Showing that a family of polynomials has positive and real roots. - MathOverflow

Showing that a family of polynomials has positive and real roots.

2012-07-16T19:56:50Z

Hi everybody, for my research I am dealing with the following function:

$$\alpha_n(x):=\left.\frac{\partial^{2n+1}}{\partial z^{2n+1}}\frac{\sinh(z)}{\cosh(z)-1+x}\right|_{z=0},\quad n\in \mathbb{N},$$

It is possible to show that $$\alpha_n(x)=\frac{P_n(x)}{x^{n+1}},$$ where $P_n(\cdot)$ is a polynomial of order $n$ in $x$, having integer coeffients.

To make few concrete examples $$\alpha_0(x)=\frac{1}{x}$$ $$\alpha_1(x)=\frac{-3+x}{x^2}$$ $$\alpha_2(x)=\frac{30-15 x+x^2}{x^3}$$ $$\alpha_3(x)=\frac{-630+420 x-63 x^2+x^3}{x^4}$$ $$\alpha_4(x)=\frac{22680-18900 x+4410 x^2-255 x^3+x^4}{x^5}$$ and so on.

What I would need to show (and it is veryfied for all the special cases I was able to compute, like those above) is that all the roots of $P_n(x)$ (and therefore those of $\alpha_n(x)$) are real and strictly greater than 2.

An explicit albeit complicated expression for $\alpha_n(x)$ can be obtained, namely:

\begin{equation*} \begin{array}{ll} \alpha_n(x)=&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j+1}^{n} (2k)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k}\sum_{i=j}^{k}\left[ {i+1\choose j+1}\binom{2k}{2 i+1}-{i\choose j+1}\binom{2k}{2 i}\right](-2)^{j+1-2k}+\\ &x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j}^{n}{(2k+1)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k+1}}\sum_{i=j}^{k} \left[ {i+1\choose j+1}\binom{2k+1}{2 i+1}-{i\choose j+1}\binom{2k+1}{2 i}\right](-2)^{j-2k}, \end{array} \end{equation*}

where the number between the curly brakets are the Stirling number of the second kind; Moreover,

$$\alpha_n(2)=-\sum_{k=1}^{2n+1}{k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}}(-2)^{-k}\neq0.$$

If someone is interested, I can post more on how I got these expressions. Thanks in advance to everybody that will try to help me.

Best Regards

Enzo

Answer by John Mangual for Showing that a family of polynomials has positive and real roots.

2012-07-16T23:43:48Z

How about re-writing with Cauchy Residue formula?

$$ \alpha_n(x) = \frac{(2n+1)!}{2\pi i} \oint \frac{dz}{z^{2n+2}}\cdot \frac{\sinh z }{\cosh z -1 + x} $$

Not sure how it helps you find roots or establish they are real.

Answer by David Speyer for Showing that a family of polynomials has positive and real roots.

2012-07-16T23:44:32Z

I'm noticing a pattern in the numerators. To avoid getting indices wrong, I'll write out an example: $$(22680, 18900, 4410, 255, 1) = (36 \times 630, 16 \times 630+21 \times 420, 9 \times 420 + 10 \times 63, 4 \times 63 + 3 \times 1, 1)$$ Here $(36, 21, 10, 3)$ is the odd indexed triangular numbers and $(16, 9, 4,1)$ are the squares.

The above recursion translates into the following relation between the $\alpha$'s: $$\alpha_{n+1} = (x-2) D( x D \alpha_n) + D\alpha_n$$ where $D$ denotes differentiation with respect to $x$. As Victor points out in the comments below, this recursion is equivalent to the relation $$D_z D_z f = (x-2) D_x ( x D_x f) + D_x f$$ where $f = \sinh(z)/(\cosh(z)-1+x)$. This latter relation can be verified, without much insight, by typing it into a computer algebra system.

Inductively, suppose that $\alpha_n$ has $n$ real roots, $2 < q_1 < q_2 < \cdots < q_n$. Set $\beta = D \alpha_n$, then by Rolle's theorem $\beta$ has $n-1$ real roots $2 < r_1 < q_1 < r_2 < q_2 < \cdots < r_{n-1} < q_{n}$. Moreover, $\lim_{x \to \infty} \alpha(x)=0$, forcing another root of $\beta$ at $r_n > q_n$.

Using Rolle's theorem again and the fact that $\lim_{x \to \infty} x \beta =0$ again, we see that $D(x \beta)$ has $n$ real roots, $2 < s_1 < r_1 < s_2 < r_2 < \cdots < s_n < r_n$. We make a little chart of the signs of our functions: $$\begin{array}{rcccccccccc} x: & 2 & r_1 & s_1 & \cdots & s_{n-2} & r_{n-1} & s_{n-1} & r_n & s_n & \gg s_n \\ \beta: & \pm & 0 & \mp & \cdots & - & 0 & + & 0 & - & - \\ (x-2) D(x \beta) : & 0 & \mp & 0 & \cdots & 0 & + & 0 & - & 0 & + \\ \alpha_{n+1} : & \pm & \mp & \mp & \cdots & - & + & + & - & - & + \\ \end{array}$$ To see the bottom right sign, we note that the dominant term of $\alpha_{n+1}$ is $x^{-1}$, which is positive.

So $\alpha_{n+1}$ changes signs $n+1$ times in $[2, \infty)$, and must have $n+1$ real zeroes in that range.

UPDATE Slicker proof for the final steps: $$\alpha_{n+1}=(x-2) D x \beta + \beta = (x-2) x D \beta + (x-1) \beta = \sqrt{x(x-2)} D \left( \sqrt{x(x-2)} \beta \right)$$ Since $\beta$ has $n$ real roots in $[2, \infty)$, and is $O(x^{-2})$ as $x \to \infty$, Rolle's theorem applied to $\sqrt{x(x-2)} \beta$ shows that $\alpha_{n+1}$ has $n+1$ real roots in $[2, \infty)$.

Answer by Robert Israel for Showing that a family of polynomials has positive and real roots.

2012-07-17T01:05:59Z

It looks to me like the ordinary generating function of $P_n(x)$ is $$ \sum_{n=0}^\infty P_n(x) s^n = - \sum_{k=0}^\infty \dfrac{(2k+1)! s^k}{2^k \prod_{j=1}^{k+1} (j^2 x s - 1)}$$

Answer by matthias beck for Showing that a family of polynomials has positive and real roots.

2012-08-01T03:43:10Z

This paper by P. Bränden and this paper by M. Visontai & N. Williams give a somewhat general approach to proving real rootedness of polynomials (especially those coming up combinatorially).