A textbook of Computer Based Numerical and Statiscal Techniques part 21 docx

Use Gauss’s forward formula to find a polynomial of degree four which takes thefollowing values of the function fx: Sol.. Use Gauss’s forward formula to find y30 for the following data..

Take the mean of equation (1) and (2)

f u = + + − + ∆f + − ∆ + ∆ − 

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

( 1)

+

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

u+ u u− u−  − ∆ − + + ∆ −  +

1 ( 1)

−  − 

5

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

( 1) ( 1)( 2)

( 2)

f

This formula is very useful when 1

2

u= and gives best result when 1 3

4< <u 4

4.4.5 Laplace-Everett’s Formula

Gausss forward formula is given by

( 2)( 1) ( 1)( _ 2) ( 1) ( 1)( 2)( 3)

We know

(0) (1) (0) ( 1) (0) ( 1) ( 2) ( 1) ( 2)

∆ − = ∆ − − ∆ − Therefore, using this in equation (1), we get

f u = f +u f −f + − ∆ f − + + − ∆ f − ∆ f −

(1 ) (0) (1) ( 1) 2 ( 1) ( 1) ( 1) 2 ( 1)

+( 1) ( 1)( 2) 4 ( 2) ( 2)( 1) ( 1)( 2) 4 ( 2)

(1 ) (0) (1) ( 1)(2 ) 2 ( 1) ( 1) ( 1) 2 (0)

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

Substitute 1 – u = w in second part of equation (2)

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

This is called Laplace-Everett’s formula It gives better estimate value when 1

2

u>

Example 1 From the following table, find the value of e 1.17 using Gauss forward formula:

x

x

e = f x

( ) 2.7183 2.8577 3.0042 3.1582 3.3201 3.4903 3.6693 Sol The difference table is as given:

1 2.7183

0.1394

0.179 1.30 3.6693

Now, let taking origin at 1.15, here h = 0.05

Then, 1.17 1.15 0.4

0.05

On applying Gauss’s forward interpolation, we have

( ) (0) (0) ( 1) 2 ( 1) ( 1) ( 1) 3 ( 1) ( 1) ( 1)( 2) 4 ( 2)

f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

5 ( 2)( 1) ( 1)( 2)

( 2) 5!

f

(0.4)( 0.6) (1.4)(0.4)( 0.6)

(1.4)(0.4)( 0.6)( 1.6)(2.4)

120

(0.4) 3.1582 0.06476 0.000948 0.0000224 0.0000010752

(0.4) 3.22199 (Approx.)

Example 2 Given that

x

x

log 1.39794 1.47712 1.54407 1.60206 1.65321

log 3.7 = ?

Sol

2

1

0

1

2

25 1.39794

0.07918

0.05115

45 1.65321

x

x

x

x

x

−

−

−

−

x = a + hu, x = 37, a = 35, h = 5

37 35

0.4

x a− = − =

2 3 4

1.54407 0.4 0.05798 (0.4)(0.4 1) ( 0.00896) (0.4)(0.4 1)(0.4 1) (0.00212)

(0.4)(0.4 1)(0.4 1)(0.4 2)

( 0.00115) 24

( ) 1.54407 0.023192 0.0010752 0.00011872 0.00002576

Since log 3.7 = log3.7 10 log37

⇒ log 3.7 = log 37 – log 10 = 1.56819272 – 1 = 0.56819272 Ans

Example 3 From the following table find y when x = 1.45.

x

Sol

−

−

−

0.112

1.008

u u u

f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

u=1.45 1.40.2− =0.25

(0.25)( 0.75) (0.25)( 0.75)(1.25)

= 0.047875

Example 4 Use Gauss’s forward formula to find a polynomial of degree four which takes the

following values of the function f(x):

Sol Taking center at 3 i.e., x0 = 3 and h = 1

h

−

Now, for the given data difference table becomes:

−

−

−

−

−

−

2

2

Gauss forward formula is

f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −

=1 2 6 2 2 14 24 4 3 12 2 104 32 2 4 28 3 142 2 308 80

Example 5 Use Gauss’s forward formula to find y30 for the following data.

18.4708 17.8144 17.1070 16.3432 15.5154 Sol Let us take the origin at x = 29

30 29 1

0.25

Now, for the given data difference table is:

−

−

−

−

21 18.4708

0.6564

0.8278

37 15.5154

Putting these values in Gauss forward interpolation formula, we have

0.25

(0.25)( 0.750) 17.1070 (0.25) ( 0.7638) ( 0.0564)

2

+ (0.25)( 0.750)(1.25) ( 0.0076) (1.25)(0.25)( 0.750)( 1.75) ( 0.0022)

y0.25 = 17.1070 – 0.19095 + 0.0052875 + 0.00002968 – 0.00000375

y0.25 = 16.9216 Ans

PROBLEM SET 4.3

1 The values of e –x at x = 1.72 to x = 1.76 are given in the following table:

( )

