civil-iii-engineering mathematics - iii [10mat31]-notes

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Engineering Mathematics III

Unit 1: Fourier Series 1-25

Unit 2: Fourier Transforms26-38

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 Transforms120-136 CitS

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UNIT- I

FOURIER SERIES

CONTENTS: Introduction

Periodic function

Trigonometric series and Eulers 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 Ci

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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) = x x = 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 functions

2. h(x) = xsinx is even, since x and sinx are odd functions

3. h(x) = x2

sinx is odd, since x2

is even and sinx is odd.

a a

2. If f(x) is even, then f ( x)dx 2 f ( x)dx a 0

a

3. If f(x) is odd, then f ( x)dx 0 a

For example, a a

cos xdx 2 cos xdx, as cosx is even a 0

a

and sin xdx 0, a

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 that

f(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 since

Sin(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.

FOURIER SERIES

Series with

arbitrary period

Half-range series Complex series Harmonic Analysis

FORMULA FOR FOURIER SERIES

Consider a real-valued function f(x) which obeys the following conditions called

Dirichlets conditions :

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l

l

n

n

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

1 a 2l

a0 l f ( x)dx a

(1)

1 a 2l n

an a

f ( x) cos l

xdx,

n 1,2,3,..... (2)

1 a 2l n

bn a

f ( x) sin l

xdx,

n 1,2,3,...... (3)

Then, the infinite series

a n

n

0 a 2 n 1

cos l

x bn

sin x l

(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 Eulers 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

a n

n

f(x) = 0 a 2 n 1

cos l

x bn

sin l

x . (5)

Suppose f(x) is discontinuous at x, then the sum of the series (4) would be

1 f ( x ) f ( x )2

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 Cases Case (i) Suppose a=0. Then f(x) is defined over the interval (0,2l). Formulae (1), (2), (3) reduce

to

1 2l 1

2l n a0 f ( x)dx an f ( x) cos xdx, l

0 l

0 l

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0

a

l

l

l

n

l

1 2l

n

bn l

f ( x) sin l xdx,

n 1,2,......

(6)

0

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

1 2

a0 =

f ( x)dx 0

1 2

an

f ( x) cos nxdx , n=1,2,.. (7)

1 2

bn

f ( x) sin nxdx 0

n=1,2,..

Also, in this case, (5) becomes

f(x) =

0 a 2 n 1

cos nx bn

sin nx

(8)

Case (ii)

Suppose a=-l. Then f(x) is defined over the interval (-l , l). Formulae (1), (2) (3) reduce

to

1 l

a0 f ( x)dx l

b 1

(9)

f ( x) sin n

xdx,

1 l

n n

l l

an f ( x) cos l xdx

l 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

1

a0 = f ( x)dx

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n

an 1

f ( x) cos nxdx

, n=1,2,.. (10)

1 bn

f ( x) sin nxdx

n=1,2,..

Putting l = in (5), we get f(x) = a0 a 2 n 1

cos nx bn

sin nx

Some useful results :

1. The following rule called Bernoullis generalized rule of integration by parts is useful in

evaluating the Fourier coefficients. ' ''

uvdx uv1 u v2 u v3 ....... Here u, u ,.. are the successive derivatives of u and

v1 vdx, v2 v1dx,...... We illustrate the rule, through the following examples :

2 2 cos nx sin nx cos nx

x sin nxdx x 2x 2 2 3 n n n

e2 x e2 x e2 x e2 x

x3e

2 x dx x3 3x2 6 x 6

2 4 8 16

2. The following integrals are also useful :

eax

cos bxdx a

eax

2 b2

eax

a cos bx b sin bx

eax

sin bxdx

a2 b2

a sin bx b cos bx

3. If n is integer, then

sin n = 0 , cosn = (-1)n , sin2n = 0, cos2n=1

Problems

1. Obtain the Fourier expansion of

f(x) = 1 x in - < x <

2

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2

2

2

a

2

2

2

n

We have,

1 1

1 a0

f ( x)dx

( x)dx

1 = x

x2

2

1

1

1 an

f ( x) cos nxdx

( x) cos nxdx

Here we use integration by parts, so that

1 sin nx

cos nx

an 2

x n

(1)

n

1

0 0 2

b 1

n

1

( x) sin nxdx

1

x cos nx

(1) sin nx

2

n n2

(1)n

n

Using the values of a0 , an and bn in the Fourier expansion

f ( x) 0 a

cos nx bn

sin nx

we get,

2 n 1 n 1

(1)n

f ( x) 2 n 1

sin nx n

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 (1)n

cos ech

n 1 n 2

Here,

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a n2 a

a0

1

1 e ax e ax

dx

a

ea e a

a

2 sinh a

a

a 1

n

e ax

cos nxdx

1 e ax

an

2 2 a cos nx n sin nx a n

2a (1)n sinh a

a

2 n2

1

bn = e ax

sin nxdx

1 e ax =

a sin nx n cos nx

a2 n2

2n (1)n sinh a

= 2 a n2

sinh a

2a sinh a

(1)n

2 n(1)n

Thus,f(x) = a

2 n 1

cos nx sinh a 2 n 1

sin nx n2

For x=0, a=1, the series reduces to

sinh

2 sinh

(1)n

f(0)=1 =

n2 1

n 1

or

sinh

2 sinh 1

(1)n

1 =

2

n2 1

2 sinh

(1)n

n 2

or 1 =

n2 1

Thus, n 2

(1)n

cos ech 2 n2 1

This is the desired deduction.

n 2

3. Obtain the Fourier expansion of f(x) = x2

over the interval (-, ). Deduce that

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2

2

2

1

n

6

1

1 6 2

2

1 ......

32

1

The function f(x) is even. Hence a0

f ( x)dx 2

f ( x)dx

0

2

2 x3 = x

2 dx

or a0

0

2 2

3

3 0

1 an

f ( x) cos nxdx

2

= f ( x) cos nxdx,

0

since f(x)cosnx is even

= 2

x2

0

cos nxdx

Integrating by parts, we get

a 2

x2

sin nx 2 x

cos nx 2

sin nx

n

n

n2 n3

4(1)n

n2

0

Also, 1

bn

f ( x) sin nxdx 0 since f(x)sinnx is odd.

Thus

2 f ( x) 4

3

n 1

(1)n cos nx

n2

2 4

1 3

2

2 1

n 1 n

2

Hence,

1 1

1 6 22 32

.....

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0

n

2

2 2

2


x,0 x f ( x)

2 x, x 2

Deduce that

1 1

1 ......

8 32

52

1

2

Here, a0 = f ( x)dx = f ( x)dx

0

2

xdx

0

since f(x)cosnx is

1

2

an f ( x) cos nxdx

f ( x) cos nxdx

even. 0

2

= x cos nxdx

2 x

sin nx 1

cos nx

n n

2 =

2 (1)n 1

0

Also,

n2

b 1

f ( x) sin nxdx 0 , since f(x)sinnx is odd

Thus the Fourier series of f(x) is

f ( x)

2

1 (1)n 1cos nx

For x= , we get

2

f ( )

n 1 n

1 (1)n 1cos n2 n 1 n

2 2 cos(2n 1)

or

2 n 1

(2n

1)2

Thus,

2 1 2 8 n 1 (2n 1)

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1

1 (1) 1cos

2

n 2

2

2

n

n

2

n

2

or 1

1

1 ......

8 32

52

This is the series as required.


