Let $x_1 < x_2 < . . . < x_m$, and let $y_1 , y_2 , . . . , y_m$ be real numbers. There exists exactly one polynomial $p$ of degree $≤ m − 1,$ such that $p(x_i) = y_i, 1 ≤ i ≤ m.$ This is equivalent to showing that the Vandermonde matrix $V_m$ is invertible. Prove it using the following steps.

  1. Explain: It suffices to prove that $p(x) = 0$ is the only solution when all the $y_i$ are zero.
  2. Assume that $p$ were a polynomial of degree $≤ m − 1$ with $p(x_i ) = 0$ for all $i$. Show that there are $m − 1$ different points in which the derivative $p'$ is zero.

My attempt:
1. $$p(x)=\sum_{j=1}^mc_jx^{j-1}=V_m\vec{c}=\vec{y}.$$ This is a system of linear questions. So, $V_m$ is invertible if columns are linearly independent. In other words the only solution to $V_m\vec{c}=\vec{0}$ is $\vec{c}=\vec{0}.$
I'm stuck on the second point, any help would be appreciated!

  • 2
    $\begingroup$ Rolle's theorem.: If $p$ is differentiable and If $p(x_i)=p(x_{i+1})=0$ (and $x_i<x_{i+1}$) there is some $\xi\in(x_i,x_{i+1})$ such that $p'(\xi)=0$. $\endgroup$ – Jens Schwaiger Mar 2 at 4:59
  • 1
    $\begingroup$ @JensSchwaiger Pleaes your comment an answer. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 2 at 5:02

Following a request I transform my comment above into an answer:

Rolle's theorem: If $p$ is differentiable and if $p(x_i)=p(x_{i+1})=0 $ (and $x_i<x_{i+1}$) there is some $ξ∈(x_i,x_{i+1}$) such that $p′(ξ)=0$.

  • $\begingroup$ So, the polynomial of the degree $m-1$ has at most $m-1$ roots, therefore, shouldn't it have $m-2$ different points where the derivative is 0 (since between any two zeros of the function there is one point where the derivative is 0)? $\endgroup$ – dxdydz Mar 2 at 17:00
  • $\begingroup$ We want to show with induction that $p$ as above is$\equiv 0$. The induction hypothesis and Rolle show that $p'\equiv 0$. Thus $p$ is constant and therefore $\equiv 0$ since $p$ by assumption has the zeroes $x_i$. $\endgroup$ – Jens Schwaiger Mar 2 at 22:19
  • $\begingroup$ Can you clarify it a little bit more? So from what I get, we want to prove with induction that polynomial $p(x)=0$. So base case is trivial $(n=0):$ 0=0. Now we make the induction hypothesis that polynomial of degree $m-1 \ \ p(x_i)=0$ for all $i$.We want to show that polynomial of degree $m \ \ p(x_i)=0$ for all $i$. By Rolle's theorem there exist $m-1$ different points where $p'=0$. How do we know that $p'$ is 0 everywhere (so that p is constant)? Correct me if I'm wrong $\endgroup$ – dxdydz Mar 3 at 4:48
  • $\begingroup$ @dxdydz: $p'$ has degree $m-2$ and vanishes at $m-2$ points $\xi_i\in(x_i,x_{i+1})$. Thus by induction hypothesis $p'$ has to vanish everywhere. $\endgroup$ – Jens Schwaiger Mar 3 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.