Ebook Advanced Engineering Mathematics Gujarat Technological University 2016 PDF Full Chapter PDF
Ebook Advanced Engineering Mathematics Gujarat Technological University 2016 PDF Full Chapter PDF
Ebook Advanced Engineering Mathematics Gujarat Technological University 2016 PDF Full Chapter PDF
Ravish R Singh
Vice Principal
Shree L R Tiwari College of Engineering
Thane, Maharashtra
Mukul Bhatt
Assistant Professor
Department of Humanities and Sciences
Thakur College of Engineering and Technology
Mumbai, Maharashtra
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This book is designed for the 2nd year GTU engineering students pursuing the
course Advanced Engineering Mathematics, SUBJECT CODE: 2130002 in their
3rd Semester. It covers the complete GTU syllabus for the course on Advanced
Engineering Mathematics, which is common to all the engineering branches.
Objective
The crisp and complete explanation of topics will help students easily understand the
basic concepts. The tutorial approach (i.e., teach by example) followed in the text will
enable students develop a logical perspective to solving problems.
Features
Each topic has been explained from the examination point of view, wherein the theory
is presented in an easy-to-understand student-friendly style. Full coverage of concepts
is supported by numerous solved examples with varied complexity levels, which is
aligned to the latest GTU syllabus. Fundamental and sequential explanation of topics
are well aided by examples and exercises. The solutions of examples are set follow-
ing a ‘tutorial’ approach, which will make it easy for students from any background
to easily grasp the concepts. Exercises with answers immediately follow the solved
examples enforcing a practice-based approach. We hope that the students will gain
logical understanding from solved problems and then reiterate it through solving simi-
lar exercise problems themselves. The unique blend of theory and application caters to
the requirements of both the students and the faculty. Solutions of GTU examination
questions are incorporated within the text appropriately.
xii Preface
Highlights
∑ Crisp content strictly as per the latest GTU syllabus of Advanced Engineering
Mathematics (Regulation 2014)
∑ Comprehensive coverage with lucid presentation style
∑ Each section concludes with an exercise to test understanding of topics
∑ Solutions of GTU examination papers from 2012 to 2014 present appropriately
within the chapters
∑ Solution to Summer and Winter 2015 GTU question papers placed at the end of the
book
∑ Rich exam-oriented pedagogy:
Non GTU solved examples within chapters: 531
Solved GTU questions within chapters: 105
Unsolved exercises: 571
Chapter Organization
The content spans the following six chapters which wholly and sequentially cover
each module of the syllabus.
Chapter 1 introduces Some Special Functions.
Chapter 2 discusses Fourier Series and Fourier Integral.
Chapter 3 presents Ordinary Differential Equations and Applications.
Chapter 4 covers Series Solution of Differential Equations.
Chapter 5 deals with Laplace Transforms and Applications.
Chapter 6 presents Partial Differential Equations and Applications.
Acknowledgements
We are grateful to the following reviewers who reviewed various chapters of the script
and generously shared their valuable comments:
Manokamna Agrawal Silver Oak College of Engineering and Technology,
Ahmedabad, Gujarat
JC Prajapati Marwadi Education Foundation Group of
Institutions, Rajkot
Shailesh Patel SPB Patel Engineering College, Gujarat
Kinnari Sutaria AD Patel Institute of Engineering, Karamsad
Prakash Kumar Patel Babaria Institute of Technology, Vadodara, Gujarat
We would also like to thank all the staff at McGraw Hill Education (India), especially
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Baveja, Anuj Shrivastava and Atul Gupta for coordinating with us during the editorial,
copyediting, and production stages of this book.