, x 0 f(x) =

x,0 x

Deduce that

1 1

1 ......

8 32

52

Here,

1 0

a0 dx xdx 0

0 2

a 1 cos nxdx x cos nxdx

0

1 (1)n 1

n2

1 0

bn sin nxdx x sin nxdx

Fourier series is

0

1 1 2(1)n n

f(x) =

nx

1 2(1) sin nx 4 n 1 n n 1 n

Note that the point x=0 is a point of discontinuity of f(x). Here f(x+) =0, f(x

-)=- at x=0.

Hence 1

[ f ( x ) f ( x )]

1 0

2 2 2

The Fourier expansion of f(x) at x=0 becomes

1

1 [(1)n 1]

2 4

2

n 1 n

1

or 4

Simplifying we get,

n 1

[(1)n 1]

1

1

1 ......

8 32

52

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1

1

1

1

3 2

n

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

1

1 1

Here a0 = f ( x)dx = 2 f ( x)dx 1 0

1 x3

= 2 (1 x2 )dx 2 x

0 3 0

4 3

1 1

an f ( x) cos(nx)dx 1

1

2 f ( x) cos(nx)dx 0

1

as f(x) cos(nx) is even

= 2 (1 x2 )

0


cos(nx)dx

nx

nx 1

nx

a 21 x2 sin

(2x) cos (2) sin

n

n (n )2

(n )3 0

4(1)n 1 =

n2 2

1 1

bn f ( x) sin(nx)dx 1

=0, since f(x)sin(nx) is odd.

The Fourier series of f(x) is

2 4 (1)n 1

f(x) = 2 n 1

cos(nx)


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2

2

2

3

2

2

4x 1 3

in 3 x 0

2

f(x) = 1 4x

in0 x 3

3 2

Deduce that

1 1

1 ......

8 32

52

The period of f(x) is 3

3 3

2 2

Also f(-x) = f(x). Hence f(x) is even

1 a0

3 / 2

f ( x)dx 2

3 / 2

f ( x)dx 3 / 2 3 / 2

3 / 2 0

4 3 / 2

1

4 x dx 0

3 0

1

3 3 / 2

nx

an f ( x) cos dx 3 / 2

3 / 2

2 3 / 2

3 / 2

2nx

f ( x) cos dx 3 / 2 0 3

3 / 2

nx

nx

sin 2

cos 2

4

1 4x 3

4 3 3 3

2n 3

2n 2

3

0

= 4

n2 2

1 (1)n

3

2

1 nx

Also, bn 3

3 2

f ( x) sin

3 dx 0

2

Thus

4 1

n 2nx

f(x) =

putting x=0, we get

n 1 n

1 (1) cos 3

4 1 n

f(0) = n 1 n

1 (1)

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2 2 2

2

8 1 1 or 1 = 1 ......

3 5

Thus,

1

1

1 ......

8 32

52

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

nx

f ( x) bn sin

l (11)

n 1

2 l

nx where bn

l f ( x) sin

l dx

0

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

f ( x) bn sin nx n 1

2

bn f ( x) sin nxdx 0

Cosine series :

Let us define

( x) f ( x)

in (0,l) .....given

( x)

a

in (-l,0) ..in order to make the function even. Then the Fourier series of f(x) is given by

nx

f ( x) 0 a n cos l

(12)

where,

2 n 1

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l

3

n

0

n

a0 2

l

f ( x)dx 0

2 l

nx an

l f ( x) cos

l dx

0

The series (12) is called half-range cosine series over (0,l)

Putting l = in (12), we get

f ( x) a0 a

cos nx 2

where

n 1

a 2

0

f ( x)dx

a 2 f ( x) cos nxdx

n 1,2,3,.. Problems :

0

1. Expand f(x) = x(-x) as half-

range sine series over the interval (0,).

We have,

2

bn f ( x) sin nxdx 0

2 2

(x x

0


) sin nxdx

b 2 x x2

cos nx sin nx

2x ( 2) cos nx

n

n

n2

n3

4

n3

1 (1)n

0

The sine series of f(x) is

f ( x) 4

1 1 (1)n

sin nx n 1 n

2. Obtain the cosine series of

x,0 x 2

f ( x)

over(0, )

x,

x

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2

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4

c

c

2

2

n

2 2

Solution : a0

xdx ( x)dx 2 0 2

2 2

an x cos nxdx ( x) cos nxdx 0 2

Performing integration by parts and simplifying, we get

n

a 2

1 (1)n 2 cos

n

8

n2

, n 2,6,10,.....

2

n2

Thus, the Fourier cosine series is

2 cos 2 x cos 6x cos10 x f(x) =

12

32

52

......

3. Obtain the half-range cosine series of f(x) = c-x in 0



l

n

n

l

l

c

c 2

c 0

then we obtain a complex exponential Fourier series

n

f ( x) i

cn e x ..(*)

where n

1 2l

cn 2l

f ( x)e

i n

x

l dx

n = 0, 1, 2, 3, .

Note:-(1)

Putting l in (*), we get Fourier series valid in (l, l ) as n

f ( x) i

cn e x ..(**)

where n

1 l n

2

f ( x ) e

i

n x

d x

Note:-(2)

Putting = 0 and l

l l

in (*), we get Fourier series valid in (0, 2 ) as

f ( x) c einx

where

Note:-(3)

n

1 n

f ( x ) e

i n x

d x

Putting l in (**), we get Fourier series valid in ( , ) as

f ( x) c einx

where n

1 2 n

2

f ( x ) e in x

d x

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

n

f ( x)

cn e inx

(1)

where

cn

n

1

f ( x)einx dx

2

1

xe inx

dx

n = 0, 1, 2, 3, .

2

Integrating by parts and substituting the limits, we get n

cn

Using this in (1), we get

i(1)

n

, n 0

f ( x) i ( 1)

n

einx

, n 0

n n

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 c which is given by

1 c

2

f ( x)einx

dx 1

2 e

ax einx

dx

1 1 e( a in ) x e

( a in ) x dx

2

2 (a in)

(1)n (a in) sinh a

(a2 n2 )

Hence the complex Fourier series for f(x) is

f ( x) cn e inx n

sinh a

(1)n (a in)

n

a2 n2

einx

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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y = f(x) cosx cos2x sinx sin2x ycosx yco

x

s2 ysinx ysin2x

1.0

1

1

0

0

1 1

0

0

a

3 2

points. In such a case, the integrals in Eulers 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 Eulers formulae, (1), (2), (3) may be redefined as

1 b 1

a 2l

f ( x) f ( x)dx b a

a

a0 2 2l

f ( x)dx 2[ f ( x)] .

