# When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research?

A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is

\begin{align} x\quad&=\quad \sqrt[3]{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) + \sqrt{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) ^2 + \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right) ^3 } }\\ &+\quad \sqrt[3]{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) - \sqrt{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) ^2 + \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right) ^3 } } \;-\;\frac{b}{3a} \end{align}

Unlike quadratic, cubic, and quartic polynomials, the general quintic cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions, as rigorously demonstrated by Abel (Abel's impossibility theorem) and Galois. However, certain classes of quintic equations can be solved [...] Source: http://mathworld.wolfram.com/QuinticEquation.html

At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:

If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?

• I once did some experimenting with a 3 dimensional polynomial of degree 15, but it escaped my lab and neighbouring departments started looking strangely at me when I met them so I put that research on pause for a bit. No but seriously in signal processing digital causal, linear and time invariant filters are polynomials in the $\mathcal Z$ transformdomain and have as large degree as the number of filter taps - 1 which can be anything ranging from 2 to thousands. – mathreadler Mar 21 '17 at 20:40
• Just because the formula for the roots of a cubic polynomial might be complicated, doesn't mean that cubic polynomials themselves should necessarily become more rare in nature. Same goes for higher order polynomials. – Morgan Sherman Mar 21 '17 at 21:48
• You get polynomials of all kinds of degrees in research. But often what you want to do then is not to explicitly express the roots in terms of the coefficients. Often you want instead to prove some structural result about the roots: when are they real vs complex, what region must they lie in, etc. – Sasho Nikolov Mar 22 '17 at 0:10
• Several of the comments and answers address your question focused on solving higher order polynomials by noting that in most important applications (and there are many) you don't usually need to find roots. The polynomials are more useful for things like modeling and interpolation. – Ethan Bolker Mar 22 '17 at 0:13
• I once found 3rd grade polynoms. It was about the approximative Van der Waals formula for real gases. I solved it numerically (more exactly, I wrote a function which finds the real roots of any degree polynoms numerically). – peterh Mar 22 '17 at 2:00

It's actually (somewhat) rare in research that you are lucky enough to have a polynomial that is only degree one or two. Some examples I haven't seen mentioned already: cyclotomic polynomials, irreducible polynomials used to construct field extensions, all sorts of polynomials used to construct algebraic curves used in cryptography.

While we don't always have nice formulas to explicitly find the roots, we have other ways to work with them. For example, we can choose nice polynomials that have a special form, or construct a polynomial so we already know the roots. We can use computational techniques to approximate the roots. We can use them in applications where we don't care what the roots are.

The characteristic polynomial of an $n \times n$ matrix has degree $n$. We often care about matrices larger than $2 \times 2$.

• The example I immediately thought of - linear recurrences that go more than $2$ steps back - is a special case of this. – Misha Lavrov Mar 21 '17 at 21:08
• And to find that characteristic polynomial will require, usually, about $c n^3$ operations, which itself is a cubic polynomial in the variable $n$. – Morgan Sherman Mar 21 '17 at 21:50
• @MorganSherman In fact it's worse, with $\lambda$ as a variable in there you can't compute the determinant by triangular reduction anymore, so you're stuck with the $O(n!)$ cofactor expansion algorithm. This is essentially never done for even $n=3$ except in special cases. – Ian Mar 21 '17 at 23:53
• @Ian You don't need to do cofactor expansion. See comments below this answer for example: math.stackexchange.com/a/405922/21512 – Morgan Sherman Mar 22 '17 at 0:06
• @alephzero (Cont.) Thus characteristic polynomials of larger matrices are important (because eigenvalues of such matrices are important), but we rarely see any particular ones. – Ian Mar 22 '17 at 3:18

Cubic polynomials are ubiquitous in computer-aided design and computer graphics.

They also are the basis for computer fonts.

Finite element analysis is based on polynomial functions. Isogeometric analysis uses NURBS.

• I'd like to add that other methods for numerically solving PDEs also use polynomial functions. High-order finite element, discontinuous galerkin, and even finite difference methods all involve polynomials of degree > 2 at some point in their analysis. – NoseKnowsAll Mar 22 '17 at 7:02
• For more about graphics and fonts, see this question: math.stackexchange.com/questions/825699/… – bubba Mar 24 '17 at 10:23
• @NoseKnowsAll Not to mention spectral methods... – ekkilop Jul 19 '17 at 19:43

From the applied math side of things, I come across higher order polynomials all the time. Here are some examples of when:

Eigenvalue problems for large matrices

Linear constant coefficient higher-order ODEs (think characteristic equation, or Laplace transform to transform ODE to solving a polynomial)

Numerical analysis - interpolation, cubic splines, higher accuracy integration schemes

Many times we expand analytic functions by their Taylor series, which as you know are polynomials of whatever degree we choose to stop expanding at.

