434
Chapter 8. Fourier and Laplace Transforms
8.1 Fourier Series This section explains three Fourier series: sines, cosines, and exponentials e 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 sin x. It has period 2 since sin.x C 2/ D sin x. It is an odd function since sin. x/ D sin x, and it vanishes at x D 0 and x D . Every function sin nx has those three properties, and Fourier looked at infinite combinations of the sines: Fourier sine series S.x/ D b1 sin x C b2 sin 2x C b3 sin 3x C D
1 X
bn sin nx (1)
nD1
If the numbers b1 ; b2 ; b3 ; : : : drop off quickly enough (we are foreshadowing the importance of their decay rate) then the sum S.x/ will inherit all three properties: Periodic S.x C 2/ D S.x/
Odd S. x/ D S.x/
S.0/ D S./ D 0
200 years ago, Fourier startled the mathematicians in France by suggesting that any odd periodic function S.x/ could be expressed as an infinite series of sines. This idea started an enormous development of Fourier series. Our first step is to find the number bk that multiplies sin kx. The function S.x/ is “transformed” to a sequence of b’s. P Suppose S.x/ D bn sin nx. Multiply both sides by sin kx. Integrate from 0 to : Z Z Z S.x/ sin kx dx D b1 sin x sin kx dx C C bk sin kx sin kx dx C (2) 0
0
0
On the right side, all integrals are zero except the highlighted one with n D k. This property of “orthogonality” will dominate the whole chapter. For sines, integral D 0 is a fact of calculus : Z
Sines are orthogonal
0
Zero comes quickly if we integrate Product of sines
R
sin nx sin kx dx D 0
cos mx dx D
sin nx sin kx D
sin mx m
1 cos.n 2
0
D0 k/x
if n ¤ k :
(3)
0. So we use this: 1 cos.n C k/x : 2
(4)
Integrating cos .n k/x and cos .n C k/x gives zero, proving orthogonality of the sines. The exception is when n D k. Then we are integrating .sin kx/2 D 21 12 cos 2kx: Z Z Z 1 1 sin kx sin kx dx D dx cos 2kx dx D : (5) 2 0 0 2 0 2 The highlighted term in equation (2) is .=2/bk. Multiply both sides by 2= to find bk .
435
8.1. Fourier Series
Sine coefficients
S. x/ D
bk D
S.x/
2
Z
S.x/ sin kx dx D
0
1
Z
S.x/ sin kx dx:
(6)
Notice that S.x/ sin kx is even (equal integrals from to 0 and from 0 to ). I will go immediately to the most important example of a Fourier sine series. S.x/ is an odd square wave with S W .x/ D 1 for 0 < x < . It is drawn in Figure 8.1 as an odd function (with period 2) that vanishes at x D 0 and x D . S W .x/ D 1
0
2
x
Figure 8.1: The odd square wave with S W .x C 2/ D S W .x/ D f1 or 0 or 1g. Find the Fourier sine coefficients bk of the odd square wave S W .x/.
Example 1 Solution
For k D 1; 2; : : : use formula (6) with S.x/ D 1 between 0 and :
bk D
2
Z
0
sin kx dx D
2
cos kx k
0
D
2
2 0 2 0 2 0 ; ; ; ; ; ;::: 1 2 3 4 5 6
(7)
The even-numbered coefficients b2k are all zero because cos 2k D cos 0 D 1. The oddnumbered coefficients bk D 4 =k decrease at the rate 1=k. We will see that same 1=k decay rate for all functions formed from smooth pieces and jumps. Put those coefficients 4=k and zero into the Fourier sine series for S W .x/: Square wave
SW.x/ D
4
sin x sin 3x sin 5x sin 7x C C C C 1 3 5 7
(8)
Figure 8.2 graphs this sum after one term, then two terms, and then five terms. You can see the all-important Gibbs phenomenon appearing as these “partial sums” include more terms. Away from the jumps, we safely approach S W .x/ D 1 or 1. At x D =2, the series gives a beautiful alternating formula for the number : 1 1 1 1 1 1 4 1 1 1D C C so that D 4 C C : (9) 1 3 5 7 1 3 5 7 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. This overshoot is the one greatest obstacle to calculation of all discontinuous functions (like shock waves). We try hard to avoid Gibbs but sometimes we can’t.
