A textbook of Computer Based Numerical and Statiscal Techniques part 19 pptx

Following are the marks obtained by 492 candidates in a certain examination: Marks No of Candidates Find out a No.. Use Newton’s forward difference formula to obtain the interpolating po

∆ ∆2 ∆3 ∆4 ( )

1

8

From Newton’s formula,

P(n) (0) (0) ( 1) 2 (0) ( 1)( 2) 3 (0) ( 1)( 2)( 3) 4 (0)

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

1

=

4 3

2

1

− + + + Ans

Example 6 Find the value of sin 52 ° from the given table:

θ

Sol Here a = 45°, h = 5, x = 52 therefore u = 1.4

°

−

°

589

468

By forward interpolation formula

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

2!

u u

f a hu+ = f a + ∆u f a + − ∆ f a +

⇒ 104 f x( ) = (1.4)(0.4) ( )1.4 (0.4)( 0.6)

−

= 7880

∴ f(52) = 0.7880 or sin 52° = 0.7880

Example 7 Following are the marks obtained by 492 candidates in a certain examination:

Marks

No of Candidates

Find out

(a) No of candidates, if they secure more than 48 but less than 50 marks.

(b) Less than 48 but not less than 45 marks.

Sol First we form the difference table as follows:

−

−

−

43

79

Marks less than No of Candidates

Here, h = 5,

For (a), x = 48, a = 40, 48 40 1.6

5

Now on applying Newton’s forward difference formula, we have

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

1.6 (0.6)( 0.4)( 1.4)( 2.4) 222

120

= 210 + 68.8 + 5.28 – 1.5904 – 0.576 – 2.38694

= 284.08 – 4.553344 = 279.52 ≅ 280

∴ No of students secured more than 48 marks but less than 50 marks = 307 – 280 = 27

for (b), x = 48, a = 40, ∴ u = 1.6

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

1.6 (0.6)( 0.4)( 1.4)( 2.4) 222

120

× − − −

279.52 280

∴ No of students secured more than 45 marks but less than 48 marks = 280 –253 = 27

Example 8 Use Newton’s forward difference formula to obtain the interpolating polynomial f(x) satisfying the following data:

x

f x

If another point x = 5, f(x) = 26 is added to the above data, will the interpolating polynomial be the same as before or different Explain why?

Sol The difference table is

−

−

−

−

8

25

As we seen here that its third difference is being constant either on taking fifth entry or on

not taking fifth entry (i.e., 26) So the interpolating polynomial should be same as before, and no

change will occur

Here, h = 1, x = 5, a = 1

h

− =

f(x) = 26 + (x – 1)(x – 8) ( 1)( 2) ( 1)( 2)( 3)

= 26 – 8x + x – 3x2 + 9x – 6 – ( 3 2 )17

6

x − x + x−

= 28 3 2 17 3 17 2 187 17

= 176 x3−20x2+1936 x+11

which is the required interpolating polynomial

Example 9 If p , q, r, s be the successive entries corresponding to equidistant arguments in a table Show that when third differences are taken into account, the entry corresponding to the argument half way between the arguments at q and r is A+ B ,

24 where A is the arithmetic mean of q and r and B is arithmetic mean of 3q – 2p – s and 3r – 2s – p.

Sol Given A is the arithmetic mean of q and r

2

q r+

Also, B is the arithmetic means of 3q – 2p – s and 3r – 2s – p.

q− p s− + − −r s p q− p− +s r

=

q r+ − p s+

Let quantities p, q, r and s corresponds to argument a, a + h, a + 2h and a + 3h respectively then the value of the argument lying half way between a + h and a + 2h will be

a + h +

2

h

i.e., a + 3

2

h

Hence, a + mh = a + 3

2h

2 Now, difference table as follows:

−

− +

2

3

q p

s r

Using Newton’s forward interpolation formula up to third difference only and taking

m = 3/2, we get

3 2

f a + h

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

p+ q p− + r− + −q p s− + −r q p

= 1 ( 9 9 ) 9( )

p s

p q r s q r  + 

A −  − = A− A+

= 24

B

A+

Example 10 The table below shows value of tan x for 0.10 ≤ x ≤ 0.30

x x

Evaluate tan 0.12 using Newton’s forward difference table.

Sol Let us, first form the difference table

tan 0.10 0.1003

0.0508

0.0540 0.30 0.3093

a = 0.10, x = 0.12

∴ u = 0.12 0.10 0.02 0.4

0.05 0.05

x a h

− = − = =

∴ By using Newton’s forward difference formula, we get

f(u) = tan (0.12) = f(a) + ( ) ( 1) 2 ( ) ( 1)( 2) 3 ( ) ( 1)( 2)( 3) 4 ( )

= 0.1003 0.4 0.0508 0.4( 0.6) (0.0008) 0.4( 0.6)( 1.6) (0.0002)

+0.4( 0.6)( 1.6)( 2.6) (0.0002)

24

= 0.1003 + 0.02032 – 0.000096 + 0.0000128 – 0.00000832

= 0.1206328 – 0.00010432

= 0.12052848 Approx

PROBLEM SET 4.1

1 The population of a town in the decimal census was as given below Estimate the population for the year 1895

93

Year x Population y in thousands

[Ans 54.8528 thousands]

2 If l x represents the number of persons living at age x in a life table, find as accurately as the data will permit l x for values of x = 35, 42 and 47 Given l20 = 512, l30 = 390, l40 = 360,

l50 = 243

[Ans 394, 326, 274]

