Ebook Advanced engineering mathematics (7th edition) Part 2

(BQ) Part 2 book Advanced engineering mathematics has contents: Fourier series, the fourier integral and transforms, special functions and eigenfunction expansions, the wave equation, the heat equation, the potential equation, complex integration, singularities and the residue theorem,...and other contents.

P A R T

4

Fourier Analysis, Special Functions, and Eigen- function

In 1807, Joseph Fourier submitted a paper to the French Academy of Sciences in competition for

a prize offered for the best mathematical treatment of heat conduction In the course of this workFourier shocked his contemporaries by asserting that “arbitrary” functions (such as might specifyinitial temperatures) could be expanded in series of sines and cosines Consequences of Fourier’swork have had an enormous impact on such diverse areas as engineering, music, medicine, andthe analysis of data

A Fourier series is a representation of a function as a series of constant multiples of sine and/orcosine functions of different frequencies To see how such a series might arise, we will look at aproblem of the type that concerned Fourier

Consider a thin homogeneous bar of metal of lengthπ, constant density and uniform cross section Let u (x, t) be the temperature in the bar at time t in the cross section at x Then (see Section 12.8.2) u satisfies the heat equation

∂u

∂t = k

∂2u

∂x2for 0< x < π and t > 0 Here k is a constant depending on the material of the bar If the left and

right ends are kept at temperature zero, then

Fourier found that the functions

u n (x, t) = b nsin(nx)e −kn2t satisfy the heat equation and the boundary conditions, for every positive integer n and any number

b n However, there is no choice of n and bn for which this function satisfies the initial condition,which would require that

This function will still satisfy the heat equation and the boundary conditions u (x, 0) = u(π, 0) =

0 To satisfy the initial condition, the problem is to choose the numbers bnso that

for 0≤ x ≤ π was too much for Fourier’s contemporaries to accept, and the absence of rigorous

proofs in his paper led the Academy to reject its publication (although they awarded him theprize) However, the implications of Fourier’s work were not lost on natural philosophers ofhis time If Fourier was right, then many functions would have expansions as infinite series oftrigonometric functions

Although Fourier did not have the means to supply the rigor his colleagues demanded, thiswas provided throughout the ensuing century and Fourier’s ideas are now seen in many importantapplications We will use them to solve partial differential equations, beginning in Chapter 16.This and the next two chapters develop the requisite ideas from Fourier analysis

Construct graphs of S N (x) and x(π − x), for 0 ≤ x ≤ π,

for N =2 and then N =10 This will give some sense of

the correctness of Fourier’s claim that this polynomial

could be exactly represented by the infinite series

2 Let p (x) be a polynomial Prove that there is no number

k such that p(x) = k sin(nx) on [0, π] for any positive

integer n.

3 Let p (x) be a polynomial Prove that there is no finite

sumN

n=1bnsin(nx) that is equal to p(x) for 0≤ x ≤π,

for any choice of the numbers b1, · · · , bN

Let f (x) be defined on [−L, L] We want to choose numbers a0, a1, a2· · · and b1, b2, · · ·

[akcos(kπx/L) + b ksin(kπx/L)]. (13.1)

This is a decomposition of the function into a sum of terms, each representing the influence of adifferent fundamental frequency on the behavior of the function

To determine a0, integrate equation (13.1) term by term to get

To solve for the other coefficients in the proposed equation (13.1), we will use the following three

facts, which follow by routine integrations Let m and n be integers Then

Now let n be any positive integer To solve for an, multiply equation (13.1) by cos(nπx/L) and

integrate the resulting equation to get

Because of equations (13.3) and (13.4), all of the terms on the right are zero except the coefficient

of an , which occurs in the summation when k = n The last equation reduces to

This expression contains a0if we let n= 0

Similarly, if we multiply equation (13.1) by sin(nπx/L) instead of cos(nπx/L) and

con-Recall that f is piecewise continuous on [a, b] if f is continuous at all but perhaps finitely many points of this interval, and, at a point where the function is not continuous, f has finite

limits at the point from within the interval Such a function has at worst jump discontinuities, orfinite gaps in the graph, at finitely many points Figure 13.1 shows a typical piecewise continuousfunction

If a < x0< b, denote the left limit of f (x) at x0 as f (x0−), and the right limit of f (x) at x0

FIGURE 13.1 A piecewise tinuous function.

