Is there a closed form for these polynomials?

Question

Let $P_0(x)=1, P_{-1}(x)=0$ and define via recursion $P_{n+1}(x)=xP_{n}(x)-P_{n-1}(x)$.

The first few polynomials are

$$ P_0(x)= 1\\ P_1(x) = x \\ P_2(x) = x^2-1 \\ P_3(x)= x^3 -2 x\\ P_4(x) = x^4 - 3 x^2 +1 \\ P_5(x) = x^5 - 4 x^3 +3 x$$

It appears the polynomials are always the form:

$$P_n(x)=\sum_{k=0}^{\lfloor n/2\rfloor+1} (-1)^k\ \ f(n,k)\ x^{n-2k}$$

Where for example one can calculate

$$f(n,0)=1 \\ f(n,1)=n-1 \\ f(n,2)=\frac{(n-2)(n-3)}{2}$$

Is there a closed form for these polynomials? Aside from that I am specifically interested in whether or not $\sum_{n=0}^\infty |P_n(x)|^2$ diverges for every $x \in \mathbb{C}$.

Answer 1

$P_n(x)=U_n(x/2)$ where $U_n$ is the $n$-th Chebyshev polynomial of the second kind, with (very close) recurrence relationship $U_{n+1}(x)=2x U_n(x)-U_{n-1}$.

A formula that could be of interest for the aim you follow is:

$\dfrac{1}{1-2tx+t^2}=\sum_{k=0}^{\infty}U_k(x)t^k$

(Gradshteyn and Ryzhik formula 8.945)

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(New edit) In connection with the series you are looking for.

1) Your problem is connected with the Darboux-Christoffel kernel.

$K_n(s,t)=\sum_{k=0}^{n}U_k(x)U_k(t)$

Notice that it is not the $P_n$ but the $U_n$.

Let $K_n(x,x)=W_n(x)=\sum_{k=0}^{n}U_k^2(x)$.

One can show that $W_n(x)$ can be expressed as a closed-form formula:

$W_n(x)=U_{n+1}^{\prime}(x)U_{n}(x)-U_{n}^{\prime}(x)U_{n+1}(x)$ (prime for derivative wrt $x$).

(letter $W$ for "Wronskian", a classical tool in ordinary differential equations).

By its definition, $W_n$ is an even function of $x$. It can be proven that the (dominant) degree $2n$ coefficients of $U_{n+1}^{\prime}(x)U_{n}(x)$ and $U_{n}^{\prime}(x)U_{n+1}(x)$ do not cancel. Thus degree($W_n(x)$)=$2n$.

For example, for $n=6$ : $W_6(x)=4-6x^2+30x^4-45x^6+30x^8-9x^{10}+x^{12}$.

I have plotted several curves $y=W_n(x)$ on $[-1,1]$. All of them make small variations above or below a second degree curve, which has its minimum in

$(0,\left \lfloor{\dfrac{n+2}{2}}\right \rfloor)$ and passes through the points $(\pm 1,\left \lfloor{\dfrac{2n+4}{3}}\right \rfloor)$ (where $\left \lfloor a \right \rfloor$ designate the integer part of $a$).

Here is the curve of $W_{30}$:

Thus, experimentaly, one can be fairly confident that for any fixed $x$, the sequence $W_n(x)$ tends to $\infty$ when $n$ tends to $\infty$. But, this is not a proof...

2) In addition, one can read the online article

https://www.ma.utexas.edu/mp_arc/c/08/08-107.pdf

by Barry Simon, a recognized specialist of orthogonal polynomials, especially the $U_n$.

3) Have a look to the similar question (What is the name for the polynomials of the form : $ P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$?)

Answer 2

As @JeanMarie states, the recurrence is for $U_n(x/2)$. A cool closed form for these Chebyshevs is

$$U_n(\cos{\theta}) = \frac{\sin{(n+1) \theta}}{\sin{\theta}}$$

so in your case such an expression would look like

$$P_n(x) = \left (1-\frac{x^2}{4} \right )^{-1/2} \sin{\left [(n+1) \arccos{\left ( \frac{x}{2} \right )} \right ]}$$

Answer 3

Here is part of an answer for the second part of the question, namely a funny little argument I found why $\sum_n |P_n(\lambda)|^2$ must diverge for any $\lambda \in \mathbb{C-R}$.

To see this consider on $\mathscr l^2 (\mathbb N)$ the left shift operator $L$ and its adjoint, the right shift operator $R$.

Note that since $L+R$ is hermitian its point-spectrum must be a subset of $\mathbb R$. In trying to find the eigenvectors of this operator is where these polynomial initially came up.

Suppose $(L+R)(x)=\lambda x$ for an $x$ in $\mathscr l^2 (\mathbb N)$. Write $x= \sum_n x_n e_n$. $L(x)=\sum_n x_{n+1} e_n$ and $R(x)=\sum_n x_n e_{n+1}$. $(L+R)(x)=\lambda x$ is equivalent to

$$\sum_n (x_{n+1}+x_{n-1}-\lambda x_n) e_n = 0 $$

Or

$$x_{n+1}= \lambda x_n - x_{n-1} \tag{1}$$

where $x_{-1}=0$.

If we have that $x_0=0$, then the only way to satisfy equation (1) is $x_n=0$ for all $n$, so we can suppose $x_0\neq0$, by rescaling the vector wlog $x_0=1$, and equation (1) is satisfied only by the polynomials given in the original question.

If for some $\lambda$: $\sum_n |P_n(\lambda)|^2 < \infty$, then with $x:=\sum_n P_n(\lambda) e_n$ we have a well defined vector in $\mathscr l^2 (\mathbb N)$ whose components satisfy (1) and for that reason is an eigenvector of $L+R$ with eigenvalue $\lambda$.

Since as stated before $L+R$ is hermitian it cannot have nonreal eigenvalues and for that reason $\sum_n |P_n(\lambda)|^2$ must diverge if $\lambda \notin \mathbb R$.

The remaining question of whether or not the sum diverges for the other $\lambda$ is equivalent to $L+R$ having empty point spectrum.