A textbook of Computer Based Numerical and Statiscal Techniques part 13 doc

Also find the value of y at x = 6 by extending the table and verify that the same value is obtained by substitution... Since we have five observations, therefore the 4th differences will

Example 1 Construct a forward difference table for the following values:

0 5 10 15 20 25 ( ) 7 11 14 18 24 32

x

f x

Sol Forward difference table for given data is:

−

−

−

0 7

4

8

25 32

Example 2 If y = x 3 + x 2 – 2x + 1, calculate values of y for x = 0, 1, 2, 3, 4, 5 and form the difference table Also find the value of y at x = 6 by extending the table and verify that the same value

is obtained by substitution.

Sol For x = 0, 1, 2, 3, 4, 5, we get the values of y are 1, 1, 9, 31, 73, 141 Therefore, difference

table for these data is as:

0

100

6 241

x y ∆y ∆ y ∆ y

Because third differences are zero therefore

∆3y3 = 6 ⇒ ∆2y4 – ∆2y3 = 6

⇒ ∆2y4–26 = 6 ⇒ ∆2y4 = 32

Now, ∆2y4 = 32 ⇒ ∆y5 – ∆y4 = 32

⇒ ∆y5 – 68 = 32 ⇒ ∆y5 = 100

Further, ∆y5 = 100 ⇒ y6 – y5 = 100

Verification: For given function x3 + x2 – 2x + 1, at x = 6, y(6) = (6)3 + (6)2 – 2(6) + 1 = 241

Hence Verified

Example 3 Given f(0) = 3, f(1) = 12, f(2) = 81, f(3) = 200, f(4) = 100 and f(5) = 8 From the difference table and find ∆ 5 f(0).

Sol The difference table for given data is as follows:

9

92

x f x ∆f x ∆ f x ∆ f x ∆ f x ∆ f x

−

−

−

−

−

−

Hence, ∆5f(0) = 755

Example 4 Construct the forward difference table, given that:

9962 9848 9659 9397 9063 8660

x y and point out the values of ∆ 2 y 10 , ∆4 y 5

Sol For the given data, forward difference table is as:

5 9962

114

403

30 8660

x y ∆y ∆ y ∆ y ∆ y

−

−

−

−

−

−

−

−

From the table, ∆2y10,∆4y5 is as ∆2y10 = –73 and ∆4y5 = –1

Example 5 Find f(6) given that f(0) = –3, f(1) = 6, f(2) = 8, f(3) = 12, the third differences being constant.

Sol For given data we construct the difference table:

−

−

( ) ( ) ( ) ( )

9

4

3 12

We have, f(6) = f(0+6) = E6 f(0) = (1+ ∆)6 f(0)

= (1 6+ ∆ + ∆ + ∆15 2 20 3) (0)f [Higher differences being zero]

= f(0) 6+ ∆f(0) 15+ ∆2f(0) 20+ ∆3f(0)

= –3 + 6 × 9 + 15 × (–7) + 20 × 9

= –3 + 54 – 105 + 180

= 126

Example 6 Prove that:

(a) f(4) = f(3) + ∆f(2) + ∆ 2 f(1) + ∆ 3 f(1).

(b) f(4) = f(0) + 4∆f(0) + 6∆ 2 f(–1) + 10∆ 3 f(–1)

Sol

(a) We have, f(4) – f(3) = ∆f(3)

= ∆[f(2) + ∆f(2)] [Because ∆f(2) = f(3) – f(2)]

= ∆f(2) + ∆2f(2)

= ∆f(2) + ∆2[f(1) + ∆f(1)] [Because ∆f(1) = f(2) – f(1)]

= ∆f(2) + ∆2f(1) + ∆3f(1)

Therefore, f(4) = f(3) + ∆f(2) + ∆2f(1) + ∆3f(1)

(b) We have, f(4) = E5 f(–1) = (1 + ∆)5 f(–1)

= {1 + 5C1∆+5C2∆2 + 5C3∆3} f(–1)

(On taking up to third differences)

= f(–1) + 5∆f(–1) + 10∆2f (–1) + 10∆3f(–1)

= [ f(–1) + ∆f(–1)] +4[∆f(–1)+ ∆2f(–1)] + 6∆2 f(–1) + 10∆3f(–1)

= [ ( 1)f − + ∆ − + ∆ − + ∆ − + ∆f( 1)] 4 [ ( 1)f f( 1)] 6 2f( 1) 10− + ∆3f( 1)−

= f(0) 4+ ∆f(0) 6+ ∆2f( 1) 10− + ∆3f( 1)−

Because, f( 1)− + ∆ − = − +f( 1) f( 1) f(0)− − =f( 1) f(0)

Example 7 Find the function whose first difference is e x

Sol We know that ∆ =e x e x h− −e x =e e x( h−1), where h is the interval of differencing.

