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I read an essay by a mathematician (whose name I forgot) who visited TIFR in India and he was writing about his mathematical experiences during the visit. In his essay the mathematician had written that he was stunned by the elegance and beauty of the following theorem.

Let $a_i$ be integers and

$$ f(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0, $$

then $$\displaystyle\max_{x\in(-2,2)} |f(x)| \ge 2.$$

Questions:

  • What is so stunning or elegant about it?
  • I am looking for a proof
  • Any reference to this theorem in literature?

Note: I am trying to recall and search the essay. I shall update my post with its link if I am able to find it.

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Nilotpal Sinha
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    What's stunning is that the maximum does not depend on the choice of $a_i$. You can't find a polynomial as close to the null function as you wish. – E. Joseph Oct 27 '16 at 08:35
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    Presumably, the proof goes as follows: we may evaluate $f(x)$ at some $x_1,\ldots,x_n\in(-2,2)$ (Chebyshev nodes?). Assuming $|f(x_i)|<2$, there is a matrix $M$ sending the vector $(a_0,a_1,a_{n-1})$ into a vector with small norm. Given $a_i\in\mathbb{Z}$ and a good lower bound on the eigenvalues of $M$, that should give a contradiction. – Jack D'Aurizio Oct 27 '16 at 08:48
  • @RobArthan: the theorem is true for $n=1$. –  Oct 27 '16 at 09:02
  • Cedric Villani had recently visited India. Are you referring to him? – StubbornAtom Oct 27 '16 at 10:52
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    No, this person was most likely a number theorist and he visited India in the 1970s or 1980s and met K. Ramachandra and R. Balasubramanian. Could me Michael Wladschmidt but I am not sure? – Nilotpal Sinha Oct 27 '16 at 11:02
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    @NilotpalSinha : in this document, Michel Waldschmidt says: "I visited TIFR from the end of October to end of December 1976". – Watson Oct 27 '16 at 11:41
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    @NilotpalSinha Bhai! I am quite sure, that Ross Honsberger has this theorem mentioned in his famous Mathematical Gems series. I am not sure though in which one. I remember seeing this theorem, while scrolling his book in a library :). – C.S. Dec 11 '16 at 16:35

1 Answers1

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By Chebyshev's Theorem (see Chebyshev's Theorem regarding real polynomials: Why do only the Chebyshev polynomials achieve equality in this inequality?), if $p(x)$ is a real polynomial of degree $n\geq 1$ with leading coefficient $1$ then $$\max_{x\in [-1,1]}|p(x)|\geq \frac{1}{2^{n-1}}.$$ Now $p(x):=f(2x)/2^n$ is a polynomial with leading coefficient $1$ and it follows that $$\max_{x\in [-2,2]}|f(x)|=\max_{x\in [-1,1]}|f(2x)|=2^n\cdot\max_{x\in [-1,1]}|p(x)|\geq \frac{2^n}{2^{n-1}}=2.$$

P.S. As far as I can see, here we do not need that the coefficients $a_i$ have to be integers.

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Robert Z
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