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I have a question here:

A finite sequence of real numbers $c_1, c_2, \dots, c_{n−1}$ is called saw– like if we have $(−1)^k(c_k − c_{k+1}) \leq 0$ for all $k = 1, \dots , n − 2$ or if we have $(−1)^k(c_k − c_{k+1}) \geq 0$ for all $k = 1, \dots , n − 2$. Prove that $c_1, c_2, \dots , c_{n−1}$ is a saw–like sequence if and only if there exist a polynomial $f (x)$ of degree $n$ with real coefficients such that

(1) $x_1 \leq \dots \leq x_{n−1}$ are critical points of $f$ ;

(2) $f (x_k) = c_k$ for $k = 1, \dots , n − 1$.

I've proven one direction,which is "if we have such a polynomial $f$, then the finite sequence {$c_k$} is saw-like." But I don't know how to prove the other direction. I think I need to construct a polynomial $f$ then prove that it satisfies all the properties. But I don't know how to construct it? Use Lagrange interpolation formula? Thanks!

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I don't know the answer, but maybe I can offer a small insight: You have $2n-2$ equations: $f(x_k)=c_k$ and $f'(x_k)=0$ with $2n$ unknowns: The $n+1$ coefficients of $f$ plus the $x_k$. That should give you enough wiggle room, but it is quite clear that determining the $x_k$ must be a part of the solution. You can't just pick any set of points $x_k$. In fact, I would expect the solution to be essentially unique, up to a change of variables of the form $x\mapsto f(ax+b)$ for the polynomial. –  Harald Hanche-Olsen Feb 22 '12 at 15:15

1 Answer 1

There is a generalization of Lagrange interpolation, called Hermite interpolation, which matches both values of a function and values of its derivative(s). Looks like just the thing. http://en.wikipedia.org/wiki/Hermite_interpolation should get you started.

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In Hermite interpolation, just as in Lagrange nerpolation, the nodes (the $x_k$) are given. In th OP problem, the nodes are unknwon. –  Julián Aguirre Feb 25 '12 at 9:06

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