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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ has such property that for every distinct $x_0,x_1,...,x_n\in\mathbb{R}$ Lagrange interpolating polynomial for $f$ in these points has degree at most $n-1$. Prove that $f$ is a polynomial.

Completely don't know how to start.

xan
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1 Answers1

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To condense my comments: Consider the Lagrange interpolation polynomials

  • $q$ for the $n$ points $x_1,x_2,\dots,x_n$ and
  • $p_x$ for the $n+1$ points $x,x_1,\dots,x_n$,

under the assumption that the points $x,x_1,x_2,\dots,x_n$ are all different. Then what difference is there between the coefficients of $q$ and $p_x$ and what does that tell about the values of $p_x$, $q$, and $f$ at the point $x$?

Lutz Lehmann
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  • Honestly, I've been trying to move from this point a few hours and still nothing.. So I know that $\forall_{1\le i \le n} q(x_i)=p_x(x_i)=f(x_i)$ but I don't know what can I say about coefficients from it. I was also trying to use the formula: $p_x(x)=\sum_{i=0}^{n}f(x_i)\prod_{0\le j \le n, \ j\neq i} \frac{x-x_j}{x_i-x_j}$ but also no conclusion. – xan Jan 04 '14 at 14:54
  • $q$ and $p_x$ are both polynomials that have degree $n-1$ and coincide on $n$ different points. So $q=p_x$ independent of $x$. So $q(x)=p_x(x)=f(x)$, and if you cut out the middle, $q(x)=f(x)$ for all $x$. – Lutz Lehmann Jan 04 '14 at 15:27
  • OK, now I see. I had problems with understanding what you've written in the comment above, but finally understood. Thank you very much! – xan Jan 04 '14 at 15:45