0 1

–1

–3 –2

x

2 –2

f is piecewise continuous on [−3, 4], having a single discontinuity at x = 2 Furthermore,

f (2−) = 2 and f (2+) = 1/2 A graph of f is shown in Figure 13.2. 

f is piecewise smooth on [a, b] if f is piecewise continuous and fexists and is continuous

at all but perhaps finitely many points of(a, b).

This derivative is itself piecewise continuous Therefore f is piecewise smooth on [−3, 4]. 

We can now state a convergence theorem

THEOREM 13.1 Convergence of Fourier Series

Let f be piecewise smooth on [−L, L] Then, for each x in (−L, L), the Fourier series of f on [−L, L] converges to

1

2( f (x+) + f (x−)).

x y

FIGURE 13.3 Convergence of a Fourier series at a jump discontinuity.

At both−L and L this Fourier series converges to

1

2( f (L−) + f (−L+)). 

At any point in(−L, L) at which f (x) is continuous, the Fourier series converges to f (x), because then the right and left limits at x are both equal to f (x) At a point interior to the interval where f has a jump discontinuity, the Fourier series converges to the average of the left

and right limits there This is the point midway in the gap of the graph at the jump discontinuity(Figure 13.3) The Fourier series has the same sum at both ends of the interval

EXAMPLE 13.4

Let f (x) = x − x2for−π ≤ x ≤ π In Example 13.1 we found the Fourier series of f on [−π, π] Now we can examine the relationship between this series and f (x).

f(x) = 1 − 2x is continuous for all x, hence f is piecewise smooth on [−π, π] For −π <

x < π, the Fourier series converges to x − x2 At bothπ and −π, the Fourier series converges to

Figures 13.4, 13.5, and 13.6 show the fifth, tenth and twentieth partial sums of this Fourier

series, together with a graph of f for comparison The partial sums are seen to approach the

function as more terms are included 

–4

–12

–2 –3

–6

–10 –8

0 –3

–10 –8

0 –3

–10 –8

1.5 2

1

0.5

x

1 0.5

0 –0.5

f is piecewise smooth on [−2π, 2π] The Fourier series of f on [−2π, 2π] converges to:

4π for x = 2π and x = −2π.

This conclusion does not require that we write the Fourier series 

13.2.1 Even and Odd Functions

A function f is even on [−L, L] if its graph on [−L, 0] is the reflection across the vertical

axis of the graph on[0, L] This happens when f (−x) = f (x) for 0 < x ≤ L For example,

x 2n

and cos(nπx/L) are even on any [−L, L] for any positive integer n.

A function f is odd on [−L, L] if its graph on [−L, 0) is the reflection through the

origin of the graph on(0, L] This means that f is odd when f (−x)=− f (x) for 0< x ≤ L.

For example, x 2n+1and sin(nπx/L) are odd on [−L, L] for any positive integer n.

Figures 13.8 and 13.9 show typical even and odd functions, respectively

A product of two even functions is even, a product of two odd functions is even, and aproduct of an odd function with an even function is odd

is odd and the integrals defining the sine coefficients will be zero If the function is odd

then f (x) cos(nπx/L) is odd and the Fourier series will contain only the sine terms, since

the integrals defining the constant term and the coefficients of the cosine terms will bezero

0

x

6 2

–2 –4

–40 –20

4

FIGURE 13.9 A typical odd function.

We will compute the Fourier series of f (x) = x on [−π, π].

