††thanks: $\dagger$ Corresponding author: A. Acus

# Coordinate-free exponentials of general multivector (MV) in
$\mathit{Cl}_{p,q}$ algebras for $p+q=3$

A. Acus  Institute of Theoretical Physics and Astronomy,Vilnius
University,Saulėtekio 3, LT-10257 Vilnius, Lithuania<EMAIL_ADDRESS>A. Dargys  Center for Physical Sciences and Technology, Saulėtekio 3, LT-10257
Vilnius, Lithuania<EMAIL_ADDRESS>
###### Abstract.

Closed form expressions in a coordinate-free form in real Clifford geometric
algebras (GAs) $\mathit{Cl}_{0,3}$, $\mathit{Cl}_{3,0}$, $\mathit{Cl}_{1,2}$
and $\mathit{Cl}_{2,1}$ are found for exponential function when the exponent
is the most general multivector (MV). The main difficulty in solving the
problem is connected with an entanglement or mixing of vector and bivector
components. After disentanglement, the obtained formulas simplify to the well-
known Moivre-type trigonometric/hyperbolic function for vector or bivector
exponentials. The presented formulas may find wide application in solving GA
differential equations, in signal processing, automatic control and robotics.

###### Key words and phrases:

Clifford (geometric) algebra, exponentials of Clifford numbers, computer-aided
theory

###### 1991 Mathematics Subject Classification:

Primary 15A18; Secondary 15A66

##  1\. Introduction

In the complex number algebra, which is isomorphic to $\mathit{Cl}_{0,1}$
Clifford geometric algebra (GA), the exponential can be expanded into
trigonometric function sum (de Moivre’s theorem). In 3D GA algebras similar
formulas are known in special cases only. In particular, when the square of
the blade is equal to $\pm 1$, the GA exponential can be also expanded in de
Moivre-type sum, in either trigonometric or hyperbolic functions respectively
[1, 2]. However, expansion of GA exponential in case of 3D algebras
$\mathit{Cl}_{3,0}$, $\mathit{Cl}_{1,2}$, $\mathit{Cl}_{2,1}$ and
$\mathit{Cl}_{0,3}$, when the exponent is a general multivector (MV), as we
shall see is more complicated and as far as the authors know has not been not
analyzed fully as yet. Authors of papers [3, 4] have considered general
properties of the functions of MV variable for Clifford algebras $n=p+q\leq
3$, including the exponential function. For this purpose they have used the
property that the pseudoscalar $I$ in these algebras commutes with all MV
elements and $I^{2}=\pm 1$. This has allowed to introduce more general
functions related to a polar decomposition of MVs. However, the analysis is
not full enough. A different approach to resolve the problem is to factor, if
possible, the exponential into product of simpler exponentials, for example,
in the polar form [5, 6, 7, 8]. General bivector exponentials in
$\mathit{Cl}_{4,1}$ algebra were analyzed in [9] in connection with 3D
conformal GA. In [10], exact closed form expressions for coefficients at basis
elements to calculate GA exponentials in coordinate form are presented for all
3D GAs. The final formulas written in some orthogonal basis appeared rather
complicated and inconvenient in carrying a detailed analysis of the
exponentials, although they may be useful in some cases, for example, for all-
purpose computer programs to calculate GA exponentials with numerical
coefficients.

In this paper the exact exponential formulas [10] are transformed to
coordinate-free form what brings in a clear geometric interpretation to the
problem and to carry a detailed analysis. Also, special cases when various
conditions and relations are imposed upon GA elements are considered that may
be useful in applications of formulas in the paper. In Sec. 2 the notation is
introduced. In Sec. 3 the exponential of the simplest, namely
$\mathit{Cl}_{0,3}$ algebra is considered. Since algebras $\mathit{Cl}_{0,3}$
and $\mathit{Cl}_{1,2}$ are isomorphic, in Sec. 4 the exponentials in the both
algebras are investigated simultaneously. In Sec. 5 the exponential of
$\mathit{Cl}_{2,1}$ is presented. In Sec. 6 possible application of
exponentials to solve GA linear differential equations are presented. Finally,
in Sec. 7 we discuss further development of the problem, including the inverse
function, viz. the GA logarithm.

##  2\. Notation and general properties of GA exponential

The calculation have been done with GA program written for Mathematica package
[11]. In GA space endowed with orthonormal basis we expand a general 3D MV in
inverse degree lexicographic ordering:
$\\{1,\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3},\mathbf{e}_{12},\mathbf{e}_{13},\mathbf{e}_{23},\\\
\mathbf{e}_{123}\equiv I\\}$, where $\mathbf{e}_{i}$ are basis vectors,
$\mathbf{e}_{ij}$ are the bivectors and $I$ is the pseudoscalar.111An
increasing order of digits in basis elements is used, i.e., we write
$\mathbf{e}_{13}$ instead of $\mathbf{e}_{31}=-\mathbf{e}_{13}$. This
convention is reflected in opposite signs of some terms in the obtained
formulas. The number of subscripts indicates the grade. The scalar is a
grade-0 element, the vectors $\mathbf{e}_{i}$ are the grade-1 elements, etc.
In the orthonormalized basis used here the geometric product of basis vectors
satisfy the anti-commutation relation,

$\mathbf{e}_{i}\mathbf{e}_{j}+\mathbf{e}_{j}\mathbf{e}_{i}=\pm 2\delta_{ij}.$
(2.1)

For $\mathit{Cl}_{3,0}$ and $\mathit{Cl}_{0,3}$ algebras the squares of basis
vectors, correspondingly, are $\mathbf{e}_{i}^{2}=+1$ and
$\mathbf{e}_{i}^{2}=-1$, where $i=1,2,3$. For mixed signature algebras such as
$\mathit{Cl}_{2,1}$ and $\mathit{Cl}_{1,2}$ the squares are
$\mathbf{e}_{1}^{2}=\mathbf{e}_{2}^{2}=1$, $\mathbf{e}_{3}^{2}=-1$ and
$\mathbf{e}_{1}^{2}=1$, $\mathbf{e}_{2}^{2}=\mathbf{e}_{3}^{2}=-1$,
respectively.

