Department of Informatics Engineering Faculty of Industrial Technology

  Department of Informatics Engineering Faculty of Industrial Technology Universitas Gunadarma 2018

  C H A P T E R 1 3 Fourier Series

  

Mathematicians of the eighteenth century, including Daniel Bernoulli and Leonhard Euler, expressed the

problem of the vibratory motion of a stretched string through partial differential equations that had no solu-

tions in terms of “elementary functions.” Their resolution of this difficulty was to introduce infinite series of

sine and cosine functions that satisfied the equations. In the early nineteenth century, Joseph Fourier, while

studying the problem of heat flow, developed a cohesive theory of such series. Consequently, they were

named after him. Fourier series and Fourier integrals are investigated in this chapter and Chapter 14. As you

explore the ideas, notice the similarities and differences with the chapters on infinite series and improper

integrals.

  Periodic Functions

A function f (x) is said to have a period T or to be periodic with period T if for all x, f (x + T) = f (x), where

T is a positive constant. The least value of T > 0 is called the least period or simply the period of f (x).

  EXAMPLE 1. The function sin x has periods 2 π, 4π, 6π, . . . , since sin(x + 2π), sin( + 4π), sin(x + 6 π), . . . all equal sin x. However, 2π is the least period or the period of sin x. EXAMPLE 2. The period of sin nx or cos nx, where n is a positive integer, is 2 π/n. EXAMPLE 3. The period of tan x is π. EXAMPLE 4. A constant has any positive number as period.

  Other examples of periodic functions are shown in the graphs of Figure 13.1(a), (b), and (c). f (x)

  Period x

  (a)

Figure 13.1

CHAPTER 13 Fourier Series

  350 ∞ a n x π n x π

   

  a cos b sin (1)

•  n n 

  ∑

  2 _{n =  } ^{1 L L}

  where the Fourier coefficients a and b are n n _L

  1 n x π

  a f x ( ) cos dx

   = _n

  − L ∫

   L L

  n 0,1, 2,... (2)

  =  _L 1 n x π

  b f x ( ) sin dx

   = _n

  − L ∫

  L L

   To correlate the coefficients with the expansion, see the following Examples 1 and 2.

  Orthogonality Conditions for the Sine and Cosine Functions

Notice that the Fourier coefficients are integrals. These are obtained by starting with the series (1) and em-

ploying the following properties called orthogonality conditions: _L m x π n x π

  (a) _L cos cos dx = 0 if m ≠ n and L if m = n ∫ − _L L L m x π n x π

  (b) (3) _L sin sin dx = 0 if m ≠ n and L if m = n

  ∫ − _L L L m x π n x π (c) Where m and n assume any positive integer values. _L sin cos dx = 0.

  ∫ − L L

  

An explanation for calling these orthogonality conditions is given on Page 355. Their application in de-

termining the Fourier coefficients is illustrated in the following pair of examples and then demonstrated in

detail in Problem 13.4.

  EXAMPLE 1. To determine the Fourier coefficient a , integrate both sides of the Fourier series (1) and em- ploy the orthogonality conditions (2). _{L L L} ∞ a n x n x

  π π  _{n n} + + f x dx ( ) = dx a cos b sin dx _{L L L ∑}  }

  ∫ − ∫ − ∫ −

  2 n ^{1 L L}

  

= 

Now _{L L L} a n x n x

  π π _{L L L} dx = a L , sin dx = 0, cos dx = 0,

  ∫ − ∫ − ∫ −

  2 L L

  therefore

_L

  1

  a = f x dx ( )

_L

∫ −

  L x

  π

  EXAMPLE 2. To determine a , multiply both sides of series (1) by cos and then integrate. Using the or- _{1 L} L

  1 x π thogonality conditions (3) and (3) , we obtain Now see Problem 13.4. _{a c} a = f x ( ) cos dx . _{1 L}

  ∫ − L L

  If L = π, the series (1) and the coefficients (2) or (3) are particularly simple. The function in this case has the period 2 π.