Because x cos (nx) is an odd function on [−π, π] for n = 0, 1, 2, · · · , each a n= 0 We need

only compute the bns:

x sin (nx) dx

=

2

n2π sin(nx) −

2x

n π cos(nx)

π0

= −2

ncos(nπ) =2

n (−1) n+1 The Fourier series of x on [−π, π] is

This converges to x for −π < x < π, and to 0 at x = ±π Figure 13.10 shows the twentieth

partial sum of this Fourier series compared to the function 

x4

d x=25and

a n= 2

 1 0

x4cos(nπx)dx

= 2



(nπx)4sin(nπx) + 4(nπx)3

cos(nπx) (nπ)5

1 0

0.6 0.8

0.4

0

x

1 0.5

0 –1

By Theorem 13.1, this series converges to x4 for −1 ≤ x ≤ 1 Figure 13.11 shows the

twentieth partial sum of this series, compared with the function 

13.2.2 The Gibbs Phenomenon

A.A Michelson was a Prussian-born physicist who teamed with E.W Morley of Case-WesternReserve University to show that the postulated “ether,” a fluid which supposedly permeated all ofspace, had no effect on the speed of light Michelson also built a mechanical device for construct-ing a function from its Fourier coefficients In one test, he used eighty coefficients for the series

of f (x) = x on [−π, π] and noticed unexpected jumps in the graph near the endpoints At first,

he thought this was a problem with his machine It was subsequently found that this behavior ischaracteristic of the Fourier series of a function at a point of discontinuity In the early twentiethcentury, the Yale mathematician Josiah Willard Gibbs finally explained this behavior

To illustrate the Gibbs phenomenon, expand f in a Fourier series on [−π, π], where



= 0 = f (0).

FIGURE 13.12 The Gibbs phenomenon.

The Fourier series therefore converges to the function on(−π, π) This series is

y-axis as N increases This is the Gibbs phenomenon.

Postscript We will add two comments on the ideas of this section, the first practical and thesecond offering a broader perspective

1 Writing and graphing partial sums of Fourier series are computation intensive activities.Evaluating integrals for the coefficients is most efficiently done in MAPLE using theint command, and partial sums are easily graphed using the sum command to enter thepartial sum and then the plot command for the graph

2 Partial sums of Fourier series can be viewed from the perspective of orthogonal

pro-jections onto a subspace of a vector space (Sections 6.6 and 6.7) Let PC[−L, L] be

the vector space of functions that are piecewise continuous on[−L, L] and let S be the

subspace spanned by the functions

C0(x) = 1, C n (x) = cos(nπx/L) and S n (x) = sin(nπx/L) for n = 1, 2, · · · , N.

A dot product can be defined on P L [−L, L] by

f · g =

 L

−L

f (x)g(x) dx.

Using this dot product, these functions form an orthogonal basis for S If f if piecewise

continuous on[−L, L], the orthogonal projection of f onto S is

Compare these coefficients in the orthogonal projection with the Fourier coefficients of f on [−L, L] First

of f on [−L, L].

This broader perspective of Fourier series will provide a unifying theme when we considergeneral eigenfunction expansions in Chapter 15

In each of Problems 1 through 12, write the Fourier series

of the function on the interval and determine the sum of

the Fourier series Graph some partial sums of the series,

compared with the graph of the function

21 Carry out the program of Problem 20 for the function

of Problem 16

If f is piecewise continuous on [−L, L], we can represent f (x) at all but possibly finitely many

points of[−L, L] by its Fourier series This series may contain just sine terms, just cosine terms,

or both sine and cosine terms We have no control over this

If f is defined on the half interval [0, L], we can write a Fourier cosine series (containing just cosine terms) and a Fourier sine series (containing just sine terms) for f on [0, L].

FIGURE 13.13 Even extension of a function defined on [0, L].

for n = 0, 1, 2, · · · Notice that we can compute an strictly in terms of f on [0, L] The tion of g showed us how to obtain this cosine series for f , but we do not need g to compute the

construc-coefficients of this series

Based on these ideas, define the Fourier cosine coefficients of f on [0, L] to be the numbers

in which the an s are the Fourier cosine coefficients of f on [0, L].

