0
$\begingroup$

Let $N$ be a positive integer, and consider the equation

\begin{equation} \frac{1}{2} \sum_{n=1}^{N} \psi^{(1)} \left( \frac{x+1-n}{2} \right) = \frac{N}{x}, \end{equation} in the real unkown $x > N - 1$, where $\psi^{(1)}(x)$ is the trigamma function.

I have conjectured that this equation has no solution. Does someone have an idea of a possible proof?

Any help is welcome.

NB From the series representation of the trigamma function we know that the map $(0,\infty) \ni x \mapsto \psi^{(1)}(x)$ is strictly decreasing. So our statement is proved if we can prove the inequality \begin{equation} \psi^{(1)}(x) > \frac{1}{x} \quad (x > 0), \end{equation} but I do not know how to prove it for now.

$\endgroup$
  • $\begingroup$ The last inequality is just $$\sum_{k = 0}^\infty \frac 1 {(k + x)^2} > \int_0^\infty \frac {dk} {(k + x)^2}.$$ $\endgroup$ – Maxim May 28 at 13:04
  • $\begingroup$ @Maxim Wonderful one-life proof!!!! Thank you very very very ... much, Maxim for your help. I would never be able to find it by myself! $\endgroup$ – Maurizio Barbato May 28 at 15:25
0
$\begingroup$

Our system has actually no solution as conjectured, since the inequality stated in the note above holds true. More precisely, one can prove that

\begin{equation} \psi^{(1)}(x) > \frac{1}{x} + \frac{1}{2x^2} \quad (x > 0). \end{equation} For a proof, see e.g. Elbert and Laforgia On Some Properties of the Gamma Function, Section 2; Gordon, A Stochastic Approach to the Gamma Function, Theorem 4; Alzer, On Some Inequalities for the Gamma Function, Theorem 9.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.