436
Chapter 8. Fourier and Laplace Transforms
4 sin x sin 3x Solid curve C 1 3 4 sin x Dashed 1
x
4 sin x sin 9x 5 terms: CC 1 9 Gibbs overshoot ! SW D 1 x
2
Figure 8.2: The sums b1 sin x C C bN sin N x overshoot the square wave near jumps.
Fourier Cosine Series The cosine series applies to even functions C.x/ D C. x/. They are symmetric across 0 : Cosine series
C.x/ D a0 C a1 cos x C a2 cos 2x C D a0 C
1 X
an cos nx:
(10)
nD1
Every cosine has period 2. Figure 8.3 shows two even functions, the repeating ramp RR.x/ and the up-down train UD.x/ of delta functions. That sawtooth ramp RR is the integral of the square wave. The delta functions in UD give the derivative of the square wave. (For sines, the integral and derivative are cosines.) RR and UD will be valuable examples, one smoother than S W , one less smooth. First we find formulas for the cosine coefficients a0 and ak . The constant term a0 is the average value of the function C.x/ : Z Z 1 1 a0 D average a0 D C.x/ dx D C.x/ dx: (11) 0 2 I just integrated every term in the cosine series (10) from 0 to . On the right side, the integral of a0 is a0 (divide both sides by ). All other integrals are zero : Z sin nx cos nx dx D D 0 0 D 0: (12) n 0 0 In words, the constant function 1 is orthogonal to cos nx over the interval Œ0; . The other cosine coefficients ak come from the orthogonality of cosines. As with sines, we multiply both sides of (10) by cos kx and integrate from 0 to : Z Z Z Z C.x/ cos kx dx D a0 cos kx dxC a1 cos x cos kx dxCC ak .cos kx/2 dxC 0
0
0
0
You know what is coming. On the right side, only the highlighted term can be nonzero. For k > 0, that bold nonzero term is ak =2. Multiply both sides by 2= to find ak : Cosine coefficients C. x/ D C.x/
2 ak D
Z
0
1 C.x/ cos kx dx D
Z
C.x/ cos kx dx : (13)
437
8.1. Fourier Series
RR.x/ D jxj
0
2ı.x/
Repeating Ramp RR.x/ Integral of Square Wave
2
x
2ı.x
2/
Up-down UD.x/
0
x
2
Derivative of Square Wave
2ı.x C /
2ı.x
/
Figure 8.3: The repeating ramp RR and the up-down UD (periodic spikes) are even. The slope of RR is 1 then 1 : odd square wave S W . The next derivative is UD : ˙ 2ı. Example 2
Find the cosine coefficients of the ramp RR.x/ and the up-down UD.x/.
The simplest way is to start with the sine series for the square wave : 4 sin x sin 3x sin 5x sin 7x S W .x/ D C C C C D slope of RR 1 3 5 7
Solution
Take the derivative of every term to produce cosines in the up-down delta function : 4 Œcos x C cos 3x C cos 5x C cos 7x C : (14) 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 correct and important. At x D 0, the cosines are all 1 and their sum is C1. At x D , the cosines are all 1. Then their sum is 1. (The R downward spike is 2ı.x /.) The true way to recognize ı.x/ is by the integral test ı.x/f .x/ dx D f .0/ and Example 3 will do this. For the repeating ramp, we integrate the square wave series for S W .x/ and add a0 . The average ramp height is a0 D =2, halfway from 0 to : cos x cos 3x cos 5x cos 7x Ramp series RR.x/ D C C C C : (15) 2 4 12 32 52 72 Up-down spikes
UD.x/ D
The constant of integration is a0 . Those coefficients ak drop off like 1=k 2 . They could R be computed directly from formula (13) using x cos kx dx, and integration by parts (or an appeal to Mathematica or Maple). It was much easier to integrate every sine separately in S W .x/, which makes clear the crucial point : Each “degree of smoothness” in the function brings a faster decay rate of its Fourier coefficients ak and bk . Every integration divides those numbers by k. No decay 1=k decay 1=k2 decay 1=k4 decay r k decay with r < 1
Delta functions (with spikes) Step functions (with jumps) Ramp functions (with corners) Spline functions (jumps in f 000 ) Analytic functions like 1=.2 cos x/
438
Chapter 8. Fourier and Laplace Transforms
The Fourier Series for a Delta Function Example 3
Find the (cosine) coefficients of the delta function ı.x/, made 2-periodic.