By applying Theorem 13.1 to g, we obtain the following convergence theorem for cosine

series on[0, L].

THEOREM 13.2 Convergence of Fourier Cosine Series

Let f be piecewise smooth on [0, L] Then

1 If 0< x < L, the Fourier cosine series for f on [0, L] converges to

1

2( f (x+) + f (x−)).

2 At 0 this cosine series converges to f (0+).

3 At L this cosine series converges to f (L−). 

= 4e2(−1) n− 1

4+ n2π2 .

The cosine series for e 2x

This Fourier cosine series converges to e 2x

for 0≤ x ≤ 1 Figure 13.14 shows e 2x

and the fifthpartial sum of this cosine series 

0.2 0

x

1 0.4 0.6 0.8

FIGURE 13.14 Fifth partial sum of the cosine series in Example 13.9.

x y

FIGURE 13.15 Odd extension of a tion defined on [0, L].

func-The Fourier expansion of h on [−L, L] has only sine terms because h is odd on [−L, L] But h(x) = f (x) on [0, L], so this gives a sine expansion of f on [0, L] This suggests the following

is the Fourier sine series for f on [0, L].

Again, we have a convergence theorem for sine series directly from the convergence theoremfor Fourier series

THEOREM 13.3 Convergence of Fourier Sine Series

Let f be piecewise smooth on [0, L] Then

1 If 0< x < L, the Fourier sine series for f on [0, L] converges to

1

2( f (x+) + f (x−)).

2 At x = 0 and x = L, this sine series converges to 0. 

Condition (2) is obvious because each sine term in the series vanishes at x = 0 and at x = L,

regardless of the values of the function there

= 2nπ(1 − (−1) n e2)

4+ n2π2 The sine expansion of e 2x

This series converges to e 2xfor 0< x < 1 and to 0 at x = 0 and x = 1 Figure 13.16 shows

the function and the fortieth partial sum of this sine expansion 

4 6

x

2

1 0.6 0.8

In each of Problems 1 through 10, write the Fourier cosine

and sine series for f on the interval Determine the sum

of each series Graph some of the partial sums of these

12 Let f (x) be defined on [−L, L] Prove that f can

be written as a sum of an even function and an oddfunction on this interval

13 Determine all functions on[−L, L] that are both even

and odd

Term by term differentiation of a Fourier series may lead to nonsense

n (−1) n+1sin(nx)

for−π < x < π Differentiate this series term by term to get

∞



n=1

2(−1) n+1cos(nx).

This series does not converge on(−π, π). 

However, under fairly mild conditions, we can integrate a Fourier series term by term

THEOREM 13.4 Integration of Fourier Series

Let f be piecewise continuous on [−L, L], with Fourier series

on(−π, π) Term by term differentiation results in a series that does not converge on the interval.

However, we can integrate this series term by term:

n (−1) n+1(cos(nx) − (−1) n ). 



b n (−1) n

Upon substituting these expressions for A0, An, and Bn into the Fourier series for F (x), we obtain

the conclusion of the theorem 

Valid term by term differentiation of a Fourier series requires stronger conditions

THEOREM 13.5 Differentiation of Fourier Series

Let f be continuous on [−L, L] and suppose that f (−L) = f (L) Let f be piecewisecontinuous on[−L, L] Then the Fourier series of f on [−L, L] converges to f (x) on [−L, L]:

for−L ≤ x ≤ L Further, at each x in (−L, L) at which f(x) exists, the term by term derivative

of the Fourier series converges to the derivative of the function:

L (−a nsin(nπx/L) + b ncos(nπx/L)). 

The idea of a proof of this theorem is to begin with the Fourier series for f(x), noting that this series converges to f(x) at each point where f exists Use integration by parts to

relate the Fourier coefficients of f(x) to those for f (x), similar to the strategy used in proving

for−2 ≤ x ≤ 2 Only cosine terms appear in this series because x2 is an even function Now,

f(x) = 2x is continuous and f is twice differentiable for all x Therefore, for −2 < x < 2,

This can be verified by expanding 2x in a Fourier series on [−2, 2]. 