1

x

e

Hence, required function is given by

1

x h

e

e − .

Example 8 Find the first term of the series whose second and subsequent terms are 8, 3,

0, –1, and 0.

Sol If the interval of differencing is unity, then

f(1) = E–1f(2)

= (1+ ∆)–1 f(2)

= (1 – ∆ + ∆2 – ∆3 + )f(2).

Since we have five observations, therefore the 4th differences will be constant and 5th differences will be zero

2

( ) ( ) ( )

5

1

1

x f x ∆f x ∆ f x

−

−

−

−

Hence, f(1) = f(2) – ∆f(2) + ∆2f(2) [Higher order differences are 0]

f(1) = 8 – (–5) + 2 = 15

3.4.2 Backward or Ascending Differences

If we subtract from each value of y except y0, the previous value of y, we get y1 – y0,

y2 –y1, y3 – y2, y n – y n–1 These differences are called first backward differences of y and are

denoted by ∇y The symbol ∇ denotes the backward difference operator That is,

∇y1 =y1−y0

∇y2 =y2−y1

∇yn = y n – y n-1

Also it can be written as,

∇f x h( + ) = f x h( + −) f x( ) Similarly, second forward difference is given by,

∇2f x h( + ) = ∇f x h( + − ∇) f x( )

In general,

+

r

y = ∇ −1 + − ∇ −1

1 , or

∇n f x h( + ) = ∇n−1f x h( + − ∇) n−1f x( )

Backward Difference Table:

0 0

1 2

3

3

2

4

4 4

y

y

∇

∇

∇

∇

Example 9 Construct the backward difference table for y = log x given that:

1 1.3010 1.4771 1.6021 1.6990

x y and find the values of ∇3 log 40 and ∇4 log 50.

Sol For the given data, backward difference table as:

−

−

0.3010

20 1.3010 0.1249

0.1761 0.0738

0.1250 0.0230

40 1.6021 0.0281

0.0969

50 1.6990

Hence,Hnc∇3 log 40 = 0.0738 and ∇4 log 50 = 0.0508 −

Example 10 Given that:

1 8 27 64 125 216 343 512

x y Construct backward difference table and obtain ∇4 ( )f 8

Sol Backward difference table for given data is as:

7

169

8 512

x f x ∇f x ∇ f x ∇ f x ∇ f x

Hence, ∇4f(8) 0.=

Example 11 Construct the backward difference table from the data:

sin 30 o = 0.5, sin 35 o = 0.5736, sin 40 o = 0.6428, sin 45 o = 0.7071

Assuming third difference to be constant, find the value of sin 25°.

Sol Backward difference table for given data is as:

−

−

−

−

−

25 0.4225

0.0775

30 0.5000 0.0039

35 0.5736 0.0044

40 0.6428 0.0049

0.0643

45 0.7071

Since third differences are constant therefore

∇3y40 = – 0.0005

⇒ ∇2y40 – ∇2y35 = 0.0005

⇒ –0.0044 – ∇2y35 = –0.0005

⇒ ∇2y35 = –0.0039

Again, ∇y35 – ∇y30 = –0.0039

⇒ 0.0736 – ∇y30 = –0.0039

Again, y30 – y25 = 0.0775

⇒ 0.50 – y25 = 0.0775

Therefore, sin 25° = 0.4225

3.4.3 Central Differences

The central difference operator is denoted by the symbol δ and central differences is given by,

( ) ( ) ( ) or

x

y

,

1/2

y

δ = y1 – y0

3/2

y

δ =y2 – y1

−

δ 1 2

n

y = y n−y n−1

Central Difference Table:

δ

δ

δ δ

1/2 2

3

3

2

7/2

y

y

3.4.4 Other Difference Operators

(a) The Operator E: The operator E is called shift operator or displacement or translation

operator It shows the operation of increasing the argument value x by its interval of differencing

h so that.

Ef (x) = f(x + h) or Ey x = y x+h

Similarly, Ef(x + h) = f(x + 2h)

In general, E y n x = y x nh+ or E f x n ( )= f x nh( + )

In the same manner, E–1 f(x) = f(x – h)

Also, E–2 f(x) = f(x – 2h)

E –n f(x) = f(x – nh)

This is called inverse of shift operator

(b) Differential Operator D: The differential operator for a function y = f(x) is defined by

( )

dx

2 ( )

D f x =

2

2 ( )

d

f x

dx and so on.

