A textbook of Computer Based Numerical and Statiscal Techniques part 25 ppsx

Show that the maximum is obtained by Sol... Applying Lagrange’s formula, find a cubic polynomial which approximate the following data Sol... Hence the result... Find the unique polynomia

Example 1 Using Lagrange’s formula, find the value of

(i) y x if y 1 = 4, y 3 = 120, y 4 = 340, y 5 = 2544,

(ii) y 0 if y –30 = 30, y –12 = 34, y 3 = 38, y 18 = 42,

Sol (i) Here, x0 = 1, x1 = 3, x2 = 4, x3 = 5,

f(x0) = 4, f(x1) = 120, f(x2) = 340, f(x3) = 2544,

Now using Lagrange’s interpolation formula, we have

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

+

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

+

(1 3 1 4 1 53)( 4)( 5) 4 ( ( )(3 1 31)( 4 34)( )( 55) ) 120

( )( )( ) ( ( ) ( )( ) ( )( ) )

y x = f(x) = –1

6 (x – 3) (x – 4) (x – 5) + 30 (x – 1) (x – 4) (x – 5)

– 340

3 (x – 1) (x – 3) (x – 5) + 318 (x – 1) (x – 3) (x – 4)

(ii) Here, x0 = – 30, x1 = – 12, x2 = 3, x3 = 18,

y0 = 30, y1 = 34, y2 = 38, y3 = 42, Now from Lagrange’s interpolation formula, we have

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

+

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

+

(3 3030 3 12 3 18)( 12)( 18) 38 (18( 30 18 12 18 330)( )( 12)( )( 3) ) 42

y x = – 0.001052188 (x + 12) (x – 3) (x – 18) + 0.00419753 (x + 30) (x – 3) (x – 18)

– 0.005117845 (x + 30) (x + 12) (x – 18) + 0.001944444 (x + 30) (x + 12) (x – 3)

for x = 0

y0 = – 0.001052188 (12) (–3) (–18) + 0.00419753 (30) (–3) (–18)

– 0.005117895 (30) (12) (–18) + 0.001944444 (30) (12) (–3)

y0 = – 0.681817824 + 6.7999986 + 33.1636356 – 2.09999952

y0 = 39.9636546 – 2.781817344

y0 = 37.1818 Ans

Example 2 If y 0 , y 1 , y 2 , y 3 y 9 are consecutive terms of a series Prove that y 5 = 1

70

[56(y 4 + y 6 ) – 28 (y 3 + y 7 ) + 8 (y 2 + y 8 ) – (y 1 + y 9 )]

Sol Here, the arguments are 1, 2, 3, 9 so for these values Lagrange’s formula is given by

( 1)( 2)( 3)( 4)( 6)( 7)( 8) ( 9)

x

y

=

( 1) ( )( )( ) ( ) ( )( )( )1 2 31 5 6 7 8

y

2

2 1 1 2 4 5 6 7

y x

+

( −3 2 1) ( ) ( ) ( ) ( )( )( )−1 −33 −4 −5 −6

y

x + ( 4 3.2.1) ( ) ( ) ( ) ( )24 3 4 5

y

+

( 6 5.4.3.2) ( )( )( )6 1 2 3

y

x− − − − + ( 7)6.5.4.3.1( 2)( 1)7

y

x− − − + ( 8 7.6.5.4.2.1) 8 ( )1

y

( 9 8.7.6.5.3.2.1) 9 ( 1) ( 2)( 3)( 4)( 6)( 7)( 8)( 9)

x

+

=

( 1 10080)1( ) ( 2 1680)2( ) 720(3 3) 720(4 4) 720(6 6)

y

x−

Now for y5, put x = 5

4 3 2 1 1 2 3 4 4 10080 3 1680 720 2

5

576

y

40320 5040 1440 720 720 1440 5040 40320

5

576

y

720 y +y −1440 y −y + 5040 y +y − 40320 y +y

720 y +y −1440 y +y + 5040 y +y − 40320 y −y

1

y5 = 1

70 [56 (y4 + y6) – 28 (y3 + y7) + 8 (y2 + y8) – (y1 + y9)]. Proved

Example 3 Find f(x) as a polynomial of x if

Sol Now from Lagrange’s interpolation formula,

(7 11 7)( 00 7)( 63 7)( 63) 1611

×

P(x) = 3

224 x(x – 3) (x – 6) (x – 7) +

6

126 (x + 1) (x – 3) (x – 6) (x – 7) + 39

144 (x + 1)x (x – 6) (x – 7) –

822

126 (x + 1)x (x – 3) (x – 7) +

1611

224 (x + 1)x (x – 6) (x – 3)

