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ALLC1-1

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1

Introduction

The limitations of analytical methods in practical applications have led mathematicians to evolve numerical methods.

We know that exact methods often fail in drawing plausible inferences from a given set of tabulated data or in finding roots of transcendental equations or in solving non-linear differential equations.

Even if analytical solutions are available, they are not amenable to direct numerical interpretation.

The aim of numerical analysis is therefore to provide constructive methods for obtaining answers to such problems in a numerical form. With the advent of high speed computers and increasing demand for numerical solution to various problems, numerical techniques have become indispensible tools in the hands of engineers and scientists. e.g., we can solve equations x2 – 5x + 6 = 0, ax2 + bx + c = 0, y″ + 3y′ + 2y = 0 by analytical methods, but transcendental equations such as a cos2 x + bex = 0 cannot be solved by analytical methods or is impossible. Such equations are solved by numerical analysis.

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ALLC2-2

Manish Goyal Laxmi Publications PDF

43

ERROR ANALYSIS AND ESTIMATION

Rounding off all other numbers to two decimal digits, we have 0.15, 15.45, 0.00, 8.12,

0.02, 0.64 and 0.17.

The sum S is given by

S = 305.1 + 143.3 + 0.15 + 15.45 + 0.00 + 8.12 + 0.02 + 0.64 + 0.17

= 472.59 = 472.6.

To determine absolute error, we note that first two numbers have each an absolute error of 0.05 and remaining seven numbers have an absolute error of 0.005 each.

∴ Absolute error in all 9 numbers

= 2(0.05) + 7(0.005) = 0.1 + 0.035 = 0.135 ≈ 0.14.

In addition to the above absolute error, we have to take into account rounding error which is 0.01. Hence the total absolute error in S = 0.14 + 0.01 = 0.15

Thus,

S = 472.6 ± 0.15.

Example 17. 5.5 = 2.345 and 6.1 = 2.470 correct to four significant figures. Find the relative error in taking the difference of these numbers.

1

× 0.001 = 0.0005

2

∆x1

∆x2

∆x1

0.0005

+

∴ Relative error <

=2

=2

= 0.008 .

X

X

0.125

X

Example 18. 10 = 3.162 and e ~

– 2.718 correct to three decimal places. Find the percentage error in their difference.

0.0005

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TIT-ALLC

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Contents

Chapter

Pages

1.

1�24

INTRODUCTION

1.1.

1.2.

1.3.

1.4.

1.5.

1.6.

1.7.

1.8.

1.9.

1.10.

1.11.

1.12.

1.13.

1.14.

1.15.

1.16.

1.17.

1.18.

2.

Introduction to Computers ........................................................................................ 4

Definitions ................................................................................................................... 4

Introduction to ‘‘C’’ Language ................................................................................... 6

Advantages/Features of ‘C’ Language ...................................................................... 6

‘C’ Character Set ......................................................................................................... 6

‘C’ Constants ............................................................................................................... 7

‘C’ Variables ................................................................................................................ 8

‘C’ Key Words .............................................................................................................. 9

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ALLC4-3

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192

NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’

Sol. R.H.S. =

=

LM

MN

FG IJ − FG 1 ∆IJ

H K H2 K

1

1

1

1− ∆ +

2

2

2

1

.

2

FG

H

2

+ ...... u0

FG

H

1

1

1

1+ ∆ u0 =

1

2

2

1+ ∆

2

IJ

K

OP

PQ

3

IJ

K

−1

u0 = (2 + ∆)–1 u0 = (1 + E)–1 u0

= (1 – E + E2 – E3 + ...) u0 = u0 – u1 + u2 – u3 + ...... = L.H.S.

Example 4. Using method of separation of symbols, show that :

n(n − 1) ux–2 + ...... + (– 1)n ux–n.

2 n(n − 1) –2

R.H.S. = ux – nE–1 ux +

E ux + ...... + (– 1)n E–n ux

2

∆n ux–n = ux – nux–1 +

Sol.

LM

N

F 1I

= G1 − J

H EK

−1

= 1 − nE + n

ux =

Example 5. Show that :

F

GH

FG E − 1IJ

H E K

ex u0 + x ∆ u0 +

Sol.

F

GH

L.H.S. = ex 1 + x∆ +

F

GH

= 1 + xE +

OP

Q

n(n − 1) −2

E + ...... + (− 1) n E − n ux = (1 – E–1)n ux

2 n

ux =

∆n

En

ux = ∆n E–n ux = ∆n ux–n = L.H.S.

I

JK

x2 2 x2

∆ u0 + ....... = u0 + u1x + u2

+ ...... .

2!

2!

I

JK

x 2 ∆2

+ ...... u0 = ex . ex∆ u0 = ex(1+∆) u0 = exE u0

2!

I

JK

I

JK

F

GH

x2 x2E2 u2 + ....... = R.H.S.

+ ...... u0 = u0 + xu1 +

2!

2!

Example 6. Prove the following identity :

x2 x u1 +

∆u1 + .....

1− x

(1 − x) 2

L.H.S. = xu1 + x2 E u1 + x3 E2u1 + ...... = x (1 + xE + x2E2 + ......) u1 u1x + u2x2 + u3x3 + ...... =

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ALLC4-2

Manish Goyal Laxmi Publications PDF

181

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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