Maths - 2D multivector Exponent

Infinite Series

The exponent is given by the series:

e(m) =

∞ ∑ n=0

(v)n n!

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

 

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