The operator ∆ is an analogous to the operator D of differential calculus In finite differences,

we deal with ratio of simultaneous increments of mutually dependent quantities where as in differential calculus, we find the limit of such ratios when the increment tends to 0

(c) The Unit Operator 1: The unit operator 1 has a property that 1 f(x) = f(x) It is also called

identity operator

(d) Averaging Operator µµµµµ: The operator µ is a averaging operator and is defined by,

µyx = + −

+

1

2 y x h y x h

i.e., µf x( ) =  + + − 

1

3.4.5 Properties of Operators

1 The operators ∆ ∇, , , , and E δ µ D are all linear operators

i.e., ∇ (af (x + h) + bφ(x + h) = [af (x + h)+bφ(x + b)] – [af (x) + bφ(x)]

= a[f(x + h) – f(x)] + b[φ(x + h) – φ(x)]

= a∇f(x + h) + b∇φ(x+h) Hence, ∇ is a linear operator

On substituting a = 1, b = 1, we get

∇[f(x + h) + φ(x + h)] = ∇f(x + h) + ∇φ(x + h)

Also on substituting b = 0, we get

∇[af(x + h)], = a∇f(x + h)

2 The operator is distributive over addition

3 All the operators follows the law of indices i.e.,

∆p∆q f(x) = ∆p+q f(x) = ∆ q∆p f(x)

Also, ∆[ f(x) + φ(x)] = ∆[φ(x) + f(x)]

4 E and ∆ are not commutative with respect to variables.

5 If f(x) = 0, then it does not mean that either ∆ = 0 or f(x) = 0.

6 Operators E and ∆ cannot stand without operands.

3.4.6 Relation between Different Operators

There are few relations defined between these operators Some of them are:

1 ∇ = 1 – E–1 or E = (1 – ∇)–1

2 ∆ = E – 1 or E = 1 + ∆

3 E∇ = ∇E = ∆

4 E = e hD = 1 + ∆, where D is the differential operator

5 δ = E1/2 – E–1/2

6 µ = 1

2 (E 1/2 + E–1/2)

7 δE1/2 = ∆

Proof:

3 (E∇) f(x) = E{∇f(x)} = E{f(x) – f (x – h)}

= Ef (x) – Ef (x – h)

Also, (∇E) f(x) = ∇ {Ef (x)} = ∇f (x + h)

From (1) and (2), we get E∇ = ∆ and ∇E = ∆

4 Ef (x) = f (x + h)

= f(x) + h f′(x) +

2 2!

h f′′(x) + (By using Taylor’s theorem)

= 1.f(x) + hDf (x) +

2

2 !

h

D2 f (x) +

= ehD f (x)

Ef (x) = e hD f(x) or E = e hD

Since, E = 1 + ∆, therefore ∆ = e hD – 1

2

h

x – y x –

2

h = E1/2 y x – E–1/2 y x

= (E1/2 – E–1/2)y x

Therefore, δ = E1/2 – E–1/2

2 y x 2h y x 2h

+

1

2 (E 1/2y x + E–1/2 y x)

= 1

2 (E 1/2 + E–1/2)y x

Therefore, µ = 1

2 (E 1/2 + E–1/2)

7 δE1/2 y x = δy x+h =

2

h x y

+ – y x = ∆y x

Therefore, δE1/2 = ∆

Example 12 Show that:

(a) (E 1/2 + E –1/2 ) (1 + ∆) 1/2 = 2 + ∆

(b) ∆ = 1

2 δ2 + δ 1+ δ2 4 Sol (a) Since 1 + ∆ = E therefore

(E1/2 + E–1/2) E1/2 = E + 1 = 1 + ∆ + 1 = ∆ + 2

(b) 1 2 1 2/ 4

= 1

2 (E 1/2 – E–1/2)2 + (E1/2 – E–1/2) 1( 1/ 2 1/ 2)2

1

−

= 1

2 (E + E

–1–2) + (E1/2 – E–1/2)

1/ 2 1/ 2

2

= 1

2 (2E – 2) = E – 1 = ∆ Example 13 Prove that (1) ∆ + ∇ = ∆ ∇−

∇ ∆ (2) (1 + ∆) (1 – ∇) ≡ 1

Where ∆ and ∇are forward and backward difference operators respectively.

Sol (1) ∆ ∇− 

∇ ∆

1 1

1 1

E E

−

−

−

=

1 1

E

E

y x = E 1

E

 y x = (E – E –1 )y x

= {(1 + ∆) – (1 – ∇)} y x = (∆ + ∇) y x

(2) (1 + ∆) (1 – ∇) y x = (1 + ∆) [y x – ∇y x]

= (1 + ∆) [y x –{y x – y x–h}] = (1 + ∆) [y x –h]

= E(y x – h ) = EE–1y x = 1 y x (the interval of differencing being 1) Hence, (1 + ∆) (1 – ∇) ≡ 1