2 length intervals [This is because c+2 csinmxdx= 0,..are true for anyc].Result: So whenever we take a functionfdefined from[c,c+ 2 ](anyinterval of length2 ) toR, satisfying the Dirichlet conditions, we have3f(x) =a02+ n=1ancosnx+bnsinnx,wherea0=1 c+2 cf(x)dxan=1 c+2 cf(x)cosnxdxbn=1 c+2 cf(x)sinnxdx3 Derivation of Fourier Series Expansion of afunction defined in an arbitrary period[a,b]:Now suppose thatf(x) is defined in an arbitrary interval [a,b] and satisfy theDirichlet conditions.

6 Let us takeb a2=l, half the length of the interval. Nowdefine the new variablez= this simple transformation, we can convert functions on any finite interval(say, [a,b]) to functions in the new variablez, whose domain is an interval of2 length. This is becausex=a z= laandx=b z= lb=2 b a(b a+a) = 2 + when the variablexinf(x) moves fromatob, the new variablezinthe new functionF(z) (which is the same functionfin the new variable) movesfromctoc+ 2 , wherec= la. Hence the Fourier Series Expansion is applicableforF(z).

7 Thusf(x) =F(z) =a02+ n=1ancosnz+bnsinnz,wherea0=1 c+2 cF(z)dzan=1 c+2 cF(z)cosnz dzbn=1 c+2 cF(z)sinnz dzand changing back to the original variablex(note thatdz= ldx), we have4f(x) =a02+ n=1ancosn lx+bnsinn lx,wherea0=1l baf(x)dxan=1l baf(x)cosn lxdxbn=1l baf(x)sinn lxdx,which is the general form of Fourier Series Expansion for functions on anyfinite interval. Also note that this is applicable to the first case of our discussion,where we need to takea= ,b= ,l= and then everything becomes thesame as in the previous IllustrationWe now take a simple problem to demonstrate the evaluation of Fourier the functionfdefined byf(x) = 10if 2 x 1,xif 1< x <1,10,if1 x shall find the Fourier Series Expansion of this function.

8 Here, note thatthe length of the interval is 4. So 2l= 4 andl= 2. We need to writef(x) =a02+ n=1ancosn 2x+bnsinn 2x,wherea0=12 2 2f(x)dxan=12 2 2f(x)cosn 2xdxbn=12 2 2f(x)sinn 2xdx,Nowa0=12( 1 2 10dx+ 1 1xdx+ 2110dx)=12( 10 + 0 + 10) = 0an=12( 1 2 10cosn 2xdx+ 1 1xcosn 2xdx+ 2110cosn 2xdx)5=12( 10 1 2cosn 2xdx+ 1 1xcosn 2xdx+ 10 21cosn 2xdx)=12(20n (sinn 2 sinn ) + 0 +20n (sinn sinn 2)) = 0bn=12( 1 2 10sinn 2xdx+ 1 1xsinn 2xdx+ 2110sinn 2xdx)=12{20n (cos(n 2) cos(n )) + 2[ 2n cos(n 2) +4n2 2sin(n 2) 0] 20n (cos(n ) cos(n 2)}=18n cos(n 2) 20n cos(n ) +4n2 2sin(n 2).)

9 So whenn= 1 b1=4 2, n= 2 b2=19 ,..Thus the Fourier Expansion off(x) isf(x) =02+ 0cos 2x+4 2sin 2x+ 0cos2 2x+19 sin2 2x+..=4 2sin 2x+19 sin2 2x+..,which is valid at all points in [ 2,2] except at 1 and 1, since the functionis continuous at all points except 1 and 1. Whenx= 1, the sum of the serieswill be equal to the value 10+ 12= and atx= 1, it is10+12= Some special cases:Suppose the function is an odd/even function in a symmetric interval [ c,c].That isf( x) =f(x) for allx R[ Even] orf( x) = f(x) for allx R[ Odd].

10 Then from the results, c cf(x)dx= 0 whenf(x) is odd c cf(x)dx= 2 c0f(x)dxwhenf(x) is even,we have some specialities in the Even functions:Supposef(x) is even in a domain [ c,c]. Then it can be observed thatbn= 0and so the Fourier Series becomesf(x) =a02+ n=1ancosn cxwherea0=2c c0f(x)dxan=2c c0f(x)cosn Odd functions:Similarly whenf(x) is odd in a domain [ c,c]. Thena0=an= 0 and theFourier Series becomesf(x) = n=1bnsinn cxwherebn=2c c0f(x)sinn cxdxNote: If you observe carefully, in the illustration problem, the function isactually odd and the domain is [ 2,2].