The general MV that belongs to real Clifford algebras $\mathit{Cl}_{p,q}$ for
$n=p+q=3$ can be expressed as

$\begin{split}\mathsf{A}=&\,a_{0}+a_{1}\mathbf{e}_{1}+a_{2}\mathbf{e}_{2}+a_{3}\mathbf{e}_{3}+a_{12}\mathbf{e}_{12}+a_{23}\mathbf{e}_{23}+a_{13}\mathbf{e}_{13}+a_{123}I\\\
\equiv&\,a_{0}+\mathbf{a}+\mathcal{A}+a_{123}I\equiv
a_{0}+A+a_{123}I,\end{split}$ (2.2)

where $a_{i}$, $a_{ij}$ and $a_{123}$ are the real coefficients, and
$\mathbf{a}=a_{1}\mathbf{e}_{1}+a_{2}\mathbf{e}_{2}+a_{3}\mathbf{e}_{3}$ and
$\mathcal{A}=a_{12}\mathbf{e}_{12}+a_{23}\mathbf{e}_{23}+a_{13}\mathbf{e}_{13}$
is, respectively, the vector and bivector. $I$ is the pseudoscalar,
$I=\mathbf{e}_{123}$. In accord with [10] the sum of vector and bivector is
denoted by symbols $A\equiv\mathsf{A}_{12}=\mathbf{a}+\mathcal{A}$. Similarly,
the exponential of $\mathsf{A}$ is denoted as

$\begin{split}\mathrm{e}^{\mathsf{A}}=&\,\mathsf{B}=b_{0}+b_{1}\mathbf{e}_{1}+b_{2}\mathbf{e}_{2}+ba_{3}\mathbf{e}_{3}+b_{12}\mathbf{e}_{12}+b_{23}\mathbf{e}_{23}+ba_{13}\mathbf{e}_{13}+a_{123}I\\\
\equiv&\,b_{0}+\mathbf{b}+\mathcal{B}+b_{123}I\equiv\,b_{0}+B+b_{123}I.\end{split}$
(2.3)

The main involutions, namely the reversion, grade inversion and Clifford
conjugation are denoted, respectively, by tilde, circumflex and their
combination,

$\begin{split}&\widetilde{\mathsf{A}}=a_{0}+\mathbf{a}-\mathcal{A}-a_{123}I,\\\
&\wideparen{\mathsf{A}}=a_{0}-\mathbf{a}+\mathcal{A}-a_{123}I,\\\
&\widetilde{\wideparen{\mathsf{A}}}=a_{0}-\mathbf{a}-\mathcal{A}+a_{123}I.\end{split}$
(2.4)

###  2.1. General properties of GA exponential

The exponential of MV is another MV that belongs to the same geometric
algebra. Therefore, we shall assume that the defining equation for exponential
is $\mathrm{e}^{\mathsf{A}}=\mathsf{B}$, where
$\mathsf{A},\mathsf{B}\in\mathit{Cl}_{p,q}$ and $p+q=3$. The following
properties hold for MV exponential:

$\begin{split}&\exp(\mathsf{A}\mathsf{B})=\exp(\mathsf{A})\exp(\mathsf{B})\quad\text{if
and only if\ }\mathsf{A}\mathsf{B}=\mathsf{B}\mathsf{A},\\\
&\widetilde{\mathrm{e}^{\mathsf{A}}}=\mathrm{e}^{\widetilde{\mathsf{A}}},\quad\wideparen{\mathrm{e}^{\mathsf{A}}}=\mathrm{e}^{\wideparen{\mathsf{A}}},\quad\widetilde{\wideparen{\mathrm{e}^{\mathsf{A}}}}=\mathrm{e}^{\widetilde{\wideparen{\mathsf{A}}}},\\\
&\mathsf{V}\,\exp(\mathsf{A})\mathsf{V}^{-1}=\exp(\mathsf{V}\mathsf{A}\mathsf{V}^{-1}).\end{split}$
(2.5)

From first formula an exponential identity
$\exp\mathsf{A}=(\exp\mathsf{A/m})^{m}$, $m\in\mathbb{N}$ follows. In the last
expression the transformation $\mathsf{V}$, for example the rotor, has been
lifted to exponent.

The GA exponential $\mathrm{e}^{\mathsf{A}}$ can be expanded in a series that
has exactly the same structure as a scalar exponential [2], from which GA
trigonometric and hyperbolic GA functions as well as various other relations
that are analogues of respective scalars functions follow [10]. For example,

$\begin{split}&\cos^{2}\mathsf{A}+\sin^{2}\mathsf{A}=1,\quad\cosh^{2}\mathsf{A}-\sinh^{2}\mathsf{A}=1,\\\
&\sin(2\mathsf{A})=2\sin\mathsf{A}\cos\mathsf{A}=2\cos\mathsf{A}\sin\mathsf{A},\\\
&\cos(2\mathsf{A})=\cos^{2}\mathsf{A}-\sin^{2}\mathsf{A}.\end{split}$ (2.6)

Also, it should be noted that GA functions of the same argument commute. Thus,
the sine and cosine functions as well as hyperbolic GA sine and cosine
functions satisfy: $\sin\mathsf{A}\cos\mathsf{A}=\cos\mathsf{A}\sin\mathsf{A}$
and $\sinh\mathsf{A}\cosh\mathsf{A}=\cosh\mathsf{A}\sinh\mathsf{A}$.

In sections 3-5 the exact formulas for GA exponentials are presented. If the
MV is in a numerical form, a finite series expansion may be useful as well to
know or just to check the GA formula with the exponential. To minimize the
number of multiplications it is convenient the exponential to represent in a
nested form (aka Horner’s rule)

$\mathrm{e}^{\mathsf{A}}=1+\frac{\mathsf{A}}{1}(1+\frac{\mathsf{A}}{2}(1+\frac{\mathsf{A}}{3}(1+\frac{\mathsf{A}}{4}(1+\dots)))),$
(2.7)

which requires fewer MV products to calculate the truncated series than
working out each power of $\mathsf{A}$. If numerical coefficients in
$\mathsf{A}$ are not too large the exponential $\mathrm{e}^{\mathsf{A}}$ can
be approximated to high precision by (2.7). The series may be programmed as a
simple iterative procedure repeated $k$-times that begins from the end (dots)
with the initial value at $\mathsf{A}/k=1$ and then iteratively moving to
left.222In Mathematica the algorithm reads:
expMVHorner$[\mathsf{A}_{-},n_{-}]:=$
$\text{Module}[\\{\mathsf{B}=1,s=n+1\\},\text{While}[(s=s-1)>0,$
$\mathsf{B}=1.+\text{GP}[\mathsf{B},\mathsf{A}/s]];\mathsf{B}]$, where $n$ is
the number of iterations and GP is the geometric product. If MV coefficients
are large ($a_{J}\geq 3$), in addition, the formula
$(\exp(\mathsf{A}/m))^{m}$, where $m$ is the integer, may be applied at first
to accelerate the convergence and then to raise the result to the $m$th power.

