Finding the value of a polynomial at zero Given a polynomial $p_n(x)\,$ with $\, n \geq 0 \,$  such  that $\, p_0(x)=1 \;,\; p_1(x)=x\;$ and $\; p_n(x)=xp_{n-1}(x)-p_{n-2}(x)\;$ ,  how can I find  $\;p_{10}(0)\; $  ?
 A: Since $p_n(0)=-p_{n-2}(0), \ n\geq2,$ it follows that the sequence $p_n(0)$ is $1,0,-1,0,1,0,\ldots$
A: Since $p_n(x)=xp_{n-1}(x)-p_{n-2}(x)$, it's trivial to see that $p_n(0)=-p_{n-2}(0)$. Then you can prove by induction that there is a simple closed-form expression for $p_n(0)$ when $n$ is even (and an even simpler one when $n$ is odd).
A: $$p_n(0) = 0\cdot p_{n-1}(0) - p_{n-2}(0) = - p_{n-2}(0) = + p_{n-4}(0) = -p_{n-6}(0) = \cdots$$
A: Writing $x$ as $2 \cos t = e^{it} + e^{-it}$, 
$$p_n(x) = x p_{n-1}(x) - p_{n-2}(x) \iff  p_{n}(x) = (e^{it}+e^{-it})p_{n-1}(x) - p_{n-2}(x)$$
has solutions of the form: 
$$p_n(x) = A e^{int} + B e^{-int}$$
Throw in the initial conditions, $p_0(x) = 1$ and $p_1(x) = x$, it is not hard to see
$$A = \frac{e^{it}}{e^{it}-e^{-it}}\quad\text{and}\quad B = \frac{-e^{-it}}{e^{it}-e^{-it}}$$
This implies:
$$p_n(t) = \frac{e^{i(n+1)t}-e^{-i(n+1)t}}{e^{it}-e^{-it}} = \frac{\sin (n+1)t}{\sin t} = U_n(\cos t) = U_n(\frac{x}{2})
$$
where $U_n(x)$ is the Chebyshev polynomails of the second kind. In particular, this means 
$$p_{10}(x) = U_{10}(\frac{x}{2}) = x^{10}-9x^8+28x^6-35x^4+15x^2-1$$
and hence $p_{10}(0) = -1$. If you want $p_n(0)$ in general, you have:
$$p_n(0) = U_n(0) = \frac{\sin (n+1)\frac{\pi}{2}}{\sin \frac{\pi}{2}} = \begin{cases}1,& n \equiv 0 \pmod 4\\-1, & n\equiv 2 \pmod 4\\0,&\text{otherwise}\end{cases}$$
