convert/0F1 converts, when possible, hypergeometric, MeijerG, and special functions admitting a 1F1 hypergeometric representation into special functions admitting a 0F1 hypergeometric representation; that is, into one of

FunctionAdvisor( `0F1` );

The 26 functions in the "0F1" class are particular cases of the hypergeometric function and are given by:

$\left[\mathrm{AiryAi}\,\mathrm{AiryBi}\,\mathrm{BesselI}\,\mathrm{BesselJ}\,\mathrm{BesselK}\,\mathrm{BesselY}\,\mathrm{HankelH1}\,\mathrm{HankelH2}\,\mathrm{KelvinBei}\,\mathrm{KelvinBer}\,\mathrm{KelvinHei}\,\mathrm{KelvinHer}\,\mathrm{KelvinKei}\,\mathrm{KelvinKer}\,\cos \,\cosh \,\cot \,\coth \,\csc \,\mathrm{csch}\,\sec \,\mathrm{sech}\,\sin \,\sinh \,\tan \,\tanh \right]$

convert/0F1 accepts as optional arguments all those described in convert/to_special_function.

$\frac{\frac{\frac{1}{\mathrm{\Gamma }\left(1+a\right)}{z}^a}{2^a}\mathrm{WhittakerM}\left(0\,a\,2\mathrm{I}z\right)}{{\left(2\mathrm{I}z\right)}^{\frac{1}{2}+a}}$

$\frac{z^a\mathrm{WhittakerM}\left(0\,a\,2\mathrm{I}z\right)}{\mathrm{\Gamma }\left(1+a\right)2^a{\left(2\mathrm{I}z\right)}^{\frac{1}{2}+a}}$

$\mathrm{convert}\left(\,\mathrm{`0F1`}\right)$

$\frac{z^a\mathrm{BesselI}\left(a\,\mathrm{I}z\right)}{{\left(\mathrm{I}z\right)}^a}$

$\frac{2^{-a-1}\mathrm{\pi }{z}^a}{\sin \left(\mathrm{\pi }\left(1+a\right)\right)}\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[0\right]\,\left[-a\right]\right]\,-\frac{1}{4}{z}^2\right)+\frac{2^{-1+a}\mathrm{\pi }}{\sin \left(\mathrm{\pi }a\right)}{z}^{-a}\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[0\right]\,\left[a\right]\right]\,-\frac{1}{4}{z}^2\right)$

$\frac{2^{-a-1}\mathrm{\pi }{z}^a\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[0\right]\,\left[-a\right]\right]\,-\frac{z^2}{4}\right)}{\sin \left(\mathrm{\pi }\left(1+a\right)\right)}+\frac{2^{-1+a}\mathrm{\pi }{z}^{-a}\mathrm{MeijerG}\left(\left[\left[\right]\,\left[\right]\right]\,\left[\left[0\right]\,\left[a\right]\right]\,-\frac{z^2}{4}\right)}{\sin \left(\mathrm{\pi }a\right)}$

$\mathrm{convert}\left(\,\mathrm{`0F1`}\right)$

$\frac{2^{-a-1}\mathrm{\pi }\mathrm{BesselI}\left(a\,z\right)2^a}{\sin \left(\mathrm{\pi }\left(1+a\right)\right)}+\frac{2^{-1+a}\mathrm{\pi }{z}^{-a}\mathrm{BesselI}\left(-a\,z\right)z^a}{\sin \left(\mathrm{\pi }a\right)2^a}$

$\mathrm{LaguerreL}\left(-\frac{1}{2}-a\,2a\,2\mathrm{I}z\right)-\mathrm{WhittakerM}\left(0\,a\,2\mathrm{I}z\right)$

$\mathrm{LaguerreL}\left(-\frac{1}{2}-a\,2a\,2\mathrm{I}z\right)-\mathrm{WhittakerM}\left(0\,a\,2\mathrm{I}z\right)$

$\mathrm{convert}\left(\,\mathrm{`0F1`}\right)$

$\frac{2^a\mathrm{\Gamma }\left(1+a\right)\left(\left(\genfrac{}{}{0ex}{}{-\frac{1}{2}+a}{-\frac{1}{2}-a}\right){\ⅇ}^{\mathrm{I}z}-{\left(2\mathrm{I}z\right)}^{\frac{1}{2}+a}\right)\mathrm{BesselI}\left(a\,\mathrm{I}z\right)}{{\left(\mathrm{I}z\right)}^a}$