A textbook of Computer Based Numerical and Statiscal Techniques part 17 ppsx

x–n – 1, n > 0 Because of their properties, this function play an important role in the theory of finite differences and also it helps in finding the various order differences of a polyn

=

x

=

[1 ( 1)]

x

x x

E

−

 − + 

= (1 + ∆E) x u0 = (1 + x C1∆E + x C2∆2E2 + )u0

0 x 1 0 x 2 0

=

2



3

6 7/2 (1/2)(3/2)(5/2) 1

 



1



3

2 1 ( 1/2)( 3/ 2)( 5/2) 1

= 1 1/2[1 1 2 1]1/2

=

1/2 2

2 [4(1 ) ]

E

E

−

= [(2+ ∆) ]2 −1/2u x= + ∆(2 )−1u x = +(1 E)−1u x

= [1− +E E2−E3+E4−E5+ ]u x

= u x−u x+1+u x+2−u x+3+u x+4−u x+5+ = L.H.S

3 R.H.S = (1 + x) n u0 + n C1 (1 + x) n–1 x∆u0 + n C 2 (1 + x) n–2 x 2 2

0

u

∆ +

= {(1 + x) + x∆} n u 0 = (1 + x (1 + ∆)} n u 0

= (1 + xE) n u 0 = [1 + n C 1 xE + n C 2 x 2 E 2 + n C 3 x 3 E 3 + ]u 0

= u 0 + n C 1 u 1 x + n C 2 u 2 x 2 + n C 3 u 3 x 3 + = L.H.S.

Example 10 Prove that x n 1 2 n x 1 3.. 3 n x 1 3 5 . . 4 n x

∆ − ∆ + ∆ − ∆ + = x+ 1n−x−1n

n x

 −  −  

   

1

2

n

 + −  − −  = +  − − 

3.9 FACTORIAL NOTATIONS

The product of n consecutive factors each at a constant difference and the first factor being x is called a factorial function or a factorial polynomial of degree n and is defined by

x (n) = x(x – h) (x – 2h) (x – 3h) (x – (n – 1)), n > 0

If interval of differencing being unity then

x (n) = x(x – 1) (x – 2) (x – 3) (x–(n – 1)), n > 0

Because of their properties, this function play an important role in the theory of finite differences and also it helps in finding the various order differences of a polynomial directly by simple rule of differentiation

Example 11 Obtain the function whose first difference is 9x 2 + 11x + 5.

Sol Let f(x) be the required function so that

( )

f x

∆ = 9x2 + 11x + 5

Let 9x2 + 11x + 5 = 9 [x]2 + A [x] + B

= 9x (x – 1) + Ax + B

On substitution x = 0, we get B = 5 and for x = 1, we get A = 20.

Therefore we have ∆f x( ) = 9[x]2 + 20 [x] + 5

On integrating, we have

f (x) = [ ]3 + [ ]2 + [ ]+

= 3x(x – 1) (x – 2) + 10x (x – 1) + 5x + c

= 3x3 + x2 + x + c, where c is the constant of integration Example 12 Find a function f x for which ∆f x = x(x – 1).

Sol We have ∆f x = x(x – 1) = x(2)

( ) 3 3

x

+ C, where C is an arbitrary constant

3x x− x− +C

Example 13 Prove that if m is a positive integer then ( )( ) ( ) ( )

.

− +

− Sol To prove this we start from R.H.S and use factorial notations to simplify

1

− +

− =

+

−

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

!

m

 − + + 

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

!

m

= ( 1)

!

m x m

+

Example 14 Prove that 2 ( ) ( 2 )

∆ = − ; where m is a positive integer and interval of differencing being unity.

Sol We know, x( )m = x x( −1)(x−2) (x m− −1)

Therefore, ∆x( )m = {(x+1) (x x−1)(x−2) (x+ − −1 m 1)} {− x x( −1) (x m− −1)}

= x x( −1)(x−2) (x m− −2){x+ − − −1 (x m 1)}

= mx (m–1)

Also, ∆2 ( )x m = ∆ ∆( x( )m = ∆{mx(m−1)}

= m x∆ (m−1) = m m( −1)x(m−2) Hence proved

Example 15 Evaluate 1

ax bx −

Sol ∆n(ax n+bx n−1) = a∆n(x n)+ ∆b n(x n−1)

= an!+b(0)=an!

Example 16 Denoting ( )x

n

x(x 1) (x n 1)

, n!

