Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 19D / a

Table of contents
- Solution a

Solution

The Lagrange interpolation polynomial basis for nodes

0, 2, 3

gives

a (x) = \frac{( x - 2 ) ( x - 3 )}{6}, b (x) = - \frac{x ( x - 3 )}{2}, c (x) = \frac{x ( x - 2 )}{3} .

Each polynomial is one at its own node and zero at the other two. For a polynomial

f

of degree at most two, the difference between

f

and this interpolation polynomial has degree at most two and three distinct roots, so it is identically zero. This also proves uniqueness of the quadratic coefficients.

 Ancestors (10)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version