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I'm pretty sure that if the function $f$ is continuous on $[x_1,x_3]$ and twice-differentiable on $(x_1,x_3)$, with $x_1 < x_2 < x_3$, then there must exist $x$ in $(x_1,x_3)$ for which $f''(x)$ equals $$\dfrac{((x_2-x_1)(f(x_3)-f(x_1)) - (f(x_2)-f(x_1))(x_3-x_1))}{ ((x_2-x_1)(x_3-x_2)(x_3-x_1))}.$$ (This complicated expression gives the constant second-derivative of the unique polynomial of degree 2 passing through the points $(x_i,f(x_i))$ for $1 \leq i \leq 3$.)

A simple proof would be nice, but even nicer would be a pointer to the literature. Surely this result is known, and indeed is just a special case of a known result for curves passing through $n$ prescribed points.

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Let $q(x)$ be the quadratic through the $(x_i, f(x_i))$. Set $g(x) = f(x) - q(x)$. Then $g(x_1) = g(x_2) = g(x_3) = 0$ and you want to show $g''(x)=0$ for some $x$. Apply Rolle's theorem to get that $g'$ has zeroes $y_1$ and $y_2$ in $(x_1, x_2)$ and $(x_2, x_3)$ respectively, then apply it again to get that $g''$ has a zero in $(y_1, y_2)$.


I can sort of give you a pointer to the literature. Rename your points $x_1$, $x_2$, ..., $x_{n}$ and $z$ with $x_1 < x_2 < \ldots < x_{n}$. Let $q(t)$ be the degree $n-1$ Lagrange interpolating polynomial for $f$ through $(x_i, f(x_i))$. Then the degree $n$ Lagrange interpolating polynomial through the $(x_i, f(x_i))$ and $(z, f(z))$ is $$r(t) = q(t) + \frac{f(z)-q(z)}{\prod (z-x_i)} \prod (t-x_i).$$ You want to know that there is some $c \in (\min(x_1,z), \max(x_n,z))$ where $f^{(n)}(c) = r^{(n)}(c)$. In other words, where $f^{(n)}(c) = \frac{f(z) - q(z)}{\prod (z-x_i)} n!$. You can rearrange this to give $$f(z) = q(z) + \frac{f^{(n)}(c)}{n!} \prod (z-x_i).$$ This is a known formula for the error in Lagrange interpolation. Here is a Wikipedia article. The case where $x_1=x_2=\cdots = x_n$ is Taylor's theorem with the Lagrange form of the remainder; Tim Gowers has a nice blog post. Comments on Prof. Gowers' post suggest that Atkinson’s Intro to Numerical Analysis would give you a more official source; I haven't read it.

I don't know a source where the $n+1$ points are treating on equal footing, but my ignorance of analysis literature is no sign that it doesn't exist.

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The stated result is a special case of the Schwarz mean value theorem, which plays a crucial role in Dörge's proof of the Hilbert irreducibility theorem. References I know of are the books Diophantine Geometry by Lang (p. 148), Selected Topics on Polynomials by Schinzel (p. 174), and Generic Polynomials by Jensen, Ledet and Yui (p. 69).

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  • $\begingroup$ I think the $n$ point version IS the Schwarz mean value theorem. (I'm using Lemma 2.6 in Lang "Fundamentals of Diophantine Geometry" for my reference. books.google.com/… ) By Cramer's rule, that ratio of determinants is the leading coefficient of the interpolating polynomial. $\endgroup$ – David E Speyer Dec 4 '14 at 14:04
  • $\begingroup$ By stated result I meant the OP's $3$-point version, not the one from your answer which is a different formulation of the Schwarz version. $\endgroup$ – Peter Mueller Dec 4 '14 at 15:34

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