A textbook of Computer Based Numerical and Statiscal Techniques part 15 doc

3.5 FUNDAMENTAL THEOREM ON DIFFERENCES OF POLYNOMIAL Statement: If fx be the nth degree polynomial in x, then the nth difference of fx is constant and ∆n+1 fx and all higher differences

Similarly, ∆n 1

x

 

 

(−+ )(− +− ) (−+ )

1 2 3

1 2

n

( 1)( 1 2 ) (! )

n

n

−

(2) Similarly, ∆(ab cx ) = a∆b cx = a{b c(x + 1) – b cx}

= a{b cx b c – b cx } = a(b c –1)b cx

2(ab cx)

∆ = ∆∆(ab cx)= ∆{a b( c−1)b cx}

= a b( c− ∆1) b cx=a b( c−1)2b cx

Proceeding in the same manner, we get

( )

n ab cx

∆ = a(b c – 1)n b cx

Example 29 If p, q, r and 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 of q and r is A+ 1 B

24 , where A is the arithmetic mean of q, r and B is

the arithmetic mean of 3q – 2p – s and 3r – 2s – p.

Sol On taking h being the interval of differencing the difference table is as:

−

− +

2

3 3

3

q p

s r

The argument half way between the arguments of q and r is 1

2 (a + h + a + 2h) i.e.,

3 2

a+ h Hence, the required entry is given by,

u a+(3/2)h = E3/2u a = (1 + ∆)3/2 u a

=  + ∆ + ∆ + −  ∆ 

 

(Higher order differences being neglected)

p+ q p− + r− q p+ − s− r+ q p−

= p1− + +32 83 161+q32− −34 163 +r38+163 −161 s

16p 16q 16r 16s

− + + −

= −161 p+ +(q r)161 +12−161 s

Again A = arithmetic mean of q and r = 1( )

2 q r+

B = Arithmetic mean of 3q – 2p – s and 3r – 2s – p is

2 q− p s− + r− − =s p 2 q r s p+ − −

24

q r

q r s p

+ + + − −

Substituting this value in (1), we get u a+(3/2)h = A + 1

24B.

Example 30 Given u 0 , u 1 , u 2 , u 3 , u 4 and u 5 Assuming that, fifth order differences to be constant Show that: = + − + −

1 2

2 256 where a = u0 + u5, b = u1 + u4, c = u2 + u3

2

u = E5/2u0 = (1 + ∆)5/2u0

= 0 5 0 15 2 0 5 3 0 5 4 0 3 5 0

u + ∆ +u ∆ u + ∆ u − ∆ u + ∆ u

= 0 5( 1 0) 15( 2 2 1 0) 5 ( 3 3 2 3 1 0)

3

= 0 5 1 4 2 3

256 u +u −256 u +u +128 u +u

256a−256b+128c

= 2563 a−25625 b+12+12811 c

c+ a c− + c b−

(R.H.S.)

Example 31 Given:

u 0 + u 8 = 1.9243, u 1 + u 7 = 1.9590, u 2 + u 6 = 1.9823, u 3 + u 5 = 1.9956 Find u 4

Sol Since 8 entries are given, therefore we have ∆8 u0 = 0

i.e (E – 1)8 u0 = 0

i.e (E8 – 8C1E7 + 8C2E6 – 8C3E5 + 8C4E4 – 8C5E3 + 8C6E2 – 8C7E1 + 1)u0 = 0

i.e (E8 – 8E7 + 28E6 – 56E5 + 70E4 – 56E3 + 28E2 – 8E + 1)u0 = 0

i.e u8 – 8u7 + 28u6 – 56u5 + 70u4 – 56u3 + 28u2 – 8u1 + u0 = 0

i.e (u8 + u0) – 8(u7 + u1) + 28(u6 + u2) – 56(u5 + u3) + 70u4 = 0

On putting the given values, we get

1.9243 – 8(1.9590) + 28(1.9823) – 56 (1.9956) + 70u4 = 0

or –69.9969 + 70u4 = 0

Example 32 Sum the following series 1 3 + 2 3 + 3 3 + + n 3 using the calculus of finite differences.

