Quaternion Numbers

Date: 01/23/97 at 16:28:40
From: Terry W. Gintz
Subject: Quaternion division and transcendental functions

Hi,

Presumably, when William Hamilton discovered quaternions (sometime in 
the 1830s), he must have gone further than just multiplying them.  I'm 
very interested in how quaternion division is accomplished (from my 
experience with complex numbers, first you have to find the inverse of 
the divisor, then multiply the dividend by that. But how to find the 
inverse of a quaternion number?  Can you do this in basic algebraic 
terms, please?  My knowledge of calculus is mainly differential.)

I'm also interested in whether quaternion math can be extended to
transcendental functions like trig and exponential functions.

My main use of the above info will be to implement extended 3D
quaternion plotting in my program ZPlot.  It already plots formulas 
like q^2+c and cq^2-1 very nicely, but I'd like to see what q^e+c 
looks like, etc.

Terry W. Gintz

Date: 01/23/97 at 16:58:05
From: Doctor Wilkinson
Subject: Re: Quaternion division and transcendental functions

I am not an expert on quaternions, and don't know the answer to your 
second question, but I do know how do divide quaternions.  It's just 
like with complex numbers.  If you multiply a + bi + cj + dk 
by a - bi -cj - dk, you get a^2 + b^2 + c^2 +d^2, as you can verify 
quite easily,  The rules ij = -ji = k, etc. make everything work out 
beautifully.  You can use this to clear quaternions out of the 
denominators.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 01/24/97 at 14:36:25
From: Doctor Toby
Subject: Re: Quaternion division and transcendental functions

Another doctor has already told you about dividing quaternions, so I 
will talk about trigonometric and exponential functions.

If you continue to take calculus, you'll probably learn about these 
functions in the context of infinite series (if you haven't already).
To calculate exp x = e^x, add up x^n/n! for all natural numbers n
(including n = 0), where n! = the factorial of n = n*(n-1)*...*2*1.
That is, exp x = x^0/0! + x^1/1! + x^2/2! + ....  (Note that 0! = 1, 
and x^0 = 1 no matter what x is, even if x = 0.)  The series for cos x 
is the sum of (-1)^n x^2n/(2n)!, and the series for sin x is the sum 
of (-1)^n x^(2n+1)/(2n+1)!.  Using these series, you can calculate for 
yourself that exp ix = cos x + i sin x in the complex numbers.

(There are many technical issues connected with infinite series that 
you will learn about if you take a course on them.  It so happens that 
the series I will discuss here have no technical difficulties 
whatsoever in the real, complex, or quaternion numbers.  So you don't 
have to worry about the technical issues; any manipulation you 
normally would do with a finite sum is perfectly legitimate with these 
infinite sums.)

It is traditional to use the infinite series for exp, cos, and sin as 
*definitions* of these functions in nonreal contexts. In particular, 
we may take these series as definitions in the quaternions.  I will 
examine exp in the quaternions in some detail.

Using the series for exp and the binomial theorem, you can prove the 
identity: exp (a + b) = exp a * exp b for real and complex numbers, 
because these numbers commute. (x * y = y * x for all real or complex 
x and y.)  This identity does *not* hold in the quaternion numbers.
For example, exp (i + j) = cos sqrt 2 + (i + j)(sin sqrt 2)/sqrt 2,
while exp i * exp j = (cos 1)^2 + (i + j)(sin 2)/2 + k(sin 1)^2, and 
exp j * exp i = (cos 1)^2 + (i + j)(sin 2)/2 - k(sin 1)^2.  However, 
whenever ab = ba (for example, whenever a is real), then 
exp (a + b) = exp a * exp b does hold in the quaternions.

That fact is the secret to calculating exp (a + ib + jc + kd).  Since 
a commutes with ib + jc + kd (when a, b, c, and d are real),
exp (a + ib + jc + kd) = exp a * exp (ib + jc + kd). Now, you can't 
break exp (ib + jc + kd) into exp ib * exp jc * exp kd, but you don't 
need to.  Simply divide this vector by its magnitude. The quantity 
L = (ib + jc + kd)/sqrt (b^2 + c^2 + d^2) satisfies L^2 = -1 (check it 
for yourself).  And you can go back to the infinite series to see that
exp Lx = cos x + L sin x, whenever L^2 = -1 and x is real.  So let 
M = sqrt (b^2 + c^2 + d^2), so ib + jc + kd = LM.  Then 
exp (a + ib + jc + kd) = exp a (cos M + L sin M).