  Dirichlet Conditions Suppose that

CHAPTER 13 Fourier Series

  351 Then the series (1) with Fourier coefficients converges to (a) f (x) if x is a point of continuity f x ( + + 0) f x ( − 0)

  (b) if x is a point of discontinuity

  2 Here f (x + 0) and f (x – 0) are the right- and left-hand limits of f (x) at x and represent lim f (x + ⑀) and lim

  ε → + ε → + f (x – ⑀), respectively. For a proof, see Problems 13.18 through 13.23.

  The conditions 1, 2, and 3 imposed on f (x) are sufficient but not necessary, and are generally satisfied in

practice. There are at present no known necessary and sufficient conditions for convergence of Fourier series.

It is of interest that continuity of f (x) does not alone ensure convergence of a Fourier series.

  Odd and Even Functions

  

3

  5

  3 A function f (x) is called odd if f (–x) = –f (x). Thus, x , x – 3x + 2x, sin x, and tan 3x are odd functions.

  4

  6 2 x –x

A function f (x) is called even if f (–x) = f (x). Thus, x , 2x – 4x + 5, cos x, and e + e are even func-

tions.

  The functions portrayed graphically in Figure 13.1(a) and (b) are odd and even, respectively, but that of Figure 13.1(c) is neither odd nor even. In the Fourier series corresponding to an odd function, only sine terms can be present. In the Fourier series

corresponding to an even function, only cosine terms (and possibly a constant, which we shall consider a

cosine term) can be present.

  Half Range Fourier Sine or Cosine Series

A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms are present,

respectively. When a half range series corresponding to a given function is desired, the function is generally

defined in the interval (0, L) [which is half of the interval (–L, L), thus accounting for the name half range]

and then the function is specified as odd or even, so that it is clearly defined in the other half of the interval,

namely, (–L, 0). In such case, we have _L

  2 n x π

   = a 0, b = f x ( ) sin dx for half range sine series _{n n L}

  ∫ −

   L L

  (4)

   _L 2 n x π

   = b 0, a = f x ( ) cos dx for half range consine series _{n n}

  ∫ L L

  

  Parseval’s Identity

If a and b are the Fourier coefficients corresponding to f (x) and if f (x) satisfies the Dirichlet conditions,

n n then

₂

_L ∞

  1 a ²

  { ( )} f x dx ( a b ) (5) ² = _{n n} ²

  ∑ − L

  ∫ L

  2 _{n = 1}

  (See Problem 13.13.)

CHAPTER 13 Fourier Series

  352

Theorem The Fourier series corresponding to f (x) may be integrated term by term from a to and the result-

_x

ing series will converge uniformly to f x dx ( ) provided that F(x) is piecewise continueus in –L < x < L

_a

  ∫ and both a and x are in this interval.

  Complex Notation for Fourier Series Using Euler’s identities, _{i – i}

  θ θ e = cos = cos

  θ + i sinθ, e θ – i sin θ (6) where i = −

  1 (see Problem 11.48), the Fourier series for f (x) can be written as

  

∞

_{in x L /} π (7) f x ( ) = c e _{n n}

∑

  

=−∞

where _L

  1 _{in x L /}

  − π (8)

c = f x e ( ) dx

_{n L}

  ∫ −

  2 L

  

In writing the equality (7), we are supposing that the Dirichlet conditions are satisfied and, further, that

(x) is continuous at x. If f (x) is discontinuous at x, the left side of (7) should be replaced by

f

  ( ( f x 0) f x ( 0)

• − ) .

2 Boundary-Value Problems

  

Boundary value problems seek to determine solutions of partial differential equations satisfying certain

prescribed conditions called boundary conditions. Some of these problems can be solved by use of Fourier

series (see Problem 13.24).

  

EXAMPLE. The classical problem of a vibrating string may be idealized in the following way. See Figure

13.2.

  Suppose a string is tautly stretched between points (0, 0) and (L, 0). Suppose the tension F is the same at

every point of the string. The string is made to vibrate in the xy plane by pulling it to the parabolic position

  2

g (x) = m(Lx – x ) and releasing it (m is a numerically small positive constant). Its equation will be of the form

y = f (x, t). The problem of establishing this equation is idealized by (a) assuming that the constant tension F

is so large as compared to the weight wL of the string that the gravitational force can be neglected, (b) the

displacement at any point of the string is so small that the length of the string may be taken as L for any of

its positions, and (c) the vibrations are purely transverse.