Polynomials are ubiquitous throughout probably every field of higher mathematics :)

• Also orthogonal polynomials are used quite often to calculate special functions and for other purposes, and the order of them is usually higher than $2$. – Ruslan Mar 22 '17 at 5:47

Well a lot of modern cryptographic algorithms involve finite fields, and a practical way to construct a finite field of size $q = p^n$ where $p$ is a prime is as the set of polynomials over $\def\ff{\mathbb{F}}$$\ff_p of degree less than n, with addition and multiplication modulo some fixed irreducible polynomial over \ff_p of degree n. Note also that your question is (seemingly) based on a false dichotomy, namely that either you do not encounter a polynomial of degree higher than two, or you need to find its roots. As my above example application shows, this dichotomy is false with a certain irony; the irreducible polynomial has a trivial root in the constructed field, just X where X is the indeterminate, and we do not care about the other roots! (Just for kicks, consider \ff_8 constructed from \ff_2 using the irreducible cubic (X^3+X+1). The roots are X and X^2 and X^2+X, which you can prove easily by evaluating the cubic on each of them.) To expand the comment, in digital signal processing a digital filter can be described by it's \mathcal Z-transform, which is a polynomial in as many dimensions (variables) as the signal has (in this case N):$${\mathcal Z}\{f\}(z) = \sum_{(i_1,\cdots,i_N )\in {\mathbb Z}^N}c_{i_1,i_2,\cdots,i_N}\prod_{k=1}^N{(z_k)}^{i_k}$$where the coefficients c_{i_1,i_2,\cdots,i_N} are elements of an N index tensor. Polynomials of higher dimension than 1 are in the general case really difficult to factor because there are so many ways to factor them. Their theory is much more complicated than linear algebras ( "straight lines equations" in many dimensions ). But they occur almost everywhere. Geometric surfaces like found in graphics and computer aided design ("CAD" models) in mechanical engineering and filter factorization in signal processing are two examples. Vaughan Jones won the Fields Medal for finding a polynomial, which is usually not of degree two. An example in engineering: solving a thermal problem involving radiation. Heat transfer is a combination of conduction, convection and radiation. Simplifying things a lot, conduction and convection correspond to a first-degree polynomial in temperature T, but radiation corresponds to a term T^4 according to the Stefan-Boltzmann law. The equation for finding the temperature of a solid can therefore be written as (again, simplifying a lot)$$T^4+a T + b = 0,$$where a and b are determined by material properties, external temperatures, geometry, and so on. From memory: a is always positive and b is always negative. This equation has 4 complex roots, but only one positive real, and that is the temperature of the solid. I encountered this example in my real work, and suprised my coworkers by solving this explicitly without a need for approximations. This was five years ago, and I still remember it; this shows that at least for me personally, higher-order polynomials are a rarity in engineering. In engineering, some of the input always has some inaccuracy (from measurement errors or because they are specified within a range). If polynomials are encountered, it therefore makes no sense to try to find exact solutions; good approximations are ok. Numerical analysis has given tools to find roots for equations quickly. One of the equations below has a closed-form solution in elementary functions, the other one not.$$x^3-3x^2+16x-4=0e^{-x}-x-\arctan(x)=0$$If x represents an unkown in engineering (e.g. the optimal length in meters for your machine), you don't care about a closed form: if you can get the solution to within 0.0001, it is enough. • To add to your final point, nobody who needs the roots of a real cubic for engineering applications care about the closed form in radicals! In part this is because there is no way to simplify the closed form except to numerically compute it approximately! Worse still, if one wishes to avoid complex numbers one would have to use the radical form in one case and the trigonometric form in the other. Finally, if the coefficients are based on empirical measurements, there is absolutely no reason to ask for the closed form because the cubic is approximate from the beginning. =) – user21820 Mar 24 '17 at 9:34 Two more examples (that are both post-1950s mathematics): • In CAD (Computer Assisted Design), there is a growing concern about "offset curves", also named "wavefront curves", "dilated curves"... issued from the (mathematical morphology) dilation of curves defined by polynomials with degree \leq n, i.e., with parametric equations$$x=p(t); y=q(t) \ \ \ \text{with} \ \ \ \max(deg(p),deg(q))=n.