Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Factoring a rational function as:

$$\frac{(x-a)(x-b)\cdots(x-z)}{(x-A)(x-B)\cdots(x-Z)}$$

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.

share|improve this question
3  
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.
http://dx.doi.org/10.1006/jnth.1999.2473

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

share|improve this answer
    
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 $a_0$, $$R(x) = a_0 + a_1 \left( x + \frac{80100}{647(x-1)} - \frac{47232}{647 (x-2)} + \frac{3555}{647(x-3)}\right)$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.