We start from the $\mathit{Cl}_{0,3}$ geometric algebra (GA) where the
expanded exponential in the coordinate form has the simplest MV coefficients.

##  3\. MV exponentials in $\mathit{Cl}_{0,3}$ algebra

###  3.1. Exponential in coordinate-free form

Symbolic formulas in GA may be written in coordinate and coordinate-free
forms. The latter presentation is compact and carries clear geometrical
interpretation and therefore is preferred. However, the formulas written in
the coordinate form sometimes may be helpful too, in particular, in GA
numerical calculations with non-symbolic programmes. In [10] we have found a
general MV exponentials in coordinate form in all three-dimensional (3D) GAs.
Although the expressions are rather involved, however, they acquire a simple
form if coordinate-free notation (second line in Eqs. (2.2) and (2.3)) is
used. Moreover, geometrical analysis of GA formulas becomes simpler and more
evident when formulas are rewritten in a coordinate-free form.

In a case of $\mathit{Cl}_{0,3}$ algebra, after multiplication of coordinates
by respective basis elements and then collection to vector, bivector,
trivector and their products one can transform the exponential components in
[10] to a generic coordinate-free form,

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{1}{2}\mathrm{e}^{a_{0}}\Bigl{(}\mathrm{e}^{a_{123}}(1+I)\bigl{(}\cos
a_{+}+\frac{\sin a_{+}}{a_{+}}(\mathbf{a}+\mathcal{A})\bigr{)}$ (3.1)
$\displaystyle\qquad+\mathrm{e}^{-a_{123}}(1-I)\bigl{(}\cos a_{-}+\frac{\sin
a_{-}}{a_{-}}(\mathbf{a}+\mathcal{A})\bigr{)}\Bigr{)},\ $

where $a_{-}$ and $a_{+}$ are the scalars,

$\displaystyle a_{-}=$
$\displaystyle\sqrt{-(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})+2I\mathbf{a}\mathbin{\wedge}\mathcal{A}}$
(3.2) $\displaystyle=$
$\displaystyle\sqrt{(a_{3}+a_{12})^{2}+(a_{2}-a_{13})^{2}+(a_{1}+a_{23})^{2}},$
$\displaystyle a_{+}=$
$\displaystyle\sqrt{-(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})-2I\mathbf{a}\mathbin{\wedge}\mathcal{A}}$
(3.3) $\displaystyle=$
$\displaystyle\sqrt{(a_{3}-a_{12})^{2}+(a_{2}+a_{13})^{2}+(a_{1}-a_{23})^{2}}\,,$

They show how the vector and bivector components are mixed up. The appearance
of trigonometric functions in Eq. (3.1) indicates that the exponential in
$\mathit{Cl}_{0,3}$ has an oscillatory character. When the denominator in the
formula (3.1), either $a_{+}$ or $a_{-}$, reduces to zero we will have a
special case. The generic formula (3.1) then should be modified by replacing
the corresponding ratios by their limits, $\lim_{a_{\pm}\to 0}\frac{\sin
a_{\pm}}{a_{\pm}}=1$.

If either vector $\mathbf{a}$ or bivector $\mathcal{A}$ in (3.1)-(3.3) is
absent then $a_{+}=a_{-}=a$, where $a$ is a magnitude of the vector
$a=\lvert\mathbf{a}\rvert=\bigl{(}\mathbf{a}\widetilde{\wideparen{\mathbf{a}}}\bigr{)}^{\frac{1}{2}}=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\,$,
or of the bivector
$a=\lvert\mathcal{A}\rvert=\bigl{(}\mathcal{A}\widetilde{\wideparen{\mathcal{A}}}\bigr{)}^{\frac{1}{2}}=\sqrt{a_{12}^{2}+a_{13}^{2}+a_{23}^{2}}$
. If, in addition, the scalar and pseudoscalar are absent, $a_{0}=a_{123}=0$,
the formula (3.1) reduces to the well-known trigonometric expressions for
exponential of vector and bivector in a polar form [2],

$\mathrm{e}^{\mathbf{a}}=\cos\lvert\mathbf{a}\rvert+\frac{\mathbf{a}}{\lvert\mathbf{a}\rvert}\sin\lvert\mathbf{a}\rvert,\quad\mathrm{e}^{\mathcal{A}}=\cos\lvert\mathcal{A}\rvert+\frac{\mathcal{A}}{\lvert\mathcal{A}\rvert}\sin\lvert\mathcal{A}\rvert.$
(3.4)

The trigonometric functions appear in (3.4) because both the vector and the
bivector in $\mathit{Cl}_{0,3}$ satisfy $\mathbf{a}^{2}<0$ and
$\mathcal{A}^{2}<0$. If exponential consists of scalar and pseudoscalar only
then $a_{+}=a_{-}=0$ and the Eq. (3.1) simplifies to hyperbolic functions

$\mathrm{e}^{a_{0}+Ia_{123}}=\mathrm{e}^{a_{0}}(\cosh a_{123}+I\sinh
a_{123}),\quad I^{2}=1.$ (3.5)

In the following we shall distinguish two kinds of coordinate-free formulas
for exponential functions, namely, generic and special. The formula (3.1) is
an example of generic formula since it is valid for almost all real
coefficient $a_{J}$ values, where $J$ is a compound index: $J=i$, $ij$, or
$ijk$. The expression (3.5) represents the special formula, since in the case
$a_{+}=0$ and/or $a_{-}=0$, and as a result we have division by zero in (3.1)
and therefore should use a modified formula (which, in this case can be
obtained by computing limit of (3.1) when $a_{+}\to 0$, and/or $a_{-}\to 0$).
On the other hand the Eq. (3.4) represents an important in practice case of
generic solution (obtained by simply equating the coefficients at scalar and
pseudoscalar and, respectively, at bivector and vector, by zero). For
completeness, in a case of logarithmic functions it would be interesting to
remark that one may add an additional free MV to the generic or special
symbolic solution [12]. A formula with a free parameters included could be
referred to as general solution. The latter kind of solution with a free MV is
absent for an exponential function.