Preface xiii
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ROADMAP TO THE SYLLABUS
This text is useful for
GO TO
CHAPTER 1: Introduction to Some Special Functions
GO TO
CHAPTER 2: Fourier Series and Fourier Integral
GO TO
CHAPTER 3: Ordinary Differential Equations and Applications
xvi Roadmap to the Syllabus
GO TO
CHAPTER 4: Series Solution of Differential Equations
GO TO
CHAPTER 5: Laplace Transforms and Applications
GO TO
CHAPTER 6: Partial Differential Equations and Applications
CHAPTER
1
Introduction to Some
Special Functions
Chapter Outline
1.1 Introduction
1.2 Gamma Function
1.3 Beta Function
1.4 Bessel Function
1.5 Error Function and Complementary Error Function
1.6 Heaviside’s Unit Step Function
1.7 Pulse of Unit Height and Duration Function
1.8 Sinusoidal Pulse Function
1.9 Rectangle Function
1.10 Gate Function
1.11 Dirac’s Delta Function
1.12 Signum Function
1.13 Sawtooth Wave Function
1.14 Triangular Wave Function
1.15 Half-Wave Rectified Sinusoidal Function
1.16 Full-Wave Rectified Sinusoidal Function
1.17 Square-Wave Function
1.1 INTRODUCTION
There are some special functions which have importance in mathematical analysis,
functional analysis, physics, or other applications. In this chapter, we will study
different special functions such as gamma, beta, Bessel, error, unit step, Dirac delta
functions, etc. The study of these functions will help in solving many mathematical
problems encountered in advanced engineering mathematics.
1.2 Chapter 1 Introduction to Some Special Functions
denoted by n .
Hence, n e x x n 1 dx , n > 0
0
(iv) n1 n
sin n
(v) 1
2
B(m, n) is also known as Euler’s integral of the first kind. The beta function can also
be defined by
B(m, n) 2 2 sin 2 m 1
x cos2 n 1
x dx
0
1.5 Error Function and Complementary Error Function 1.3
xn x2 x4
1
2n n 1 2(n 2) 2 4(2 n 2)(2 n 4)
defined by –3 –2 –1 0 1 2 3
–0.25
2 2
erfc( x ) e t dt –0.5
x
–0.75
where x may be a real or complex variable. –1
Relation between error function and the com- Fig. 1.2 Error function
plementary error function is given by
2 2
erfc( x ) e t dt
x
2 t2 2 x t2
e dt e dt
0 0
2
erf ( x )
2
1 erf ( x )
f (t ) a sin at 0 t
a
0 t
a
Fig. 1.6 Sinusoidal pulse
function
1.9 RECTANGLE FUNCTION
[Summer 2013]
The rectangle function (Fig. 1.7) is defined by
f (t ) 1 a t b
0 otherwise
Fig. 1.7 Rectangle function
In terms of unit step function, the rectangle function
can be expressed as
f(t) = u(t – a) – u(t – b)
1.6 Chapter 1 Introduction to Some Special Functions
0 t
Fig. 1.9 Any function
The area enclosed by the function f(t) and the t-axis is f(t)
given by
0
f (t ) dt f (t ) dt f ( t ) dt f (t ) dt
0
1
0 dt 0
0
1
t 0
1
As 0, the height of the rectangle increases indefinitely in such a way that its area
is always equal to 1. This function is known as Dirac’s delta function or unit impulse
function and is denoted by (t).
(t ) lim f (t )
0
The displaced (delayed) delta or displaced impulse
function (t – a) (Fig. 1.10) represents the function
(t) displaced by a distance a to the right.
0 t a Fig. 1.10 Delayed function
1
(t a ) lim f (t ) lim a t a
0 0
0 a t
1.15 1.7
(iii) f (t ) (t ) t f (0)
(iv) f (t ) (t a ) dt f (a)
Chapter Outline
2.1 Introduction
2.2 Fourier Series
2.3 Trigonometric Fourier Series
2.4 Fourier Series of Functions of Any Period
2.5 Fourier Series of Even and Odd Functions
2.6 Half-Range Fourier Series
2.7 Fourier Integral
2.1 INTRODUCTION
Fourier series is used in the analysis of periodic functions. Many of the phenomena
studied in engineering and sciences are periodic in nature, e.g., current and voltage
in an ac circuit. These periodic functions can be analyzed into their constituent
components by a Fourier analysis. The Fourier series makes use of orthogonality
relationships of the sine and cosine functions. It decomposes a periodic function into a
sum of sine-cosine functions. The computation and study of Fourier series is known
as harmonic analysis. It has many applications in electrical engineering, vibration
analysis, acoustics, optics, signal processing, image processing, etc.
n x n x
f ( x) a0 an cos bn sin
n 1 l n 1 l
This series is known as a trigonometric Fourier series or simply, a Fourier series. For
example, a square function can be constructed by adding orthogonal sine components
(Fig. 2.1).
