Yet another alternate transformation idea.

[matsoljare]

The conventional way of squaring complex numbers can be described as squaring the distance from zero, while doubling the angle. The conventional Mandelbrot and Julia sets are of course generated by the simpler, more linear way of adding a value to the horizontal and vertical position value. It's the conflict between these two different geometries, the two different ways of representing the location of a point, which generates the amazing complexities we all know.

What if we instead add a constant to the angle and distance, and square or multiply the real and imaginary parts?

[JosLeys]

The conventional Mandelbrot and Julia sets are of course generated by the simpler, more linear way of adding a value to the horizontal and vertical position value.

That is not correct. The value is added to z², not to z...

[Timeroot]

@JosLeys: I think he meant regular, linear addition after the angle doubling and radius squaring.

@Matsoljare: At first I was about to say that this would be equivalent to working in polar coordinates, but no...

In the regular Mset:

xNew=xOld*xOld-yOld*yOld+Cx
yNew=2*xOld*yOld+Cy

In your idea:

radiusNew=radiusOld+Cradius
angleNew=angleOld+Cangle
which means:
xMiddle=cos(angleNew)*radiusNew
yMiddle=sin(angleNew)*radiusNew      -->

xMiddle=cos(atan(zOld)+atan(C))*cabs(zOld)
yMiddle=sin(atan(zOld)+atan(C))*cabs(zOld) -->

xMiddle=[cos(atan(zOld))*cos(atan(C))-sin(atan(zOld))*sin(atan(c))]*cabs(zOld)
yMiddle=[sin(atan(zOld))*cos(atan(C))-cos(atan(zOld))*sin(atan(c))]*cabs(zOld) -->

xMiddle=[1/[sqrt((1 + zOld^2)(1+C^2))]  + zOld*C/[sqrt((1 + zOld^2)(1+C^2))] ]*cabs(zOld)
yMiddle=[zOld/[sqrt((1 + zOld^2)(1+C^2))]-C/[sqrt((1 + zOld^2)(1+C^2))]*cabs(zOld)

Then for xNew, square it, and for yNew, double it (or vice versa - they will give different fractals, I think).

So the most efficient UF code would be:

init:
 c=#pixel
 c1=(1+c^2)
 z=0
loop:
 sqrtMult=1/sqrt(c1*sqrt(1+z^2))
 z=((sqrtMult+z*c*sqrtMult)^2,2*(z*sqrtMult+c*sqrtMult))
final:
 //find a bailout condition that works!

Good luck! 

[matsoljare]

So can anyone give me an example to see if this actually works?

[Krumel]

So you replace the standard formula

with the following formula:

Well, I tried it out, and it turns out, that it gives pretty lame results for real n.
Then I tried a complex n, and it turns out, that the results are quite interesting.
For example, for n = -0.86 -0.52i you get the attached image.

ps: Bailout conditions are the same as in the standard mandelbrot.

[kram1032]

shouldn't it rather be abs(z)*n?

* -> ^
and
+ -> *
and such?

[Krumel]

That's right, but if you do that, then you'll get just a boring dot...

[kram1032]

ah, I see

I experimented a bit with the properties of ln to reduce or e to increase the grade of an operation...

like: ln(x)+ln(y)=ln(x*y) and ln(x)-ln(y)=ln(x/y) and lnx^y)=y*ln(x) and e^x*e^y=e^(x+y) and (e^x)^y=e^(x*y) and x^ln(y)=y^ln(x)
and with those I wondered if I could somehow manufacture a "fake" ternation and the operation that would be beyond + and -

The lower grade operation would be ln(e^x+e^y) and ln(e^x-e^y), respectively.
The higher grade operation would be something like (e^x)^(e^y) or something like that... - I'm not exactly sure about that.

Maybe, two of those in combination would lead to interesting variations...

[matsoljare]

I'm sure many have tried substituting x*(x+constant) for x*x in the normal Mandelbrot/Julia formula, but can anyone remind me what the result looks like?