Solution The spike in ı.x/ occurs at x D 0. All the integrals are 1, because the cosine of 0 is 1. We divide by 2 for a0 and by for the other cosine coefficients ak .
Average a0 D
1 2
Z
ı.x/ dx D
1 2
Cosines ak D
1
Z
ı.x/ cos kx dx D
1
Then the series for the delta function has all cosines in equal amounts : No decay. Delta function
ı.x/ D
1 1 C Πcos x C cos 2x C cos 3x C : 2
(16)
This series cannot truly converge (its terms don’t approach zero). But we can graph the sum after cos 5x and after cos 10x. Figure 8.4 shows how these “partial sums” are doing their best to approach ı.x/. They oscillate faster while going higher. There is a neat formula for the sum ıN that stops at cos N x. Start by writing each term 2 cos x as e ix C e ix . We get a geometric progression from e iN x up to e iN x . ıN D
1 h 1 C e ix C e 2
ix
C C e iN x C e
iN x
i
D
1 sin.N C 12 /x : 2 sin 12 x
(17)
This is the function graphed in Figure 8.4. ı10 .x/
height 21=2
ı5 .x/
height 11=2
0
height 1=2 height 1=2
Figure 8.4: The sums ıN .x/ D .1 C 2 cos x C C 2 cos N x/=2 try to approach ı.x/.
439
8.1. Fourier Series
Complete Series: Sines and Cosines Over the half-period Œ0; , the sines are not orthogonal to all the cosines. In fact the integral of sin x times 1 is not zero. So for functions F .x/ that are not odd or even, we must move to the complete series (sines plus cosines) on the full interval. Since our functions are periodic, that “full interval” can be Œ ; or Œ0; 2. We have both a’s and b’s.
Complete Fourier series F .x/ D a0 C
1 X
nD1
an cos nx C
1 X
bn sin nx :
(18)
nD1
On every “2 interval” the sines and cosines are orthogonal. We find the Fourier coefficients ak and bk in the usual way: Multiply (18) by 1 and cos kx and sin kx. Then integrate both sides from to to get a0 and ak and bk . a0 D
1 2
Z
F .x/ dx ak D
1
Z
F .x/ cos kx dx bk D
1
Z
F .x/ sin kx dx
Orthogonality kills off infinitely many integrals and leaves only the one we want. Another approach is to split F .x/ D C.x/ C S.x/ into an even part and an odd part. Then we can use the earlier cosine and sine formulas. The two parts are C.x/ D Feven .x/ D
F .x/ C F . x/ 2
S.x/ D Fodd .x/ D
F .x/
F . x/ : 2
(19)
The even part gives the a’s and the odd part gives the b’s. Test on a square pulse from x D 0 to x D h—this one-sided thin box function is not odd or even. 1=h for 0 < x < h Example 4 Find the a’s and b’s if F.x/ D tall box D 0 for h < x < 2 The integrals for a0 and ak and bk stop at x D h where F .x/ drops to zero. The coefficients decay like 1=k because of the jump at x D 0 and the drop at x D h : Solution
Coefficients of square pulse
ak D
1 h
Z
h 0
cos kx dx D
a0 D
1 2
sin kh kh
Z
h 0
1= h dx D
bk D
1 h
Z
1 D average 2
h 0
sin kx dx D
1
cos kh : kh
Important As h approaches zero, the box gets thinner and taller. Its width is h and its height is 1= h and its area is 1. The box approaches a delta function ! And its Fourier coefficients approach the coefficients of the delta function as h ! 0 : a0 D
1 2
ak D
sin kh 1 approaches kh
bk D
1
cos kh approaches 0: (20) kh
440
Chapter 8. Fourier and Laplace Transforms
Energy in Function = Energy in Coefficients There is an extremely important equation (the energy identity) that comes from integrating .F .x//2 . When we square the Fourier series of F .x/, and integrate from to , all the “cross terms” drop out. The only nonzero integrals come from 12 and cos2 kx and sin2 kx. Those integrals give 2 and and , multiplied by a02 and ak2 and bk2 : Energy
R
.F.x//2 dx D 2a02 C .a12 C b21 C a22 C b22 C /:
(21)
The energy in F .x/ equals the energy in the coefficients. The left side is like the length squared of a vector, except the vector is a function. The right side comes from an infinitely long vector of a’s and b’s. The lengths are equal, which says that thepFourier transp form from function to vector is like an orthogonal matrix. Normalized by 2 and , sines and cosines are an orthonormal basis in function space.
