[PDF] CHAPTER 4 FOURIER SERIES AND INTEGRALS

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CHAPTER 4

FOURIER SERIES AND INTEGRALS

4.1 FOURIER SERIES FOR PERIODIC FUNCTIONS

This section explains three Fourier series:sines, cosines, and exponentialse ikx Square waves (1 or 0 or-1) are great examples, with delta functions in the derivative. We look at a spike, a step function, and a ramp"and smoother functions too. Start with sinx.Ithasperiod2since sin(x+2)=sinx. It is an odd function since sin(-x)=-sinx, and it vanishes atx=0andx=. Every function sinnx

has those three properties, and Fourier looked atin“nite combinations of the sines:Fourier sine seriesS(x)=b

1 sinx+b 2 sin2x+b 3 sin3x+···= n=1 b n sinnx(1)

If the numbersb1

,b 2 ,...drop offi 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()=0

200 years ago, Fourier startled the mathematicians in France by suggesting thatany

functionS(x) with those properties could be expressed as an infinite series of sines. This idea started an enormous development of Fourier series. Our first step is to compute fromS(x)thenumberbk that multiplies sinkx.

SupposeS

(x)=bn sinnx.Multiply both sides bysinkx.Integrate from0to: 0

S(x)sinkxdx=

0 b 1 sinxsinkx dx+···+ 0 b k sinkxsinkxdx+···(2) On the right side, all integrals are zero except the highlighted one withn=k. This property of "orthogonality" will dominate the whole chapter. The sines make 90
angles in function space, when their inner products are integrals from 0 to:Orthogonality 0 sinnxsinkxdx=0 ifn?=k.(3) 317

318Chapter 4 Fourier Series and Integrals

Zero comes quickly if we integrate

cosmxdx= sinmx m 0 =0Š0. So we use this:

Product of sinessinnxsinkx=1

2cos(nŠk)xŠ12cos(n+k)x.(4)

Integrating cosmxwithm=nŠkandm=n+kproves orthogonality of the sines. The exception is whenn=k. Then we are integrating (sinkx) cos2kx: sinkxsinkxdx= 1

2dxŠ

1

2cos2kxdx=π2.(5)

The highlighted term in equation (2) isb

k

Γ/2. Multiply both sides of (2) by 2/π:

Sine coeffcients

S(Šx)=ŠS(x)

b k =2

S(x)sinkxdx=1

S(x)sinkxdx.(6)

Notice thatS(x)sinkxiseven(equal integrals fromŠπto 0 and from 0 toπ). I will go immediately to the most important example of a Fourier sine series.S(x) is anodd square wavewithSW(x)=1for0Šπ0π2π Figure 4.1: The odd square wave withSW(x+2π)=SW(x)={1or0orŠ1}.

Example 1Find the Fourier sine coefficientsb

k of the square waveSW(x). SolutionFork=1,2,...use the first formula(6)withS(x)=1between0andπ: b k =2 sinkxdx=2

Šcoskxkffl

0 =2

21,02,23,04,25,06,..."

(7)

The even-numbered coecientsb

?k are all zero because cos2kπ= cos0 = 1. The odd-numbered coecientsb k =4/πkdecrease at the rate 1/k. We will see that same

1/kdecay rate for all functions formed fromsmooth pieces and jumps.

Put those coecients 4/πkand zero into the Fourier sine series forSW(x):

Square waveSW(x)=4

(8) Figure 4.2 graphs this sum after one term, then two terms, and then five terms. You can see the all-importantGibbs phenomenonappearing as these "partial sums"

4.1 Fourier Series for Periodic Functions319

include more terms. Away from the jumps, we safely approachSW(x)=1orŠ1. Atx=Γ/2, the series gives a beautiful alternating formula for the numberΓ: 1= 4

11Š13+15Š17+···?

so that≈=4?11Š13+15Š17+···? .(9) The Gibbs phenomenon is the overshoot that moves closer and closer to the jumps. Its height approaches 1.18...and it does not decrease with more terms of the series! 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"t. xxŠΓΓDashed 4

Γsinx1Solid curve

4 sinx1+sin3x3?

5terms:4Γ?

sinx1+···+sin9x9? overshootŠ?SW=1 2

Figure 4.2:Gibbs phenomenon: Partial sums?

N 1 b n sinnxovershoot near jumps.

