2. Find Solution using Lagrange's Interpolation formula
| x | f(x) |
| 2 | 0.69315 |
| 2.5 | 0.91629 |
| 3 | 1.09861 |
x = 2.7
Solution:
The value of table for `x` and `y`
| x | 2 | 2.5 | 3 |
|---|
| y | 0.69315 | 0.91629 | 1.09861 |
|---|
Lagrange's Interpolating Polynomial
The value of x at you want to find `P_n(x) : x = 2.7`
Lagrange's formula is
`f(x) = ((x - x_1)(x - x_2))/((x_0 - x_1)(x_0 - x_2)) xx y_0 + ((x - x_0)(x - x_2))/((x_1 - x_0)(x_1 - x_2)) xx y_1 + ((x - x_0)(x - x_1))/((x_2 - x_0)(x_2 - x_1)) xx y_2`
`y(2.7) = ((2.7 - 2.5)(2.7 - 3))/((2 - 2.5)(2 - 3)) xx 0.69315 + ((2.7 - 2)(2.7 - 3))/((2.5 - 2)(2.5 - 3)) xx 0.91629 + ((2.7 - 2)(2.7 - 2.5))/((3 - 2)(3 - 2.5)) xx 1.09861`
`y(2.7) = ((0.2)(-0.3))/((-0.5)(-1)) xx 0.69315 + ((0.7)(-0.3))/((0.5)(-0.5)) xx 0.91629 + ((0.7)(0.2))/((1)(0.5)) xx 1.09861`
`y(2.7) = (-0.06)/(0.5) xx 0.69315 + (-0.21)/(-0.25) xx 0.91629 + (0.14)/(0.5) xx 1.09861`
`y(2.7) = 0.994116`
Solution of the polynomial at point `2.7` is `y(2.7) = 0.994116`
This material is intended as a summary. Use your textbook for detail explanation.
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