Complex Exponentials ck e ikx This is a small step and we have to take it. In place of separate formulas for a0 and ak and bk , we will have one formula for all the complex coefficients ck . And the function F .x/ might be complex (as in quantum mechanics). The Discrete Fourier Transform will be much simpler when we use N complex exponentials for a vector. We practice with the complex infinite series for a 2-periodic function : Complex Fourier series F .x/ D c0 C c1 e ix C c
1e
ix
C D
1 X
cn e inx
(22)
nD 1
If every cn D c n , we can combine e i nx with e i nx into 2 cos nx. Then (22) is the cosine series for an even function. If every cn D c n , we use e i nx e i nx D 2i sin nx. Then (22) is the sine series for an odd function and the c’s are pure imaginary. To find ck , multiply (22) by e ikx (not e i kx ) and integrate from to : Z
F .x/e
i kx
dx D
Z
c0 e
i kx
dx C
Z
ix
c1 e e
i kx
dx C C
Z
ck e ikx e
ikx
dx C
The complex exponentials are orthogonal. Every integral on the right side is zero, except for the highlighted term (when n D k and e i kx e i kx D 1). The integral of 1 is 2. That surviving term gives the formula for ck : Fourier coefficients
Z
F.x/e
ikx
dx D 2ck
for k D 0; ˙1; : : : l
(23)
Notice that c0 D a0 is still the average of F .x/. The orthogonality of e i nx and e i kx is checked by integrating e i nx times e i kx . Remember to use that complex conjugate e i kx .
441
8.1. Fourier Series
Example 5 Example 6
For a delta function, all integrals are 1 and every ck is 1=2. Flat transform ! 1 for s x s C h Find ck for the 2-periodic shifted box F .x/ D 0 elsewhere in Π;
The integrals (23) have F D 1 from s to s C h : " #s C h Z sCh i kx 1 1 e ck D 1 e ikx dx D De 2 s 2 ik
Solution
iks
s
1
e i kh 2 i k
!
:
Notice above all the simple effect of the shift by s. It “modulates” each ck by e The energy is unchanged, the integral of jF j2 just shifts, and je i ks j D 1. Shift F.x/ to F.x
s/
! Multiply every ck
by e
iks
.
(24) i ks
.
(25)
Example 7 A centered box has shift s D h=2. It becomes balanced around x D 0. This even function equals 1 on the interval from h=2 to h=2: Centered by s D
h 2
ck D e i kh=2
1
e i kh 1 sin.kh=2/ D : 2 i k 2 k=2
Divide by h for a tall box. The ratio of sin.kh=2/ to kh=2 is called the “sinc” of kh=2. 1 1 X kh Fcentered 1= h for h=2 x h=2 D sinc e i kx D Tall box 0 elsewhere in Œ ; h 2 1 2 1 That division by h produces area D 1. Every coefficient approaches 2 as h ! 0. The Fourier series for the tall thin box again approaches the Fourier series for ı.x/.
The Rules for Derivatives and Integrals The derivative of e i kx is i ke i kx . This great fact puts the Fourier functions e i kx in first place for applications. They are eigenfunctions for d=dx (and the eigenvalues are D i k). Differential equations with constant coefficients are naturally solved by Fourier series. Multiply by ik The derivative of F.x/ D
X
ck e ikx is dF=dx D
X
ikck e ikx
The second derivative has coefficients .i k/2 ck D k2 ck . High frequencies are growing stronger. And in the opposite direction (when we integrate), we divide by i k and high frequencies get weaker. The solution becomes smoother. Please look at this example : Response 1=.k2 C 1/ to frequency k
d 2y e i kx i kx C y D e is solved by y.x/ D dx 2 k2 C 1
This was a typical problem in Chapter 2. The transfer function is 1=.k 2 C 1/. There we learned : The forcing function e i kx is exponential so the solution is exponential.
442
Chapter 8. Fourier and Laplace Transforms
All we are doing now is superposition. Allow all the exponentials at once ! X d 2y C y D ck e i kx dx 2
is solved by
y.x/ D
X ck e i kx : k2 C 1
(26)
1. Derivative rule dF =dx has Fourier coefficients ikck (energy moves to high k). 2. Shift rule F.x
s/ has Fourier coefficients e
iks
ck (no change in energy).
