Answering a recent question I came across the family of polynomials:


with numerical evidence of the following interesting properties:

  • $P_n(2)=\begin{cases}+1,& n=0,3\mod4\\-1,&n=1,2\mod4\end{cases}$
  • $P_n(4)=(-1)^n(2n+1)$
  • all roots are simple and real and belong to the interval $(0,4)$
  • the absolute value of the polynomial on $(0,2)$ seems to be bounded by a value close to $\sqrt2$

The first two properties can be easily proved using the recurrence relation $$ P_{n+1}(x)=P_n(x)-x\sum_{j=0}^n P_j(x), $$ but I have no idea how to approach the other two. Any hint is appreciated. Additionally I would like to estimate the absolute value of the largest extremum (it is situated between the last two zeros).

  • 1
    $\begingroup$ After a few minutes of messing around this, Chebyshev-like identity seems to be true: $(-1)^n\sin(\tfrac{x}{2})P_n(2\cos(x)+2) = \sin(\frac{2n+1}{2}x)$. Maybe this helps. $\endgroup$
    – Lee Fisher
    Feb 20 at 23:31

2 Answers 2


Let $S_n(x) = \sum_{k=0}^n P_k(x)$, and $U_n(y) = S_n(2 - 2y)$, it is easy to prove that $U_0(y) = 1$, $U_1(y) = 1 + 1 - (2-2y) = 2y$ and $$U_{n+1}(y) = 2y U_n(y) - U_{n-1}(y)$$ $U_n$ is then the Chebychev Polynomials of the second kind

\begin{align} P_n(2 - 2\cos(x)) &= S_{n}(2 - 2\cos(x)) - S_{n-1}(2 - 2\cos(x))\\ &= U_{n}\left(\cos(x)\right) - U_{n-1}\left(\cos(x)\right)\\ &= \frac1{\sin(x)} \left(\sin((n+1)x) - \sin(nx)\right)\\ &= \frac{2}{\sin x}\sin\left(\frac x2\right)\cos\left(\frac{2n+1}2x\right)\\ &= \frac{\cos\left(\frac{2n+1}{2}x\right)}{\cos\left(\frac x2\right)} \end{align}

This proves that $t_k = 2 - 2 \cos\left(\frac{2k+1}{2n+1}\pi \right)$, $k = 0,\ldots,n-1$ are the roots of $P_n$.

For the fourth question:

\begin{align} \sup_{t\in (0,2)} \left|P(t)\right| &= \sup_{x\in \left(0,\frac{\pi}2\right)}\left|P(2-2\cos(x))\right|\\ &= \sup_{x\in\left(0,\frac\pi2\right)} \left|\frac{\cos\left(\frac{2n+1}{2}x\right)}{\cos\left(\frac x2\right)}\right|\\ &\le \sup_{x\in\left(0,\frac\pi2\right)} \left|\frac{1}{\cos\left(\frac x2\right)}\right|\\ &= \frac1{\cos\left(\frac\pi 4\right)} = \sqrt 2 \end{align}

Note: $$P_n(x) = \begin{cases} 1 & \text{if $n = 0$}\\ U_{n}\left(1-\frac x2\right) - U_{n-1}\left(1-\frac x2\right) &\text{if $n \ge 1$} \end{cases} $$

this may help in any study of $P_n$.

  • $\begingroup$ An excellent answer! $\endgroup$
    – user
    Feb 21 at 9:13

We have the identity: $$\sin\left(\frac{x}{2}\right)P_n(2\cos(x) +2) = (-1)^{n+1}\sin\left(\frac{2n+1}{2}x\right)$$

This holds by induction on the recursion you provided.

Now this should answer your third point, yes all the roots are in the interval $[0,4]$. It should not be difficult to find the value at the largest extrema using this identity.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .