A textbook of Computer Based Numerical and Statiscal Techniques part 27 ppt

Divided differences are symmetric with respect to the arguments i.e., independent of the order of arguments.. The nth divided differences of a polynomial of nth degree are constant.. The

The second divided difference for x0, x1, x2 is given by

f(x0, x1, x2) = [x0, x1, x2] = [ 1 2] [ 0 1]

−

−

The third divided difference for x0, x1, x2, x3 is given by

f(x0, x1, x2, x3) = [x0, x1, x2, x3] = [ 1 2 3] [ 0 1 2]

−

f(x0, x1, x2 ,x n ) = [x0, x1, x2 ,x n] = [ 1 2 ] [ 0 1 1]

0

n

−

−

−

5.4.1 Properties of Divided Differences

1 Divided differences are symmetric with respect to the arguments i.e., independent of the

order of arguments

i.e., [x0, x1] = [x1, x0]

Also, [x0, x1, x2] = [x2, x0, x1] or [x1, x2, x0]

2 The nth divided differences of a polynomial of nth degree are constant

Let f(x) = A0x n + A1x n–1 + + A n–1 x + A n by a polynomial of degree n provided A0 ≠ 0 and arguments be equally spaced so that

x1 – x0 = x2 – x1 = x n – x n–1 = h

Then first divided difference

[x0, x1] = 1 0 0

− =∆

−

Second divided difference

[x0, x1, x2] = 12

2 0

y

∆

[x0, x1, x2 x n–1 , x n] = 1

! n

n h = 0

n y

∆

Since, function is a nth degree polynomial

Therefore, ∆n y0 = constant

∴ nth divided difference will also be constant

3 The nth divided difference can be expressed as the quotient of two determinants of order

(n + 1) i.e.,

[x0, x1] = 1 0

−

− = 1−1 0 − 1−0 0

⇒ [x0, x1] = 1 0 1 0

÷

Similarly, [x0, x1, x2] =

x x x ÷ x x x and so on

4 The nth divided difference can be expressed as the product of multiple integral

5.4.2 Relation Between Divided Differences and Ordinary Differences

Let the arguments x0, x1, x2, x n, be equally spaced such that

x1 – x0 = x2 – x1 = = x n – x n–1 = h

∴ x1 = x0 + h

x2 = x0 + 2h

x n = x0 + nh Now, first divided difference for arguments (x0, x1) be given by

( )

x f x

f x f x

x x

−

( 0 ) ( )0

h

+ −

= f x( )0

h

∆

(1) Similarly, second divided difference be given as

( )

1 2

2 0

x x∆ f x = ( 1 2) ( 0 1)

1

−

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

−

−

−

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

−

= 12

2h [f x( 0−2 ) 2 (h − f x0+ +h) f x( 0)] = 2 ( )

0 2 2!

f x h

∆

(2)

( )

1 2 3 0

x x x∆ f x =

1

x −x [f x x x( ,1 2, 3)−f x x x( 0, 1, 2)]

= 1

3h

( ) 2 ( )

2

0 1

f x

f x

−

( ) ( )

3 6

h

[From (1)

= 3 ( )

0 3 3!

f x h

∆

In general, we have

1 n

n

x∆x f(x0) = ( )0

!

n n

f x

n h

∆

which shows the relation between divided difference and ordinary difference

5.5 NEWTON’S DIVIDED DIFFERENCE FORMULA

Let y0, y1, y n , be the values of y = f(x) corresponding to the arguments x0, x1, x n, then from the definition of divided differences, we have

0

0

x x

x x

−

−

So, that, y=y0 + (x − x0) [x x, 0] (1) Again, [ ] [ 0] [ 0 1]

1

= x, x x , x

x, x , x

x x

−

−

which gives, [x, x0] = [x , x0 1] + (x x− 1) [x, x , x0 1] (2) From (1) and (2), y=y0 + [x x− 0] (x x0, 1) (+ x −x0)(x x− 1) [x x x, 0, 1] (3)

2

x x x x

x x

−

=

−

which gives, [x x x, 0, 1] = [x x x0, 1, 2] + (x x− 2) [x x x x, 0, 1, 2] (4) From (3) and (4)

y = y0 + (x x− 0) [x x0, 1] + (x x− 0)(x x− 1) [x x x0, 1, 2] + (x x− 0)(x x− 1)(x x− 2) [x x x x, 0, 1, 2]

Proceeding in this manner, we get

y = f(x) = y0 + (x – x0) [x0, x1] + (x – x0) (x – x1) [x0, x1, x2]

+ (x – x0) (x – x1) (x – x2) [x0, x1, x2, x3] + + (x – x0) (x – x1) (x – x2)

(x – x n–1 ) [x0, x1, x n ] + (x – x0) (x – x1) (x – x n ) [x, x0, x1, x2, x n] This is called Newton’s general interpolation formula with divided differences, the last term

being the remainder term after (n + 1) terms.

