Here we calculate the exponent of a multivector. When we looked at the result for pure vectors (on this page) we saw that it depends on whether the dimensions commute or anticommute and whether the dimensions square to positive or negative. A summary of the results is given in the following table:
commutative  square to  result  example  derivation 

anticommute  all positive  exp(v) = cosh(v) + norm(v)*sinh(v)  Euclidean vector  see below 
anticommute  all negative  exp(v) = cos(v) + norm(v)*sin(v)  bivector  see below 
commute  two positive  exp(x + Dy) = cosh(v) + D sinh(v)  double number  double number pages 
commute  one positive, one negative  r e^{iθ} = r (cos(θ) + i sin(θ))  complex number  complex number pages 
where:
 v = √(v•v)) = magnitude scalar value
 norm(v) = v*(1/√(v•v)) = normalised (unit length) vector
To summarise and interpret the results, if the dimensions commute (as they do for complex numbers for example) then the result is a pure vector but, if the dimensions anticommute (as they do for vectors in euclidean space for example) then the result is a scalar plus a vector.
in order to generalise this result to any multivector we will use infinite series:
Infinite Series
The exponent is given by the series:
e^{(m)} = 


Where:
 m = multivector
 n= integer
 e = 2.71828
We have to be careful with multivectors because they are not in general commutative. I think the above series applies but I'm not absolutely sure.
We now need to plug in a value for (m)^{n} which we have calculated on this page.
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 
We start with the series:
exp(m) = 1 + m ^{1}/1! + m ^{2}/2! + m ^{3}/3!+ m ^{4}/4! + m ^{5}/5! + …
When we look at powers of multivectors (here) then we made the assumption that the multivector squares to a pure scalar value (does not apply if scalar part is non zero so be warned: these results are not general):
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 