  The force acting on a segment PQ is ₂ w y

  ∂ ₂

  x , x < x < x +

  ∆ ₂ ^{1 ∆x, g ≈ 32 ft per sec}

  g t

  ∂

  If

α and β are the angles that F makes with the horizontal, then the vertical difference in tensions is

F (sin α – sin β). This is the force producing the acceleration that accounts for the vibratory motion.

CHAPTER 13 Fourier Series

  353

Figure 13.2

  Now

    tan tan y y  α β   ∂ ∂ 

  F {sin α – sin β} = F − ≈ F{tan α – tan β} = F ( x + ∆ x t , ) − ( , ) x t

      ∂ x ∂ x

    ^{2 2}

• 1 tan α
• 1 tan β     where the squared terms in the denominator are neglected because the vibrations are small.

  Next, equate the two forms of the force, i.e., ₂ y y w y

   ∂ ∂  ∂

  F ( x + ∆ x t , ) − ( , ) x t = ∆ x

    ² ∂ x ∂ x g ∂ t

   

  Fg divide by the resulting equation is

  α = ,

  ∆x, and then let ∆x → 0. After letting ₂

w

₂

y y

  ∂ ^{2 ∂} ₂ = α ₂ ∂ t ∂ x This homogeneous second partial derivative equation is the classical equation for the vibrating string.

  Associated boundary conditions are

y (0, t) = 0, y(L, t) = 0, t > 0

  The initial conditions are _{2 ∂}

y

y x ( , 0) m Lx ( x ), ( , 0) x 0, 0 x L

  = − = < <

  

t

  ∂

  The method of solution is to separate variables, i.e., assume

y (x, t) = G(x)H(t)

  Then, upon substituting, ₂ G (x) H G ″ (t) = α ″ (x)H(t)

  Separating variables yields _{" "} ″

  G ″ H ₂ k , k , where k is an arbitrary constant

  = = α

  G H Since the solution must be periodic, trial solutions are

  G x ( ) = c sin − k x c cos − k x , 0 < + _{1 2}

CHAPTER 13 Fourier Series

  354 Thus,

y = [ sin c − k x c ][ sin α − k t c cos α + − k t ]

_{1 3 4} Now impose the boundary condition y = 0 at x = L; thus, 0 = [c sin − k L ] [c sin − k t + c cos

  α α

  1

  3

  4 − k t ].

  ⫽ c 0, as that would imply y = 0 and a trivial solution. The next-simplest solution results from the

  1 n π n π n π n π

      − = + choice k , since y = c sin x c sin α t c cos α t and the first factor is zero when _{1 3 4}

     

  L  L   l L  x = L. y

  ∂ With this equation in place, the boundary condition = ( , 0) x = 0, 0 < < x L can be considered.

  ∂ t

  y n n n n n

  ∂ π π π π π    

  = c sin x c α cos α t − c α sin α t _{1 3 4}    

  ∂ t  L   L L L L 

  At t = 0, n n

  π π  

  

c sin x c α

_{1 3}

  =    L  L

  n

  π

  ⫽ Since c 0 and sin is not identically zero, it follows that c = 0 and that x

  1

  3 L n n n

   π   π π 

  

y c sin x c α cos α t

₁

₄

  =      L   L L 

  The remaining initial condition is ₂

y (x, 0) = m(Lx – x ), 0 < x < L

  When it is imposed, _{2 π π} n n

m Lx ( − x ) = c c α sin x

_{1 4}

L L

  However, this relation cannot be satisfied for all x on the interval (0, L). Thus, the preceding extensive analysis of the problem of the vibrating string has led us to an inadequate form: n n n

  π π π

  

y = c c α sin x cos α t

_{1 4} L L L

and an initial condition that is not satisfied. At this point the power of Fourier series is employed. In particu-

lar, a theorem of differential equations states that any finite sum of a particular solution also is a solution.

