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 Author Topic: Triplex Algebra  (Read 27519 times) Description: Triplex Algebra 0 Members and 1 Guest are viewing this topic.
David Makin
Global Moderator
Fractal Senior

Posts: 2269

 « Reply #60 on: December 15, 2009, 03:12:30 AM »

Here is the solution for the exponential function:

Awesome!

I had Mathematica sum this in terms of x, y, z instead of r, theta, phi, and I got results that numerically match yours. I tested my results with 100k points where x, y, and z were randomly chosen between -4 and 4, and our answers always agreed to at least ~12 digits.

In terms of x, y, z:
$e^{(x,y,z)}=\left(\frac{1}{2} e^{x-t y} \left(\cos (t x+y)+e^{2 t y} \cos (t x-y)\right),\frac{1}{2} e^{x-t y} \left(\sin (t x+y)-e^{2 t y} \sin (t x-y)\right),e^{\sqrt{x^2+y^2}} \sin (z)\right)$ where $t=\frac{z}{\sqrt{x^2+y^2}}$

Most excellent !

log() anybody ?
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The meaning and purpose of life is to give life purpose and meaning.

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Safarist

Posts: 85

 « Reply #61 on: December 15, 2009, 07:36:15 AM »

Here's what I get for the natural log:
$\ln (x,y,z)=(a,b,c)$
where
$a=\frac{1}{4} \ln \left(\frac{2 z^2 ((x-1) x+(y-1) y) \left((x-1) x+y^2+y\right)}{t}+\left(x^2+y^2\right)^2+z^4\right)$
$b=\frac{1}{4} \left(-\text{atan2}\left((x-1) z-\sqrt{t} y,\sqrt{t} x+y z\right)+\text{atan2}\left(\sqrt{t} y+(x-1) z,\sqrt{t} x-y z\right)-\text{atan2}\left(-\sqrt{t} y-x z+z,\sqrt{t} x-y z\right)+\text{atan2}\left(\sqrt{t} y-x z+z,\sqrt{t} x+y z\right)\right)$
$c=\text{atan}\left(\frac{z}{\sqrt{t}+1}\right)$
$t=(x-1)^2+y^2$

I derived this by analogy with the series expansion for ln(z) about z=1, which is $\ln (z)=\sum _{k=1}^{\infty } \frac{(-1)^{k-1} (z-1)^k}{k}$. For triplex numbers I did the following:
o $(z-1)$ became $(x-1,y,z)$.
o Raising to power $k$ used the regular integer power formula.
o For multiplying by $\frac{(-1)^{k-1}}{k}$, I multiplied each component of the triplex by $\frac{(-1)^{k-1}}{k}$.
o Sum each component of the triplex separately.
o Feed the whole mess to Mathematica and have it simplify the result as much as possible.

I tested these results by comparing a large number of random points within 1/4 unit of (1,0,0) to this triplex series truncated to 1000 terms. The results all matched to 16 digits. I only used sample points near (1,0,0) because this series has a limited radius of convergence.

Using this formula, we get the following:

$\ln (x,0,0)=(\ln (|x|),0,0)$.

$\ln (1,0,0)$ is undefined, but $\lim_{\epsilon \to 1} \, \ln (\epsilon ,0,0)=(0,0,0)$. This is analogous to $\ln (1)=0$.

$\ln (e,0,0)=(1,0,0)$. Analogous to $\ln (e)=1$.

$\ln (-1,0,0)$ is 0 (it should be pi), but $\lim_{\epsilon \to 0^+} \, \ln (-1,\epsilon ,0)=(0,\pi ,0)$ and $\lim_{\epsilon \to 0^-} \, \ln (-1,\epsilon ,0)=(0,-\pi ,0)$. This is similar-ish to $\ln (-1)=i \pi$.

Unfortunately, for most triplex numbers, this log formula and the exp formulas above don't appear to be inverses of one another.
 « Last Edit: December 15, 2009, 07:42:42 AM by BradC » Logged
bugman
Conqueror

Posts: 122

 « Reply #62 on: December 15, 2009, 07:52:32 PM »

Here is the solution for the exponential function:

Awesome!

I had Mathematica sum this in terms of x, y, z instead of r, theta, phi, and I got results that numerically match yours. I tested my results with 100k points where x, y, and z were randomly chosen between -4 and 4, and our answers always agreed to at least ~12 digits.

In terms of x, y, z:
$e^{(x,y,z)}=\left(\frac{1}{2} e^{x-t y} \left(\cos (t x+y)+e^{2 t y} \cos (t x-y)\right),\frac{1}{2} e^{x-t y} \left(\sin (t x+y)-e^{2 t y} \sin (t x-y)\right),e^{\sqrt{x^2+y^2}} \sin (z)\right)$ where $t=\frac{z}{\sqrt{x^2+y^2}}$

Good work BradC. I just tried to further simplify my formula and indeed it exactly reduces to your formula. I don't know why I didn't see this before. Here is the simplified formula using the notation from my original formula:
 Exponential-Simplified.gif (1.27 KB, 498x63 - viewed 492 times.) « Last Edit: December 16, 2009, 07:03:58 AM by bugman » Logged
Safarist

Posts: 85

 « Reply #63 on: December 16, 2009, 02:23:53 AM »

I've done some numerical experiments regarding a triplex^triplex power function...