Fourier coefficients, and Fourier sine and cosine coefficients, satisfy an important set of

inequalities called Bessel’s inequalities.

THEOREM 13.6 Bessel’s Inequalities

2 The Fourier cosine coefficients an of g (x) on [0, L] satisfy

n πx

L

sin

n πx

L

sin

We will use Bessel’s inequality to derive an upper bound for the sum of a series Let f (x) = x2

on[−π, π] The Fourier series is



2π

3

2+

5−29

which is approximately 1.0823 Infinite series are generally difficult to sum, so it is sometimes

useful to be able to derive an upper bound 

With stronger assumptions than just existence of the integral over the interval, we can derive

an important equality satisfied by the Fourier coefficients of a function on[−L, L] or by the

Fourier sine or cosine coefficients of a function on[0, L] We will state the result for f (x)

defined on[−L, L].

THEOREM 13.7 Parseval’s Theorem

Let f be continuous on [−L, L] and let f be piecewise continuous Suppose that f (−L) =

f (L) Then the Fourier coefficients of f on [−L, L] satisfy

 L

−L

f (x)2d x. 

Proof Begin with the fact that, from the Fourier convergence theorem,

4

π

2+

∞



n=1

4

π

(−1) n 4n2− 1

A proof is outlined in Problem 6

(a) Write the Fourier series of f (x) on [−π, π] and

show that this series converges to f (x) on (−π, π).

(b) Use Theorem 12.5 to show that this series can be

integrated term by term

(c) Use the results of (a) and (b) to obtain a

trigono-metric series expansion ofx

−π f (t)dt on [−π, π].

2 Let f (x) = |x| for −1 ≤ x ≤ 1.

(a) Write the Fourier series for f on [−1, 1].

(b) Show that this series can be differentiated term by

term to yield the Fourier expansion of f(x) on

[−π, π].

(c) Determine f(x) and expand this function in a

Fourier series on[−π, π] Compare this result with

that of (b)

3 Let f (x) = x sin(x) for −π ≤ x ≤ π.

(a) Write the Fourier series for f on [−π, π]

(b) Show that this series can be differentiated term

by term and use this fact to obtain the Fourier

expansion of sin(x) + x cos(x) on [−π, π].

(c) Write the Fourier series for sin(x) + x cos(x) on

[−π, π] and compare this result with that of (b).

4 Let f (x) = x2for−3 ≤ x ≤ 3.

(a) Write the Fourier series for f on [−3, 3].

(b) Show that this series can be differentiated term by

term and use this to obtain the Fourier expansion

of 2x on [−3, 3].

(c) Expand 2x in a Fourier series on [−3, 3] and

compare this result with that of (b)

5 Let f and fbe piecewise continuous on[−L, L] Use

Bessel’s inequality to show that

This result is called Riemann’s lemma.

6 Prove Theorem 13.8 by filling in the details of the lowing argument Denote the Fourier coefficients of

fol-f (x) by lower case letters, and those of f(x) by upper

case Show that

Thus show by comparison that

∞



n=1

(|an | + |b n |)

converges Finally, show that

|a ncos(nπx/L) + bnsin(nπx/L)| ≤ |an | + |b n|and apply a theorem of Weierstrass on uniform conver-gence

A function f has period p if f (x + p) = f (x) for all x The smallest positive p for which this holds is the fundamental period of f For example sin (x) has fundamental period 2π.

The graph of a function with fundamental period p simply repeats itself over intervals of length p We can draw the graph for −p/2 ≤ x < p/2, then replicate this graph on p/2 ≤ x <

3 p /2, 3p/2 ≤ x < 5p/2, −3p/2 ≤ x < −p/2, and so on.

Now suppose f has fundamental period p Its Fourier series on [−p/2, p/2], with L = p/2,

Now look for numbers cnandδ n so that

a ncos(nω0x ) + b nsin(nω0x ) = c ncos(nω0x + δn ).

