Vector exponential

An exponential is usually defined according to the traditional power series. Consider first the exponential of a vector:

$$\exp( {\bf v}) {\bf 1}+ {\bf v}+ \frac{1}{2!}{{\bf v}}^2 + \frac{1}{3!}{\bf v}^3 + \ldots$$ = (60)

$$[ {\bf 1}+ \frac{1}{2!}({\bf v}\cdot{\bf v}) + \frac{1}{4!}({\bf v}\cdot{\bf v})^2 + . . . ] +$$ = (61)

$$\frac{{\bf v}}{\surd({\bf v}\cdot{\bf v})} [\surd({\bf v}\cdot{\bf v})+\frac{1}{3!}\surd({\bf v}\cdot{\bf v})^3+\ldots ]$$ (62)

$${\bf 1}\cosh(\surd({\bf v}\cdot{\bf v})) + \frac{{\bf v}}{\surd({\bf v}\cdot{\bf v})} \sinh(\surd({\bf v}\cdot{\bf v})),$$ = (63)

$${\bf 1}\cosh(\vert{\bf v}\vert) + \frac{{\bf v}}{\vert{\bf v}\vert} \sinh(\vert{\bf v}\vert),$$ = (64)

a result analogous to Euler's formula. The exponential of a quaternion is not much more complicated, since any scalar which could be added would commute with the quaternion, so its exponential could just be set aside as a multiplying scalar factor. How much to set aside in the general case depends on satisfying the identity $\cosh^2(x) - \sinh^2(x) = 1$, but in general there is much to be said in favor of working with vectors of unit norm and treating norms separately.

  

Figure: Arc length along the rectangular hyperbola defines hyperbolic trigonometry when the Lorentz metric is used.

The interesting properties of the exponential lie in its law of exponents. Notice that the angle, $\surd({\bf v}\cdot{\bf v})$, is the norm of ${\bf v}$, and that imaginary quantities can be avoided by using trigonometric functions, such as should be done in association with the quaternion ${\bf i}$.

Consider, for unit vectors ${\bf u}$ and ${\bf v}$,

$$\begin{eqnarray*}\exp(\alpha {\bf u}) \exp (\beta {\bf v}) & = & ( {\bf 1}\cos... ...+ \\ & & ({\bf u}\times {\bf v}) \sinh(\alpha) \sinh(\beta), \end{eqnarray*}$$

and the prospects for seeing this as

$$\begin{eqnarray*}\exp(\gamma {\bf w}) & = & {\bf 1}\cosh(\gamma) + {\bf w}\sinh(\gamma). \end{eqnarray*}$$

Just define a new angle, $\cosh(\theta) = ({\bf u}\cdot{\bf v})$; then copy the two parts of the previous result:

$$\begin{eqnarray*}\cosh(\gamma)&=& \cosh(\alpha)\cosh(\beta)+\sinh(\alpha)\sinh... ...+ \\ & & ( {\bf u}\times {\bf v}) \sinh(\alpha) \sinh(\beta). \end{eqnarray*}$$

  

Figure: Quaternion exponentials and sequences of quaternion exponentials can be visualized in a nomogram based on the unit one-sheeted hyperboloid of radius squared -1.