Question

Using Lagrange’s interpolation formula find a polynomial 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

\(= {\frac{(x - 1)(x -3)(x - 4)}{(0 - 1) (0 - 3)(0 - 4) }\times(-12) + \frac{(x - 0)(x - 3)(x - 4)}{(1 - 0)(1 - 3)(1 - 4)} \times 0 + \frac{(x - 0)(x - 1)(x - 4)}{(3 - 0)(3 - 1)(3 - 4)} \times 6+ \frac{(x - 0)(x - 1)(x - 3)}{(4 - 0)(4 - 1)(4 - 3)} \times 12}\)

\(= \frac{(x - 1)(x - 3)(x - 4)}{(-1) (-3)(-4)} (-12) + 0 + \frac{(x)(x - 1)(x - 4)}{(3) (2)(-1)} \times 6 + \frac{(x)(x - 1)(x - 3)}{(4)(3)(1)} \times 12\)

\(= \frac{(x -1)(x^2 - 7x + 12)}{(-12)} \times (-12) + \frac{x(x^2 - 5x + 4)}{-6} \times 6 + \frac{x(x^2 - 4x + 3)}{12} \times 12\)

= (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

Answer 2

The given values are 

x₀ = 0, y₀ = -12 

x₁ = 1, y₁ = 0 

x₂ = 3, y₂ = 6 

x₃ = 4, y₃ = 12

By Lagrange’s interpolaiton formula,

⇒ y = (x – 1) (x – 3) (x – 4) – x (x – 1) (x – 4) + x (x – 1) (x – 3) 

⇒ y = (x – 1) (x – 4)[(x – 3) – x] + x (x – 1) (x – 3)

⇒ y = (x – 1) (x – 4) (-3) + x (x – 1) (x – 3) 

⇒ y = (x – 1) [-3x + 12 + x² – 3x] 

⇒ y = (x – 1) (x² – 6x + 12) 

⇒ y = x³ – 6x² + 12x – x² + 6x – 12 

y = x³ – 7x + 18x – 12 is the required polynomial which passes through the given points