Example 1. Exponential of MV in $\mathit{Cl}_{0,3}$. Let’s compute the
exponential of
$\mathsf{A}=-8-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23}-4\mathbf{e}_{123}$
using the coordinate-free expression (3.1). We find $a_{-}=\sqrt{53}$ and
$a_{+}=\sqrt{353}\,$. The exact numerical answer then is

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\tfrac{\mathrm{e}^{-8}}{2}\biggl{(}\mathrm{e}^{4}(1-\mathbf{e}_{123})$
$\displaystyle\hphantom{\frac{\mathrm{e}^{-8}}{2}\mathrm{e}^{4}}\times\Bigl{(}\cos\sqrt{53}+\tfrac{\sin\sqrt{53}}{\sqrt{53}}(-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23})\Bigr{)}$
$\displaystyle\hphantom{\frac{\mathrm{e}^{-8}}{2}}+\mathrm{e}^{-4}(1+\mathbf{e}_{123})$
$\displaystyle\hphantom{\frac{\mathrm{e}^{-8}}{2}}\times\Bigl{(}\cos\sqrt{353}+\tfrac{\sin\sqrt{353}}{\sqrt{353}}(-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23})\Bigr{)}\biggr{)}.$

For comparison, the calculation of the exponential series (2.7) by Mathematica
v.12 in a floating point regime gives six significant figures of the exact
solution after summation of 70 series terms or iterations. For larger
iterations the number of significant figures increases. However it should be
noted that the convergence is not monotonic and to get the first significant
figure for all basis MV elements no less than 64 series terms (iterations) are
needed in this particular case.

###  3.2. Vector-bivector mixing in $\mathit{Cl}_{0,3}$

In equations (3.2) and (3.3), the outer product
$\mathbf{a}\mathbin{\wedge}\mathcal{A}$ in general case is a trivector. It
entangles or mixes up vector and bivector components in the exponential (3.1).
This is easy to see if we equate to zero either $\mathbf{a}$ or $\mathcal{A}$.
Then, $a_{+}=a_{-}=\lvert\mathbf{a}\rvert$ if $\mathcal{A}=0$ and
$a_{+}=a_{-}=\lvert\mathcal{A}\rvert$ if $\mathbf{a}=0$, where
$\lvert\mathbf{a}\rvert=\bigl{(}\mathbf{a}\widetilde{\wideparen{\mathbf{a}}}\bigr{)}^{\frac{1}{2}}$
and
$\lvert\mathcal{A}\rvert=\bigl{(}\mathcal{A}\widetilde{\wideparen{\mathcal{A}}}\bigr{)}^{\frac{1}{2}}$.
The trivector also vanishes if $\mathbf{a}$ and $\mathcal{A}$ are unequal to
zero but the vector $\mathbf{a}$ lies in the plane $\mathcal{A}$. In this
case333Such a situation is encountered in classical electrodynamics where
magnetic field bivector and electric field vector lie in a the same plane. the
components satisfy the condition
$I\mathbf{a}\mathbin{\wedge}\mathcal{A}=a_{1}a_{23}-a_{2}a_{13}+a_{3}a_{12}=0$.
Then the entanglement (mixing) coefficients become $a_{+}=a_{-}\to
a_{m}=\sqrt{\lvert\mathbf{a}\rvert^{2}+\lvert\mathcal{A}\rvert^{2}}$ and the
exponential (3.1) reduces to

$\mathrm{e}^{\mathsf{A}}=\mathrm{e}^{a_{0}}\mathrm{e}^{Ia_{123}}\big{(}\cos
a_{m}+\frac{\mathbf{a}+\mathcal{A}}{a_{m}}\sin
a_{m}\big{)},\quad\mathbf{a}\|\mathcal{A}.$ (3.6)

Thus, the exponential $\mathrm{e}^{\mathsf{A}}$ in this case can be
factorized. It is interesting that the last multiplier represents an entangled
(mixed up) Moivre-type formula for a sum of vector and bivector, where the
magnitude of $(\mathbf{a}+\mathcal{A})$ is

$a_{m}=\lvert\mathbf{a}+\mathcal{A}\rvert=\sqrt{(\mathbf{a}+\mathcal{A})\widetilde{\wideparen{(\mathbf{a}+\mathcal{A})}}}=\sqrt{\lvert\mathbf{a}\rvert^{2}+\lvert\mathcal{A}\rvert^{2}}.$
(3.7)

We shall remind that in Eq. (3.6) the vector $\mathbf{a}$ lies in the plane
$\mathcal{A}$. A similar formula can be obtained in opposite case if we
assumes that, for example, the vector $\mathbf{a}\|\mathbf{e}_{3}$ and
$\mathcal{B}\|\mathbf{e}_{12}$, i.e., the vector and bivector are
characterized by a single scalar coefficient.444This approach reminds a
popular method in physics where a judicious choice of mutual orientation of
the fields and coordinate vectors allows to simplify the problem
substantially. Then the expression (3.1) gives

$\mathrm{e}^{\mathsf{A}}=-\big{(}\cos(a_{12}-a_{3})+\frac{a_{12}\mathbf{e}_{12}+a_{3}\mathbf{e}_{3}}{a_{12}-a_{3}}\sin(a_{12}-a_{3})\big{)}.$
(3.8)

In conclusion, apart from the pure Moivre-type expressions (see Eqs. (3.4) and
(3.5)), the generic GA exponential (3.1) also contains mixed or entangled MVs
which may be disentangled as shown by Eqs (3.6) and (3.8)) if additional the
conditions are imposed.

