Question

Using interpolation, find the value of f(x) when x = 15

x	3	7	11	19
f(x)	42	43	47	60

Answer 1

Here the intervals are unequal

By Lagrange’s in-terpolation formula we have,

x₀ = 3

x₁ = 7

x₂ = 11

x₃ = 19

y₀ = 42

y₁ = 43

y₂ = 47

y₃ = 60 and x = 15.

y = `"f"(x) = ((x - x_1)(x - x_2)(x - x_3))/((x_0 - x_1)(x_0 - x_2)(x_0 - x_3)) xx y_0 + ((x_0 - x_1)(x - x_2)(x - x_3))/((x_1 - x_0)(x_1 - x_2)(x_1 - x_3)) xx y_1 + ((x - x_0)(x - x_1)(x - x_3))/((x_2  - x_0)(x_2 - x_1)(x_2 - x_3))xx y_2 + ((x - x_0)(x - x_1)(x - x_2))/((x_3- x_0)(x_3 - x_1)(x_3 - x_2)) xx y_3`

y(15) = `"f"(15) = ((15 - 7)(15 - 11)(15 - 19))/((3 - 7)(3 - 11)(3 - 19)) xx 42 + ((15 - 3)(15 - 11)(15 - 19))/((7 - 3)(7 - 11)(7 - 19)) xx 43 + ((15 - 3)(15 - 7)(15 - 19))/((11 - 13)(11 - 7)(11 - 19)) xx 47 + ((15 - 3)(15 - 7)(15 - 11))/((19 - 3)(19 - 7)(19 - 11)) xx 60`

= `((8)(4)(-4))/((-4)(-8)(-4)) xx 42 + ((12)(4)(-4))/((4)(-4)(-12)) xx 43 + ((12) xx (8) xx (-4))/((8) xx (4) xx (-8)) xx 47 + ((12) xx (8) xx (4))/((16) xx (12) xx (8)) xx 60`

= 10.5 – 43 + 70.5 + 15

= 53