1 a 2l

n x

n x

an 2 f ( x) cos dx 2 f ( x) cos 2l a l l

1 a 2l

n x

n x

bn 2 f ( x) sin dx 2 f ( x) sin 2l a 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 a2 b2 and that

of second harmonic is

a

2 b2

1 1

and so on. 2 2

Problems:

1. Find the first two harmonics of the Fourier series of f(x) given the following table

:

x 0 2 3

4 53 3

f(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

0

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n

n

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

We have

a 2

f ( x) cos n

l

x 2[ y cos nx]

as the length of interval= 2l = 2 or

l=

b 2

f ( x) sin n

l

x 2[ y sin nx]

Putting, n=1,2, we get

2 y cos x 2(1.1) a 2[ y cos x] 0.367 1

6 6

2 y cos 2 x 2(0.3) a 2[ y cos 2 x] 0.1 2

b1 [ y sin x]

6

2 y sin x 6

6

1.0392

b2 [ y sin 2x] 2 y sin 2 x

6

0.0577

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 :

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x 0 1 2 3 4 5

y 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

a x x 2x 2x 3x 3x y 0 a cos b sin a cos b sin a cos b sin

2 1

or

l 1

l 2

l 2

l 3

l 3

l

a x x 2x 2x 3x 3x y 0 a cos b sin a cos b sin a cos b sin

2 1

3 1

3 2

3 2

3 3

3 3

3

Put x

, then 3

y a0 a cos b sin a

cos 2 b

sin 2 a

cos 3 b sin 3 (1) 2

1 1 2 2 3 3

We prepare the following table using the given values :

x = x 3

y 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 0

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2 y 1 a 2[ f ( x)] 2[ y] (42) 14 0

6 3

a 2[ y cos ] 2

(8.5) 2.833 1

6

b 2[ y sin ] 2

(12.99) 4.33 1

6

a 2[ y cos 2 ] 2

(4.5) 1.5 2

6

b 2[ y sin 2 ] 2

(2.598) 0.866 2

6

a 2[ y cos 3 ] 2

(8) 2.667 3 6

b3 2[ y sin 3 ] 0

Usingthesein(1),we get

y 7 2,833cos x

(4.33) sin x

1.5 cos 2x

0.866 sin 2x

2.667 cosx

3

3

3

3

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

2 of period T. Let us denote

T t . We have

a0 2[ A]

a 2

A cos 2

t 2[ A cos ]

(1)

1

T

b 2 Asin

2 t 2[ Asin ]

1 T

We prepare the following table:

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t

0

2t

T

0

A

1.98

cos

1

sin

0

Acos

1.98

Asin

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

2 A 4.5 a 1.5 0

6 3

2 A cosa1

6

2 Asinb1

6

1.12

0.3733 3

3.0137

1.0046 3

The Fourier expansion upto the first harmonic is

a 2t 2t

A 0 a cos b sin 2

1

T 1

T

2t 2t 0.75 0.3733cos 1.0046 sin

T T

The expression shows that A has a constant part 0.75 in it. Also the amplitude of the first

harmonic is a2 b2

= 1.0717. 1 1

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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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Fourier cosine and Fourier sine transforms & Inverse Fourier

cosine and sine transforms

FOURIER TRANSFORMS

Introduction

Fourier Transform is a technique employed to solve ODEs, PDEs,IVPs, BVPs and Integral equations.

The subject matter is divided into the following sub topics :

FOURIER TRANSFORMS

Infinite Sine Cosine Convolution

Fourier Transform Transform Theorem &

Transform Parsevals Identity

Infinite Fourier Transform

Let f(x) be a real valued, differentiable function that satisfies the following conditions:

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1) f(x) and its derivative f x are continuous, or have only a finite number of simple discontinuities in every finite interval, and

2) the integral f x dx exists. -

Also, let be non - zero real parameter. is defined by

The infinite Fourier Transform of f(x)

f F f x f xeix

dx

provided the integral exists.

The infinite Fourier Transform is also called complex Fourier Transform or

just the Fourier Transform. The inverse Fourier Transform of f denoted by F

-1 f is defined by F

1 f f x 1 2

f e ix d

Note : The function f(x) is said to be self reciprocal with respect to Fourier transform

if f f .

Basic Properties

1. Linearity Property

For any two functions f(x) and (x) (whose Fourier Transforms exist) and any two constants a and b,

F af x b x aF f x bF x

Proof :

By definition, we have

F af x b x

af x b x eix dx

a

f xe ix dx b

x e ix dx

aF f x bF x

This is the desired property.

In particular, if a = b = 1, we get

F f x x F f x F xAgain if a= -b = 1, we get

F f x x F f x F x

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i u

2. Change of Scale Property

If f Ff x , then for any non - zero constant a, we have

1

Ff x f

Proof : By definition, we have

a a

F f ax f ax e ix dx (1)

Suppose a > 0. let us set ax = u. Then expression (1) becomes

F f ax

f u e a du

a

1

f

(2) a a

Suppose a < 0. If we set again ax = u, then (1) becomes

i u

F f ax

f u e

a du

a

1

f u e a du

a

i u

1 a

f a

(3)

Expressions (2) and (3) may be combined as

1

F f ax f

3. Shifting Properties

a a

This is the desired property

For any real constant a,

(i)

F f x a eia f

Proof : (i) We have

(ii) F eiax f x

F f x

f a

f

f x e ix dx

Hence, F f x a f x a e ix

dx

Set x-a = t. Then dx = dt.Then,

F f x a

f t ei (t a ) dt

= eia

f t e it

dt

ii) We have

= eia

f

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2

f a

f x ei a x

dx

f x e iax e i x dx

g x e iax

dx ,

where

g ( x )

f ( x )e iax

This is the desired result.

F g x

F eiax f x

4. Modulation Property

If F f x

f ,

then, F f xcos ax 1 f a 2

where a is a real constant.

f a

Proof : We have

cos ax

Hence

e iax e iax

2

F f xcos ax F f x 2

1 f a f a , by using linearity and shift properties. 2

This is the desired property.

Note : Similarly

F f xsin ax 1 f a f a

Problems:

1. Find the Fourier Transform of the function f(x) e-a x

where a 0

For the given function, we have

F f x

e a x

e ix

dx

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0

e e dx

0

a

e a x

eix

dx

a x ix 0

Using the fact that x x, 0 x and x - x, - x 0, we get

F f x

e ax

ix

ax

ix

e

dx 0 e e dx0

a i x

a i x

e

dx e dx

e a i x 0

0

e a i x

a i a i 0

1 1 a i

a i

2a a 2 2

2. Find the Fourier Transform of the function

1, f(x)

0,

x a

x a

where a is a positive constant. Hence evaluate

(i) sin a

cos x d

(ii) sin

d0


ix

F f x

f ( x)e

dx

a

f ( x)eix

dx

f ( x)eix

dx f ( x)e ix dx

a

e a

ix

a

dx

a

2 sin a

a

Thus F f x f 2 sin

(1)

Inverting f by employing inversion formula, we get

f x 1

2 sin a

e ix d

2 1 sin acosx i sin x

d

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2 2

1 sin acos x

sin a sin x

d i d

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

1 sin a cosx

f ( x) d

or

sin a cosx d f ( x)

, 0,

x a

x a

For x 0, a 1, this yields

sin

d

Since the integrand is even, we have sin

2 d 0 or

sin d

0 2

3. Find the Fourier Transform

x 2

of f(x) e - a

2 x

2

where ' a' is a positive constant.

Deduce that f(x) e 2 is self reciprocal with respect to Fourier Transform.

Here

F f x

e a x

e ix

dx

2 2

e a x i x dx

i

ax

2

2

2

e

2 a 4 a dx

2

2

ax

i

2

e 4 a e

2 a dx

Setting t ax - i

, we get 2a

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F f x e

a 2

e t 2

s

s

s

s

s s

dt

2

2 -

s

2

4

1 2

a

2

e a

2

4 a 2 0

e t

dt

1

e a

4 a 2 , using gamma function.

f

This is the desired Fourier Transform of f(x).

2

e a

4 a 2

For

a 2 1 2

in f(x) e- a

-x 2

2 x

2

2

we get f(x) e

2

and hence,

f 2 e

2

- x2

Also putting x in f(x) e 2 , we get f( ) e - 2

2 .

