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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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Engineering Mathematics – III
Unit 1: Fourier Series ……………………………1-25
Unit 2: Fourier Transforms…………………………26-38
Unit 3: Applications Of Partial Differential Equations….39-48
Unit 4: Curve Fitting & Optimization ……………….49-66
Unit 5: Numerical Methods I………………………….67-83
Unit 6: Numerical Methods II …………….………….84-105
Unit 7: Numerical Methods III……………………….106-119
Unit 8: Difference Equations & Z – Transforms……120-136
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UNIT- I
FOURIER SERIES
CONTENTS:
Introduction
Periodic function
Trigonometric series and Euler’s formulae
Fourier series of period 2
Fourier series of even and odd functions
Fourier series of arbitrary period
Half range Fourier series
Complex form of Fourier series
Practical Harmonic Analysis0
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UNIT- I
FOURIER SERIES
DEFINITIONS :
A function y = f(x) is said to be even, if f(-x) = f(x). The graph of the even function is always symmetrical about the y-axis.
A function y=f(x) is said to be odd, if f(-x) = - f(x). The graph of the odd function is always symmetrical about the origin.
For example, the function f(x) = x in [-1,1] is even as f(-x) = xx = f(x) and the
function f(x) = x in [-1,1] is odd as f(-x) = -x = -f(x). The graphs of these functions are shown below :
Graph of f(x) = x Graph of f(x) = x
Note that the graph of f(x) = x is symmetrical about the y-axis and the graph of f(x) = x
is symmetrical about the origin.
1. If f(x) is even and g(x) is odd, then
h(x) = f(x) x g(x) is odd h(x) = f(x) x f(x) is even h(x) = g(x) x g(x) is even
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For example,1. h(x) = x2 cosx is even, since both x2 and cosx are even functions2. h(x) = xsinx is even, since x and sinx are odd functions3. h(x) = x2 sinx is odd, since x2 is even and sinx is odd.
2. If f(x) is even, then
a
a
a
dxxfdxxf0
)(2)(
3. If f(x) is odd, then
a
a
dxxf 0)(
For example,
a
a
a
xdxxdx0
,cos2cos as cosx is even
and
a
a
xdx ,0sin as sinx is odd
PERIODIC FUNCTIONS :-
A periodic function has a basic shape which is repeated over and over again. The fundamental range is the time (or sometimes distance) over which the basic shape is defined. The length of the fundamental range is called the period.
A general periodic function f(x) of period T satisfies the condition
f(x+T) = f(x)
Here f(x) is a real-valued function and T is a positive real number.
As a consequence, it follows thatf(x) = f(x+T) = f(x+2T) = f(x+3T) = ….. = f(x+nT)
Thus,f(x) = f(x+nT), n=1,2,3,…..
The function f(x) = sinx is periodic of period 2 sinceSin(x+2n) = sinx, n=1,2,3,……..
The graph of the function is shown below :
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Note that the graph of the function between 0 and 2 is the same as that between 2 and 4 and so on. It may be verified that a linear combination of periodic functions is also periodic.
FOURIER SERIES
A Fourier series of a periodic function consists of a sum of sine and cosine terms. Sines and cosines are the most fundamental periodic functions.
The Fourier series is named after the French Mathematician and Physicist Jacques Fourier (1768 – 1830). Fourier series has its application in problems pertaining to Heat conduction, acoustics, etc. The subject matter may be divided into the following sub topics.
FORMULA FOR FOURIER SERIES
Consider a real-valued function f(x) which obeys the following conditions called Dirichlet’s conditions :
FOURIER SERIES
Series with arbitrary period
Half-range series Complex series Harmonic Analysis
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1. f(x) is defined in an interval (a,a+2l), and f(x+2l) = f(x) so that f(x) is a periodic function of period 2l.
2. f(x) is continuous or has only a finite number of discontinuities in the interval (a,a+2l).
3. f(x) has no or only a finite number of maxima or minima in the interval (a,a+2l).
Also, let
la
a
dxxfl
a2
0 )(1
(1)
la
a
n xdxl
nxf
la
2
,cos)(1
,.....3,2,1n (2)
la
a
n xdxl
nxf
lb
2
,sin)(1
,......3,2,1n (3)
Then, the infinite series
1
0 sincos2 n
nn xl
nbx
l
na
a (4)
is called the Fourier series of f(x) in the interval (a,a+2l). Also, the real numbers a0, a1, a2, ….an, and b1, b2 , ….bn are called the Fourier coefficients of f(x). The formulae (1), (2) and (3) are called Euler’s formulae.
It can be proved that the sum of the series (4) is f(x) if f(x) is continuous at x. Thus we have
f(x) =
1
0 sincos2 n
nn xl
nbx
l
na
a ……. (5)
Suppose f(x) is discontinuous at x, then the sum of the series (4) would be
)()(2
1 xfxf
where f(x+) and f(x-) are the values of f(x) immediately to the right and to the left of f(x) respectively.Particular CasesCase (i) Suppose a=0. Then f(x) is defined over the interval (0,2l). Formulae (1), (2), (3) reduce to
2 2
0
0 0
1 1( ) ( )cos ,
l l
n
na f x dx a f x xdx
l l l
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2
0
1( )sin ,
l
n
nb f x xdx
l l
,......2,1n (6)
Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (0,2l).
If we set l=, then f(x) is defined over the interval (0,2). Formulae (6) reduce to
a0 =
2
0
)(1
dxxf
2
0
cos)(1
nxdxxfan , n=1,2,….. (7)
2
0
sin)(1
nxdxxfbn n=1,2,…..
Also, in this case, (5) becomes
f(x) =
1
0 sincos2 n
nn nxbnxaa
(8)
Case (ii)
Suppose a=-l. Then f(x) is defined over the interval (-l , l). Formulae (1), (2) (3) reduce to
(9)
n =1,2,……
Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (-l , l).
If we set l = , then f(x) is defined over the interval (-, ). Formulae (9) reduce to
a0 =
dxxf )(
1
l
l
n
l
l
xdxl
nxf
la
dxxfl
a
cos)(
1
)(1
0
l
l
n xdxl
nxf
lb ,sin)(
1
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nxdxxfan cos)(
1
, n=1,2,….. (10)
nxdxxfbn sin)(
1 n=1,2,…..
Putting l = in (5), we get f(x) =
1
0 sincos2 n
nn nxbnxaa
Some useful results :
1. The following rule called Bernoulli’s generalized rule of integration by parts is useful in evaluating the Fourier coefficients.
.......3''
2'
1 vuvuuvuvdx
Here uu , ,….. are the successive derivatives of u and
,......, 121 dxvvvdxv
We illustrate the rule, through the following examples :
166
86
43
2
cos2
sin2
cossin
2222
2323
3222
xxxxx ee
xe
xe
xdxex
n
nx
n
nxx
n
nxxnxdxx
2. The following integrals are also useful :
bxbbxaba
ebxdxe
bxbbxaba
ebxdxe
axax
axax
cossinsin
sincoscos
22
22
3. If ‘n’ is integer, thensin n = 0 , cosn = (-1)n , sin2n = 0, cos2n=1
Problems
1. Obtain the Fourier expansion of
f(x) = x2
1in - < x <
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We have,
dxxdxxfa )(2
11)(
10
=
22
1 2xx
nxdxxnxdxxfan cos)(2
11cos)(
1
Here we use integration by parts, so that
002
1
2cos
)1(sin
2
1
n
nx
n
nxxna
n
n
nx
n
nxx
nxdxxb
n
n
)1(
sin)1(
cos
2
1
sin)(2
11
2
Using the values of a0 , an and bn in the Fourier expansion
1 1
0 sincos2
)(n n
nn nxbnxaa
xf
we get,
1
sin)1(
2)(
n
n
nxn
xf
This is the required Fourier expansion of the given function.
2. Obtain the Fourier expansion of f(x)=e-ax in the interval (-, ). Deduce that
2
2 1
)1(2cos
n
n
nech
Here,
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0
1 1
2sinh
axax
a a
ea e dx
a
e e a
a a
2 2
2 2
1cos
1cos sin
2 ( 1) sinh
axn
ax
n
n
a e nxdx
ea a nx n nx
a n
a a
a n
bn =
nxdxe ax sin
1
=
nxnnxa
na
e ax
cossin1
22
=
22
sinh)1(2
na
an n
Thus,f(x) =
1
221
22sin
)1(sinh
2cos
)1(sinh2sinh
n
n
n
n
nxna
nanx
na
aa
a
a
For x=0, a=1, the series reduces to
f(0)=1 =
1
2 1
)1(sinh2sinh
n
n
n
or
1 =
22 1
)1(
2
1sinh2sinh
n
n
n
or 1 =
22 1
)1(sinh2
n
n
n
Thus,
2
2 1
)1(2cos
n
n
nech This is the desired deduction.
3. Obtain the Fourier expansion of f(x) = x2 over the interval (-, ). Deduce that
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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......3
1
2
11
6 22
2
The function f(x) is even. Hence 0
0
1 2( ) ( )a f x dx f x dx
=
0 0
32
3
22 xdxx
or 3
2 2
0
a
nxdxxfan cos)(
1
=
0
,cos)(2
nxdxxf since f(x)cosnx is even
=
0
2 cos2
nxdxx
Integrating by parts, we get
Also,
0sin)(
1nxdxxfbn since f(x)sinnx is odd.
Thus2
21
( 1) cos( ) 4
3
n
n
nxf x
n
22
21
2
21
14
3
1
6
n n
n
Hence, .....3
1
2
11
6 22
2
2
032
2
)1(4
sin2
cos2
sin2
n
n
nx
n
nxx
n
nxxa
n
n
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4. Obtain the Fourier expansion of
2,2
0,)(
xx
xxxf
Deduce that ......5
1
3
11
8 22
2
Here, a0 =
dxxf )(
1=
0
)(2
dxxf
0
0
cos)(2
cos)(1
2
nxdxxfnxdxxfa
xdx
n
since f(x)cosnx is
even.
=
0
cos2
nxdxx
=
1)1(2
cos1
sin2
2
02
n
n
n
nx
n
nxx
Also,
0sin)(
1nxdxxfbn , since f(x)sinnx is odd
Thus the Fourier series of f(x) is
nxn
xfn
n cos1)1(12
2)(
12
For x= , we get
)(f
n
nn
n cos1)1(12
2 12
or
1
2)12(
)12cos(22
2 n n
n
Thus,
12
2
)12(
1
8 n n
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or ......5
1
3
11
8 22
2
This is the series as required.
5. Obtain the Fourier expansion of
f(x) =
xx
x
0,
0,
Deduce that ......5
1
3
11
8 22
2
Here,
2
1 0
0
0
xdxdxa
1)1(1
coscos1
2
0
0
n
n
n
nxdxxnxdxa
n
n
n
nxdxxnxdxb
)1(211
sinsin1 0
0
Fourier series is
f(x) =
112
sin)1(21
cos1)1(11
4 n
n
n
n nxn
nxn
Note that the point x=0 is a point of discontinuity of f(x). Here f(x+) =0, f(x-)=- at x=0.
Hence 2
02
1)]()([
2
1 xfxf
The Fourier expansion of f(x) at x=0 becomes
12
2
12
]1)1[(1
4
]1)1[(11
42
n
n
n
n
nor
n
Simplifying we get,
......5
1
3
11
8 22
2
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as f(x) cos(nx) is even
6. Obtain the Fourier series of f(x) = 1-x2 over the interval (-1,1).
The given function is even, as f(-x) = f(x). Also period of f(x) is 1-(-1)=2
Here a0 =
1
1
)(1
1dxxf =
1
0
)(2 dxxf
= 2
1
0
1
0
32
32)1(
xxdxx
3
4
1
0
1
1
)cos()(2
)cos()(1
1
dxxnxf
dxxnxfan
= 1
0
2 )cos()1(2 dxxnx
Integrating by parts, we get
1
0
322
)(
sin)2(
)(
cos)2(
sin12
n
xn
n
xnx
n
xnxan
= 22
1)1(4
n
n
1
1
)sin()(1
1dxxnxfbn =0, since f(x)sin(nx) is odd.
The Fourier series of f(x) is
f(x) =
12
1
2)cos(
)1(4
3
2
n
n
xnn
7. Obtain the Fourier expansion of www.vtuc
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f(x) =
2
30
3
41
02
3
3
41
xinx
xinx
Deduce that ......5
1
3
11
8 22
2
The period of f(x) is 32
3
2
3
Also f(-x) = f(x). Hence f(x) is even
2/3
0
2
2/3
0
2/3
2/3
2/3
0
2/3
2/3
2/3
0
0
3
2
3
2cos
3
4
3
23
2sin
3
41
3
4
3
2cos)(
2/3
2
2/3cos)(
2/3
1
03
41
3
4
)(2/3
2)(
2/3
1
n
xn
n
xnx
dxxn
xf
dxxn
xfa
dxx
dxxfdxxfa
n
= n
n)1(1
422
Also,
23
23
0
23
sin)(3
1dx
xnxfbn
Thus
f(x) =
122 3
2cos)1(1
14
n
n xn
n
putting x=0, we get
f(0) =
1
22)1(1
14
n
n
n
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or 1 =
......
5
1
3
11
8222
Thus, ......5
1
3
11
8 22
2
HALF-RANGE FOURIER SERIES
The Fourier expansion of the periodic function f(x) of period 2l may contain both sine and cosine terms. Many a time it is required to obtain the Fourier expansion of f(x) in the interval (0,l) which is regarded as half interval. The definition can be extended to the other half in such a manner that the function becomes even or odd. This will result in cosine series or sine series only.
Sine series :Suppose f(x) = (x) is given in the interval (0,l). Then we define f(x) = -(-x) in (-l,0). Hence f(x) becomes an odd function in (-l , l). The Fourier series then is
1
sin)(n
n l
xnbxf
(11)
where
l
n dxl
xnxf
lb
0
sin)(2
The series (11) is called half-range sine series over (0,l).
Putting l= in (11), we obtain the half-range sine series of f(x) over (0,) given by
1
sin)(n
n nxbxf
0
sin)(2
nxdxxfbn
Cosine series :
Let us define
in (0,l) given..... in (-l,0) …..in order to make the function even.Then the Fourier series of f(x) is given by
1
0 cos2
)(n
n l
xna
axf
(12)
where,
)(
)()(
x
xxf
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l
n
l
dxl
xnxf
la
dxxfl
a
0
0
0
cos)(2
)(2
The series (12) is called half-range cosine series over (0,l)
Putting l = in (12), we get
Problems :
1. Expand f(x) = x(-x) as half-range sine series over the interval (0,).
We have,
0
2
0
sin)(2
sin)(2
nxdxxx
nxdxxfbn
Integrating by parts, we get
n
n
n
n
nx
n
nxx
n
nxxxb
)1(14
cos)2(
sin2
cos2
3
032
2
The sine series of f(x) is
1
3sin)1(1
14)(
n
n nxn
xf
2. Obtain the cosine series of
xx
xxxf
2,
20,
)( ),0( over
..1,2,3,ncos)(2
)(2
cos2
)(
0
0
0
1
0
nxdxxfa
dxxfa
where
nxaa
xf
n
nn
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2
0
02
2
02
2: ( )
2
2cos ( ) cosn
Solution a xdx x dx
a x nxdx x nxdx
Performing integration by parts and simplifying, we get
,.....10,6,2,8
2cos2)1(1
2
2
2
nn
n
na n
n
Thus, the Fourier cosine series is
f(x) =
......
5
10cos
3
6cos
1
2cos2
4 222
xxx
3. Obtain the half-range cosine series of f(x) = c-x in 0<x<c
Here
c
cdxxcc
a0
0 )(2
c
n dxc
xnxc
ca
0
cos)(2
Integrating by parts and simplifying we get,
nn n
ca )1(1
222
The cosine series is given by
f(x) =
122
cos)1(112
2 n
n
c
xn
n
cc
COMPLEX FORM OF FOURIER SERIES
The standard form of Fourier series of f(x) over the interval (α, α+2l) is :
1
0 sincos2
)(n
nn l
xnb
l
xna
axf
If we use Euler formulae
sincos ie i
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then we obtain a complex exponential Fourier series
n
xl
ni
necxf
)( ………..(*)
where21
( )2
nl i xl
nc f x e dxl
n = 0, 1, 2, 3, …….
Note:-(1)
Putting l in (*), we get Fourier series valid in ( , )l l as
n
xl
ni
necxf
)( ………..(**)
where
1( )
2
nl i xl
nl
c f x e dxl
Note:-(2)
Putting α = 0 and l in (*), we get Fourier series valid in (0,2 ) as
( ) inx
nn
f x c e
where
1( )
2inx
nc f x e dx
Note:-(3)
Putting l in (**), we get Fourier series valid in ( , ) as
( ) inx
nn
f x c e
where
2
0
1( )
2inx
nc f x e dx
Problems:
1. Obtain the complex Fourier series of the function f(x) defined by f(x) = x over the interval (- , ).Here f(x) is defined over the interval (- , ). Hence complex Fourier series of f(x) is
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n
inxnecxf )( (1)
where
dxxe
dxexfc
inx
inxn
2
1
)(2
1
n = 0, 1, 2, 3, …….
