convert/Airy - Maple Programming Help

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convert/Airy

convert special functions admitting 1F1 or 0F1 hypergeometric representation into Airy functions

 Calling Sequence convert(expr, Airy)

Parameters

 expr - Maple expression, equation, or a set or list of them

Description

 • convert/Airy converts, when possible, special functions admitting a 1F1 or 0F1 hypergeometric representation into Airy functions. The Airy functions are
 The 2 functions in the "Airy" class are:
 $\left[{\mathrm{AiryAi}}{,}{\mathrm{AiryBi}}\right]$ (1)

Examples

 > $\mathrm{BesselK}\left(\frac{1}{3},z\right)$
 ${\mathrm{BesselK}}{}\left(\frac{{1}}{{3}}{,}{z}\right)$ (2)
 > $\mathrm{convert}\left(,\mathrm{Airy}\right)$
 ${\mathrm{π}}{}\sqrt{\frac{{{3}}^{{1}{/}{3}}{}{{2}}^{{2}{/}{3}}}{{{z}}^{{2}{/}{3}}}}{}{\mathrm{AiryAi}}{}\left(\frac{{1}}{{2}}{}{{3}}^{{2}{/}{3}}{}{{2}}^{{1}{/}{3}}{}{{z}}^{{2}{/}{3}}\right)$ (3)
 > $\mathrm{BesselJ}\left(\frac{1}{3},z\right)$
 ${\mathrm{BesselJ}}{}\left(\frac{{1}}{{3}}{,}{z}\right)$ (4)
 > $\mathrm{convert}\left(,\mathrm{Airy}\right)$
 $\frac{{1}}{{2}}{}\sqrt{\frac{{{3}}^{{1}{/}{3}}{}{{2}}^{{2}{/}{3}}}{{{z}}^{{2}{/}{3}}}}{}\left(\sqrt{{3}}{}{\mathrm{AiryAi}}{}\left({-}\frac{{1}}{{2}}{}{{3}}^{{2}{/}{3}}{}{{2}}^{{1}{/}{3}}{}{{z}}^{{2}{/}{3}}\right){-}{\mathrm{AiryBi}}{}\left({-}\frac{{1}}{{2}}{}{{3}}^{{2}{/}{3}}{}{{2}}^{{1}{/}{3}}{}{{z}}^{{2}{/}{3}}\right)\right)$ (5)
 > $\mathrm{BesselI}\left(-\frac{1}{3},\frac{2{z}^{\frac{3}{2}}}{3}\right)$
 ${\mathrm{BesselI}}{}\left({-}\frac{{1}}{{3}}{,}\frac{{2}}{{3}}{}{{z}}^{{3}{/}{2}}\right)$ (6)
 > $\mathrm{convert}\left(,\mathrm{Airy}\right)$
 $\frac{{1}}{{2}}{}\frac{\sqrt{{3}}{}\left(\sqrt{{3}}{}{\mathrm{AiryAi}}{}\left({\left({{z}}^{{3}{/}{2}}\right)}^{{2}{/}{3}}\right){+}{\mathrm{AiryBi}}{}\left({\left({{z}}^{{3}{/}{2}}\right)}^{{2}{/}{3}}\right)\right)}{{\left({{z}}^{{3}{/}{2}}\right)}^{{1}{/}{3}}}$ (7)
 > $\mathrm{KummerM}\left(\frac{5}{6},\frac{5}{3},2Iz\right)$
 ${\mathrm{KummerM}}{}\left(\frac{{5}}{{6}}{,}\frac{{5}}{{3}}{,}{2}{}{I}{}{z}\right)$ (8)
 > $\mathrm{convert}\left(,\mathrm{Airy}\right)$
 $\frac{{1}}{{9}}{}\frac{{{ⅇ}}^{{I}{}{z}}{}{\mathrm{π}}{}\sqrt{{3}}{}{{2}}^{{1}{/}{3}}{}\sqrt{\frac{{{3}}^{{1}{/}{3}}{}{{2}}^{{2}{/}{3}}}{{{z}}^{{2}{/}{3}}}}{}\left(\sqrt{{3}}{}{\mathrm{AiryAi}}{}\left({-}\frac{{1}}{{2}}{}{{3}}^{{2}{/}{3}}{}{{2}}^{{1}{/}{3}}{}{{z}}^{{2}{/}{3}}\right){-}{\mathrm{AiryBi}}{}\left({-}\frac{{1}}{{2}}{}{{3}}^{{2}{/}{3}}{}{{2}}^{{1}{/}{3}}{}{{z}}^{{2}{/}{3}}\right)\right)}{{\mathrm{Γ}}{}\left(\frac{{2}}{{3}}\right){}{{z}}^{{1}{/}{3}}}$ (9)
 > $\mathrm{MeijerG}\left(\left[\left[\right],\left[\frac{1}{6},\frac{2}{3}\right]\right],\left[\left[\frac{1}{3},0\right],\left[\frac{2}{3},\frac{1}{6}\right]\right],z\right)$
 ${\mathrm{MeijerG}}{}\left(\left[\left[{}\right]{,}\left[\frac{{1}}{{6}}{,}\frac{{2}}{{3}}\right]\right]{,}\left[\left[\frac{{1}}{{3}}{,}{0}\right]{,}\left[\frac{{2}}{{3}}{,}\frac{{1}}{{6}}\right]\right]{,}{z}\right)$ (10)
 > $\mathrm{convert}\left(,\mathrm{Airy}\right)$
 $\frac{{1}}{{4}}{}\frac{\left(\sqrt{{3}}{}\left({{z}}^{{1}{/}{3}}{+}{\left({-}{z}\right)}^{{1}{/}{3}}\right){}{\mathrm{AiryAi}}{}\left({-}{{3}}^{{2}{/}{3}}{}{\left({-}{z}\right)}^{{1}{/}{3}}\right){+}\left({-}{{z}}^{{1}{/}{3}}{+}{\left({-}{z}\right)}^{{1}{/}{3}}\right){}{\mathrm{AiryBi}}{}\left({-}{{3}}^{{2}{/}{3}}{}{\left({-}{z}\right)}^{{1}{/}{3}}\right)\right){}\sqrt{\frac{{{3}}^{{1}{/}{3}}}{{\left({-}{z}\right)}^{{1}{/}{3}}}}}{{\left({-}{z}\right)}^{{1}{/}{6}}{}{\mathrm{π}}}$ (11)