# duality of Gudermannian and its inverse function

There are a lot of formulae concerning the Gudermannian function and its inverse function containing a hyperbolic function or a trigonometric function or both, such that if we change functions of one kind to the corresponding functions of the other kind, then the new formula also is true.

Some exemples:

 $\displaystyle\mbox{gd}\,x\;=\;\int_{0}^{x}\!\frac{dt}{\cosh{t}},\qquad\mbox{gd% }^{-1}x\;=\;\int_{0}^{x}\!\frac{dt}{\cos{t}}$ (1)
 $\displaystyle\frac{d}{dx}\mbox{gd}\,x\;=\;\frac{1}{\cosh{x}},\qquad\frac{d}{dx% }\mbox{gd}^{-1}\,x\;=\;\frac{1}{\cos{x}}$ (2)
 $\displaystyle\tan(\mbox{gd}\,x)\;=\;\sinh{x},\qquad\tanh(\mbox{gd}^{-1}x)\;=\;% \sin{x}$ (3)
 $\displaystyle\sin(\mbox{gd}\,x)\;=\;\tanh{x},\qquad\sinh(\mbox{gd}^{-1}x)\;=\;% \tan{x}$ (4)
 $\displaystyle\tan\frac{\mbox{gd}\,x}{2}\;=\;\tanh\frac{x}{2},\qquad\tanh\frac{% \mbox{gd}^{-1}x}{2}\;=\;\tan\frac{x}{2}$ (5)

For proving (5) we can check that

 $\frac{d}{dx}[2\arctan(\tanh\frac{x}{2})]\;=\;\frac{1}{\cosh{x}},$

and since both the expression in the brackets and the http://planetmath.org/node/11997Gudermannian vanish in the origin, we have

 $\mbox{gd}\,x\;\equiv\;2\arctan(\tanh\frac{x}{2}).$

This equation implies (5).

The duality (http://planetmath.org/DualityInMathematics) of the formula pairs may be explained by the equality

 $\displaystyle\mbox{gd}\,ix\;=\;i\,\mbox{gd}^{-1}x.$ (6)
Title duality of Gudermannian and its inverse function DualityOfGudermannianAndItsInverseFunction 2013-03-22 19:06:41 2013-03-22 19:06:41 pahio (2872) pahio (2872) 7 pahio (2872) Topic msc 33B10 msc 26E05 msc 26A09 msc 26A48 InverseGudermannianFunction IdealInvertingInPruferRing