It is well-known that the Chebyshev polynomials of the first kind satisfy the recurrence relation $$ \begin{cases} T_{n}(x)=2xT_{n-1}(x)-T_{n-2}(x) \qquad n \geq 2 \\ T_{0}(x)=1, \ \ T_{1}(x)=x \\ \end{cases} $$

The perturbed Chebyshev polynomials are defined by the recurrence $$f_0(x)=b,\ \ f_1(x)=x−c,\ \ f_{n+1}(x)=(x−d)f_n(x)−af_{n−1}(x),\ n\geq1, $$ where $a,b,c,d\in \mathbb R$ and $a>0$. These polynomials generalize the Chebyshev polynomials, which are obtained by setting $a=1/4$, $c=d=0$ and $b\in\{1,2\}$. See, for example:

Stoll, Thomas. "Decomposition of perturbed Chebyshev polynomials." Journal of Computational and Applied Mathematics 214.2 (2008): 356-370.

Let us consider the sequence $\{T_n\}_{n\in\mathbb N}$. Perturb this sequence could mean also consider $\{T_{\alpha_n}\}_{n\in\mathbb N}$ with $\alpha_n$ is a sequence of real numbers. I wonder if a similar sequence of Chebyshev polynomials is known, and if it is used in some applications. I thought this question by reading some articles on Non-harmonic Fourier series, where $\{e^{i n t}\}_{n\in\mathbb N}$ is replaced by $\{e^{i \alpha_n t}\}_{n\in\mathbb N}$.

Thanks in advance.

  • 1
    $\begingroup$ $T_n$ and $f_n$ are polynomials of degree $n$. Is your $T_{\alpha_n}$ supposed to be a polynomial, and if so of what degree? $\endgroup$ – Robert Israel Aug 31 '15 at 19:19
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    $\begingroup$ To me, the reason that the Chebyshev polynomials are so important lies in the identities $T_n((x+x^{-1})/2)=(x^n+x^{-n})/2$ and the consequence that they commute under composition. Neither of these conditions holds when you perturb, although I guess that if $a-\frac14,b-1,c,d$ are all small, you'll have small perturbations of these fundamental properties. A nice way to think of the Chebyshev polynomials is they tell you what happens to the power map $x^n$ when you identify $\mathbb R^*/(x=x^{-1})$ with $\mathbb R$ via $x\to (x+x^{-1})/2$. $\endgroup$ – Joe Silverman Aug 31 '15 at 20:29
  • $\begingroup$ @RobertIsrael, Yes, doing so $\{T_{\alpha_n}\}_{n\in\mathbb N}$ is no longer a polynomials sequence. $\endgroup$ – Mark Sep 1 '15 at 11:29
  • $\begingroup$ Related to what @JoeSilverman wrote, $T_n(\cos \theta) = \cos(n \theta)$. You could define for non-integer $\alpha$, $T_\alpha(z) = \cos(\alpha \arccos(z))$, an analytic function on the complement of the branch cuts for $\arccos$. $\endgroup$ – Robert Israel Sep 2 '15 at 19:15

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