Fourier Coefficients are Best

Let me look again at the first termb

1 sinx=(4/Γ)sinx.Thisistheclosest possible approximationto the square waveSW, by any multiple of sinx(closest in the least squares sense). To see this optimal property of the Fourier coecients, minimize the error over allb 1

The error is

0 (SWŠb 1 sinx) 2 dxTheb 1 derivative isŠ2? 0 (SWŠb 1 sinx)sinxdx.

The integral of sin

2 xisΓ/2. So the derivative is zero whenb 1 =(2/Γ)? 0

S(x)sinxdx.

This is exactly equation (6) for the Fourier coecient. Eachb k sinkxis as close as possible toSW(x). We can find the coecientsb k one at a time,because the sines are orthogonal. The square wave hasb 2 = 0 because all other multiples of sin2xincrease the error. Term by term, we are "projecting the function onto each axis sinkx."

Fourier Cosine Series

The cosine series applies toeven functionswithC(Šx)=C(x):

Cosine seriesC(x)=a

0 +a 1 cosx+a 2 cos2x+···=a 0 n=1 a n cosnx.(10)

320Chapter 4 Fourier Series and Integrals

Every cosine has period 2Γ. Figure 4.3 shows two even functions, therepeating rampRR(x)andtheup-down trainUD(x) of delta functions. That sawtooth rampRRis the integral of the square wave. The delta functions inUDgive the derivative of the square wave. (For sines, the integral and derivative are cosines.) RRandUDwill be valuable examples, one smoother thanSW, one less smooth.

First we find formulas for the cosine coecientsa

anda k . The constant terma is theaverage valueof the functionC(x): a 0 =Averagea =1 ffi

C(x)dx=1

2Γ ffi

Šffi

C(x)dx.(11)

I just integrated every term in the cosine series (10) from 0 toΓ.Ontherightside, the integral ofa isa Γ(divide both sides byΓ). All other integrals are zero: ffi cosnxdx=sinnx n ffi =0Š0=0.(12) In words, the constant function 1 is orthogonal to cosnxover the interval [0,Γ].

The other cosine coecientsa

k come from theorthogonality of cosines.Aswith sines, we multiply both sides of (10) by coskxand integrate from 0 toΓ: ffi

C(x)coskxdx=

ffi a coskxdx+ ffi a cosxcoskxdx+··+ 0 a k (coskx) 2 dx+·· You know what is coming. On the right side, only the highlighted term can be nonzero. Problem 4.1.1 proves this by an identity for cosnxcoskx"now (4) has a plus sign. The bold nonzero term isa k

σ/2and we multiply both sides by 2/Γ:

Cosine coe?cients

C(Šx)=C(x)

a k =2 ffi

C(x)coskxdx=1

ffi

Šffi

C(x)coskxdx.(13)

Again the integral over a full period fromŠΓtoΓ(also 0 to 2Γ) is just doubled. xŠΓ0Γ2ΓRR(x)=|x|

Repeating RampRR(x)

Integral of Square Wave

xŠΓ0Γ2Γ

Š2Σ(x+Γ)2Σ(x)

Š2Σ(xŠΓ)2Σ(xŠ2Γ)

Up-downUD(x)

Figure 4.3: The repeating rampRRand the up-downUD(periodic spikes) are even. The derivative ofRRis the odd square waveSW.The derivative ofSWisUD.

4.1 Fourier Series for Periodic Functions321

Example 2Find 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+sin3x3+sin5x5+sin7x7+···? Take the derivative of every term to produce cosines in the up-down delta function:

Up-down seriesUD(x)=4

Those coefficients don"t decay at all. The terms in the series don"t approach zero, so officially the series cannot converge. Nevertheless it is somehow correct and important. Unofficially this sum of cosines has all1"s atx=0and all-1"s atx=Γ.Then+∞ and-∞are consistent with2Σ(x)and-2Σ(x-Γ). The true way to recognizeΣ(x)is by the test?Σ(x)f(x)dx=f(0)and Example 3 will do this. For the repeating ramp, we integrate the square wave series forSW(x)and add the average ramp heighta 0 =Γ/2, halfway from0toΓ:

Ramp seriesRR(x)=Γ

2-Γ4?

cosx1 2 +cos3x 3 2 +cos5x 5 2 +cos7x 7 2 .(15)

