Tanhc function: Difference between revisions

Revision as of 22:46, 16 March 2015

Revision as of 22:46, 16 March 2015

Line 37:

\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 \bf WhittakerM}\left(0,\,1/2,\,2\,z\right)}}{

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

==Gallery==