2.4 Fourier Series of Functions of Any Period 2.3
f (x)
f (x)
x Square function
O
Determination of a0
Integrating both the sides of Eq. (2.1) w.r.t. x in the interval (c, c + 2l),
c 2l c 2l c 2l n x c 2l n x
f ( x )dx a0 dx an cos dx bn sin dx
c c c
n 1 l c
n 1 l
= a0(c + 2l – c) + 0 + 0
= 2la0
1 c 2l
Hence, a0 f ( x )dx …(2.2)
2l c
2.4 Chapter 2 Fourier Series and Fourier Integral
Determination of an
n x
Multiplying both the sides of Eq. (2.1) by cos and integrating w.r.t. x in the
l
interval (c, c + 2l),
c 2l n x c 2l n x c 2l n x n x
f ( x ) cos dx a0 cos dx an cos cos dx
c l c l c
n 1 l l
c 2l n x n x
bn sin cos dx
c
n 1 l l
0 lan 0
= l an
1 c 2l n x
Hence, an f ( x ) cos dx …(2.3)
l c l
Determination of bn
n x
Multiplying both the sides of Eq. (2.1) by sin and integrating w.r.t. x in the
interval (c, c + 2l ), l
c 2l n x c 2l n x c 2l n x n x
f ( x )sin dx a0 sin dx an cos sin dx
c l c l c
n 1 l l
c 2l n x n x
bn sin sin dx
c
n 1 l l
0 0 lbn
= l bn
1 c 2l n x
Hence, bn f ( x )sin dx …(2.4)
l c l
The formulae (2.2), (2.3), and (2.4) are known as Euler’s formulae which give the values
of coefficients a0, an, and bn. These coefficients are known as Fourier coefficients.
Corollary 1 When c = 0 and 2l = 2
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1 2
where a0 f ( x ) dx
2 0
1 2
an f ( x ) cos nx dx
0
1 2
bn f ( x )sin nx dx
0
2.4 Fourier Series of Functions of Any Period 2.5
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1
where a0 f ( x ) dx
2
1
an f ( x ) cos nx dx
1
bn f ( x )sin nx dx
Corollary 3 When c = 0
n x n x
f ( x) a0 an cos bn sin
n 1 l n 1 l
1 2l
where a0 f ( x ) dx
2l 0
1 2l n x
an f ( x ) cos dx
l 0 l
1 2l n x
bn f ( x )sin dx
l 0 l
Corollary 4 When c = – l
n x n x
f ( x) a0 an cos bn sin
n 1 l n 1 l
1 l
where a0 f ( x )dx
2l l
1 l n x
an f ( x ) cos dx
l l l
1 l n x
bn f ( x )sin dx
l l l
Example 1
Find the Fourier series of f (x) = x in the interval (0, 2 ).
