Analogue of the Chebyshev polynomials over C? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:54:08Z http://mathoverflow.net/feeds/question/8912 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8912/analogue-of-the-chebyshev-polynomials-over-c Analogue of the Chebyshev polynomials over C? Scott Aaronson 2009-12-14T21:31:55Z 2009-12-15T08:19:47Z <p>I've been driven up a wall by the following question: let p be a complex polynomial of degree d. Suppose that |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ (for some small δ>0). Then what's the best upper bound one can prove on |p(1)|? (I only care about the asymptotic dependence on d and δ, not the constants.)</p> <p>For the analogous question where p is a degree-d <i>real</i> polynomial such that |p(x)|&le;1 for all 0&le;x&le;1-&delta;, I know that the right upper bound on |p(1)| is |p(1)|≤exp(d&radic;δ). The extremal example here is p(x)=T<sub>d</sub>((1+&delta;)x), where T<sub>d</sub> is the d<sup>th</sup> Chebyshev polynomial.</p> <p>Indeed, by using the Chebyshev polynomial, it's not hard to construct a polynomial p in z <i>as well as its complex conjugate z*</i>, such that</p> <p>(i) |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ, and</p> <p>(ii) p(1) ~ exp(dδ).</p> <p>One can also show that this is optimal, for polynomials in both z and its complex conjugate.</p> <p>The question is whether one can get a better upper bound on |p(1)| by exploiting the fact that p is really a polynomial in z. The fastest-growing example I could find has the form p(z)=C<sub>d,&delta;</sub>(1+z)<sup>d</sup>. Here, if we choose the constant C<sub>d,&delta;</sub> so that |p(z)|≤1 whenever |z|=1 and |z-1|≥δ, we find that</p> <p>p(1) ~ exp(dδ<sup>2</sup>)</p> <p>For my application, the difference between exp(dδ) and exp(dδ<sup>2</sup>) is all the difference in the world!</p> <p>I searched about 6 approximation theory books---and as often the case, they answer every conceivable question except the one I want. If anyone versed in approximation theory can give me a pointer, I'd be incredibly grateful.</p> <p>Thanks so much! --Scott Aaronson</p> <p>PS. The question is answered below by David Speyer. For anyone who wants to see explicitly the polynomial implied by David's argument, here it is:</p> <p>p<sub>d,&delta;</sub>(z) = z<sup>d</sup> T<sub>d</sub>((z+z<sup>-1</sup>)(1+&delta;)/2+&delta;),</p> <p>where T<sub>d</sub> is the d<sup>th</sup> Chebyshev polynomial.</p> http://mathoverflow.net/questions/8912/analogue-of-the-chebyshev-polynomials-over-c/8915#8915 Answer by David Speyer for Analogue of the Chebyshev polynomials over C? David Speyer 2009-12-14T21:43:22Z 2009-12-14T21:43:22Z <p>I may be missing something obvious here. Let $f(z, z^{*})$ be the polynomial in $z$ and $z^{*}$ of degree $d$ which achieves $\exp(d \delta)$. Let $g(z)$ be the Laurent polynomial obtained from $f$ by replacing $z^{*}$ by $z^{-1}$. On the unit circle, we have $f=g$.</p> <p>Now, let $h$ be the polynomial $z^d g$. This is an honest polynomial, because we multiplied by a high enough power of $z$ to clear out all the denominators and, for $z$ on the unit circle, we have $|h|=|f|$.</p> <p>Doesn't this mean that $h$ is a polynomial of degree $2d$, obeying your conditions, with $|h(1)| \sim \exp(d \delta)$?</p>