Integrating by parts and substituting the limits, we get
n
ic
n
n
)1( , n 0
Using this in (1), we get
n
inxn
en
ixf)1(
)( , n 0
This is the complex form of the Fourier series of the given function
2. Obtain the complex Fourier series of the function f(x) = eax over the interval (- , ).
As the interval is again (- , ) we find n
c which is given by
( )
( )
1 1( )
2 2
1 1
2 2 ( )
inx ax inxn
a in xa in x
c f x e dx e e dx
ee dx
a in
2 2
( 1) ( )sinh
( )
n a in a
a n
Hence the complex Fourier series for f(x) is
n
inxnecxf )(
n
inxn
ena
inaa22
)()1(sinh
HARMONIC ANALYSIS
The Fourier series of a known function f(x) in a given interval may be found by finding the Fourier coefficients. The method described cannot be employed when f(x) is not known explicitly, but defined through the values of the function at some equidistant
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points. In such a case, the integrals in Euler’s formulae cannot be evaluated. Harmonic analysis is the process of finding the Fourier coefficients numerically.
To derive the relevant formulae for Fourier coefficients in Harmonic analysis, we employ the following result : The mean value of a continuous function f(x) over the interval (a,b) denoted by [f(x)] is defined as The Fourier coefficients defined through Euler’s formulae, (1), (2), (3) may be redefinedas
b
a
dxxfab
xf )(1
)( 2
0
12 ( ) 2[ ( )]
2
a l
a
a f x dx f xl
.
212 ( )cos 2 ( ) cos
2
a l
n
a
n x n xa f x dx f x
l l l
212 ( )sin 2 ( )sin
2
a l
n
a
n x n xb f x dx f x
l l l
Using these in (5), we obtain the Fourier series of f(x). The term a1cosx+b1sinx is called the first harmonic or fundamental harmonic, the term a2cos2x+b2sin2x is called the
second harmonic and so on. The amplitude of the first harmonic is 21
21 ba and that
of second harmonic is 22
22 ba and so on.
Problems:
1. Find the first two harmonics of the Fourier series of f(x) given the following table :
x 0 3
32 3
43
5 2f(x) 1.0 1.4 1.9 1.7 1.5 1.2 1.0
Note that the values of y = f(x) are spread over the interval 0 x 2 and f(0) = f(2) = 1.0. Hence the function is periodic and so we omit the last value f(2) = 0. We prepare the following table to compute the first two harmonics.
x0 y = f(x) cosx cos2x sinx sin2x ycosxycos2
xysinx ysin2x
0 1.0 1 1 0 0 1 1 0 0
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 22
We have
]sin[2sin)(2
]cos[2cos)(2
nxyl
xnxfb
nxyl
xnxfa
n
n
as the length of interval= 2l = 2 or
l=
Putting, n=1,2, we get
1.06
)3.0(2
6
2cos2]2cos[2
367.06
)1.1(2
6
cos2]cos[2
2
1
xyxya
xyxya
0577.06
2sin2]2sin[
0392.16
sin2]sin[
2
1
xyxyb
xyxyb
The first two harmonics are a1cosx+b1sinx and a2cos2x+b2sin2x. That is (-0.367cosx + 1.0392 sinx) and (-0.1cos2x – 0.0577sin2x)
2. Express y as a Fourier series upto the third harmonic given the following values :
60 1.4 0.5 -0.5 0.866 0.866 0.7 -0.7 1.2124 1.2124
120 1.9 -0.5 -0.5 0.866 -0.866 -0.95 -0.95 1.6454 -1.6454
180 1.7 -1 1 0 0 -1.7 1.7 0 0
240 1.5 -0.5 -0.5 -0.866 0.866 -0.75 -0.75 1.299 1.299
300 1.2 0.5 -0.5 -0.866 -0.866 0.6 -0.6 -1.0392 -1.0392
Total -1.1 -0.3 3.1176 -0.1732
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 23
x 0 1 2 3 4 5y 4 8 15 7 6 2
The values of y at x=0,1,2,3,4,5 are given and hence the interval of x should be 0 x < 6. The length of the interval = 6-0 = 6, so that 2l = 6 or l = 3.The Fourier series upto the third harmonic is
l
xb
l
xa
l
xb
l
xa
l
xb
l
xa
ay
3sin
3cos
2sin
2cossincos
2 3322110
or
3
3sin
3
3cos
3
2sin
3
2cos
3sin
3cos
2 3322110 x
bx
ax
bx
ax
bx
aa
y
Put 3
x , then
3sin3cos2sin2cossincos2 332211
0 bababaa
y ……… (1)
We prepare the following table using the given values :
x =3
xy ycos ycos2 ycos3 ysin ysin2 ysin3
0 0 04 4 4 4 0 0 0
1 600 08 4 -4 -8 6.928 6.928 0
2 1200 15 -7.5 -7.5 15 12.99 -12.99 0
3 1800 07 -7 7 -7 0 0 0
4 2400 06 -3 -3 6 -5.196 5.196 0
5 3000 02 1 -1 -2 -1.732 -1.732 0
Total 42 -8.5 -4.5 8 12.99 -2.598 0www.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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0]3sin[2
667.2)8(6
2]3cos[2
866.0)598.2(6
2]2sin[2
5.1)5.4(6
2]2cos[2
33.4)99.12(6
2]sin[2
833.2)5.8(6
2]cos[2
14)42(3
1
6
2][2)]([2
3
3
2
2
1
1
0
yb
ya
yb
ya
yb
ya
yyxfa
Usingthesein(1),we get
xxxxx
y cos667.2
3
2sin866.0
3
2cos5.1
3sin)33.4(
3cos833,27
This is the required Fourier series upto the third harmonic.
3. The following table gives the variations of a periodic current A over a period T :
t(secs) 0 T/6 T/3 T/2 2T/3 5T/6 T
A (amp) 1.98 1.30 1.05 1.30 -0.88 -0.25 1.98
Show that there is a constant part of 0.75amp. in the current A and obtain the amplitude of the first harmonic.
Note that the values of A at t=0 and t=T are the same. Hence A(t) is a periodic function
of period T. Let us denote tT
2
. We have
]sin[22
sin2
]cos[22
cos2
][2
1
1
0
AtT
Ab
AtT
Aa
Aa
(1)
We prepare the following table:
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 25
tT
t 2 A cos sin Acos Asin
0 0 1.98 1 0 1.98 0
T/6 600 1.30 0.5 0.866 0.65 1.1258
T/3 1200 1.05 -0.5 0.866 -0.525 0.9093
T/2 1800 1.30 -1 0 -1.30 0
2T/3 2400 -0.88 -0.5 -0.866 0.44 0.7621
5T/6 3000 -0.25 0.5 -0.866 -0.125 0.2165
Total 4.5 1.12 3.0137
Using the values of the table in (1), we get
0046.13
0137.3
6
sin2
3733.03
12.1
6
cos2
5.13
5.4
6
2
1
1
0
Ab
Aa
Aa
The Fourier expansion upto the first harmonic is
T
t
T
t
T
tb
T
ta
aA
2sin0046.1
2cos3733.075.0
2sin
2cos
2 110
The expression shows that A has a constant part 0.75 in it. Also the amplitude of the first
harmonic is 21
21 ba = 1.0717.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 26
UNIT-II
FOURIER TRANSFORMS
CONTENTS:
Introduction
Finite Fourier transforms and Inverse finite Fourier transforms
Infinite Fourier transform (Complex Fourier transform) and
Inverse Fourier transforms
Properties [Linearity, Change of scale, Shifting and Modulation
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 27
Fourier cosine and Fourier sine transforms & Inverse Fourier
cosine and sine transforms
FOURIER TRANSFORMS
Introduction
Fourier Transform is a technique employed to solve ODE’s, PDE’s,IVP’s, BVP’s and Integral equations.The subject matter is divided into the following sub topics :
FOURIER TRANSFORMS
Infinite Sine Cosine ConvolutionFourier Transform Transform Theorem &Transform Parseval’s
Identity
Infinite Fourier Transform
Let f(x) be a real valued, differentiable function that satisfies the following conditions:
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 28
andinterval,finiteevery in itiesdiscontinusimple
ofnumber finiteaonly haveor ,continuousarexfderivativeitsandf(x)1)
bydefinedisf(x)ofTransformFourier infiniteThe parameter.realzero-nonbelet Also,
exists.xfintegral the2)-
dx
dxexfxfFf xiˆ
provided the integral exists.
bydefinedis f̂F
by denotedf̂ofTransformFourier inverseThe Transform.Fourier just theor TransformFourier complex calledalsoisTransformFourier infiniteThe
1-
defxffF xi
ˆ2
1ˆ1
Note : The function f(x) is said to be self reciprocal with respect to Fourier transform
ff ˆif .
Basic Properties
1. Linearity Property
For any two functions f(x) and (x) (whose Fourier Transforms exist) and any two constants a and b,
xbFxfaFxbxafF
Proof :
By definition, we have
dxexbxafxbxafF xi
xbFxfaF
dxexbdxexfa xixi
This is the desired property.
In particular, if a = b = 1, we get xFxfFxxfF
Again if a= -b = 1, we get xFxfFxxfF
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 29
2. Change of Scale Property
have wea,constant zero-nonany for then ,xfFf̂If
aa
f̂
1xfF
Proof : By definition, we have
)1(dxeaxfaxfF xi
Suppose a > 0. let us set ax = u. Then expression (1) becomes
a
dueufaxfF
ua
i
)2(ˆ1
af
a
Suppose a < 0. If we set again ax = u, then (1) becomes
a
dueufaxfF a
ui
dueufa
ua
i
1
)3(ˆ1
af
a
Expressions (2) and (3) may be combined as
af
aaxfF
ˆ1
This is the desired property3. Shifting Properties
For any real constant ‘a’,
feaxfFi ai ˆ)(
afxfeFii iax ˆ)(Proof : (i) We have
dxexffxfF xi
ˆ
Hence, dxeaxfaxfF xi
Set x-a = t. Then dx = dt.Then,
dtetfaxfF ati )(
= aie dtetf ti
= aie f̂ii) We have
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 30
dxexfaf xai
ˆ
dxeexf xiiax
iaxiax exfxgwheredxexg )()(,
xgF
xfeF iaxThis is the desired result.
4. Modulation Property
,ˆ fxfFIf
afafaxxfFthen ˆˆ2
1cos,
where ‘a’ is a real constant.
Proof : We have
2
cosiaxiax ee
ax
Hence
2cos
iaxiax eexfFaxxfF
property.desired theisThis
.propertiesshift andlinearity usingby ,ˆˆ2
1afaf
Note : Similarly
afafaxxfF ˆˆsin2
1
Problems:0whereef(x)function theofTransformFourier theFind1. x-a a
For the given function, we have
dxeexfF xixa
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 31
dxeedxee xixaxixa
0
0
get we0, x -x,-xand x 0 x,xfact that theUsing
dxeedxeexfF xiaxxiax
0
0
dxedxexiaxia
0
0
0
0
ia
e
ia
e xiaxia
iaia
11
22
2
a
a
2. Find the Fourier Transform of the function
ax
ax
0,
1,f(x)
where ‘a’ is a positive constant. Hence evaluate
d
xai
cossin)(
dii
0
sin)(
For the given function, we have
dxexfxfF xi)(
dxexfdxexfdxexf xi
a
xia
a
xia )()()(
dxe
a
a
xi
asin
2
)1(sin
2ˆ
a
fxfFThus
get weformula,inversion employingby f̂Inverting
de
axf xi
sin
22
1
d
xixa sincossin1
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 32
d
xaid
xa sinsincossin1
Here, the integrand in the first integral is even and the integrand in the second integral is odd. Hence using the relevant properties of integral here, we get
d
xaxf
cossin)( 1
or
)(
cossinxfd
xa
ax
ax
0,
,
d
sinyields this1,a0,For x
Since the integrand is even, we have
0
sin2 d
or
20
sin
d
Transform.Fourier respect to with reciprocalselfis22x
ef(x) that Deduce
constant.positiveaisa'' where2x2a-ef(x)ofTransformFourier theFind3.
Here
dxeexfF xixa 22
dxe xixa 22
dxeaa
iax
2
22
42
dxee a
iax
a
2
2
2
24
getwe,2a
i-ax t Setting
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 33
a
dteexfF ta
22
2
4
dteea
ta
0
422
2
21
function.gammausing,1 2
2
4
aea
2
2
4ˆ aea
f
This is the desired Fourier Transform of f(x).
2
2x-
2
2
2
2f̂
hence,andef(x)get we
2x2a-ef(x)in 21aFor
e
getwe,ef(x)in xputtingAlso 2x2- 2
2-e)f(
.
Hence, )f( and f̂ are same but for constant multiplication by 2 .
f̂)f(Thus
reciprocalselfis e)f( that followsIt 22
-x
x
FOURIER SINE TRANSFORMS
Let f(x) be defined for all positive values of x.
ThusxfFor
xdxxf
s .f̂by
denotedisThis f(x).ofTransformSineFourier thecalledis sinintegralThe
s
0
dxxxfxfFs sinf̂0s
dxf s sinˆ2integralethrough th
definedisf̂ofTransformsineFourier inverseThe
0
s
dxff
Thusf
ss
s
sinˆ2Ff(x)
.For f(x)by denotedisThis
0
1-s
-1s
Properties
The following are the basic properties of Sine Transforms.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 34
(1) LINEARITY PROPERTYIf ‘a’ and ‘b’ are two constants, then for two functions f(x) and (x), we
have xgbFxfaFxbxafF sss
Proof : By definition, we have
dxxxbxafxbxafFs sin0
xbFxfaF ss This is the desired result. In particular, we have
xFxfFxxfF sss and
xFxfFxxfF sss
(2) CHANGE OF SCALE PROPERTY
have we0,afor then ,f̂xfFIf s s
aa s
f̂
1axfFs
Proof : We have
dxxaxf sinaxfF0s
Setting ax = t , we get
a
dtt
atf
sinaxfF
0s
aa s
f̂
1
(3) MODULATION PROPERTY
have we0,afor then ,ˆxfFIf s sf
afafaxxfF sss ˆˆcos2
1
Proof : We have
0
sincoscos dxxaxxfaxxfFs
dxxaxaxf sinsin
2
10
property.Linearity usingby ,ˆˆ
2
1 afaf ss
Problems:
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 35
1. Find the Fourier sine transform of
ax
ax
,0
0,1xf
For the given function, we have
a
adxxdxx sin0sinf̂
0s
ax
0
cos
acos1
x
ef(x)of transformsineFourier theFind2.
-ax
Here
0s
sinf̂
x
dxxe ax
Differentiating with respect to , we get
0s
sinf̂
x
dxxe
d
d
d
d ax
0
sin dxxx
axe
performing differentiation under the integral sign
0
cos dxxxx
e ax
022
sincos xxaa
e ax
22
a
a
Integrating with respect to , we get
ca
1s tanf̂
00f̂But s when
c=0
a
1s tanf̂
3. Find f(x) from the integral equation
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 36
2,0
21,2
10,1
sinxf0
xdx
Let () be defined by
2,0
21,2
10,1
Given
0
ˆsin s
fxdxxf
Using this in the inversion formula, we get
dxx
sin2
xf0
dxdxdx sinsinsin
2
2
1
1
02
2
1
1
0
2 0sin2sin
dxdx
xxx
2cos2cos12
FOURIER COSINE TRANSFORMS
dxxxf cosintegralThe x.of valuespositivefor definedbef(x)Let 0
Thus.For f̂by denotedisandf(x)ofTransformCosineFourier thecalledis cc xf
xdxxfxf
cos2
Ff̂0cc
Thus .or xfby denotedisThis .2
integralthe
through definedisofTransformCosineFourier inverseThe
cf̂
1-cFcosˆ
cf̂
0
dxfc
0
1-c cosˆ2ˆFxf
dxff c
Basic Properties
The following are the basic properties of cosine transforms :
(1) Linearity property
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 37
xxfxafFhave we(x),andf(x)functionsfor two then constants, twoareb''anda''If
c
cc bFaFxb
(2) Change of scale property
aa
cc
cc
f̂1
axfF
have we0,afor then ,f̂xfFIf
(3) Modulation property
aaax
ccc
cc
f̂f̂2
1cosxfF
have we0,afor then ,f̂xfFIf
The proofs of these properties are similar to the proofs of the corresponding properties of Fourier Sine Transforms.
Problems: 1) Find the cosine transform of the function
2,0
21,2
10,
x
xx
xx
xf
We have
xdxxdxxxdxx
xdxxf
cos0cos2cos
cosf̂
2
2
1
1
0
0c
Integrating by parts, we get
2
12
1
0
2c
cos1
sin2
cossinf̂
xx
xxx
x
2
12coscos2
2) -ax Find the cosine transform of f(x) e , a 0. Hence evaluate 2 20
cos kx
x adx
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 38
Here
22c
0
22
0c
f̂
sincosa
cosf̂
a
a
Thus
xxae
xdxe
ax
ax
Using the definition of inverse cosine transform, we get
xda
x cosa
2f
0 22
or
0 22
cos
2
da
xe
aax
Changing x to k, and to x, we get
a
edx
a
kx ax
2x
cos0 22
3) Solve the integral equation
aedxxxf cos
0
Let () be defined by() = e-a
c0
f̂cosGiven
dxxxf
Using this in the inversion formula, we get
22
00 22
0
0
2
sincos2
cos2
cos2
f
xa
a
xxaxa
e
xde
xdx
a
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 39
UNIT III Applications of Partial Differential Equations
CONTENTS:
Introduction
Various possible solutions of the one dimensional wave equation
Various possible solutions of the one dimensional heat equation
Various possible solutions of the two dimensional Laplace’s equation
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 40
D’Alembert’s solutions of the one dimensional wave equation
APPLICATION OF PARTIAL DIFFERENTIAL EQUATIONS
Introduction
A number of problems in science and engineering will lead us to partial differential equations. In this unit we focus our attention on one dimensional wave equation ,one dimensional heat equation and two dimensional Laplace’s equation.Later we discuss the solution of these equations subject to a given set of boundary conditions referred to as boundary value problems.Finally we discuss the D’Alembert’s solution of one dimensional wave equation.