= prove that for any polynomial φ(x) of degree

=

=∑k x i

i

i 0

Sol We have E n f(a) = f(a + nh)

( ) n ( ) n ( ) n n n ( )

Substitute a = 0, n = x we have for h = 1

(0) x (0) x (0) x x x (0)

Again f(x) = φ(x) is the given polynomial of degree k therefore we have ∆ φk ( )x =L (constant) Therefore higher order differences become zero

∴ 2

( )x (0) x C (0) x C (0) x C k k (0)

= ( )

1

(0)

k

x i i

i=

∆ φ

∑

3.10 RECIPROCAL FACTORIAL NOTATION

The reciprocal factorial is denoted by x(−n) and is defined by

( n)

(x h x)( 2 ) (h x nh)=(x nh)n

If the interval of differencing being unity, then

( n)

(x 1)(x 2) (x n)=(x n)n

Example 17 Prove that ∆2 ( m) x− =m(m 1)x+ ( m 2)− −

Sol We have, x(−m) = (x+1)(x+12) (x m+ )

(x 2)(x 3) (x m 1) (− x 1)(x 2) (x m)

(x 2)(x 3) (x m) (x m 1) (x 1)

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

m

−

= −mx[ (− +m 1)]= −mx(− −m 1) Therefore, ∆2 (x−m) = ∆ ∆( x(−m))= ∆ −( mx(− −m 1))

= − ∆m x(− −m 1)= − − −m m( 1)x(− −m 2)

= m (m + 1) x (–m–2)

Example 18 Express f(x) = ( + ) (− + )

x 1

x 1 x 3 in terms of negative factorial polynomials.

( 1)(1 3)

x

−

Multiply and divide by (x + 2)

= ( (1)(1)(2)(2) 3)

= x11−(x 1)(4x 2) (+ x 1) (x42)(x 3)

= x(–1) – 4x(–2) + 4x(–3)

Corollary:ollary:

To show that, ∆n x( )n = n! h n and ∆n+ 1 x (n) = 0

Since, (x + h) (n) = (x + h) x (x – h) (x – 2h) {x – (n–2)h}

(x) (n) = x(x – h) (x – 2h) {x – (n – 1)h}

Therefore from (1), we have

⇒ (x + h) (n) – (x) (n) = (x + h) x (x – h) (x – 2h) {x – (n – 2)h}

– x (x – h) ( x – 2h) {x – (n – 1)h}

or (x + h) (n) – (x) (n) = x (x – h) (x – 2h) {x – (n – 2)h} [(x + h) –{x – (n – 1)h}]

This is a polynomial of degree (n – 1) in factorial notation.

Similarly, 2 ( )n

x

∆ = ∆ ∆ x( )n 

  = ∆ nhx(n−1 ) [using (a)]

or nh∆x(n−1 ) = nh ( )(n 1 ) (n 1)

= nh[(x + h) x (x – h) {x – (n – 3)h} – x (x – h) {x – (n – 2)h}]

= nh[x(x–h) {x – (n – 3) h}] (nh – h)

= nh (n – 1) h [x (x – h) (x – 2h) {x – (n – 3)h}]

= n(n – 1) h2 x (n–2)

This is a polynomial of degree (n – 2) in factorial notation Proceeding in the same way,

we get

( )n

n x

∆ = n (n – 1) (n – 2) 3.2.1.h n x n–n

= n! h n x0= n! h n This term is a constant

Again, ∆n+1x( )n = ∆ ∆ n x( )n 

= ∆ n h! n

= 0 Because n!h n is constant

3.11 METHOD OF REPRESENTING POLYNOMIAL IN FACTORIAL NOTATIONS

3.11.1 Direct Method

Example 19 Express 2x 3 – 3x 2 + 3x – 10 and its differences in factorial notation, consider the interval of differencing being unity.

Sol Let 2x3 – 3x2 + 3x – 10 = Ax(3) + Bx(2) + Cx(1) + D

= Ax (x – 1) (x – 2) + Bx (x – 1) + Cx + D (1) Where A, B, C and D are constants to be found

Now substitute x = 0 on both the sides of equation (1), we get D = – 10.

Again, substituting x = 1 on both sides of equation (1), we get

2 – 3 + 3 – 10 = C + D i.e., C = 2

Again substituting x = 2, we get

16 – 12 + 6 – 10 = 2B + 2C + D

or 0 = 2B – 6 i.e., B = 3.

By equating the coefficient of x2 on both sides of (1), we get A = 2

Hence, the required polynomial in factorial notation will be

f(x) = 2x(3) + 3x(2) + 2x(1) – 10 Again by the simple rule of differentiation, we have

∆f x( ) 6= x(2)+6x+2

2f x( ) 12x 6

3f x( ) 12

Example 20 Find the relation between α, β and γ in order that α β+ x+γx 2 may be expressible

in one term in the factorial notation.

Sol Let, f x( )= α + β + γ = +x x2 (a bx)(2) where a and b are certain unknown constants.

Now, (a bx+ )(2) = (a bx a b x+ )[ + ( −1)]

= (a bx a b bx+ )( − + ) (= +a bx)2− −ab b x2

= (a2−ab) (2+ ab b x b x− 2) + 2 2= α + β + γx x2

Comparing the coefficients of various powers of x we get

α = − β = − γ =

Eliminating a and b from the above equations, we get

γ + αγ = β This is the required relation

3.11.2 Method of Synthetic Division

It is also called the method of detached coefficient

Example 21 Represent 2x 3−3x 2+3x 10− in factorial notation using synthetic division method.

