Find Solution using Lagrange's Interpolation formula
x = 1
Finding f(2)
Solution:
The value of table for `x` and `y`
Lagrange's Interpolating Polynomial
The value of x at you want to find `P_n(x) : x = 1`
Lagrange's Interpolation formula is
`f(x) = ((x - x_1)(x - x_2)(x - x_3)(x - x_4))/((x_0 - x_1)(x_0 - x_2)(x_0 - x_3)(x_0 - x_4)) xx y_0 + ((x - x_0)(x - x_2)(x - x_3)(x - x_4))/((x_1 - x_0)(x_1 - x_2)(x_1 - x_3)(x_1 - x_4)) xx y_1 + ((x - x_0)(x - x_1)(x - x_3)(x - x_4))/((x_2 - x_0)(x_2 - x_1)(x_2 - x_3)(x_2 - x_4)) xx y_2 + ((x - x_0)(x - x_1)(x - x_2)(x - x_4))/((x_3 - x_0)(x_3 - x_1)(x_3 - x_2)(x_3 - x_4)) xx y_3 + ((x - x_0)(x - x_1)(x - x_2)(x - x_3))/((x_4 - x_0)(x_4 - x_1)(x_4 - x_2)(x_4 - x_3)) xx y_4`
`y(1) = ((1 - 0)(1 - 3)(1 - 6)(1 - 7))/((-1 - 0)(-1 - 3)(-1 - 6)(-1 - 7)) xx 3 + ((1 - -1)(1 - 3)(1 - 6)(1 - 7))/((0 - -1)(0 - 3)(0 - 6)(0 - 7)) xx -6 + ((1 - -1)(1 - 0)(1 - 6)(1 - 7))/((3 - -1)(3 - 0)(3 - 6)(3 - 7)) xx 39 + ((1 - -1)(1 - 0)(1 - 3)(1 - 7))/((6 - -1)(6 - 0)(6 - 3)(6 - 7)) xx 822 + ((1 - -1)(1 - 0)(1 - 3)(1 - 6))/((7 - -1)(7 - 0)(7 - 3)(7 - 6)) xx 1611`
`y(1) = ((1)(-2)(-5)(-6))/((-1)(-4)(-7)(-8)) xx 3 + ((2)(-2)(-5)(-6))/((1)(-3)(-6)(-7)) xx -6 + ((2)(1)(-5)(-6))/((4)(3)(-3)(-4)) xx 39 + ((2)(1)(-2)(-6))/((7)(6)(3)(-1)) xx 822 + ((2)(1)(-2)(-5))/((8)(7)(4)(1)) xx 1611`
`y(1) = (-0.2679) xx 3 + 0.9524 xx -6 + 0.4167 xx 39 + (-0.1905) xx 822 + 0.0893 xx 1611`
`y(1) = -3`
Solution of the polynomial at point `1` is `y(1) = -3`
This material is intended as a summary. Use your textbook for detail explanation.
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