|
Q. 10. Using any one of the interpolation
formulae (e.g., Lagrange's/Newton's etc.) and the
following table, find the value of f(x) at x = 3.5
: (Dec. 2000)
| x |
1 |
2 |
5 |
6 |
| f(x) |
1.2 |
4.8 |
26 |
39 |
Solution. Given, x = 3.5, x0 =
1, x1 = 2, x2 = 5, x3
= 6, f(x0) = 1.2, f(x1) = 4.8,
f(x2) = 26, f(x3) = 39
Using Lagrange’s Formula
| |
(3.5 - 2)(3.5 - 5)(3.5 - 6)
|
|
(3.5 - 1)(3.5 - 5)(3.5 - 6)
|
|
| f(3.5) = |
|
X (1.2) + |
|
X (4.8)
+ |
| |
(1 - 2)(1 - 5)(1 - 6)
|
|
(2 - 1)(2 - 5)(2 - 6)
|
|
| |
|
|
|
|
| |
(3.5 - 1)(3.5 - 2)(3.5 - 6)
|
|
(3.5 - 1)(3.5 - 2)(3.5 - 5)
|
|
| |
|
X (26) + |
|
X (39) |
| |
(5 - 1)(5 - 2)(5 - 6)
|
|
(6 - 1)(6 - 2)(6 - 5)
|
|
or f(3.5) = 12.7563
Q. 11. The following values of the function
f(x) for value of x are given :
f(1) = 4, f(2) = 5, f(7) = 5, f(8) = 4
Using Newton's Divided Difference Formula or Lagrange’s
Formula, find the value of f(6). (Dec.
99)
Solution. Given, x = 6, x0 = 1,
x1 = 2, x2 = 7, x3
= 8, f(x0) = 4, f(x1) = 5, f(x2)
= 5, f(x3) = 4
Using Newton's Divided Difference Formula
The difference table is shown below:
| x |
f(x) |
Df(x) |
D2f(x) |
D3f(x) |
| 1 |
4 |
(5 -
4)/(2 - 1) = 1
(5 - 5)/(7 - 2) = 0
(4 - 5)/(8 - 7) = -1
|
(0 -
1)/(7 - 1) = -1/6
(-1 - 0)/(8 - 2) = -1/6 |
0 |
| 2 |
5 |
| 7 |
5 |
| 8 |
4 |
Here, Df(x0)
= 1, D2f(x0)
= -1/6, D3f(x0)
= 0
f(x) = f(x0) + (x - x0)Df(x0)
+ (x - x0)(x - x1)D2f(x0)
+ (x - x0)(x - x1)(x - x2)D3f(x0)
f(6) = 4 + [(6 - 1) X 1] + [(6 - 1) X (6 - 2) X (-1/6)]
+ [(6 - 1) X (6 - 2) X (6 - 7) X 0]
or f(6) = 5.66
|
