A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx

Find the value of an annuity at 5 3 8%, given the following table: Rate Annuity Value... In an examination, the number of candidates who obtained marks between certain limits are as foll

Sol The difference table for the given date is as follows:

3.6

21.1

To obtain first term, we use Newton’s forward interpolation formula,

Here, a = 3, h = 1, x = 1 ∴ u = –2

Hence we have

On putting the subsequent values, we get

f(1) = ( ) ( )2 ( 3) ( 2)( 3)( 4)

Similarly, to obtain tenth term, we use Newton’s backward interpolation formula So

a + nh = 9, h = 1, a + nh + uh = 10

= 73.9 + 21.1 + 4.5 + 0.5 = 100

Example 4 Find the value of an annuity at 5 3

8%, given the following table:

Rate

Annuity Value

Sol Difference table,

9.4014 1

2

1

2

7.6892

−

−

−

−

x = 538= 438 ,a=6,n=12

y = ( ) ( 1.25)( 0.25) 2 ( 1.25)( 0.25)(0.75) 3

= 137.6483 ( 1.25)( 7.6892) ( 1.25)( 0.25)(0.6978)

2!

( 1.25)( 0.25)(0.75) ( 1.25)( 0.25)(0.75)(1.75)

= 147.2251 Approx

Example 5 In an examination, the number of candidates who obtained marks between certain limits are as follows:

Marks

No of candidates

Find no of candidates who obtained fewer than 70 marks.

Sol First, we form the difference table

Marks less than x No of candidates y ∇ ∇ ∇ ∇

−

−

−

−

62

17

Here, we have h = 20, a = 99

∴ u = 70 99 1.45

20

− = − Now on applying ‘Newton’s backward difference formula, we get

f(70) = ( ) ( ) ( 1) 2 ( ) ( 1)( 2) 3 ( ) ( 1)( 2)( 3) 4 ( )

f a + ∇u f a + + ∇ f a + + + ∇ f a + + + + ∇ f a

= 235 + (–1.45)(17) + ( 1.45)( 0.45) ( 33) ( 1.45)( 0.45)(0.55) ( 18)

= 235 – 24.65 – 10.76625 – 1.076625

= 235 – 36.492875

= 198.507 { 198

∴ Total no of candidates who obtained fewer than 70 marks are 198

Example 6 The area A of a circle of diameter d is given for the following values:

d A

Find A for 105.

Sol First of all we form the difference table as follow:

−

648

766

100 7854

Here, h = 5, a = 100, x = 105

∴ u = 105 100 1

5

− = Now on applying Newton’s backward difference formula, we have

(105)

f = f a( )+ ∇u f a( )+u u(2!+1)∇2f a( )+u u( +1)(3!u+2)∇3f a( )+u u( +1)(u4!+2)(u+3)∇4f a( )

= 7854 1 766 40 2 1 2 3 2 1 2 3 4 4

= 7854 + 766 + 46

= 8666

Which is the required area for the given diameter of circle

Example 7 The probability integral 1 t 2

2 0

2

P= π∫πe− dt has the following values:

x

P

0.682689 0.706282 0.728668 0.749856 0.769861 0.788700

Calculate P for x = 1.235

Sol First we form the difference table

−

−

−

−

−

( ) 1.00 0.682689

0.023593

0.018839 1.25 0.788700

Here, h = 0.05 a = 1.20

∴ u = 1.235 1.20 0.3

0.05

( )

f x = ( ) ( ) ( 1) 2 ( ) ( 1)( 2) 3 ( )

f a + ∇u f a + + ∇ f a + + + ∇ f a +

= 0.788700 ( 0.3)(0.018839) ( 0.3)(0.7)( 0.001166) ( 0.3)(0.7)(1.7)(0.000017)

( 0.3 (0.7)(1.7)(2.7)) ( 0.3)(0.7)(1.7)(2.7)(3.7)

= 0.788700 – 0.0056517 + 0.0012243 – 0.0000010115 – 0.00000008032

= 0.7888225488 – 0.00566189532

= 0.78316065356

Example 8: Calculate the value of tan 48°15′ from the following table:

1.00000 1.03553 1.07237 1.11061 1.15037 1.19175

x

tan x

Sol Given that a + nh = 50

h = 1

a + nh + uh = 48°15′ = 48.25°

The difference table for given data is as follows:

°

°

°

−

°

°

3553

4138

a nh a nh a nh a nh

y+ + ∇u y + + + ∇ y+ + + + ∇ y + +

5

48.25

( 1.75)( 0.75)(0.25)(1.25) ( 1.75)( 0.75)(0.25)(1.25)(2.25)

5

48.25

10 y = 112040.2867

⇒ y48.25 = tan 48°15′ = 1.120402867

PROBLEM SET 4.2

1 The population of a town is as follows:

Year Population in lakhs

Estimate the increase in population during the period 1955 to 1961

[Ans 621036.8 lakhs.]

