Tanhc函數

Tanhc函數定義如下^[1]

$tanhc(z)={\frac {\tanh \left(z\right)}{z}}$

復域虛部

${\it {Im}}\left({\frac {\tanh \left(x+iy\right)}{x+iy}}\right)$

復域實部

${\it {Re}}\left({\frac {\tanh \left(x+iy\right)}{x+iy}}\right)$

復域絕對值

$\left|{\frac {\tanh \left(x+iy\right)}{x+iy}}\right|$

一階微商

${\frac {1-\left(\tanh \left(z\right)\right)^{2}}{z}}-{\frac {\tanh \left(z\right)}{{z}^{2}}}$

微商實部

$-{\it {Re}}\left(-{\frac {1-\left(\tanh \left(x+iy\right)\right)^{2}}{x+iy}}+{\frac {\tanh \left(x+iy\right)}{\left(x+iy\right)^{2}}}\right)$

微商虛部

$-{\it {Im}}\left(-{\frac {1-\left(\tanh \left(x+iy\right)\right)^{2}}{x+iy}}+{\frac {\tanh \left(x+iy\right)}{\left(x+iy\right)^{2}}}\right)$

微商絕對值

$\left|-{\frac {1-\left(\tanh \left(x+iy\right)\right)^{2}}{x+iy}}+{\frac {\tanh \left(x+iy\right)}{\left(x+iy\right)^{2}}}\right|$

積分函數

$\int _{0}^{z}\!{\frac {\tanh \left(x\right)}{x}}{dx}$

用其他特殊函數表示

$tanhc(z)=2\,{\frac {{\rm {KummerM}}\left(1,\,2,\,2\,z\right)}{\left(2\,iz+\pi \right){{\rm {KummerM}}\left(1,\,2,\,i\pi -2\,z\right)}{{\rm {e}}^{2\,z-1/2\,i\pi }}}}$
$tanhc(z)=2\,{\frac {{\it {HeunB}}\left(2,0,0,0,{\sqrt {2}}{\sqrt {z}}\right)}{\left(2\,iz+\pi \right){\it {HeunB}}\left(2,0,0,0,{\sqrt {2}}{\sqrt {1/2\,i\pi -z}}\right){{\rm {e}}^{2\,z-1/2\,i\pi }}}}$
$tanhc(z)={\frac {i{{\rm {\ WhittakerM}}\left(0,\,1/2,\,2\,z\right)}}{{{\rm {WhittakerM}}\left(0,\,1/2,\,i\pi -2\,z\right)}z}}$
$tanhc(z)={\frac {i\left({{\rm {e}}^{2\,z}}-1\right)}{\left({{\rm {e}}^{i\pi -2\,z}}-1\right){{\rm {e}}^{2\,z-1/2\,i\pi }}z}}$

級數展開

$tanhc\approx (1-{\frac {1}{3}}{z}^{2}+{\frac {2}{15}}{z}^{4}-{\frac {17}{315}}{z}^{6}+{\frac {62}{2835}}{z}^{8}-{\frac {1382}{155925}}{z}^{10}+{\frac {21844}{6081075}}{z}^{12}-{\frac {929569}{638512875}}{z}^{14}+O\left({z}^{16}\right))$

$\int _{0}^{z}\!{\frac {\tanh \left(x\right)}{x}}{dx}=(z-{\frac {1}{9}}{z}^{3}+{\frac {2}{75}}{z}^{5}-{\frac {17}{2205}}{z}^{7}+{\frac {62}{25515}}{z}^{9}-{\frac {1382}{1715175}}{z}^{11}+O\left({z}^{13}\right))$