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 anti-commute 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 |
|---|---|---|---|---|
| anti-commute | all positive | exp(v) = cosh(|v|) + norm(v)*sinh(|v|) | Euclidean vector | see below |
| anti-commute | 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 eiθ = 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 anti-commute (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 - x3/3! + x5/5! ... +(-1)rx2r+1/(2r+1)! | all values of x |
| cos(x) | 1 - x2/2! + x4/4! ... +(-1)rx2r/(2r)! | all values of x |
| ln(1+x) | x - x2/2! + x3/3! ... +(-1)r+1xr/(r)! | -1 < x <= 1 |
| exp(x) | 1 + x1/1! + x2/2! + x3/3! ... + xr/(r)! | all values of x |
| exp(-x) | 1 - x1/1! + x2/2! - x3/3! ... | all values of x |
| e | 1 + 1/1! + 2/2! + 3/3! | =2.718281828 |
| sinh(x) | x + x3/3! + x5/5! ... +x2r+1/(2r+1)! | all values of x |
| cosh(x) | 1 + x2/2! + x4/4! ... +x2r/(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):
m2 = s = positive or negative scalar
so substituting gives:
exp(m) = 1 + m + s/2! + s*m/3!+ s2/4! + s2*m/5! + …
splitting up into real and vector parts gives:
exp(m) = 1 + s/2! + s2/4! +…+ m*( 1+ s/3! + s2/5! + …)
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 |
