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INTERPOLATION

Sol. (i) The difference table is : x

f(x)

0

–3

1

6

∆2f(x)

∆f(x)

∆3f(x)

9

–7

2

2

8

3

12

9

2

4

f(0 + 6) = E6f(0) = (1 + ∆)6f(0) = (1 + 6∆ + 15∆2 + 20∆3) f(0)

= – 3 + 6 (9) + 15 (– 7) + 20 (9) = – 3 + 54 – 105 + 180 = 126.

(ii) Max. power of x in polynomial will be 10 and coeff. of x10 will be abcd.

Here k = abcd, h = 1, n = 10

∴ Expression = k hn n ! = abcd 10 !.

Example 20. (i) Prove that if m is a (+)ve integer then

x (m) x (m−1)

(x + 1)(m)

+

= m!

(m − 1) ! m!

(ii) Given u0 + u8 = 1.9243, u1 + u7 = 1.9590 u2 + u6 = 1.9823, u3 + u5 = 1.9956. Find u4.

Sol. (i)

R.H.S. =

=

x( x − 1) ...... ( x − m + 1) x( x − 1) ...... ( x − m + 2)

+ m!

(m − 1) ! x( x − 1) ( x − 2) ...... ( x − m + 2)

[(x – m + 1) + m] m!

( x + 1) x( x − 1)( x − 2) ...... ( x − m + 2) ( x + 1) ( m)

=

= L.H.S. m! m!

(ii) Taking ∆8 u0 = 0

(E – 1)8 u0 = 0

⇒ u8 – 8c1u7 + 8c2u6 – 8c3u5 + 8c4u4 – 8c5u3 + 8c6u2 – 8c7u1 + 8c8u0 = 0

(u0 + u8) – 8(u1 + u7) + 28(u2 + u6) – 56(u3 + u5) + 70 u4 = 0

⇒ u4 = 0.99996.

(After putting the values)

Example 21. Evaluate :

=

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NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’

d =

Test statistic, t=

Σd − 7

Σd 2

=

= – 0.7778 ; s2 =

− (d ) 2 = 2.617 n

9 n d s/ n − 1

− 0.7778

=

2.6172 / 8

= – 1.359

The tabulated value of t0.05 at 8 d.f. is 2.31.

Conclusion. Since | t | = 1.359 < t0.05, H0 is accepted i.e., training was not effective in improving performance.

ASSIGNMENT

1.

2.

The mean life of 10 electric motors was found to be 1450 hrs with S.D. of 423 hrs. A second sample of 17 motors chosen from a different batch showed a mean life of 1280 hrs with a S.D. of

398 hrs. Is there a significant difference between means of the two samples ?

The marks obtained by a group of 9 regular course students and another group of 11 part time course students in a test are given below :

Regular

:

56

62

63

54

60

51

67

69

58

Part time

:

62

70

71

62

60

56

75

64

72

68

66

Examine whether the marks obtained by regular students and part time students differ significantly at 5% and 1% level of significance.

3.

A group of 10 boys fed on diet A and another group of 8 boys fed on a different diet B recorded the following increase in weight (kgs) :

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MATRIX INVERSION

Operating R3

FG 11025 IJ

H 64 K

LM

MN

− 15 / 4

1 0 − 3 / 35

9/4

0

− 15 / 4

6/7

45 / 4

0

~ 0 1

0 0

1

945 / 64 − 4725 / 32 11025 / 64

Operating R13

FG 3 IJ , R FG − 6 IJ

H 35 K H 7 K

L1 0

~ M0 1

MN0 0

OP

PQ

23

A–1 =

OP

PQ

− 525 / 32

0 225 / 64

945 / 64

0 − 525 / 32 2205 / 16 − 4725 / 32 ~ [I : A–1]

1 945 / 64 − 4725 / 32 11025 / 64

LM

MN

225

1

− 1050

64

945

OP

PQ

− 1050

8820

− 9450

945

− 9450 .

11025

ASSIGNMENT

1.

Find the inverse of the following matrices using Gauss-Jordan elimination method.

LM

MN

LM

MN

OP

PQ

1.