Newton’s divided difference formula can also be written as

y = y0 + (x – x0)∆y0 + (x – x0) (x – x1) 2

0

y

∆ + (x – x0) (x – x1) (x – x2) 3

0

y

∆

+ (x – x0) (x – x1) (x – x2) (x – x3)∆4y0 + + (x – x0) (x – x1) (x – x n–1) ∆n y0 Example 1 Apply Newton’s divided difference formula to find the value of f(8) if

f(1) = 3, f(3) = 31, f(6) = 223, f(10) = 1011, f(11) = 1343,

Sol The divided difference table is given by

28 2 14

332 1 332

=

=

=

=

=

On, applying Newton’s divided difference formula, we have

f(x) or y x = y0 + (x – x0)∆y0 + (x – x0) (x – x1) 2

0

y

∆ + (x – x0) (x – x1) (x – x2) 3

0

y

∆ +

f(x) or y x = 3 + (x – 1) × 14 + (x – 1) (x – 3) × 10 + (x – 1) (x – 3) (x – 6) × 1

for f(8), we put x = 8 in above equation, we get

f(8) = 3 + (7) (14) + (7) (5) (10) + (7) (5) (2)

f(8) = 3 + 98 + 350 + 70 f(8) = 521 Ans

Example 2 Find the function u x in powers of x – 1 given that

u 0 = 8, u 1 = 11, u 4 = 68, u 5 = 123,

Sol Here,

x0 = x0, x1 = 1, x2 = 4, x3 = 5,

y0 = 8, y1 = 11, y2 = 68, y3 = 123, The divided difference table is given by

( )

3

55

From Newton’s divided difference formula, we have

y x = 8 + (x – 0)× (3) + (x – 0)(x–1)× (4) + (x – 0) (x – 1) (x – 4)×1

y x = 8 + 3x + 4x2 – 4x + x(x2 – 5x + 4)

y x = x3 – x2 + 3x + 8

In order to express it in powers of (x – 1), we use synthetic division method as,

1

x3 – x2 + 3x + 8 = (x – 1)3 + 2 (x – 1)2 + 4 (x – 1) + 11 Ans

Example 3 Find the Newton’s divided difference interpolation polynomial for:

( )

: 1.625 5.875 31.0 131.0 282.125 521.0

x

f x

Sol The divided difference table is given as, if here

( )

: 1.625 5.875 31.0 131.0 282.125 521.0

x

f x

The table for divided difference is as:

( )

4.25

159.25

By using Newton’s divided difference formula, we have

y x = 1.625 + (x – 0.5)× 4.25 + (x – 0.5) (x – 1.5)× 5 + (x – 0.5) (x – 1.5) (x – 3)× 1

y x = 1.625 + 4.25x – 2.125 + 5x2 – 10x + 3.75 + (x – 0.5) (x2 – 4.5x + 4.5)

y x = 1.625 + 4.25x – 2.125 + 5x2 – 10x + 3.75 + x3 – 4.5 x2 + 4.5x – 0.5x2 + 2.25 x – 3.25

∴ y x = x3 + 11x – 10x + 1

y x = x3 + x + 1 Ans

Example 4 Construct a divided difference table for the following:

( )

: 22 30 82 106 216

x

f x

Sol The divided difference table is given as,

8

2 1

26 8

4 1

22 8

216 106

22 5

− =

−

− =

−

−

=

−

−

=

Example 5 (i) Prove that 3

bcd

1 a

 

∆    = – 1

abcd (ii) Show that the nth divided differences [x 0 , x 1 , x n ] for u x = 1

x is

( )n

1

x , x , x

( )

( ) ( )

2

3

2

1

1

1

1

1

1 1

a

a

b a

b

c b

c

d c

−

= −

−

−

−

−

−

−

= −

−

From the table, we observe that

3

bcd∆ 1

a

 

  = – 1

(ii) From (1), we see that

3

bcd∆ 1

a

 

  = – 1

abcd = (– 1)

3 f(a, b, c, d)

In general,

0 , 1 , n

n

x x∆ x

0

1

x

 

 

  = (–1)

n f(x0, x1, x2, x n) = ( )

1 , , ,

n n

x x x x

−

Example 6 (i) Find the third divided difference with arguments 2, 4, 9, 10 of the function f(x) =

x 3 – 2x.

