The usual Chebyshev polynomials of the first kind are defined by the recursive relation: $$T_0(x)=1,\\ T_1(x)=x,\\ T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$$ One can use the same recursive relation to define the Chebyshev Polynomials of the second kind, by slightly changing the initial condition: $$U_0(x)=1,\\ U_1(x)=2x,\\ U_{n+1}(x)=2xU_n(x)-U_{n-1}(x)$$
An alternative (but equivalent) way to define the Chebyshev Polynomials is through the trigonometric identities: $$T_n(\cos(\theta))=\cos(n\theta)\\ U_n(\cos(\theta))\sin(\theta)=\sin((n+1)\theta)$$
I am interested in considering "generalized" Chebyshev Polynomials, which are defined through the same recursive relation: $$P_{n+1}(x)=2xP_n(x)-P_{n-1}(x)$$ where this time the initial condition can be more general ($P_0$ would be some constant and $P_1$ would be some polynomial of degree one). Given a generalized Chebyshev Polynomial, I am interested to know whether it can also be defined through some nice trigonometric identity, similar to $T_n$ and $U_n$. My question is:
Does anyone have an idea on how to approach such a problem? I can just plug in $\cos(\theta)$ as $x$ and try to see what I can get, but perhaps there is a smarter (less ad hoc) way, which does not require me to fix a specific initial condition.
Has this been done somewhere? Even for specific cases (other than the two I stated). It seems likely that this problem has been visited in the past, so it'd be good to know if there are some relevant existing results.
Thanks in advance.
In general, if you have a recurrence relation of the form $$ a_0, \cdots, a_{n-1} \text{ arbitrary }, \quad a_n = \sum_{i=1}^{n} c_i a_{n-i}, \quad c_i \in R $$ (where $R$ is some ring (commutative unital let's say)), the set of sequences $a_n$ satisfying that recurrence relation is a free $R$-module of dimension $n$; that's just because there's always a unique solution given any set of initial parameters.
I'm saying this because you're using the same recurrence relation in both sequences $T_n$ and $U_n$, the difference being that you're working with the ring $\mathcal R$ of (let's say) analytic functions in one variable $x$ over the real numbers (polynomials and trigonometric functions are analytic). By analytic I mean admitting a convergent Taylor expansion everywhere.
For initial parameters $a_0, a_1 \in \mathcal R$, let $C(a_0,a_1)_n$ denote the corresponding solution to your recurrence relation, so that $C(a_0,a_1)_n$ is characterized by $$ C(a_0,a_1)_0 = a_0, \quad C(a_0,a_1)_1 = a_1, \quad C(a_0,a_1)_n = 2x C(a_0,a_1)_{n-1} - C(a_0,a_1)_{n-2}. $$ Notice that for $a_0,a_1,b_0,b_1 \in \mathcal R$, we have $$ C(a_0,a_1)_n + C(b_0,b_1)_n = C(a_0 + b_0, a_1 + b_1)_n $$ and for $\lambda \in \mathcal R$, $$ C(\lambda a_0, \lambda a_1)_n = \lambda C(a_0,a_1)_n $$ by $\mathcal R$-linearity of the condition imposed on $C$ (this is what I meant by $\mathcal R$-linearity; I do not mean $\mathbb R$-linearity!).
We have $$ C(1,x)_n = T_n, \qquad C(1,2x) = U_n. $$ If $f,g \in \mathcal R$ are arbitrary, the $\mathcal R$-linearity shows that $$ C(f,g)_n = f C(1,0)_n + g C(0,1)_n, $$ so it suffices to compute only those.
We see that as vectors in $\mathcal R^{\oplus 2}$, $$ 2(1,x) - (1,2x) = (1,0) $$ so $$ C(1,0)_n = 2 T_n - U_n. $$ Now let us have a look at $C(0,1)_n$. We have $$ a_0 = 0 \\ a_1 = 1 \\ a_2 = 2x(1) - 0 = 2x $$ and the recurrence relation continues afterwards exactly like $U_n$ since it has the same initial conditions, just shifted by an index. That means $$ \forall n \ge 1, \quad C(0,1)_n = U_{n-1}, \quad C(0,1)_0 = 0. $$ Therefore, if $f, g \in \mathcal R$ are arbitrary analytic functions, you will have $$ C(f,g)_n = f C(1,0)_n + g C(0,1)_n = f(2 T_n - U_n) + g U_{n-1}. $$ For instance, if $f=1$ and $g=x$, you have $$ T_n = C(1,x)_n = (2 T_n - U_n) + x U_{n-1}. $$ (and for $n=0$, we have $T_0 = C(1,0)_0 + C(0,1)_0 = 1 + 0 = 1$; I couldn't use the above formula in the general case because $C(0,1)$ is defined separately for $n=0$).
How come is that true? Well let's put U's on one side and T's on the other: $$ U_n - x U_{n-1} = T_n $$ This is true for $n=1$: $$ U_1 - x U_0 = 2x - x(1) = x = T_1 $$ and for $n=2$: $$ U_2 - x U_1 = (4x^2-1) - x(2x) = 2x^2 - 1 = T_2 $$ and just because you might be worried about the apparition of $n-3$ in the following, it's also true for $n=3$: $$ U_3 - x U_2 = \\ (2x(4x^2-1) - 2x) - x(4x^2 - 1) \\ = 4x^3 - 3x \\ = 2x(2x^2-1) - x \\ = T_3 $$
and if we substitute in the recurrence relation: $$ U_n - x U_{n-1} \\ = (2x U_{n-1} - U_{n-2}) - x(2x U_{n-2} - U_{n-3}) \\ = 2x(U_{n-1} - x U_{n-2}) - (U_{n-2} - x U_{n-3}) \\ = 2x T_{n-1} - T_{n-2} \\ = T_n. $$ So yeah, I didn't do any sorcery apparently. Note that by re-arranging the previous equation: $$ T_n = (2 T_n - U_n) + x U_{n-1}. $$ You get $$ U_n \\ = (2T_n - T_n) + x U_{n-1} \\ = T_n + x U_{n-1} \\ = ((2 T_n - U_n) + x U_{n-1}) + x U_{n-1} \\ = (2 T_n - U_n) + 2x U_{n-1} \\ = (1) C(1,0)_n + 2x C(0,1)_n. $$ I guess in some sense $T_n$ and $U_n$ were considered fundamental to this system since $U_n = C(0,1)_{n+1}$ and $C(0,1)_0 = 0$, so people just dismissed the "zero" party of $C(0,1)$.
That's what I have to say. I guess there could be more fundamental stuff to discuss about this, but I'm out for now!
Hope that helps,