3 From the following table, find the value of e0.24

x

x

e

[Ans e0.24=1.271249]

4 From the table, Estimate the number of students who obtained marks between 40 and 45

35

Marks

No of students

[Ans 17]

5 Find the cubic polynomial which takes the following values

10

x

f x

[Ans 2x3-7x2+6x+1]

6 Find the number of men getting wages between Rs 10 and Rs.15 from following table:

42

Wages in Rs Frequency

[Ans 15]

7 The following are the numbers of deaths in four successive ten year age groups Find the number of deaths at 45-50 and 50-55

31496

Age Deaths

[Ans 11278, 12947]

8 The following table give the marks secured by 100 students in Mathematics:

11

Range of marks

No of students

Use Newton’s forward difference interpolation formula to find

(1) the number of students who got more than 55 marks

(2) the number of students who secured marks in the range from 36 to 45

[Ans (i) 28, (ii) 36]

9 From the following table of half-yearly premium for policies maturing at different ages, estimate the premium for policies maturing at age of 46

74.48

Age Premium in Rupees

[Ans Rs 110.52]

4.3 NEWTON’S GREGORY FORMULA FOR BACKWARD INTERPOLATION

Let y = f(x) be a function of x which takes the values f(a), f(a + h), f(a + 2h), f(a+ nh) for (n + 1) equally spaced values a, a + h, , a + nh of the independent variable x Let us assume f(x) be a nth degree polynomial given by

f x =A +A x a nh− − +A x a nh x a n− − − − − h + A x a hn x a n− − − − − h

(x – a – h) (1)

Where A0, A1, A2, A n are to be determined

Put x = a + nh, a + n−1 , h a in (1) respectively.

Put x = a+ (n – 1)h, then

f a n+ − h =A −hA = f a nh+ −hA (from 2)

⇒ 1

f a nh A

h

∇ +

Put x = a + (n – 2)h, then

f a n+ − h =A − hA + − h −h A

⇒ 2!h A2 2 = f a n( + −2 )h − f a nh( + ) 2+ ∇f a nh( + )= ∇2f a nh( + )

2

2!

f a nh A

h

Proceeding in similar way, ( )

!

n

f a nh A

n h

Substituting the vaues in (1), we get

( ) ( ) ( ) f a nh ( )( 1 )

h



( )

!

n n

f a nh

x a h

n h



 .(6)

Put x = a +nh + uh, then x – a – nh = uh

and x – a – (n – 1)h = (u +1)h

x – a – h = (u n+ −1)h

∴ Equation (6) becomes

+ +

= + + ∇ + + ( 1)∇2 + + + + + − ∇ ( )

n n

f a nh

u u

n h

n

n

Which in required Newton’s Gregory formular for backward interpolation This formula is used when we want to interpolate the value near the end of the table

Example 1 The table below gives the value of tan x for 0.10 ≤ x ≤ 0.30

x x

Find (a) tan 0.50 (b) tan 0.26 (c) tan 0.40

Sol First of all we construct the difference table:

0.10 0.1003

0.0508

0.0540 0.30 0.3093

(a) Here, h = 0.05, a = 0.30, x = 0.50

0.05

tan (0.50) = ( ) ( ) ( 1) 2 ( ) ( 1)( 2) 3 ( ) ( 1)( 2)( 3) 4 ( )

= 0.3093 4 0.0540 4 5 0.0014 4 5 6 0.0004 4 5 6 7 0.0002

= 0.3093 + 0.216 + 0.014 + 0.008 + 0.007

= 0.5543

(b) Here, h = 0.05, a = 0.30, x = 0.26

∴ u = 0.26 0.30 0.8

0.05

( 0.8) (0.2) (1.2) (2.2)

0.0002 24

= 0.3093 – 0.0432 – 0.000112 – 0.0000128 – 0.00000352

= 0.3093 – 0.04332882≅0.2662

(c) Here, h = 0.05, a = 0.30, x = 0.40

0.05

− =

(0.40) tan(0.40) ( ) ( ) ( 1) 2 ( ) ( 1)( 2) 3 ( )

4 ( 1)( 2)( 3)

( ) 4!

f a

On putting the subsequent values, we get

f(0.40) = 0.3093 + 2 × 0.054 + 2 3 0.0014 2 3 4 0.0004 2 3 4 5 0.0002

× × + × × × + × × × =

= 0.3093 + 0.108 + 0.0042 + 0.0016 + 0.0001

= 0.4241

Example 2 Using Newton’s backward difference formula find the value of e –1.9 from the following table of value of e –x

x

x

e−

Sol Difference table for the given data as follows:

0.0814

0.0385 2.00 0.1353

x

−

−

Here, u = 1.9 2 0.4

0.25− = −

using Newton’s backward difference formula

( x)

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

On putting the subsequent values, we get

( x)

f e− = 0.1353 + (–0.4) × (–0.0385) + (−0.4)× −( 0.441)×

0.0108 2

+( 0.4) ( 0.4 1)( 0.4 2)×(0.0033)

6

( 0.4) ( 0.4 1)( 0.4 2)( 0.4 3)

0.0006 24

= 0.1353+0.0154 – 0.001296 + 0.0002112 + 0.000024

= 0.14959

Example 3 In the following table, values of y are consecutive terms of a series of which 23.6 is the 6th term Find the first and tenth terms of the series.

x y