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a pole, $\psi^{(n)}(x) = (-1)^{n+1} n! \sum_{j \geq 0} (x+j)^{-n-1}$; when $n$ is odd the exponent $n+1$ is even, every term has the same sign, and the sum cannot- vanish. So the odd orders are absent from this table rather than missing from- it.+ vanish. comment-one-zero-per-gap: For even $n$ the function $\psi^{(n+1)}$ has no real zeros by CITE{comment-only-even-n}, and so has constant sign. Hence $\psi^{(n)}$ is
which $\Gamma$ attains its minimum on the positive reals, since $\Gamma' = \psi \Gamma$ and $\Gamma > 0$ there. CITE{OEIS}- comment-index-is-proven: 'Unlike the neighbouring tables of Airy and Bessel extrema,- the index here is proven and not merely computed. Those record that they cannot- establish a value to be the $n$-th zero rather than a neighbour, which would need- a count of the zeros below it. Here CITE{comment-one-zero-per-gap} supplies that- count: one zero per gap, so counting the gaps counts the zeros.'+ comment-half-integers: For even $n\geq 2$ the $k$-th largest zero tends to $-k+\tfrac12$+ as $n\to\infty$, which is why so many entries here are half-integers to many digits.+ In the series of CITE{comment-only-even-n} the terms $j=k-1-m$ and $j=k+m$ cancel+ in pairs at the midpoint of $(-k,-k+1)$, since $n+1$ is odd; what is left are+ the terms $j\geq 2k$, which have no partner because the sum starts at $j=0$. Balancing+ that remainder against the two nearest poles puts the zero above the midpoint+ by about $\dfrac{(2k+1)^{-(n+1)}}{4(n+1)}$. Links: Wiki:
index is proven too, by the monotonicity in the comments. Checked against mpmath at 75 zeros and against OEIS A030169 for the positive zero of the digamma.+ complete-note: every even $n\leq 40$ is here, with the $50$ largest zeros of each Display properties: number-header: $k$<sup>th</sup> largest zero of $\psi^{(n)}$
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