If we identify the complex number $x+i y$ with the triplex number $(x,y,0)$, and if we define $(a,b,c)^{(x,y,z)}=e^{(x,y,z) \ln (a,b,c)}$ using the above series-based definitions of exp and ln, then the equation $(a,b,0)^{(c,d,0)}=(a+i b)^{c+i d}$ appears to hold for$(b\neq 0\text{ or }a>0)\text{ and }(c\neq 0\text{ or }d\neq 0)$. In other words, this triplex power operation is equivalent to the complex power operation almost everywhere, so it's basically a generalization of the complex power operation.

Note that since triplex multiplication is commutative, we don't have the same ambiguity when defining the power function that quaternions have (see http://www.fractalforums.com/theory/rasing-a-quaternion-to-a-quaternionic-power/).
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Safarist

Posts: 85

 « Reply #64 on: December 16, 2009, 02:31:33 AM »

A picture of the triplex exponential map:
 ComplexExponentialMap.jpg (59.03 KB, 763x432 - viewed 344 times.)  TriplexExponentialMap.jpg (155.26 KB, 1156x800 - viewed 675 times.) Logged
cbuchner1
Fractal Phenom

Posts: 440

 « Reply #65 on: December 16, 2009, 02:38:22 AM »

A picture of the triplex exponential map:

Beautiful!
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kram1032
Fractal Senior

Posts: 1575

 « Reply #66 on: December 16, 2009, 04:56:31 PM »

very nice map which looks pretty much analogous
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bugman
Conqueror

Posts: 122

 « Reply #67 on: December 16, 2009, 06:24:37 PM »

That is really interesting. I would like to examine your formulas when I have more time. I also found a formula for ln({x, y, z}} and sin({x, y, z}) but they were very long and I haven't had time to see if I could simplify them.
 « Last Edit: December 16, 2009, 09:52:58 PM by bugman » Logged
kram1032
Fractal Senior

Posts: 1575

 « Reply #68 on: December 16, 2009, 07:33:36 PM »

ln is the only thing that would be missing to have a nice formula for "true" exponentiation
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David Makin
Global Moderator
Fractal Senior

Posts: 2269

 « Reply #69 on: December 16, 2009, 08:10:06 PM »

ln is the only thing that would be missing to have a nice formula for "true" exponentiation
I think you missed this ?
http://www.fractalforums.com/theory/triplex-algebra/msg9962/#msg9962
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kram1032
Fractal Senior

Posts: 1575

 « Reply #70 on: December 16, 2009, 08:34:26 PM »

wait...
Quote
Unfortunately, for most triplex numbers, this log formula and the exp formulas above don't appear to be inverses of one another.
That could be a problem...

Or was that an error?
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David Makin
Global Moderator
Fractal Senior

Posts: 2269

 « Reply #71 on: December 16, 2009, 08:41:40 PM »

wait...
Quote
Unfortunately, for most triplex numbers, this log formula and the exp formulas above don't appear to be inverses of one another.
That could be a problem...

Or was that an error?

I think it's probably correct, given that for fully triplex numbers (z*z)*z != z^3
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The meaning and purpose of life is to give life purpose and meaning.

http://www.fractalgallery.co.uk/
"Makin' Magic Music" on Jango
Safarist

Posts: 85

 « Reply #72 on: December 16, 2009, 08:52:21 PM »

Quote
Unfortunately, for most triplex numbers, this log formula and the exp formulas above don't appear to be inverses of one another.
That could be a problem...
Or was that an error?

I think that's just the way it comes out. If you use power series to define exp and ln for triplex numbers, then the resulting functions don't turn out to be inverses of each other. For example, $e^{\ln (0.1,0.2,0.3)}=(0.224322,0.00416262,0.679246)$ and $\ln \left(e^{(0.1,0.2,0.3)}\right)=(0.124986,0.217693,0.293778)$. I'm not sure what to think about this. It would be nice if they were inverses, but I don't know how important it is that they're not. Or it's possible that I made a mistake somewhere. I'm interested to see what bugman comes up with if he looks at it some more.
 « Last Edit: December 16, 2009, 08:55:27 PM by BradC » Logged
kram1032
Fractal Senior

Posts: 1575

 « Reply #73 on: December 16, 2009, 09:05:48 PM »

hmmm...

could you do the plane-analogy you did for e^z with ln(z), ln(e^z) and e^(ln(z)) ?

this could visually express the error
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bugman
Conqueror

Posts: 122

 « Reply #74 on: December 16, 2009, 09:59:28 PM »

I expect the log of the triplex to have an infinite number of valid solutions because that is true of the complex case:

"every nonzero complex number z has infinitely many logarithms"
http://en.wikipedia.org/wiki/Complex_logarithm

"If z = re^iθ with r > 0 (polar form), then w = ln r + iθ is one logarithm of z; adding integer multiples of 2πi gives all the others"
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