Application: Laplace’s Equation in a Circle Our first application is to Laplace’s equation uxx C uyy D 0 (Section 7.4). The idea is to construct u.x; y/ as an infinite series, choosing its coefficients to match u0 .x; y/ along the boundary. The shape of the boundary is crucial, and we take a circle of radius 1. Begin with the solutions 1, r cos , r sin , r 2 cos 2, r 2 sin 2, ... to Laplace’s equation. Combinations of these special solutions give all solutions in the circle: u.r; / D a0 C a1 r cos C b1 r sin C a2 r 2 cos 2 C b2 r 2 sin 2 C
(27)
It remains to choose the constants ak and bk to make u D u0 on the boundary. For a circle, and C 2 give the same point. This means that u0 ./ is periodic : Set r D 1
u0 ./ D a0 C a1 cos C b1 sin C a2 cos 2 C b2 sin 2 C
(28)
This is exactly the Fourier series for u0 . The constants ak and bk must be the Fourier coefficients of u0 ./. Thus Laplace’s boundary value problem is completely solved, if an infinite series (27) is acceptable as the solution. Example 8 Point source u0 D ı./. The boundary is held at u0 D 0, except for the source at x D 1, y D 0 (where D 0). Find the temperature u.r; / inside the circle. Delta function
u0 ./ D
1 1 1 X i n 1 C .cos C cos 2 C cos 3 C / D e 2 2 1
Inside the circle, each cos n is multiplied by r n to solve Laplace’s equation : Inside the circle
u.r; / D
1 1 C .r cos C r 2 cos 2 C r 3 cos 3 C / 2
(29)
Poisson managed to sum this infinite series ! It involves a series of powers .re i /n . His sum gives the response at every .r; / to the point source at r D 1, D 0: Temperature inside circle
u.r; / D
1 1 2 1 C r 2
r2 2r cos
(30)
At the center r D 0, this produces the average of u0 D ı./ which is a0 D 1=2. On the boundary r D 1, this gives u D 0 except u D 1 at the point where cos 0 D 1.
443
8.1. Fourier Series
u0 ./ D 1 on the top half of the circle and u0 D 1 on the bottom half.
Example 9
The boundary values u0 are a square wave S W . We know its sine series : 4 sin sin 3 sin 5 C C C (31) Square wave for u0 ./ S W ./ D 1 3 5
Solution
Inside the circle, multiplying by r, r 3 , r 5 , : : : gives fast decay of high frequencies : 4 r sin r 3 sin 3 r 5 sin 5 Rapid decay inside u.r; / D C C C 1 3 5
(32)
Laplace’s equation has smooth solutions inside, even when u0 ./ is not smooth.
Problem Set 8.1 1
(a) To prove that cos nx is orthogonal to cos kx when k ¤ n, use the formula 1 .n C k/x C 12 cos .n k/x. Integrate from x D 0 .cos nx/ .cos kx/ D R 2 cos 2 to x D . What is cos kx dx ? (b) From 0 to , cos x is not orthogonal to sin x. The period has to be 2 : Z Z Z2 Find .sin x/ .cos x/ dx and .sin x/ .cos x/ dx and .sin x/ .cos x/ dx:
2
Suppose F .x/ D x for 0 x . Draw graphs for 2 x 2 to show three extensions of F : a 2-periodic even function and a 2-periodic odd function and a -periodic function.
3
Find the Fourier series on x for
0
4
5
0
(a) f1 .x/ D sin3 x, an odd function (sine series, only two terms) (b) f2 .x/ D j sin xj, an even function (cosine series) (c) f3 .x/ D x for x (sine series with jump at x D ) P Find the complex Fourier series e x D ck e i kx on the interval x . 1 The even part of a function is 2 .f .x/ C f . x//, so that feven .x/ D feven . x/. Find the cosine series for feven and the sine series for fodd . Notice the jump at x D . From the energy formula (21), the square wave sine coefficients satisfy Z Z 2 2 2 .b1 C b2 C / D jS W .x/j dx D 1 dx D 2:
2
Substitute the numbers bk from equation (8) to find that D 8.1 C 6
If a square pulse is centered at x D 0 to give f .x/ D 1 for jxj < ; f .x/ D 0 for 2 draw its graph and find its Fourier coefficients ak and bk .
