Question

Estimate the ⍋4 f(x) value by constructing divided difference table?

x : 0 12 2 5 6

y : 1 19 55 241 415

A)0 B) 2 C) 9 ​

Answer 1

Answer:

n

Step-by-step explanation:

WORKED OUT PROBLEMS

The problems in this section are solved using Newton's divided difference formula and Lagrange's formula. Since By Sheperd's Zig-Zag rule any aritrary path from function values to its highest divided difference to compute the value of f(x) in all these examples first fuction value and its higher divided differences are been used to compute f(x).

1. Find f(2) for the data f(0) = 1, f(1) = 3 and f(3) = 55.

x

0

1

3

f

1

3

55

Solution :

By Newton's divided difference formula :

Divided difference table

xi

fi

0

1

2

1

3

8

26

3

55

Now Newton's divided difference formula is

f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2]

f(2) = 1 + (2 - 0) 2 + (2 - 0)(2 - 1) 8

= 21

By Lagrange's formula :

(x - x1) (x - x2)

(x - x0)(x - x1)

f(x) =

f0+ . . . +

f2

(x0 - x1) (x0 - x2)

(x2 - x0)(x2 - x1)

(2 - 1)(2 - 3)

(2 - 0)(2 - 3)

(2 - 0)(2 - 1)

f(2) =

1+

3 +

55

(0 - 1) (0 - 3)

(1 - 0)(1 - 3)

(3 - 0)(3 - 1)

f(2) = 21

2. Find f(3) for

x

0

1

2

4

5

6

f

1

14

15

5

6

19

Solution :

By Newton's divided difference formula :

Divided difference table

xi

fi

0

1

13

1

14

-6

1

1

2

15

-2

0

-5

1

0

4

5

2

0

1

1

5

6

6

13

6

19

Now Newton's divided difference formula is

f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2] + (x - x0) (x - x1) (x - x2)f [x0, x1, x2, x3]

+ (x - x0) (x - x1) (x - x2)(x - x3)f [x0, x1, x2, x3, x4]

+ (x - x0) (x - x1) (x - x2)(x - x3)(x - x4)f [x0, x1, x2, x3, x4, x5]

f(3) = 1 + (3 - 0) 13 + (3 - 0)(3 - 1) -6 + (3 - 0) (3 - 1) (3 - 2) 1

= 10

By Lagrange's formula :

(3 - 1)(3 - 2)(3 - 4)(3 - 5)(3 - 6)

(3 - 0)(3 - 2)(3 - 4)(3 - 5)(3 - 6)

f(3) =

1+

14+

(0 - 1)(0 - 2)(0 - 4)(0 - 5)(0 - 6)

(1 - 0)(1 - 2)(1 - 4)(1 - 5)(1 - 6)

(3 - 0)(3 - 1)(3 - 4)(3 - 5)(3 - 6)

(3 - 0)(3 - 1)(3 - 2)(3 - 5)(3 - 6)

15+

5+

(2 - 0)(2 - 1)(2 - 4)(2 - 5)(2 - 6)

(4 - 0)(4 - 1)(4 - 2)(4 - 5)(4 - 6)

(3 - 0)(3 - 1)(3 - 2)(3 - 4)(3 - 6)

(3 - 0)(3 - 1)(3 - 2)(3 - 4)(3 - 5)

6+

19

(5 - 0)(5 - 1)(5 - 2)(5 - 4)(5 - 6)

(6 - 0)(6 - 1)(6 - 2)(6 - 4)(6 - 5)

f(2) = 10

3. Find f(0.25) for

x

0.1

0.2

0.3

0.4

0.5

f

9.9833

4.9667

3.2836

2.4339

1.9177

Solution :

By Newton's divided difference formula :

Divided difference table

xi

fi

0.1

9.9833

-50.166

0.2

4.9667

166.675

-16.83

-416.68

0.3

3.2836

41.67

833.42

-8.497

-83.32

0.4

2.4339

16.675

-5.162

0.5

1.9177

Now Newton's divided difference formula is

f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2] + (x - x0) (x - x1) (x - x2)f [x0, x1, x2, x3]

+ (x - x0) (x - x1) (x - x2)(x - x3)f [x0, x1, x2, x3, x4]

f(3) = 9.9833 + (0.25 - 0.1) -50.166 + (0.25 - 0.2)(0.25 - 0.3) 166.675 +

(0.25 - 0.1) (0.25 - 0.2) (0.25 - 0.3) -416.68 + (0.25 - 0.1) (0.25 - 0.2) (0.25 - 0.3)(0.25 - 0.4) 833.42

= 3.912

By Lagrange's formula :

f(0.25) =

(.25 - .2)(.25 - .3)(.25 - .4)(.25 - .5)

(.25 - .1)(.25 - .3)(.25 - .4)(.25 - .5)

9.9833+

4.9667 +

(.1 - .2)(.1 - .3)(.1 - .4)(.1 - .5)

(.2 - .1)(.2 - .3)(.2 - .4)(.2 - .5)

(.25 - .1)(.25 - .2)(.25 - .4)(.25 - .5)

(.25 - .1)(.25 - .2)(.25 - .3)(.25 - .5)

3.2836+

2.4339 +

(.3 - .1)(.3 - .2)(.3 - .4)(.3 - .5)

(.4 - .1)(.4 - .2)(.4 - .3)(.4 - .5)

(.25 - .1)(.25 - .2)(.25 - .3)(.25 - .4)

1.9177

(.5 - .1)(.5 - .2)(.5 - .3)(.5 - .4)

f(0.25) = 3.912