##  4\. MV exponentials in $\mathit{Cl}_{3,0}$ and $\mathit{Cl}_{1,2}$
algebras

After multiplication of the scalar coefficients the full expressions of which
are given in [10] by respective basis elements and collection into sum, and
finally combining the resulting expression into a coordinate-free form we find
the following generic exponential of MV $\mathsf{A}$,

$\displaystyle\exp(\mathsf{A})=$ $\displaystyle\mathrm{e}^{a_{0}}(\cos
a_{123}+I\sin a_{123})\Bigl{(}\cos a_{-}\cosh a_{+}+I\sin a_{-}\sinh a_{+}$
$\displaystyle+\frac{1}{a_{\smash{+}}^{2}+a_{\smash{-}}^{2}}(\cosh a_{+}\sin
a_{-}-I\cos a_{-}\sinh a_{+})$ (4.1)
$\displaystyle\hphantom{+\frac{1}{\sqrt{a_{\smash{+}}^{2}+a_{\smash{-}}^{2}}}\quad}\times\bigl{(}a_{-}(\mathbf{a}+\mathcal{A})+a_{+}I(\mathbf{a}+\mathcal{A})\bigr{)}\Bigr{)},\allowdisplaybreaks$
where scalar coefficients $a_{\pm}$ are $\displaystyle
a_{-}=\frac{-2I\mathbf{a}\mathbin{\wedge}\mathcal{A}}{\sqrt{2}\sqrt{\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}+\sqrt{(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})^{2}-4(\mathbf{a}\mathbin{\wedge}\mathcal{A})^{2}}}},$
(4.2) $\displaystyle
a_{+}=\frac{\sqrt{\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}+\sqrt{(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})^{2}-4(\mathbf{a}\mathbin{\wedge}\mathcal{A})^{2}}}}{\sqrt{2}}\,$
$\displaystyle\qquad\textrm{when}\quad\mathbf{a}\mathbin{\wedge}\mathcal{A}\neq
0,\quad\textrm{and}\allowdisplaybreaks$
$\displaystyle\begin{cases}a_{+}=\sqrt{\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}},\quad
a_{-}=0,&\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}>0\\\
a_{+}=0,\quad
a_{-}=\sqrt{-(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})},&\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}<0,\end{cases}$
$\displaystyle\qquad\textrm{when}\quad\mathbf{a}\mathbin{\wedge}\mathcal{A}=0.$

Since the exponential (4) is in a coordinate-free form the above formulas are
valid for both mutually isomorphic $\mathit{Cl}_{3,0}$ and $\mathit{Cl}_{1,2}$
algebras. If one expands the formulas into coordinates, of course, the
resulting expressions will differ by signs of some terms. Note that
determinant555In 3D algebras the determinant of MV $\mathsf{A}$ is defined by
$\operatorname{Det}(\mathsf{A})=\mathsf{A}\widetilde{\wideparen{\mathsf{A}}}\wideparen{\mathsf{A}}\widetilde{\mathsf{A}}$
[13, 14].
$\operatorname{Det}(\mathbf{a}+\mathcal{A})=(a_{\smash{+}}^{2}+a_{\smash{-}}^{2})^{2}$.
When $\operatorname{Det}(\mathbf{a}+\mathcal{A})=0$ we have special case
$\exp(\mathsf{A})=\mathrm{e}^{a_{0}}(\cos a_{123}+I\sin a_{123})$ which again
can be straightforwardly obtained by computing limit of (4), when both
$a_{+}\to 0$ and $a_{-}\to 0$. Simultaneous vanishing of $a_{+}$ and $a_{-}$
means vanishing of both the inner
$\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}$ and outer
$\mathbf{a}\mathbin{\wedge}\mathcal{A}$ products. Then the well-known Moivre-
type formulas from Eq. (4) are

$\displaystyle\exp{\mathsf{A}}$ $\displaystyle=\mathrm{e}^{a_{0}}(\cos
a_{123}+\sin a_{123}I)$ $\displaystyle\mathbf{a}$
$\displaystyle=\mathcal{A}=0,$ (4.3) $\displaystyle\exp{\mathsf{\mathcal{A}}}$
$\displaystyle=\cos\lvert\mathcal{A}\rvert+\frac{\mathcal{A}}{\lvert\mathcal{A}\rvert}\sin\lvert\mathcal{A}\rvert$
$\displaystyle a_{0}$ $\displaystyle=a_{123}=\mathbf{a}=0,$ (4.4)
$\displaystyle\exp{\mathsf{\mathbf{a}}}$
$\displaystyle=\cosh\lvert\mathbf{a}\rvert+\frac{\mathbf{a}}{\lvert\mathbf{a}\rvert}\sinh\lvert\mathbf{a}\rvert$
$\displaystyle a_{0}$ $\displaystyle=a_{123}=\mathcal{A}=0.$ (4.5)

The equation (4.3) represents a special case when $a_{+}=a_{-}=0$.

Similarly to $\mathit{Cl}_{0,3}$ algebra (see Subsec. 3.2), in the exponential
(4) the vector and bivector may be disentangled if we assume that
$\mathbf{a}\|\mathcal{A}$, i.e. the vector $\mathbf{a}$ lies in the plane
$\mathcal{A}$. Then vector-bivector sum can be expressed by trigonometric and
hyperbolic functions in the both $\mathit{Cl}_{3,0}$ and $\mathit{Cl}_{1,2}$
algebras,

$\exp{\mathsf{A}}=\begin{cases}&\cos\sqrt{\lvert\mathcal{A}\rvert^{2}-\lvert\mathbf{a}\rvert^{2}}+\frac{\mathbf{a}+\mathcal{A}}{\sqrt{\lvert\mathcal{A}\rvert^{2}-\lvert\mathbf{a}\rvert^{2}}}\sin\sqrt{\lvert\mathcal{A}\rvert^{2}-\lvert\mathbf{a}\rvert^{2}}\qquad\text{if\
}\mathbf{a}^{2}<\mathcal{A}^{2},\\\
&\cosh\sqrt{\lvert\mathbf{a}\rvert^{2}-\lvert\mathcal{A}\rvert^{2}}+\frac{\mathbf{a}+\mathcal{A}}{\sqrt{\lvert\mathbf{a}\rvert^{2}-\lvert\mathcal{A}\rvert^{2}}}\sinh\sqrt{\lvert\mathbf{a}\rvert^{2}-\lvert\mathcal{A}\rvert^{2}}\quad\text{if\
}\mathbf{a}^{2}>\mathcal{A}^{2}.\end{cases}$ (4.6)

