User talk:Physicist137

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SwisterTwister talk 04:16, 27 February 2017 (UTC)

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Cohn's theorem

The new article titled Cohn's theorem says:

An nth-degree polynomial,

$p(z)=p_{0}+p_{1}z+\cdots +p_{n}z^{n}$

is called self-inversive if

$p(z)=\omega p^{*}(z),\qquad |\omega |=1,$

where

$p^{*}(z)=z^{n}{\bar {p}}\left({\frac {1}{z}}\right)={\bar {p}}_{n}+{\bar {p}}_{n-1}z+\cdots +{\bar {p}}_{0}z^{n}$

is the reciprocal polynomial associated with $p(z)$ and the bar means complex conjugation.

I changed something in it: You had the notation p_n referring BOTH to the polynomial itself and to one of the coefficients.

One problem in the present form of that article is that I cannot tell whether the passage above means

• for EVERY complex number ω for which |ω| = 1, or
• for SOME complex number ω for which |ω| = 1, or
• something else.

Can you clarify that? Michael Hardy (talk) 20:37, 12 April 2018 (UTC)

Hi Michael Hardy,

The correct is for SOME ω. Perhaps a more precise definition would be something like: "p(x) is called self-inversive if there exists a $\omega \in C:|\omega |=1$ so that: $p(z)=\omega p^{*}(z)$."