A textbook of Computer Based Numerical and Statiscal Techniques part 23 pot

Apply Stirling’s formula to find a polynomial of degree four which takes the values of yas given below: x [Ans.. Using Bessel’s formula find the value of y at x = 3.75 for the data given

7 Apply Stirling’s formula to find a polynomial of degree four which takes the values of y

as given below:

x

[Ans 2 4 8 2

1

3u −3u + ]

8 Apply Stirling’s formula to interpolate the value of y at x = 1.91 from the following data: x

y

[Ans 6.7531]

4.5 BESSEL’S

Example 1 Using Bessel’s formula find the value of y at x = 3.75 for the data given below:

Sol Difference table for the given data is as:

2.102

1.215

−

−

−

−

Here h = 0.5

∴ u = x a h− =3.75 3.50.5− =0.5

Now from Bessel’s formula, we have

1 ( 1)

−  − 

3 f (–1)

5

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

( 2)

f

18.644 20.225 (0.5 0.5)( 1.581) 1.99 2.37

16 9

0 (0.5 1)(0.5)( 0.5)(2.5) 0

2

+

= 19.407 (Approx.)

Example 2 Following table gives the values of e x for certain equidistant values of x Find the value

of e x at x = 0.644 using Bessel’s formula:

x

x

e

1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237

Sol Given h = 0.01, take it origin as 0.64

0.644 0.64 0.004

0.4 0.01 0.01

x a

u

h

u = 0.4

Difference table for the given data is as:

−

−

−

−

( )

0.61 1.840431

0.018497

0.019445 0.67 1.954237

Bessel’s formula

1 ( 1)

u u

 

− − 

3 f (–1)

5

1 ( 1)( 2) ( 1) ( 2) ( 1) ( 1)( 2)( 1/2)

( 2)

f

1.896481 1.915541 ( 0.1)(0.01906) 0.4( 0.6) 0.000191 0.000189

−

(0.4)( 0.6)( 0.1)(0.000002) (1.4)(0.4)( 0.6)( 1.6) 0.000001 0.000002

(1.4)(0.4)( 0.6)( 0.1)( 1.6)

( 0.000001) 120

= 1.906011–0.001906 – 0.0000228 + 0.000000008 + 0.000000033 + 1.68 × 10–10

= 1.904082

Example 3 Find the value of y 25 from the following data using Bessel’s formula Data being y 20 =

2854, y 24 = 3162, y 28 = 3544, y 32 = 3992.

Sol The difference table for the data is as:

308

448

−

25 24 0.25

4

(0.25) ( 0.25) 382 (0.25)( 0.75)

= 3353 – 95.5 – 6.5625 – 0.0625

= 3250.875

Example 4 The pressure p of wind corresponding to velocity v is given by following data Estimate pressure when v = 25.

v P

Sol The difference table for the given data is as:

−

0.9

3.4

Let origin = 20, h = 10,

25 20 0.5

10

Bessel’s formula for interpolation is:

2

1 ( 1)

u u

 −  −

1

6.4 0 0.16250 0 2

= 3.2 – 0.16250

P25 = 3.03750

Example 5 Probability distribution function values of a normal distribution are given as follows:

x

p x

( ) 0.39104 0.33322 0.24197 0.14973 0.07895

Find the value of p(x) for x = 1.2.

Sol Taking the origin at 1.0 and h = 0.4

x = a + uh ⇒ 1.2 = 1.0 + u × 0.4

1.2 1.0 1

The difference table is:

−

−

−

−

−

5782

7078

Bessel‘s formula is

3

1 ( 1)

−  − 

105f (0.5) =

1 1 1

24197 14973 2 2 2146 99

  − 

  

= 19457.0625

∴ f (0.5) = 0.194570625

Hence, p(1.2) = 0.194570625.

