Kenneth Kuttler

650 CHAPTER 33. BOUNDARY VALUE PROBLEMS, FOURIER SERIES

∣∣∣∣∫ L

−L

(f (x+)+ f (x−)

2− f (x− y)

)Dn (y)dy

∣∣∣∣=

∣∣∣∣∫ L

0( f (x+)+ f (x−))Dn (y)dy−

∫ L

0( f (x− y)+ f (x+ y))Dn (y)dy

∣∣∣∣≤

∣∣∣∣∣∫ L

f (x+)− f (x+ y)2Lsin

(π

2L y) sin

((n+

)π

Ly)

∣∣∣∣∣+∣∣∣∣∣∫ L

f (x−)− f (x− y)2Lsin

(π

2L y) sin

((n+

)π

Ly)

∣∣∣∣∣Both of these converge to 0 thanks to Lemma 33.4.3. To use this lemma, it is only necessaryto verify that the functions

y→ f (x−)− f (x− y)2Lsin

(π

2L y) , y→ f (x+)− f (x+ y)

2Lsin(

2L y)

are each Riemann integrable on [−L,L].I will show this now. Each is continuous except for finitely many points of discontinuity.

The only remaining issue is whether the functions are bounded as y→ 0. However, thereexists a constant K such that∣∣∣∣∣ f (x+)− f (x+ y)

2Lsin(

2L y) ∣∣∣∣∣≤ K |y|∣∣2Lsin

(π

2L y)∣∣

and this expression converges to K/π , so the function is Riemann integrable. The otherfunction is similar. ■

Example 33.5.4 Let f (x) = |x| for x ∈ [−1,1) and let f be periodic of period 2. Find theFourier series of f .

Here you need L = 1. Then

a0 =12

∫ 1

−1|x|dx =

ak =∫ 1

−1|x|cos(kπx)dx =

2π2k2

((−1)k−1

)Note that ak = 0 if k is even and it equals −4/

(π2k2

)when k is odd.

Since the function is even, the bk = 0. Therefore, the Fourier series equals

12−

∞

∑k=1

π2 (2k−1)2 cos(2k−1)πx

Now here is the graph of the function between −1 and 1 along with the sum up to 2 in theFourier series. You will notice that after only three terms the Fourier series appears to bevery close to the function on the interval [−1,1]. This also shows how the Fourier seriesapproximates the periodic extension of this function off this interval.

650 CHAPTER 33. BOUNDARY VALUE PROBLEMS, FOURIER SERIES_ [ ar Dx(v)dh]—Lvensee y)dy— [ov (x—y)+f(x+y))D no}dy— rates sin( (+5) ry) “yt1\ aP f(x-)—f(x-y) sin n + _ —y dy2Lsin (FF OT y) 2) LBoth of these converge to 0 thanks to Lemma 33.4.3. To use this lemma, it is only necessaryto verify that the functionsf(x) —f (x-y) f+) —f (x+y)2Lsin(fy) ” yo 2Lsin (37 y)yoare each Riemann integrable on [—L, L].I will show this now. Each is continuous except for finitely many points of discontinuity.The only remaining issue is whether the functions are bounded as y + 0. However, thereexists a constant K such thatae ___ Kb2Lsin (sry) ~ |2Lsin (+y)|and this expression converges to K/7, so the function is Riemann integrable. The otherfunction is similar.Example 33.5.4 Let f (x) = |x| for x € [—1,1) and let f be periodic of period 2. Find theFourier series of f.Here you need L = 1. Then1/7) 1=_ dx=—ao 5 | bles 5a= |x| cos (kx) dx = = ((-1)'-1)Note that a, = 0 if k is even and it equals —4/ (ak?) when k is odd.Since the function is even, the by = 0. Therefore, the Fourier series equals— 4— y so 608 (2k — 1) xk=1 1? (2k — 1)Now here is the graph of the function between —1 and | along with the sum up to 2 in theFourier series. You will notice that after only three terms the Fourier series appears to bevery close to the function on the interval [—1,1]. This also shows how the Fourier seriesapproximates the periodic extension of this function off this interval.