Various possible solutions of standard p.d.es by the method of separation of variables.We need to obtain the solution of the ODEs by taking the constant k equal to i) Zero ii)positive:k=+p2 iii)negative:k=-p2
Thus we obtain three possible solutions for the associated p.d.e
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 41
Various possible solutions of the one dimensional wave equation utt =c2uxx by the method of separation of variables.
Consider 2 2
22 2
u uc
t x
Let u= XT where X=X(x),T=T(t) be the solution of the PDEHence the PDE becomes
2 2 2 22 2
2 2 2 2
XT XT d T d Xc or X c
t x dt dx
Dividing by c2XT we have 2 2
2 2 2
1 1d T d X
c T dt X dx
Equating both sides to a common constant k we have2
2
1 d X
X dx=k and
2
2 2
1 d T
c T dt=k
2
20
d XkX
dx and
22
20
d Tc kT
dt
2 0D k X and 2 2 0D c k T
Where D2 =2
2
d
dxin the first equation and D2 =
2
2
d
dtin the second equation
Case(i) : let k=0AEs are m=0 amd m2=0 amd m=0,0 are the roots Solutions are given byT = 0 0
1 1 2 3 2 3t xc e c and X c x c e c x c
Hence the solution of the PDE is given byU= XT= 1 2 3c c x cOr u(x,t) =Ax+B where c1c2=A and c1c3=BCase(ii) let k be positive say k=+p2
AEs are m –c2p2=0 and m2-p2=0m= c2p2 and m=+psolutions are given by
2 2' ' '1 2 3
c p t px pxT c e andX c e c e Hence the solution of the PDE is given by
2 2'1
c p tu XT c e .( ' '2 3
px pxc e c e )
Or u(x,t) =2 2'
1c p tc e (A’ pxe +B pxe ) where c1’c2’=A’ and c1’c3’=B’
Case(iii): let k be negative say k=-p2
AEs are m+ c2p2=0 and m2+p2=0m=- c2p2 and m=+ipsolutions are given by
2 2'' '' ''1 2 3cos sinc p tT c e and X c px c px
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 42
Hence the solution of the PDE is given by2 2'' '' ''
1 2 3.( cos sin )c p tu XT c e c px c px 2 2 '' ''( , ) ( cos sin )c p tu x t e A px B px
Various possible solutions of the two dimensional Laplace’s equation uxx+uyy=0 by the method of separation of variables
Consider 2 2
2 20
u u
x y
Let u=XY where X(x), Y=Y(y) e the solution of the PDEHence the PDE becomes
2 2
2 2
( )0
XY XY
x y
2 2
2 2
( )0
d X d YY X and dividing by XY we have
dx dy
2 2
2 2
1 1 ( )d X d Y
X dx Y dy
Equating both sides to a common constant k we have
2
2
1 d X
X dx=k and
2
2
1 ( )d Y
Y dy =k
Or (D2-k)X=0 and (D2+k)Y=0
Where D =d
dxin the first equation and D =
d
dyin the second equation
Case(i) Let k=0AE arem2=0 in respect of both the equationsM=0,0 and m=0,0Solutions are given byX= c1x+c2 and Y= c3y+c4
Hence the solution of the PDE is given byU=XY=( c1x+c2 ) (c3y+c4)
Case(ii) : let k be positive ,say k=+p2
m2-p2=0 and m2+p2=0m=+p and m=+ipsolutions are given by
' ' ' '1 2 3 4cos sinpx pxX c e c e and Y c py c py
Hence the solution of the PDE is given by
' ' ' '1 2 4( 3cos sin )px pxu XY c e c e c py c py
Case(iii) Let k be negative say k= -p2
AEs are m2+p2=0 and m2-p2=0
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 43
M=+ip and m=+pSolutions are given by
'' ''1 2( cos sin )X c px c px and '' ''
3 4( )py pyY c e c e Hence the solution of the PDE is given by
'' '' '' ''1 2 3 4( cos sin )( )py pyu XY c px c px c e c e
EXAMPLES
1.Solve the wave equation utt=c2uxx subject to the conditions
u(t,0)=0 ,u(l,t)=0, ,0 0u
xt
and u(x,0) =u0sin3( x/l)
Soln: 1
, sin cosnn
n x n ctu x t b
l l
Consider u(x,0) =u0sin3( x/l)
1
,0 sinnn
n xu x b
l
30
1sin sinn
n
x n xu b
l l
30
1
3 1 3sin sin sin
4 4 nn
x x n xu b
l l l
comparing both sides we get
01
3
4
ub , 2 0b , 0
3 4 5, 0 04
ub b b
,
Thus by substituting these values in the expanded form we get
0 03 3 3( , ) sin cos sin cos
4 4
u ux ct x ctu x t
l l l l
2.Solve the wave equation utt=c2utt subject to the conditions
u(t,0)=0 ,u(l,t)=0, ,0 0u
xt
when t=0and u(x,0) =f(x)
Soln: : 1
, sin cosnn
n x n ctu x t b
l l
Consider u(x,0)=f(x) then we have
Consider u(x,0) = 1
sinnn
n xb
l
F(x) = 1
sinnn
n xb
l
The series in RHS is regarded as the sine half range Fourier series of f(x) in (0,l) and hence
0 01 2 3
3 3 2 3sin sin sin sin sin
4 4
u ux x x x xb b b
l l l l l
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 44
0
2( )sin
l
n
n xb f x dx
l l
Thus we have the required solution in the form
1
, sin cosnn
n x n ctu x t b
l l
3. Solve the Heat equation 2
22
u uc
t x
given that u(0,t)=0,u(l,0)=0 and u(x,0)=
100x/l
Soln: 0
2 100sin
l
n
x n xb dx
l l l
=
20
200sin
l n xx dx
l l
22
0
. cos sin2001
/ /
l
n
n x n xx
l lbl n l n l
1
2
200 1 200 1200 1. cos .
n n
nb l nl n n n
The required solution is obtained by substituting this value of nb
Thus 2 2 21
21
200 1( , ) sin
n n c t
n
n xu x t e
n l l
4.Obtain the solution of the heat equation2
22
u uc
t x
given that u(0,t)=0,u(l,t)and
u(x,0) =f(x)where
20
2( )2
2
Tx lin x
lf xT l
l x in x ll
Soln: 0
2( )sin
l
n
n xb f x dx
l l
2
02
2 2 2sin ( )sin
l
l
nl
Tx n x Tx n xb dx l x dx
l l l l l
2
02
4sin ( )sin
l
l
l
T n x n xx dx l x dx
l l l
2 2
8sin
2n
T nb
n
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 45
The required solution is obtained by substituting this value of bn
Thus 2 2 2
2 2 21
8 1( , ) sin sin
2
n c t
n
T n n xu x t e
n l l
5. Solve the heat equation 2
2
u u
t x
with the boundary conditions u(0,t)=0,u(l,t)and
u(x,0) =3sin x
Soln: 2
( , ) ( cos sin )............................(1)p tu x t e A px B px Consider u(0,t)=0 now 1 becomes
0=2p te (A) thus A=0
Consider u(1,t)=0 using A=0 (1) becomes
0=2p te (Bsinp)
Since B≠0,sinp=0or p=n2 2 2
( , ) ( sin )n c t
u x t e B n x
In general 2 2 2
1
( , ) sinn c t
nn
u x t b e n x
Consider u(x,0)= 3sin n x and we have
3 1 2 3sin sin sin 2 sin 3n x b x b x b x
Comparing both sides we get 1 2 33, 0, 0b b b We substitute these values in the expanded form and then get
2
( , ) 3 (sin )t
u x t e x
6.Solve 2 2
2 20
u u
x y
subject to the conditions u(0,y)=0,u( ,y)=0 and u(x,∞) =0 and
u(x,0)=ksin2xSoln: The befitting solution to solve the given problem is given by
( , ) ( cos sin ) py pyu x y A px B px Ce De
( , ) 0 ( sin ) 0,py pyu y gives B p Ce De
Since A=0,B≠0 and we must havesin p =0,therefore p=n where n is a integer
( , ) ( sin ) ny nyu x y B nx Ce De
The condition u(x,∞) =0 means that u→as y→∞
(0, ) 0 ( ) 0, 0py pyu y gives A Ce De A
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 46
0=(Bsinnx)( nyCe )since 0nye asy Since B≠0 we must have C=0
We now have u(x,y) =BDsinnx nye
Taking n= 1,2,3,……… and BD =b1 ,b2 ,b3 ,b4…………………
We obtain a set of independent solutions satisfying the first three conditions Their sum also satisfy these conditions hence
1
( , ) sin nyn
n
u x y b n xe
Consider u(x,0)=ksin2x and we have
1
( ,0) sinnn
u x b nx
Ksin2x=b1sinx+b2sin2x+b3sin3x+…………….Comparing both sides we get b1=0, b2=0, b3=0Thus by substituting these values in the expanded form we get
2( , ) sin 2 yu x y k xe
D’Alembert’s solution of the one dimensional wave equation
We have one dimensional wave equation
2 2
22 2
u uc
t x
Let v=x+ct and w=x-ctWe treat u as a function of v and w which are functions of x and tBy chain rule we have,
. .u u v u w
x v x w x
Since v=x+ct and w=x-ct, 1 1v w
andx x
.(1) .(1)u u u u u
x v w v w
2
2
u u u u
x x x x v w
Again by applying chain rule we have,
2
2. .
u u u v u u w
x v v w x w v w x
2 2 2 2 2
2 2 2.(1) .(1)
u u u u u
x v v w w v w
But 2u
v w
=2u
w v
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 47
But 2 2 2 2
2 2 22
u u u u
x v w v w
Similarly . .u u v u w
x v t w t
Since v=x+ct and w=x-ct, v w
c and ct t
.( ) .( ) ,u u u u u u
c c or ct v w t v w
Again applying chain rule we have,
2
2. .
u u u u v u u wc c
r t t v w v w t w v w t
2 2 2 2 2
2 2 2.( ) .( )
u u u u uc c c c
r v v w w v w
2 2 2 2 22 2
2 2 2
u u u u uc c
t v v w w v w
2 2 2 22
2 2 22
u u u uc
t v w v w
2 2 22
2 22
u u uc
v w v w
=2 2 2
22 2
2u u u
cv w v w
2
4u
w v
=0 or
2u
w v
=0
We solve this PDE by direct integration, writing it in the form,
0u
w v
Integrating w.r.t w treating v as constant we get ( )u
f vv
Now integrating w.r.t, we get ( ) ( )u f v dv G w U=F(v)+G(w) where F(v)= ( )f v dvBut v=x+ct and w=x-ctThus u=u(x,t)=F(x+ct)+G(x-ct)This is the D’Alembert’s solution of the one dimensional wave equation
EXAMPLES
1. Obtain the D’Alembert’s solution of the wave equation utt=c2uxx subject to the
conditions u(x,0)=f(x) and ( ,0) 0u
xt
Soln: D’Alembert’s solution of the wave equation is given by
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 48
u(x,t)=F(x+ct0=G(x-ct)………………………………………..(1)Consider u(x,0)=f(x) (10 becomesu(x,0)=F(x)+G(x)f(x)=F(x)+G(x)…………………………………………………….(2)Differentiating partially wrt t we have
u
t
(x,t) =F’(x+ct).(c)+G’(x-ct).(-c)
u
t
(x,)=c[F’(x)-G’(x)]
0= c[F’(x)-G’(x)][F’(x)-G’(x)]=0 integrating wrt x we have[F(x)-G(x)]=kWhere k is the constant of integration……………………………(3)By solving simultaneously the equations,[F(x)+G(x)]=f(x)[F(x)-G(x)]=k
We obtain F(x)= 1 1( ) ( ) ( )
2 2f x k andG x f x k
Thus F(x+ct)= 1( )
2f x ct k and G(x-ct)= 1
( )2
f x ct k
Substituting these in(10 we have
U(x,t)= 1 1( ) ( )
2 2f x ct k f x ct k
Thus the required solution is
u(x,t)= 1( ) ( )
2f x ct f x ct
2. Obtain the D’Alembert’s solution of the wave equation utt=c2uxx given that u(x,0)=f(x)=l2-x2and ut(x,0)=0
Soln. Assuming the D’Alembert’s solution
u(x,t)= 1( ) ( )
2f x ct f x ct and f(x)=l2-x2
u(x,t)= 2 2 2 21( ) ( )
2l x ct l x ct
2 2 2 212 2 2
2l x c t
u(x,t)= 2 2 2 2l x c t
3. . Obtain the D’Alembert’s solution of the wave equation utt=c2uxx given that u(x,0)=asin2πx and
0 0u
when tt
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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Soln: Assuming the D’Alembert’s solution
u(x,t)= 1( ) ( )
2f x ct f x ct
By data f(x)= asin2πx or f(x)= 1 cos 22
ax
u(x,t)= 1. 1 cos2 ( ) 1 cos 2 ( )
2 2
ax ct x ct
u(x,t)= 2 cos 2 2 cos 2 24
ax ct x ct
2 2cos2 cos 24
ax ct
u(x,t)= 1 1cos2 cos22
ax ct
UNIT IVCURVE FITTING AND OPTIMIZATION
CONTENTS:
Curve fitting by the method of least squares
Fitting of curves of the form
y ax b
2y ax bx c
bxy ae
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 50
by ax
Optimization :
Linear programming
Graphical method
Simplex method
CURVE FITTING AND OPTIMIZATION
CURVE FITTING [BY THE METHOD OF LEAST SQUARE]:
We can plot ‘n’ points ( , )i ix y where i=0,1,2,3,………At the XY plane. It is difficult to draw a graph y=f(x) which passes through all these points but we can draw a graph which passes through maximum number of point. This curve is called the curve of best fit. The method of finding the curve of best fit is called the curve fitting.
FITTING A STRAIGHT LINE Y = AX + B
We have straight line that sounds as best approximate to the actual curve y=f(x)
passing through ‘n’ points ( , )i ix y , i=0,1.2………..n equation of a straight line is (1)y a bx
Then for ‘n’ points (2)i iY a bx
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 51
Where a and b are parameters to be determined; iY is called the estimated value. The
given value iY corresponding to ix .
2
2
2
(3)
=
S = (4)
i i
i i
i i
Let S y Y
y a bx
y a bx
We determined a and b so that S is minimum (least). Two necessary conditions for this S
0 and 0b
S
a
.differentiate (4) w.r.t a and b partially
i
i
i
0
0 = y
0= y
y
or y
i i
i i
i
i
i
Sy a bx
a
y a bx
a b x
na b x
na b x
na b x
2
2
0 2
i i i
i i i i
Sy a bx x
b
x y ax bx
2
i0 = yi i ix a x b x
2i
2
y
or
where n = number of points or value.