Example 6 Given y 0 , y 1 , y 2 , y 3 , y 4 , y 5 (fifth difference constant), prove that

1 2

1 25(c b) 3(a c)

where a = y 0 + y 5 , b = y 1 + y 4 , c = y 2 + y 3

Sol Put u= 12 in bessel’s formula, we get

y = y +y − ∆ y + ∆ y− + ∆ y− + ∆ y−

Shifting origin to 2, we have

2

y = y +y − ∆ y + ∆ y + ∆ y + ∆ y

c

1 2

c

c

1 2

25( ) 3( )

2 256

c

Example 7 If third differences are constant, prove that

x 2

+ = + − ∆ + ∆ · Sol Put u=12 in Bessel’s formula, we get

2

y = y +y − ∆ y + ∆ y− Shifting the origin to x.

2

x

+ = + − ∆ + ∆

Example 8 Given that:

x

f x

Apply Bessel’s formula to find the value of f(9).

Sol Taking the origin at 8, h = 2,

9 = 8 + 2u or u = 1

2 The difference table is:

15293

10353

−

−

−

−

Bessel’s formula is

1

1 ( 1)

u

u u

−  − 

−

1

 

4

1/2

y

1/2

10 y =71078.27344

∴ y1/2 =7.107827344

Hence, f(9) = 7.107827344

Example 9 Find a polynomial for the given data using Bessel’s formula f(2) = 7, f(3) = 9, f(4) =

12, f(5) = 16.

Sol Let us take origin as 3 therefore,

u = x – 3

2

4

Bessel’s formula:

u

P P

u u u

  + − × +

=

2

3

u

=

9

2 2

u

u

Put u = x – 3

y =

2

( 3) 5

( 3) 9

x

x

2 x − x+ +2 x− +

y = 1 2 1 6

2x −2x+ Ans

PROBLEM SET 4.6

1 Find y (0.543) from the following values of x and y:

x

y x

[Ans 6.303]

2 Apply Bessel’s formula to find the value of y2.73 from the data given below:

2.5 0.4938, 2.6 0.4953, 2.7 0.4965, 2.8 0.4974

2.9 0.4981, 3.0 0.4987

[Ans 0.496798]

3 Find y25 by using Bessel’s interpolation formula from the data:

20 24, 24 32, 28 35, 32 40

[Ans 32.9453125]

4 Apply Bessel’s formula to evaluate y62.5 from the data:

x

x

y

[Ans 7957.1407]

5 Apply Bessel’s formula to obtain the value of f (27.4) from the data:

x

f x

[Ans 3.649678336]

6 Apply Bessel’s formula to find the value of f(12.2) from the following data:

( ) 0 0.19146 0.34634 0.43319 0.47725 0.49379 0.49865

x

f x

[Ans 0.39199981]

7 Apply Bessel’s interpolation formula, show that tan160° = 0.2867, Given that:

tan 0 0.0875 0.1763 0.2679 0.3640 0.4663 0.5774

x

x

8 Apply Bessel’s formula to find a polynomial of degree 3 from the data:

x y

x

u + u + u+ u= −

]

9 From the following table find the value of f (0.5437) by Gauss and Bessel’s formula:

( ) 0.529244 0.537895 0.546464 0.554939 0.663323 0.571616 0.579816

x

f x

[Ans 0.558052]

10 Apply Bessel’s formula to obtain a polynomial of degree three:

x

f x

[Ans

3

x

4.6 LAPLACE EVERETTS

Example 1 Find the value of f(27.4) from the following table:

u

f x

Sol Here, u = 27.4 27.0 0.4,

1

origin is at 27.0 h = 1

Also, w = 1 – u = 0.6

Difference table is:

−

−

−

−

−

−

4

154

115

By Laplace Everett’s formula,

f (0.4) = (1.4)(0.4)( 0.6) (2.4)(1.4 (0.4)( 0.6)( 1.6))

( ) { (1.6)(0.6)( 0.4) (2.6)(1.6)(0.6)( 0.4)( 1.4) }

= 3649.678336

Hence f (27.4) = 3649.678336.

Example 2 Using Laplace Everett’s formula, find f(30), if f(20) = 2854, f(28) = 3162, f(36) = 7088, f(44) = 7984.

Sol Take origin at 28, h = 8

⇒ 28 + 8u = 30 ⇒ u = 0.25

Also, w = 1 – u = 1 – 25 = 0.75