3
$\begingroup$

The Chebyshev T-polynomials have at least two natural definitions, either via the characterizing property $\cos(n\theta) = T_n(\cos(\theta))$, which I will call the geometric definition, or via a recurrence relation $T_{n+1} = 2x\ T_n - T_{n-1}$. My question concerns the relationship between these two definitions, and specifically asks if other families of polynomials defined by similar recurrence relations have a natural geometric interpretation, similarly to how $T_n$ expresses the relation between x-coordinates of particular points on the circle.

Starting from the geometric definition of $T_n$, it is straightforward to derive the recurrent definition. Is there a natural way of going the other direction? One thought I have had is as follows. If we treat the $T_n$ as elements of the coordinate ring of the circle, then $T_n$ expresses the relationship between the two parameterizations of the x-coordinates given by $\theta \mapsto \cos(\theta)$ and $\theta \mapsto \cos(n\theta)$. Can we do the same sort of thing with other families of polynomials defined by similar recurrence relations (say, for simplicity, second-order linear polynomial recurrences with coefficients of degree $\leq 1$)?

One potential obstacle I have encountered is that $\cosh(n\theta) = T_n(\cosh(\theta))$ as well, so that the hyperbola $X^2 - Y^2 = 1$ is just as natural a choice as the circle for a geometric object associated to $T_n$.

Here's my main question: given a particular family of polynomials $P_n$ related by a polynomial recurrence of an appropriately "nice" type, can one associate one or more algebraic curves (or other geometric objects) so that $P_n$ expresses some relationship between various points on the curve?

Improve this question
$\endgroup$
1
  • $\begingroup$ Now that I think about it... don't the other three kinds of Chebyshev polynomials, due to their being representable as compositions of trigonometric functions and their inverses, admit similar geometric interpretations as well? $\endgroup$
    – J. M. isn't a mathematician
    Commented Aug 29, 2010 at 3:59

2 Answers 2

Reset to default
5
$\begingroup$

You have to realize that you are asking a really vague question. Specifically, you haven't defined "similar (nice) recurrence relation" and "some relationship between various points on the curve".

When you say similar recurrence relation, I believe the next step to generalizing Chebyshev polynomials is to consider orthogonal polynomials, which satisfy a recurrence relation, but at the same time they come from certain differential equations which sometimes allow us to express them using Rodrigues' formula. For Chebyshev polynomials, this connects their recurrence relation $$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$$ to the formula $$T_n(x)=\frac{\Gamma(1/2)\sqrt{1-x^2}}{(-2)^n \Gamma(n+1/2)}\frac{d^n}{dx^n}\left([1-x^2]^{n-1/2}\right).$$ Similar formulas apply to most families of orthogonal polynomials, however the formula $$T_n(x)=cos(n \arccos(x))$$ is a bit special and characteristic of Chebyshev polynomials. In fact if you had a family of polynomials $\{P_k\}_{k\geq 1}$ so that you could express them as $$P_n(x)=f(nf^{-1}(x))$$ for some function $f$, then they would trivially satisfy $P_n(P_m(x))=P_m(P_n(x))=P_{mn}(x)$ and by a result of Block and Thielman, these polynomials must coincide with a linear transformation of either the power sequence $\{x^k\}$ or the Chebyshev polynomials $\{T_k\}$.

Another perspective which sheds a little more light on the geometric properties of Chebyshev polynomials is Dessin d'enfant. The Belyi functions corresponding to trees are the Shabat polynomials which are generalizations of Chebyshev polynomials. I believe this is the right context to observe the symmetries of functional composition of Chebyshev functions. See these articles.

Improve this answer
$\endgroup$
1
$\begingroup$

I'll just focus on your first question (from the recursive relation to the geometric definition), in the hope the answer may give you hints for the other ones. If you think $x$ as a parameter, you want to solve the linear difference equation $T_{n+1}-2xT_{n}+T_{n-1}=0$ with the initial conditions $T_0=1$, $T_1=x$. By the usual methods (general solution as a linear combination of exponential sequences, or also, via generating series) you easily find the solution: $$T_n(x)=\frac{1}{2}\left[\big(x+\sqrt{x^2-1}\big)^n+\big(x-\sqrt{x^2-1}\big)^n\right].$$ The presence of the term $\sqrt{x^2-1}$ suggests the position $x:=\cos\theta$, so that the above writes $$T_n(\cos\theta)=\frac{e^{in\theta}+e^{-in\theta}}{2}=\cos(n\theta). $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you. I wonder if this can be extended to address the second question. If you take $y = \sqrt{x^2 - 1}$ then we have the relation $x^2 - y^2 = 1$, from which the hyperbola arises naturally as the geometric object underlying the Chebyshev polynomials. It seems like a similar technique can be used to introduce a $y$ element that generates a finite extension of $\mathbb{C}[x]$ - an algebraic curve! $\endgroup$
    – Nick Salter
    Commented Aug 28, 2010 at 23:51
  • $\begingroup$ Nick: You might want to recall that the definition $T_n(x)=\cos(n \arccos(x))$. Remember also that $\arccos$ is real only for x in the interval $[-1,1]$ (which anyway is the usual interval of interest for Chebyshev). The key here is that for arguments outside $[-1,1]$, $\arccos(x)=i\mathrm{arcosh}(x)$, and $\cos(ix)=\cosh(x)$. Thus it is no surprise that $T_n(x)=\cosh(n \mathrm{arcosh}(x))$ is a valid representation for the Chebyshev polynomial outside the usual support interval. $\endgroup$
    – J. M. isn't a mathematician
    Commented Aug 29, 2010 at 0:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .