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*a4)*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*a4)/5! + … )
| 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 |
When we look at powers of multivectors (here) then we made the assumption that the multivector squares to a pure scalar value:
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 |