$$(if n=2 for example it is a parabola, that can be studied as a quadratic Bezier curve). These curves are still polynomially defined but undergo an important degree elevation, for example have degree 8 when n=2. For explanations, I refer you to the very nice introduction here. Remark: these types of curves are generally studied as circle envelopes. [I had the privilege to meet their author at his home in Strasbourg in 1976]. Well, since any polynomial of order N with real coefficients can be factored into N_1 first-order binomial factors and N_2 quadratic factors where$$ N = N_1 + 2 N_2 $$then one might make a case that they never have to come across polynomials of higher degree than 2. In practice, I seldom come across polynomials of higher order than 3 (I sometimes use cubic Hermite splines for better-than-linear interpolation with look-up tables). I have used 7th-order polynomials for approximating vacuum-tube distortion curves (with 4x oversampling) in audio. Once I used a function$$ f(x) = \alpha x + x^9 $$it was meant to be just a linear feedback term, but the x^9 was meant to be a !!wake up!! factor that didn't kick in until x got large. But otherwise, it's pretty much all biquads for me. If I need roots it's pretty easy:$$ f(x) = x^2 \ - \ 2b\, x \ + c = (x-r_1)(x-r_2)$$where$ r_n = b + (-1)^n \sqrt{b^2-c} $or if the roots are complex conjugate$ r_n = b + i (-1)^n \sqrt{c-b^2} $. It almost never needs to be more complicated than that. • The linear / quadratic factorability is true only for single variate polynomials. Multi variate polynomials would be much less interesting if was always possible to do the same with them. – mathreadler Mar 22 '17 at 15:46 For the implementation of some Jacobi polynomial integration code I had to prove $$2^{-(\sigma+\tau+1)} \int\limits_{-1}^{+1} p_{n+r}^{(\lambda,\tau)} (t) p_n^{(\sigma,\tau)} (t) (1-t)^{\sigma} (1+t)^{\tau} \: dt\\ \ = \ \frac{\Gamma(\sigma+n+1)\Gamma(\tau+(n+r)+1)}{\Gamma(\sigma+\tau+n+(n+r)+2)} \cdot \frac{(\lambda+\tau+(n+r)+1)_n}{n!} \cdot \frac{(\sigma-\lambda-r+1)_r}{r!}$$ with$n,r \ge 0$integers,$\sigma,\tau > -1$and$\lambda$real numbers. Essentially this boiled down showing $$(\lambda+\tau+2n+2r) (\sigma+\tau+n+1) (\sigma+n+1) (\lambda+\tau+2n+r) r\\ + (\lambda+\tau+2n+2r) (\tau+n) (\sigma+\tau+2n+r+1) n (\sigma-\lambda-r+1)\\ - (\lambda+n+r) (\sigma+\tau+2n+1) (\sigma+\tau+2n+r+1) (\lambda+\tau+n+r) r \\ \ = \ (n+r) (\sigma+\tau+2n+1) (\tau+n+r) (\lambda+\tau+2n+r) (\sigma-\lambda-r+1),$$ which is an equality of fifth degree polynomials. A more general formula can be found in Erdelyi's Tables of Integral Transforms vol. 2, p. 287, 16.4.16. Jacobi polynomials are hypergeometric polynomials, a kind of special functions. There is a vast literature finding proofs for special functions or combinatorical identities (semi-)automatically using the Wilf-Zeilberger method, see https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair. Many proof certificates contain polynomials of high degree. Gaussian quadrature is a class of methods of numerical integration based on polynomial approximation. Typically, polynomials of degree$5$or$10$are employed. For many of the functions that arise in applied mathematics, these give very accurate results. Polynomials of arbitrarily high degree can in principle be used, but the additional accuracy is seldom relevant or worth the extra complexity entailed. In combinatorics and linear algebra and adjacent fields, finding eigenvalues of matrices is extremely important. Eigenvalues of matrices are solutions to polynomials with degree equal to the dimension of the matrix. Note that we can say "the dimension" because eigenvalues are only defined for square matrices. This is sufficiently important that it's its own topic in graph theory, known a spectral graph theory. I'm currently working on an applied problem in spectral graph theory where I am dealing with matrices with dimensions ranging from$3$to$96$(most are around$45\$).