Hence, f( ) and f are same but for constant multiplication by

Thus f( ) f

2 .

x

It follows that f( x) e 2

is self reciprocal

FOURIER SINE TRANSFORMS

Let f(x) be defined for all positive values of x.

The integral 0 f xsin xdx is called the Fourier Sine Transform of f(x). This is denoted

by fs or F f x.

Thus

f Fs f x

f xsin x dx

s 0 The inverse Fourier sine Transform of f is defined

through the integral 2

f sin x d0

This is denoted by f(x) or F-1 f . Thus

f(x)

F-1

f s 2

f sin x d 0

Properties

The following are the basic properties of Sine Transforms.

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s s

s

(1) LINEARITY PROPERTY

If a and b are two constants, then for two functions f(x) and (x), we

have

F af x b x aF f x bF g xs s s

Proof : By definition, we have

Fs af x b x 0 af x b xsin x dx

aF f x bF xs s

This is the desired result. In particular, we have

F f x x F f x F xs s s

and

Fs f x x Fs f x Fs x

(2) CHANGE OF SCALE PROPERTY

If F f x f , then for a 0, we have s s

1

F f ax f

Proof : We have

F f ax

a a

f axsin x dx

Setting ax = t , we get

s 0

dt

F f ax f t sin t s 0

1

a a

f a a

(3) MODULATION PROPERTY

If F f x f , then for a 0, we have s s

F f xcos ax 1 f a f as Proof : We have

2 s s

1

Fs f xcos ax 0

f xcos ax sin x dx

2 0

f xsin a x sin axdx

1 f a f a, by using Linearity property. 2 s s

Problems:

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f

s

2

s

1. Find the Fourier sine transform of

1,

0 x a


f x 0,

x a

a

f s 0 sin

a

x dx a 0 sin x dx

cosx

0

1 cos a

2. Find the Fourier sine transform of f(x) e -ax

x

Here

s e

ax sin x dx

0 x

Differentiating with respect to , we get

d f

d e ax sin x dx d

e ax

d 0 x

0 sin x dx x performing differentiation under the integral sign

e ax

0

x

e ax

x cos x dx

a

2

a

a cosx sin x 0

a 2 2

Integrating with respect to , we get

f s

tan 1

c a

But f 0 when 0 c=0

f tan 1 s a

3. Find f(x) from the integral equation

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f x sin xdx 2,

2

1

c

1,

0 1

1 2

Let () be defined by

0

1,

0,

0 1

2

Given

2,

0,

0

1 2

2

f xsin xdx

f s

Using this in the inversion formula, we get

f x 2

sin x dx

0

2 sin x d

2 sin x d

sin x d

0

1 2

2

1sin x d

2 2 sin x d 0

0 1

2 1 cos x 2 cos 2 x

x

FOURIER COSINE TRANSFORMS

Let f(x) be defined for positive values of x. The integral 0

f xcosx dx

is called the Fourier Cosine Transform of f(x) and is denoted by f or F f x. Thus

f F f x 2

c c

f xcosxdx

c c 0

The inverse Fourier Cosine Transform of fc is defined through

the integral 2

f cos x d . This is denoted by f x or F - 1 f .

Thus

0

f x F -1 f

f cos

c

x d

c

Basic Properties

c 0

c

The following are the basic properties of cosine transforms :

(1) Linearity property

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c

c

If ' a' and ' b' are two constants, then for two functions f(x) and (x), we have Fc af x b x aFc f x bFc x

(2) Change of scale property

If Fc f x f , then for 1

a 0, we have

Fc f ax fc a a

(3) Modulation property

If Fc f x f , then for a 0, we have

F f x cos ax 1 f a f ac 2

c c

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

x, 0 x 1

f x 2 x,

1 x 2

0,

We have

x 2

f c 0

f xcosxdx

1 2

0

x cosxdx 1 2 xcosxdx 2 0 cosxdxIntegrating by parts, we get

sin x 1 cosx

sin x 2

cosx

f c x 2 2 x

1 2

0 1

2 cos cos 2 1 2

2) Find the cosine transform of f(x) e-ax , a 0. Hence evaluate cos kx

dx

0 x 2 a 2

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2

a

e

a

x

Here

ax

fc e cosxdx 0

e ax

Thus

a

2 a cosx sin x

0

f a

c

a 2 2

Using the definition of inverse cosine transform, we get

f x 2 a

cosxd 0 a 2 2

or

eax

2a

0

cosx

d 2 a 2

Changing x to k, and to x, we get

cos kx dx

e ax

0 x 2 a 2 2a

3) Solve the integral equation

f x cosx dx e 0

Let () be defined by

() = e-a

Given 0

f xcosx dx f c

Using this in the inversion formula, we get

f x 2

0 cos

xd

2 a 0

cos

xd

2 e

a cos

x sin

x

0 2 2 0 2 a

a 2 x 2

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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 Laplaces equation

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DAlemberts 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 Laplaces 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 DAlemberts 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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2

1 1 2 3 2 3

2 2

2 2

2

Various possible solutions of the one dimensional wave equation

utt =c2uxx by the method of separation of variables.

2 2

Consider u

c2 u

t 2 x2 Let u= XT where X=X(x),T=T(t) be the solution of the PDE

Hence the PDE becomes

2 XT 2 XT c2

or X d 2T d 2 X

c2 t 2 x2 dt 2 dx2

2 2

Dividing by c2XT we have

1 d T

1 d X

c2T dt

2 X dx

2

Equating both sides to a common constant k we have

1 d 2 X

X dx2

=k and 1 d 2T

=k c2T dt 2 2

d 2 X kX 0

dx2

d T and dt

2

c2 kT 0

D2 k X 0

Where D2

= d

and D2 c2 k T 0

in the first equation and D2

= d

in the second equation

dx2 dt

2

Case(i) : let k=0

AEs are m=0 amd m2=0 amd m=0,0 are the roots

Solutions are given by

T = c e0t c and X c x c e0 x c x c

Hence the solution of the PDE is given by

U= XT= c1 c2 x c3 Or u(x,t) =Ax+B where c1c2=A and c1c3=B

Case(ii) let k be positive say k=+p2

AEs are m c2p2=0 and m2-p2=0 m= c

2p

2 and m=+p

solutions are given by

T c ' ec p t andX c ' e px c ' e px 1 2 3 Hence the solution of the PDE is given by

u XT c ' ec p t .( c ' e px c ' e px ) 1 ' c2 p2t

2 3

px px

Or u(x,t) = c 1e (A e +B e ) where c1c2=A and c1c3=B

Case(iii): let k be negative say k=-p2

AEs are m+ c2p

2=0 and m

2+p

2=0

m=- c2p

2 and m=+ip

solutions are given by '' c2 p2t

'' ''

T c 1e and X c 2 cos px c 3 sin px

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1 2 3

2 2 ''

1 2 4


u XT c '' e c2 p2t .(c

'' cos px c '' sin px)

u( x, t) e c p t ( A

cos px B ''

sin px)

Various possible solutions of the two dimensional Laplaces equation uxx+uyy=0 by the method of separation of variables

2 2

Consider u

u

0

x2 y 2

Let u=XY where X(x), Y=Y(y) e the solution of the PDE

Hence the PDE becomes

2 XY x2

2 ( XY ) 0

y 2

d 2 X

Y X dx

2

d 2 (Y )

dy2

0 and dividing by XY we have

1 d 2 X 1 d 2 (Y )

