convert/Hankel - Maple Help

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convert/Hankel

convert special functions admitting 1F1 or 0F1 hypergeometric representation into Hankel (Bessel of third kind) functions

 Calling Sequence convert(expr, Hankel)

Parameters

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

Description

 • convert/Hankel converts, when possible, special functions admitting a 1F1 or 0F1 hypergeometric representation into Hankel (Bessel of third kind) functions. The Hankel functions are
 The 2 functions in the "Hankel" class are:
 $\left[{\mathrm{HankelH1}}{,}{\mathrm{HankelH2}}\right]$ (1)

Examples

 > $\mathrm{BesselJ}\left(a,z\right)$
 ${\mathrm{BesselJ}}{}\left({a}{,}{z}\right)$ (2)
 > $\mathrm{convert}\left(,\mathrm{Hankel}\right)$
 $\frac{{1}}{{2}}{}{\mathrm{HankelH1}}{}\left({a}{,}{z}\right){+}\frac{{1}}{{2}}{}{\mathrm{HankelH2}}{}\left({a}{,}{z}\right)$ (3)
 > $\mathrm{BesselK}\left(a,z\right)$
 ${\mathrm{BesselK}}{}\left({a}{,}{z}\right)$ (4)
 > $\mathrm{convert}\left(,\mathrm{Hankel}\right)$
 $\frac{{I}{}\left(\left({-}{\left({{z}}^{{a}}\right)}^{{2}}{}{{ⅇ}}^{{I}{}{a}{}{\mathrm{π}}}{+}{\left({\left({I}{}{z}\right)}^{{a}}\right)}^{{2}}{}{{ⅇ}}^{{2}{}{I}{}{a}{}{\mathrm{π}}}\right){}{\mathrm{HankelH1}}{}\left({a}{,}{I}{}{z}\right){+}\left({\left({\left({I}{}{z}\right)}^{{a}}\right)}^{{2}}{-}{\left({{z}}^{{a}}\right)}^{{2}}{}{{ⅇ}}^{{I}{}{a}{}{\mathrm{π}}}\right){}{\mathrm{HankelH2}}{}\left({a}{,}{I}{}{z}\right)\right){}{\mathrm{π}}}{{\left({I}{}{z}\right)}^{{a}}{}{{z}}^{{a}}{}\left({2}{}{{ⅇ}}^{{2}{}{I}{}{a}{}{\mathrm{π}}}{-}{2}\right)}$ (5)
 > $\mathrm{KummerU}\left(a+\frac{1}{2},2a+1,z\right)$
 ${\mathrm{KummerU}}{}\left({a}{+}\frac{{1}}{{2}}{,}{2}{}{a}{+}{1}{,}{z}\right)$ (6)
 > $\mathrm{convert}\left(,\mathrm{Hankel}\right)$
 $\frac{\left(\left({-}\frac{{1}}{{8}}{}\frac{{{ⅇ}}^{{I}{}{a}{}{\mathrm{π}}}{}{\left(\frac{{1}}{{2}}{}{I}{}{z}\right)}^{{a}}{}{{2}}^{{2}{}{a}{+}{1}}}{{{z}}^{{2}{}{a}}{}{{2}}^{{a}}}{+}\frac{{1}}{{8}}{}\frac{{{2}}^{{a}}}{{\left(\frac{{1}}{{2}}{}{I}{}{z}\right)}^{{a}}{}{{2}}^{{-}{1}{+}{2}{}{a}}}\right){}{\mathrm{HankelH1}}{}\left({a}{,}\frac{{1}}{{2}}{}{I}{}{z}\right){+}\left({-}\frac{{1}}{{8}}{}\frac{{\left(\frac{{1}}{{2}}{}{I}{}{z}\right)}^{{a}}{}{{2}}^{{2}{}{a}{+}{1}}}{{{z}}^{{2}{}{a}}{}{{ⅇ}}^{{I}{}{a}{}{\mathrm{π}}}{}{{2}}^{{a}}}{+}\frac{{1}}{{8}}{}\frac{{{2}}^{{a}}}{{\left(\frac{{1}}{{2}}{}{I}{}{z}\right)}^{{a}}{}{{2}}^{{-}{1}{+}{2}{}{a}}}\right){}{\mathrm{HankelH2}}{}\left({a}{,}\frac{{1}}{{2}}{}{I}{}{z}\right)\right){}\sqrt{{\mathrm{π}}}{}{{ⅇ}}^{\frac{{1}}{{2}}{}{z}}}{{\mathrm{sin}}{}\left({\mathrm{π}}{}\left({a}{+}{1}\right)\right)}$ (7)
 > $\mathrm{LaguerreL}\left(-\frac{1}{2}-a,2a,2Iz\right)-\mathrm{WhittakerM}\left(0,a,2Iz\right)$
 ${\mathrm{LaguerreL}}{}\left({-}\frac{{1}}{{2}}{-}{a}{,}{2}{}{a}{,}{2}{}{I}{}{z}\right){-}{\mathrm{WhittakerM}}{}\left({0}{,}{a}{,}{2}{}{I}{}{z}\right)$ (8)
 > $\mathrm{convert}\left(,\mathrm{Hankel}\right)$
 $\frac{{1}}{{2}}{}\frac{{\mathrm{binomial}}{}\left({a}{-}\frac{{1}}{{2}}{,}{-}\frac{{1}}{{2}}{-}{a}\right){}{{ⅇ}}^{{I}{}{z}}{}{\mathrm{Γ}}{}\left({a}{+}{1}\right){}\left({\mathrm{HankelH1}}{}\left({a}{,}{-}{z}\right){+}{\mathrm{HankelH2}}{}\left({a}{,}{-}{z}\right)\right){}{{2}}^{{a}}}{{\left({-}{z}\right)}^{{a}}}{-}\frac{{1}}{{2}}{}\frac{{\left({2}{}{I}{}{z}\right)}^{{a}{+}\frac{{1}}{{2}}}{}{\mathrm{Γ}}{}\left({a}{+}{1}\right){}\left({\mathrm{HankelH1}}{}\left({a}{,}{-}{z}\right){+}{\mathrm{HankelH2}}{}\left({a}{,}{-}{z}\right)\right){}{{2}}^{{a}}}{{\left({-}{z}\right)}^{{a}}}$ (9)
 > $\mathrm{MeijerG}\left(\left[\left[\right],\left[\right]\right],\left[\left[\frac{1a}{2}\right],\left[-\frac{1a}{2}\right]\right],z\right)$
 ${\mathrm{MeijerG}}{}\left(\left[\left[{}\right]{,}\left[{}\right]\right]{,}\left[\left[\frac{{1}}{{2}}{}{a}\right]{,}\left[{-}\frac{{1}}{{2}}{}{a}\right]\right]{,}{z}\right)$ (10)
 > $\mathrm{convert}\left(,\mathrm{Hankel}\right)$
 $\frac{{1}}{{2}}{}\frac{{{z}}^{\frac{{1}}{{2}}{}{a}}{}\left({\mathrm{HankelH1}}{}\left({a}{,}{-}{2}{}\sqrt{{z}}\right){+}{\mathrm{HankelH2}}{}\left({a}{,}{-}{2}{}\sqrt{{z}}\right)\right){}{{2}}^{{a}}}{{\left({-}{2}{}\sqrt{{z}}\right)}^{{a}}}$ (11)