Is the reciprocal of polygamma functions of odd order convex on $ \mathbb{R^+}$, while that of even order above 0 concave? Plotting the functions suggest so, but I've been trying for days to come up with a proof to no avail. The 2nd derivative of the reciprocal is:

$$ \frac{d ^2\psi ^{(m)}(z)^{-1}}{d z^2}=\frac{2 \psi ^{(m+1)}(z)^2-\psi ^{(m)}(z) \psi ^{(m+2)}(z)}{\psi ^{(m)}(z)^3} $$

I can't show the numerator is always positive for $z>0$. Curiously, $\psi ^{(m+1)}(z)^2-\psi ^{(m)}(z) \psi ^{(m+2)}(z)<0$, which I can show. Positivity of the numerator is equivalent to

$$ \frac{(m+2)}{(m+1)}\frac{ \zeta (m+1,z) \zeta (m+3,z)}{ \zeta (m+2,z)^2}<2 $$

so showing this would suffice. Alternatively, since the $z\rightarrow0$ and $z\rightarrow\infty$ limits are not negative, showing the numerator is monotonic also works.

  • $\begingroup$ The numerator looks like a Turan-type inequality; see mscand.dk/article/viewFile/10396/8417. Have you searched the literature? A quick google search for "Turan-type inequalities for polygamma function" turns up this paper: emis.de/journals/JIPAM/images/198_05_JIPAM/198_05.pdf $\endgroup$ – parsiad Nov 8 '16 at 18:50
  • $\begingroup$ Thanks @parsiad, I wasn't aware of such inequalities. Eq 2.4 in your 2nd link corresponds to what I mentioned I could show. The problem is the factor of 2 behind $\psi^{(m+1)}(z)^2$ makes the inequality differ from Turan-like. $\endgroup$ – obsolesced Nov 8 '16 at 19:07

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