To solve for these constants, use a trigonometric identity to write this equation as

a ncos(nω0x) + b nsin(nω0x) = c ncos(nω0x) cos(δ n ) − c nsin(nω0x) sin(δ n ).

One way to satisfy this equation is to put

c ncos(δ n ) = a n and cnsin(δ n ) = −b n

If we square both sides of these equations and add the results, we obtain

When each an = 0, these equations enable us to write the phase angle form of the Fourier series of f (x) on [−p/2, p/2]:

This phase angle form is also called the harmonic form of the Fourier series for f (x)

on[−p/2, p/2] The term cos(nω0x + δn ) is called the nth harmonic of f , c n is the nth harmonic amplitude, and δ n is the nth phase angle of f

If f has fundamental period p, then in the expressions for the coefficients an and bn, we can

compute the integrals over any interval[α, α + p], since any interval of length p carries all of the information about a p-periodic function.

This means that the Fourier coefficients of p-periodic f can be obtained as

Since f is 3-periodic, and we are given an algebraic expression for f (x) only on [0, 3), we will use this interval to compute the Fourier coefficients of f That is, use p = 3 and α = 0 in the

preceding discussion We also haveω o = 2π/p = 2π/3.

The Fourier coefficients are

a0=23

 3 0

x2

d x = 6,

a n=23

 3 0

x2cos

8

4 6

0

FIGURE 13.17 f (x) in Example 13.16.

and

b n=23

 3 0

x2sin

also on[0, 3/2] But from Figure 13.17, f (x) = (x + 3)2on[−3/2, 0).

This Fourier series converges to

FIGURE 13.18 Amplitude spectrum of f

in Example 13.16.

The amplitude spectrum of a periodic function f is a plot of points (nω0, c n /2) for n =

1, 2, · · · , and also the point (0, |c0|/2) For the function of Example 13.14, this is a plot of

points(0, 3) and, for nonzero integer n, points



2n π

3 , 92n2π2

In Problems 1, 2, and 3, let f be periodic of period p.

1 If g is also periodic of period p, show that α f + βg is

periodic of period p, for any numbers α and β.

2 Letα be a positive number Show that g(t) = f (αt) has

period p /α and h(t) = f (t/α) has period αp.

3 If f is differentiable, show that fhas period p.

In each of Problems 4 through 12, find the phase angle

form of the Fourier series of the function and plot some

points of the amplitude spectrum Some of these functions

are specified by a graph

4 Let f (x)= x for 0≤ x <2, with fundamental period 2.

and let f has fundamental period 2.

6 Let f (x) = 3x2for 0≤ x < 4 and let f have

8 f (x) = cos(πx) for 0 ≤ x < 1 and f has fundamental

period 1

9 f has the graph of Figure 13.19.

x y

–1

1

–1 –2

FIGURE 13.21 f (x) in Problem 11, Section 13.5.

12 f has the graph of Figure 13.22.

y

x k

–2 –1

FIGURE 13.22 f (x) in Problem 12, Section 13.5.

There is a complex form of Fourier series that is sometimes used As preparation for this, recallthat, in polar coordinates, a complex number (point in the plane) can be written

z = x + iy = r cos(θ) + ir sin(θ)

where

r = |z| =x2+ y2andθ is an argument of z This is the angle (in radians) between the positive x− axis and the

line from the origin through(x, y), or this angle plus any integer multiple of 2π Using Euler’s formula, we obtain the polar form of z:

z = r[cos(θ) + i sin(θ)] = re i θ

Now

e i θ = cos(θ) + i sin(θ),

and by replacingθ with −θ, we get

e −iθ = cos(θ) − i sin(θ).