Example 2. Exponential of MV in $\mathit{Cl}_{3,0}$. Let’s take the same MV
$\mathsf{A}=-8-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23}-4\mathbf{e}_{123}$
as in Example 1 and calculate the exponential using the coordinate-free
expression (4). We find
$\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A}=31$,
$-2I\mathbf{a}\mathbin{\wedge}\mathcal{A}=-150$. Then
$a_{-}=-75\sqrt{\frac{2}{31+\sqrt{23461}}}$ and
$a_{+}=\sqrt{\tfrac{31+\sqrt{23461}}{2}}$. The exact numerical answer
therefore is

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{1}{\mathrm{e}^{8}}\bigl{(}\cos(4)-\sin(4)I\bigr{)}\biggl{(}\cos\Bigl{(}75\sqrt{\tfrac{2}{31+\sqrt{23461}}}\Bigr{)}\cosh\Bigl{(}\sqrt{\tfrac{31+\sqrt{23461}}{2}}\Bigr{)}$
$\displaystyle\hphantom{\frac{1}{\mathrm{e}^{8}}\bigl{(}\cos(4)-\sin(4)I\bigr{)}\biggl{(}}-\sin\Bigl{(}75\sqrt{\tfrac{2}{31+\sqrt{23461}}}\Bigr{)}\sinh\Bigl{(}\sqrt{\tfrac{31+\sqrt{23461}}{2}}\Bigr{)}I\allowdisplaybreaks$
$\displaystyle+\frac{1}{\sqrt{23461}}\biggl{(}\Bigl{(}-75\sqrt{\tfrac{2}{31+\sqrt{23461}}}(-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23})$
$\displaystyle\hphantom{+\frac{1}{\sqrt{23461}}\biggl{(}\Bigl{(}}+\sqrt{\tfrac{31+\sqrt{23461}}{2}}(-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23})I\Bigr{)}$
$\displaystyle\quad\times\Bigl{(}-\sin\Bigl{(}75\sqrt{\tfrac{2}{31+\sqrt{23461}}}\Bigr{)}\cosh\Bigl{(}\sqrt{\tfrac{31+\sqrt{23461}}{2}}\Bigr{)}$
$\displaystyle\quad\qquad-\cos\Bigl{(}75\sqrt{\tfrac{2}{31+\sqrt{23461}}}\Bigr{)}\sinh\Bigl{(}\sqrt{\tfrac{31+\sqrt{23461}}{2}}\Bigr{)}I\Bigr{)}\biggr{)}\biggr{)}.$
(4.7)

Example 3. Exponential in $\mathit{Cl}_{1,2}$ with disentanglement included.
Let’s take a simple MV, $\mathsf{A}=3-\mathbf{e}_{1}+2\mathbf{e}_{12}$, which
represents disentangled case $\mathbf{a}\mathbin{\wedge}\mathcal{A}=0$. Then
we have $a_{+}=\sqrt{5}$ and $a_{-}=0$. The answer is expressed in hyperbolic
functions:
$\exp(\mathsf{A})=\mathrm{e}^{3}\Bigl{(}\cosh\sqrt{5}+(-\mathbf{e}_{1}+2\mathbf{e}_{12})\frac{\sinh\sqrt{5}}{\sqrt{5}}\Bigr{)}$.

##  5\. MV exponential in $\mathit{Cl}_{2,1}$

After assembling $\mathit{Cl}_{2,1}$ coefficients given in [10] into MV and
then regrouping them in a coordinate-free form we have

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{1}{2}\mathrm{e}^{a_{0}}\Bigl{(}\mathrm{e}^{a_{123}}(1+I)\bigl{(}\mathop{\mathrm{co}}(a_{+}^{2})+\mathop{\mathrm{si}}(a_{+}^{2})(\mathbf{a}+\mathcal{A})\bigr{)}$
(5.1)
$\displaystyle\qquad+\mathrm{e}^{-a_{123}}(1-I)\bigl{(}\mathop{\mathrm{co}}(a_{-}^{2})+\mathop{\mathrm{si}}(a_{-}^{2})(\mathbf{a}+\mathcal{A})\bigr{)}\Bigr{)}$
where scalar coefficients $a_{\pm}$ are $\displaystyle
a_{-}^{2}=-(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})+2I\mathbf{a}\mathbin{\wedge}\mathcal{A},\quad
a_{-}^{2}\gtrless 0,$ (5.2) $\displaystyle
a_{+}^{2}=-(\mathbf{a}\cdot\mathbf{a}+\mathcal{A}\cdot\mathcal{A})-2I\mathbf{a}\mathbin{\wedge}\mathcal{A},\quad
a_{+}^{2}\gtrless 0.$

To simplify notation in Eq. (5.1) we have introduced $\mathop{\mathrm{si}}$
and $\mathop{\mathrm{co}}$ functions that depending on sign under square root
go over to trigonometric or hyperbolic functions. In $\mathop{\mathrm{si}}$
and $\mathop{\mathrm{co}}$ functions, either lower or upper signs should be
included. All in all, this gives four cases for both si and co,

$\displaystyle\mathop{\mathrm{si}}(a_{+}^{2})=$
$\displaystyle\begin{cases}\frac{\sin\sqrt{a_{+}^{2}}}{\sqrt{a_{+}^{2}}},&a_{+}^{2}>0\\\
\frac{\sinh\sqrt{-a_{+}^{2}}}{\sqrt{-a_{+}^{2}}},&a_{+}^{2}<0\\\
\end{cases};\qquad\mathop{\mathrm{co}}(a_{+}^{2})=\begin{cases}\cos\sqrt{a_{+}^{2}},&a_{+}^{2}>0\\\
\cosh\sqrt{-a_{+}^{2}},&a_{+}^{2}<0\\\ \end{cases}$ (5.4)
$\displaystyle\mathop{\mathrm{si}}(a_{-}^{2})=$
$\displaystyle\begin{cases}\frac{\sin\sqrt{a_{-}^{2}}}{\sqrt{a_{-}^{2}}},&a_{-}^{2}>0\\\
\frac{\sinh\sqrt{-a_{-}^{2}}}{\sqrt{-a_{-}^{2}}},&a_{-}^{2}<0\\\
\end{cases};\qquad\mathop{\mathrm{co}}(a_{-}^{2})=\begin{cases}\cos\sqrt{a_{-}^{2}},&a_{-}^{2}>0\\\
\cosh\sqrt{-a_{-}^{2}},&a_{-}^{2}<0\\\ \end{cases}$ (5.5)

When $a_{-}=0$ and/or $a_{+}=0$ we have special cases, which again can be
easily included taking already mentioned limits, i.e., by putting
$\mathop{\mathrm{co}}(0)=1$ and $\mathop{\mathrm{si}}(0)=1$.

