Csdrhrt

Can anyone tell me if the following set of polynomials have a special name?
$$P_{0}(x)=1,P_{1}(x)=x$$
$$P_{n}(x)=xP_{n-1}-P_{n-2}$$
The above gives:
$$P_{2}(x)=x^2-1;P_{3}(x)=x^3-2x;P_{4}(x)=x^4-3x^2+1;P_{5}(x)=x^5-4x^3+3x$$
So $P_{n}(x)$ has parity $(-1)^n$. I was trying to find out whether they are orthogonal, but couldn't find a suitable weight function. My main concern is to prove that $P_{n}(x)$ has n distinct real roots, all larger than or equal to -2.

Your polynomials are indeed a special case of classical orthogonal polynomials.
According to Abramowitz/Stegun 22.7.6 you have
$$P_n(x) = S_n(x)= U_ nleft(frac{x}{2}right)$$
where $U_n(x)$ is the well-known Chebyshev polynomial of the second kind.

The weight function for the interval $(-2,2)$ is
$$w(x)=left(1-frac{x^2}{4}right)^{1/2}$$

And of course this means that the root are simple, distinct and located in the interval $(-2,2).$ For a proof see e.g. my answer
Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros .

The orthogonality of $P_n$ follows from the correspending property of $U_n$ and $sin$ , see e.g. https://www.sciencedirect.com/science/article/pii/0377042793901485:

With $x=costheta$ and $sin theta = (1-x^2)^{1/2}$ you have
$$U_n(cos theta) = frac{sinbig((n{+}1)thetabig)}{sintheta}$$
so $$(1-x^2)^{1/2}U_n(x) = sinbig((n{+}1)thetabig)$$

(Although I did not see any fully formulated proof yet, maybe a direct proof from the recursion can be modelled after https://planetmath.org/orthogonalityofchebyshevpolynomialsfromrecursion)

In what follows the explicit expression for the roots of the polynomials $P_n(x)$ will be derived. The statement "the roots are all distinct, real and less than 2 by absolute value" follows immideately.

Consider a family of $ntimes n$ bidiagonal matrices:
$$begin{align}
A^{(n)}_{ij}=&delta_{i-j,1}+delta_{j-i,1},
end{align}tag{1}$$
given below for $n=5$ as example:
$$
A^{(5)}=begin{pmatrix}
0&1&0&0&0\
1&0&1&0&0\
0&1&0&1&0\
0&0&1&0&1\
0&0&0&1&0
end{pmatrix}.
$$

Lemma 1. The eigenvalues of the matrix (1) are:
$$
begin{align} lambda_m=2cosfrac{pi m}{n+1},&
text{with associated eigenvectors } u_{mk}=sinfrac{pi m}{n+1}k,
end{align}tag{2}
$$
where $m$ and $k$ run from 1 to $n$.

Though it would suffice for the proof to let the matrix $A$ act on the given vectors, we present below an extended "constructive" version.

Assume the elements of an eigenvector $u$ have the form:
$$
u_k=e^{alpha k}+ae^{-alpha k},tag{3}
$$
with some parameters $a$ and $alpha$, which are to be found.

Obviously for all $k=2dots(n-1)$
$$
(Au)_k=left(e^{alpha k}+ae^{-alpha k}right)left(e^alpha+e^{-alpha}right)
=left(e^alpha+e^{-alpha}right)u_k.tag{4}$$

Thus it remains only to find such $a$ and $alpha$ that the equation (4) is satisfied for $k=1$ and $k=n$ as well.

For $k=1$:
$$
e^{alpha 2}+ae^{-alpha 2}=left(e^alpha+e^{-alpha}right)left(e^alpha+a e^{-alpha}right)
Leftrightarrow
1+a=0.tag{5}
$$

For $k=n$:
$$
e^{alpha (n-1)}+ae^{-alpha(n-1)}=left(e^alpha+e^{-alpha}right)left(e^{alpha n}+a e^{-alpha n}right)
Leftrightarrow e^{alpha (n+1)}+ae^{-alpha(n+1)}=0.tag{6}
$$

It follows: $a=-1$, $alpha=frac{pi m}{n+1}i$, where $m$ is an integer number. Plugging the values into (3) and (4) one obtains (2).

As all $n$ eigenvalues are distinct, Lemma 1 is proved.

Lemma 2. The characteristic polynomials of negated matrix (1):
$$
Q_n(x)equivleft|A^{(n)}+x I^{(n)}right|,
$$
where $I^{(n)}$ is $ntimes n$ dimensional identity matrix, are the polynomials in question:
$$Q_n(x)=P_n(x)tag{7}.$$

For $n=1$ and $n=2$ the equality (7) is obvious. Assume that (7) is valid for all $n<N$. Then it is valid for $n=N$ as well.

Indeed, applying the Laplace expansion to matrix $A^{(N)}$ (with $N>2$) one readily obtains:
$$
Q_N(x)=x Q_{N-1}(x)-Q_{N-2}(x)stackrel{I.H.}{=}x P_{N-1}(x)-P_{N-2}(x)=P_N(x).tag{8}
$$

Thus, by induction Lemma 2 is proved.

Now, as the eigenvalues of a matrix are exactly the roots of its characteristic polynomial,

Lemma 3. The roots of $P_n(lambda)$ are:
$$
lambda^{(n)}_m=2cosfrac{pi m}{n+1}, quad m=1dots n
$$
is a simple Corollary of Lemmas 1 and 2.