Solve these equations for cos(θ) and sin(θ) to obtain the complex exponential forms of the

This follows from Euler’s formula, since

e i x = cos(x) + i sin(x) = cos(x) − i sin(x) = e −ix Now let f be a piecewise smooth periodic function with fundamental period 2L To derive a complex Fourier expansion of f (x) on [−L, L], begin with the Fourier series of f (x) With



e i n ω0x + e −inω0x

+ bn 12i

Because of the periodicity of f , the integral defining the coefficients can be carried out over

any interval[α, α + 2L] of length 2L The Fourier convergence theorem applies to this complex

Fourier expansion, since it is just the Fourier series in complex form

0 0.5

FIGURE 13.23 Graph of f in Example 13.17.

All the terms sin(nπ) = 0 In−1

n=−∞, replace n with −n and sum from n = 1 to ∞, then combine

the two summations from 1 to∞ to write

This is the Fourier series for f (x) = x on [−1, 1]. 

The amplitude spectrum of a complex Fourier series of a periodic function is a graph of the

points(nω0, |d n|) Sometimes this graph is also referred to as a frequency spectrum.

In each of Problems 1 through 7, write the complex Fourier

series of f , determine the sum of the series, and plot some

points of the frequency spectrum

13.7 Filtering of Signals

A periodic signal f (t) of period 2L is sometimes filtered to cancel out or diminish unwanted

effects, or perhaps to enhance other effects We will briefly examine one way this is done

Suppose f has complex Fourier series

series is in general better behaved than the sequence of partial sums of the series itself If SN is

the N th partial sum of the series, this average has the form

0 f (t) dt exists, then σ N (t) → f (t) for any t at which f is continuous, a much

stronger result than holds for partial sums of Fourier series

Inserting the summation for Sk (t), we have



d n e πint/L

This is of the form of equation (13.15) with the Cesàro filter function

is called the sequence of filter factors for the Cesàro filter.

This “averaging” filter damps out the Gibbs effect in the convergence of a Fourier series To

observe this, let f have fundamental period 2 π, and, on [−π, π],

Figure 13.24 shows graphs of S10(t) and σ10(t), and Figure 13.25 graphs of S30(t) and σ30(t),

showing the Gibbs effect in the partial sums of the Fourier series, and this effect damped out inthe smoother Cesàro sums The Cesàro filter also damps out the higher frequency terms in theFourier series because 1− |n/N| tends to zero as n increases toward N.

0 0.5

The Hamming filter is named for Richard Hamming, who was a senior research scientist

at Bell Labs, and is defined by

In each of Problems 1 through 5, graph the function, the

fifth partial sum of its Fourier series on the interval, and

the fifth Cesàro sum, using the same set of axes Repeat this

process for the tenth and twenty-fifth partial sums Notice

in particular the graphs at points of discontinuity of the

function, where the Gibbs phenomenon appears

on [−2, 2], together with the fifth Cesàro sum, the

fifth Hamming and Gauss filtered partial sums, using

the same set of axes Repeat this with the tenth and

twenty-fifth partial sums

on [−2, 2], together with the fifth Cesàro sum, the

fifth Hamming and Gauss filtered partial sums, usingthe same set of axes Repeat this with the tenth andtwenty-fifth partial sums

C H A P T E R 14

The Fourier Integral and Transforms

F O U R I E R T R A N S F O R M F O U R I E R C O S I N E

If f (x) is defined for −L ≤ x ≤ L, we may be able to represent f (x) as a Fourier series on this interval However, Fourier series are tied to intervals If f is defined over the entire line and is

not periodic, then the idea of a Fourier series representation is replaced with the idea of a Fourierintegral representation, in which the role of∞

n=0is played by∞

0

We will give an informal argument to suggest the form that the Fourier integral should take

Assume that f is absolutely integrable, which means that ∞

−∞| f (x)| dx converges We also assume that f is piecewise smooth on every interval [−L, L].

Write the Fourier series of f (x) on an arbitrary interval [−L, L] With the formulas for the

coefficients included, this series is

+

1

.

We want to let L → ∞ to obtain a representation of f (x) over the entire real line It is not clear what this quantity approaches, if anything, as L→ ∞, so we will rewrite some terms First,let