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The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 8: Fourier series II

MATH3068 Analysis Semester 1, 2007

Web Page: http://www.maths.usyd.edu.au:8000/u/UG/SM/MATH3068/Lecturer: Donald Cartwright

1. In the last tutorial, a function f (x) was dened on [0, 2] by the following formula, then extendedby periodicity:

f (x) = x

2 if 0 < x < 2,0 if x = 0 or x = 2.

We found that its Fourier series is

k=1

sin kxk

.

What does the Parseval Identity tell us in this case?

Solution: One side of the Parseval Identity is

12

|f (x)|2 dx.

Now f (x) is odd, and so |f (x)|2 = ( f (x))2 is even. So we can integrate over [0 , ] and multiplyby 2:

12

|f (x)|2 dx =

1

0(f (x))2 dx =

1

0

( x)2

4 dx =

1

( x)3

12

x =

x =0

= 2

12.

The other side of the Parseval Identity is

14

|a0 |2 + 12

k=1

|ak |2 + |bk |2 ,

which in this case, when all the ak s are 0, and bk = 1 /k for k = 1 , 2, . . . , is

12

k=1

|bk |2 = 12

k=1

1k2

.

So Parsevals Identity tells us the famous formula

k=1

1

k2 =

2

6 .

2. In the last tutorial, a function f (x) was dened on [0, 2] by the following formula, then extendedby periodicity:

x2

4

x2

+ 2

6We found that its Fourier series is

k=1

cos kxk2

.

What does the Parseval Identity tell us in this case?

Solution: This time the function f (x) is even. So |f (x)|2 = ( f (x))2 is even, and

12

|f (x)|2 dx =

1

0(f (x))2 dx =

1

0

x2

4

x2

+ 2

62

dx.

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(b) The formula we need to prove may be written

1 + 2n

k=1

cos(kx) sin12

x = sin n + 12

x . ( )

Using 2 cos(kx)sin 12 x = sin k + 12 x sin k

12 x , the left hand side becomes

sin12

x +n

k=1

sin k + 12

x sin k 12

x .

This is a collapsing sum

a0 + ( a1 a0) + ( a2 a1) + + ( an an 1)

for ak = sin k + 12 x . So it equals an = sin n + 12 x , and this is the right hand side of

the formula ( ) we wanted to show.(c) Using cos = ( ei + e i )/ 2, we have 2 cos(kx) = eikx + e ikx , and so

D n (x) =n

k= n

eikx

This equals e inx + e (n 1) ix + + ei (n 1) x + einx , and so equals

e inx 1 + eix + e2ix + + e2nix

= e inxei (2 n +1) x 1

eix 1 (sum of a geometric progression)

= ei (n +1) x e inx

eix 1

= ei (n +

12 )x e i (n +

12 )x

eix

2 e ix

2multiplying top and bottom by e ix/ 2

= 2i sin n + 12 x2i sin 12 x

= sin n + 12 x

sin 12 x .

(b) Check that D n (x) = 2 n + 1 if x is a multiple of 2. Also, check that 12 Dn (t) dt = 1.

Solution: Just observe that cos( kx) equals 1 when x = 2m , m an integer, to see thatD n (2m ) = 2 n + 1. Also,

12

D n (t) dt =

12

1 dt +

1

n

k=1

cos(kt ) dt = 1 +

1

n

k=1

sin(kx)k

x =

x =

= 1 .

5. Let f (x) be a 2-periodic piecewise continuous function.(a) Show that if f (x) is odd (i.e., f ( x) f (x)), then all the Fourier cosine coefficients ak of

f (x) are zero.Solution: If f (x) is odd, then so is f (x)cos(kx) for k = 0 , 1, 2, . . .. Hence

ak = 12

f (x)cos(kx) dx = 0 .

(b) Show that if f (x) is even (i.e., f ( x) f (x)), then all the Fourier sine coefficients bk of f (x) are zero.Solution: If f (x) is even, then f (x)sin( kx) is odd for k = 1 , 2, . . . . Hence

bk = 12

f (x)sin( kx) dx = 0 .

