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,


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


  • 7
    $\begingroup$ It appears that, furthermore, the roots of your polynomials are interlaced. For example, the roots of your quadratic are 2.37652, 12.6235; the roots of your cubic are 2.20173, 5.14109, 55.6572 and $2.20 < 2.37 < 5.14 < 12.6 < 55.6$. This suggests that there might be some sort of recursion relating the polynomials, as in Rolle's theorem or Sturm's theorem. $\endgroup$ Jul 16, 2012 at 20:40
  • $\begingroup$ Hi David, thanks for the comment. I have noticed that as well, and it reminded me the proof that the Hermite polynomial have real roots. Unfortunately, the same procedure does not seems to work here, since we have that the derivative is taken in $z$ and the roots are in $x$. $\endgroup$
    – Enzo
    Jul 16, 2012 at 21:43
  • 1
    $\begingroup$ An early thing to try (although it doesn't work in this case) is to input the sequence of coefficients, (3,1,15,30,1,63,420,630 and 3,30,15,1,630 leaving off the first couple of terms and the signs, into the OEIS. Then, you might also try submitting them to superseeker, which you can learn about at the OEIS website (I haven't done this step with your sequence). If that doesn't work, you should then submit the sequence to the OEIS, so the next researcher doesn't start from scratch. $\endgroup$ Jul 17, 2012 at 12:37

4 Answers 4


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 $[2, \infty)$.

  • 3
    $\begingroup$ It is worth noting there is nothing special about taking $z=0$ in the definition of $\alpha$ besides (possibly) the base case. Indeed, we can directly verify that $D_{zz}f(x,z)=(x-2)D_x(xD_xf(x,z))+D_xf(x,z)$, where $f(x,z) = \sinh(z)/(\cosh(z)-1+x)$, which clearly implies the above relation between $\alpha_{n+1},\alpha_n$ after we differentiate both sides $2n-1$ times w.r.t. $z$ and plug in $z=0$ (here the relevant partial derivatives are commutative). Mathematica code: D[D[Sinh[z]/(Cosh[z] - 1 + x), z], z] - D[Sinh[z]/(Cosh[z] - 1 + x), x] - (x-2)D[x*D[Sinh[z]/(Cosh[z] - 1 + x), x], x] $\endgroup$ Jul 17, 2012 at 2:58
  • $\begingroup$ Hi again David. Nice suggestion! I will check the details, but it seems to be right. Regards, Enzo $\endgroup$
    – Enzo
    Jul 17, 2012 at 6:42
  • $\begingroup$ If we take power series around $z=z_0$, then $\alpha_n(x)$ is a rational function with numerator of degree $2n$ and denominator $(\cosh(z_0)-1+x)^{2n+1}$. It's possible that this proof might show that the numerator has $n$ roots (I haven't checked) but I don't think it will get you to $2n$. $\endgroup$ Jul 17, 2012 at 11:02
  • $\begingroup$ @David: You're right; I wasn't careful about that. BTW, a minor point: the recursion is not "equivalent," as the definition of $\alpha_n$ involves taking $z=0$ at the end. $\endgroup$ Jul 17, 2012 at 13:48

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).


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


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.

  • $\begingroup$ Hi John, that is how I got the explicit formula. Since $$f(z,x)=\left.\frac{\sinh(z)}{\cosh(z)-1+x}=\frac{(y+1)(y-1)}{y^2+1+2y(x-1)}\right|_{y=e^z}$$ one can use the Faa di Bruno's formula $$\alpha_n(x)=\sum_{k=1}^{2n+1}k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}\frac{1}{2\pi i}\int_\rho{z+1}{(z^2+1+2z(x-1))(z-1)^{k}}$$ where $\rho$ is a closed curve encircling $z=1$. Then the integral can be obtained by Cauchy Residue formula. $\endgroup$
    – Enzo
    Jul 17, 2012 at 6:48
  • $\begingroup$ Sorry the integral should be $$\int_{\rho}\frac{z+1}{(z^2+1+2z(x-1))(z-1)^k}dz$$ $\endgroup$
    – Enzo
    Jul 17, 2012 at 6:51

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