About multiplicative inverses, Birkhoff and Mac Lane's A Survey of
Modern Algebra says quaternions have them. Their explanation: if
x = a + b i + c j + d k (conventional notation, a b c d are real)
and you define
x* = a - b i - c j - d k
then
x x* = x* x = a^2 + b^2 + c^2 + d^2
so as long as one of a b c d is not zero, x has the inverse
x*/(a^2 + b^2 + c^2 + d^2) .
(It is easy to see that
x (inverse of x) = (inverse of x) x = 1 .)
Kip Murray
On Sun, 27 Apr 2008, Don Guinn wrote:
Well, the problem with quaternions is that multiplication is not
commutative. Therefore there is no such thing as a multiplicative inverse.
You can have a left inverse or a right inverse. Fortunately, they are
conjugates, depending on how you define conjugate. In calculating
transcendentals, computers typically pick a "principle" solution. There are
many solutions to something like the ArcSin of x. Where is the principle
solution in quaternions?
Take "a*b" is equivalent to "a(+&.^.)b" in real and complex numbers. Not
necessarily in quaternions. Picking principle solutions in the typical way
does not work. But, interestingly, there are solutions for ^. in quaternions
where that relationship is true.
I defined the sin(x) as follows. Right now I don't remember where I found
that definition. Long time ago. I slept since then.
Qsin=: 3 : 0"0
't V mV'=. QtVm y
Qcloseq((sin t)*(cosh mV)),((cos t)*(sinh mV)*(V%mV))
)
where
QtVm=: (({.;}.),[:<[:+/&.:*:}.)&>
NB.*QtVm - Break a quaternion into real, imaginary and magnitude.
NB. Parts consist of t,V and |y|
and
sin=: 1&o.
cos=: 2&o.
sinh=: 5&o.
cosh=: 6&o.
Say something like
1 o. 1i2j3k4
91.7837i21.8865j32.8297k43.773
Is this a good principle value for 1i2j3k4? I'm not sure.
It would be interesting to see if a Taylor series gave the same answer. I
haven't tried that yet. Will have to.
On Sun, Apr 27, 2008 at 7:43 PM, Raul Miller <[EMAIL PROTECTED]> wrote:
> On Sun, Apr 27, 2008 at 8:42 AM, Don Guinn <[EMAIL PROTECTED]> wrote:
> > I took a shot at defining the circular functions in quaternions. I'm not
> > sure that they are correct.
>
> Taylor Series?
>
> --
> Raul
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>
>
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Kip Murray
[EMAIL PROTECTED]
http://www.math.uh.edu/~km
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