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(c) Show that if f (x) is real-valued, then all the Fourier cosine and sine coefficients ak and bkof f (x) are real.Solution: If f (x) is real-valued, then f (x)cos(kx) and f (x)sin( kx) is real-valued for all k,and so k = 1 , 2, . . .. Hence

ak = 12

f (x)sin( kx) dx and bk = 12

f (x)sin( kx) dx are real numbers .

(d) Find corresponding conditions, expressed in terms of the complex Fourier coefficients ck of f (x), which must hold f (x) is even, odd and real-valued, respectively.Solution: If f (x) is odd, then

ck = 12

f (x)e ikx dx =

12

( f ( x))e ikx dx =

12

f (u)eiku du = c k .

If f (x) is even, then

ck = 12

f (x)e ikx dx = 12

f ( x)e ikx dx = 12

f (u)eiku du = c k .

If f (x) is real-valued, then

ck = 12

f (x)e ikx dx =

12

f (x)eikx dx =

12

f (x)eikx dx = c k .

(e) Show that if 12

a0 +

k=1

ak cos(kx) + bk sin(kx)

is the Fourier series of f (x), then

12 a0 +

k=1ak cos(kx)

is the Fourier series of the even function g(x) = f (x )+ f ( x )2 , and that

k=1

bk sin(kx)

is the Fourier series of the odd function h(x) = f (x ) f ( x )2 .Solution: Let Ak , k = 0 , 1, . . . , and Bk , k = 1 , 2, . . . , be the Fourier coefficients of g(x).Then Bk = 0 for all k, because g(x) is even. On the other hand,

Ak = 1

g(x)cos(kx) dx = 1

f (x) + f ( x)2 cos(kx) dx

= 12

f (x) cos(kx) dx +

12

f ( x) cos(kx) dx

= 12

f (x) cos(kx) dx +

12

f (u) cos(ku ) du

= 1

f (x)cos(kx) dx

= ak .

So12 a0 +

k=1 ak cos(kx)

is the Fourier series of g(x).

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Let C k , k = 0 , 1, . . . , and D k , k = 1 , 2, . . . , be the Fourier coefficients of h(x). Then C k = 0for all k, because h(x) is odd. On the other hand,

D k = 1

g(x)sin( kx) dx =

1

f (x) + f ( x)2

sin(kx) dx

= 12

f (x)sin( kx) dx + 12

f ( x)sin( kx) dx

= 12

f (x)sin( kx) dx +

12

f (u)sin( ku ) du

= 1

f (x)sin( kx) dx

= bk .

So

k=1

bk sin(kx)

is the Fourier series of h(x).

(f) (harder) Show that the converses of each of the assertions (a), (b) and (c) hold if f (x)is continuous . Do these converses still hold if f (x) is merely assumed to be piecewisecontinuous? [Hint: You cannot assume that the Fourier series of f (x) converges to f (x) ateach point x. Instead, in (a), for example, consider g(x) = ( f (x) + f ( x)) / 2 and apply theuniqueness theorem.]Solution: (i) Suppose that the cosine coefficients of a function f (x) are all 0. If we knewthat the Fourier series of f (x) converged to f (x) at each point x, then we would know that

f (x) =

k=1

bk sin(kx) for all x.

This implies that f (x) is odd, because

f ( x) =

k=1

bk sin( kx) =

k=1

bk sin(kx) = f (x) for all x.

The trouble is, we dont know that the Fourier series of f (x) converges to f (x) at eachpoint x, because we are not assuming that f (x) is differentiable. To avoid assuming differ-entiability, consider the function g(x) = ( f (x) + f ( x)) / 2. Its Fourier series is just the zeroseries, by the rst half of part (e). By the Parseval identity, applied to g(x), we see that

12

|g(x)|2 dx = 0 .

When f (x) is continuous at every point, so that g(x) is continuous everywhere too, thisimplies that g(x) = 0 for all x, and so f ( x) = f (x) for all x. When f (x) is merelyassumed to be piecewise continuous, all we can say is that g(x) = 0 at each point whereg(x) is continuous. Therefore, we can only say that f ( x) = f (x) at each point x0 forwhich f (x) is continuous at both x0 and x0 .(ii) Suppose that the sine coefficients of a function f (x) are all 0. If we knew that the Fourierseries of f (x) converged to f (x) at each point x, then we would know that

f (x) = 12

a0 +

k=1

ak cos(kx) for all x.

