A textbook of Computer Based Numerical and Statiscal Techniques part 33 pot

Find f′ 5 from the following table: Sol.. So we use Newton’s divide difference formula... Here, the arguments are not equally spaced and therefore we shall apply Newton’s divided differ

Here, x n 2.04,h 0.02u x x n du 1atx 2.03

−

∴ u=2.03 2.040.02− = −0.010.02= −12

Then, by Newton’s backward formula, we have

Differentiating w.r.t x, we have

( ) 1 2 1 2 3 2 6 2 3 4 3 18 2 22 6 4

h

( )

2

y









24



+





 =50 0.0090 0.000008 0.000017[− + + ]= −0.44875 Ans

Again differentiating equation (2) w.r.t x,

2

h

2

2

(0.02)

y

=2500[−0.0002 0.0001 0.00012− − ]

y′′(2.03)= −1.05. Ans

Example 9 Find f′( )5 from the following table:

Sol Here the arguments are not equally spaced So we use Newton’s divide difference formula

Difference table

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

1

3

3

1 144

−

16

−

6 3

Newton’s divided difference formula is given by:

2

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 x x f x

Differentiating (1) w.r.t x, we get

f x′ = ∆f x + x x− −x ∆ f x + x x− x x− + −x x x x− + −x x x x− ∆ f x +

At x = 5

1 (5) 1 (10 1 2) [(5 2)(5 4) ((5 1)(5 4) (5 1)(5 2)] 0

3

[(5 2)(5 4)(5 8) (5 1)(5 4)(5 8) (5 1)(5 2)(5 8) (5 1)(5 2)(5 4)] 1

144

−

1 7 1 1 [ 9 12 36 12] 1 7 45

  Hence f′(5) 3.6458.= Ans

Example 10 Find f′′′(5) from the data given below:

Sol Here, the arguments are not equally spaced and therefore we shall apply Newton’s divided

difference formula

2

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 x x f x

x f (x) ∆f x( ) ∆ 2f x( ) ∆ 3f x( ) ∆ 4f x( ) ∆ 5f x( ) ∆ 6f x( )

644

2119372

Substituting values in eqn (1), we get

f x( )=57 (+ −x 2)(644) (+ −x 2)(x−4)(1765) (+ −x 2)(x−4)(x−9)(556)

+ −(x 2)(x−4)(x−9)(x−13)(45) (+ −x 2)(x−4)(x−9)(x−13)(x−16)(1)

=57 644(+ x− +2) 1765(x2−6x+ +8) 556(x3−15x2+62x−72)

+45(x4−28x3+257x2−878x+936)+x5−44x4+705x3−4990x2

+14984x−14976

f x′ = + x− + x − x+ + x − x + x−

+5x4−176x3+2115x2−9980x+14984

2

f′′′x = + x− + x − x+ =60x2+24x+6

Where x = 5;

2

f′′′ = + + = Ans

Example 11 Find f′ (4) from the following data:

( )

Sol Though this problem can be solved by Newton’s divided difference formula, we are giving here, as an alternative, Lagrange’s method Lagrange’s polynomial, in this case, is given by

e j e j

( 0)( 2)( 1)(125) ( 0)( 2)( 5)(1)

x− x− x− x− x− x−

4( 3 6 2 5 ) 25( 3 3 2 2 ) 1( 3 7 2 10 ) 3

∴ f x′( ) 3= x2

When x = 4, f′(4) = 3(4)2 = 48 Ans

Example 12 Find f′(0.6)and f′′(0.6)from the following table:

( )

Sol Here, the derivatives are required at the central point x = 0.6, so we use Stirling’s

formula

Difference table

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

0.2138

0.3235

Here we have x0 0.6,h 0.1,u x x0 at x 0.6,u 0

h

−

Stirling’s formula is

( )

Differentiating (1) w.r.t x, we get

h

e j e j

Using difference table, we have

∆ =y0 0.2468,∆y−1=0.2833, ∆2y−1=0.0365, ∆3y−1=0.0035,

∆3y−2=0.0037, ∆4y−2=0.0002,

0.2468 0.2833 1 0.0035 0.0037 1

=10[0.26505 0.0006]−

= f’(0.6) = 2.6445 Ans

Again differentiating (1), we get

2

h

2

12 (0.1)

=100[0.0365 0.000016]− Hence, f′′(0.6)=3.6484 Ans

Example 13 Find f′(93)from the following table:

( )

