A textbook of Computer Based Numerical and Statiscal Techniques part 28 potx

Evaluate f9 using Newton’s divided difference formula... Express the function as a sums of partial functions.. The difference is that in the Lagrange’s interpolation the interpolating po

Example 9 Using Newton’s divided difference formula, prove that

f(x) = f(0) + x∆f (–1) +(x 1 x)

2!

+

( )

2 f −1

∆ +(x 1 x x) ( 1)

3!

f

∆ (–2) +

Sol Taking the arguments, 0, –1, 1, –2, the Newton’s divided difference formula is

f(x) = f(0) + x−∆1 f(0) + x(x + 1) 2

1,1

−∆ f(0) + x(x + 1) (x – 1) 3

1,1 2

− −∆ f(0) + (1)

f(x) = f(0) + x∆0 f(–1) + x(x + 1) 2

0,1

∆ f(–1) + x(x + 1) (x – 1) 3

0, 1,1∆− f(–2) +

Now, ∆0 f(–1) = ( ) ( )

( )

f − −f

− − =∆f(–1)

2

0,1∆ f(–1) =

( )

1

∆ − ∆ − 

= 1

2

3

1,0,1

−∆ f(–2) =

( )

1

−

∆ − − ∆ − 

= 1

3

∆ − −∆ − 

= 3 ( )2

3 2

f

× = 3 ( )2

3!

f

and so on

Substituting these values in (1)

f(x) = f(0) + x∆f (–1) +( 1)

2!

x+ x 2

f

∆ (–1) +( 1) ( 1)

3!

x+ x x− 3

f

∆ (–2)+

Example 10 Using Newton’s divided difference formula, calculate the value of f(6) from the following data:

( )

x : 1 2 7 8

f x : 1 5 5 4

Sol The divided difference table is

4

2

3

1 0

14 1

6 1

−

−

−

Applying Newton’s divided difference formula,

f(x) = 1 + (x – 1) (4) + (x – 1) (x – 2)−23

  + (x – 1) (x – 2) (x – 7)

1 14

 

 

 

f(6) = 1 + 20 + (5) (4) 2

3

− 

 

  + (5) (4) (–1)

1 14

 

 

 

= 6.2381

Example 11 Find the value of log 10 656 using Newton’s divided difference formula from the data given below:

10

log x : 2.8156 2.8182 2.8189 2.8202

Sol Divided difference table for the given data is as:

260 65 4

70 _ 65

5

65 70

3 130

65 2

=

=

− = −

=

For the given argument, the divided difference formula is,

f(x) = y0 + (x – x0)∆y0 + (x – x0) (x – x1)∆2y0 + (x – x0) (x – x1) (x – x2)∆3y0

105 f(x) = 281560 + (x – 654) 65 + (x – 654) (x – 658).1 + (x – 654) (x – 658) (x – 659) (0.38)

On Substituting x = 656, we get

105 f(x) = 281560 + 2×65 + (–4) + (–4) (–3)×(0.38)

= 281560 + 130 – 4 + 4.56

105 f(x) = 281690 56

⇒ f(x) = 2.8169056

⇒ log10 656 = 2.8169056

PROBLEM SET 5.3

1 By means of Newton’s divided difference formula,

Find the value of f(8) and f(15) from the following table:

( )

: 48 100 294 900 1210 2028

x

2 Using Newton’s divided difference formula, find a polynomial function satisfying the following data:

( ) :: 12454 31 0 25 9 13355

x

f x

Hence find f(3) [Ans f (x) = 3x4 – 5x3+ 6x2 – 14x + 5, f (3) = 89]

3 From the given data, find f(x) as a polynomial in powers of (x – 5)

( )

: 4 26 58 112 466 922

x

f x

[Ans (x – 5)3 + 17 (x – 5)2 + 98 (x – 5) + 194]

4 Find the value of log10 33 from the data given below:

log10 2 = 0.3010, log10 3 = 0.4771, log10 7 = 0.8451, [Ans log10 33 = 1.5184]

5 Find f’(10) from the given data:

( )

: 13 23 899 17315 35606

x

6 Find approximately the real root of the equation x3 – 2x – 5 = 0 [Ans 2.0945595]

7 Find the value of f(x) at point x = 2 and 5 from the data:

( )

: 1.5 3 6 : 0.25 2 20

x

8 Evaluate f(9) using Newton’s divided difference formula.

( ) :: 150 392 1452 2366 52025 7 11 13 17

x

9 Use Newton’s divided difference formula to find f(7) if f(3) = 24, f(5) = 120, f(8) = 504

10 There is a data be given

( )

: 2 3 12 147

x

f x

What is the form of the function? [Ans x 3 + x 2 – x + 2]

11 Find yx in powers of x – 4 where y0 = 8, y1 = 11, y4 = 68, y5 = 125

12 Express the function as a sums of partial functions

( ) ( )( )

2 2

( 1 ) (3 ) (13 ) (71 )

13 If f(x) = u (x) v(x), find the divided difference f(x0, x1) in terms of u (x0), v (x1) and the divided

differences u(x0, x1), v (x0, x1)