X dx

2 Y dy

2

Equating both sides to a common constant k we have

1 d 2 X

X dx2

=k and

1 d 2 (Y )

=k Y dy

2

Or (D2-k)X=0 and (D

2+k)Y=0

Where D = d

dx

in the first equation and D = d

dy

in the second equation

Case(i) Let k=0

AE arem2=0 in respect of both the equations

M=0,0 and m=0,0


X= c1x+c2 and Y= c3y+c4


U=XY=( c1x+c2 ) (c3y+c4)

Case(ii) : let k be positive ,say k=+p2

m2-p

2=0 and m

2+p

2=0

m=+p and m=+ip

solutions are given by

X c ' e px c ' e px and Y c ' cos py c ' sin py 1 2 3 4 Hence the solution of the PDE is given by

u XY c ' e px c ' e px (c ' 3 cos py c ' sin py) Case(iii) Let k be negative say k= -p

2

AEs are m2+p

2=0 and m

2-p

2=0

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1 2 3 4

1 2 3 4

0

n

M=+ip and m=+p


X (c '' cos px c ' ' sin px)

and Y (c '' e py

c '' e

py

)


u XY (c '' cos px c '' sin px)(c '' e py c '' e py )

EXAMPLES

1.Solve the wave equation utt=c2uxx subject to the conditions

u(t,0)=0 ,u(l,t)=0, u

x, 0 0 t

and u(x,0) =u0sin3( x/l)

n x n ct

Soln: u x, t bn sin cos n 1 l l

Consider u(x,0) =u0sin3( x/l)

u x, 0 bn sin n 1

n x

l

u sin3 x b

sin

n x

l n1 n

l

u 3

sin 3 x

1 sin

3 x b

sin n x

3u0 sin

x

u0 sin

3 x b sin

x b

sin 2 x

b sin 3 x

0 4 l 4 l

n1 l

4 l 4 l 1

l 2

l 3

l

comparing both sides we get

b 3u0 , b 0 , b

u

0 , b

0 b

0 , 1 4

2

3 4

4 5

Thus by substituting these values in the expanded form we get

u( x, t) 3u0 sin

x cos

ct

u0 sin 3 x

cos 3 ct

4 l l 4 l l

2.Solve the wave equation utt=c2utt subject to the conditions

u(t,0)=0 ,u(l,t)=0, u

x, 0 0 t

when t=0and u(x,0) =f(x)

n x n ct

Soln: : u x, t bn sin cos n 1 l l

Consider u(x,0)=f(x) then we have

Consider u(x,0) =

bn

sin n 1

n x

l

F(x) =

bn

sin n 1

n x

l

The series in RHS is regarded as the sine half range Fourier series of f(x) in (0,l) and

hence

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0

l

n 2

l

n

2 l

0

l l l

2 l

n x bn

l f ( x) sin

l dx

Thus we have the required solution in the form n x n ct

u x, t bn sin cos n 1 l l

2

3. Solve the Heat equation u

c2 u

given that u(0,t)=0,u(l,0)=0 and u(x,0)=

100x/l

l

t x2

l

Soln: b 2

100 x sin

n x dx =

200 x sin

n x dx

l 0

l l l 0

l

l

n x n x

200 x. cos

b

sin

1 l

n l 2 n / l n / l

2

0

b 200

. 1 l cos n

200 1 n

200 .

1n 1

n l

2 n n n

The required solution is obtained by substituting this value of b

n

Thus u( x, t )

200 1n 1

e

n2 2c2t

2

sin

n x

n 1 n l l

u 2u 4.Obtain the solution of the heat equation c2

given that u(0,t)=0,u(l,t)and

u(x,0) =f(x)where

2Tx in 0 x

l

t x2

f ( x) l 2

2T l x in l x l

l 2 l

Soln: b 2

f ( x) sin n x

dx l

0 l

l

b 2

2Tx sin

n x dx

2Tx (l x) sin

n x dx

n

l l l

l l

2

l

2 l

4T

x sin n x

dx

(l x) sin n x

dx

0 l

b 8T

2

sin n

n n

2 2 2

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2

n2

2

2

The required solution is obtained by substituting this value of bn 2 2 2

8T 1

n n c t

n x

Thus u( x, t ) sin e n 1 2 l

sin l

5. Solve the heat equation

u(x,0) =3sin x

u 2u

t x2

with the boundary conditions u(0,t)=0,u(l,t)and

Soln: u( x, t) e p t ( A cos px B sin px)............................(1)

Consider u(0,t)=0 now 1 becomes p2t

0= e (A) thus A=0 Consider u(1,t)=0 using A=0 (1) becomes

p2t

0= e (Bsinp)

Since B0,sinp=0or p=n n2 2c2t

u( x, t ) e (B sin n x)

In general u( x, t ) bn e n1

n2 2c2t

sin n x

Consider u(x,0)= 3 sin n x and we have

3 sin n x b1 sin x b2 sin 2 x b3 sin 3 x

Comparing both sides we get b1 3, b2 0, b3 0

We substitute these values in the expanded form and then get

2 t

u( x, t ) 3e (sin x)

6.Solve 2u 2u

0 subject to the conditions u(0,y)=0,u( ,y)=0 and u(x,) =0 and x2

u(x,0)=ksin2x

y 2

Soln: The befitting solution to solve the given problem is given by

u( x, y) ( A cos px B sin px) Ce py De py u(0, y) 0 gives( A) Ce py De py 0, A 0

u( , y) 0 gives(B sin p ) Ce py De py 0, Since A=0,B0 and we must have

sin p =0,therefore p=n where n is a integer

u( x, y) (B sin nx) Ce ny De ny The condition u(x,) =0 means that uas y

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0=(Bsinnx)( Ce ny

)since e ny

Since B0 we must have C=0

0 asy

We now have u(x,y) =BDsinnx e ny

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

u( x, y) bn n1

sin n xe ny

Consider u(x,0)=ksin2x and we have

u( x, 0) bn sin nx n 1

Ksin2x=b1sinx+b2sin2x+b3sin3x+. Comparing both sides we get b1=0, b2=0, b3=0

Thus by substituting these values in the expanded form we get

u( x, y) k sin 2xe2 y

DAlemberts solution of the one dimensional wave equation

We have one dimensional wave equation

2u 2u c2

t 2 x2 Let v=x+ct and w=x-ct We treat u as a function of v and w which are functions of x and t

By chain rule we have,

u u

. v

u

. w

x v x w x v

Since v=x+ct and w=x-ct, x

1 and w

1 x

u

x

u .(1)

u v w

.(1) u v

u w

2u

x2 x

u x x

u v

u w

Again by applying chain rule we have, 2

u u

u

. v

u

u

. w

x2

v v

w

x

w

v

w

x

2 2 2 2 2

u u

u u

.(1)

u

.(1)

x2

v2

vw

wv

w2

But 2u

= vw

2u

wv

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v 0

w

2u 2u 2u 2u But

x2 2 v2 wv

w2

Similarly u

u

. v

u

. w

x v t w t

Since v=x+ct and w=x-ct, v

c and w

c t t

u

t

u .(c)

u

v w

.(c) , or u

c u

t v

u

w

Again applying chain rule we have,

2

u u

c u

u

. v

c u

u

. w

r 2

t t

v w

v

w

t

w

v

w t

2 2 2 2 2

u c

u u u

.(c) c

u

.(c)

r 2

v2

vw

wv

w2

2 2 2 2 2

u c2

u u u

c2

u

t 2

v2

vw

wv

w2

2 2 2 2

u c2

u 2

u u

t 2

v2

wv

w2

2 2 2

2 2 2

c2 u 2

v2

u

wv

u =

w2 c

2 u v2

2 u

wv

u

w2

2 2u

4 u

wv

=0 or

=0 wv

We solve this PDE by direct integration, writing it in the form,

u

u Integrating w.r.t w treating v as constant we get

v

f (v)