The constant of integration isa

0 .Those coe⎷cientsa k drop o? like1/k 2 .Theycouldbe computed directly from formula(13)using?xcoskxdx, but this requires an integration by parts (or a table of integrals or an appeal toMathematicaorMaple). It was much easier 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 its

Fourier coefficientsa

k andb k

No decay Deltafunctions (with spikes)

1/kdecay Stepfunctions (with jumps)

1/k 2 decay Rampfunctions (with corners) 1/k 4 decay Splinefunctions (jumps inf ffffff r k decay withr<1Analyticfunctions like 1/(2-cosx) Each integration divides thekth coeωcient byk. So the decay rate has an extra

1/k. The Riemann-Lebesgue lemmaŽ says thata

k andb k approach zero for any continuous function (in fact whenever?|f(x)|dxis “nite). Analytic functions achieve a new level of smoothness"they can be di∂erentiated forever. Their Fourier series and Taylor series in Chapter 5 convergeexponentially fast. The poles of 1/(2-cosx) will be complex solutions of cosx= 2. Its Fourier series converges quickly becauser k decays faster than any power 1/k p . Analytic functions are ideal for computations"the Gibbs phenomenon will never appear. Now we go back toΣ(x) for what could be the most important example of all.

322Chapter 4 Fourier Series and Integrals

Example 3Find the (cosine) coefficients of thedelta functionδ?x?,made?π-periodic. SolutionThe spike occurs at the start of the interval[?,π]so safer to integrate from -πtoπ.Wefinda 0 =?/?πand the othera k =?/π(cosines becauseδ?x?is even):

Averagea

0

δ?x?dx=1

2Cosinesa

k

δ?x?coskxdx=1

Then the series for the delta function has all cosines in equal amounts:

Delta functionδ?x?=?

Again this series cannot truly converge (its terms don"t approach zero). But we can graph the sum aftercos?xand aftercos??x. Figure 4.4 shows how these partial sumsŽ are doing their best to approachδ?x?. They oscillate faster and faster away fromx=?. Actually there is a neat formula for the partial sumδ N ?x?that stops atcosNx.Start by writing each term?cosθase iθ ?e

Šiθ

N ?π[? ? ?cosx?···??cosNx]=??π◦??e ix ?e

Šix

?···?e iNx ?e

ŠiNx

This is a geometric progression that starts frome

ŠiNx

and ends ate iNx .Wehavepowers of the same factore ix . The sum of a geometric series is known:

Partial sum

up tocosNxδ N ?x?=? ?πe i(N+ 1 2 )x -e

Ši(N+

1 2 )x e ix/2 -e

Šix/2

?πsin?N? 1 2 ?x sin 1 2 x.??7? This is the function graphed in Figure 4.4. We claim that for anyNthe area underneath N ?x?is?. (Each cosine integrated from-πtoπgives zero. The integral of?/?πis ?.) The central lobeŽ in the graph ends whensin?N? 1 2 ?xcomes down to zero, and that happens when?N? 1 2 ?x=±π. I think the area under that lobe (marked by bullets) approaches the same number?.?8...that appears in the Gibbs phenomenon.

In what way doesδ

N ?x?approachδ?x??Thetermscosnxin the series jump around at each pointx?=?, not approaching zero. Atx=πwe see 1 2π [?-???-??···]and the sum is?/?πor-?/?π. The bumps in the partial sums don"t get smaller than?/?π. The right test for the delta functionδ?x?is to multiply by a smoothf?x?=a k coskx and integrate, becausewe only knowδ?x?from its integralsδ?x?f?x?dx=f???:

Weak convergence

of N (x)to(x) N (x)f(x)dx=a 0 +···+a N f(0).??8?

In this integrated sense (weak sense)thesumsδ

N ?x?do approach the delta function!

The convergence ofa

0 ?···?a N is the statement that atx=?the Fourier series of a smoothf?x?=a k coskxconverges to the numberf???.

4.1 Fourier Series for Periodic Functions323

Šffff0?

5 (x)Δ 10 (x) height 11?2?height 21/2ff heightŠ1?2?height 1/2ff

Figure 4.4: The sumsΔ

N (x)=(1+2cosx+···+2cosNx)/2fftry to approachΔ(x).

Complete Series: Sines and Cosines

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