i i ix n x b x
xy a x b x
FITTING A SECOND DEGREE PARABOLA Y=AX2 + BX + C
Let us take equation of parabola called parabola of best fit in the form2 (1)y a bx cx www.vt
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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Where a, b, c are parameters to be determined. Let be the value of corresponding to the v
2
2
i
22
i
2i
(2)
Also
S= y (3)
y
y (4)
i i i
i
i i
i i
Y a bx cx
Y
a bx cx
S a bx cx
We determine a, b, c so that S is least (minimum).The necessary condition for this are
0, 0 & 0S S S
a b c
diff (4) w.r.t ‘a’ partially
2
2
2
2
2
0 2
0
0
i i i
i i i
i i i
i i i
Sy a bx cx
a
y a bx cx
y a bx cx
y a b x c x
2
2
2
i i i
i i i
y a b x c x
y na b x c x
or y na b x c x
diff (4) w.r.t ‘b’ partially
2
2 3
2
0 2
i i i i
i i i i i
Sy a bx cx x
b
x y ax bx cx
2 3
2 3
0
0
i i i i i
i i i i i
x y ax bx cx
x y a x b x c x
2 3
2 3
i i i i ix y a x b x c x
or xy a x b x c x
diff (4) w.r.t ‘c’ partiallywww.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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2 2
2 2 3 4
2 2 3 4
2 2 3 4
2 2 3 4
2 2 3 4
2
0 2
0
0
i i i i
i i i i i
i i i i i
i i i i i
i i i i i
Sy a bx cx x
c
x y ax bx cx
x y ax bx cx
x y a x b x c x
x y a x b x c x
or x y a x b x c x
Hence the normal equation for second degree parabola are 2y na b x c x
2 3xy a x b x c x 2 2 3 4x y a x b x c x
PROBLEMS:
1) Fit a straight line y a bx to the following data
: 5 10 15 20 25
: 16 19 23 26 30
x
y
Solution: Let (1)y a bx
Normal equation
2
(2)
(3)
y na b x
xy a x b x
2
5 16 25 80
10 19 100 190
15 23 225 345
20 26 400 520
25 30
x y x xy
2
625 750
75 =114 =1375 =1885
x y x xy
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 54
(1) & (2)
114 15 75
1885 75 1375
12.3
0.7
(1) 12.3 0.7
a b
a b
a
b
y x
2) Fit equation of straight line of best fit to the following data
: 1 2 3 4 5
: 14 13 9 5 2
x
y
Solution:
Let (1)y a bx
Normal equation
2
2
(2)
(3)
1 14 1 14
2 13 4 26
3 9 9
y na b x
xy a x b x
x y x xy
2
27
4 5 16 20
5 2 25 10
15 =43 =55 =97
x y x xy
(2) & (3)
43 15 15
97 15 56
a b
a b
Solving above equations we get
18.2
3.2
(1) 18.2 3.2
a
b
y x
3) The equation of straight line of best fit find the equation of best fit
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 55
: 0 1 2 3 4
: 1 1.8 3.3 4.5 6.3
x
y
Solution: let (1)y a bx
Normal equation
2
2
(2)
(3)
0 1 0 0
1 1.8 1 1.8
2 3.3
y na b x
xy a x b x
x y x xy
2
4 6.6
3 4.5 9 13.5
4 6.3 16 25.2
10 =16.9 =30 =47.1
(2) & (3)
16.9 5 10
47.1 10 30
0.72
1.33
(1)
x y x xy
a b
a b
a
b
0.72 1.33y x
4) If p is the pull required to lift a load by means of pulley block. Find a linear block of the form p=MW+C Connected p &w using following data
: 50 70 100 120
: 12 15 21 25p
Compute p when W=150. Solution: Given p=y & W=x equation of straight line is:
let (1 )y a bx
Normal equationswww.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 56
2
2
(2)
(3)
=
50 12 2500 600
70 15
y na b x
xy a x b x
x p y x xy
2
4900 1050
100 21 10000 2100
120 25 144000 3000
10 =16.9 =30 =x y x xy 47.1
(2) & (3)
73 4 340
6750 340 31800
2.27
0.187
(1) 2.27 0.187
put 150
30.32
a b
a b
a
b
y x
y
5) Fit a curve of the form xy ab
Solution: Consider
y
(1)
Take log on both side
logy=log
=loga+log
logy=loga+log
(2) ; logy=Y y=e
Corresponding normal equation loga=
x
x
x
x
y ab
ab
b
b
y A Bx
A
B
2
A a=e
(3) logb=B =e
(4)
Y nA B x b
xY A x B x
Solving the normal equation (3) & (4) for a & b . Substitute these values in (1) we get
curve of best fit of the form xy ab
6) Fit a curve of the curve xy ab for the data
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 57
: 1 2 3 4 5 6 7 8
: 1.0 1.2 1.8 2.5 3.6 4.7 6.6 9.1
x
y
Solution:
2
2
(1)
Normal equations are
(2); A=loga
(3) Y=logy, B=logb
Y=logy
xlet y ab
Y nA B x
xY A x B x
x y x
1 1.0 0 1 0
2 1.2 0.182 4 0.364
3 1.8 0.587
xY
9 1.761
4 2.5 0.916 16 3.664
5 3.6 1.280 25 6.4
6 4.7 1.547 36 9.282
7 6.6 1.887 49 13.209
8 9.1 2.208 64 17.664
36 Y =8.607 x
2
A
B
=204 =52.34
(1) & (2)
8.607 8 36
52.34 36 203
0.382
0.324
then e 0.682
e 1.382
(1) 0.682 1.382x
x xY
A B
A B
A
B
a
b
y
7) Fit a curve of the form by ax for the following data
: 1 1.5 2 2.5
: 2.5 5.61 10.0 15.6
x
y Solution:: Consider
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 58
2
2
(1)
Normal equations are
(2); Y=logy
(3) A=loga, X=logx
X= log
blet y ax
Y nA b x
XY A X B X
x y x X
log XY
1 2.5 0 0 0.916 0
1.5 5.62 0.405 0.164 1.7
Y y
26 0.699
2 10.0 0.693 0.480 2.302 1.595
2.5 15.6 0.916 0.839 2.747 2.5
2
A
2
16
X=2.014 X =1.483 Y=7.691 XY=4.81
(1) & (2)
7.691 4 2.014
4.81 2.014 1.483
A=0.916, e 2.499 2.5
1.999 2
(1) 2.5
A b
A b
a
b
y x
8) Fit a parabola 2y a bx cx for the following data
: 1 2 3 4
: 1.7 1.8 2.3 3.2
x
y Sol:
2
2
2 3
2 2 3 4
(1)
Normal equation
(2)
(3)
(4)
y a bx cx
y na b x c x
xy a x b x c x
x y a x b x c x
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 59
2 3 4 2
1 1.7 1 1 1 1.7
x y x x x xy x y
1.7
2 1.8 4 8 16 3.6 7.2
3 2.3 9 27 81
2 3 4 2
6.9 20.7
4 3.2 16 64 256 12.8 51.2
9 30 100 354 25 80.8
(1), (3) &
x y x x x xy x y
2
(4)
9 4 10 30
25 10 30 100
80.8 30 100 354
2
-0.5
0.2
(1) 2 - 0.5 0.2
a b c
a b c
a b c
a
b
c
y x x
9) Fit a curve of the form bxy ae for the following
: 0 2 4
: 8.12 10 31.82
x
y Sol:
2
2
(1)
Normal equation
Y=nA+b (2); Y=logy
(3) A=loga
y Y=logy
0
bxy ae
x
xy A x b x
x x xY
8.12 2.094 0 0
2 10 2.302 4 4.604
4 31.82 3..46 16 2
0..341
13.84
6 Y=7.86 20 18.444
(2) & (3)
7.856=3A+6b
18.444=6A+20b
A=1.935, 6.924
b=0.341
(1) 6.924
A
x
x x xY
a e
y e
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 60
10) Fit a II degree parabola 2ax bx c to the least square method & hence
find y when x=6
: 1 2 3 4 5
: 10 12 13 16 19
x
y
Sol:
2
2
2
2 3
2 2 3 4
(1)
(1)
Normal equation
(2)
(3)
(4)
y ax bx c
y c bx ax
y nc b x a x
xy c x b x a x
x y c x b x a x
2 3 4 2
1 10 1 1 1 10 10
2 12 4 8 16 24
x y x x x xy x y
14
3 13 9 27 81 39 117
4 16 16 64 256 64 256
5 19 25 125 625 95
2
475
15 70 55 225 979 232 906
70 5 15 55
232 15 55 225
906 55 225 979
0.285
0.485
9.4
(1) 0.285 0.485 9.4
at 6, 22.6
c b a
c b a
c b a
a
b
c
y x x
x y
OPTIMIZATION:Optimization is a technique of obtaining the best result under the given conditions. Optimization means maximization or minimization
LINEAR PROGAMMING:Linear Programming is a decision making technique under the given constraints on the condition that the relationship among the variables involved is linear. A general relationship among the variables involved is called objective function. The variables involved are called decision variables.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 61
The optimization of the objective function Z subject to the constraints is the mathematical formulation of a LPP.
A set real values ),........,( 21 nxxxX which satisfies the constraint BAX )( is called solution.
A set of real values xi which satisfies the constraints and also satisfies non negativity
constraints 0ix is called feasible solution.
A set of real values xi which satisfies the constraints along with non negativity restrictions and optimizes the objective function is called optimal solution.
An LPP can have many solutions.
If the optimal value of the objective function is infinity then the LPP is said to have unbounded solutions. Also an LPP may not possess any feasible solution.
GRAPHICAL METHOD OF SOLVING AN LPP
LPP involved with only two decision variables can be solved in this method. The method is illustrated step wise when the problem is mathematically formulated.
The constraints are considered in the form of 1
b
y
a
x
which graphically represents straight lines passing through the points (a,0) and (0,b) since there are only two decision variables.These lines along with the co-ordinate axes forms the boundary of the region known as the feasible region and the figure so formed by the vertices is called the convex polygon.
The value of the objective function nn xcxcxcZ ......2211 is found at all these vertices.The extreme values of Z among these values corresponding the values of the decision variables is required optimal solution of the LPP.
PROBLEMS:
1) Use the graphical method to maximize Z = 3x + 4y subject to the constraints .0,0,18052,402 yxyxyx
Solution: Let us consider the equations
)2.(..........1
3690)1...(..........1
4020
18052402
yxyx
yxyxwww.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 62
Let (1) and (2) represent the straight lines AB and CD respectively where we have A = (20, 0), B = (0, 40) ; C = (90 , 0), D = (0,36)We draw these lines in XOY plane.
Shaded portion is the feasible region and OAED is the convex polygon. The point E being the point of intersection of lines AB and CD is obtained by solving the equation:
)35,5.2(),(.18052,402 yxEyxyx
The value of the objective function at the corners of the convex polygin OAED are tabulated.
Corner
Value of Z = 3x + 4y
O(0,0) 0A(20, 0) 60E(2.5, 35) 147.5D(0,36) 144
35,5.25.147)( yxwhenZThus Max
2) Use the graphical method to maximize 21 53 xxZ subject to the constraints .0,,600,1500,20002 2122121 xxxxxxx
Solution: Let us consider the equations
)3.........(600)2.(..........115001500
)1.....(110002000
600,1500,2000
22121
22121
xxxxx
xxxxx
Let (1) and (2) represent the straight lines AB and CD respectively where we have A = (2000, 0), B = (0, 1000) ; C = (1500 , 0), D = (0,1500)
.6002 x is a line parallel to the 1x axis.We draw these lines in XOY plane.www.vt
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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On solving
)600,800(,600,20002
)600,900(,600,1500
)500,1000(,1500,2000
221
221
2121
Ggetwexxx
Fgetwexxx
Egetwexxxx
Also we have C = (1500 , 0), H= (0, 600)
The value of the objective function at the corners of the convex polygin OAED are tabulated.
Corner Value of 21 53 xxZ
O(0,0) 0C(1500, 0) 4500E(1000, 500) 5500F(900, 600) 5700G(800,600) 5400
600,9005700)( 21 xxwhenZThus Max
3) Use the graphical method to minimize 21 1020 xxZ subject to the constraints .0,,6034,303,402 21212121 xxxxxxxx
Solution: Let us consider the equations
)3.........(12015
)2.(..........13010
)1.....(12040
60334,303,402
212121
212121
xxxxxx
xxxxxx
Let (1) ,(2) and (3) represent the straight lines AB ,CD and EF respectively where we have A = (40, 0), B = (0,20) ; C = (10 , 0), D = (0,30) ; E = (15,0), F = (0, 20)We draw these lines in XOY plane.www.vt
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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Shaded portion is the feasible region and EAHG is the convex polygon. The point G being the point of intersection of lines EF and CD. The point H being the point of intersection of lines AB and CDis obtained by solving the equation:
The value of the objective function at the corners of the convex polygin OAED are tabulated.
Corner Value of 21 53 xxZ
A(15 , 0) 300A(40, 0) 800H(4 , 18) 260G (6, 12) 240
12,6240)( 21 xxwhenZThus MIN
4) Show that the following LPP does not have any feasible solution.Objective function for maximization:Z = 20x+30y.
Constraints: .0,0,6397,2443 yxyxyx
Solution: Let us consider the equations
)2.(..........1
79)1...(..........1
68
63972443
yxyx
yxyx
Let (1) and (2) represent the straight lines AB and CD respectively where we have A = (8, 0), B = (0,6) ; C = (9 , 0), D = (0,7)We draw these lines in XOY plane.
It is evident that there is no feasible region.Thus we conclude that the LPP does not have any feasible solution.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 65
SIMPLEX METHOD
Simplex method is an efficient algebraic method to solve a LPP by systematic procedure and hence an algorithm can be evolved called the simplex algorithm.
In this method it is necessary that all the constraints in the inequality formn is converted into equality form thus arriving at a system of algebraic equations.
If the constraint is involved with we add a non zero ariables 0)(1 says to the LHS to make it an equality and the same variable is called slack variable.
LPP with all constraints being equalities is called a standard form of LPP.
A minimizing LPP is converted into an equivalent maximization problem. Minimizing the given objective function P is equivalent to maximizing –P under the same constraints and Min.P = -(Max value of –P )
PROBLEMS:
1) Use Simplex method to maximize z= 2x + 4y subject to the constraints 0,0,2432,223 yxyxyx
Solution: Let us introduce slack variables 1s and 2s to the inequalities to write them in
the following form.
1 2
1 2
1 2
3 1. 0. 22
2 3 0. 1. 24
2 4 0. 0. is the objective function
x y s s
x y s s
x y s s z
Solution by the simplex method is presented in the following table.
Thus the maximum value of Z is 32 at x=0 and y=8
NZV x y1s 2s Qty Ratio
1s 3 1 1 0 22 22/1=22 8 is least, 3 is pivot,
2s is replaced by y.
Also 1/3.R2 2s 2 3 0 1 24 24/3=8
Indicators(∆)
-2 -4 0 0 0
1s 3 1 1 0 221 2 1R R R
y 2/3 1 0 1/3 8∆ -2 -4 0 0 0
3 2 34R R R
1s 7/3 0 1 -1/3 14
y 2/3 0 0 1/3 8∆ 2/3 0 0 4/3 32 No negative indicators
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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2) Use Simplex method to maximize Z =2x + 3y + z subject to the constraints 3 2 11, 2 5 19,3 4 25 0, 0, 0x y z x y z x y z x y z
Solution: Let us introduce slack variables 1s , 2s , 3s to the inequalities to write them in
the following form.
3
3
3
3
1 2
1 2
1 2
1 2
3 2 1. 0. 0. 11
2 5 0. 1. 0. 19
3 4 0. 0. 1. 25
2 3 0. 0. 0. is theobjectivefunction
x y z s s s
x y z s s s
x y z s s s
x y z s s s P
Solution by the simplex method is presented in the following table.
Thus the maximum value of P is 19 at x = 8 and y = 1, z = 0
NZV x y z1s 2s 3s Qty Ratio
1s 1 3 2 1 0 0 11 11/3=3.33 3.33 is least, 3 is pivot 1s is replaced
by y. Also 1/3.R1 2s 1 2 5 0 1 0 19 19/3=9.5
3s 3 1 4 0 0 1 35 25/1=25
Indicators(∆)
-2 -3 -1 0 0 0 0
y 1/3 1 2/3 1/3 0 0 11/3
2s 1 2 5 0 1 0 192 1 22R R R
3s 3 1 4 0 0 1 253 1 3R R R
∆ -2 -3 -1 0 0 0 04 1 43R R R
Y 1/3 1 2/3 1/3 0 0 11/3 11/3÷1/3=11 8 is least, 8/3 is pivot 3s is replaced
by x. Also 3/8.R3 2s 1/3 0 11/3 -2/3 1 0 35/3 35/3÷1/3=35
s3 8/3 0 10/3 -1/3 0 1 64/3 64/3÷8/3=8∆ -1 0 1 1 0 0 11y 1/3 1 2/3 1/3 0 0 11/3
1 3 11/ 3R R R
2s 1/3 0 11/3 -2/3 1 0 35/32 3 21/ 3R R R
x 1 0 10/8 -1/8 0 3/8 8∆ -1 0 1 1 0 0 1
4 3 4R R R y 0 1 1/4 3/8 0 -
1/81
2s 0 0 13/4 -5/8 1 -1/8
9
x 1 0 10/8 -1/8 0 3/8 9∆ 0 0 9/4 7/8 0 3/8 19 No negative
indicatorswww.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 67
3) Use Simplex method to minimize P =x - 3y + 2z subject to the constraints 3 2 7, 2 4 12, 4 3 8 10 0, 0, 0x y z x y x y z x y z
Solution: The given LPP is equivalent to maximizing the objective function –P subject to the same constraintsThat is –P=P’=-x + 3y - 2z is to be maximizedLet us introduce slack variables 1s , 2s , 3s to the inequalities to write them in the
following form.
3
3
3
3
1 2
1 2
1 2
1 2
3 2 1. 0. 0. 7
2 4 0 0. 1. 0. 12
4 3 8 0. 0. 1. 10
3 2 0. 0. 0. is theobjectivefunction
x y z s s s
x y z s s s
x y z s s s
x y z s s s P
Solution by the simplex method is presented in the following table.
Thus the maximum value of P is 19 at x = 8 and y = 1, z = 0
NZV x y z1s 2s 3s Qty Ratio
1s 3 -1 2 1 0 0 7 7/-1=-7 3 is least, 4 is pivot 2s is
replaced by y. Also 1/4.R2
2s -2 4 0 0 1 0 12 12/4=3
3s -4 3 8 0 0 1 10 10/3=3.3
Indicators(∆)
1 -3 2 0 0 0 0
s1 3 -1 2 1 0 0 71 1 2R R R
y -1/2 1 0 0 1/4 0 3
3s -4 3 8 0 0 1 103 2 33R R R
∆ 1 -3 2 0 0 0 04 2 43R R R
s1 5/2 0 2 1 1/4 0 10 10÷5/2=4 4 is least, 5/2 is pivot, 1s is
replaced by x. Also 2/5.R1
y -1/2 1 0 0 1/4 0 3 3÷-1/2=-6
3s -5/2 0 8 0 -3/4 1 1 1÷-5/2=-2/5
∆ -1/2 0 2 0 3/4 0 9x 1 0 4/5 2/5 1/10 0 4y -1/2 1 0 0 1/4 0 3
2 1 21/ 2R R R
3s -5/2 0 8 0 -3/4 1 1 3 1 35 / 2R R R
∆ -1/2 0 2 0 3/4 0 94 1 41/ 2R R R
x 1 0 4/5 2/5 1/10 0 4y 0 1 2/5 1/5 3/10 0 5
3s 0 0 10 1 -1/2 1 11
∆ 0 0 12/5 1/5 4/5 0 11 No negative indicators
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 68
UNIT VNUMERICAL METHODS - 1
CONTENTS:
Introduction
Numerical solution of algebraic and transcendental equations
Regula –falsi method
Newton –Raphson method
Iterative methods of solution of a system of equation
Gauss –Seidel method
Relaxation method
Largest eigen value and the corresponding eigen vector by
Rayeigh’s power method
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 69
NUMERICAL METHODS-1Introduction:Limitations of analytical methods led to the evolution of Numerical methods. Numerical Methods often are repetitive in nature i.e., these consist of the repeated execution of the same procedure where at each step the result of the proceeding step is used. This process known as iterative process is continued until a desired degree of accuracy of the result is obtained.
