Tanc函数

Tanc 函数 定义如下^[1]

<math>\operatorname{Tanc}(z)=\frac {\tan(z) }{z}</math>

File:Tanc 2D plot.png

Tanc 2D plot

File:Tanc'(z) 2D plot.png

Tanc'(z) 2D plot

File:Tanc integral.png

Tanc integral 2D plot

File:Tanc integral 3D plot.png

Tanc integral 3D plot

虚域虚部

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

虚域实部

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

绝对值

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

一阶导数

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

导数实部

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

   </math>

导数虚部

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

    </math>

导数绝对值

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

与其他特殊函数的关系[编辑]

<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>

<math>\operatorname{Tanc}(z)={\frac {2\,i{\it HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {iz} \right) }{

\left( 2\,z+\pi  \right) {\it HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {1

/2\,i \left( 2\,z+\pi \right) } \right) }} </math>

<math>\operatorname{Tanc}(z)={\frac {{{\rm WhittakerM}\left(0,\,1/2,\,2\,iz\right)}}{

{{\rm WhittakerM}\left(0,\,1/2,\,i \left( 2\,z+\pi \right) \right)}z}}

</math>

级数展开[编辑]

<math>\operatorname{Tanc} z \approx (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 \left( {z}^{16} \right) )</math>

<math>\int _{0}^{z}\!{\frac {\tan \left( x \right) }{x}}{dx}=(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 \left( {z}^{ 17} \right) )</math>