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Factoring a rational function as:


Clearly displays the x-intercepts (roots) and vertical asymptotes (poles) of a rational function.

Similarly, dividing the denominator D into the numerator N to get N = DQ+R and writing N/D as Q + R/D clearly displays the end-behavior (horizontal or even diagonal asymptotes and so forth).

Is there some similar factorization that displays the local minimum and maximums of the real-valued real-input rational function?

In other words, if the local extrema occur at a,b,…,z then the expression for the function has the a,b,…,z explicit.

The only example I can think of is "completing the square" exhibits the vertex of a parabola.

Edit: After a few hours of work, I've also managed to easily express a cubic in terms of its two vertices (I had no idea cubics were so symmetrical). I'm hoping similar ideas will work for any polynomial.

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a stupid joke but I got to say it: your function is 0. –  J. J. Nov 24 '10 at 17:03
on more serious side; a polynomial is a rational function and it's minimums/maximums are located at the zeros of its derivative, which may be a quite high-degree polynomial whose roots are not apparent at all –  J. J. Nov 24 '10 at 17:05
Cubics have "two vertices"? It can have two extrema, yes (for the all-real roots case), but... –  J. M. Nov 25 '10 at 4:02

2 Answers 2

up vote 3 down vote accepted

I don't think there is a nice answer for the general case of your question. However, there is nice answer to a closely related question - see the review below.

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:

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

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

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Thanks for the interesting paper. I'm not too worried about the derivatives having weird roots. I assume we factor at least over R (if not C), and am ok if the result does not make perfect sense if the polynomial is not R-derived. I've tried writing cubics in some sort of "vertex form", but have had little success. –  Jack Schmidt Nov 24 '10 at 19:03
@Jack: The nearest to "vertex form" you can get for polynomials of degree 3 or higher is their "depressed" form: briefly, if you have $a_n x^n+a_{n-1} x^{n-1}+\dots+a_0$, you then make the substitution $u=x-\frac{a_{n-1}}{na_n}$ so that your polynomial in $u$ contains no $u^{n-1}$ term. –  J. M. Nov 24 '10 at 22:51

Suppose $R(x) = P(x)/Q(x)$ is a rational function where $P(x)$ has degree $n$ and $Q(x)$ has degree $m$. To keep things simple, I'll suppose $Q(x)$ has $m$ distinct roots $r_j,\ j=1\ldots m$. Then we can express $R(x)$ in partial fractions as $R(x) = A(x) + \sum_{j=1}^m \frac{c_j}{x - r_j}$: if $m \le n$, $A(x)$ is a polynomial of degree $n-m$, while if $m > n$, $A(x) = 0$. Then $R'(x) = A'(x) - \sum_{j=1}^m \frac{c_j}{(x - r_j)^2}$. In general $R'(x)$ can have up to $m+n-2$ zeros $s_k$, and $n-1$ of them will determine the coefficients of $A'$ and the $c_j$ up to a multiplicative constant by the linear equations $$ 0 = A'(s_k) + \sum_{j=1}^m \frac{c_j}{(s_k - r_j)^2} $$ For example, take $n=4$, $m=3$, and suppose you want the poles at $1, 2, 3$ and three of the zeros of $R'$ at $4, 5, 6$. The three equations $$ a_1 - \frac{c_1}{(s-1)^2} - \frac{c_2}{(s-2)^2} - \frac{c_3}{(s-3)^2} = 0 \ \text{for } s = 4,5,6 $$ have solution $$ c_{{1}}={\frac {80100}{647}}\,a_{{1}},c_{{2}}=-{\frac {47232}{647}}\,a_{{1}},c_{{3}}={\frac {3555}{647}}\,a_{{1}} $$ where $a_1$ is arbitrary. Thus, for some constants $a_1$ and $a_0$, $$R(x) = a_0 + a_1 \left( x + \frac{80100}{647(x-1)} - \frac{47232}{647 (x-2)} + \frac{3555}{647(x-3)}\right)$$

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The gist of this is "If you have chosen the mins, maxes, zeros, and poles, then you can write such a function easily in the standard form.", right? This looks very useful, and gets at the pedagogical purpose. I think my original question (a year ago, so I am little fuzzy) was wanting to have a different standard form that made things nice. I suspect it was for writing more intuitive web-apps: ms.uky.edu/~jack/2010-08-MA109/RationalFunctions.html and ms.uky.edu/~jack/2010-08-MA109/CubicVertexFormula.html –  Jack Schmidt Jan 17 '12 at 18:24

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