Solution of Algebraic and Transcendental Equations:
The equation f(x) = 0 said to be purely algebraic if f(x) is purely a polynomial in x.If f(x) contains some other functions like Trigonometric, Logarithmic, exponential etc. then f(x) = 0 is called a Transcendental equation.
Ex: (1) x4 - 7x3 + 3x + 5 = 0 is algebraic(2) ex - x tan x = 0 is transcendental
Method of false position or Regula-Falsi Method:This is a method of finding a real root of an equation f(x) = 0 and is slightly an improvisation of the bisection method.Let x0 and x1 be two points such that f(x0) and f(x1) are opposite in sign.
Let f(x0) > 0 and f(x1) < 0The graph of y = f(x) crosses the x-axis between x0 and x1
Root of f(x) = 0 lies between x0 and x1www.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 70
Now equation of the Chord AB is
)1...()xx(xx
)x(f)x(f)x(fy 0
01
010
When y =0 we get x = x2
)2...()x(f)x(f)x(f
xxxx.e.i 0
01
0102
Which is the first approximationIf f(x0) and f(x2) are opposite in sign then second approximation
)x(f)x(f)x(f
xxxx 0
02
0203
This procedure is continued till the root is found with desired accuracy.
Poblems:
1. Find a real root of x3 - 2x -5 = 0 by method of false position correct to three decimal places between 2 and 3.
Answer:Let f(x) = x3 - 2x - 5 = 0
f(2) = -1f(3) = 16
a root lies between 2 and 3Take x0 = 2, x1 = 3 x0 = 2, x1 = 3
)x(f)x(f)x(f
xxxxNow 0
01
0102
)1(116
232
= 2.0588 f(x2) = f(2.0588) = -0.3908
Root lies between 2.0588 and 3Taking x0 = 2.0588 and x1 = 3
f(x0) = -0.3908, f(x1) = 16
)x(f.)x(f)x(f
xxxxgetWe 0
01
0103
)3908.0(3908.16
9412.00588.2
= 2.0813f(x3) = f(2.0813) = -0.14680
Root lies between 2.0813 and 3
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 71
Taking x0 = 2.0813 and x1 = 3f(x0) = 0.14680, f(x1) =16
0897.2)14680.0(1468.16
9187.00813.2x 4
Repeating the process the successive approximations arex5 = 2.0915, x6 = 2.0934, x7 = 2.0941, x8 = 2.0943
Hence the root is 2.094 correct to 3 decimal places.
2. Find the root of the equation xex = cos x using Regula falsi method correct to three decimal places.
Solution: Let f(x) = cosx - xex
Observe f(0) = 1 f(1) =cos1 - e = -2.17798
root lies between 0 and 1Taking x0 = 0, x1 = 1f(x0) = 1, f(x1) = -2.17798
)x(f.)x(f)x(f
xxxx 0
01
0102
31467.0)1(17798.3
10
f(x2) = f(0.31467) = 0.51987 +ve Root lies between 0.31467 and 1
x0 = 0.31467, x1 = 1f(x0) = 0.51987, f(x1) = -2.17798
44673.0)51987.0(51987.017798.2
31467.0131467.0x3
f(x3) = f(0.44673) = 0.20356 +ve Root lies between 0.44673 and 1
49402.020356.038154.2
55327.044673.0x 4
Repeating this process
x5 = 0.50995, x6 = 0.51520, x7 = 0.51692, x8 = 0.51748x9 = 0.51767, etc
Hence the root is 0.518 correct to 4 decimal places
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 72
Newton Raphson Method
This method is used to find the isolated roots of an equation f(x) = 0, when the derivative of f(x) is a simple expression.
Let m be a root of f(x) = 0 near a. f(m) = 0We have by Taylor's series
.....)a(f!2
)ax()a(f)ax()a(f)x(f ''
2'
.....)a(f)am()a(f)m(f ' Ignoring higher order termsf(m) = f(a) + (m - a) f' (a) = 0
)a(f
)a(famor
'
)a(f
)a(famor
'
Let a = x0, m = x1
ionapproximatfirst theis)x(f
)x(fxxthen
0'
001
.
.
.
ionapproximatsecond theis)x(f
)x(fxx
1'
112
MethodRaphsonNewtonforformulaiterativetheis)x(f
)x(fxx
k'
kk1k
1. Using Newton's Raphson Method find the real root of x log10 x = 1.2 correct to four decimal places.Answer:Let f(x) = x log10 x - 1.2 f(1) = -1.2, f(2) = -0.59794, f(3) = 0.23136
10log
xlog1)x(f2.1
10log
xlogx)x(fhaveWe
e
e'
e
e
xlogelog 1010
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 73
K1010
k10kk1k xlogelog
2.1xlogxxx
Let x0 = 2.5 (you may choose 2 or 3 also)
7465.25.2logelog
2.15.2log5.25.2x
1010
101
7406.27465.2logelog
2.17465.2log7465.27465.2x
10102
Repeating the procedure7406.2x3
equationgiventheofroottheis7406.2x
2.Using Newton's Method, find the real root of xex = 2. Correct to 3 decimal places.Answer:Let f(x) = xex - 2 f(0) = -2 f(1) = e - 2 = 0.7182Let x0 = 1 f' (x) = (x + 1) ex
We have
kxk
kxk
k1ke)1x(
2exxx
8678.0e2
2e1x1
8527.0e)8678.1(
2e)8678.0(8678.0x
8678.0
8678.0
2
8526.0e)8527.1(
2e)8527.0(8527.0x
8527.0
8527.0
3
placesdecimal3toCorrect.rootrequiredtheis,8526.0x
3. Find by Newton's Method the real root of 3x = cosx + 1 near 0.6, x is in radians. Correct for four decimal places.
Answer:Let f(x) = 3x - cosx - 1 f'(x) = 3 + sinx
k
1kkk1k xsin3
xcosx3xx
6071.0)6.0(sin3
1)6.0(cos)6.0(36.0x6.0xWhen 10
6071.0)6071.0(sin3
1)6071.0(cos)6071.0(36071.0x 2
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 74
Since x1 = x2
The desired root is 0.6071
4. Obtain the iterative formula for finding the square root of N and find 41Answer:
NxLet or x2 - N = 0 f(x) = x2 - N f'(x) = 2xNow
k
2k
k1k x2
Nxxx
k
kk x2
N
2
xx
k
k1k x
Nx
2
1x.e.i
41findTo
4136thatObserve
6xChoose 0
4166.66
416
2
1x1
4031.64166.6
414166.6
2
1x 2
4031.64031.6
414031.6
2
1x3
Since x2 = x3 = 6.4031
4031.641ofvalueThe
5. Obtain an iterative formula for finding the p-th root of N and hence find (10)1/3
correct to 3 decimal places.Answer:Let xp = Nor xp - N = 0Let f(x) = xp - N
1p' px)x(f
1pk
pk
k1kpx
NxxxNow
Observe that 8 < 10
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 75
3/13/1 108 3/1)10(2.e.i
Use x0 = 2, p = 3, N=10
1666.2)2(3
1022x
2
3
1
1545.2)1666.2(3
10)1666.2(1666.2x
2
3
2
1544.2)1545.2(3
10)1545.2(1545.2x
2
3
3
1544.2)10( 3/1
6. Obtain an iterative formula for finding the reciprocal of p-th root of N. Find (30)-1/5 correct to 3 decimal places.
Answer:Let x -p = Nor x -p - N = 0 f(x) = x -p - N f'(x) = -px -p - 1
Now
1p
k
pk
k1kxp
Nxxx
5.02
1)32(cesin 5/1
We use x0 = 0.5, p = 5, N = 30
processthepeatingRe,50625.0)5.0(5
30)5.0(5.0x
6
5
1
506495.0x,506495.0x 32
5065.0)30( 5/1
GAUSS-SEIDAL ITERATION METHOD (to solving the system linear simultaneous equations.)
Example 1.Use Gauss-Seidal iteration method to solve the following system of equations. 3x + 20y - 2z = -18 20x + y - 2z = 17 2x - 3y + 20z = 25
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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Solution:. Rearranging thegiven system of equations 20x + y - 2z = 17 3x + 20y - 2z = -18 ………………….(1) 2x - 3y + 20z =25
The above system equations is arranged such that, diagonally dominant. System (1)
x =
201 [ 17- y +2z ]
y =
20
1 [-18-3x + z ] ………… (2)
z =
201 [25 –2x + 3y ]
Let the initial approximations to the solution of the system (A) be
,0)0( x ,0)0( y 0)0( zFirst Iteration:Using ( 2 )
)1(x201 [ 17- y
)0(
+2z)0(
]
)1(x201 [ 17- 0 + 2(0) ] = 8500.0
y)1(
=
20
1 [-18-3x)1(
+ z)0(
]
y)1(
=
201 [ -18-3(0.8500 ) + 0 ] = -1.0275
z)1(
=
201 [25 –2x
)1(
+ 3y)1(
]
z)1(
=
201 [25 –2(0.8500) + 3(- 1.0275)]
= 1.0109
Second Iteration: Usining ( 2 )
)2(x201 [ 17- y
)1(
+2z)1(
]
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 77
)2(x201 [ 17- (-10275) + 2(1.0109 ) ]
= 1.0025
y)2(
=
201 [-18-3x
)2(
+ z)1(
]
y)2(
=
201 [ -18-3(1.0025 ) + 1.0109 ]
= - 0.99928
z)2(
=
20
1 [25 –2x)2(
+ 3y)2(
]
z)2(
=
201 [25 –2(1.0025) + 3(- 0.99928)]
= 0.9998
Third Iteration: Usining ( 2 )
)3(x201
[ 17- y)2(
+2z)2(
]
)3(x201 [ 17- (- 0.99928)+ 2(0.9998) ]
= 1.0000
y)3(
=
201 [-18-3x
)3(
+ z)2(
]
y)3(
=
20
1 [ -18-3(1.0000) +0.9998 ]
= -1.0000
z)3(
=
201 [25 –2x
)3(
+ 3y)3(
] www.vtuc
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 78
z)3(
=
201 [25 –2(1.00) + 3(-1.0000)]
= 1.0000
Answer after three iterations x =1.000, y = -1.0000 and z = 1.000
2) Use Gauss-Seidal iteration method to solve the following system of equations. x + y +54z=110 27x +6 y - z = 85 …………(A) 6x+15y + 2z =72Solution :. Rearranging the system of equations (A)
27x +6 y - z = 85
6x+15y + 2z =72 (B) x + y +54z=110
The above system equations is arranged such that, the system is diagonally dominant.
System (B) x =
271 [ 85-6 y +z ]
y =
151 [72-6x -2z ] (C)
z =
541 [110 –x -y ]
Let the initial approximations to the solution of the system (A) be
,1)0( x ,0)0( y 0)0( zFirst Iteration:Using ( C )
)1(x271 [ 85-6 y
)0(
+z)0(
]
)1(x271
[ 85- 6 (0) +0] =3.148148
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 79
y)1(
=
151
[72- 6x)1(
-2 z)0(
] y)1(
=
151 [72- 6(3.148148) -2 (0) ]
=3.54074
z)1(
=
541 [110 –x
)1(
-y)1(
]
z)1(
=
541 [110 –3.148148 -3.54074 ]=1.913168
Second Iteration: Usining ( C )
)2(x27
1 [ 85-6 y)1(
+z)1(
]
)2(x271 [ 85- 6 (3.54074) +1.913168]
=2.432175
y)2(
=
151 [72- 6x
)2(
-2 z)1(
]
y)2(
=
151 [72- 6(2.432175) -2( 1.913168) ]
= 3.57204
z)2(
=
541 [110 –x
)2(
-y)2(
]
z)2(
=
541 [110 –2.432175 -3.57204 ]
=1.925837
ert45t
Third Iteration: Usining ( C )
)3(x271 [ 85-6 y
)2(
+z)2(
]
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 80
)3(x27
1 [ 85- 6 (3.57204) +2(1.925837)]
=2.425689
y)3(
=
151 [72- 6x
)3(
-2 z)2(
]
y)3(
=
15
1 [72- 6(2.425689) -2( 1.913168) ]
=3.57313
z)3(
=
541 [110 –x
)2(
-y)2(
]
z)3(
=
541 [110 –2.432175 -3.57204 ]
=1.925947
Answer after three iterations x =2.425689, y = 3.57313 and z =
RELAXATION METHOD:
The method is illustrated for the following diagonally dominant system of three independent equations in three unknowns.
11 1 12 2 13 3 1
21 1 22 2 23 3 2
31 1 32 2 33 3 3
a x a x a x b
a x a x a x b
a x a x a x b
From these equations the residuals 1 2 3, ,R R R are defined as follows
1 11 1 12 2 13 3 1
2 21 1 22 2 23 3 2
3 31 1 32 2 33 3 3
R a x a x a x b
R a x a x a x b
R a x a x a x b
The values of 1 2 3, ,x x x that make the residuals 1 2 3, ,R R R simultaneously zero
constitutes the exact solution of the system of equations. If this is not possible we need to make the residuals as close to zero as possible.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 81
The values of 1 2 3, ,x x x at that stage constitutes an approximate solution. The
relaxation method aims in reducing the values of residuals as close to zero as possible by modifying the value of the variables at every stage.
At every step the numerically largest residual is reduced to zero by choosing an appropriate integral value for the corresponding variable.
These integral values are called the increments in 1 2 3, ,x x x denoted by 1 2 3, ,x x x respectively. If the system possess exact solution then 1 2 3, ,R R R becomes zero
simultaneously and the required solution will be the sum of all these increments.i.e., 1 1 2 2 3 3, ,x x x x x x
When the system does not possess exact solution, we need to continue the process b y magnifying the prevailing residuals on multiplication with 10. Later we divide by 10 to obtain the solution correct to one decimal place.
The process will be contined to meet the requirement of the solution to the desired accuracy.
PROBLEMS:1) Solve the following system of equations by relaxation method.
1 2 3
1 2 3
1 2 3
12 31
2 8 24
3 4 10 58
x x x
x x x
x x x
Solution: The given system of equations are diagonally dominant.
1 1 2 3
2 1 2 3
3 1 2 3
12 31
2 8 24
3 4 10 58,
Let R x x x
R x x x
R x x x
be the residuals.
Let 1 2 3, ,x x x respectively represent the increments for 1 2 3, ,x x x in the following
relaxation table.
1x 2x 3x 1R 2R 3R
0 0 0 -31 -24 -580 0 6 -25 -30 20 4 0 -21 2 182 0 0 3 6 240 0 -2 1 8 40 -1 0 0 0 0
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 82
1 2 3, ,R R R are all zero and the exact solution of the given system is
1 1 2 2 3 32, 4 1 3, 6 2 4x x x x x x Thus ( 1 2 3, ,x x x ) = (2, 3, 4 ) is the solution of the given system of equations.
2) Solve the following system of equations by relaxation method.
10 2 3 205
2 10 2 154
2 10 120
x y z
x y z
x y z
Solution: The given system of equations are diagonally dominant.
1
2
3
10 2 3 205
2 10 2 154
2 10 120,
Let R x y z
R x y z
R x y z
be the residuals.
Let , ,x y z respectively represent the increments for , ,x y z in the following relaxation table.
x y z 1R 2R 3R
0 0 0 -205 -154 -12021 0 0 5 -196 -1620 20 0 -35 4 -1820 0 18 -89 -32 -29 0 0 1 -50 -200 5 0 -9 0 250 0 3 -18 -6 52 0 0 2 -10 10 1 0 0 0 0
1 2 3, ,R R R are all zero and the exact solution of the given system is
32, 26, 21x x y y z z Thus ( , ,x y z ) = (32 , 26, 21 ) is the solution of the given system of equations.
Rayleigh’s power method:
Given square matrix A, if there exist a scalar λ and a non zero column matrix X, such that AX = λ X, then λ is called an eigen value of A and X is called an eigen vector of A corresponding to an eigen value λ.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 83
Rayleigh’s power method is an iterative method to determine the numerically largest eigen value and the corresponding eigen vector of a square matrix.
Working procedure: Suppose A is the given square matrix, we assume initially an eigen vector 0X in a
simple for like [1,0,0] [0,1,0] [0,0,1] [1,1,1]or or or and find the matrix
product 0AX which will also be a column matrix.
We take out the largest element as the common factor to obtain 110AX X .
We then find 1
AX and again put in the form 1 22AX X by normalization.
The iterative process is continued till two consecutive iterative values of λ and X are same upto a desired degree of accuracy.
The values so obtained are respectively the largest eigen value and the corresponding eigen vector of the given square matrix A.
Problems:
1) Using the Power method find the largest eigen value and the corresponding eigen vector starting with the given initial vector.
Tgiven 001
201
020
102
0 1 1
1 2 2
2 0 1 1 2 1
: 0 2 0 0 0 2 0
1 0 2 0 1 0.5
2 0 1 1 2.5 1
0 2 0 0 0 2.5 0
1 0 2 0.5 2 0.8
Solution AX X
AX X
2 3 3
2 0 1 1 2.8 1
0 2 0 0 0 2.8 0
1 0 2 0.8 2.6 0.93
AX X
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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3 4 4
4 5 5
5
2 0 1 1 2.93 1
0 2 0 0 0 2.93 0
1 0 2 0.93 2.86 0.98
2 0 1 1 2.98 1
0 2 0 0 0 2.98 0
1 0 2 0.98 2.96 0.99
2 0 1 1
0 2 0 0
1 0 2 0.99
AX X
AX X
AX
6 6
6 7 7
2.99 1
0 2.99 0
2.98 0.997
2 0 1 1 2.997 1
0 2 0 0 0 2.997 0
1 0 2 0.997 2.994 0.999
X
AX X
Thus the largest eigen value is approximately 3 and the corresponding eigen vector is [1 ,0, 1]’
2) Using the Power method find the largest eigen value and the corresponding eigen vector starting with the given initial vector.