###  5.1. Special cases

If both the vector $\mathbf{a}$ and bivector $\mathcal{A}$ are equal to zero
the exponential (5.1) simplifies to
$\exp{\mathsf{A}}=\exp(a_{0}+Ia_{123})=\mathrm{e}^{a_{0}}(\cosh a_{123}+I\sinh
a_{123})$.

The exponential of vector, when $a_{0}=a_{123}=\mathcal{A}=0$, is

$\exp(\mathbf{a})=\begin{cases}&\cos\sqrt{-\mathbf{a}^{2}}+\frac{\mathbf{a}}{\sqrt{-\mathbf{a}^{2}}}\sin\sqrt{-\mathbf{a}^{2}}\quad\text{if
\ }\mathbf{a}^{2}<0,\\\
&\cosh\sqrt{\mathbf{a}^{2}}+\frac{\mathbf{a}}{\sqrt{\mathbf{a}^{2}}}\sinh\sqrt{\mathbf{a}^{2}}\qquad\text{if
\ }\mathbf{a}^{2}>0.\end{cases}$ (5.6)

The exponential of bivector, when $a_{0}=a_{123}=\mathbf{a}=0$, is

$\exp(\mathcal{A})=\begin{cases}&\cos\sqrt{-\mathcal{A}^{2}}+\frac{\mathcal{A}}{\sqrt{-\mathcal{A}^{2}}}\sin\sqrt{-\mathcal{A}^{2}}\quad\text{if
\ }\mathcal{A}^{2}<0,\\\
&\cosh\sqrt{\mathcal{A}^{2}}+\frac{\mathcal{A}}{\sqrt{\mathcal{A}^{2}}}\sinh\sqrt{\mathcal{A}^{2}}\qquad\text{if
\ }\mathcal{A}^{2}>0.\end{cases}$ (5.7)

If $a_{0}$ and $a_{123}$ are not equal to zero then $\exp(\mathbf{a})$ and
$\exp(\mathcal{A})$ should be multiplied by $\mathrm{e}^{a_{0}}(\cosh
a_{123}+I\sinh a_{123})$.

Example 4. Exponential of MV in $\mathit{Cl}_{2,1}$.
1\. Case $a_{-}^{2}<0,a_{+}^{2}>0$. Using the same MV
$\mathsf{A}=-8-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23}-4\mathbf{e}_{123}$
for $\mathit{Cl}_{2,1}$ we have $a_{-}^{2}=-141$, $a_{+}^{2}=159$. The answer
then is

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{1}{2\mathrm{e}^{8}}\biggl{(}\frac{1}{\mathrm{e}^{4}}(1+I)\Bigl{(}\tfrac{\sin\sqrt{159}}{\sqrt{159}}\bigl{(}-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23}\bigr{)}$
$\displaystyle\hphantom{\frac{1}{2\mathrm{e}^{8}}\biggl{(}\frac{1}{\mathrm{e}^{4}}(1+I)\Bigl{(}\tfrac{\sin\sqrt{159}}{\sqrt{159}}\bigl{(}-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}}+\cos\sqrt{159}\Bigr{)}$
$\displaystyle+\mathrm{e}^{4}(1-I)\Bigl{(}\tfrac{\sinh\sqrt{141}}{\sqrt{141}}(-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}+6\mathbf{e}_{23})$
$\displaystyle\hphantom{\frac{1}{2\mathrm{e}^{8}}\biggl{(}\frac{1}{\mathrm{e}^{4}}(1+I)\Bigl{(}\tfrac{\sin\sqrt{159}}{\sqrt{159}}\bigl{(}-6\mathbf{e}_{2}-9\mathbf{e}_{3}+5\mathbf{e}_{12}-5\mathbf{e}_{13}}+\cosh\sqrt{141}\Bigr{)}\biggr{)}.$

Example 5. Exponential in $\mathit{Cl}_{2,1}$. Case $a_{-}^{2}<0,a_{+}^{2}<0$.
Exponentiating $\mathsf{A}=-6\mathbf{e}_{2}+5\mathbf{e}_{12}+\mathbf{e}_{123}$
of $\mathit{Cl}_{2,1}$ we have $a_{-}^{2}=-11$, $a_{+}^{2}=-11$. The answer
then is

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{1}{2}\biggl{(}\bigl{(}\mathrm{e}(1+I)+\mathrm{e}^{-1}(1-I)\bigr{)}\Bigl{(}\tfrac{\sinh\sqrt{11}}{\sqrt{11}}\bigl{(}-6\mathbf{e}_{2}+5\mathbf{e}_{12}\bigr{)}+\cosh\sqrt{11}\Bigr{)}\biggr{)}.$

Example 6. Exponential in $\mathit{Cl}_{2,1}$. Case $a_{-}^{2}>0,a_{+}^{2}>0$.
Exponentiating
$\mathsf{A}=2+\mathbf{e}_{3}+6\mathbf{e}_{12}+3\mathbf{e}_{123}$ of
$\mathit{Cl}_{2,1}$ we have $a_{-}^{2}=49$, $a_{+}^{2}=25$. The answer then is