2.6 Chapter 2 Fourier Series and Fourier Integral
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1 2
a0 f ( x )dx
2 0
1 2
x dx
2 0
2
1 x2
2 2
0
2
1 4
2 2
1 2
an f ( x ) cos nx dx
0
1 2
x cos nx dx
0
2
1 sin nx cos nx
x (1)
n n2 0
1 cos 2 n cos 0
sin 2 n sin 0 0
n2 n2
0 cos 2 n cos 0 1
1 2
bn f ( x )sin nx dx
0
1 2
x sin nx dx
0
2
1 cos nx sin nx
x (1)
n n2 0
1 cos 2 n
2 sin 2 n sin 0 0
n
2
cos 2 n 1
n
1
Hence, f ( x) 2 sin nx
n 1n
2.4 Fourier Series of Functions of Any Period 2.7
1 1
x 2 sin x sin 2 x sin 3 x
2 3
Example 2
Find the Fourier series of f (x) = x2 in the interval 0, 2 and, hence,
2
1 1 1
deduce that
12 12 22 32
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1 2
a0 f ( x )dx
2 0
1 2
x 2 dx
2 0
2
1 x3
2 3
0
3
1 8
2 3
2
4
3
1 2
an f ( x ) cos nx dx
0
1 2
x 2 cos nx dx
0
2
1 sin nx cos nx sin nx
x2 2x 2
n n2 n3 0
1 cos 2 n
4 sin 2 n sin 0 0
n2
1 4
cos 2 n 1
n2
4
n2
2.8 Chapter 2 Fourier Series and Fourier Integral
1 2
bn f ( x )sin nx dx
0
1 2
x 2 sin nx dx
0
2
1 cos nx sin nx cos nx
x2 2x 2
n n2 n3 0
1 2 cos 2 n cos 2n
n cos 0
4 2 3
2
n n n3
2
1 4
cos 2 n cos 0 1
n
4
n
2
4 1 1
Hence, f ( x) 4 2
cos nx 4 sin nx
3 n 1n n 1n
2
4 1 1 1
x2 4 2
cos x 2
cos 2 x cos3 x ... (1)
3 1 2 32
Putting x = in Eq. (1),
2
2 4 1 1 1
4 2
cos 2
cos 2 cos 3
3 1 2 32
2
4 1 1 1
4
3 12 22 32
2
1 1 1
12 12 22 32
Example 3
1
Find the Fourier series of f ( x ) ( x ) in the interval (0, 2 ).
2
1 1 1
Hence, deduce that 1 [Winter 2013]
4 3 5 7
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
2.4 Fourier Series of Functions of Any Period 2.9
1 2
a0 f ( x )dx
2 0
1 2 1
( x )dx
2 0 2
2
1 x2
x
4 2
0
1 2 2
(2 2 )
4
0
1 2
an f ( x ) cos nx dx
0
1 2 1
( x ) cos nx dx
0 2
2
1 sin nx cos nx
( x) ( 1)
2 n n2 0
1 cos 2 n cos 0
2
2 n n2
0 cos 2 n cos 0 1
1 2 1
bn ( x )sin nx dx
0 2
2
1 cos nx sin nx
( x) ( 1)
2 n n2 0
1 cos 2 n cos 0
( ) sin 2 n sin 0 0
2 n n
1
cos 2 n cos 0 1
2 n n
1
n
1
Hence, f ( x) sin nx
n 1n
1 1 1 1 1
( x) sin x sin 2 x sin 3 x sin 4 x sin 5 x
2 2 3 4 5
1 1
sin 6 x sin 7 x ...(1)
6 7
2.10 Chapter 2 Fourier Series and Fourier Integral
Example 4 2
x
Obtain the Fourier series of f ( x ) in the interval 0 x 2 .