Tgiven 8.08.01
512
132
114
0 1 1
1
4 1 1 1 5.6 1
: 2 3 1 0.8 5.2 5.6 0.93
2 1 5 0.8 5.2 0.93
4 1 1 1 5.86 1
2 3 1 0.93 5.72 5.86 0.98
2 1 5 0.93 5.72 0.98
Solution AX X
AX
2 2X
2 3 3
4 1 1 1 5.96 1
2 3 1 0.98 5.92 5.96 0.99
2 1 5 0.98 5.92 0.99
AX X
3 4 4
4
4 1 1 1 5.98 1
2 3 1 0.99 5.96 5.98 0.997
2 1 5 0.99 5.96 0.997
4 1 1 1 5.994 1
2 3 1 0.997 5.988 5.994 0.999
2 1 5 0.997 5.988 0.999
AX X
AX
5 5X
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 85
Thus after five iterations the numerically largest eigen value is 5.994 and corresponding eigen vector is [1, 0.999, -0,999]’
UNIT VI
NUMERICAL METHOD -2
CONTENTS:
Introduction
Finite difference:
Forward and Backward difference
Newtons forward and backward interpolation formula
Newtons divided difference formula
Lagranges interpolation formula
Numerical integration
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 86
NUMERICAL METHODS-2
Finite Differences
Let y = f(x) be represented by a table
x : x0 x1 x2 x3 …. xn
y : y0 y1 y2 y3 … yn
where x0, x1,x2….xn are equidistant. (x1 - x0 = x2 - x1 = x3 - x2 =….=xn - xn-1 = h)
We now define the following operators called the difference operators.
Forward difference operator (∆)
)x(f)hx(f)x(f
1n,...,2,1,0r,yyy r1rr
sdifferenceforwardfirst
yyy
.
.
yyy
yyy
1nn1n
121
010
sdifferenceondsecthecalledare,....,y,y,y 22
12
02
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 87
)yy()y(yNow 01002
)yy()yy(yy 011201
012 yy2y
12312ly yy2yy|||
r1r2rr2 yy2yy
012303 yy3y3yy:Note
rk
2kr2k
1kr1k
krrk C)1(....yCyCyy
Difference Tablex y y 2y 3y 4y 5y
x0 y0
y0
x1 y1 2y0
y1 3y0
x2 y2 2y1 4y0
y2 3y0
x3 y3 2y2
y3
x4 y4
.sdifferenceleadingthecalledare,....y,y,y 03
02
0
Ex: The following table gives a set of values of x and the corresponding values ofy = f(x)
x : 10 15 20 25 30 35
y : 19.97 21.51 22.47 23.52 24.65 25.89
Form the difference table and find )15(f),20(f),10(f),10(f 432
x y 2 3 4 5
10 19.97
1.54
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 88
15 21.51 -0.58
0.96 0.67
20 22.47 0.09 -0.68
1.05 -0.01 0.72
25 23.52 0.08 0.04
1.13 0.03
30 24.65 0.11
1.24
35 25.89
04.0)15(f,03.0)20(f,58.0)10(f,54.1)10(f 432
Note: The nth differences of a polynomial of n the degree are constant.
Backward difference operator ()
Let y = f(x)
We define f(x) = f(x) - f(x - h)
i.e. y1 = y1 - y0 = ∆ y0
y2 = y2 - y1 = ∆ y1
y3 = y3 - y2 = ∆ y2
'
'
yn = yn - yn-1 = ∆ yn - 1
1r1rrr yyyy
Note:
1. f(x + h) = f(x + h) - f(x) = ∆ f(x)
2. 2 f(x + 2h) = (f(x + 2h))
= {f(x + 2h) - f(x + h)}
= f(x + 2h) - f(x + h)
= f(x + 2h) - f(x + h) - f(x + h) + f(x)
= f(x + 2h) -2f(x + h) + f(x)
= ∆2 f(x)
|||ly n f(x + nh) = ∆n f(x)
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 89
Backward difference tablex y y 2y 3y
X0 y0
y1
X1 y1 2y2
y2 3y3
X2 y2 2y3
y3
X3 y3
1. Form the difference table for
x 40 50 60 70 80 90
y 184 204 226 250 276 304
and find y (30), 2y (70), 5y (90)
Soln:
x y y 2y 3y 4y 5y
40 184
20
50 204 2
22 0
60 226 2 0
24 0 0
70 250 2 0
26 0
80 276 2
28
90 304
y (80) = 26, 2y (70) = 2, 5y (90) = 0
2. Given
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 90
x 0 1 2 3 4
f(x) 4 12 32 76 156
Construct the difference table and write the values of f (4), 2f (4), 3f (3)
x y y 2y 3y
0 4
8
1 12 12
20 12
2 32 24
44 12
3 76 36
80
4 156
3) Find the missing term from the table:
x 0 1 2 3 4
y 1 3 9 - 81
Explain why the value obtained is different by putting x = 3 in 3x.
Denoting the missing value as a, b, c ..etc. Construct a difference table and solve.
x y y 2y 3y 4y
0 1 2
1 3 6 4
2 9 a - 9 a - 15 a - 19 -4a + 124
3 a 81 - a 81 - a -3a +105
4 81
Put 4y = 0 (assuming f(x) its be a polynomial of degree 3)
i.e., -4a + 124 = 0
a = 31
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 91
Since we have assumed f(x) to be a polynomial of degree 3 which is not 3x we obtained a
different value.
4) Given u1 = 8, u3 = 64, u5 = 216 find u2 and u4
x u u 2u 3u
x1 8
x2 a a - 8 -2a + 72 b + 3a - 200
x3 64 64 - a b + a - 128 -3b - a + 408
x4 b b - 64 -2b + 280
x5 216 216 -b
We carryout upto the stage where we get two entries ( 2 unknowns) and equate each of
those entries to zero. (Assuming) to be a polynomial of degree 2.
b + 3a - 200 = 0
-3b - a + 408 = 0 We get a = 24 b = 128
Interpolation:
The word interpolation denotes the method of computing the value of the function y =
f(x) for any given value of x when a set (x0, y0), (x1, y1),…(xn, yn) are given.
Note:
Since in most of the cases the exact form of the function is not known. In such cases the
function f(x) is replaced by a simpler function (x) which has the same values as f(x) for
x0, x1, x2….,xn.
....y!3
)2u()1u(uy
!2
)1u(uyuy)x( 0
30
200
0n y
!n
)1nu...()2u()1u(u
is called the Newton Gregory forward difference formula
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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Note :
1. Newton forward interpolation is used to interpolate the values of y near the beginning
of a set of tabular values.
2. y0 may be taken as any point of the table but the formula contains those values of y
which come after the value chosen as y0.
Problems:
1) The table gives the distances in nautical miles of the visible horizon for the given
heights in feet above the earths surface.
x = height 100 150 200 250 300 350 400
y = distance 10.63 13.03 15.04 16.81 18.42 19.90 21.27
Find the values of y when i) x = 120, ii) y =218
Solution:
x y 2 3 4 5 6
100 10.632.40
150 13.03 -0.392.01 0.15
200 15.04 -0.24 -0.071.77 0.08 0.02
250 16.81 -0.16 -0.05 0.021.61 0.03 0.04
300 18.42 -0.13 -0.011.48 0.02
350 19.90 -0.111.37
400 21.27
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 93
Choose x0 = 100
i) 4.050
100120u,120x
)39.0(!2
)14.0()4.0()40.2(
!1
4.063.10)120(f
)15.0(!3
)24.0()14.0()4.0(
)07.0(!4
)34.0()24.0()14.0()4.0(
)02.0(!5
)44.0()34.0()24.0()14.0()4.0(
649.11)02.0(!6
)54.0()44.0()34.0()24.0()14.0()4.0(
ii) Let x = 218, x0 = 200, 36.050
18
50
200218u
)16.0(2
)64.0(36.0)77.1(36.004.15)218(f
...)03.0(6
)64.1()64.0(36.0
= 15.7
2) Find the value of f(1.85).
x y y 2y 3y 4y 5y 6y1.7 5.474
0.5751.8 6.049 0.062
0.637 0.0041.9 6.686 0.066 0.004
0.703 0.008 -0.0042.0 7.389 0.074 0 0.004
0.777 0.008 02.1 8.166 0.082 0
0.859 0.0082.2 9.025 0.090
0.94923 9.974
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 94
5.01.0
8.185.1
h
xxu85.1x,8.1xChoose 0
0
)066.0(2
)5.0()5.0()637.0()5.0(049.6)85.1(f
)008.0(6
)5.1()5.0)(5.0(
0.00050.0008-0.31856.049
= 6.359
3) Given sin 45o = 0.7071, sin 50o = 0.7660, sin 55o =0.8192, sin 60o = 0.8660.
Find sin 48o.
x y 2 3
45 0.7071
0.589
50 0.7660 -0.0057
0.0532 0.0007
55 0.8192 -0.0064
0.0468
60 0.8660
6.0h
xxu5h;45x,48x 0
0
)0589.0()6.0(7071.048sin o
)0057.0(2
)4.0)(6.0(
7431.0)0007.0(
6
)4.1()4.0()6.0(
4) From the following data find the number of students who have obtained 45
marks. Also find the number of students who have scored between 41 and 45 marks.
Marks 0 - 40 41 - 50 51 - 60 61 -70 71 - 80
No. of students 31 42 51 35 31
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 95
x y 2 3 4
40 31
42
50 73 9
51 -25
60 124 -16 37
35 12
70 159 -4
31
80 190
2
9)5.0()5.0()42()5.0(31)45(f
!3
)25()5.1()5.0()5.0(
488672.47)37(!4
)5.2()5.1()5.0()5.0(
f(45) - f(40) = 70 = Number of students who have scored between 41 and 45.
5) Find the interpolating polynomial for the following data:
f(0) = 1, f(1) = 0, f(2) = 1, f(3) = 10. Hence evaluate f(0.5)
xy 2 3
0 1
-1
1 0 2
1 6
2 1 8
9
3 10
x1
0xu
)2(!2
)1x(x)1(x1)x(f
1x2x6
!3
)2x()1x(x 23
6) Find the interpolating polynomial for the following data:
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 96
x: 0 1 2 3 4
f(x) : 3 6 11 18 27
x y 2 3 4
0 33
1 6 25 0
2 11 2 07 0
3 18 29
4 27
x1
0xu
)2(2
)1x(x)3(x3)x(f
)0(
!x
)1x(x 2xx23
Newton Gregory Backward Interpolation formula
....y!3
)2u()1u(uy
!2
)1u(uyuyy n
3n
2nn
h
xxuwhere n
1) The values of tan x are given for values of x in the following table. Estimate
tan (0.26)
x 0.10 0.15 0.20 0.25 0.30
y 0.1003 0.1511 0.2027 0.2553 0.3093
x y 2 3 4
0.10 0.10030.0508
0.15 0.1511 0.0008
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 97
0.0516 0.00020.20 0.2027 0.0010 0.0002
0.0526 0.00040.25 0.2553 0.0014
0.05400.30 0.3093
8.005.0
3.026.0u
)054.0)(8.0(3093.0)26.0(f )0014.0()2.0(2
)8.0( 2659.0)0004.0(
6
)2.1()2.0()8.0(
2) The deflection d measured at various distances x from one end of a cantilever is
given by the following table. Find d when x = 0.95
95.0xwhen3308.0d25.02.0
195.0u
x d 2 3 4 5
0 00.0347
0.2 0.0347 0.04790.0826 -0.0318
0.4 0.1173 0.0161 0.00030.0987 -0.0321 -0.0003
0.6 0.2160 -0.016 00.0827 -0.032
0.8 0.2987 -0.04810.0346
1.0 0.3333
3) The area y of circles for different diameters x are given below:
x : 80 85 90 95 100
y : 5026 5674 6362 7088 7854
Calculate area when x = 98
x y y 2y 3y 4y80 5026
648
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 98
85 5674 40688 -2
90 6362 38 4726 2
95 7088 40766
100 7854
Answer:
4.0n
xxu n
y = 7542
4) Find the interpolating polynomial which approximates the following data.
x 0 1 2 3 4
y -5 -10 -9 4 35
x y 2 3 4
0 -5-5
1 -10 61 6
2 -9 12 013 6
3 4 1831
4 35
1
4xu
!2
18)3x()4x()31()4x(35)x(f
!3
)6()2x()3x()4x(
f(x) 5-6x 2x x 23
Interpolation with unequal intervalsNewton backward and forward interpolation is applicable only when x0, x1,…,xn-1 are
equally spaced. Now we use two interpolation formulae for unequally spaced values of x.
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 99
i) Lagranges formula for unequal intervals:
If y = f(x) takes the values y0, y1, y2,….,yn corresponding to x = x0, x1, x2,…,xn then
)x(f)xx)...(xx()xx(
)xx)...(xx()xx()x(f 0
n02010
n21
)x(f)xx)...(xx()xx()xx(
)xx)...(xx()xx()xx(1
n1312101
n320
....)x(f)xx)...(xx()xx()xx(
)xx)...(xx()xx()xx(2
n2321202
n310
)x(f)xx)...(xx()xx()xx(
)xx)...(xx()xx()xx(n
1nn2n1n0n
1n210
is known as the lagrange's
interpolation formula
ii) Divided differences ()
]x,x[xx
yyy)x(f 10
01
0100
]x,x[xx
yyy 12
12
121
]x,x[xx
yyy n1n
1nn
1nn1n
second divided difference
02
010
20
2
xx
yyy)x(f
]x,x,x[xx
]x,x[]x,x[210
02
0112
]x,x,x[xx
]x,x[]x,x[
xx
yyy|| 321
13
1223
13
121
2ly
similarly definedbecan,....y03
Newton's divided difference interpolation formula
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 100
03
21002
10000 y)xx)(xx()xx(y)xx()xx(y)xx(y)x(fy
0n
n10 y)xx...()xx()xx(...
is called the Newton's divided difference formula.
Note:Lagrange's formula has the drawback that if another interpolation value were
inserted, then the interpolation coefficients need to be recalculated.
Inverse interpolation: Finding the value of y given the value of x is called interpolation
where as finding the value of x for a given y is called inverse interpolation.
Since Lagrange's formula is only a relation between x and y we can obtain the inverse
interpolation formula just by interchanging x and y.
0n02010
n21 x.)yy)...(yy()yy(
)yy)....(yy()yy(x
...x)yy...()yy()yy()yy(
)yy)...(yy()yy()yy()yy(1
n1312101
n3200
n1nn1n0n
1n10 x.)yy)...(yy()yy(
)yy)...(yy()yy(...
is the Lagranges formula for inverse interpolation
1) The following table gives the values of x and y
x : 1.2 2.1 2.8 4.1 4.9 6.2
y : 4.2 6.8 9.8 13.4 15.5 19.6
Find x when y = 12 using Lagranges inverse interpolation formula.
Using Langrages formula
05040302010
54321 x)yy()yy()yy()yy()yy(
)yy()yy()yy()yy()yy(x
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
DEPT OF MATHEMATICS/SJBIT Page 101
445251505
43210 x)yy)...(yy()yy()yy(
)yy()yy()yy()yy()yy(....
055.0964.0419.3252.1234.0022.0
= 3.55
2) Given the values
x : 5 7 11 13 17
f(x) : 150 392 1452 2366 5202
Evaluate f(9) using (i) Lagrange's formula (ii) Newton's divided difference formula.
i) Lagranges formula
392.)177()137()117()57(
)179()139()119()59()150(
)175()135()115()75(
)179()139()119()79()9(f
)2366()1713()1113()713()513(
)179()119()79()59()1452(
)1711()1311()711()511(
)179()139()79()59(
810)5202()1317()1117()717()517(
)139()119()79()59(
f(9) = 810
ii)
5 150
121
7 392 24
265 1
11 1452 32 0
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457 1
13 2366 42
709
17 5202
f(9) = 150 + 121 (9 - 5) + 24 (9 - 5) (9 - 7) + 1(9 - 5) (9 - 7) (9 - 11) = 810
3) Using i) Langranges interpolation and ii) divided difference formula. Find the
value of y when x = 10.
x : 5 6 9 11
y : 12 13 14 16
i) Lagranges formula
13)116)(96)(56(
)1110)(910)(510(12
)115()95()65(
)1110()910()610()10(fy
16)911)(611)(511(
)910)(610)(510(14
)119)(69)(59(
)1110)(610)(510(
3
44
ii)Divided differencex y 2 3
5 12
1
6 13
6
1
4
3/2
3
1
20
1
6
10/3
690
27
5116
1
15
2
9 14
15
2
5
3/2
12
2
11 16
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20
1)910)(610)(510(
6
1).610)(510()510(12)10(f
3
44
4) If y(1) = -3, y(3) = 9, y(4) =30, y(6) = 132 find the lagranges interpolating
polynomial that takes the same values as y at the given points.
Given:
x 1 3 4 6
y -3 9 30 132
9.)63)(43)(13(
)6x)(4x)(1x()3(.
)61)(41)(31(
)6x()4x()3x()x(f
132.)46)(36)(16(
)4x)(3x)(1x(30.