P(x) = 3

224 (x4 – 16x3 + 81x2 – 126x) +

39

144 (x4 – 12x3 + 29x2 + 42x)

+ 6

126 (x4 – 15x3 + 23x2 – 45x – 126) – 822

126 (x

4 – 9x3 + 11x2 + 21x) + 1611

224 (x

4 – 8x3 + 9x2 + 18x)

+ x2 3 81 39 29 23 6 822 11 9 1611

+ x− ×45 6126 −126 3224× +39 42144× −822 21126× −1611 18224× −6

P(x) = x4(1) + x3 (–3) + x2(5) – 6

P(x) = x4 – 3x 3 + 5x2 – 6 which is the required polynomial

Example 4 Value of f(x) are given at a, b, c Show that the maximum is obtained by

Sol Here, x0 = a, x1 = b, x2 = c,

f(x0) = f(a), f(x1) = f(b), f(x2) = f(c)

By applying Lagrange’s formula, we have

f(x) = ( )( )

(x a b b)(x a c c) f a( ) ( (x b a a b)( )(x c c) ) f b( ) ( (x c a x a c) ( ) ( b b) )f c( )

f(x) = ( )

For maximum, we have

f’(x) = 0

( )( ) ( ( )( ) ( ) ) ( ( )( ) ( ) )

= {2x – (b + c)} (b – c) f(a) + {2x – (a + c)} (c – a) f(b) + {2x – (a + b)}(a–b) f (c) = 0

= 2x [(b – c) f(a) + (c – a) f(b) + (a – b) f(c)] = (b2 – c 2 ) f(a) + (c2 – a2) f(b) + (a2 – b2) f(c)

(3 2 ≠ 0)

Example 5 Applying Lagrange’s formula, find a cubic polynomial which approximate the following

data

Sol Now, using Lagrange’s formula, we have

+

(222 2 1 2)( 1 )( 33)3 ( (32)(2 3 1 3 2)( 1)( )( 2) )5

+

5 x − x + −x − 3x − x − x+ − 4x − x− +4 x +x − x− 

∴ f(x) = – 1 3 3 2 241 3.9

15x −20x + 60 x−

Example 6 (i) Determine by Lagrange’s formula, the percentage number of criminals under 35

years:

(ii) Find a Lagrange’s interpolating polynomial for the given data:

x 0 = 1, x 1 = 2.5, x 2 = 4, and x 3 = 5.5 f(x 0 ) = 4, f(x 1 ) = 7.5, f(x 2 ) = 13 and f(x 3 ) = 17.5 also, find the value of f(5)

Sol (i) Here x0 = 25, x1 = 30, x2 = 40 and x3 = 50

f(x0) = 52, f(x1) = 67.3, f(x 2 ) = 84.1 and f(x 3) = 94.4

By using Lagrange’s interpolation formula, we have

(25 30 2530)( 4040 25)( 5050) 52 (30( 25 3025) ( )( 4040 30) ( )( 5050) ) 67.3

(40 25 4030)( 2530)(405050) 84.1 (50( 25 5025)( )( 3030 50)( ) ( 4040) ) 94.4

Now for f(35), put x = 35

f(35) = ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ( ) ( ) ( ) ) ( ) ( ) ( ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( ) ( )

f(35) = – 10.5 + 50.475 + 42.05 + 4.72

f(35) = 77.405

(ii) By using Lagrange’s formula, we have

(1 2.5 1 4 1 5.52.5)( 4)( 5.5) 4 (2.5 1 2.5( 1)( ) ( 44 2.5)( )( 5.55.5) ) 7.5

4 1 4 2.5 4 5.5 5.5 1 5.5 2.5 5.5 4

Put x = 5

f(5) = ( )( )( )

( )( )( ) ( )( )( )

20.25−6.75+6.75+20.25 = 0.246913 – 2.2222 + 9.62962 + 8.641975

f(5) = 18.51850831 – 2.222222 = 16.296296. Ans

+ ( 13)

6.75

−

(x – 1) (x – 2.5) (x – 5.5) + 17.5

20.25 (x – 1) (x – 2.5) (x – 4)

P(x) = – 0.1481x3 + 1.5555x2 – 1.6666x + 4.2592

which is a required polynomial

At x = 5, f(x) = 16.3012 (Approx.)