2 From the following table find the value of tan 17°

tan 0 0.0699 0.1405 0.2126 0.2867 0.3640 0.4402

θ

θ

[Ans 0.3057]

3 From the given table find the value of log 5875

log 1.60206 1.65321 1.69897 1.74036 1.77815 1.81291

x

x

[Ans 3.7690058]

4 From the following table, find y when x = 1.84 and 2.4

5.474 6.050 6.686 7.389 8.166 9.025 9.974

x

x

e

[Ans 6.36, 11.02]

5 From the following table of half yearly premium for policies maturing at different ages, estimate the premium for policy maturing at the age of 63:

( .) 114.84 96.16 83.32 74.48 68.48

Age Premium in Rs

[Ans 70.585152]

6 The values of annuities are given for the following ages Find the value of annuity at the age of 271

2.

Age

Annuity

16.195 15.919 15.630 15.326 15.006

[Ans 15.47996]

7 Show that Newton’s Gregory interpolation formula can be written in the form as

x

u =u + ∆ − ∆x u xa u +xab∆ u −xabc∆ u +

where a = 1 1( 1), = 1 1 ( +1), = 1 1( 1)

4.4 CENTRAL DIFFERENCE FORMULAE

As earlier we study formulae for leading terms and differences These formulae are fundamental and are applicable to nearly all cases of interpolation, but they do not converge as rapidly as central difference formulae The main advantage of central difference formulae is that they give more accurate result than other method of interpolation Their disadvantages lies in complicated calculations and tedious expression, which are rather difficult to remember These formulae are used for interpolation near the middle of a argument values In this category we use the following formulae:

4.4.1 Gauss Forward Difference Formula

We know Newton’s Gregory forward difference formula is given by

f a hu+ = ( ) ( ) ( 1) 2 ( ) ( 1)( 2) 3 ( )

f a + ∆u f a + − ∆ f a + − − ∆ f a

+ ( 1)( 2)( 3) 4 ( )

4!

f a

Substitute a = 0, h = 1 in (1), we get

f u( ) = (0) (0) ( 1) 2 (0) ( 1)( 2) 3 (0) ( 1)( 2)( 3) 4 (0)

f + ∆u f + − ∆ f + − − ∆ f + − − − ∆ f +

.(2) Now obtain the values of ∆2f(0),∆3f(0),∆4f(0)

To get these values,

3f( 1)

∆ − = ∆2f(0)− ∆2f( 1)−

⇒ ∆2f(0) = ∆3f( 1)− + ∆2f( 1)−

Also, ∆4f( 1)− = ∆3f(0)− ∆3f( 1)−

⇒ ∆3f(0) = ∆4f( 1)− + ∆3f( 1)−

5f( 1)

∆ − = ∆4f(0)− ∆4f( 1)−

⇒ ∆4f(0) = ∆5f( 1)− + ∆4f( 1)−

6f( 1)

∆ − = ∆5f(0)− ∆5f( 1)−

⇒ ∆5f(0) = ∆6f( 1)− + ∆5f( 1)− and so on

Substituting these values in equation (2)

( )

f u = (0) (0) ( 1) 3 ( 1) 2 ( 1) ( 1)( 2)[ 4 ( 1) 3 ( 1)]

f + ∆u f + − ∆ f − + ∆ f − + − − ∆ f − + ∆ f −

( )

f + ∆u f + − ∆ f − + − + − ∆ f −

6 ( 1)( 2)( 3)( 4)

( 1) 120

f

( )

f u = f(0)+ ∆u f(0)+u u(2!−1)∆2f( 1)− +(u+1) (3!u u−1)∆3f( 1)− +(u+1) (u u4!−1)(u−2)∆4f( 1)−

5 ( 1) ( 1)( 2)( 3)

( 1)

5!

f

But ∆5f (–2) =∆4f (–1) – ∆4f (–2)

⇒ ∆4f (–1) =∆4f (–2) + ∆5f (–2)

and ∆6f (–2) =∆5f (–1) + ∆5f (–2)

⇒ ∆5f (–1) =∆5f (–2) + ∆6f (–2)

The equation (3) becomes

f u( ) = (0) (0) ( 1) 2 ( 1) ( 1) ( 1) 3 ( 1) ( 1) ( 1)( 2) 4 ( 2)

f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)

