Tanhc函数

Tanhc函数定义如下^[1]

File:Tanhc 2D plot.png

Tanhc 2D plot

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

Tanhc'(z) 2D

File:Tanhc integral 2D plot.png

Tanhc 积分图

File:Tanhc integral 3D plot.png

Tanhc integral 3D plot

<math> tanhc(z)={\frac {\tanh \left( z \right) }{z}} </math>

复域虚部

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

复域实部

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

复域绝对值

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

一阶微商

<math>{\frac {1- \left( \tanh \left( z \right) \right) ^{2}}{z}}-{\frac {

\tanh \left( z \right) }{{z}^{2}}}

    </math>

微商实部

<math> \it Re} \left( \frac {1- \left( \tanh \left( x+iy \right)

\right) ^{2}}{x+iy}}+{\frac {\tanh \left( x+iy \right) }{ \left( x+iy
\right) ^{2}}} \right)
   </math>

微商虚部

积分函数

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

用其他特殊函数表示[编辑]

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

\right) {{\rm KummerM}\left(1,\,2,\,i\pi -2\,z\right)}{{\rm e}^{2\,z-1/2\,i\pi }}}}</math>

<math>tanhc(z)=2\,{\frac {{\it HeunB} \left( 2,0,0,0,\sqrt {2}\sqrt {z} \right) }{

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

1/2\,i\pi -z} \right) {{\rm e}^{2\,z-1/2\,i\pi }}}}</math>

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

{{\rm WhittakerM}\left(0,\,1/2,\,i\pi -2\,z\right)}z}}</math>

<math>tanhc(z)={\frac {i \left( {{\rm e}^{2\,z}1 \right) }{ \left( {{\rm e}^{i\pi -

2\,z}1 \right) {{\rm e}^{2\,z-1/2\,i\pi }}z}}</math>

级数展开[编辑]

<math>tanhc \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 {\tanh \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}+O \left( {z}^{13}