14 Find f(3), using Newton’s divided difference formula from the given data

( )

: 0 1 2 4 5 6 : 1 14 15 5 6 19

x

Hermite’s interpolation is similar to as Lagrange’s interpolation The difference is that in the

Lagrange’s interpolation the interpolating polynomial consider with f(x) at the interpolation points

x0, x1, x2, x n where as Hermite’s interpolation formula interpolate both the function as well as its derivative at each of the points Sometimes it is also called osculating interpolation formula

Let the set of data points (x i , y i , y’ i), 0≤ i≤ n be given A polynomial of the least degree say H(x) is to be determined such that

H(x i ) = y i and H’ (x i ) = y’ i ; i = 0, 1, 2, n (1)

H(x) is called Hermite’s interpolating polynomial

Since there are 2n + 2 conditions to be satisfied, H(x) must be a polynomial of degree

≤ 2n + 1

The required polynomial may be written as

H(x) =

0

n i i

u

=

∑ (x) y i +

0

n i i

v

=

Where u i (x) and vi(x) are polynomials in x of degree≤ 2n + 1 and satisfy.

(i) u i (x i) =

0,

1 ,

≠

 = 

(ii) v i (x j) = = 0 ∀i, j (3b)

(iv) v′i (x j) =1 ,0,i i≠j j

 = 

Using the Lagrange fundamental polynomials L i (x), we choose

2

2

u x A x L x

v x B x L x



′ =   



where L i (x) is defined as

( 00) ( 11) ( 11) ( 1 1) ( )

Since L i 2 (x) is a polynomial of degree 2n, A i (x) and B i (x) must be linear polynomials.

Let A i (x) = a i x + b i and B i (x) = c i x + d i

therefore from (4),

2

u (x) = (a x + b )[L (x)]

v x c x d L x





(5)

using conditions (3a) and (3b) in (5), we get

Again, using conditions (3c) and (3d) in (5), we get

From equations 6(a, b, c, & d), we deduce

( ) ( )

2

1 2 1

i

c

λ

′



′



Hence, from (5)

u i (x) = [– 2xL i ’ (x i ) + 1 + 2x i L i ’ (x i )] [L i (x)]2

= [1 – 2 (x – x i ) L i ’ (x i )] [L i (x)]2 and v i (x) = (x – x i ) [L i (x)]2

Therefore from (2),

H(x) =

0

n

i=

∑ {1 – 2 (x – x i ) L i ’ (x i )} [L i (x)]2 y i + [ ]2

0

n

i

=

′

−

∑

Which is the required Hermite’s interpolation formula

Example 1 Apply Hermite’s interpolation formula to obtain a polynomial of degree 4 from the following data.

i i

i

y′ 0 0 24

: : :

Sol Using h Hermite’s interpolation formula

2 2

( ) ( ) 1 2( i) i( )i i i + ( i) i( ) i

L x

=

We have,

H(x) = {1 – 2 (x – x0) L0’ (x0)} [L0 (x)]2 y0 + {1 – 2(x – x1) L1’ (x1)} [L1 (x)]2y1

{1 – 2 (x – x2) L2’ (x2)} [L2 (x)]2 y2 + (x – x0) [L0 (x0)]2 y’0

+ (x – x1) [L1 (x)]2 y ′1 + (x – x2) [L2 (x)]2 y’2

L0 (x) = ( )( )

( 0 11)( 0 22)

x x x x

( 1)( 2) 1( 2 )

L1 (x) = ( )( )

( 1 00)( 1 22) ( ( )( )( ) ) 2

2

x x

L2 (x) = ( )( )

( 0)( 1 ) ( ( )( )( ) ) ( 2 )

L0’ (x) = 2 3

2

x−

, L 1 ’ (x) = 2 – 2x, L2’ (x) = 2 1

2

x− ,

L0’ (x0) = 3

2

− , L1’ (x1) = 0, L2’ (x2) = 3

2,

Using these values in equation, we get

H (x) = [1 – 2 (x – 0) (– 3/2)]1

4 (x2 + 3x + 2)2 × 1

+ [1 – 2 (x – 2) (3/2)]1

4 (x2 – x)2 × 9 + (x – 2)1

4 (x2 – x)2×24

H(x) = (1 + 3x)1

4(x

2 + 3x + 2)2 + (1 – 3x + 6)9

4 (x

2 – x)2 + (x – 2)6 (x2 – x)2

∴ H(x) = x4 – 2x2 + 1

Example 2 Apply Hermite’s interpolation formula to find a cubic polynomial which meets the following specifications.

