Tanc function: Difference between revisions

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In mathematics, the '''Tanctanc function''' is defined for <math>z \neq 0</math> as<ref>{{Cite web |last=Weisstein, |first=Eric W. "|title=Tanc Function." From MathWorld--A Wolfram Web Resource. http|url=https://mathworld.wolfram.com/TancFunction |access-date=2022-11-17 |website=mathworld.htmlwolfram.com |language=en}}</ref>

<math display="block">\operatorname{Tanctanc}(z)=\frac {\tan(z) }{z}</math>

[[File:Tanc 2D plot.png|thumb|Tanc 2D plot]]

[[File:Tanc'(z) 2D plot.png|thumb|Tanc'(z) 2D plot]]

[[File:Tanc integral.png|thumb|Tanc integral 2D plot]]

[[File:Tanc integral 3D plot.png|thumb|Tanc integral 3D plotplo]]

== Properties ==

;Imaginary part in complex plane

The first-order derivative of the tanc function is given by

:<math> \operatorname{Im} \left( \frac {\tan(x+iy) }{x+iy} \right) </math>

;Real part in complex plane

:<math> \operatorname{Re} \left( \frac {\tan \left( x+iy \right) }{x+iy} \right) </math>

;absolute magnitude

:<math> \left| \frac {\tan(x+iy) }{x+iy} \right| </math>

;First-order derivative

:<math> \frac {1- (\tan(z))^2}{z} - \frac {\tan(z)}{z^2} </math>

The [[Taylor series]] expansion is<math display="block">\operatorname{Tanctanc} z \approx \left(1+ \frac {1}{3} z^2 + \frac {2}{15} z^4 + \frac {17}{315} z^6 + \frac{62}{2835} z^8 + \frac {1382}{155925} z^{10} + \frac{21844}{6081075} z^{12}+ \frac {929569}{638512875} z^{14} + O(z^{16} ) \right)</math>which leads to the series expansion of the integral as<math display="block">\int _0^z \frac {\tan(x) }{x} \, dx = \left(z+ \frac {1}{9} z^3 + \frac {2}{75} z^5 + \frac {17}{2205} z^7 + \frac {62}{25515} z^9+ \frac {1382}{1715175} z^{11}+ \frac {21844}{

;Real part of derivative

79053975} z^{13} + \frac{929569}{9577693125} z^{15}+ O (z^{17}) \right)</math>The [[Padé approximant]] is<math display="block">\operatorname{Tanctanc} \left( z \right) = \left( 1-{\frac {7}{51}}\,{z}^{2} + {\frac {1}{255}}\,{z}^{4}-{\frac {2}{69615}}\,{z}^{6}+{\frac {1}{34459425}}\,{z}^{8} \right) \left( 1-{\frac {8}{17}}\,{z}^{2}+{\frac {7}{255}}\,{z}^{4}-{\frac {4}{9945}}\,{z}^{6}+{\frac {1}{765765}}\,{z}^{8} \right) ^{-1} </math>

:<math> -\operatorname{Re} \left( -\frac {1- (\tan(x+iy))^2}{x+iy} +\frac{\tan(x+iy)}{(x+iy)^2} \right) </math>

;Imaginary part of derivative

:<math>-\operatorname{Im} \left( -\frac {1-(\tan(x+iy))^2}{x+iy} + \frac {\tan(x+iy)}{(x+iy)^2} \right) </math>

;absolute value of derivative

:<math> \left| -\frac{1-(\tan(x+iy))^2}{x+iy}+\frac {\tan(x+iy)}{(x+iy)^2} \right| </math>

==In terms of other special functions==

=== In terms of other special functions ===

* <math>\operatorname{Tanc}(z)={\frac {2\,i{{\rm KummerM}\left(1,\,2,\,2\,iz\right)}}{ \left( 2\,z+\pi

\right) {{\rm KummerM}\left(1,\,2,\,i \left( 2\,z+\pi \right) \right)}}}</math>, where <math>{\rm{KummerM}}(a,b,z)</math> is Kummer's [[confluent hypergeometric function]].

*<math>\operatorname{Tanc}(z)= \frac {2i \operatorname{HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {iz} \right) }{(2z+\pi) \operatorname{HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {(i/2) (2z+\pi) } \right) } </math>, where <math>{\rm{HeunB}}(q, \alpha, \gamma, \delta, \epsilon ,z)</math> is the biconfluent [[Heun function]].

* <math>\operatorname{Tanc}(z)= \frac {{\rm WhittakerM}(0,\,1/2,\,2\,iz)}{{\rm WhittakerM}(0,\,1/2,\,i (2z+\pi)) z} </math>, where <math>{\rm{WhittakerM}}(a,b,z)</math> is a [[Whittaker function]].

==Series expansion==

<math display="block">\operatorname{Tanc} z \approx \left(1+ \frac {1}{3} z^2 + \frac {2}{15} z^4 + \frac {17}{315} z^6 + \frac{62}{2835} z^8 + \frac {1382}{155925} z^{10} + \frac{21844}{6081075} z^{12}+ \frac {929569}{638512875} z^{14} + O(z^{16} ) \right)</math>

<math display="block">\int _0^z \frac {\tan(x) }{x} \, dx = \left(z+ \frac {1}{9} z^3 + \frac {2}{75} z^5 + \frac {17}{2205} z^7 + \frac {62}{25515} z^9+ \frac {1382}{1715175} z^{11}+ \frac {21844}{

79053975} z^{13} + \frac{929569}{9577693125} z^{15}+ O (z^{17}) \right)</math>

==Padé approximation==

<math display="block">\operatorname{Tanc} \left( z \right) = \left( 1-{\frac {7}{51}}\,{z}^{2} + {\frac {1}{255}}\,{z}^{4}-{\frac {2}{69615}}\,{z}^{6}+{\frac {1}{34459425}}\,{z}^{8} \right) \left( 1-{\frac {8}{17}}\,{z}^{2}+{\frac {7}{255}}\,{z}^{4}-{\frac {4}{9945}}\,{z}^{6}+{\frac {1}{765765}}\,{z}^{8} \right) ^{-1} </math>

==Gallery==