2
Hence, deduce that
2
1 1 1
2 2 [Winter 2014]
12 1 2 32
Solution
The Fourier series of f(x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1 2
a0 f ( x ) dx
2 0
2
1 2 x
dx
2 0 2
2
1 ( x )3
8 3 0
1
( 3 3
)
24
2
12
1 2
an f ( x ) cos nx dx
0
2
1 2 x
cos nx dx
0 2
2.4 Fourier Series of Functions of Any Period 2.11
2
1 2 sin nx cos nx sin nx
( x) 2( x )( 1) 2( 1)( 1)
4 n n2 n3 0
1 cos 2 n cos 0
2 2 sin 2 n sin 0 0
4 n2 n2
1 2 2
2
cos 2 n cos 0 1
4 n n2
1
n2
1 2
bn f ( x ) sin nx dx
0
2
1 2 x
sin nx dx
0 2
2
1 2 cos nx sin nx cos nx
( x ) 2( x )( 1) 2( 1)( 1)
4 n n2 n3 0
2 2
x 1 1 1
2
cos x 2
cos 2 x cos 3 x ....(1)
2 12 1 2 32
Putting x = in Eq. (1),
2
1 1 1
0 2 2
12 1 2 32
2
1 1 1
2 2
12 1 2 32
2.12 Chapter 2 Fourier Series and Fourier Integral
Example 5
3x2 6x 2 2
Find the Fourier series of f ( x ) in the interval (0, 2 )
12
2
1 1
Hence, deduce that 1 2
6 2 32
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1 2
a0 f ( x )dx
2 0
1 2 3x2 6x 2 2
dx
2 0 12
2
1 x3 x2 2
3 6 2 x
24 3 2
0
3 2
1 8 4 3
3 6 4
24 3 2
0
1 2
an f ( x ) cos nx dx
0
1 2 3x2 6x 2 2
cos nx dx
0 12
2
1 sin nx cos nx sin nx
(3 x 2 6x 2 2
) (6 x 6 ) 6
12 n n2 n3 0
1 cos 2 n cos 0
(6 ) 2
( 6 ) sin 2 n sin 0 0
12 n n2
1 6 6
2
cos 2n cos 0 1
12 n n2
1
n2
1 2
bn f ( x )sin nx dx
0
2.4 Fourier Series of Functions of Any Period 2.13
1 2 3x2 6x 2 2
sin nx dx
0 12
2
1 cos nx sin nx cos nx
(3 x 2 6x 2 2
) (6 x 6 ) 6
12 n n2 n3 0
1
Hence, f ( x) 2
cos nx
n 1 n
2 2
3x 6x 2 1 1
cos x cos 2 x cos 3 x … (1)
12 22 32
Putting x = 0 in Eq. (1),
2
1 1
cos 0 2
cos 0 cos 0
6 2 32
1 1
1 2
2 32
Example 6
Find the Fourier series of f(x) = e–x in the interval (0, 2 ).
1 ( 1)n
Hence, deduce that . [Summer 2014]
2 sinh n 2 n2 1
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1 2
a0 f ( x )dx
2 0
1 2
e x dx
2 0
1 x
2
e
2 0
2.14 Chapter 2 Fourier Series and Fourier Integral
2
e e0
2
2
1 e
2
1 2
an f ( x ) cos nx dx
0
1 2 x
e cos nx dx
0
2
x
1 e
( cos nx n sin nx )
n2 1 0
2
1 e 1
2
( cos 2 n ) 2
( cos 0) sin 2 n sin 0 0
n 1 n 1
1 2
(1 e ) cos 2 n cos 0 1
(n2 1)
1 2
bn f ( x )sin nx dx
0
1 2 x
e sin nx dx
0
2
x
1 e
( sin nx n cos nx )
n2 1 0
2
1 e 1
2
( n cos 2 n ) 2
( n cos 0) sin 2 n sin 0 0
n 1 n 1
n 2
2
(1 e ) cos 2 n cos 0 1
(n 1)
2 2 2
1 e 1 e 1 1 e n
Hence, f ( x ) 2
cos nx 2
sin nx … (1)
2 n 1n 1 n 1n 1
Putting x = in Eq. (1),
2 2
1 e 1 e ( 1)n
f( ) 2
cos n ( 1)n , sin n 0
2 n 1n 1
2 2
1 e 1 e 1 ( 1)n
e
2 2 n 2 n2 1
2
1 e ( 1)n
n 2 n2 1
2.4 Fourier Series of Functions of Any Period 2.15
( 1)n
2
e (1 e ) n 2 n2 1
( 1)n
e e n 2 n2 1
1 ( 1)n
Hence,
2 sinh n 2 n2 1
Example 7
Find the Fourier series of f ( x ) 1 cos x in the interval (0, 2 ). Hence,
1 1
deduce that 2
.