Example 7 By means of Lagrange’s formula, show that

(i) y 0 = 1

2 (y 1 – y –1 ) –

1

(ii) y 3 = 0.05 (y 0 + y 6 ) – 0.3 (y 1 + y 5 ) + 0.75 (y 2 + y 4 )

(iii) y 1 = y 3 – 0.3 (y 5 – y –3 ) + 0.2 (y –3 – y –5 )

Sol (i) For the arguments – 3, –1, 1, 3, the Lagrange’s formula is

+

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

+

+

−

+

Putting x = 0 in (1), we get

y0 = –161 y–3 + 169 y -1 +169 y1 – 161 y3

= 1

2 (y1 + y–1) –1

8 ( 3 1) ( 1 3)

2 y y 2 y− y−

(ii) For the arguments 0, 1, 2, 4, 5, 6, the Lagrange’s formula is

( ) ( ) ( ) ( ) ( )

+

+

+

Putting x = 3 in (2), we get

y3 = 0.05 y0 – 0.3 y1 + 0.75 y2 + 0.75 y4 – 0.3 y5 + 0.05y6

y3 = 0.05 (y0 + y6) – 0.3 (y1 + y5) + 0.75 (y2 + y4)

(iii) For the arguments –5, –3, 3, 5, the Lagrange’s formula is

+

( ) ( )( ) 3 ( ( )( )( )( )( ) ) 5

+

Putting x = 1 in equation (3), we get

y1 = (–0.2) y–5 + 0.5 y–3 + y3 – 0.3y5

= y3 – 0.3 (y5 – y–3) + 0.2 (y–3 – y–5)

Example 8 Prove that Lagrange’s formula can be expressed in the

( )

( ) ( ) ( )

2 n n

2 n

2 n

2 n

P x | x x x

= 0

where P n (x) = f(x).

Sol Let P n (x) = a0 + a1x + a2x2 + a n x n (1)

Given that P n (x) = f(x)

Substitute x = x0, x1, x2, x n Successively in equation (2)

⇒ f(x0) = a0 + a1x0 + a2x2

0+ + a n x0n

f(x1) = a0 + a1x1 + a2x12 + + a n x1n

f(x n ) = a0 + a1x n + a2x2

n + + a n x n

n Now Eliminating a0, a1, a2 , a n from above equations, we get

( ) ( ) ( ) ( )

2 2

2

2

|

|

|

|

n n

n n

n

−

−

−

−

( ) ( ) ( ) ( )

2 2

2

2

n n

n n

n

= 0

Example 9 Four equidistant values u –1 , u 0 , u 1 and u 2 being given, a value is interpolated by Lagrange’s formula show that it may be written in the form.

u x = yu 0 + xu 1 + y y( 2 1)

3!

−

∆2 u –1 + x x( 2 1)

3!

−

∆2 u 0 where x + y = 1.

Sol ∆2u1= (E – 1)2 u–1 = (E2 – 2E + 1) u–1 = u1 – 2u0 + u–1

∆2u0 = (E2 – 2E + 1) u0 = u2 – 2u1 + u0

R.H.S = (1 – x) u0 + xu1 +( ) ( { )2 }

3!

(u1 – 2u0 + u–1)

+ ( 2 1)

3!

(u2 – 2u1 + u0) where y = 1 – x

2

6

u

Applying Lagrange’s formula for the arguments – 1, 0, 1 and 2.

