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NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’

Table II-A

SIGNIFICANT VALUES OF THE VARIANCE RATIO

F-DISTRIBUTION (RIGHT TAIL AREAS)

5 PER CENT POINTS v1

1

2

3

1

2

3

4

5

161.40

18.51

10.13

7.71

6.61

199.50

19.00

9.55

6.94

5.79

215.70

19.16

9.28

6.59

5.41

6

7

8

9

10

5.99

5.59

5.32

5.12

4.96

5.14

4.74

4.46

4.26

4.10

11

12

13

14

15

4.84

4.75

4.67

4.60

4.54

16

17

18

19

20

4

12

24

5

6

8

224.60

19.25

9.12

6.39

5.19

230.20

19.30

9.01

6.26

5.05

234.00

19.35

8.94

6.16

4.95

238.90

19.37

8.84

6.04

4.82

4.76

4.35

4.07

3.865

3.71

4.53

4.12

3.84

3.63

3.48

4.39

3.97

3.69

3.48

3.33

4.28

3.87

3.58

3.37

3.22

4.15

3.73

3.44

3.23

3.07

4.00

3.57

3.28

3.07

2.91

3.84

3.41

3.12

2.90

2.74

3.67

3.23

2.93

2.71

2.54

3.98

3.88

3.80

3.74

3.68

3.59

4.49

3.41

3.34

3.29

3.365

3.26

3.18

3.11

3.06

3.20

3.11

3.02

2.96

2.90

3.09

3.00

2.92

2.85

2.79

2.95

2.85

2.77

2.70

2.64

2.79

2.69

2.60

2.53

2.48

2.61

2.50

2.42

2.35

2.29

2.40

2.30

2.21

2.13

2.07

4.49

4.45

4.41

4.38

4.35

3.63

3.59

3.55

3.52

3.49

3.24

3.20

3.16

3.13

3.10

3.01

2.96

2.93

2.90

2.87

2.85

2.81

2.77

2.74

2.71

2.74

2.70

2.66

2.63

2.60

2.59

2.55

2.51

2.48

2.45

2.42

2.38

2.34

2.31

2.28

2.24

2.19

2.15

2.11

2.08

2.01

1.96

1.92

1.88

1.84

21

22

23

24

25

4.32

4.30

4.28

4.26

4.24

3.47

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NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS

pe

slo

The eqn. of the tangent at P0(x0, y0) is y – y0

F dy I

= GH JK dx

P0

Q2

)

, y0

(x 0 f e

p

slo

( x – x0 )

= f(x0, y0) (x – x0) y = y0 + (x – x0) f(x0, y0)

)

,y1

(f x 1

g(x

)

Y

y=

In this method, we use the property that in a small interval, a curve is nearly a straight line. Thus at the point

(x0, y0), we approximate the curve by the tangent at the point (x0, y0).

Q1

P0 y0

...(2)

O

x0

y2

y1

x1

x2

X

This gives the y-coordinate of any point on the tangent. Since the curve is approximated by the tangent in the interval (x0, x1), the value of y on the curve corresponding to x = x1 is given by the above value of y in eqn. (2) approximately.

Putting x = x1(= x0 + h) in eqn. (2), we get y1 = y0 + hf(x0, y0)

Thus Q1 is (x1, y1)

Similarly, approximating the curve in the next interval (x1, x2) by a line through

Q1(x1, y1) with slope f(x1, y1), we get y2 = y1 + hf(x1, y1)

In general, it can be shown that, yn+1 = yn + hf(xn, yn)

This is called Euler’s formula.

dy changes rapidly over an dx interval, its value at the beginning of the interval may give a poor approximation as compared to its average value over the interval and thus the value of y calculated from Euler’s method may be in much error from its true value. These errors accumulate in the succeeding intervals and the value of y becomes much erroroneous ultimately.

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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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Curve-Fitting, Cubic

Spline and Approximation

10.1. CURVE FITTING

Let there be two variables x and y which give us a set of n pairs of numerical values

(x1, y1), (x2, y2).......(xn, yn). In order to have an approximate idea about the relationship of these two variables, we plot these n paired points on a graph thus, we get a diagram showing the simultaneous variation in values of both the variables called scatter or dot diagram. From scatter diagram, we get only an approximate non-mathematical relation between two variables.

Curve fitting means an exact relationship between two variables by algebraic equations, infact this relationship is the equation of the curve. Therefore, curve fitting means to form an equation of the curve from the given data. Curve fitting is considered of immense importance both from the point of view of theoretical and practical statistics.

Theoretically, it is useful in the study of correlation and regression. Practically, it enables us to represent the relationship between two variables by simple algebraic expressions e.g., polynomials, exponential or logarithmic functions.

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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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