Sol Let, f x( ) 2= x3−3x2+3x−10=Ax(3)+Bx(2)+Cx(1)+D

=Ax x( −1)(x− +2) Bx x( − +1) Cx D+

In this case we divide the function f x( ) by x Then the remainder will be –10 and the quotient will be 2x2−3x+3

Now divide the quotient by (x – 1) i.e.,

−

−

− +

− +

2 2

2 1 ( 1) 2 3 3

3 1 2

x

x x

In this case quotient is (2x – 1) and remainder is 2 = C Again divide (2x – 1) by (x – 2),

2 ( 2) 2 1

2 4 3

x

−

Therefore the quotient 2 = A and the remainder are 3 = B Hence the required polynomial

will be

(3) (2)

2x +3x +2x−10

We can also simplify the above in the following way:

Taking the coefficients of various powers of x in f(x), we have

0

2 = A

The following steps are to be followed in the method of synthetic division:

1 Put the coefficients of different powers of x in order beginning with the coefficients of higher power of x.

2 Put 1 in the left hand side column and write zero below the coefficients of highest power

of x (in this case we have written zero below 2 which is coefficient of x3)

3 Now multiply 2 by 1 and 0 by 1, add them to get the sum 2, put 2 below –3 as given in

(a) Now multiply –3 by 1 and 2 by 1 and add them to get the sum –1 which is to be written below 3 The remainder –10 is the value of D.

4 Add the terms of corresponding columns of (a) and get 2, –1 and 2 of (b)

5 Now again apply the steps (1) and (3): in this way we get (2 × 2) + (0 × 2) = 4 which

is to be written below –1 The remainder 2 of (b) is equal to C.

6 Apply step (4) on (b) and get 2 and 3 of (c).

7 Again apply the steps (2), (3) and (4) to get 2 which will be equal to A and remainder 3

of (c) which will be equal to B.

Example 22 Express f(x) = x 4 – 12x 3 + 24x 2 – 30x + 9, and its successive differences in factorial notation Hence show that ∆5f x( ) = 0.

Sol Let f(x) = A[x] 4 + B [x] 3 + C[x] 2 + D[x] + E.

Use method of synthetic division, we divide by x, x – 1, x – 2, x – 3 etc successively, then

3

1 = A

Hence, f(x) = [x] 4 – 6[x] 3 – 5[x] 2 – 17[x] + 9

∴ ∆f x( ) = 4[x]3 – 18[x]2 – 10[x] – 17

( )

2f x

∆ = 12[x]2 – 36[x] – 10

( )

3f x

∆ = 24[x] – 36 ⇒ ∆4f x( )= 24

Example 23: Find the lowest degree polynomial which takes the following values:

( )

Sol We know that

f (a + nh) = f (a) + n C1 ∆f a( ) + n C2∆2f a( ) + + n C n∆n f a( ) (1) Putting a = 0, h = 1, n = x, we get

f(x) = f(0) + x C1 ∆f( )0 + x C2∆2f( )0 + x C3∆3f( )0

= f(0) + x(1) ∆f( )0 +

( ) 2 2!

x

( )

2f 0

( ) 3 3!

x

( )

3f 0

Now form the difference table for the given data to find ∆f (0), ∆2f(0), ∆3f(0) etc.

3

11

Substituting the values of f(0),∆f(0),∆2f(0),∆3f(0) in (2), we get

f(x) =

(2) (1)

2!

x x

= 3x x x+ ( − =1) x2+2x

Example 24 A second degree polynomial passes through the points (0,1), (1,3), (2,7), (3,13) Find the polynomial.

Sol Let f(x) = Ax2 + Bx + C.

Now form the difference table for given points:

2

4

6

Since ∆f x( ) = A x∆ + ∆ + ∆2 B x C

Therefore, ∆f x( ) = A x{( +1)2−x2}+B x( + − +1 x) 0

= A x(2 + +1) B

Put x = 0, ∆f(0) = A B+ ⇒ + =A B 2

Again, ∆2f x( ) = 2A ⇒ ∆2f(0)= =2 2A⇒ =A 1

Therefore polynomial for the given point is f(x) = x2 + x + 1.

Example 25 Write down the polynomial of lowest degree which satisfies the following set of numbers 0, 7, 26, 63, 124, 215, 342, 511.

Sol We know

By forming the difference table we calculatef(0),∆f(0),∆2f(0),∆3f(0) etc

7

169

Now substituting the values off(0),∆f(0),∆2f( )0 ,∆3f(0)from the difference table, we obtain

f(x) = 0 7 (1) 12 (2) 6 (3)

= 7x+6 (x x− +1) x x( −1)(x−2)

= 7x+6x2−6x x+ 3−3x2+2x

= x3+3x2+3x