Quaternionic powers of quaternions?

1. Jan 14, 2005

Aphex_Twin

Can such things be defined?

I know there are complex powers to complex numbers (using polar forms), but what about quaternions (or perhaps complex powers of quaternions...).

Any ideas?

2. Jan 14, 2005

matt grime

Well, the complex powers are defined by exp and log, and these in turn have power series expansions, so one may be able to apply them to quaternions, however, multiplication isn't commutative, so the sums are going to be slightly odd to work out.

3. Jan 14, 2005

Aphex_Twin

A little something more concrete?

4. Jan 14, 2005

matt grime

Like what? Fractional, irrational, and complex powers are defined with exp and log which are defined as power series for the real numbers (for those numbers possesing a valid log expansion, obviously - what are you going to replace Arg with?), for heaven's sake, so how can a generalization to some other non-commutative ring possibly be more concrete? Some particular numbers may have nice exponentiation relations, eg i^i=-1, but even in the complex numbers, most do not (what is sqrt(2) ^(sqrt(5)-isqrt(3))?)

It's just a formal definition, it may not even make sense - after all, in what topology are you considering convergence issues?

Last edited: Jan 14, 2005
5. Jan 14, 2005

dextercioby

Maybe he meant "concrete",as in the relative of "cement"???

Daniel.

6. Jan 14, 2005

Aphex_Twin

Please, be gentle, I am not a math guru and I'm not familiar with power series. So if you can spell it out in more layman's terms? ;)

7. Jan 14, 2005

matt grime

Power series aka mclaurin or taylor series. (not a guru topic: taught in high school in the UK, or it used to be), but a simple way of trying to work out what f(x) is for some x, and some function.

$$e^x = \sum_{r=0}^{\infty} \frac{x^r}{r!}$$

is the definition (strictly speaking, one of them) of e^x.

log(x) is the inverse function (thought of as e^x a function from R to R^+) there are various ways of working out log(y) in terms of power series too (that is polynomials in y)

x^r is defined to be
$$e^{rlogx}$$

e, or exp has a nice obvious continuation to the complex numbers, log doesn't have such a good extension (it is a "many valued function")

some times, in some cases you may find x^y for complex arguments in nice ways, but not always. So surely you'd be better off finding out what the complex version of your question is before thinking about the quarternionic one.

8. Jan 14, 2005

Aphex_Twin

Ah, yes, Taylor ;) English is not my first language so I don't know all Math English names. Though Taylor is not my strong point.

I already figured out p^q where p and q are complex. It goes something like this:

I write the first in polar form and the second in cartesian form:

p=r*e^i*t
q=a+b*i

p^q = (r*e^i*t)^(a+b*i) = r^(a+b*i) * e^[i*t*(a+b*i)]

r^(a+b*i) = r^a * r^(b*i) = r^a * e^[b*i*ln(r)]

e^[i*t*(a+b*i)] = e^(i*t*a+i*t*b*i) = e^(i*t*a-t*b)

p^q = r^a * e^[i*b*ln(r)]*e^(i*t*a) * e^(-t*b) = r^a/(e^t*b) * e^[i*b*ln(r)]*e^(i*t*a) ...

And it boils down to multiplying complex numbers.

How do you do this with quaternions?

9. Jan 14, 2005

matt grime

you need to show that that is well-defined and leads to no contradictions first, that is that it is independent of the representation of p and q that you've chosen. (if i replace t by t+2pi then what happens?) Moreover, if I take p=1 and choose t to be 2pi, and raise to the power sqrt(2) I get an complex number with strictly nonzero imaginary part. is that right?

then you'd have to decide if the quarternions have a nice choice of r, theta style norm and argument. They are a 2-d vector space over C for instance. But you'd also need to check that algebraic operations on H, have a nice reflection in the geometric properties of this vector space.

Last edited: Jan 14, 2005
10. Jan 14, 2005

Aphex_Twin

Am I missing something here?

11. Jan 14, 2005

Hurkyl

Staff Emeritus
Let's consider p = q = i as an example.

There are lots of ways to wrote p in polar form. Two examples are:

i = 1 * e^(1/2 pi i)
i = 1 * e^(5/2 pi i)

Now, if we apply your method, using the first polar form we get:

i^i = 1^i * e^((1/2) pi i * i)
= e^(i ln 1) * e^(-pi/2)
= e^0 * e^(-pi/2)
= e^(-pi/2)

Using the second form,

i^i = 1^i * e^((5/2) pi i * i)
= e^(i ln 1) * e^(-5pi/2)
= e^0 * e^(-5pi/2)
= e^(-5pi/2)

Which are clearly different numbers. The problem here is that your procedure is not well-defined. That means that when given two different representations of the same number, your method gives two different results.

Just looking at the end result, Quaternions raise another problem -- their addition is commutative, but their multiplication is not. So, if we naively try to apply the identities for log and exp:

a * b = e^(ln a) * e^(ln b) = e^(ln a + ln b) = e^(ln b + ln a) = e^(ln b) * e^(ln a) = b * a

Which "proves" that multiplication is commutative... but we know multiplication is not, which is a contradiction.

12. Jan 15, 2005

Aphex_Twin

What if I make r and t dependent on a c and d, where p=c+d*i
r = |p| = sqrt(c^2 + d^2)
t = arg (p) = atan(d/c)

This way there is a unique representation. I'm looking more into it...

I was also looking at the formula here:
http://mathworld.wolfram.com/ComplexExponentiation.html

And trying to figure it out (without success thus far).

Last edited: Jan 15, 2005
13. Jan 15, 2005

dextercioby

I see u have lost connection with the quaternions.You're speaking about old,rusty complex numbers...
What do you mean unique representation??
You mean from
$$p=c+id \Rightarrow p=re^{it}$$ through a one to one mapping (understood in terms of pairs/elements of R^{2}) ??If so,think again and use the famous formula due to Leonhard Euler.

Daniel.

14. Jan 15, 2005

Hurkyl

Staff Emeritus
Technically, that is not true. arg and atan are multivalued functions. To get a unique representation, you have to make an (arbitrary) choice of a branch cut. There is a standard choice of branch cut that we denote Arg and Atan, which does give a unique representation (called the principal value)

15. Jan 15, 2005

Aphex_Twin

Alright, I'm getting it. I presume there are more such "branch cuts" to be made when talking of quaternionic powers of quaternions.

This means for quaternions I can use the formula for complex exponentiation, replace the real numbers with complex numbers (chosing an appropriate norm and atan representations) and make more restrictions. I'll see what I can get.

16. Jan 17, 2005

matt grime

I think Hurkyl's proof that any reasonalbe extension of log and exp implies the quartenions are commutative ought to be the final nail in the coffin, don't you?