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

$$P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$$

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).

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    $\begingroup$ 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

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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 $P_n$.

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

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