CHAPTER 4 FOURIER SERIES AND INTEGRALS

1 CHAPTER 4 FOURIER SERIES AND FOURIER SERIES FOR PERIODIC FUNCTIONSThis section explains three FOURIER SERIES :sines, cosines, and waves (1 or 0 or 1) are great examples, with delta functions in the look at a spike, a step function, and a ramp and smoother functions with since sin(x+2 )=sinx. It is an odd functionsince sin( x)= sinx, and it vanishes atx=0andx= . Every function sinnxhas those three properties, and FOURIER looked atinfinite combinations of the sines: FOURIER sine seriesS(x)=b1sinx+b2sin 2x+b3sin 3x+ = n=1bnsinnx(1)If the numbersb1,b2,..drop off quickly enough (we are foreshadowing the im-portance of the decay rate) then the sumS(x) will inherit all three properties:PeriodicS(x+2 )=S(x)OddS( x)= S(x)S(0) =S( )=0200 years ago, FOURIER startled the mathematicians in France by suggesting thatanyfunctionS(x) with those properties could be expressed as an infinite SERIES of idea started an enormous development of FOURIER SERIES .

2 Our first step is tocompute fromS(x)thenumberbkthat multiplies (x)= both sides from0to : 0S(x)sinkx dx= 0b1sinxsinkx dx+ + 0bksinkxsinkx dx+ (2)On the right side, all INTEGRALS are zero except the highlighted one withn= property of orthogonality will dominate the whole CHAPTER . The sines make90 angles in function space, when their inner products are INTEGRALS from 0 to :Orthogonality 0sinnxsinkx dx=0 ifn =k.(3)317318 CHAPTER 4 FOURIER SERIES and IntegralsZero comes quickly if we integrate cosmx dx= sinmxm 0=0 0. So we use this:Product of sinessinnxsinkx=12cos(n k)x 12cos(n+k)x.(4)Integrating cosmxwithm=n kandm=n+kproves orthogonality of the exception is whenn=k. Then we are integrating (sinkx)2=12 12cos 2kx: 0sinkxsinkx dx= 012dx 012cos 2kx dx= 2.

3 (5)The highlighted term in equation (2) isbk /2. Multiply both sides of (2) by 2/ :Sine coefficientsS( x)= S(x)bk=2 0S(x)sinkx dx=1 S(x)sinkx dx.(6)Notice thatS(x)sinkxiseven(equal INTEGRALS from to 0 and from 0 to ).I will go immediately to the most important example of a FOURIER sine (x)is anodd square wavewithSW(x)=1for0<x< . It is drawn in Figure asanoddfunction(withperiod2 )thatvanishesatx=0andx= . xSW(x)=1 0 2 Figure : The odd square wave withSW(x+2 )=SW(x)={1or0or 1}.Example 1 Find the FOURIER sine coefficientsbkof the square waveSW(x).SolutionFork=1,2,..use the first formula(6)withS(x)=1between0and :bk=2 0sinkx dx=2 coskxk 0=2 21,02,23,04,25,06,.."(7)The even-numbered coefficientsb2kare all zero because cos 2k = cos 0 = 1.

4 Theodd-numbered coefficientsbk=4/ kdecrease at the rate 1/k. We will see that same1/kdecay rate for all functions formed fromsmooth pieces and those coefficients 4/ kand zero into the FOURIER sine SERIES forSW(x):Square waveSW(x)=4 sinx1+sin 3x3+sin 5x5+sin 7x7+ (8)Figure graphs this sum after one term, then two terms, and then five terms. Youcan see the all-importantGibbs phenomenonappearing as these partial sums FOURIER SERIES for Periodic Functions319include more terms. Away from the jumps, we safely approachSW(x)=1or /2, the SERIES gives a beautiful alternating formula for the number :1=4 11 13+15 17+ so that =4 11 13+15 17+ .(9)The Gibbs phenomenon is the overshoot that moves closer and closer to the height approaches it does not decrease with more terms of the SERIES !

5 Overshoot is the one greatest obstacle to calculation of all discontinuous functions(like shock waves in fluid flow). We try hard to avoid Gibbs but sometimes we can Dashed4 sinx1 Solid curve4 sinx1+sin 3x3 5terms:4 sinx1+ +sin 9x9 overshoot SW=1 2 Figure :Gibbs phenomenon: Partial sums N1bnsinnxovershoot near Coefficients are BestLet me look again at the first termb1sinx=(4/ )sinx. Thisistheclosest possibleapproximationto the square waveSW, by any multiple of sinx(closest in the leastsquares sense). To see this optimal property of the FOURIER coefficients, minimize theerror over allb1:The error is 0(SW b1sinx)2dxTheb1derivative is 2 0(SW b1sinx) integral of sin2xis /2. So the derivative is zero whenb1=(2/ ) 0S(x) is exactly equation (6) for the FOURIER as close as possible toSW(x).

6 We can find the coefficientsbkone at a time,because the sines are orthogonal. The square wave hasb2= 0 becauseall other multiples of sin 2xincrease the error. Term by term, we are projecting thefunction onto each axis sinkx. FOURIER Cosine SeriesThe cosine SERIES applies toeven functionswithC( x)=C(x):Cosine seriesC(x)=a0+a1cosx+a2cos 2x+ =a0+ n=1ancosnx.(10)320 CHAPTER 4 FOURIER SERIES and IntegralsEvery cosine has period 2 . Figure shows two even functions, therepeatingrampRR(x)andtheup-down trainUD(x) of delta functions. That sawtoothrampRRis the integral of the square wave. The delta functions inUDgive thederivative of the square wave. (For sines, the integral and derivative are cosines.)RRandUDwill be valuable examples, one smoother thanSW, one less we find formulas for the cosine coefficientsa0andak.

7 The constant terma0is theaverage valueof the functionC(x):a0=Averagea0=1 0C(x)dx=12 C(x)dx.(11)I just integrated every term in the cosine SERIES (10) from 0 to .Ontherightside,the integral ofa0isa0 (divide both sides by ). All other INTEGRALS are zero: 0cosnx dx= sinnxn 0=0 0=0.(12)In words, the constant function 1 is orthogonal to cosnxover the interval [0, ].The other cosine coefficientsakcome from theorthogonality of , we multiply both sides of (10) by coskxand integrate from 0 to : 0C(x)coskx dx= 0a0coskx dx+ 0a1cosxcoskx dx+ + 0ak(coskx)2dx+ You know what is coming. On the right side, only the highlighted term can benonzero. Problem proves this by an identity for cosnxcoskx now (4) has aplus sign.

8 The bold nonzero term isak /2and we multiply both sides by 2/ :Cosine coefficientsC( x)=C(x)ak=2 0C(x)coskx dx=1 C(x)coskx dx .(13)Again the integral over a full period from to (also 0 to 2 ) is just doubled. x 0 2 RR(x)=|x|Repeating RampRR(x)Integral of Square Wave x 0 2 2 (x+ )2 (x) 2 (x )2 (x 2 )Up-downUD(x)Figure : The repeating rampRRand the up-downUD(periodic spikes) are derivative ofRRis the odd square derivative FOURIER SERIES for Periodic Functions321 Example 2 Find the cosine coefficients of the rampRR(x)and the up-downUD(x).SolutionThe simplest way is to start with the sine SERIES for the square wave:SW(x)=4 sinx1+sin 3x3+sin 5x5+sin 7x7+ .Take the derivative of every term to produce cosines in the up-down delta function:Up-down seriesUD(x)=4 [cosx+cos3x+cos5x+cos7x+ ].

9 (14)Those coefficients don t decay at all. The terms in the SERIES don t approach zero, soofficially the SERIES cannot converge. Nevertheless it is somehow correct and this sum of cosines has all1 s atx=0and all 1 s atx= .Then+ and are consistent with2 (x)and 2 (x ). The true way to recognize (x)isby the test (x)f(x)dx=f(0)and Example 3 will do the repeating ramp, we integrate the square wave SERIES forSW(x)and add theaverage ramp heighta0= /2, halfway from0to :Ramp seriesRR(x)= 2 4 cosx12+cos 3x32+cos 5x52+cos 7x72+ .(15)The constant of integration coefficientsakdrop off like1 directly from formula(13)using xcoskx dx, but this requires an integrationby parts (or a table of INTEGRALS or an appeal toMathematicaorMaple).

10 It was mucheasier to integrate every sine separately inSW(x), which makes clear the crucial point:Each degree of smoothness in the function is reflected in a faster decay rate of itsFourier decayDeltafunctions (with spikes)1/kdecayStepfunctions (with jumps)1/k2decayRampfunctions (with corners)1/k4decaySplinefunctions (jumps inf )rkdecay withr<1 Analyticfunctions like 1/(2 cosx)Each integration divides thekth coefficient byk. So the decay rate has an extra1/k. The Riemann-Lebesgue lemma says thatakandbkapproach zero for anycontinuous function (in fact whenever |f(x)|dxis finite). Analytic functions achievea new level of smoothness they can be differentiated forever. Their FOURIER seriesand Taylor SERIES in CHAPTER 5 convergeexponentially poles of 1/(2 cosx) will be complex solutions of cosx= 2.