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Hermite polynomials of order $n\in\mathbb{N}_0$ can be expressed as a special case of the confluent hypergeometric function (also called Tricomi's confluent hypergeometric function): $$H_n(x) = 2^n U\left(-\tfrac12 n, \tfrac12, x^2\right)$$ for $\operatorname{Re}(x)\geq0$. This representation allows us to generalize Hermite polynomials to negative and even fractional orders $n$ which is a great help in calculating some integrals. My problem is the following: I have Hermite polynomials of negative fractional order $-r$ and want to evaluate them at negative $x$, but the above representation does not exist, i.e. I cannot write $$H_{-r}(-x) = 2^{-r} U\left(\tfrac12 r, \tfrac12, x^2\right),$$ however using the known relation $H_n(-x)=(-1)^nH_n(x)$ also won't work as I get complex numbers then (because of $(-1)^{-r}$). Do representations of Hermite polynomials using hypergeometric and other functions exist that extend the region of convergence to points on the left half-plane $\operatorname{Re}(x)\leq0$? I couldn't find any.

Caesar.tcl
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  • $U$ has a branch point at the origin and one usually arranges the branch cut to lie along the negative real axis. So, at once, there are at least two different values you might want: the one that follows a path around the branch point in the upper half-plane and the one that follows a path in the lower half-plane. Along each path $U$ is continuous, but when we arrive at the negative real axis, the two paths produce two different values for $U$. I suspect your "$(-1)^n$ formula is giving you "the other one". (continued) – Eric Towers Nov 07 '21 at 16:23
  • However, the DLMF gives a formula to relate the various values of $U$ obtained by orbiting around the origin (any integer) $m$ times at section 13.3(ii). This would allow you to start with the value of $U$ (and $\mathbf{M}$) that you have, then generate the values of $U$ at $m = \pm 1$ to get values by switching to a path that goes around once more clockwise or once more anticlockwise. Maybe one of those will be real valued. However, there may be no real-valued choice of branch on the negative real axis. – Eric Towers Nov 07 '21 at 16:27
  • Thanks Eric, I will look into it – Caesar.tcl Nov 08 '21 at 09:34
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    $x \mapsto H_r(x)$ is entire for any $r$, $$H_r(x) = \frac {2^r \sqrt \pi} {\Gamma{ \left( \frac {1 - r} 2 \right)}} \hspace {1 px} {_1 \hspace {-1 px} F_1} {\left( -\frac r 2; \frac 1 2; x^2 \right)} - \frac {2^{r + 1} x \sqrt \pi} {\Gamma{ \left( -\frac r 2 \right)}} \hspace {1 px} {_1 \hspace {-1 px} F_1} {\left( \frac {1 - r} 2; \frac 3 2; x^2 \right)},$$ with one of the terms disappearing if $r \in \mathbb N^0$. – Maxim Feb 11 '22 at 16:01
  • @Maxim Thank you! – Caesar.tcl Feb 12 '22 at 12:55

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