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{\mathrm{e}^{2}}{2}\Bigl{(}\mathrm{e}^{3}(1+I)\bigl{(}\tfrac{\sin
5}{5}\bigl{(}\mathbf{e}_{3}+6\mathbf{e}_{12}\bigr{)}+\cos 5\bigr{)}$
$\displaystyle+\mathrm{e}^{-3}(1-I)\bigl{(}\tfrac{\sin
7}{7}(\mathbf{e}_{3}+6\mathbf{e}_{12})+\cos 7\bigr{)}\Bigr{)}.$

Example 7. Exponential in $\mathit{Cl}_{2,1}$. Case $a_{-}^{2}>0,a_{+}^{2}<0$.
Exponentiating
$\mathsf{A}=2-10\mathbf{e}_{2}-10\mathbf{e}_{3}+2\mathbf{e}_{13}+\mathbf{e}_{23}+\mathbf{e}_{123}$
of $\mathit{Cl}_{2,1}$ we have $a_{-}^{2}=35$, $a_{+}^{2}=-45$. The answer
then is

$\displaystyle\exp(\mathsf{A})=$
$\displaystyle\frac{\mathrm{e}^{2}}{2}\Bigl{(}\mathrm{e}(1+I)\bigl{(}\tfrac{\sinh
3\sqrt{5}}{3\sqrt{5}}\bigl{(}-10\mathbf{e}_{2}-10\mathbf{e}_{3}+2\mathbf{e}_{13}+\mathbf{e}_{23}\bigr{)}+\cosh(3\sqrt{5})\bigr{)}$
$\displaystyle+\mathrm{e}^{-1}(1-I)\bigl{(}\tfrac{\sin\sqrt{35}}{\sqrt{35}}(-10\mathbf{e}_{2}-10\mathbf{e}_{3}+2\mathbf{e}_{13}+\mathbf{e}_{23})+\cos\sqrt{35}\bigr{)}\Bigr{)}.$

##  6\. Examples of application: Solution of GA differential equations

The exponential function plays a fundamental role in solution of linear
differential equations. For example, the solution of a homogeneous equation
for MV $\mathsf{X}$,

$\frac{\mathrm{d}\mathsf{X}(t)}{\mathrm{d}t}=\mathsf{X}(t),\quad\mathsf{X}(0)=\mathsf{X}_{0},$
(6.1)

is GA exponential $\mathsf{X}(t)=\mathrm{e}^{t\mathsf{A}}\mathsf{X}_{0}$,
where $t$ is the parameter (in physics usually it is the time). Treating
$t\mathsf{A}$ as a new MV, after expansion of exponential we will obtain the
evolution of MV in time. More generally, with suitable assumptions upon
smoothness of $\mathsf{X}(t)$, the solution of the inhomogeneous system

$\frac{\mathrm{d}\mathsf{X}(t)}{\mathrm{d}t}=\mathsf{X}(t)+f(t),\quad\mathsf{X}(0)=\mathsf{X}_{0},$
(6.2)

may be expressed by

$\mathsf{X}(t)=\mathrm{e}^{t\mathsf{A}}\mathsf{X}_{0}+\int_{0}^{t}\mathrm{e}^{(t-s)\mathsf{A}}f(s)\mathrm{d}s.$
(6.3)

Some of differential MV equations, for example,

$\frac{\mathrm{d}\mathsf{X}(t)}{\mathrm{d}t}=\mathsf{A}\mathsf{X}(t)+\mathsf{B}\mathsf{X}(t),\quad\mathsf{X}(0)=\mathsf{X}_{0},$
(6.4)

have solutions that consist of products of GA exponentials:
$\mathsf{X}(t)=\mathrm{e}^{t\mathsf{A}}\,{X}_{0}\mathrm{e}^{t\mathsf{B}}$. The
answer can be easily checked by direct substitution of $\mathsf{X}(t)$ into
Eq. (6.4) and application of Leibniz’s differentiation theorem [15] .

Trigonometric GA functions, as well as GA roots, arise in the solution of
second order differential equations. For example, the GA equation [16]

$\frac{\mathrm{d}^{2}\mathsf{X}(t)}{\mathrm{d}t^{2}}+\mathsf{A}\mathsf{X}=0,\quad\mathsf{X}(0)=\mathsf{X}_{0}\text{\
and \ }(\mathrm{d}\mathsf{X}/\mathrm{d}t)_{t=0}=\mathsf{X}^{\prime}_{0}$ (6.5)

has the solution

$\mathsf{X}(t)=\cos(\sqrt{\mathsf{A}}\,t)\mathsf{X}_{0}+\big{(}\sqrt{\mathsf{A}}\big{)}^{-1}\sin(\sqrt{\mathsf{A}}\,t)\mathsf{X}_{0}^{\prime},$
(6.6)

where $\sqrt{\mathsf{A}}$ is the square root of $\mathsf{A}$. The
trigonometric functions of MV argument can be expressed by exponentials [10].
Closed form expression for square root of MV in $p+q=3$ can be found in [17].

A concrete example of application of GA exponential in physics can be found in
[10].

##  7\. Discussion

The main results are the formulas (3.1), (4) and (5.1), where the GA
exponentials are presented in an expanded coordinate-free form. Since in 3D
algebras the scalar and pseudoscalar belong to the GA center the respective
real coefficients $a_{0}$ and $a_{123}$ appear in scalar exponentials only.
However, entanglement (mixing) of vector and bivector components takes place.
The mixing is characterized by scalar coefficients $a_{+}$ and $a_{-}$, where
the terms of the form $(a_{i}-a_{jk})^{2}$, $i\neq j\neq k$ appear. The
disentanglement can be done by equating either vector or bivector to zero what
gives the well-known trigonometric-hyperbolic Moivre-type formulas for vector
and bivector. However, more interesting is that the disentanglement also can
be achieved if the vector is parallel to bivector as a result we obtain the
exponential which consists of a sum of scalar, vector and bivector, Eqs. (3.6)
and (4.6).

The related to exponential is the logarithm function. GA logarithm problem is
more difficult to analyze [12]. Contrary to exponential the logarithm,
similarly to complex logarithm, is a not a single valued function and
therefore one must proceed with caution not to mix different branches. In
addition, one must introduce principle value

Finally, the GA exponentials with complex coefficients also need deeper
analysis, more so, since the relativity theory may be introduced by
complexified 3D rather then 4D algebras [18],

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