Now integrating w.r.t, we get u f (v)dv G(w) U=F(v)+G(w) where F(v)=

But v=x+ct and w=x-ct

f (v)dv

Thus u=u(x,t)=F(x+ct)+G(x-ct)

This is the DAlemberts solution of the one dimensional wave equation

EXAMPLES

1. Obtain the DAlemberts solution of the wave equation utt=c2uxx subject to the

conditions u(x,0)=f(x) and u

( x, 0) 0 t

Soln: DAlemberts solution of the wave equation is given by

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u(x,t)=F(x+ct0=G(x-ct)..(1) Consider u(x,0)=f(x) (10 becomes

u(x,0)=F(x)+G(x)

f(x)=F(x)+G(x).(2) Differentiating partially wrt t we have

u (x,t) =F(x+ct).(c)+G(x-ct).(-c)

t u

(x,)=c[F(x)-G(x)] t 0= c[F(x)-G(x)] [F(x)-G(x)]=0 integrating wrt x we have [F(x)-G(x)]=k

Where 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 f ( x) k andG( x)

1 f ( x) k

Thus F(x+ct)=

2 2

1 f ( x ct) k and G(x-ct)=

2

1 f ( x ct) k

2 Substituting these in(10 we have

U(x,t)= 1 f ( x ct) k

1 f ( x ct) k

2 2 Thus the required solution is

u(x,t)= 1 f ( x ct) f ( x ct)

2

2. Obtain the DAlemberts solution of the wave equation utt=c2uxx given that

u(x,0)=f(x)=l2-x

2and ut(x,0)=0

Soln. Assuming the DAlemberts solution

u(x,t)=

u(x,t)=

1 f ( x ct) f ( x ct) and f(x)=l2-x2

2

1 l 2 ( x ct)2 l 2 ( x ct)2

2

1 2l 2 2x2 2c2t 2

2

u(x,t)= l 2 x2 c2t 2

3. . Obtain the DAlemberts solution of the wave equation utt=c2uxx given that

u(x,0)=asin2x and

u 0

t

when t 0

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Soln: Assuming the DAlemberts solution

u(x,t)= 1 f ( x ct) f ( x ct)

2

By data f(x)= asin2x or f(x)=

a 1 cos 2 x

2

u(x,t)=

u(x,t)=

1 . a 1 cos 2 ( x ct) 1 cos 2 ( x ct)

2 2

a 2 cos 2 x 2 ct cos 2 x 2 ct

4

a 2 2 cos 2 x cos 2 ct

4

u(x,t)= a 1 1cos 2 x cos 2 ct

2

UNIT IV CURVE FITTING AND OPTIMIZATION

CONTENTS:

Curve fitting by the method of least squares

Fitting of curves of the form

y ax b

y ax 2 bx c

y aebx

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y axb

Optimization :

Linear programming

Graphical method

Simplex method

CURVE FITTING AND OPTIMIZATION

CURVE FITTING [BY THE METHOD OF LEAST SQUARE]:

We can plot n points ( xi , yi ) 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 ( xi , yi ) , i=0,1.2..n equation of a straight line is

y a bx

Then for n points

(1)

Yi a bxi

(2)

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2

S

Where a and b are parameters to be determined; Yi is called the estimated value. The

given value Yi

corresponding to xi .

Let S

y Y 2

(3) i i 2

= yi a bxi S = y

a bx

2

(4) i i We determined a and b so that S is minimum (least). Two necessary conditions for this

S 0 and

a

S 0

b .

differentiate (4) w.r.t a and b partially

S

a yi a bxi

0 yi a bxi 0 = yi a b xi 0= yi na b xi

yi na b xi

or y na b x

b 2 yi a bxi x i

0 2

x y ax

bx2

i i i i

0 = xi yi a xi b xi

i i

i i

x y n x b x2

or xy a x b x2

where n = number of points or value.

FITTING A SECOND DEGREE PARABOLA Y=AX2

+ BX + C

Let us take equation of parabola called parabola of best fit in the form

y a bx cx2 (1)

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i

Where a, b, c are parameters to be determined. Let be the value of corresponding to the v

Y a bx cx2 (2) i i i

Also

S= yi Y 2

(3)

2 2

yi a bxi cxi

S y

a bx cx2

(4) i i i

We determine a, b, c so that S is least (minimum).

The necessary condition for this are

S 0,

S 0 &

S 0

a b c

diff (4) w.r.t a partially

S 2

y a bx cx2

a

i i i

0 2

y a bx cx2

i i i

0 y a bx

cx2

i i i

i i i

0 y a b x c x2

i i i

y a b x c x2

i i i

y na b x c x2

or y na b x c x2 diff (4) w.r.t b partially

S 2

y a bx cx2 x

b

i i i i

0 2

x y ax bx2 cx3

i i i i i

0 x y ax

bx2 cx3

i i i i i

i i

i i i

0 x y a x b x2 c x3

xi yi a xi b xi

c xi

2 3

or xy a x b x2 c x3

diff (4) w.r.t c partially

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S 2

y a bx cx2 x2

c

i i i i

0 2

x2 y ax2 bx3 cx4

0 x2 y

i i i i i

ax2 bx3 cx4

i i i i i

i i i

i i

0 x2 y a x2 b x3 c x4

i i i

i i

x2 y a x2 b x3 c x4

or x2 y a x2 b x3 c x4

Hence the normal equation for second degree parabola are

y na b x c x2

xy a x b x2 c x

3

x2 y a x2 b x3 c x4

PROBLEMS:

1) Fit a straight line y a bx

to the following data

x : 5 10 15 20 25

y : 16 19 23 26 30

Solution: Let y a bx (1)

Normal equation

y na b x xy a x b x2

(2)

(3)

x y x

2 xy

5 16 25 80

10 19 100 190

15 23 225 345

20 26 400 520

25 30 625 750

x y x2 xy 75 =114 =1375 =1885

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(1) & (2)

114 15a b 75

1885 75a b 1375

a 12.3

b 0.7

(1) y 12.3 0.7 x

2) Fit equation of straight line of best fit to the following data

x : 1 2 3 4 5

y : 14 13 9 5 2

Solution:

Let

y a bx

Normal equation

(1)

y na b x

xy a x b x2 x y x

2 xy

(2)

(3)

1 14 1 14

2 13 4 26

3 9 9 27

4 5 16 20

5 2 25 10

x y x2 xy 15 =43 =55 =97

(2) & (3)

43 15a 15b

97 15a 56b Solving above equations we get

a 18.2

b 3.2

(1) y 18.2 3.2 x

3) The equation of straight line of best fit find the equation of best fit

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x : 0 1 2 3 4

y : 1 1.8 3.3 4.5 6.3

Solution: let y a bx (1)

Normal equation

y na b x xy a x b x2 x y x

2 xy

(2)

(3)

0 1 0 0

1 1.8 1 1.8

2 3.3 4 6.6

3 4.5 9 13.5

4 6.3 16 25.2

x y x2 xy 10 =16.9 =30 =47.1

(2) & (3)

16.9 5a 10b

47.1 10a 30b

a 0.72

b 1.33

(1) y 0.72 1.33x

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

p : 12 15 21 25

Compute p when W=150.

