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Say I wish to calculate $U(0,-1)$ which is $D_{-1/2}(-1)$ using the whitaker notation. According to http://dlmf.nist.gov/12.7.10, the identity is $$U(0,z)=\sqrt{\frac{z}{2\pi}}\mathcal{K}_{1/4}\left(\frac{1}{4}z^2\right)$$ However, the fact that z is under the square root sign gives a complex number. In wolfram alpha the value of $D_{-1/2}(-1)\approx1.83039$ which is more what I am expecting.

Am I missing something?

sachinruk
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  • I believe that equation might have some assumptions (not an expert on the area), but if you have a look at the definition of $U(a,z)$ at http://en.wikipedia.org/wiki/Parabolic_cylinder_function then your square root problem goes away – Kevin Dec 02 '13 at 05:11
  • Hi Kevin, not sure I'm following? I don't think there z>0. Infact $z\in\mathcal{R}$ from my understanding. http://dlmf.nist.gov/12.3 – sachinruk Dec 02 '13 at 06:09

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This question is very old, but I have just spent some time on the very same problem so I thought I might post my work-around. I agree the sign of the argument of $D_{-1/2}(x)$ is not mentioned in https://dlmf.nist.gov/12. Then the defining differential equation is second-order and even in $z$. So the vector space of solutions has to be of dimension two, and to contain one purely odd and one purely even solution. Then I'm afraid what mix of these is provided by one particular numerical implementation ends up being a matter of convention.

What I've figured out regarding the scipy.special.pbdv implementation (which turned to be just what I needed) goes as follows (some equalities I use are not in https://dlmf.nist.gov/12, so I use the numbering from Abramovitz & Stegun):

  • As you said, for real positive argument, (A & S, 19.3.7), $D_{-1/2}(x)$ returns $U(0,x)$, which turns out (A & S, 19.27.5), to be equal to $\sqrt{\frac{x}{2\pi}}\mathrm{K}_{1/4}\left(\frac{1}{4}x^2\right)$.
  • For real negative argument, use is made instead of (A & S, 19.3.8), which states that $U(0,-x) = \sqrt{\pi} V(0,x)$. So for $x\leq0$, $D_{-1/2}(x)$ returns $\sqrt{\pi} V(0,-x)$. Which turns out (A & S, 19.27.4), to be equal to $\frac{\sqrt{\pi |x|}}{2} \left[\mathrm{I}_{1/4}\left(\frac{1}{4}x^2\right)+\mathrm{I}_{-1/4}\left(\frac{1}{4}x^2\right)\right]$.

Then, using (A & S, 19.27.5), and noting $\mathrm{H}(x)$ the Heaviside step function, one can summarize all this as: $$ D_{-1/2}(x) = \sqrt{\frac{|x|}{2 \pi}} \mathrm{K}_{1/4}\left(\frac{x^2}{4}\right) + \mathrm{H}(-x) \sqrt{\pi |x|} \mathrm{I}_{1/4} \left(\frac{x^2}{4}\right). $$ I have no idea what happens out of the real line.

lmarie
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