2

How do I show that Lagrange's polynomial is the only one (with degree < n) that takes the given values at given points? ($f(x_{1})=y_{1}... \space f(x_{n})=y_{n}$)

Nesa
  • 1,255

1 Answers1

4

We know that degree of $f < n$.

Let us assume that there is another polynomial $g$, with degree strictly less than $n$, such that $g(x_i) = y_i$, $i=1,2,\dots , n$.

Then the polynomial $f-g$ has degree strictly less than $n$ but has $n$ zeros (namely $x_1, x_2, \dots , x_n$). Which is a contradiction

Hence, no such $g$ exists.