This implies that f (x) is even, because

f ( x) =

k=1

bk cos( kx) =

k=1

bk cos(kx) = f (x) for all x.

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The trouble is, we dont know that the Fourier series of f (x) converges to f (x) at eachpoint x, because we are not assuming that f (x) is differentiable. To avoid assuming differ-entiability, consider the function h(x) = ( f (x) f ( x)) / 2. Its Fourier series is just the zeroseries, by the second half of part (e). By the Parseval identity, applied to h(x), we see that

1

2

|h(x)|2 dx = 0 .

When f (x) is continuous at every point, so that h(x) is continuous everywhere too, thisimplies that h(x) = 0 for all x, and so f ( x) = f (x) for all x. When f (x) is merely assumedto be piecewise continuous, all we can say is that h(x) = 0 at each point where h(x) iscontinuous. Therefore, we can only say that f ( x) = f (x) at each point x0 for which f (x)is continuous at both x0 and x0 .(iii) Suppose that the cosine and sine coefficients of a function f (x) are all real. If we knewthat the Fourier series of f (x) converged to f (x) at each point x, then we would know that

f (x) = 12

a0 +

k=1

ak cos(kx) + bk sin(kx) for all x.

This implies that f (x) is real-valued.The trouble is, we dont know that the Fourier series of f (x) converges to f (x) at eachpoint x, because we are not assuming that f (x) is differentiable. To avoid assuming differ-entiability, write f (x) = u(x) + iv(x), where u(x) and v(x) are real-valued. consider thefunction v(x). Its Fourier series is just the zero series, because

1

v(x)cos( kx) dx = Im

1

f (x)cos(kx) dx = Im( ak ) = 0

and1

v(x)sin( kx) dx = Im

1

f (x)sin( kx) dx = Im( bk ) = 0 .

By the Parseval identity, applied to v(x), we see that1

2

|v(x)|2 dx = 0 .

When f (x) is continuous at every point, so that v(x) is continuous everywhere too, thisimplies that v(x) = 0 for all x, and so f (x) = u(x) for all x, so that f (x) is real-valued.When f (x) is merely assumed to be piecewise continuous, all we can say is that v(x) = 0at each point where v(x) is continuous. Therefore, we can only say that f (x) R at eachpoint x0 for which f (x) is continuous at both x0 and x0 .

6. The important integral

0sin( x )

x dx may be evaluated using the Riemann-Lebesgue Lemma andthe fact that

0 Dn (x) dx = : Check that

2

(n + 12 )

0

sin(x)x

dx = 12

0

1sin( 12 x)

112 x

sin (n + 12

)x dx

Now dene

g(x) = 1

sin( 12 x ) 11

2 x if 0 < x ,

0 if x = 0.

Check that g(x) is continuous at 0 (e.g., using LHopitals rule) and thus on [0 , ], so that theRiemann-Lebesgue Lemma is applicable. Deduce that

0sin( x )

x dx = 2 .

Solution: The function Dn (x) = 1 + 2nk=1 cos(kx) is clearly even, and as we saw in an earlier

question, it satises 12

Dn (x) dx = 1. Hence

0 Dn (x) dx = . Therefore

2

= 12

0

1sin( 12 x)

sin (n + 12

)x dx.

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because D n (x) = sin (n + 12 )x / sin12 x . Also, the change of variable x = n +

12 u shows that

(n + 12 )

0

sin(x)x

dx = 12

0

112 u

sin (n + 12

)u du

Subtracting the last two displayed equations, and we get

2

(n + 12 )

0

sin(x)x

dx = 12

0

1sin( 12 x)

112 x

sin (n + 12

)x dx

It remains to show that the function g(x) of the question is well-behaved as x 0. The clearestway to see this is to write

sin x = x x3

6 +

x5

120 ,

so thatsin x x = x3

16

+ x2

120 = x3h(x), say,

where h(x) 16 as x 0. Therefore

g(x) = 1

sin( 12 x)

112 x

= sin( 12 x)

12 x

12 x sin(

12 x)

= x 3

8 h(12 x)

12 x sin(

12 x)

= x 2

4 h(12 x)

sin( 12 x)

= x2

sin( 12 x) h(

12

x) 12

x

1 16

0 = 0 as x 0.

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