Now, in order to raise quaternions to arbitrary powers, you'll also 
have to calculate their logarithms.  You do this by accepting x as the 
logarithm of y whenever exp x = y.  Just as you can write a complex 
number in the form r(cos t + i sin t), you can write any quaternion in 
the form r(cos M + L sin M), for some real r and M and some vector L 
satisfying L^2 = -1.  The logarithm of r(cos M + L sin M) is of course 
ln r + LM, where ln is the natural logarithm in the real numbers.  Of 
course, you can add any integer multiple of 2 pi to M, so the 
logarithm is unique only up to integer multiples of 2 pi L. The 
logarithm of a real number is particularly non-unique.  Since the real 
number r equals r(cos M + L sin M) for *any* L satisfying L^2 = -1 
(when M is a multiple of 2 pi), the logarithm of r can be any 
ln r + LM.  For example, the logarithm of 2 can be ln 2 + 2i pi,
ln 2 - 4 k pi, ln 2 + 2 (i + j) pi/sqrt 2, and so on.

At any rate, the definition of x^y is exp (y log x), where log x can 
take the value of any logarithm of x.  If n is an integer, 
exp (n log x) can take only one value, no matter what value is chosen 
for log x, and this agrees with the usual definition of x^n.  Now you 
should be able to calculate q^e+c.

As for the trigonometric functions, the infinite series will give you 
the common complex formulae cos x = (exp ix + exp -ix)/2 and 
sin x = (exp ix - exp -ix)/2i.  These formulae can be derived in the 
same way for quaternions, if you replace i by the appropriate value of 
L for the quaternion x.  The other trigonometric functions can be 
defined in terms of cos and sin.

If you have any more questions, feel free to write back!

-Doctor Toby,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 01/24/97 at 20:35:14
From: Doctor Toby
Subject: Re: Quaternion division and transcendental functions

I neglected to mention that you can differentiate quaternion 
functions. Use the same definition in terms of limits as for complex 
functions. The proofs of the sum, product and chain rules go through 
unchanged. (Be sure to put the products in the right order.) You can 
differentiate the series for exp, sin, and cos to calculate 
exp' = exp, sin' = cos, and cos' = -sin; these formulas work just as 
well in the quaternions as anywhere else. (This process of 
differentiating a series by differentiating its terms is one of the 
trickiest when there are technical difficulties. Luckily, you don't 
have them with these series.)

Note also that the summation formulas for trigonometric functions
sin(a + b) = (sin a)cos b + (cos a)sin b and 
cos(a + b) = (cos a)cos b - (sin a)sin b work only when a and b 
commute.  This applies to all the formulas related to these summation 
formulas, such as half angle formulas and so on. In general, go back 
to the series to check if an identity holds. (They're usually pretty 
easy to prove from the series when they do.)

-Doctor Toby,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 01/24/97 at 20:46:14
From: Terry W. Gintz
Subject: Quaternion division

Thanks for the info.  If I understand this right, in order to divide
quaternions you multiply by the reciprocal of the denominator, which
should be (a-bi-cj-dk)/t where t = a^2+b^2+c^2+d^2.  Simplifying, the
quaternion reciprical is a/t -bi/t -cj/t -dk/t, which is clearly the
logical extention to the complex reciprocal a/t -bi/t, where 
t = a^2+b^2.

I am trying out the new quaternion formulas in my program ZPlot now, 
and seeing a lot of new shapes too! Thanks very much for the 
direction. The exponential q function seems to work well, but the 
sin(q) formula didn't seem to work right using exponentials.

I tried coding the formula for sine but the resulting 3D shape didn't 
come close to resembling the 2D Julia sine set, and the time to 
compute (exp Lx - exp -Lx)/2L was astronomical.  So I switched to an 
alternate formula, one I derived from the formula I usually use for 
computing complex sine (faster than the exponential form): 
sin x cosh M + L cos x sinh M. This gives 3D results with a shape that 
closely resembles the 2D Julia set, so I guess I got the derivation 
right.  Thanks again for your help. 

Terry W. Gintz

Date: 01/30/97 at 21:59:44
From: Doctor Toby
Subject: Re: Quaternion division and transcendental functions

You're right about how to divide quaternions.  I double-checked your 
formula, and it looks correct to me.

For a moment, I thought you were going to ask if I knew any 
alternative formulas for the sine that might compute faster, and I'd 
have had to say I don't know very well what computes quickly or 
slowly.  But now you've given me the formula! So now I can tell it to 
anyone who asks with your question again, so they can compute quickly 
right away.  Thanks and good luck!

-Doctor Toby,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/