Apply lagrange interpolation to fit the given set of data points into a function...

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We need only the x and F columns to get the datapoints.

x₀=5, x₁=10, x₂=15, x₃=20, x₄=25, x₅=30,

F₀=3.0062, F₁=7.8762, F₂=9.3662, F₃=7.1012, F₄=5.2062, F₅=12.3062.

The polynomial F(x)=

(x-x₁)(x-x₂)(x-x₃)(x-x₄)(x-x₅)F₀

(x₀-x₁)(x₀-x₂)(x₀-x₃)(x₀-x₄)(x₀-x₅)

+(x-x₀)(x-x₂)(x-x₃)(x-x₄)(x-x₅)F₁

  (x₁-x₀)(x₁-x₂)(x₁-x₃)(x₁-x₄)(x₁-x₅)

+(x-x₀)(x-x₁)(x-x₃)(x-x₄)(x-x₅)F₂

  (x₂-x₀)(x₂-x₁)(x₂-x₃)(x₂-x₄)(x₂-x₅)

+(x-x₀)(x-x₁)(x-x₂)(x-x₄)(x-x₅)F₃

  (x₃-x₀)(x₃-x₁)(x₃-x₂)(x₃-x₄)(x₃-x₅)

+(x-x₀)(x-x₁)(x-x₂)(x-x₃)(x-x₅)F₄

  (x₄-x₀)(x₄-x₁)(x₄-x₂)(x₄-x₃)(x₄-x₅)

+(x-x₀)(x-x₁)(x-x₂)(x-x₃)(x-x₄)F₅

  (x₅-x₀)(x₅-x₁)(x₅-x₂)(x₅-x₃)(x₅-x₄)

Substituting x=12.5 and other x and F values, we can compute F(12.5)=

-0.082201

+3.230473

+7.683211

-1.941734

+0.427071

-0.144213=9.1726.

F(12.5)=9.1726.

by Top Rated User (884k points)

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