Interpolation polynomial Challenge

Question

suppose $p(x)=x^k-x^t, k \neq t $ (k,t is a positive integer). function q(x) be a Interpolation polynomial from degree lower or equal n, to data $i=1,...,n+1, (x_i ,p(x_i))$. if ----------- then $p(x)=q(x)$.

my professor wrote a solution for ----------- as:

$n \leq max(k,t)$

why this is true?

i think $n \geq max(k,t)$ is True.

any hint or idea?

Answer 1

I'll prove you right using the Lagrange Interpolation error:

Let $q$ be the interpolation polynomial to a function $p$ with data points $(x_i, p(x_i))_{i=0}^{n}$. If $p$ is $n+1$ times differentiable then:

$$ p(x) - q(x) = \frac{p^{(n+1)}(\xi)}{(n+1)!} \prod_{i=0}^{n}(x-x_i)$$

Let $p(x) := x^k - x^t$. Then for $n \geq \max (k,t)$ we find that

$$ p^{(n+1)}(x) = 0$$

for all $x \in \mathbb{R}$. Using the Lagrange error we get:

\begin{align*} p(x) - q(x) &= \frac{p^{(n+1)}(\xi)}{(n+1)!} \prod_{i=0}^{n}(x-x_i)\\ &= 0 \\ &\Rightarrow p(x) = q(x) \end{align*}

Your professor confused $\leq$ with $\geq$.

Answer 2

The professor is correct. With $n+1$ points the highest degree polynomial defined is $n$ and the degree of $p(x)$ is ${\rm max}(k,t)$. So for $p(x)=q(x)$ the order of $q(x)$ must be equal or less than the order of $p(x)$, or

$$ n \le {\rm max}(k,t)$$