)64)(34)(14(
)6x)(3x)(1x(
= x3 - 3x2 + 5x - 6
5) Find the interpolating polynomial using Newton divided difference formula for
the following data:
x 0 1 2 5
y 2 3 12 147
x y 2 3
0 2
1
1 3 4
9 1
2 12 9
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45
5 147
F(x) = 2 + (x - 0)(1) + (x - 0) (x - 1) (4) + (x - 0) (x - 1) (x - 2) 1
= x3 + x2 - x + 2
Numerical Integration:-
To find the value of b
a
dxyI numerically given the set of values ),( ii yx ,
ni ,...,2,1,0 at regular intervals.(i) Simpson’s one third rule:-
2421310 .....2.....43 nnn yyyyyyyyh
I
when n is even.
(ii) Simpson’s three-eighth rule:-
363154210 .....2....38
3 nnn yyyyyyyyyy
hI
when n is a multiple of 3.
(iii) Weddle’s rule:-
......56510
36543210 yyyyyyy
hI
when n is a multiple of 6.
Problems:
1) Using Simpson’srd
3
1rule evaluate
1
021 x
dxby dividing the interval 1,0 into
4 equal sub intervals and hence find the value of correct to four decimal places.
Solution: Let us divide [0,1] into 4 equal strips (n = 4)1 0 1
length of each strip: 4 4
1 2 1 3 4The points of division are 0, , , , 1
4 4 2 4 4
h
x
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By data2
1
1y
x
Now we have the following table
x 0 1/4 1/2 3/4 1
2
1
1y
x
1 16/17 4/5 16/25 1/2
0y 1y 2y 3y 4y
Simpson’srd
3
1rule for n = 4 is given by
0 4 1 3 24 23
b
a
hydx y y y y y 1
20
1
20
1 1/ 4 1 16 16 41 4 2. 0.7854
1 3 2 17 25 5
10.7854
1
dxx
Thus dxx
To deduce the value of π: We perform theoretical integration and equate the resulting value to the numerical value obtained.
111 1 1
2 00
1tan tan (1) tan (0)
1 4
, 0.7854 4(0.7854) 3.14164
3.1416
dx xx
Wemust have
Thus
2) Given that x 4 4.2 4.4 4.6 4.8 5 5.2
xlog 1.3863 1.4351 1.4816 1.5261 1.5686 1.6094 1.6487
Evaluate 2.5
4
log dxx using Simpson’s th
8
3rule
Solution: Simpson’sth
8
3rule for n = 6 is given by
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0 6 1 2 4 5 3
5.2
4
5.2
4
33 2
8
3(0.2)log 1.3863 1.6487 3 1.4351 1.4816 1.5686 1.6094 2 1.5261
8
log 1.8279
b
a
e
e
hydx y y y y y y y
xdx
xdx
3) Using Weddle’s rule evaluate 1
20 1
xdx
x by taking seven ordinates and hence
find loge2Solution: Let us divide [0,1] into 6 equal strips ( since seven ordinates)
1 0 1 length of each strip:
6 61 2 1 3 1 4 2 5 6
The points of division are 0, , , , , , 16 6 3 6 2 6 3 6 6
h
x
By data2
1
1y
x
Now we have the following table
x 0 1/6 1/3 1/2 2/3 5/6 1
21
xy
x
0 6/37 3/10 2/5 6/13 30/61 1/2
0y 1y 2y 3y 4y 5y 6y
Weddle’s rule for n = 6 is given by
0 1 2 3 4 5 6
1
20
1
20
35 6 5
10
3(1/ 6)0 5(6 / 37) 3 /10 6(2 / 5) 6 /13 5(30 / 61) 1/ 2
1 10
0.34661
b
a
hydx y y y y y y y
xdx
x
xdx
x
To deduce the value of loge2: We perform theoretical integration and equate the resulting value to the numerical value obtained. www.vt
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112
200
1
20
1 1 1log (1 ) log 2 log 1
1 2 2 2
1log 2
1 2
1, log 2 0.3466 log 2 2(0.3466) 0.6932
2
log 2 0.6932
e e e
e
e e
e
xdx x
x
xHence dx
x
Wemust have
Thus
UNIT VIINUMERICAL METHODS -3
CONTENTS:
Numerical solution of PDE
Finite difference approximation to derivatives
Numerical solution of
One dimensional wave equation
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One dimensional heat equation
Two dimensional Laplace’s equation
UNIT-VIINUMERICAL METHODS-3
Classification of PDE s of second orderThe general second order linear PDE in two independent variables x,yis of the formAuxx+ Buxy+ Cuyy+ Dux+ Euy+ Fu=0Where A,B,C,D,E,F are in general functions of x,yThis equation is said to be
i) Parabolic if B2-4AC=0ii) Elliptic if B2-4AC<0iii) Hyperbolic if B2-4AC>0
Now let us examine the nature of three PDE which are under our discussionP.D.E A B C B24AC=0 Nature of
PDE1 One dimensional wave
equationc2uxx-utt=0C2 0 -1 4C2>0 Hyperbolic
2 One dimensional heat equationc2uxx-ut=0
C2 0 0 0 parabolic
3 two dimensional Laplace’sequationuxx+uyy=0
1 0 C2 -4<0 elliptic
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Numerical solution of a PDEConsider a rectangular region in the x—y plane. Let us divide this region into a network of rectangular of sides h and k. In other words , we draw lines x=ih,y=jk:I,j=1,2,3…….being parallel to the Y-axis and X-axis respectively resulting into a network of rectangles.the points of intersection of these lines are called mesh points or grid points.
We write u(x,y) =u(ih,jk) and the finite difference approximation for the the partial derivatives are put in the modified form
1, ,
1x i j i ju u u
h
, 1,
1y i j i ju u u
h
1, 1,
1
2x i j i ju u uh
, 1 ,
1y i j i ju u u
k
, , 1
1y i j i ju u u
k
, 1 , 1
1
2y i j i ju u uk
1, , 1,2
12xx i j i j i ju u u u
h
, 1 , , 12
12yy i j i j i ju u u u
k
The substitution of these finite diference approximation into the given PDE converts the PDE into a finite difference equation.Now we discuss numerical solution of the three important PDE namely
1) One dimensional wave equation2) One dimensional heat equation
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3) Two dimensional laplaceequation
Numerical solution of the one dimensional wave equationWe seek the numerical solution of the wave equation
2.................................(1)xx ttc u u
Subject to the boundary conditionsu(0,t)=0………………………(2)u(l,t)=0……………………….(3)and the initial conditionsu(x,0)=f(x)……………………(4)ut(x,0)=0……………………(5)
We shall substitute the finite difference approximation for the partial derivatives present in (1)
21, , 1, , 1 , , 12 2
1 1. 2 2i j i j i j i j i j i jc u u u u u uh k
22
1, , 1, , 1 , , 122 2i j i j i j i j i j i j
hc u u u u u u
k
Taking k/h= we have
2 21, , 1, , 1 , , 12 2i j i j i j i j i j i jc u u u u u u
2 2 2 2, 1 , 1, 1, , 12 1i j i j i j i j i ju c u c u u u ………………..(6)
For convenience let us choose such that 1- 2 2c =02 2
2 22 2
1 0k k h
c or k kh h c
Thus (6) reduces to the form
, 1i ju = 1, 1, , 1i j i j i ju u u
This is called the explicit formula for the solution of wave equation.
Examples
1.Solve the wave equation 2 2
2 24
u u
t x
subject to u=(0,t) =0,u(4,t)=0,ut(x,0)=0
and u(x,0)=x(4-x) by taking h=1,k=0.5 upto four steps
Soln: Below is the initial table wherein the first and last column are zero since (0,t)=0=u(4,t)
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Now consider u(x,0)=x(4-x)
1,0 2,0 3,0(1,0) 3; (2,0) 4; 3u u u u u
Next consider
,1 1,0 1,0
1,1 0,0 2,0
2,1 1,0 3,0
1,1 0,0 4,0
1
21 1
0 4 22 21 1
3 3 32 21 1
4 0 22 2
i i iu u u
u u u
u u u
u u u
We now consider the explicit formula to find the remaining values in the table
, 1 1, 1, , 1i j i j i j i ju u u u
1,2 0,1 2,1 1,0 0 3 3 0u u u u
2,2 1,1 3,1 2,0 2 2 4 0u u u u
3,2 2,1 4,1 3,0 3 0 3 0u u u u
The third row is completed
1,3 0,2 2,2 1,1 0 0 2 2u u u u
2,3 1,2 3,2 2,1 0 0 3 3u u u u
3,3 2,2 4,2 3,1 0 0 2 2u u u u
The fourth row is completed
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1,4 0,3 2,3 1,2 0 3 0 3u u u u
2,4 1,3 3,3 2,2 2 2 0 4u u u u
3,4 2,3 4,3 3,2 3 0 0 3u u u u The last row is completedThus the required values 0f ui,j are tabulated
t⁄ x
0 1 2 3 4
0 0 3 4 3 00.5 0 2 3 2 01 0 0 0 0 01.5 0 -2 -3 -2 02 0 -3 -4 -3 0
2. Solve 2 2
2 2
u u
t x
given that u(x,0) =0; u(0,t)=0 ut(x,0)=0 and u(l,t)=100sin(πt)
in the range 0≤t≤1 by taking h=1/4
Soln: Comparing the wave equation c2uxx = utt with the equation uxx= utt we have c2=1 or c=1By data h=1/4,k=h/c=1/4
We have seen that the condition ut(x,0)=0 will lead us to the formula
, 1,0 1,0
1
2i j i iu u u
1,1 0,0 2,0
1 10 0 0
2 2u u u
2,1 1,0 3,0
1 10 0 0
2 2u u u
3,1 2,0 4,0
1 10 0 0
2 2u u u
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4,1 3,0 5,0
1
2u u u
Hence
4,1u = 4 1, (1,1/ 4) 100sin( / 4) 0.707u x t u (Second row of the table is completed)We shall now conider the explicit formula to find the remaining values in the table
, 1 1, 1, , 1i j i j i j i ju u u u
1,2 0,1 2,1 1,0 0u u u u
2,2 1,1 3,1 2,0 0u u u u
3,2 2,1 4,1 3,0 70.7u u u u
4,2 3,1 5,1 4,0u u u u is inapplicable
4,2 4 2, (1,1/ 2) 100sin( / 2) 100u u x t u (Third row of the table is completed)
1,3 0,2 2,2 1,1 0u u u u
2,3 1,2 3,2 2,1 70.7u u u u
3,3 2,2 4,2 3,1 100u u u u
4,3 4 3, (1,3 / 4) 100sin(3 / 4) 70.7u u x t u (Fourth row of the table is completed)
1,4 0,3 2,3 1,2 70.7u u u u
2,4 1,3 3,3 2,2 100u u u u
3,4 2,3 4,3 3,2 70.7u u u u
4,4 4 4, (1,1) 100sin( ) 0u u x t u (Last row is completed)Thus the required values 0f ui,j are tabulated
t⁄ x
0 1/4 1/2 3/4 1
0 0 0 0 0 00.25 0 0 0 0 70.70.5 0 0 0 70.7 100.75 0 0 70.7 100 70.71 0 70.7 100 70.7 0
Numerical solution of the one dimensional heat equation
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We seek the numerical solution of heat equation ut=c2uxx..........................................(1)subject to the boundary conditions u(0,t)=0……………………………………………(2) u(l,t)=0…………………………………………….(3)and the initial condition u(x,0)=f(x)……………………………………….(4)we shall substitute the finite difference approximation for the partial derivatives present in (1)
2, 1 , 1, , 1,2
1 12i j i j i j i j i ju u c u u u
k h 2
, 1 , 1, , 1,22i j i j i j i j i j
kcu u u u u
h
Taking 2
2
kc
h=a the above equation becomes
, 1 , 1, , 1,2i j i j i j i j i ju u au au au
, 1 1, , 1,(1 2 )i j i j i j i ju au a u au …………………….(5)
This is called Schmidt explicit formula valid for 0<a<1/2For convenience let us set 1-2a=0 or a=1/2
That is 2
2
kc
h=1/2
Thus (5) becomes , 1 1, 1,
1
2i j i j i ju u u This is called Bendre Schmidt formula .
EXAMPLES
1) Find the numerical solution of the parabolic equation2
22
u u
x t
When u(0,t)=0=u(4,t) and u(x,00=x94-x0 by taking h=1
Soln: the standard form of the one dimensional heat equation is ut=c2uxx and the given equation can be put in the form ut=(1/2)uxx
c2=1/2 since h=1,k=2
21
2
h
c
the values of x=0 in 0<x<4 with h=1 are 0,1,2,3,4 and the values of t with k=1 are 0,1,2,3,4,5The initial table is given byConsider the initial condition u(x,0)=x(4-x)u1,0=u(1,0)=3; u2,0=4; u3,0=3
(First row in the table is completed)
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Now we consider the relation
, 1 1, 1,
1
2i j i j i ju u u
In particular
,1 1,0 1,0
1
2i i iu u u
1,1 0,0 2,0
1 10 4 2
2 2u u u
2,1 1,0 3,0
1 13 3 3
2 2u u u
3,1 2,0 4,0
1 14 0 2
2 2u u u
(The second row in the table is completed)
Again ,2 1,1 1,1
1
2i i iu u u
1,2 0,1 2,1
1 10 3 1.5
2 2u u u
2,2 1,1 3,1
1 12 2 2
2 2u u u
3,2 2,1 4,1
1 13 0 1.5
2 2u u u
(The third row in the table is completed)
Again ,3 1,3 1,2
1
2i i iu u u
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1,3 0,2 2,2
1 10 2 1
2 2u u u
2,3 1,2 3,2
1 11.5 1.5 1.5
2 2u u u
3,3 2,2 4,2
1 12 0 1
2 2u u u
(Fourth row in the table is completed)
Again ,4 1,3 1,3
1
2i i iu u u
1,4 0,3 2,3
1 10 1.5 0.75
2 2u u u
2,4 1,3 3,3
1 11 1 1
2 2u u u
3,4 2,3 4,3
1 11.5 0 0.75
2 2u u u
(Fifth row in the table is completed)
Again ,5 1,4 1,4
1
2i i iu u u
1,5 0,4 2,4
1 10 1 0.5
2 2u u u
2,5 1,4 3,4
1 10.75 0.75 0.75
2 2u u u
3,5 2,4 4,4
1 11 0 0.5
2 2u u u
He last row in the table is completed)Thus the required values 0f ui,j are tabulated
t⁄ x
0 1 2 3 4
0 0 3 4 3 01 0 2 03 2 02 0 1.5 02 1.5 03 0 1 701.5.7 1 04 0 0.75 1001 0.75 05 0 0.5 0.75 0.5 0
Numerical solution of the Laplace’s equation in two dimensions
Laplace’s equation in two dimensions is
2 2
2 20
u u
x y
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Here we consider a rectangular region R for which u(x,y) is know at the boundary. Let us suppose that the region is such that it can divided into a network of square mesh of side h.
We shall substitute the finite difference approximation for the partial derivatives ,hence we have
1, , 1, , 1 , , 12 2
1 12 2 0i j i j i j i j i j i ju u u u u u
h h
1, 1, , 1 , 1 ,4i j i j i j i j i ju u u u u
, 1, 1, , 1 , 1
1
4i j i j i j i j i ju u u u u ……………………(2)
This is called the standard five point formula. It may be observed that the value of ui,j at any interior mesh point is the average of its values at four neighboring points which is given in the figure below
, 1, _ 1 1, 1 1, 1 1, 1
1
4i j i j i j i j i ju u u u u ……………………………(3)
Equation (30 is called diagonal five point formula.
Examples1.solve Laplace’s uxx+uyy=0 for the following square mesh with boundary values as shown belowwww.vt
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u5 is located at the centre of the region.
5
10 21 17 12.1 12.525 .
4u by S F
Next we shall compute u7, u9, u1, u3by applying D.F
7
10 12.525 0 12.1 6.15625
4u
9
112.1 21 12.525 9 13.65625
4u
1
10 17 0 12.525 7.38125
4u
3
112.525 18.6 17 21 17.28125
4u
Finally we shall compute u2, u4, u6, u8 by applying S.F
2
17.38125 17.28125 17 12.525 13.546875
4u
4
10 12.525 7.38125 6.15625 6.515625
4u
6
112.525 21 17.28125 13.65625 16.115625
4u
8
16.15625 13.65625 12.525 12.1 11.109375
4u
Thus, u1=7.38, u2=13.55 , u3=17.28, u4=6.25, u5=12.53, u6=16.12, u7=6.16, u8=11.11, u9=13.66,
2.Solve uxx+uyy=0 in the following square region with the boundary conditions as indicated in the figure belowwww.vt
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Soln: we shall apply standard five point formula for u1, u2, u3, u4, to obtain a system of equations
1 2 3
120 100
4u u u
2 1 4
10 100
4u u u
3 4 1
130 0
4u u u
4 3 2
10 0
4u u u
Now we have a system of equations to be solved
1 2 34 120u u u ……………………………(1)
1 2 44 100u u u ………………………….(2)
1 3 44 30u u u ……………………………..(3)
2 3 44 0u u u ……………………………..(4)
Let us eliminate 1u from (1) and (2);(20 and (3)
That is 15 2u - 3u -4 4u =520………………………..(5)
4 2u -4 3u =70 …………..(6)
we shall now eliminate 4u from (4) and (5)
That is 14 2u -2 3u =520
Let us solve (6) and (7) : 2 2u -2 3u =35
14 2u -2 3u =520
Thus 2u =40.4, and 3u =22.9 hence from (1)
1u =45.825 and from(4) 4u =15.825
Thus the required values at th2 interior mesh points are
1u =45.825 , 2u =40.4, 3u =22.9 4u =15.825
3. solve the elliptic equation uxx+uyy=0 at the pivotal points for the following square mesh www.vt
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Soln: it is evident that the values at the boundary points on AB are respectively 200 and 100
Thus 1 2 3
1300 200
4u u u
2 1 4
1100 100
4u u u
3 4 1
1400 400
4u u u
4 3 2
1200 300
4u u u
1 2 34 500u u u ……………………………(1)
1 2 44 200u u u ………………………….(2)
1 3 44 800u u u ……………………………..(3)
2 3 44 500u u u ……………………………..(4)
Thus by solving simultaneously we obtain the values at the pivotal points
1u =250 , 2u =175, 3u =325 4u =250
4. Solve the elliptic partial differential equation for the following square mesh using the 5 point difference formula by setting up the linear equations at the unknown points P,Q,R,S
The linear equations for the unknownfor the 5 point difference formula are
Thus 11 1
4P Q S
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14 2
4Q P R
15 5
4R S Q
12 4
4S R P
4P-Q-S=2……………………………(1)
-P+4Q-R=6…………………………(2)
-Q+4R-S=10………………………(3)
-P-R+4S=6…………………………(4)
Thus by solving we get P=2,Q=3,S=3,R=4
UNIT VIIIDIFFERENCE EQUATIONS AND
Z- TRANSFORMSCONTENTS:
Introduction
Z-Transforms –Definition
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Damping Rule
Shifting RuleTheorem
Inverse Z-Transforms
Power Series Method
Applications Of Z-Transforms To Solve Difference Equations
DIFFERENCE EQUATIONS ANDZ- TRANSFORMS
Introduction
The Z-transform plays an important role plays in the study of communications, sample data control systems, discrete signal processing , solutions of difference equations etc.