Generalize this to infinite sum and consider

  ∞ n n

  π π

  

y b sin x cos t

  = α _n

  ∑ _{n = 1} L L with the initial condition expressed through a half range sine series, i.e.,

  ∞ n

  π ²

  

b sin x m Lx ( x ), t

_n = − = ∑ _{n = 1} L

  According to the formula on Page 351 for the coefficient of a half range sine series, _L L n x _{2 π} b ( Lx x ) sin dx _n = −

  ∫

  2 m L

  That is, _{L L} L n x π n x π ₂

CHAPTER 13 Fourier Series

  355 When integration by parts is applied to the two integrals of this expression and a little algebra is employed, the result is ₂

  4 L

  

b = (1 cos − n π )

_{n 3}

  ( n π )

  Therefore, ∞ n π n π

y = b sin x cos α t

_n

  ∑ _{n 1 L L} = with the coefficients b defined previously. n

  Orthogonal Functions

Two vectors A and B are called orthogonal (perpendicular) if A · B = 0 or or A B + A B + A B = 0, where

  1

  1

  2

  2

  3

  3 A = A i + A j + A k and B = B i + B j + B k. Although not geometrically or physically evident, these ideas

  1

  2

  3

  1

  2

  3

can be generalized to include vectors with more than three components. In particular, we can think of a

function—say, A(x)—as being a vector with an infinity of components (i.e., an infinite dimensional vector),

the value of each component being specified by substituting a particular value of x in some interval (a, b). It

is natural in such case to define two functions, A(x) and B(x), as orthogonal in (a, b) if _b

  (9) _a

A x B x dx ( ) ( ) =

∫

  2 A vector A is called a unit vector or normalized vector if its magnitude is unity, i.e., if A · A = A = 1.

  Extending the concept, we say that the function A(x) is normal or normalized in (a, b) if _b

₂

(10) _a { ( )} A x dx =

  1

  ∫

From this, it is clear that we can consider a set of functions { (x)}, k = 1, 2, 3, . . . , having the properties

_b φ k

  (11) _a φ ( ) x ) ( ) φ x dx = m ≠ n _{m n b} ∫ ₂

  { ( )} 1 1, 2, 3,... (12) _a φ x dx = m = _m

  ∫ In such case, each member of the set is orthogonal to every other member of the set and is also normalized.

  We call such a set of functions an orthonormal set.

  Equations (11) and (12) can be summarized by writing _b

  ( ) x ( ) x dx (13) _a φ φ φ = δ _{m n mn}

  ∫ where , called Kronecker’s symbol, is defined as 0 if m ⫽ n and 1 if m = n.

  δ mn

Just as any vector r in three dimensions can be expanded in a set of mutually orthogonal unit vectors i, j,

k in the form r = c i + c j + c k, so we consider the possibility of expanding a function f (x) in a set of ortho-

  1

  2

  3 normal functions, i.e.,

  ∞ f x ( ) c ( ) x a x b (14)

  = φ ≤ ≤ _{n n}

  ∑ _{n = 1}

CHAPTER 13 Fourier Series

  356

SOLVED PROBLEMS

  Fourier Series 13.1. Graph each of the following functions.

  3 x

  5  < <

  (a) f x ( ) Period

  10 = =

  3 5 x − − < < 

  

Figure 13.3

Since the period is 10, that portion of the graph in –5 < x < 5 (indicated by heavy lines in Figure 13.3) is extended periodically outside this range (indicated by dashed lines). Note that f (x) is not defined at x = 0, 5,

• –5, 10, –10, 15, –15, and so on. These values are the discontinuities of f (x).

  sin x x < < π

  

  (b) f x ( ) Period = 2

  π = 

  π < < x 2 π 

Figure 13.4 Refer to Figure 13.4. Note that f (x) is defined for all x and is continuous everywhere.