1 9
C
< jxj < ; 2
1 25
C /.
444 7
Chapter 8. Fourier and Laplace Transforms
Plot the first three partial sums and the function x. x/ : 8 sin x sin 3x sin 5x C C C ; 0 < x < : x. x/ D 1 27 125 Why is 1=k 3 the decay rate for this function? What is its second derivative?
8
Sketch the 2-periodic half wave with f .x/ D sin x for 0 < x < and f .x/ D 0 for < x < 0. Find its Fourier series.
9
Suppose G.x/ has period 2L instead of 2. Then G.x C 2L/ D G.x/. Integrals go from L to L or from 0 to 2L. The Fourier formulas change by a factor =L : ZL 1 P 1 i kx=L The coefficients in G.x/ D C ke are C k D G.x/e i kx=L dx: 2L 1 L
Derive this formula for Ck : Multiply the first equation for G.x/ by integrate both sides. Why is the integral on the right side equal to 2LCk ?
10
For Geven , use Problem 9 to find the cosine coefficient Ak from .Ck C C Geven.x/ D Geven is
1 P 0
kx Ak cos L
1 .G.x/ 2
1 has Ak D L
ZL
Geven.x/ cos
and
k /=2 :
kx dx: L
0
C G. x//. Exception for A0 D C0 : Divide by 2L instead of L.
1 .ck C c k / on the usual interval from 0 to . 2 Find a similar formula for bk from Pck and c k . In the reverse direction, find the complex coefficient ck in F .x/ D ck e i kx from the real coefficients ak and bk .
11
Problem 10 tells us that ak D
12
Find the solution to Laplace’s equation with u0 D on the boundary. Why is this the imaginary part of 2.z z 2 =2 C z 3 =3 / D 2 log.1 C z/? Confirm that on the unit circle z D e i , the imaginary part of 2 log.1 C z/ agrees with .
13
If the boundary condition for Laplace’s equation is u0 D 1 for 0 < < and u0 D 0 for < < 0, find the Fourier series solution u.r; / inside the unit circle. What is u at the origin r D 0 ?
14
With boundary values u0 ./ D 1 C 12 e i C 14 e 2i C , what is the Fourier series solution to Laplace’s equation in the circle? Sum this geometric series.
15
(a) Verify that the fraction in Poisson’s formula (30) satisfies Laplace’s equation.
16
(b) Find the response u.r; / to an impulse at x D 0; y D 1 (where D 2 ). P With complex exponentials in F .x/ D ck e i kx , the energy identity (21) changes to R P P P jF .x/j2 dx D 2 jck j2 . Derive this by integrating . ck e i kx /. c k e i kx /.
8.1. Fourier Series
17
445
A centered square wave has F .x/ D 1 for jxj =2. R (a) Find its energy jF .x/j2 dx by direct integration
(b) Compute its Fourier coefficients ck as specific numbers (c) Find the sum in the energy identity (Problem 16).
18
F .x/ D 1 C .cos x/=2 C C .cos nx/=2n C is analytic : infinitely smooth. (a) If you take 10 derivatives, what is the Fourier series of d 10 F=dx 10 ? (b) Does that series still converge quickly ? Compare n10 with 2n for n D 210 .
19
If f .x/ D 1 for jxj =2 and f .x/ D 0 for =2 < jxj < , find its cosine coefficients. Can you graph and compute the Gibbs overshoot at the jumps ?
20
Find all the coefficients ak and bk for F; I; and D on the interval x : Z x d F .x/ D ı x I.x/ D ı x dx D.x/ D ı x : 2 2 dx 2 0
21
For the one-sided tall box function in Example 4, with F D 1= h for 0 x h, what is its odd part 12 .F .x/ F . x// ? I am surprised that the Fourier coefficients of this odd part disappear as h approaches zero and F .x/ approaches ı.x/. P Find the series F .x/ D ck e i kx for F .x/ D e x on x . That function x e looks smooth, but there must be a hidden jump to get coefficients ck proportional to 1=k. Where is the jump ?
22
23
(a) (Old particular solution) Solve Ay 00 C By 0 C Cy D e i kx . P (b) (New particular solution) Solve Ay 00 C By 0 C Cy D ck e i kx .