Sol Difference table

( )

f x

( )

f x

( )

f x

∆

–3.2

Here we have x0 =90,x=93,h=15

x x u h

Now using Stirling’s formula

f x = =y y +u∆ + ∆ − + ∆ y− + − ∆ − + ∆ − + − ∆ y− +

Differentiating (1) w.r.t x, we get

h

Putting the values of x = 93, u = 0.2, h = 15 and

We get,

1.46

×

1 1.35 1.46 0.30065 0.1334 15

(93) 0.03627

Example 14 Find x for which y is maximum and find this value of y

Sol The Difference table is as follows:

1.2 0.9320

0.0316

0.0021 1.6 0.9996

Let y0=0.9320 and a = 1.2

By Newton’s forward difference formula

0 0 2 0

2

u u

y=y + ∆ +u y − ∆ y +

0.9320 0.0316 ( 1)(–0.0097)

2

u u

= + + (Neglecting higher differences)

dy du=0.0316+ 2u2−1(–0.0097)

At a maximum,

dy 0

du=

2

∴ x=x0+hu=1.2 (0.1)(3.76) 1.576+ =

To find ymax,we use backward difference formula,

x = x n + hu

(1.576) ( 1) 2 ( 1)( 2) 3

0.9996 (0.24 0.0021) (–0.24)(1 – 0.24)(–0.0099)

2

=0.9999988 0.9999= nearly

∴ Maximum y = 0.9999 (Approximately) Ans

Example 15 From the following table, for what value of x, y is minimum Also find this value of y.

Sol Difference table

0.035

–0.026

Now taking x0 = 3, we have y0 =0.205, ∆ =y0 0.035, ∆2y0= −0.016 and ∆3y0=0.

Therefore, Newton’s forward interpolation formula gives

0.205 (0.035) ( 1)( 0.016)

2

u u

(1)

Differentiating (1), w.r.t u, we get

0.035 2 1(–0.016)

2

du

−

For y to be minimum put dy 0

du=

⇒ 0.035 – 0.008(2u− =1) 0

⇒ u=2.6875

Therefore, x=x0+uh

= +3 2.6875 1 5.6875× =

Hence, y is minimum when x = 5.6875.

Putting u = 2.6875 in (1), we get the minimum value of y given by

0.205 2.6875 0.035 1(2.6875 1.6875)(–0.016)

2

= 0.2628 Ans

PROBLEM SET 6.1

1 Find f′(6) from the following table:

( )

x

[Ans –23]

2 Use the following data to find f ′(3):

( )

x

[Ans 1.8828]

3 Use the following data to find f ′(5):

( )

x

f x

[Ans 2097.69]

4 From the table, find dy

dx at x = 1, x = 3 and x = 6:

( )

x

f x

[Ans 0.3952, 0.3341, 0.2719]

5 Find f ′(5) and f ′′(5) from the following data:

( )

x

f x

6 Using Newton’s Divided Difference Formula, find f ′(10) from the following data:

( )

x

[Ans 232.869]

7 From the table below, for what value of x, y is minimum? Also find this value of y.

x

[Ans 5.6875, 0.2628]

8 A slider in a machine moves along a fixed straight rod Its distance x (in cm.) along the rod is given at various times t (in secs.)

t

x

Evaluate dx

dt at t = 0.1 and at t = 0.5 [Ans 32.44166 cm/sec.; −24.05833 cm/sec.]

9 A rod is rotating in a plane The following table gives the angle θ (radians) through which

the rod has turned for various values of the time t (seconds).

t

θ

Calculate the angular velocity and acceleration of the rod when t = 0.6 sec.

[Ans (i) 3.82 radians/sec (ii) 6.75 radians/sec2]

10 The table given below reveals the velocity ‘v’ of a body during the time ‘t’ specified Find its acceleration at t = 1.1.

t v

[Ans 44.92]

6.3 NUMERICAL INTEGRATION

Like numerical differentiation, we need to seek the help of numerical integration techniques in the following situations:

1 Functions do not possess closed from solutions Example:

2

– 0 ( )

x t

f x =C e∫ dt

2 Closed form solutions exist but these solutions are complex and difficult to use for calculations

3 Data for variables are available in the form of a table, but no mathematical relationship between them is known as is often the case with experimental data

Let y= f x( ) be a function, where y takes the values y0, y1, y2, y n for x = x0, x1, x2, x n We want to find the vaule of ( )

b

a

I=∫f x dx

Let the interval of integration (a, b) be divided into n equal subintervals of width h b a

n

−

=  so that x0=a x, 1= +x0 h x, 2= +x0 2hT, ,x n= + =x0 nh b

∴ 0

0

nh

x b

+

Newton’s forward interpolation formula is given by

Where x x0

u h

−

=

∴ du 1dx dx hdu

h

∴ Equation (1) becomes,

0

n

I=h y + ∆ +u y − ∆ y + − − ∆ y + du

∫

2

up to( 1) terms

∴

0

2

n

x

x

This is called general quadrature formula