Verify that the polynomials px 3 2x 1 4x 1x 2 qx 4x2

Verify that the polynomials p(x) = 3 + 2(x - 1) + 4(x - 1)(x + 2), q(x) = 4x^2 + 6x - 7 interpolate the data and explain why this does not violate the uniqueness part of the theorem on existence of polynomial interpolation.

Solution

Solution :

Checking for interpolating the given data:

i) p(x) = 3 + 2(x-1) + 4(x - 1)(x + 2)

   put x =1 in this we will get p(1) = 3 + 2(1-1) + 4(1 -1)(1+2) = 3

   put x = -2 in this we will get p(-2) = 3 + 2(-2-1) + 4(-2 -1)(-2+2) = -3

   put x = 0 in this we will get p(0) = 3 + 2(0-1) + 4(0 -1)(0+2) = 3 -2 -8 = - 7

   hence all given data satisfied for this polynomial and hence it interpolates the p(x) polynomial.

ii) q(x) = 4x² + 6x -7

   put x =1 in this we will get q(1) = 4.1 + 6.1 - 7 = 3

   put x = -2 in this we will get q(-2) = 4.(-2)² + 6.(-2) - 7 = -3

   put x = 0 in this we will get q(0) =4.(0)² + 6.(0) - 7 = - 7

   hence all given data satisfied for this polynomial and hence it interpolates the q(x) polynomial.

Finally we will show that this doesn\'t violate the uniqueness part of the theorem :

We will show that p(x) and q(x) are actually same :

q(x) = 4x² + 6x -7

p(x ) = 3 + 2(x-1) + 4(x - 1)(x + 2) expand it

    = 3 + 2x -2 + 4(x² +2x -x -2)

     = 3 + 2x -2 + 4x² +4x -8

     = 4x² + 6x -7

hnece p(x) = q(x) actually and uniqueness is maintained.