Solution: Given p=y & W=x equation of straight line is

:

let y a b x ( 1 )

Normal equations

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y na b x

xy a x b x2

(2)

(3)

x p= y x2 xy

50 12 2500 600

70 15 4900 1050

100 21 10000 2100

120 25 144000 3000

x 10 (2) & (3)

y =16.9

x 2 =30 xy =47.1

73 4a 340b

6750 340a 31800b

a 2.27

b 0.187

(1) y 2.27 0.187 x

put 150

y 30.32

5) Fit a curve of the form y abx

Solution: Consider

y abx

Take log on both side

logy=log abx =loga+logb

x

logy=loga+logbx

y A Bx

(1)

(2) ; logy=Y y=ey

Corresponding normal equation loga=A a=eA

Y nA B x

xY A x B x2

(3) logb=B

(4)

b=eB

Solving the normal equation (3) & (4) for a & b . Substitute these values in (1) we get

y abx

curve of best fit of the form

6) Fit a curve of the curve y abx

for the data

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x : 1 2 3 4 5 6 7 8

y : 1.0 1.2 1.8 2.5 3.6 4.7 6.6 9.1

Solution:

let y ab x

Normal equations are

Y nA B x

xY A x B x2

(1)

(2); A=loga

(3) Y=logy, B=logb

x y Y=logy x 2

xY

1 1.0 0 1 0

2 1.2 0.182 4 0.364

3 1.8 0.587 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

x 36 Y =8.607 x2 =204 xY =52.34

(1) & (2)

8.607 8 A 36B

52.34 36 A 203B

A 0.382

B 0.324

then e A a 0.682

eB b 1.382

(1)

y 0.682 1.382x

7) Fit a curve of the form y axb

for the following data x : 1 1.5 2 2.5

y : 2.5 5.61 10.0 15.6

Solution:: Consider

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let y axb

Normal equations are

Y nA b x

XY A X B X 2

(1)

(2); Y=logy

(3) A=loga, X=logx

x y X= log x X 2

Y log y XY

1 2.5 0 0 0.916 0

1.5 5.62 0.405 0.164 1.726 0.699

2 10.0 0.693 0.480 2.302 1.595

2.5 15.6 0.916 0.839 2.747 2.516

X=2.014 X2 =1.483 Y=7.691 XY=4.81

(1) & (2)

7.691 4 A 2.014b

4.81 2.014 A 1.483b

A=0.916, eA a 2.499 2.5

b 1.999 2

(1)

y 2.5 x 2

8) Fit a parabola y a bx cx2

for the following data

Sol:

x : 1 2 3 4

y : 1.7 1.8 2.3 3.2

y a bx cx2

Normal equation

(1)

y na b x c x2

xy a x b x2 c x

3

(2)

(3)

x2 y a x

2 b x3 c x

4 (4)

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x y x2 x

3 x

4 xy x

2 y

1 1.7 1 1 1 1.7 1.7

2 1.8 4 8 16 3.6 7.2

3 2.3 9 27 81 6.9 20.7

4 3.2 16 64 256 12.8 51.2

x y 9 x2 30 x3 100 x4 354 xy 25 x2 y 80.8 (1), (3) & (4)

9 4a 10b 30c

25 10a 30b 100c

80.8 30a 100b 354c

a 2

b -0.5

c 0.2

(1)

y 2 - 0.5x 0.2x2

9) Fit a curve of the form x : 0 2 4

y : 8.12 10 31.82

y aebx

for the following

Sol:

y aebx

Normal equation

Y=nA+b x

(1)

(2); Y=logy

xy A x b x2 (3) A=loga

x y Y=logy x2

xY

0 8.12 2.094 0 0

2 10 2.302 4 4.604

4 31.82 3..46 16 13.84

x 6 (2) & (3)

Y=7.86 x2 20 xY 18.444

7.856=3A+6b

18.444=6A+20b

A=1.935,

b=0.341

a e A 6.924

(1) y 6.924e0..341x

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10) Fit a II degree parabola

find y when x=6

ax 2 bx c

to the least square method & hence

x : 1 2 3 4 5

y : 10 12 13 16 19

Sol:

y ax2 bx c

y c bx ax2

Normal equation

(1)

(1)

y nc b x a x2

xy c x b x2 a x

3

(2)

(3)

x2 y c x2 b x3 a x4 (4)

x y x2 x3 x4 xy x2 y

1 10 1 1 1 10 10

2 12 4 8 16 24 14

3 13 9 27 81 39 117

4 16 16 64 256 64 256

5 19 25 125 625 95 475

15 70 55 225 979 232 906

70 5c 15b 55a

232 15c 55b 225a

906 55c 225b 979a

a 0.285

b 0.485

c 9.4

(1)

y 0.285x2 0.485x 9.4

at x 6, y 22.6

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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The optimization of the objective function Z subject to the constraints is the

mathematical formulation of a LPP.

A set real values

solution.

X ( x1 , x2 ,........xn ) which satisfies the constraint AX ()B

is called

A set of real values xi which satisfies the constraints and also satisfies non negativity

constraints xi 0 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.

x

y 1

The constraints are considered in the form of a b 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

vertices.

Z c1 x1 c2 x2 ...... cn xn

is found at all these

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

2x y 40 , 2x 5 y 180, x 0, y 0.

Solution: Let us consider the equations

2x y 40

2x 5 y 180

x

y 20 40

1.............(1) x

y 90 36

1...........(2)

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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:

2x y 40 , 2 x 5 y 180. E( x, y) (2.5,35)

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) 0

A(20, 0) 60

E(2.5, 35) 147.5

D(0,36) 144

Thus (Z)Max 147.5 when x 2.5, y 35

2) Use the graphical method to maximize Z 3x1 5x2

subject to the constraints

x1 2x2 2000 , x1 x2 1500, x2 600 , x1 , x2 0.


x1 x2 2000 , x1 x2 1500, x2 600

x1

2000

x2 1000

1.....(1) x1

1500

x2 1500

1...........(2)

x2 600.........(3)

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)

x2 600. is a line parallel to the x1 axis.


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On solving

x1 x2 2000 ,

x1 x2 1500 ,

x1 2 x2 2000 ,

x1 x2 1500,

x2 600,

x2 600,

we get

we get

we get

E (1000,500)

F (900,600)

G (800,600)

Also we have C = (1500 , 0), H= (0, 600)


tabulated.

Corner Value of

O(0,0) 0 C(1500, 0) 4500

E(1000, 500) 5500

F(900, 600) 5700

G(800,600) 5400

Z 3x1 5x2

Thus (Z )Max 5700 when x1 900, x2 600

3) Use the graphical method to minimize Z 20x1 10x2

subject to the constraints

x1 2x2 40 , 3x1 x2 30, 4x1 3x2 60 , x1 , x2 0.


x1 2x2 40 , 3 x1 x2 30, 34 x1 3x2 60

x1

x2

1.....(1) x1

x2

1...........(2) x1

x2

1.........(3) 40 20 10 30 15 20

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)


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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:


tabulated.

