Infinite Series
The exponent is given by the series:
e^{(m)} = 


Where:
 m = multivector
 n= integer
 e = 2.71828
We start with the series:
exp(m) = 1 + m ^{1}/1! + m ^{2}/2! + m ^{3}/3!+ m ^{4}/4! + m ^{5}/5! + …
We now need to plug in a value for (v)^{n} which we have calculated on this page:
exp(m) = 1 + m + (s + 2a*m)/2!
+ (2a*s + (4a²+s)*m) ^{}/3!
+ ((s² + 4*a²*s) + (4*s*a + 8*a³)*m) ^{}/4!
+ ( (4*s²*a + 8*s*a³) + (s² + 12*a²*s + 16*a^{4})*m) ^{}/5! + …
splitting up the scalars and powers of m gives:
exp(m) = 1 + s/2!
+ (2a*s ) ^{}/3!
+ (s² + 4*a²*s) ^{}/4!
+ (4*s²*a + 8*s*a³) ^{}/5! + …
+ m*( 1 + 2a)/2!
+ (4a²+s) ^{}/3!
+ (4*s*a + 8*a³) ^{}/4!
+ (s² + 12*a²*s + 16*a^{4})^{}/5! + … )
sin(x)  x  x^{3}/3! + x^{5}/5! ... +(1)^{r}x^{2r+1}/(2r+1)!  all values of x 
cos(x)  1  x^{2}/2! + x^{4}/4! ... +(1)^{r}x^{2r}/(2r)!  all values of x 
ln(1+x)  x  x^{2}/2! + x^{3}/3! ... +(1)^{r+1}x^{r}/(r)!  1 < x <= 1 
exp(x)  1 + x^{1}/1! + x^{2}/2! + x^{3}/3! ... + x^{r}/(r)!  all values of x 
exp(x)  1  x^{1}/1! + x^{2}/2!  x^{3}/3! ...  all values of x 
e  1 + ^{1}/1! + ^{2}/2! + ^{3}/3!  =2.718281828 
sinh(x)  x + x^{3}/3! + x^{5}/5! ... +x^{2r+1}/(2r+1)!  all values of x 
cosh(x)  1 + x^{2}/2! + x^{4}/4! ... +x^{2r}/(2r)!  all values of x 
When we look at powers of multivectors (here) then we made the assumption that the multivector squares to a pure scalar value:
m^{2} = s = positive or negative scalar
so substituting gives:
exp(m) = 1 + m + s/2! + s*m/3!+ s^{2}/4! + s^{2}*m/5! + …
splitting up into real and vector parts gives:
exp(m) = 1 + s/2! + s^{2}/4! +…+ m*( 1+ s/3! + s^{2}/5! + …)
In order to fit to the series above we will express this in terms of √s:
exp(m) = 1 + (√s)^{2}/2! + (√s)^{4}/4! +…+ m/(√s)*( √s+ (√s)^{3}/3! + (√s)^{5}/5! + …)
So the match to the series depends on the sign of √s as follows:
√s  exp(m) 

if √s = +ve then:  exp(m) = cosh(√s) + m/(√s)*sinh(√s) 
if √s = ve then:  exp(m) = cos(√s) + m/(√s)*sin(√s) 
if √s = 0 then:  exp(m) = 1 + m 