(ii) if f(x) = 1 2

x , find the first divided differences f(a, b), f(a, b, c,), f (a, b, c, d) (iii) If f(x) = g(x) h(x), prove that

f(x 1 , x 2 ) = g(x 1 ) h(x 1 , x 2 ) + g (x 1 , x 2 ) h (x 2 ) Sol (i)

26

131 26

9 2

269 131

980 711

269

− =

−

−

=

−

−

=

−

−

=

−

Hence third divided difference is 1

2

2 2

2 2

2

1

1

1

1

a

a

a b

ab bc ca b

abc acd abd bcd

b c

bc cd bd c

c d

c d

= −  

+ +

 

+ +

+

− 

From the above divided difference table, we observe that first divided differences,

f(a, b) = – a2 2b

a b

+

f(a, b, c) = ab 2 2 2bc ca

a b c

+ +

and f(a, b, c, d) = – abc acd2 2 2 2abd bcd

a b c d

(iii) R.H.S = g(x1) ( ) ( )2 1 ( )2 ( ) ( )1

2

h x h x g x g x

h x

+

=

1

x −x [{g (x1)h (x2) – g(x1) h(x1)} + {g(x2) h (x2) – g(x1) h (x2)}]

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

g x h x g x h x

x x

−

− = x∆2 g (x1) h (x1) =

2

x∆ f(x2) f(x1, x2) = L.H.S Hence the result

Example 7 The following are the mean temperatures (°F) on three days, 30 days apart round the pds.

of summer and winter Estimate the app dates and values of max and min temperature.

Summer Winter Day

Sol Divided difference table for summer is:

4.6

0.9

−

−

∴ f(x) = 58.8 + (x – 0) (4.6) + (x – 0) (x – 1) (–2.75)

= – 2.75x2 + 7.35 x + 58.8 For maximum and minimum of f(x), we have

f′(x) = 0

⇒ – 5.5x + 7.35 = 0⇒ x = 1.342

Again, f ’’(x) = – 5.5 < 0

∴ f(x) is maximum at x = 1.342

Since unit ≡ 30 days

1.342 = 30×1.342 = 40.26 days

∴ Maximum temperature was on 15 June + 40 days i.e., on 25 July and value of maximum

temperature is

[f (x)]max = [f(x)]1.342 = 63.711°F approximately.

Divided difference table for winter is as follows:

2.6

1.2

−

∴ f(x) = 40.7 + (x – 0) (–2.6) + x(x – 1) (1.9)

= 1.9x2 – 4.5x + 40.7 For f(x) to be maximum or minimum, we have f′(x) = 0

3.8x – 4.5 = 0 ⇒x = 1.184

Again, f′′(x) = 3.8 > 0

∴ f(x) is minimum at x = 1.184

Again, unit 1 ≡ 30 days

∴ 1.184 = 30×1.184 = 35.52 days

∴ Minimum temperature was on 16 Dec + 35.5 days i.e., on the mid night of 20th Jan and

its value can be obtained similarly

[f(x)]min = [f(x)]1.184 = 63.647 °F approximately.

Example 8 The mode of a certain frequency Curve y = f (x) is very near to x = 9 and the values

of frequency density f(x) for x = 8.9, 9.0 and 9.3 are respectively equal to 0.30, 0.35 and 0.25 Calculate the approximate value of mode.

Sol Divide difference table for given frequency density is as

50 9

3500

36 100

3

−

−

Applying Newton’s divided difference formula

100 f(x) = 30 + (x – 8.9)× 509 + (x – 8.9) (x – 9) 3500

36

− 

= – 97 222x2 + 1745.833x – 1759.7217

∴ f(x) = – 0.9722x2 + 17.45833x – 17.597217

f′(x) = – 1.9444x + 17.45833

Putting f′(x) = 0, we get

x = 17.45833

1.9444 = 8.9788

Also,f′′( )x = – 1.9444 i.e., (–) ve

∴ f(x) is maximum at x = 8.9788

Hence, mode is 8.9788