2 n 1 4n 1
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
x
f ( x) 1 cos x 2 sin
2
1 2
a0 f ( x )dx
2 0
1 2 x
2 sin dx
2 0 2
2
2 x
2 cos
2 2 0
2
( 2 cos 2 cos 0)
2
2 2
[ cos 1, cos 0 1]
1 2
an f ( x ) cos nx dx
0
1 2 x
2 sin cos nx dx
0 2
2 2 2n 1 2n 1
sin x sin x dx
2 0 2 2
2.16 Chapter 2 Fourier Series and Fourier Integral
2
2 2 2n 1 2 2n 1
cos x cos x
2 2n 1 2 2n 1 2 0
2 2 2 cos 0 2 2 cos 0
cos(2 n ) cos(2 n )
2 2n 1 2n 1 2n 1 2n 1
2 4 4
cos(2 n 1) cos(2 n 1) 1,cos 0 1
2 2n 1 2n 1
4 2 1
2
4n 1
1 2
bn f ( x )sin nx dx
0
1 2 x
2 sin sin nx dx
0 2
2 2 2n 1 2n 1
cos x cos x dx
2 0 2 2
2
2 2 2n 1 2 2n 1
sin x sin x
2 2n 1 2 2n 1 2 0
0 sin(2 n 1) sin(2n 1) sin 0 0
2 2 4 2 1
Hence, f ( x) 2
cos nx … (1)
n 1 4n 1
Putting x = 0 in Eq. (1),
2 2 4 2 1
f (0 ) 0 2
n 1 4n 1
1 1
2 2
n 1 4n 1
Example 8
Find the Fourier series of f ( x ) 1 0 x
2 x 2
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
2.4 Fourier Series of Functions of Any Period 2.17
1 2
a0 f ( x )dx
2 0
1 2
( 1)dx 2 dx
2 0
1 2
x 0 2x
2
1
( ) (4 2 )
2
1
2
1 2
an f ( x ) cos nx dx
0
1 2
( 1) cos nx dx 2 cos nx dx
0
2
1 sin nx sin nx
2
n 0 n
0 sin 2 n sin n sin 0 0
1 2
bn f ( x )sin nx dx
0
1 2
( 1)sin nx dx 2 sin nx dx
0
2
1 cos nx 2 cos nx
n 0 n
1 cos n cos 0 2 cos 2 n 2 cos n
n n n n
3
[( 1)n 1] cos 2 n cos 0 1, cos n ( 1)n
n
Hence, f ( x ) 1 3 ( 1)n 1
sin nx
2 n 1 n
1 3 2 2
2 sin x sin 3 x sin 5 x
2 3 5
1 6 1 1
sin x sin 3 x sin 5 x
2 3 5
2.18 Chapter 2 Fourier Series and Fourier Integral
Example 9
Find the Fourier series of f (x) = x + x2 in the interval (– , ), and
hence, deduce that
2
1 1 1
(i)
12 12 22 32
2
1 1 1
(ii) 2 2
[Winter 2012]
6 1 2 32
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1
a0 f ( x )dx
2
1
( x x 2 )dx
2
1 x2 x3
2 2 3
2 3 2 3
1
2 2 3 2 3
3
1 2
2 3
2
3
1
an f ( x ) cos nx dx
1
( x x 2 ) cos nx dx
4( 1)n
2
cos n ( 1)n
n
1
bn f ( x )sin nx dx
1
( x x 2 )sin nx dx
2
( 1)n ( 1)n
Hence, f ( x) 4 cos nx 2 sin nx
3 n 1 n2 n 1 n
2
1 1 1
x x2 4 2
cos x 2
cos 2 x cos3 x
3 1 2 32
1 1 1
2 sin x sin 2 x sin 3 x ...(1)
1 2 3
(i) Putting x = 0 in Eq. (1),
2
1 1 1
0 4 2
cos 0 2
cos 0 cos 0
3 1 2 32
2
1 1 1
2 2
12 1 2 32
(ii) Putting x = in Eq. (1),
2
2 1 1 1
4 2
cos 2
cos 2 cos 3
3 1 2 32
2
1 1 1
4 2 2 … (2)
3 1 2 32
2.20 Chapter 2 Fourier Series and Fourier Integral
Example 10
Find the Fourier series of f (x) = x + |x| in the interval – < x < .