  ≤ < x

  2  

  (c) f x ( ) =

  1 2 ≤ < x

  4 Period =

  6   4 ≤ < x

  6 

  

Fig.13.5

Refer to Figure 13.5. Note that f (x) is defined for all x and is discontinuous at x = ±2, ±4, ±8, ±10,

CHAPTER 13 Fourier Series

  − − − −

  ^{L L} 2 ^{L L L L L L}

     

    = − =

     

    = + =

  357

  π π π π π π

  m x n x n x dx dx L L L L m x n x n x dx dx L L L L

  2 sin sin 1 cos

  1 ( ) ( ) sin cos sin sin

  1

  2

  2 cos cos 1 cos

  1

  ∫ ∫ If m = n, we have

  = − =    

  − +  

  ∫ ∫ ∫ ∫ Note that if m = n these integrals are equal to 2L and 0, respectively. (b) We have sin A cos B = 1/2 {sin(A – B) + sin(A + B)}. Then by Problem 13.2, if m ⫽ n,

  ^{L L} 2 ^{L L}

  π π π π

  ^{L L} 2 ^{L L}

  n x n x A a b L L

   

  ∫ ∫ The results of (a) and (b) remain valid even when the limits of integration –L, L are replaced by c, c + 2L, respectively.

  = =

  − −

  π π π

  

m x n x n x

dx dx

L L L

  2 sin cos sin

  

m x n x m n x m n x

dx L L L L

  1

  If m = n,

  ∫ ∫ dx =

  = +    

  − +  

  − −

  π π π π

  − −

  m x n x m n x m n x dx dx L L L L

  ∞ =

  = = − − = −

  = 

  = = 

  ≠ 

  − −

  π π π π

  13.3. Prove (a) cos cos sin sin ^{L L} _{L L} m n m x n x m x n x

dx dx

L m n L L L L

  ∫ ∫

  = − = + − = −

  L L

  π π π π π

  π π π π π

  π π π π

  L L k x L L dx k k L k L k k

  L L L L dx k k L k k k

  L ^{L -L L -L}

  k x k x sin cos cos cos( ) L L k x cos sin sin sin( )

  ∫ ∫ (b) sin cos ^L _L m x n x dx

  π π

  ^{L L} 2 ^{L L}

  ^{L L} 2 ^{L L}

  1 ( ) ( ) sin sin cos cos

  ∫ ∫ Similarly, if m ⫽ n,

  = + =    

  − +  

  − −

  π π π π

  

m x n x m x x m n x

dx dx L L L L

  1 ( ) ( ) cos cos cos cos

  −

  Then, if m ⫽ n, by Problem 13.2,

  2 {cos(A – B) – cos (A + B)}.

  1

  2 {cos(A – B) + cos(A + B)}, sin A sin B =

  1

  (a) From trigonometry: cos A cos B =

  ∫ where m and n can assume any of the values 1, 2, 3, . . . .

  =

  π π

13.4. If the series

  ∫ (c) .

  ₁

  ∑ (1)

   

    = + +

  ∞

  π π

  n x n x f x A a b

  ( ) cos sin _{n n}

  a A = (a) Multiplying

  2

  cos sin _{n n n}

     

  −

  π

  n x b f x dx L L

  1 ( ) sin , ^L _{n L}

  ∫ and (b)

  =

  −

  π

  n x a f x dx L L

  1 ( ) cos ^L _{n L}

  ∑ converges uniformly to f(x) in (–L, L), show that for n = 1, 2, 3, . . . , (a)

  =

CHAPTER 13 Fourier Series

  358 _{L L} m x π m x π f x ( ) cos dx A cos dx _{L L} =

  ∫ − ∫ − L L

  ∞ _{L L} m x n x m x n x

  π π π π 

  

• a cos cos dx b cos sin dx _{n n}

   _{L L} 

  − ∫ − ∑ ∫ _{n 1 L L L L}

  =

   

  = a L if m ≠ _m

  Thus, _L

  1 m x π

a = f x ( ) cos dx if m = 1, 2, 3,...