Sol Hermite’s interpolation formula is

H(x) = [1 – 2(x – x0) L0’ (x0)] [L0(x)]2 y 0 +[1 – 2(x – x1) Li’(x1)] [L1(x)]2 y1

+ (x – x0) [L0 (x)]2 y′0 + (x – x1) [L i (x)] 2 y1’ (1)

Now, L0 (x) = 0

x x

x x

−

− =

0 1

0 1

x

x

− = −

−

L1 (x) = 0

0

1 0

x

x x

∴ L0’(x) = – 1 and L1’ (x) = 1

Hence, L0’(x) = –1 and L1’ (x1) = 1

∴ From (1),

H (x) = [1 – 2 (x – 0) (– 1)] (1 – x2) (0) + [1 – 2(x – 1) (1)]x2 (1)

+ (x – 0) (1 – x)2 (0) + (x – 1)x2 (1)

= x2 –2x2 (x – 1) + x2 (x – 1)

= x2 – x2 (x – 1) = x2 (2 – x)

= 2x2 – x3

Example 3 Apply Hermite interpolation to find the value of sin (1.05) from the following data

i

x Sin x Cos x

Sol Here, f(x) = sin x, f’(x) = cos x, x0 = 1 & x1 = 1.10

L0 (x) = 1

1.10

1 1.10

x x

− − = – 10x + 11

L’0 (x) = – 10

1.00 1.10 1.00

x x

L’1 (x1) = 10

f(x0) = 0.84147, f(x1) = 0.54030

f(x2) = 0.89121, f(x3) = 0.45360 Now, using Hermite’s formula, we have

H(x) ={ 1 ( ) ( ) } ( )2 ( ) { 1 ( ) ( ) } ( )2 ( )

1 2− L x x−x L x  f x + 1 2− L x x−x L x  f x

( ) ( )2 ( ) ( ) ( )2 ( )

x x L x f′ x x x L x f′ x

H(x) = {1 + 20 (x – 1)} (–10x + 11)2 (0.84147) + {1 – 20(x – 1.10)} (10x – 10)2 (0.89121)

+ (x – 1.00) (–10x + 11)2 (0.54030)2 (0.54030) + (x – 1.10) (10x – 10)2 (0.45360) Put, x = 1.05, so we get

H (1.05) = sin (1.05) = {1 + 20(1.05 – 1)} (–10(1.05) + 11)2 (0.84147)

+ {1 – 20((1.05) – 1.10)} (10(1.05) – 10)2 (0.89121) + ((1.05) – 1.00) (–10(1.05) + 11)2 (0.54030) + ((1.05) – 1.10) (10(1.05) – 10)2 (0.45360)

H(x) = 2(0.25) (0.84147) + 2(0 25) (0.89122) + (0.05) (0.25) (0.54030) – (0.5) (0.25) (0.45360)

sin (1.05) = 0.420735 + 0.445605 + 0.00675375 – 0.00567

sin (1.05) = 0.86742 Ans

Example 4 A switching path between parallel railroad tracks is to be a cubic polynomial joining positions (0,0) and (4,2) and tangent to the lines y = 0 and y = 2 as shown in the figure Apply Hermite’s interpolation formula to obtain this polynomial

Sol Since tangents are parallel to X-axis,

y′ = 0 in both the cases.

∴ We have the table of values

Hermite interpolation formula is

′

Now, L 0 (x) = 1

4

x x

− − = 1 – 4

x

L1 (x) = 0

0

x x

∴ L0′ (x) = 1 ( )

and

Hence, L0’ (x0) = – 1

4 and L1’ (x1) =

1

4.

Therefore from (1)

 − − −   −  × +  − −     ×

+ ( 0) 1 2 0 ( 4) 2 0

1

x x

 −  =

  

H(x) = 1

16 (6x

2 – x3) Ans

Example 5 Using Hermite interpolation formula, estimate the value of 1 n (3.2) from the following table:

( )

n

x

Sol Using Hermite’s interpolation formula, we have

′

H(x) = [1 –2(x – x0)L0’ (x0)] [L0(x)]2y0 + [1 – 2(x – x1) L1’ (x1)] [L1 (x)]2 y1

+ [1 – 2(x – x2) L2’ (x2)] [L2(x)]2 y2+ (x – x0) [L0 (x)]2 y0’

+ (x – x1) [L1 (x)]2 y1’ + (x – x2) [L2 (x)] 2 y2’

Now, L 0 (x) = ( )( )

( 0 11)( 0 22) ( ( ) ( )( ) )

=

= 2(x2 – 4x – 3.5 x + 14)

= 2(x2 – 7.5x + 14)

L1 (x) = ( ) ( )

( 1 00) ( 1 22) ( ( ) ( )( ) )

=

= – 4(x2 – 7x + 12)

L2 (x) = ( )( )

( 2 00)( 2 11) ( )( )

1 0.5

=

= 2(x2 – 6.5x + 10.5)

∴ L0’ (x) = 4x – 15, L1’ (x) = – 8x + 28,

L0’ (x0) = – 3, L1’ (x1) = 0

L2’ (x) = 4x – 13

i.e., L2’ (x2) = 3

2 2

2 2

On Putting x = 3.2, we get

2 2

2 2

2 2

l n (3.2) = (2.2 {4(0.0576)} × ( ) ( ) ( ) ( )

l n (3.2) = 1.6314

Example 6 Show that f a b

2

+

  = f a( ) f b( ) (b a) f ( )a f ( )b

−  ′ − ′ 

Sol By Hermite’s interpolation formula

′

′

′