Sol Let 13= u0, 23 = u1, 33 = u2, , u3 = u n–1 Therefore sum is given by

S = u0 + u1 + u2 + + u n–1

= (1 + E + E2 + E3 + + E n–1 )u0

1

n

E E

  u0 =

(1 )n 1

∆

= 1

∆

( 1) 2 ( 1)( 2) 3

n

 + ∆ + ∆ + ∆ + + ∆ − 

1

n n

We know ∆u0 = u1 – u0 = 23 – 13 = 7

∆2 u0 = u2 – 2u1 + u0 = 33 –2(2)3 + 13 = 12

Similarly we have obtained ∆3 u0 = 6 and ∆4 u0, ∆5 u0 , are all zero asu r =r3is a polynomial

of third degree

∴ S = n + ( 1)

2!

n n−

(7) + ( 1)( 2)

6

n n− n−

12 + ( 1)( 2)( 3)

24

(6)

=

2

4

n

(n2 + 2n + 1) = ( ) 2

1 2

n n

Example 33 Prove that: ∞ ∞

= =

Sol Taking right hand side of the given expression

=

2 0

2x u x 4 2 4

∞

=

= 1

2 (u0 + u1 + u2 + u3 + ) +

1 4

1

1 2

−

∆

 + 

 

  u0

= 1

2 (u0 + Eu0 ++ E

2u0 + E3u0 + ) + 1

4

1

1 2

−

∆

 + 

 

  u0

= 1

2 (1 + E + E

2 + E3+ )u0 + 1

4

1

1 2

−

∆

 + 

 

  u0

= 1

2 (1 – E)

–1u0 + 1

2 (2 + ∆)–1 u0

= 1

2 (1 – E)

–1u0 + 1

2 (1 + ∆)–1 u0

= 1

2 [(1 – E)

–1 + (1 + E)–1]u0

= 1

2 2 [1 + E

2 + E4 + E6 + ]u0

= u0 + u2 + u4 + u6 +

0

x x

u

∞

=

∑ = L.H.S

Example 34 Given that u 0 = 3, u 1 = 12, u 2 = 81, u 3 = 200, u 4 = 100, u 5 = 8 Find the value of

∆5u 0

Sol We know ∆ = E – 1, therefore,

5 0

u

∆ =(E –1) 5 u0

= (E5 – 5E4 + 10E3 – 10E2 + 5E – 1)u0

=u5 – 5u4 + 10u3 – 10u2 + 5u1 – u0

= 8 – 500 + 2000 – 810 + 60 – 3

= 755

PROBLEM SET 3.1

1 Form the forward difference table for given set of data:

X: 10 20 30 40

Y: 1.1 2.0 4.4 7.9

2 Construct the difference table for the given data and hence evaluate ∆3f(2)

X: 0 1 2 3 4

3 Find the value of E2x2 when the values of x vary by a constant increment of 2.

[Ans x2 + 8x + 16]

4 Evaluate E n e x when interval of differencing is [Ans E n e x = e x+nh]

5 Evaluate ∆3(1 –x) (1–2x) (1–3x) ; the interval of differencing being unity.

[Ans ∆3f x( )= −36]

6 If f(x) = exp (ax), evaluate ∆n f x( ) [Ans ∆n ax e =(e ah−1)n ax e ]

7 Evaluate

2 3

x E

∆ 

8 Find the value of

2 4 2

2 2 ( 1)

x x

a

9 Evaluate:

10 Prove that:

2

(c) δ = ∇ − ∇(1 )−1/2= ∆ + ∆(1 )−1/2 (d) δ = ∆E−1/2 = ∇E1/2

(e) ∇∆ = ∆∇ = δ2 (f) ∇ = ∆E−1=E−1∆ = −1 E−1

11 Show that:

(a) ∆cot(a bx+ )=sin(a bx−) sin(sinb a b bx)

(b) sin( ) (2 sin ) sin( (( )

n

(c) cos( ) (2 sin ) cos( ( )

n

12 What is the difference between 2

E

∆

 

  u x and

2 2

x x

u

E u

  If u x = x

3 and the interval of

differencing is unity Find out the expression for both [Ans 6h2 (3x – h),

2 3 3

+ + ]

13 If f(x) = e ax , show that f(0) and its leading differences form a geometrical progression.

14 A third degree polynomial passes through the points (0, –1), (1, 1), (2, 1) and (3, 2) Find

15 Prove that ∆ sin–1 x = [(x+1) 1−x2−x 1 (− +x 1) ]2

3.5 FUNDAMENTAL THEOREM ON DIFFERENCES OF POLYNOMIAL

Statement: If f(x) be the nth degree polynomial in x, then the nth difference of f(x) is constant and

∆n+1 f(x) and all higher differences are zero when the values of the independent variables are at

equal interval

Proof: Consider the polynomial f(x) = a0 + a1x + a2x2 + + a n x n (1)

Where n is a positive integer and a0, a1, a2, a n are constants

We know ∆f(x) = f(x + h) – f(x)

On applying the operator ∆ on equation (1), we get

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

⇒ f(x + h) – f(x) = [a0 + a1(x + h) + a2 (x + h)2 + + a n (x + h) n]

–[a0 + a1x + a2x2 + +a n x n]

1 2[( ) ] 3[( ) ] n[( )n n]

⇒ a1b + a2 [2C1xh + h2] + a3 [3 2 3 2 3 1 2 1 2

1 2 ] n[n 1 n nC x n n n n]

1 2 3 n 1 n n n

where b1, b2, b n–1 are constant coefficients

According to equation (2), we have the first difference of equation (1) is again a polynomial

of degree n – 1.