_{m L}

  ∫ − L L m x

  π

  (b) Multiplying Equation (1) by and integrating from –L to L, using Problem 13.3, we have

  sin _{L L} L

  m x m x

  π π

  f x ( ) sin dx A sin dx

  =

  − L − L ∫ ∫

  L L ∞ _{L L} m x n x m x n x

   π π π π 

• a sin cos dx b cos sin dx _{n n}

   _{L L} 

  − ∫ − ∑ ∫ _{n 1 L L L L}

  = 

  

  Thus, b L

  = _{m L} 1 m x π

b = f x ( ) sin dx if m = 1, 2, 3,...

_m

  − L ∫

  L L (c) Integrating Equation (1) from –L to L, using Problem 13.2, gives _{L L}

  1

  f x dx ( ) =

  2 AL or A = f x dx ( )

  − L − L ∫ ∫ _L

  2 L 1 a

  Putting m = 0 in the result of (a), we find a = f x dx ( ) and so A = .

  − L ∫

  L

  2 The above results also hold when the integration limits –L, L are replaced by c, c + 2L.

  Note that in (a), (b), and (c), interchange of summation and integration is valid because the series is as- sumed to converge uniformly to f (x) in (–L, L). Even when this assumption is not warranted, the coefficients a and b as obtained are called Fourier coefficients corresponding to f (x), and the corresponding series with _{m m} these values of a and b is called the Fourier series corresponding to f (x). An important problem in this case _{m m} is to investigate conditions under which this series actually converges to f (x). Sufficient conditions for this convergence are the Dirichlet conditions established in Problems 13.18 through 13.23.

  13.5. (a) Find the Fourier coefficients corresponding to the function

  5 x  − < <

  

f x ( ) Period = 10

  =  3 x

  5 < <

   (b) Write the corresponding Fourier series.

  (c) How should f (x) be defined at x = –5, x = 0, and x = 5 in order that the Fourier series will converge to f (x)

  < <

  for –5 x 5? The graph of f (x) is shown in Figure 13.6.

CHAPTER 13 Fourier Series

  (a) Period = 2L = 10 and L = 5. Choose the interval c to c + 2L as –5 to 5, so that c = –5. Then _{2 5 5}

  359

Figure 13.6

• −
• = = = = = = =

  1

  3

  5

  5

  3

  5

  2

  5 sin sin sin

  1

  3

  1

  6

  5

  a n x n x n n x a b L L n x x x

  2

  2

  3 3(1 cos ) cos sin sin

  ∫ ∫ ∫ ∫ (b) The corresponding Fourier series is _{1 1}

  = − =    

  −  

   

   = + =

  − −

  π π

  π π π π π

  5 ^{n n n n}

  π π π π π

  5

  3 / 2

  < <   =  ...

  = =  

  − < < 

  = −  

  x

x

f x x

x

x

  5

  5 3 / 2

  3

  5 ( ) Period = 10 3 / 2

  5

  at points of discontinuity. At x = –5, 0, and 5, which are points of discontinuity, the series converges to (3 + 0)/2 = 3/2, as seen from the graph. If we redefine f (x) as follows,

  π π π π

  f x f x

  2

  ( 0) ( 0)

  (c) Since f (x) satisfies the Dirichlet conditions, we can say that the series converges to f (x) at all points of continuity and to

  L

  ∑ ∑

     

    = + + + +

     

  −  

  ∞ ∞ = =

  dx n x n x n x dx dx dx n x n n n

  5

  1 ( ) cos ( ) cos

  5

  5

  5

  5 sin

  3

  ∫ ∫ ∫

   

   = + =

  −

  π π π

  n x n x n x dx dx dx

  5

  π π

  5

  5

  5

  3 (0) cos (3) cos cos

  1

  ∫ ∫ } ^{5 5} ₅

  = =

  π π

  n x n x a f x dx f x dx L L

  5 ^{c L n c}

  5

  n x n

    =

  3 5 3(1 cos ) cos

      ⁵

  5

  5

  5

  5

  3 (0) sin (3) sin sin

  1

  ∫ ∫ ∫ } ^{5 5 5 5} _{5 5}

  π π π

  L L

  x n a a dx dx n x n x b f x dx f x

  5 ^{n c L n c}

  5

  1 ( ) sin ( ) sin

  1

  5

  5

  5

  3 If 0, cos 3.

  3

  2

    _{5 5}

  

 = if n ≠

  5

• = +
• −

CHAPTER 13 Fourier Series

• = =
• = =

  1

  4 .