Corner Value of

A(15 , 0) 300

A(40, 0) 800

H(4 , 18) 260

G (6, 12) 240

Z 3x1 5x2

Thus (Z ) MIN 240 when x1 6, x2 12

4) Show that the following LPP does not have any feasible solution.

Objective function for maximization:Z = 20x+30y.

Constraints: 3x 4 y 24 , 7 x 9 y 63, x 0, y 0.


3x 4 y 24

7 x 9 y 63

x

y 1.............(1)

x

y 1...........(2)

8 6 9 7 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)


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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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 s1 (say) 0

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

3x y 22, 2x 3 y 24, x 0, y 0

Solution: Let us introduce slack variables

the following form.

3x y 1.s1 0.s2 22

2x 3 y 0.s1 1.s2 24

s1 and s2 to the inequalities to write them in

2x 4 y 0.s1 0.s2 z is the objective function

Solution by the simplex method is presented in the following table.

NZV x y s1 s2 Qty Ratio

s1 3 1 1 0 22 22/1=22 8 is least, 3 is pivot,

s2 is replaced by y.

Also 1/3.R2 s2 2 3 0 1 24 24/3=8

Indicators ()

-2 -4 0 0 0

s1 3 1 1 0 22 R1 R2 R1

y 2/3 1 0 1/3 8

-2 -4 0 0 0 R3 4R2 R3

s1 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

Thus the maximum value of Z is 32 at x=0 and y=8

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NZV x y z s s s Qty Ratio

s1 1 3 2 1 0 0 11 11/3=3.33 3.33 is least, 3 is

pivot s1 is replaced

by y. Also 1/3.R1 s2 1 2 5 0 1 0 19 19/3=9.5

s3 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

s2 1 2 5 0 1 0 19 R2 2R1 R2

s3 3 1 4 0 0 1 25 R3 R1 R3

-2 -3 -1 0 0 0 0 R4 3R1 R4

Y 1/3 1 2/3 1/3 0 0 11/3 11/31/3=11 8 is least, 8/3 is

pivot s3 is replaced

by x. Also 3/8.R3

s2 1/3 0 11/3 -2/3 1 0 35/3 35/31/3=35

s3 8/3 0 10/3 -1/3 0 1 64/3 64/38/3=8

-1 0 1 1 0 0 11

y 1/3 1 2/3 1/3 0 0 11/3 R1 1/ 3R3 R1

s2 1/3 0 11/3 -2/3 1 0 35/3 R2 1/ 3R3 R2

x 1 0 10/8 -1/8 0 3/8 8

-1 0 1 1 0 0 1 R4 R3 R4

y 0 1 1/4 3/8 0 - 1/8

1

s2 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 indicators

3

3

3

3

2) Use Simplex method to maximize Z =2x + 3y + z subject to the constraints

x 3 y 2z 11, x 2 y 5z 19, 3x y 4z 25 x 0, y 0, z 0

Solution: Let us introduce slack variables

the following form.

x 3 y 2z 1.s1 0.s2 0.s 11

x 2 y 5z 0.s1 1.s2 0.s 19

3x y 4z 0.s1 0.s2 1.s 25

s1 , s2 , s3 to the inequalities to write them in

2x 3 y z 0.s1 0.s2 0.s P is the objective function


1 2 3

Thus the maximum value of P is 19 at x = 8 and y = 1, z = 0

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3

3

3

3

3) Use Simplex method to minimize P =x - 3y + 2z subject to the constraints

3x y 2 z 7, 2x 4 y 12, 4x 3 y 8z 10 x 0, y 0, z 0

Solution: The given LPP is equivalent to maximizing the objective function P subject to the same constraints

That is P=P=-x + 3y - 2z is to be maximized

Let us introduce slack variables

following form.

3x y 2 z 1.s1 0.s2 0.s 7

s1 , s2 , s3 to the inequalities to write them in the

2x 4 y 0z 0.s1 1.s2 0.s 12

4x 3 y 8z 0.s1 0.s2 1.s 10

x 3 y 2 z 0.s1 0.s2 0.s P

is the objective function


NZV x y z s1 s2 s3 Qty Ratio

s1 3 -1 2 1 0 0 7 7/-1=-7 3 is least, 4 is

pivot s2 is

replaced by y.

Also 1/4.R2

s2 -2 4 0 0 1 0 12 12/4=3

s3 -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 7 R1 R1 R2

y -1/2 1 0 0 1/4 0 3

s3 -4 3 8 0 0 1 10 R3 3R2 R3

1 -3 2 0 0 0 0 R4 3R2 R4

s1 5/2 0 2 1 1/4 0 10 105/2=4 4 is least, 5/2 is

pivot, s1 is

replaced by x.

Also 2/5.R1

y -1/2 1 0 0 1/4 0 3 3-1/2=-6

s3 -5/2 0 8 0 -3/4 1 1 1-5/2=-2/5

-1/2 0 2 0 3/4 0 9

x 1 0 4/5 2/5 1/10 0 4

y -1/2 1 0 0 1/4 0 3 R2 1/ 2R1 R2

s3 -5/2 0 8 0 -3/4 1 1 R3 5 / 2R1 R3

-1/2 0 2 0 3/4 0 9 R4 1/ 2R1 R4

x 1 0 4/5 2/5 1/10 0 4

y 0 1 2/5 1/5 3/10 0 5

s3 0 0 10 1 -1/2 1 11

0 0 12/5 1/5 4/5 0 11 No negative indicators

Thus the maximum value of P is 19 at x = 8 and y = 1, z = 0

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UNIT V NUMERICAL 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

Rayeighs power method Ci

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Introduction:

NUMERICAL METHODS-1

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) < 0

The graph of y = f(x) crosses the x-axis between x0 and x1

Root of f(x) = 0 lies between x0 and x1

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1 0

Now equation of the Chord AB is

y f (x ) f (x1 ) f (x 0 )

0 x x

When y =0 we get x = x2

(x x 0 )

...(1)

i.e. x 2 x 0 x1 x 0

f (x1 ) f (x 0 )

f (x 0 )

...(2)

Which is the first approximation

If f(x0) and f(x2) are opposite in sign then second approximation

x 2 x 0

x 3 x 0

f (x 2 ) f (x 0 ) f (x 0 )

This procedure is continued till the root is found with desired accuracy.

Poblems:

1. Find a real root of x

3 - 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) = -1

f(3) = 16

a root lies between 2 and 3 Take x0 = 2, x1 = 3

x0 = 2, x1 = 3

Now x 2 x 0 x1 x 0

f (x1 ) f (x 0 )

f (x 0 )

2 3 2

(1) 16 1

= 2.0588

f(x2) = f(2.0588) = -0.3908

Root lies between 2.0588 and 3 Taking x0 = 2.0588 and x1 = 3

f(x0) = -0.3908, f(x1) = 16

x1 x 0

We get x 3 x 0 f (x1 ) f (x 0 )

.f (x 0 )

2.0588

= 2.0813

0.9412

16.3908

(0.3908)

f(x3) = f(2.0813) = -0.14680

Root lies between 2.0813 and 3

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Taking x0 = 2.0813 and x1 = 3

f(x0) = 0.14680, f(x1) =16

x 4 2.0813 0.9187

16.1468

(0.14680) 2.0897

Repeating the process the successive approximations are

x5 = 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 xe

x = cos x using Regula falsi method correct to

three decimal places.

Solution:

Let f(x) = cosx - xex

Observe

civil-iii-engineering mathematics - iii [10mat31]-notes

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