Definition
Let un = f(n) be a real-valued function defined for n=0,1,2,3,….. and un = 0 for n<0. Then the Z-transform of un denoted by Z(un) is defined by
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0
)()(n
nnn zuuZzu (1)
The transform also is referred to as the one sided Z-transform or unilateral Z-transform. Next, we define un = f(n) for n=0, 1, 2, …… ∞.
The two-sided Z-transform is defined by
nnn zuuZ )( (2)
The region of the Z-plane in which the series (1) or (2) converges is called the region of convergence of the transform.
Properties of Z-transform
1. Linearity property
Consider the sequences {un} and {vn} and constants a and b. Then
Z[aun + bvn ] = aZ(un) + bZ(vn)
Proof : By definition, we have
)()(
][][
0 0
0
nn
n n
nn
nn
n
nnnnn
vbZuaZ
zvbzua
zbvaubvauZ
In particular, for a=b=1, we get
Z[un+vn] = Z(un) + Z(vn)and for a=-b=1, we get
Z[un - vn] = Z(un) - Z(vn)
2. Damping property
Let Z(un) = )(zu . Then (i) azuuaZ n
n )( (ii) )()( azuuaZ nn
Proof : By definition, we havewww.vtuc
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a
zu
a
zuzuauaZ
n
n
nn
nn
nn
n
0 0
)()(
Thus
a
zuuaZ n
n )(
This is the result as desired. Here, we note that that if Z(un) = )(zu , then
][)( zuuaZ nn
aZZ
=
a
zu
Next,
)(
)()()(00
azu
azuzuauaZn
nn
n
nn
nn
n
Thus
)()( azuuaZ nn
This is the result as desired.
3. Shifting property
(a) Right shifting rule :
If Z(un) = )(zu , then Z(un-k) = z-k )(zu where k>0
Proof : By definition, we have
0n
nknkn zuuZ
Since un = 0 for n<0, we have un-k = 0 for n=0,1,……(k-1)
Hence
)(
].......[
.......
0
110
)1(10
zuz
zuz
zuuz
zuzu
zuuZ
k
n
nn
k
k
kk
kn
nknkn
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Thus
Z(un-k) = z-k )(zu
(b)Left shifting rule :
]......)([)( )1(1
22
110
k
kk
kn zuzuzuuzuzuZ
Proof :
0
1
0
0
)(
0
)(
0
,
n
k
n
nn
nn
k
n
nkkn
k
n
nkkn
k
n
nknkn
zuzuz
knmwherezuzzuz
zuuZ
= ]......)([ )1(1
22
110
kk
k zuzuzuuzuz
Particular cases :
In particular, we have the following standard results :
1. ])([)( 01 uzuzuZ n
2. ])([)( 110
22
zuuzuzuZ n
3. ])([)( 22
110
33
zuzuuzuzuZ n etc.
Some Standard Z-Transforms :
1.Transform of an
By definition, we have
.....1
)(
0
2
0
n
n
n
nnn
z
a
z
a
z
a
zaaZwww.vtuc
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The series on the RHS is a Geometric series. Sum to infinity of the series is
az
zor
za 1
1Thus,
az
zaZ n
)(
In particular, when a=1, we get Z(1) = 1z
z
2. Transform of ean
Here)()( nan kZeZ where k = ea
aez
z
kz
z
Thus
aan
ez
zeZ
)(
3. Transform of np , p being a positive integer
We have,
0
)1(1
0
)(
n
np
n
npp
nznz
znnZ
Also, we have by defintion
0
11)(n
npp znnZ
Differentiating with respect to z, we get
0
)1(1
0
11
)(
)(
n
np
n
npp
znn
zndz
dnZ
dz
d
Using this in (1), we get
)]([)( 1 pp nZdz
dznZ
Particular cases of )( pnZ :-
1. For p = 1, we get
2)1(1
)1()(
z
z
z
z
dz
dzZ
dz
dznZ
(1)
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Thus,
2)1()(
z
znZ
2. For p = 2, we get
3
2
22
)1()1()()(
z
zz
z
z
dz
dznZ
dz
dznZ
Thus,
3
22
)1()(
z
zznZ
3. For p = 3, we get
4
233
)1(
4)(
z
zzznZ
4. Transform of nan
By damping property, we have
22
2
)(1
)1()()(
az
az
a
za
z
z
znZnaZ
aZZ
aZZ
n
, in view of damping
propertyThus,
2)()(
az
aznaZ n
5. Transform of n2an
We have,
a
ZZ
a
ZZ
n
z
zznZanZ
3
222
)1()()(
Thus,
3
222
)()(
az
zaazanZ n
6. Transforms of coshn and sinhn
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We have
)(2
1)(cosh
2cosh
nn
nn
eeZnZ
een
)()(2
1 nn eZeZ , by using the linearity property
1cosh2
cosh
1)(2
)(
1)(2
2
1
2
22
zz
zz
eezz
eez
zeezz
ezezz
ez
z
ez
z
Next,
1cosh2
sinh
1cosh22
11
2)(sinh
2sinh
2
2
zz
z
zz
eez
ezez
znZ
een
nn
7. Transforms of cosn and sinn
We have
)(2
1)(cos
2cos
inin
inin
eeZnZ
een
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1cos2
]cos[
1)(
)(2
2
2
1
2
2
zz
zz
eezz
eezz
ez
z
ez
z
inin
inin
ii
Next,
1cos2
sin
1cos22
2
1)(sin
2sin
2
2
zz
z
zz
ee
i
z
ez
z
ez
z
inZ
i
een
ii
ii
inin
Examples :
Find the Z-transforms of the following :
1. nn
nu
4
1
2
1
We have,
nn
n ZuZ4
1
2
1)(
)14)(12(
)38(2
14
4
12
2
4
1
2
1
4
1
2
1
zz
zz
z
z
z
z
z
z
z
z
ZZnn
2. !
1
nun
Here
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z
n
n
n
nn
ezz
nz
znn
ZuZ
1
2
00
........!2
1
!1
1
1
!
1
!
1
!
1)(
,
3. nau nn cos
We have
1cos2
)cos()(cos
2
zz
zznZ
By using the damping rule, we get
22
2
2
2
)cos(
1cos2
cos
1cos2
)cos()cos(
aazz
azz
a
z
a
z
a
z
a
z
zz
zznaZ
aZZ
n
4. un = )!2(
1
n
Let us denote vn = !
1
n, so that vn+2 =
)!2(
1
n= un
Here
zn evZ
1
)( Hence
z
vvvZzvZuZ nnn
10
22 )()()( , by left shifting rule
zez
zez
z
z
11
!1
1.
1
!0
1
12
12
List of standard inverse Z-transforms
, by exponential theorem
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SUBJECT:MATHEMATICS SUBJECTCODE:10MAT31
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1. 11
1
z
zZT 2. n
T kkz
zZ
1
3.
nz
zZT
21
14.
nk
kz
kzZ n
T
21
5.
23
21
1n
z
zzZT
6.
2
3
221 nk
kz
zkkzZ n
T
7.
34
231
1
4n
z
zzzZT
8.
3
4
32231 4
nkkz
zkzkkzZ n
T
9. )2sin(12
1 nz
zZT
10. )2cos(
12
21 n
z
zZT
Evaluation of inverse Z-transforms:We obtain the inverse Z-transforms using any of the following three methods:( I ) Power series Method:- This is the simplest of all the method of finding the inverse Z-transform. If U (z) is expressed as the ratio of two polynomials which cannot be factorized, we simply divide the numerator by the denominator and take the inverse Z-transform of each term in the quotient.Problems:
(1)Find the inverse transform of log 1z
z by power series method.
Putting
2 3
1 2 3
11 11 , ( ) log log(1 ) ...
1 2 31
1 1...
2 3
yz U z y y y yt
y
z z z
1
)(,.n
nn zuzUei
0 0
( 1)nn
for nThus u
otherwisen
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(2) Find the inverse Z-transform of 2( 1)z
z by division method.
)1(
12)(
2122
2
zzz
zzz
zzU
1
1
432
14111
21
212
)1(
14
13
12
1)(
)()(3)(211
11
)1(
32
n
nn
zzz
z
z
nZ
zzzzzU
z
z
z
z
Thus nu nn
1)1( ( II ) Partial fractions Method:- This method is similar to that of finding the inverse
Laplace transforms using partial fractions. The method consists of decomposing ( )U zz
into partial fractions, multiplying the resulting expansion by z and then inverting the same.
Problems:
(1) 22 3
( 2)( 4)
z z
z z
We write 22 3
( )( 2)( 4)
z zU z
z z
as
( ) 2 3 1 116 6( 2)( 4) 2 4
U z z A Bwhere A and B
z z z z z
Therefore
1 11( )
6 2 6 4
z zU z
z z
On inversion, we have
1 11( 2) (4)
6 6n n
nu nkz
z kz )(1
(2) 3
3
20
( 2) ( 4)
z z
z z
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We write 3
3
20( )
( 2) ( 4)
z zU z
z z
as
2 2
3 3
( ) 20
4( 2) ( 4) ( 2)
U z z A Bz Cz D
z zz z z
Multiplying throughout by
we get
Comparing the coefficients of z in the above equationwe get A = 6, B=0, C = ½ and D = -½ .Thus
On inversion, we get
(3)Find the inverse Z-transform of 32,)3)(2(
)5.65(22
2
zforzz
zz
Splitting into partial fractions, we obtain
2
22
2
)3(
1
3
1
2
1)(
1)3(32)3)(2(
)5.65(2)(
zzzzU
CBAwherez
C
z
B
z
A
zz
zzzU
22
3 3
1 16 0( ) 20 2 24( 2) ( 4) ( 2)
z zU z z
z zz z z
3( 2) ( 4)z z
2 2 320 ( )( 4) ( 2) .z A Bz Cz z D z
22
3 3
1 16 0( ) 20 2 24( 2) ( 4) ( 2)
z zU z z
z zz z z
3 2 2 3
3 3
3 2 22 2 3
3 3
2 2
3
1 12 1 1 12 4 4 1( ) .
2 2 4 2 2 4( 2) ( 2)
4 4 8 48 4 4 41 1 1 1
2 2 4 2 2 4( 2) ( 2)
1 ( 2) 4 8
2 ( 2)
z z z z z z z zU z
z zz z
z z z z zz z z z z z z
z zz z
z z z z
z
2
3
1 1 2 4 12
2 4 2 2 2 4( 2)
z z z z z
z z zz
2
1 2 2 1
1(2 2. 2 ) 4
2
2 2 2
n n nn
n n n
u n
n
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3227
4
9
3
3
21
9
1
27931
3
18421
1
113
19
1
31
3
121
2
1
3232
32
32
211
zwherezzzzzz
zzzz
andthatsozz
zz
z
.0,3)2(1,2,
)1(2
3
4
3
3
3
2
3
1
3333
1222
2
1
21
2
31
00
1
31
1
1
5
3
4
2
324
3
3
2
24
3
3
2
2
nnuandnugetweinversionon
znzz
zzzzzz
zzz
nn
nn
nn
n
n
n
n
n
nn
Difference Equations
INTRODUCTIONDifference equations arise in all situations in which sequential relation exists at various discrete values of the independent variable. The need to work with discrete functions arises because there are physical phenomena which are inherently of a discrete nature. In control engineering, it often happens that the input is in the form of discrete pulses of short duration. The radar tracking devices receive such discrete pulses from the target which is being tracked. As such difference equations arise in the study of electrical networks, in the theory of probability, in statistical problems and many other fields.
Just as the subject of difference equations grew out of differential calculus to become one of the must powerful instruments in the hands of a practical mathematician when dealing with continuous processes in nature, so the subject of difference equations is forcing its way to the fore for the treatment of discrete processes. Thus the difference equations may be thought of as the discrete counterparts of the differential equations.
DefinitionA difference equation is a relation between the differences of an unknown function at one or more general values of the argument.Eg: 1) 1 2n ny y
2) 2 5 6 0n n ny y y
3) 3 2 23 2n n n ny y y y x are difference equations.
An alternative way of writing a difference equation is as follows:Putting 1E , we get(1) may be written as,
1
1 1
2 1
( 1) 2
2 [sin ]
2 (4)
n n
rn n n n n r
n n n
E y y
Ey y y ce E y y
y y y
(2) may be written as,
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)5(023
052
06)55()12(
06)1(5)1(
12
112
2
2
nnn
nnnnn
nnn
nnn
yyy
yyyyy
yyEyEE
yyEyE
(3) may be written as
)6(5116
2236333
)1(2)12(3)133(
)1(2)1(3)1(
2123
2112123
2223
223
xyyyy
xyyyyyyyyyy
xyyEyEEyEEE
xyyEyEyE
nnnn
nnnnnnnnnn
nnnn
nnnn
The equations (4), (5) and (6) can also be written in terms of the operator E. i.e,
223
2
2
)5116(
0)23(
2)1(
xyEEE
yEE
yEE
n
n
n
Order of a difference equation:The order of a difference equation is the difference between the largest and the smallest arguments occurring in the difference equation divided by the unit of increment.
(interval)increment ofunit
argumentsmallest -argument largest .eqdifferenceaoforder . ei
Note: To find the order, the equation must be expressed in a form free of s . [because the highest power of does not give the order of difference equation]
therefore the order of the difference equation
(4) is = 21
)()2(
nn
(5) is = 21
)()2(
nn
(6) is = 31
)()3(
nn
The order of a difference equation can also be obtained by considering the highest power of the operator E involved in the equation.
Solution of a difference Equation is an expression for ( )ny which satisfies the given
difference equation.The general solution of a difference equation is that in which the number of arbitrary constants is equal to the order of the difference equation.
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A particular solution (or particular integral) is that solution which is obtained from the general solution by giving particular values to the constants.
APPLICATION OF Z-TRANSFORM TO SOLVE DIFFERENCE EQUATIONS
Working Procedure to solve a linear difference equation with constant coefficient by Z-transforms:1) Take the Z-transform of both sides of the difference equations using the Z-transforms formulae and the given conditions.2) Transpose all terms without )(zU to the right.3) Divide by the coefficient of )(zU , getting )(zU as a function of z.4) Express this function in terms of Z-transforms of known functions and take the inverse Z-transform of both sides. This gives nu as a function of n which is the desired solution.
Problems:(1) Using the Z-transform, solve 1,0334 1012 uuwithuuu n
nnn
)2()2(
)()(
,)()(),()(1
102
2
01
zzZAlso
zuuzUzuZ
uzUzuZthenzUuZIfthatnoteWe
n
n
nn
Taking the Z-transforms of both sides of the above difference equation, we get
)3()(3)(4)( 01
102
zzzUuzUzzuuzUz
Using the given conditions, it reduces to
,3
1
12
5
3
1
24
1
1
1
8
3
)3)(1)(3(
1
)3)(1(
1)(
)3()34)(( 2
zzz
zzzzzz
zU
zzzzzzU
on breaking into partial fractions, then
312
5
324
1
18
3)(
z
z
z
z
z
zzU
On inversion, we obtain
nnn
n z
zZ
z
zZ
z
zZu
)3(12
53
24
1)1(
8
3
312
5
324
1
18
3 111
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(2) Solve ,0296 1012 yywithyyy nnnn using Z-transform.
If 110
2201 )()(,)()(),()(
zyyzYzyZyzYzyZthenzYyZ nnn
Also )2()2( zzZ n
Taking Z-transforms of both sides, we get
)2()(9)(6)( 01
102
zzzYyzYzzyyzYz
Since )2()96)((,0,0 210 z
zzzzYhaveweyandy
Or ,)3(
5
3
1
2
1
25
1
)3)(2(
1)(22
zzzzzz
zYon splitting into partial
fractions.
Or
2)3(
53225
1)(
z
z
z
z
z
zzY
On taking inverse Z-transform of both sides, we obtain
nnnn
zz
zz
zz
n
naaz
azZn
ZZZy
21
35
)3(31
35
31
21
)()3()3(2
25
1
25
12
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