  4

  At x = 0, the Fourier series of Problem 13.6 reduces to ² _{2 1}

  =

  6 π

  3 + + + ... ² .

  2

  1

  1

  n

  1

  _{2 2 2}

  ∑ This is valid for 0 < x < 2 π. At x = 0 and x = 2π the series converges to 2π ² . (b) If the period is not specified, the Fourier series cannot be determined uniquely in general.

     

    = = + −

  ∞ =

  π π

  f x x nx nx n n

  3 _n

  π

  4 Then ( ) cos sin .

  6 _{n n}

  < < 

  x

  3

  2

  ∑ ∑ Odd and even functions, half range Fourier series 13.8. Classify each of the following functions according to whether they are even, odd, or neither even nor odd.

  ∞ ∞ = =

  π π π

  n n

  3

  ∞ =

  1 2 , and so .

  4

  4

  Then ^{2 2 2} _{2 2 1 1}

  (0 + 4 π ² ) = 2 π ² .

  2

  1

  By the Dirichlet conditions, the series converges at x = 0 to

  3 _n

  4

  n a x dx π

  4 ( ) (2 ) 2 ,

  8 If 0, .

  1

   _{2 2 2}

      

  − + = ≠       

   − −      

  π

  

nx nx nx

x x n n n n n

  sin cos sin

  4

  { ^{2 2 2 3 2}

  =   

  1 π

  ∫ ∫

  π π

  

π

  n x a f x dx x nx dx L L

  1 ( ) cos cos ^{c L} _{n c}

  3

  π π

  360 Period = 2L = 2 π and L = π. Choosing c = 0, we have ₂

₂

₂

  { ^{2 2 2 3}

   _{2 2 2 1}

      

  − +       

   − −      

  π

  nx nx nx x x n n n

  2

  sin cos sin ( ) (2 )

  = =   

  = = =

  1 π

  ∫ ∫

  π π

  π

  n x b f x dx x nx dx L L

  1 ( ) sin sin ^{c L} _{n c}

  1

  ∫ _{2 2 2}

  1

13.7. Using the results of Problem 13.6, prove that

• ∑
• = =

CHAPTER 13 Fourier Series

  361

Figure 13.8 From Figure 13.9, it is seen that the function is neither even nor odd.

  

Figure 13.9

(c) f (x) = x(10 –x), 0 < x < 10 Period = 10 From Figure 13.10 below the function is seen to be even.

Figure 13.10 13.9. Show that an even function can have no sine terms in its Fourier expansion.

  Method 1: No sine terms appear if b = 0, n = 1, 2, 3, . . . . To show this, let us this, let us write _{L L n}

  1 n x 1 n x 1 n x π π π

  (1)

b = + f x ( ) sin dx = f x ( ) sin dx f x ( ) sin dx

_{n L L}

  ∫ − ∫ − ∫

L L L L L L

  If we make the transformation x = –u in the first integral on the right of Equation (1), we obtain _{L L}

  1

  1

  1

  n x π n u π n u π

   

  f x ( ) sin dx = f ( − u ) sin − du = − f ( − u ) sin du _L   ∫ − ∫ ∫

  L L L  L  L L _{L L} (2)

  1 n u 1 n u π π

  = − f u ( ) sin du = − f x ( ) sin dx

  ∫ ∫ L L L L

CHAPTER 13 Fourier Series

  362 Method 2: Assume

  ∞ a n x π n x π

   

  f x ( ) a cos sin b

  = + +   ^{n n}

  ∑

  2 _{n =  } ^{1 L L}

  Then ∞ a n x π n x π

   

  − = −  n N 

• f ( x ) a cos b sin

  ∑

  2 n ^{1  L L }

  = If f (x) is even, f (–x) = f (x). Hence,

  ∞ ∞ a a n x π n x π n x π n x π

     

  a cos b sin a cos b sin

  = − + + +  n n   n n 

  ∑ ∑

  2 n ^{1  L L } 2 n ^{1  L L }

  = = and so