(i)

LM

MN

OP

PQ

LM

MN

1 3 3

(ii) 1 4 3

1 3 4

1 1

1

(i) 4 3 − 1

3 5

3

OP

PQ

14

2 −4

1

− 15

0

5

10

11 − 2 − 1

OP

PQ

1 1 2

(iii) 1 2 4 .

2 4 7

Answer

LM

MN

OP

PQ

7 −3 −3

1

0

(ii) − 1

−1

0

1

LM

MN

OP

PQ

2 −1

0

2 .

(iii) − 1 − 3

0

2 −1

7.2. TRIANGULARIZATION METHOD

In this method, we write the given matrix as

A = LU where L is a lower triangular matrix and U is an upper triangular matrix.

The inverse of A can be determined as

A–1 = U–1 L–1

If lii = 1, the method is called Doolittle’s method.

If uii = 1, the method is called Crout’s method.

EXAMPLES

Example 1. Given the matrix A =

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NUMERICAL INTEGRATION AND DIFFERENTIATION

Example 3. Evaluate

z

6

dx

0

1 + x2

by using

(i) Simpson’s one-third rule

(ii) Simpson’s three-eighth rule

(iii) Trapezoidal rule

(iv) Weddle’s rule.

Sol. Divide the interval (0, 6) into six parts each of width h = 1.

The values of f(x) =

1

1 + x2

are given below :

x :

0

1

2

3

f(x) :

1

0.5

0.2

0.1

y0

y1

y2

y3

4

5

6

1

17 y4

1

26 y5

1

37 y6

(i) By Simpson’s one-third rule,

z

dx

6

1+ x

0

2

=

h

[(y0 + y6) + 4(y1 + y3 + y5) + 2(y2 + y4)]

3

=

1

3

LMFG 1 + 1 IJ + 4 FG0.5 + 0.1 + 1 IJ + 2 FG0.2 + 1 IJ OP = 1.366173413.

H 17 K Q

26 K

NH 37 K H

(ii) By Simpson’s three-eighth rule,

z

6

0

dx

1+ x

2

=

3h

[(y0 + y6) + 3(y1 + y2 + y4 + y5) + 2y3]

8

=

3

8

LMFG 1 + 1 IJ + 3 FG.5 + .2 + 1 + 1 IJ + 2(.1)OP = 1.357080836.

17 26 K

NH 37 K H

Q

(iii) By Trapezoidal rule,

z

6

0

dx

1+ x

2

=

h

[(y0 + y6) + 2(y1 + y2 + y3 + y4 + y5)]

2

=

1

2

(iv) By Weddle’s rule,

z

6

0

dx

1+ x

2

LMFG 1 + 1 IJ + 2 FG.5 + .2 + .1 + 1 + 1 IJ OP = 1.410798581.

17 26 K Q

NH 37 K H

=

3h

[y + 5y1 + y2 + 6y3 + y4 + 5y5 + y6]

10 0

=

1

3

1

1

+

1 + 5(.5) + .2 + 6(.1) +

+5

= 1.373447475.

17

10

26

37

LM

N

FG IJ

H K

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Solution of a System of Simultaneous

Linear Algebraic Equations

6.1. INTRODUCTION

The systems of simultaneous linear equations arise, both directly in modelling physical situations and indirectly in the numerical solution of other mathematical models.

Problems such as determining the potential in certain electrical networks, stresses in a building frame, flow rates in a hydraulic system etc., are all reduced to solving a set of algebraic equations simultaneously.

Linear algebraic systems are also involved in the optimization theory, least squares fitting of data, numerical solution of boundary value problems for ordinary and partial differential equations, statistical inference etc.

Consider a non-homogeneous system of n simultaneous linear algebraic equations in n unknowns as a11x1 + a12x2 + a13x3 + ... + a1nxn = b1 a21x1 + a22x2 + a23x3 + ... + a2nxn = b2 a31x1 + a32x2 + a33x3 + ... + a3nxn = b3

...(1)

M

M

M

M an1x1 + an2x2 + an3x3 + ... + annxn = bn

Using matrix notation, the above system (1) can be written as

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