0.17907 0.17728 0.17552 0.17377 0.17204

x

f x

Find the values of e–1.7425 using Gauss forward difference formula

[Ans 0.1750816846]

2 Apply Gauss’s forward formula to find the value of f(x) at x = 3.75 from the table:

( )

24.145 22.043 20.225 18.644 17.262 16.047

x

f x

[Ans 19.407426]

3 Apply Central difference formula to obtain f(32) Given that :

f(25) = 0.2707, f(30) = 0.3027, f(35) = 0.3386, f(40) = 0.3794. [Ans 0.316536]

4 Apply Gauss forward formula to find the value of U9, if

5 Apply Gauss forward formula to find a polynomial of degree three which takes the values

of y as given on next page:

x

3

6 x 2x 3x

6 Use Gauss’s forward formula to find the annuity value for 27 years from the following data:

10.3797 12.4622 14.0939 15.3725 16.3742 17.1591

Year

Annuity

[Ans 14.643]

7 Use Gauss’s forward formula to find the value of f(x), when 1,

2

x= given that:

( )

x

f x

−

GAUSS BACKWARD

Example 1 Given that

tan 1.1918 1.2349 1.2799 1.3270 1.3764

x

x

°

°

Using Gauss’s backward formula, find the value of tan 51° 42′

Sol Take the origin at 52° and given h = 1

h

Now using Gauss backward formula

f(u) = f(0) + u∆f (–1) + ( 1) 2 ( )

1 2!

f

+

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

2

f

Difference table for given data is:

50 1.1918

0.0431

0.0494

54 1.3764

From equation (1)

f(–0.3 °) = 1.2799 + (–0.3) (0.045) + ( 0.3 0.7)( ) ( 0.3 0.7 1.7) ( ) ( )

= 1.2799 – 0.0135 – 0.0002205 – 0.0000119

= 1.266167 (Approx.)

Example 2 Apply Gauss backward formula to find sin 45 ° from the following table

Sin

θ°

θ°

0.34202 0.502 0.64279 0.76604 0.86603 0.93969 0.98481 Sol Difference table for given data is:

20 0.34202

0.15998

70 0.93

sinθ°

−

−

−

−

−

0.04512

80 0.98481

−

∴ u = x a

h

− = 45 40 10

− = 5

10 = 0.5 Now using Gauss backward formula

f(u) = f(0) + u∆f (–1) + ( 1)

2!

u+ u ∆2f( )−1 + ( 1) ( 1)

3!

u+ u u−

( )

3f 2

+ ( 2)( 1) ( 1)

4!

u+ u+ u u−

( )

4f 2

∆ − + ( 2)( 1) ( 1)( 2)

5!

u+ u+ u u− u−

( )

5f 2

f(0.5) = 0.64279 + 0.5 × 0.14079 + 0.5 1.5

2

×

× (–0.01754) + 0.5 1.5( 0.5)

6

× (0.00165)

+ 0.5 1.5( 0.5 2.5)( )

24

× (–0.00737)

= 0.64279 + 0.070395 – 0.0065775 – 0.000103125 + 0.00028789

= 0.706792 Ans

Example 3 Apply Gauss backward formula to find the value of (1.06) 19 if

(1 06) 10 = 1.79085, (1.06) 15 = 2.39656, (1 06) 20 = 3.20714, (1.06) 25 = 4.29187, (1.06) 30 = 5.74349

Sol The difference table is given by

10 1.79085

0.60571

1.45162

30 5.74349

h

−

= 19 20 5

−

= – 0.2 From Gauss backward formula

f(u) = f(0) + u∆f (–1) + ( 1)

2!

u+ u

( )

2f 1

∆ − + ( 1) ( 1)

3!

u+ u u−

( )

3f 2

+ ( 2)( 1) ( 1)

4!

u+ u+ u u−

∆4f( )−2 +

f(u) = 3.20714 – 0.2 × 0.81058 – 0.2 0.8( )

2 × 0.27415 – 0.2 0.8( )( 1.2)

6

0.06928 + 0.2 0.8( ) (−1.2 1.8) ( )

× 0.02346

f(u) = 3.20714 – 0.162116 – 0.021932 + 0.002216 + 0.00033782

= 3.0256458 (Approx.)

Example 4 Using Gauss backward formula, Estimate the no of persons earning wages between Rs.

60 and Rs 70 from the following data:

( )

No of Persons in thousands

Sol Difference table for the given data is as:

−

−

−

−

120

50

∴ u = x a

h

− = 70 80 20

− = 10 20

− = – 0.5 From Gauss backward formula

4 ( 2)( 1) ( 1)

( 2) 4!

f

= 470 ( 0.5) ( 100) ( 0.5)(0.5) ( 30) ( 0.5)(0.5)( 1.5) ( 10)

+ ( 0.5)(0.5)( 1.5)(1.5) (20)

24

= 470 50 3.75 0.625 0.46875− + − +

= 423.59375

Hence No of Persons earning wages between Rs 60 to 70 is 423.59375 – 370 = 53.59375 or

54000 (Approx.)