+ ( 1) ( 1)( 2)( 3) 6 ( 2)

5!

f

+ − − − ∆ − ( )

f u

∴ = (0) (0) ( 1) 2 ( 1) ( 1) ( 1) 3 ( 1) ( 1) ( 1)( 2) 4 ( 2)

f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

( 2)( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)

This formula is known as Gauss forward difference formula

This formula is applicable when u lies between 0 and1

2

4.4.2 Gauss Backward Difference Formula

This formula is also solved by using Newton’s forward difference formula

Now, we know Newton’s formula for forward interpolation is

f a hu+ = f a + ∆u f a + − ∆ f a + − − ∆ f a + − − − ∆ f a +

(1)

Put a = 0, and h = 1, in equation (1), we get

f u = f + ∆u f + − ∆ f + − − ∆ f + − − − ∆ f +

(2)

(0) ( 1) ( 1) (0) ( 1) ( 1) (0) ( 1) ( 1) (0) ( 1) ( 1) and so on

∆ = ∆ − + ∆ −

On substituting these values in (2), we get

f u = f +u∆ − + ∆f f − + − ∆ f − + ∆ f − + − − ∆ f − + ∆ f − 

( 1)( 2)( 3)

( 1) ( 1) 4!

∴ 2 ( 1) ( 1) 3 ( 2)

f u = f + ∆ − + ∆u f u f −  + − + − ∆ f −  + − 

+ ( 1)( 2) 4 ( 1) 1 ( 3)

f

= (0) ( 1) ( 1) 2 ( 1) ( 1) ( 1) 3 ( 1)

f + ∆ − +u f + u∆ f − + + − ∆ f −

+( 1) ( 1)( 2) 4 ( 1)

4!

u u u u

f

Again ∆3f( 1)− = ∆3f( 2)− + ∆4f( 2)−

4f( 1) 4f( 2) 5f( 2)

5f( 1) 5f( 2) 6f( 2)

∆ − = ∆ − + ∆ − and so on

Therefore, equation (3) becomes

f u = f + ∆ − +u f + ∆ f − + + − ∆ f − + ∆ f − 

+( 1) ( 1)( 2) 4 ( 2) 5 ( 2) ( 1) ( 1)( 2)( 3) 5 ( 2) 6 ( 2)

f u = f + ∆ − +u f + ∆ f − + + − ∆ f − + + −  + − ∆ f −

5 ( 1) ( 1)( 2) ( 3)

f

f u = f + ∆ − +u f + ∆ f − + + − ∆ f − + + + −

4 ( 2)( 1) ( 1)( 2) 5

5!

This is known as Gauss Backward difference formula and useful when u lies between

1

and 0

2

4.4.3 Stirling’s Formula

This is another central difference formula and useful when | | 1 1 1

u < or− < <u It gives best estimation when 1 1

4 u4

− < This formula is obtained by taken mean of Gauss forward and Gauss backward difference formula

Gauss forward formula for interpolating central difference is,

f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

( 2)( 1) ( 1)( 2) 5 ( 2) ( 1) ( 1)( 2)( 3) 6 ( 2)

Gauss Backward difference is,

4 ( 2) ( 2)( 1) ( 1)( 2) 5 ( 2)

5!

Take mean of Equation (1) and (2)

f u f ∆ + ∆ − f  − + + 

4 ( 1) ( 2)

( 2)

f

∆ − + ∆ − 

5 ( 2)( 1) ( 1)( 2)

( 2) 5!

f

2 2

2

This is called Stirling’s formula

4.4.4 Bessel’s Interpolation Formula

This is one of the another type of central difference formula and obtained by (1) shifting the origin

by 1 in Gauss backward difference and then (2) replacing u by (u – 1), (3) take mean of this

equation with Gauss forward formula

Gauss backward difference formula is,

( )

f u = f + ∆ − +u f + ∆ f − + + − ∆ f − + + + − ∆ f −

5 ( 2)( 1) ( 1)( 2)

( 2) 5!

f

Now shift the origin by one, we get

( )

f u = f + ∆u f + + ∆ f + + − ∆ f − + + + − ∆ f −

5 ( 2)( 1) ( 1)( 2)

( 1) 5!

f

On replacing u by (u – 1)

( )

f u = f + − ∆u f + − ∆ f + − − ∆ f − + + − − ∆ f −

5 ( 1)(( 1) ( 2)( 3)

( 1) 5!

f

Gauss forward difference formula is,

f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

( 2)( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)