From this we say that ∆f(x) is one degree less than the degree of original polynomial

Again, on taking a difference of equation (2) i.e second difference of equation (1), we get

∆2f x( )=C2+C x C x3 + 4 2+ +n n( −1)h a x2 n n−2 …(3)

This is a polynomial of degree n – 2.

Thus, on continuing this process up to nth difference we get a polynomial of degree zero.

Such that:

( )

n f x

∆ = n n( −1)(n−2) 1.h a x n n n n−

=n h a x! n n 0

= n h a! n n

Hence, we have nth difference of the polynomial is constant and so all higher differences are each zero i.e.

1 ( ) 2 ( ) 0

n+ f x n+ f x

3.6 ESTIMATION OF ERROR BY DIFFERENCE TABLE

Let y0, y1, y2, yn be the exact values of a function y = f(x) corresponding to arguments

x0, x1, x2 , , x n Now to determine error in such a case and to correct the functional values,

let an error δ is made in entering the value of y3 in the table so that erroneous value of y3 is

y3 + δ

2

2

2

2

2

4

2

y

∆

∆

∆

From the above difference table we noted that:

1 The error in column y affects two entries in column ∆y, three entries in column ∆2y and

so on i.e the error spreads in triangular form.

2 The error increases with the order of differences

3 The coefficients of δ’s are binomial coefficients with alternative signs +, –,

4 In various difference columns of the above table the algebraic sum of the errors is zero.,

5 The errors in the column ∆i y are given by the coefficients of the binomial expansion

(1 –δ)i

6 In even differences columns of ∆2y, ∆4y, , the maximum error occurs in a horizontal line

in which incorrct value of y lies.

7 In odd difference columns of ∆1y, ∆3y, , the maximum error lies in the two middle terms and the incorrect value of y lies between these two middle terms.

Example 1 Find the error and correct the wrong figure in the following functional values:

Sol

x y

∆ ∆ ∆

−

−

3

13

Here the sum of all the third differences is zero and the adjacent values –3, 3 are equal in magnitude Also horizontal line between –3 and 3 points out the incorrect functional value 18

Therefore coefficient of first middle term on expansion of (1 – p)3 = –3

∴ Correct functional value = 18 – 1 = 17

Example 2 Find and correct by means of differences the error in the following table:

20736, 28561, 38416, 50625, 65540, 83521, 104976, 130321, 160000

Sol For the given data we form the following difference table:

−

−

20736

7825

29679 160000

→

←

From this table we have the third differences are quite iregular and the irregularity starts

around the horizontal line corresponding to the value y = 65540.

Since the algebraic sum of the fifth differences is 0, therefore –5ε = –20 ⇒ ε = 4

Therefore the true value of y5 = 65540 – 4 = 65536

Example 3 Locate the error in following entries and correct it.

1.203, 1.424, 1.681, 1.992, 2.379, 2.848, 3.429, 4.136

Sol Difference table for given data is as follows:

−

−

1203

221

707 4136

Sum of all values in column of fourth difference is –0.004 which is very small as compared

to sum of values in other columns

Errors in this column are e, –4e, 6e, –4e and e.

Term of Maximum value = 24 ⇒ 6e = 24 ⇒ e = 4

Error lies in 2379

Hence, required correct entry = 2379 – 4 = 2375

Hence, correct value = 2.375

→

←

Example 4 One number in the following is misprint Correct it.

1 2 4 8 16 26 42 64 93.

Sol Difference table for given data it as follows:

−

−

−

−

−

1

29

In the above table, the fourth difference column have algebraic sum of all the values is 0 The middle term of this difference column is 6

∴ 6e = 6 or e = 1.

∴ Correct value is given by 16 – 1 = 15

Example 5 Locate the error in the following table:

1.0000 1.5191 2.0736 2.6611 3.2816 3.9375 4.6363 5.3771 6.1776 7.0471 8.0000

x

y