[Winter 2014]
Solution
The Fourier series of f (x) with period 2 is given by
f ( x) a0 an cos nx bn sin nx
n 1 n 1
1
a0 f ( x ) dx
2
1
(x | x |) dx
2
1
x dx | x | dx
2
a a
1 f ( x ) dx 2 f ( x ) dx, if f ( x ) is even function
0 2 | x | dx a 0
2 0
0, if f ( x ) is odd function
1
x dx
0
1 x2
2 0
2
1
2
2
2.4 Fourier Series of Functions of Any Period 2.21
1
an f ( x ) cos nx dx
1
( x | x |) cos nx dx
1
x cos nx dx | x | cos nx dx
2 sin nx cos nx
x (1)
n n2 0
2 cos n cos 0
[ sin n sin 0 0]
n2 n2
2
( 1)n 1 cos n ( 1)n , cos 0 1
n2
1
bn f ( x ) sin nx dx
1
( x | x |)sin nx dx
1
x sin nx dx | x | sin nx dx
2 cos nx sin nx
x (1)
n n2 0
2 cos n
[ sin n sin 0 0]
n
2
( 1)n cos n ( 1)n
n
2 ( 1)n 1 ( 1)n
Hence, f ( x ) cos nx 2 sin nx
2 n 1 n2 n 1 n
2 2 2 2
x |x| 2
cos x 2
cos3 x cos 5 x
2 1 3 52
1 1 1
2 sin x sin 2 x sin 3 x
1 2 3
Another random document with
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sein werden. Der erste Finder des Rechter Goldes, Bergverwalter
Jung aus Bliesenbach, hat dasselbe auch bei Montenau
nachgewiesen und hat in dem ganzen Gelände, in dem die
kleinen Hügel vorkommen, Mutungen auf Gold bei der
Bergbehörde eingelegt. In der Nähe der vielen Hügel in der
Umgebung der Rechter Mühle sah ich viele schachtförmige
Vertiefungen (alte Pingen) mit hohem Baumwuchs bestanden, die
sicherlich von früherm Bergbau herstammen. Herr Jung hat die
Gegend schon in den 70er Jahren durchsucht und vermutet, daß
die vielen Hügel Halden von Seifen seien. v. Dechen, mit dem
Jung damals in Briefwechsel trat, war anderer Ansicht, indem er
ihm unter dem 27. Januar 1876 schrieb: „Die kleinen Hügel von
Montenau habe ich hier und westlich von Recht an der belgischen
Grenze gesehen. Es sind wohl keine Halden. Mit Halden von
Goldwäschen, die ich bei Goldberg, Löwenberg, Bunzlau in
Schlesien gesehen, haben dieselben keine Aehnlichkeit.
Ebensowenig weist der Bestand derselben auf irgend ein sonst
bekanntes Goldvorkommen hin. Ich habe sie für alte Grabhügel
gehalten, obgleich bei einigen, die aufgeworfen worden sind,
nichts gefunden worden ist.“ Derselben Ansicht, daß sie alte
Grabhügel seien, war früher auch der Altertumsforscher Dr. Esser
in Malmedy, er hält sie aber jetzt für Halden von Erzseifen, wie er
mir unlängst mitteilte. Dafür spricht ihre geringe Ausdehnung —
sie sind kaum 1 m hoch — und namentlich der Umstand, daß sie
nur in Thalgründen unmittelbar an Bächen (Amelsbach,
Emmelsbach, Rechter Bach u. s. w.) vorkommen. Es mögen im
Altertum hier viele Arbeiter beschäftigt und eine größere
Gewinnung von Metall im Gange gewesen sein. Nun, die Römer
verfügten sicherlich über ganz billige Arbeitskräfte und es stand
damals das Gold in weit höherm Werte als heute. Ob gegenwärtig
noch eine Rentabilität zu erzielen ist, werden die weiteren
Versuchsarbeiten ergeben.“
[24] Die Ortschaft wird seit 888 in Urkunden genannt: Nova villa,
la neuve ville, auf wallonisch: li nouve veie, dann: Lignonville, im
11. Jahrhundert: Langeneuville.
VI.
Die Bewohner von Malmedy und die
Sprachen-Verhältnisse in der
Wallonie.