# roots of recursive polynomials

I have recursive polynomials

$$Q_{n}(t)=tQ_{n-1}(t)+\frac{1-t^{2}}{n+1}Q'_{n-1}(t)$$

and

$$Q_{0}(t)=1$$

Is there a theory for finding a factorisation of recursive polynomials?

It is possible to show that $$\sum_{s=-\infty}^{\infty}sinc(\pi(x+s))^{p+1}=Q_{p-1}[cos(\pi x)]$$ where $$sinc(x)=sin(x)/x$$

Maybe you have some ideas for my case?

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What is "$sinc$"? – Wadim Zudilin Jun 14 '10 at 9:13
And what does $s$ vary over in the summation? – Robin Chapman Jun 14 '10 at 9:23

I have realised that your recurrent relation is exactly the one which appears in Eq. (2.4) in [Izvestiya: Mathematics 66:3 (2002) 489--542] (see also here). The properties of the corresponding polynomials are expressed in Lemma 2.2 and around (they seem to be exactly the ones you are trying to establish).

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Wadim, it looks to me that the denominator $n+1$ here should be $n-1$ to match the recurrence relation (2.4) from your paper. I don't know if that matters much though. – Vladimir Dotsenko Jun 14 '10 at 9:46
Vladimir, are you the author? :-) Yes, you are right but shifting the index of the sequence does not influence on its properties, does it? – Wadim Zudilin Jun 14 '10 at 9:56
Wadim, the author of what? I am just naturally curious :-) I think that indices are shifted in different directions though. Where the OP has $n$, you have $b+1$, but where he has $n+1$, you have $b$, so either one of you made a misprint, or these two sequences are different... – Vladimir Dotsenko Jun 14 '10 at 9:59
thank you, Wadim. I knew that this is derivative of cotangent, but I still don't know if it is possible to find its roots :(....I am reading your paper... – vilvarin Jun 14 '10 at 10:04
Wadim: OK, I understand what you mean - indeed, shifting by 2 seems to do the trick. Sorry for being slow. – Vladimir Dotsenko Jun 14 '10 at 10:08

To give a solution from scratch, start with the identity $$\sum_{s=-\infty}^\infty\frac1{(x+s)^2}=\frac{\pi^2}{\sin^2\pi x}.$$ We want to prove that this extends to an identity $$\sum_{s=-\infty}^\infty\frac1{(x+s)^n}=\frac{\pi^n P_n(\cos\pi x)}{\sin^n\pi x}$$ for each integer $n\ge2$, where the $P_n$ are polynomials satisfying some nice recursion. If this holds for some $n$ then differentation yields $$\sum_{s=-\infty}^\infty\frac1{(x+s)^{n+1}} =\frac{\pi^{n+1}\cos\pi x\\, P_n(\cos\pi x)}{\sin^{n+1}\pi x} +\frac{\pi^{n+1}\sin^2\pi x\\,P_n'(\cos\pi x)}{n\sin^{n+1}\pi x}$$ giving $$P_{n+1}(t)=tP_n(t)+\frac{(1-t^2)}{n}P_n'(t).$$ So, in your notation, $P_n=Q_{n-2}$.

However $$\sum_{s=-\infty}^\infty\mathrm{sinc}(\pi(x+s))^n =\sum_{s=-\infty}^\infty\frac{(-1)^{sn}\sin^n\pi x}{\pi^n(x+s)^n} =\frac{\sin^n\pi x}{\pi^n}\sum_{x=-\infty}^\infty\frac{(-1)^{ns}}{(x+s)^n}$$ which is the above sum when $n$ is even but not when $n$ is odd. Unless I have made some sign error (quite likely!) then your assertion holds for odd $p$ but needs to be modified for even $p$. Note that $$\sum_{s=-\infty}^\infty\frac{(-1)^n}{(x+s)^n}= 2^{1-n}\sum_{s=-\infty}^\infty\frac1{(x/2+s)^n} -\sum_{s=-\infty}^\infty\frac1{(x+s)^n}$$ so this is feasible.

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yes, this formula is indeed for odd p. For even p $Q_{0}(t)=t$ – vilvarin Jun 14 '10 at 10:41