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



Showing that a family of polynomials has positive and real roots.
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 $[2, \infty)$.


