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Class notes Advanced Mathematics in Chemical Engineering (CH050P)
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This document provides useful notes in the course Advanced Mathematics in Chemical Engineering, with sample problems and solutions.
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1. 1 Fourier SeriesFourier Series -
series of cosine and sine terms and arise in the Exercise: Find the resulting period of :
important practical task of representing general a) cos 12×7 c) cos 13×7
periodic functions b) sin (2×7 d) sin 13×7
that involve
very important tool in solving problems
-
differential equations Familiar periodic functions are the sine and cosine functions
ordinary and partial
theory is rather complicated but the application From flxtp) f- (x)
we can say that
-
=
,
of these series is simple f- 1×+2 p) = f- ( ( xtp) 1- p] = flxtp) = flx)
-
more universal than Taylor series etc .
.
and for any integer n
-
many discontinuous periodic functions of practical flxtnp ) -11×7 =
interest can be developed in Fourier series but do
,
not have Taylor series representations Hence , 2p 3p 4p
. . ,
. . . are also periods of f- 1×1 .
Furthermore if flx ) and , glx) have period p ,
then the func
Periodic Functions h(✗ I = aflxltbglx) la b constant)
,
A function flx) is called periodic if it is defined also has the period p .
for all real ✗ and if there is some positive number p such f- 1×+2 p) = fllxtp) 1- p] = flxtp) = flx)
that
flxtp )=f / X) It a periodic function flx) has the smallest period pl > 01 ,
th
This number p is called a period of -11×1 The graph of
. is called the fundamental period of f- 1×7 .
such a function is obtained by periodic repetition of its For cos ✗ and sin × the fundamental period
,
is 21T .
for cos 12×7 an
graph in any interval of length p Periodic phenomena . and sin 12×7 it is I. and
,
so on .
functions have many applications A function without a fundamental period is f- constant .
Examples of periodic functions with period p=2ñ
① 1. 2 Fourier series
-
trend is
repeating
-
looking at to
3¥ .
Trigonometric Series
A number of functions of period p= 21T can be
there is a basic pattern represented in terms of simple functions .
being repeated all The series that will arise in this connection will be of th
throughout form
-
solving for p Ao t A. COS ✗ + b. sin ✗ + Az cos 2×-1 basin2x . . .
Where
f-¥ )
ao ai bi.bz real constants
3¥ 21T are
p= = . . . .
.. .
- , , . .
Such a series is called a
trigonometric series and ,
the An a
Using the
②
bn are called coefficients of the series .
summation sig
this series may be written as
ao t Ein:( ancosnx + bnsinnx )
The set of functions from which the series has been made up
is often called the trigonometric system .
Each term of the series has the period p= 21T If the series
.
converges its .
sum will be a function of period 21T .
trigonometric series can be used for representing
The any practic
reference point : 0 to 21T important periodic function f. simple or complicated of any perio
-
.
-
shape of plot is repeated This series will then be called the Fourier series of f.
-
the basic unit w/ c is repeated is
equal to 21T Wtc is the
period of the function sin 1×7 .
, Fourier Series Solve for Ao :
# f. ]
"
Arise from the task of
given
flx) in terms of cosine and sine functions
representing a
.
periodic function
go =
fix) dx =
¥/ 1- ) DX
K + (K ) DX
periodic
ao=¥ /1-1<4×1;) k( I ;) ) 2¥ (-10+1171-(11--
Let fix) be a function of period 21T and is integrable + ✗ =
over a period .
let f- 1×1 be represented by a trigonometric series
-11×7=90 % ( an cos nx + bnsinnx )
# 1-
+ ,
do =
IT -11T ]
The assumption is that this series and has flx)
converges as
its sum .
do =D
The coefficients of the trigonometric series are also termed as
Euler Formulas or Fourier coefficients .
Solve for An:
ao=¥f ¥ f.
"
flx)d✗ An = flx) cos nxdx
,
¥/ )
"
An =
flx) cos nxdx n = 1.2 . . . .
=
1- K) cos nxdx + (K) cos nxdx
bn=¥f fix) sin nxdx n -
- i. a. .
. .
=n¥[ sin nxl! ,
+ sin nx
/F)
¥ (0-0-0-10)
sin nx =D at -
IT , 0,1T for a
=
N = 1 2,
, . . .
An =D
Solve for bn :
bn =
# fifth sin nxdx
f. ]
"
=
¥ c-Hsin nxdx + (Hsin nxdx
=
¥ / cosnxl? ,
-
cos nx / F)
=n¥( cos 0 -
cost MT) -
cos Init) + cos 0
]
¥/ ] cost A) cos (a)
-
=
= -
Zcoslnit) -11
, cos 101=1
=n¥[ 2-2 cos 1nA ]
=
2¥11 -
cos 1nA ]
Analyze bn for first few n's :
n cos NIT [1- cos CMT)]
l -
l l -
C- 1) = 2
2 1 I - I = 0
3 -
l l -
C- 1) =
2
4 l l -
l = 0
Summarizing :
{ n's
-
l odd
cos (nlt) =
1 even n 's
?⃝
1- cos MT ( ) =
{? odd
even n's
n 's
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