Maths - Exponent of Vectors

Here we calculate the exponent of a Vector. The result 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. If this idea of adding scalars and vectors is a problem then see the pages about clifford (geometric) algebra, what this is saying is that the exponent does not have a solution within vector space but it does have a solution in clifford algebra.

in order to derive this result we will try two approaches:

• polar form
• infinite series

However, before we start, it will be useful to review the formula for infinite series.

Infinite Series

I have not found a version of Euler's equation which applies to vectors so we need to calculate it ourselves.

The only method that I can think of is to calculate the exponent using the series:

e(v) =

∞ ∑ n=0

(v)n n!

Where:

• v = vector
• n= integer
• e = 2.71828

We have to be careful with vectors 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 (v)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

Case 1: all dimensions square to positive

We start with the series:

exp(v) = 1 + v¹/1! + v²/2! + v³/3!+ v⁴/4! + v⁵/5! + …

So if all dimensions square to positive then:

v² = v•v = positive scalar

so substituting gives:

exp(v) = 1 + v + v•v/2! + v•v*v/3!+ v•v*v•v/4! + v•v*v•v*v/5! + …

splitting up into real and vector parts gives:

exp(v) = 1 + v•v/2! + v•v*v•v/4! +…+ v*( 1+ v•v/3! + v•v*v•v/5! + …)

So we can see that it is a scalar + vector*another scalar

exp(v) = cosh(√(v•v)) + v*(1/√(v•v))*sinh(√(v•v))

Case 2: all dimensions square to negative

We start with the series:

exp(v) = 1 + v¹/1! + v²/2! + v³/3!+ v⁴/4! + v⁵/5! + …

So if all dimensions square to negative then:

v² = -v•v = negative scalar

so substituting gives:

exp(v) = 1 + v - v•v/2! - v•v*v/3!+ v•v*v•v/4! + v•v*v•v*v/5! + …

splitting up into real and vector parts gives:

exp(v) = 1 - v•v/2! + v•v*v•v/4! +…+ v*( 1- v•v/3! + v•v*v•v/5! + …)

So we can see that it is a scalar + vector*another scalar

exp(v) = cos(√(v•v)) + v*(1/√(v•v))*sin(√(v•v))

Case 3: dimensions square to mixture of positive and negative

The difference between the above two cases depends on whether v² = a positive or negative scalar.

So if we have a vector 'v' made up of the vectors x,y and z then to determine the nature of the exponent we calculate

s = v•v = x² + y ²+ z²

as follows:

√(v•v)	exp(v)
if √(v•v) = +ve then:	exp(v) = cosh(√(v•v)) + v/(√(v•v))*sinh(√(v•v))
if √(v•v) = -ve then:	exp(v) = cos(√(v•v)) + v/(√(v•v))*sin(√(v•v))
if √(v•v) = 0 then:	exp(v) = 1 + v