( )( ) ( )

From (1) and (2), we observe that R.H.S = L.H.S

Hence the result

Example 10 Find the cubic Lagrange’s interpolating polynomial from the following data

( )

: 0 1 2 5 : 2 3 12 147

x

f x

Sol Here, ( )00 ( )11 ( )22 ( )33

Lagrange’s formula is

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

+

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

f(x) = ( ) ( ) ( )

( )( ) ( ) ( )

+

( )( )( ) ( ) ( ( )( )( ) ( )( ) ) ( )

= 1

5

−

(x – 1) (x – 2) + 3

4x (x – 2) (x – 5) – 2x (x – 1) (x – 5) +

49

20x (x – 1) (x – 2)

= – 1

5 (x

3 – 8x2 + 17x – 10) +3

4 (x

3 – 7x2 + 10) – 2 (x3 – 6x2 + 5x)

+ 49

20 (x

3 – 3x2 + 2x)

⇒ f(x) = x3 + x2 – x + 2

which is the required Lagrange’s interpolating polynomial

Example 11 The function y = f(x) is given at the points (7, 3), (8, 1), (9, 1) and (10, 9) Find

the value of y for x = 9.5 using Lagrange’s interpolation formula.

Sol We are given

( )

: 7 8 9 10 : 3 1 1 9

x

f x

Here, ( )00 ( )11 ( )22 ( )33

Lagrange’s interpolation formula is

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

+

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

+

( ) ( ) ( ) ( ) ( ( ) ( ) ( ) ( ) ( ) ) ( ) ( ( ) ( ) ( ) ( ) ( ) ) ( )

9

10 7 10 8 10 9

+

= – 1

2 (x – 8) (x – 9) (x – 10) +

1

2 (x – 7) (x – 9) (x – 10) –

1

2 (x – 7) (x – 8) (x – 10) + 3

2 (x – 7) (x – 8) (x – 9) .(1)

Putting x = 9.5 in eqn (1), we get

f (9.5) = –1

2 (9.5 – 8) (9.5 – 9) (9.5 – 10) +

1

2 (9.5 – 7) (9.5 – 9) (9.5 – 10) – 1

2 (9.5 – 7) (9.5 – 8) (9.5 – 10) +

3

2 (9.5 – 7) (9.5 – 8) (9.5 – 9)

= 3.625

Example 12 Find the unique polynomial P(x) of degree 2 such that:

Use Lagrange’s method of interpolation.

Lagrange’s interpolation formula is

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

(1 3 1 43)( 4) ( )1 ( (3 1 31)( )( 44) ) ( )27 ( (4 1 41)( )( 33) ) ( )64

= 1

6 (x

2 – 7x + 12) – 27

2 (x

2 – 5x + 4) + 64

3 (x

2 – 4x + 3)

= 8x2 – 19x + 12

Hence the required polynomial is,

P(x) = 8x2 – 19x + 12

PROBLEM SET 5.1

1 Using Lagrange’s interpolation formula, find y(10) from the following table.

: 5 6 9 11 : 12 13 14 16

X

2 Use Lagrange’s interpolation formula to fit a polynomial for the data

: 1 0 2 3 : 8 3 1 12

x

x u

−

−

Hence or otherwise find the value of u1 [Ans ux = 2x3 – 6x2 + 3x + 3, u1 = 2]

3 Compute the value of f(x) for x = 2.5 from the following data:

( ) : 1 2: 1 8 27 643 4

x

f x

4 If y(1) = – 3, y(3) = 9, y(4) = 30 and y(6) = 132, find the four point Lagrange’s interpolation polynomial which takes the same values as the function y at the given points.

[Ans x3 – 3x2 + 5x – 6]

5 Find the polynomial of degree three which takes the values given below:

: 0 1 2 4 : 1 1 2 5

x

4x

2 – 2

3x + 1]

6 Find f(x) by using Lagrange’s interpolation formula:

( )

: 659 705 729 804

x

f x

Also find maximum value of f(x) [Ans 151

24 x

3 – 249

8 x

2 + 721

12 x + 659]

[No real value of x exist for which f(x) is max.]

7 Given log10 654 = 2.8156, log10 658 = 2.8182, log10 659 = 2.8189 and log10 661 = 2.8202 Find log10 656 by Lagrange’s interpolation formula [Ans 2.8169]

8 Compute Sin 15° by Lagrange’s Method from the data given below:

: 0.0000 0.50000 0.70711 0.86603 1.0000

x

y

[Ans 0.25859]

9 The percentage of Criminals for different age group are given below:

Apply Lagrange’s formula to find the percentage of criminals under 35 years of age

[Ans 77]