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Coonsider the polynomial $p(x) = x^2(4x^2-3)^2$.

This polynomial is special because:

  1. All of the local maxima $M_i$ are of the form $p(M_i)=1$
  2. All of the local minima $m_i$ are of the form $p(m_i)=0$
  3. and $p(-1)=p(1)=1$
  4. The polynomial has no other zeros, local maxima or local minima.

My questions are:

  • what conditions must a polynomial satisfy in order to have the above characteristics?
  • do they have a specific name?



These are the elements of the family I have so far:

$$p_1(x)=x^2 (4x^2-3)^2$$ $$p_2(x)=x^2 (16 x^4-20 x^2+5)^2$$ $$p_3(x)=x^2 (64 x^6-112 x^4+56 x^2-7)^2$$

Can someone find the pattern? What is $p_4(x)$?

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This is the square of $T_3(x)$, the third Chebyshev (many spellings) polynomial. You might visit for more information. The square of all the other ones will satisfy your criteria, as will $(2T_n(x)+1)/2$ as the original $T_n$ range from -1 to 1. What is wrong with $p_0(x)=ax^2$ for any positive $a$? All (the only) minima are 0, All (there are none) the maxima are 1, and $p_0(-1)=p_0(1)$ – Ross Millikan Mar 17 '11 at 4:40
Right, but $p_0(x)=x^2$ because of condition 3. – Neves Mar 17 '11 at 4:45
up vote 4 down vote accepted

I think you will find the results and techniques of the following paper useful in this regard.

MR1752251 (2001c:11035) 11D25 (11D41 11G05 11G30)
Buchholz, Ralph H.; MacDougall, James A.(5-NEWC)
When Newton met Diophantus: a study of rational-derived polynomials
and their extension to quadratic fields.

J. Number Theory 81 (2000), no. 2, 210–233.

This is an interesting paper, which surveys the problem of determining the set $D(n)$ of all "$k$-derived'' univariate polynomials of degree $n$ (where a polynomial $f \in k[x]$ is $k$-derived if $f$ and each of its successive derivatives has all roots in the ground field $k$). Define two polynomials $f_1,f_2\in k[x]$ to be equivalent if $f_1(x)=r f_2(s x+t)$ for $r,s,t\in k$, $r,s \neq 0$. Then up to equivalence, the following is known about $\mathbb Q$-derived polynomials:

$$D(1)=\{x\};\quad D(2)=\{x^2,x(x-1)\};$$ $$D(3)=\{x^3\}\cup\bigg\{x(x-1)(x-a)\ :\ a=\frac{w(w-2)}{w^2-1},w\in \mathbb Q\bigg\};$$ $$ D(4)\supseteq \{x^4\}\cup\left\{x^2(x-1)(x-a)\ \left|\ \begin{array}{l}a=\frac{9(2w+z-12)(w+z)}{(z-w-18)(8w+z)}, (w,z)\in E(\mathbb Q),\\ E\colon z^2=w(w-6)(w+18)\end{array}\right.\right\};$$ $$ D(n)\supseteq \{x^n, x^{n-1}(x-1)\}\ {\rm for}\ n\geq 5.$$

The authors prove that determining $D(n)$ in general devolves into two conjectures: (1) that no quartic with four distinct roots is $\mathbb Q$-derived; (2) that no quintic of type $x^3(x-a)(x-b)$, $a\neq b,\ a,b\neq0$, is $\mathbb Q$-derived. The first conjecture can be solved by determining all rational points on a hyperelliptic surface of degree 10. The second conjecture can be solved by determining all rational points on a curve of genus 2 (E. V. Flynn ["On $\mathbb Q$-derived polynomials'', Preprint; per revr.] has now proved this second conjecture). The authors also discuss briefly the situation of $K$-derived polynomials for quadratic extensions $K$ of $\mathbb Q$; there is, for example, the quartic $y^2=x^2(x-1)(x-\frac{37-20\sqrt{3}}{13})$ which is a ${\mathbb Q}(\sqrt{3})$-derived polynomial.

Reviewed by Andrew Bremner

share|cite|improve this answer
Thanks, very interesting. – Neves Mar 16 '11 at 23:36

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