Question

Using Lagrange’s interpolation formula find a polynominal which passes through the points (0, –12), (1, 0), (3, 6) and (4, 12)

Answer 1

We can construct a table using the given points.

x	0	1	3	4
y	– 12	0	6	12

Here x₀ = 0

x₁ = 1

x₂ = 3

x₃ = 4

y₀ = – 12

y₁ = 0

y₂ = 6

y₃ = 12

= `((x - 0)(x - 3)(x - 4))/((1 - 0)(1 - 3)(1 - 4)) xx 0  +`

= `((x - 0)(x - 1)(x - 4))/((3 - 0)(3 - 1)(3 - 4)) xx 6 + ((x - 0)(x - 1)(x - 3))/((4 - 0)(4 - 1)(4 - 3)) xx 12`

= `((x - 1)(x - 3)(x - 4))/((-1)(-3)(-4)) (-12) + 0 +`

= `((x)(x - 1)(x - 4))/((3)(2)(-1)) xx 6 + ((x)(x - 1)(x - 30))/((4)(3)(1)) xx 12`

= `((x - 1)(x^2 - 7x + 12))/((-12)) (-12) + (x(x^2 - 5x + 4))/(-6) xx (6) + (x(x^2 - 4x + 3))/12 xx 12`

= (x³ – 7x² + 12x – x² + 7x – 12) – (x³ – 5x² + 4x) + (x³ – 4x² + 3x)

= (x³ – 8x² + 19x – 12) – (x³ – 5x² + 4x) + (x³ – 4x² + 3x)

= x³ – 8x² + 19x – 12 – x³ + 5x² – 4x + x³ – 4x² + 3x

∴ y = x³ – 7x² + 18x – 12