A textbook of Computer Based Numerical and Statiscal Techniques part 32 pps

6.2.3 Derivatives Using Stirling’s FormulaIf we want to determine the values of the derivatives of the function near the middle of the given set of arguments.. We may apply any central d

Aliter: We know that

+ ∆ = = hD, = d

dx

∴ hD =log(1+ ∆)

= ∆ −∆2 +∆3 −∆4 +∆5 −∆6 +

1

D h

(1)

2

2 2

1

D h

= ∆ − ∆ + ∆ 

2

12

3

3 3

1

D h

= ∆ − ∆ + 

3

2

Applying equations (1), (2) and (3) for y0, we get

=

x x

dy

and

=

 

2

12

x x

d y

=

3

2

y

x x

d

6.2.2 Derivatives Using Newton’s Backward Difference Formula

Newton’s backward interpolation formula is given by

n n (2!1) 2 n ( 1)(3! 2) 3 n

y=y + ∇ +u y + ∇ y + + + ∇ y

(1)

where x x n

u

n

−

Differentiating both sides of equation (1) with respect to x, we get

2

At x=x u n, =0 Therefore putting u = 0 in (2), we get

n

x x

dy

Again differentiating both sides of equation (2) w.r.t x, we get

At x=x u n, =0 Therefore putting u = 0 in (3), we get

2

12

n

x x

d y

Similarly,

3

2

n

x x

d y

 

Aliter: We know that

1

1− ∇ =E− =e−hD

∴ −hD=log (1− ∇)

1

D h

and D2 12 2 12 3 1112 4

h

2

D h

Applying these identities to y n, we get

n

x x

dy

 

=

 

 

2

12

n

x x

d y

and

3

2

x x

d y

 

 

6.2.3 Derivatives Using Stirling’s Formula

If we want to determine the values of the derivatives of the function near the middle of the given set of arguments We may apply any central difference formula Therefore using Stirling’s formula, we get

n

y =y +u∆ + ∆ − + ∆ y− + − ∆ − + ∆ − + − ∆ y− +

where u x x0

h

−

Now, differentiating w.r.t x, we get

Since u x x0

h

−

=

dx= h

∴

At x=x u0, =0,therefore, putting u = 0 in (2), we get

1

0

x x

−

∆ + ∆

Again differentiating, we get

At x=x u0, =0 therefore, putting u=0 in (3), we get

0

2

12

x x

d y

 

6.2.4 Derivative Using Newton’s Divided Difference Formula

Newton’s divided difference formula for finding the successive differentiation at the given

value of x Let us consider a function f(x) of degree n, then

y= f x = f x + −x x ∆f x + −x x x x− ∆ f x + −x x x x− x x− ∆ f x

(x x )(x x ) (x x n− ) n f x( )

Differentiate this equation w.r.t ‘x’ as many times as we require and put x=x, we get the required derivatives

Example 1 Find dy dx at x = 0.1 from the following table:

Sol Difference table:

–0.0075

–0.0172

Here, x0 =0.1,h=0.1 and y0 =0.9975 we know that, Newton’s forward difference formula

=

 

0.1

x

dy

0.0075 0.0049 0.0001

Example 2 Using following table.

Find dy

dx and

2 2

d y

dx at x = 1.1.

Sol Since the values are at equidistant and we want to find the value of y at x = 1.1.

Therefore, we apply Newton’s forward difference formula

Difference table:

0.414

0.281

We have,

=

x x

dy

Putting x0 =1.1,∆ =y0 0.378,∆2y0=0.030,and so on and h = 0.1, we get

1.1

0.378 0.030 0.004 0.001 0.001

dy dx

  =10 0.378 0.015 0.0013 0.00025 0.0002[ + + + − ]

=10 0.39435[ ]=3.9435

and

 

  0

2

x

d y

2

1.1

0.030 0.004 0.001 0.001

(0.1)

d y dx

=100[−0.030 0.004 0.0009 0.0008− − + ]=100(−0.0341)

= −0.341 Ans

Example 3 Using the given table, find dy/dx at x = 1.2

0.6018

1.6359

We have

=

x x

dy

0 1.2, 0 0.7351, 0 0.1627,

x = ∆ =y ∆ y = and so on and h=0.2, we get

1.2

0.7351 0.1627 0.0361 0.0080 0.0014

dy dx

  =5 0.7351 0.08135 0.0120 0.002 0.00028[ − + − + ]

=3.32015 Ans

Example 4 Find dy dx and

2 2

d y

dx of

1/3

y=x at x = 50 from the following table:

1/3

0.0244

0.0229

Here, x0 =50,h=1.Then, we have Newton’s forward difference formula

0

x x

dy

 

50

0.0244 0.0003 0

dy dx

=(0.0244 0.00015+ )=0.02455

0

2

1

x x

d y

12[ ]

( 0.0003) (1)

=

2 2 50 0.0003

d y dx

= −

Example 5 The table given below reveals the velocity v of a body during the time t Find its,

acceleration at t = 1.1.

Sol DifDifDifferferference table:ence table:

4.6

4.4

We have, t0 =1.1,v0 =4.77 and h=0.1

Then the acceleration at t=1.1 is given by

0

t t

dv

 

1 1( ) ( )1

1 4.4 0.5 0.2

3

=45.167 (approx.)Ans

Example 6 Find f′(1.1) and f″(1.1) from the following table :

( )

Sol Difference table

x f x( ) ∆f x( ) ∆2f x( ) ∆3f x( ) ∆4f x( )

0.1280

1.5680

Here we have to find the derivatives at x=1.1 which lies between given arguments 1.0 and 1.2.

So apply Newton’s forward formula, we have

0 1

5( 1) 0.2

h

0

4(4 1)

f x = f x + ∆u f x + − ∆ f x + − − ∆ f x (2)

Differentiating w.r.t x, we get

( ) ( ) ( ) 2 ( ) 2 3 ( )

dx

Also, from (1) du 5

dx=

5

At x =1.1, u = 5(1.1 – 1) = 0.5

2 0.5 1 3 0.5 6 0.5 2

=5 0.128 0 0.002[ + − ]

( )1.1 0.63

Differentiating equation (2) again w.r.t ‘x’, we get

6 6 5

6

dx

−

∴ f′′( )1.1 =25 0.288 +(0.5 1− ×) 0.04815 =25 0.288 0.024[ − ]

Hence, f′′( )1.1 =6.6 Ans

Example 7 Find f′(1.5) from the following table:

Sol Here we want to find the derivatives of f (x) at x = 1.5, which is near the end of the

arguments Therefore, apply Newton’s backward formula

Difference Table

x f x( ) ∇f x( ) ∇ 2f x( ) ∇ 3f x( ) ∇ 4f x( )

–0.0468

0.0755

Here, x n = 1.5, h = 0.5

n

x x

dy

1.5 0.1125 0.0024 0.0609

2 0.1125 0.0012 0.0152251 2 0.098475

⇒ f′( )1.5 = −0.19695 Ans

Example 8 From the following table, find the values of dy dx and

2 2

d y

dx at x = 2.03.

Sol Difference table

–0.0086

–0.0090