Hello

[Steve]

Right negative coefficients are allowed. I haven't been thinking clearly about the problem. What I meant is there is no way of getting negative coefficients if they all start out positive.

This is what is now obvious in retrospect. For those coordinates where the coefficients a, b, and c are all positive, then the square has all positive coefficients:

Ie., if x = a +bi +cj then x² = 1(a² +2bc) +i( c²+2ab) +j(b²+2ac), which remain positive.

And they will always be positive when squaring over many iterations, so all orbits of x are positive..

This is unlike the complex plane where, when squared, the real part is a² -b² (potentially becoming negative even though both coefficients start out positive.)

If at least one coefficient in C3 is negative, then upon iterated squaring, negative values can eventually mix into all coefficients.  In this case orbits can perhaps wander almost anywhere in the space.

Now I think I have answered my question:
 

The top two images are that of C3 sliced through the i and j axes respectively. These shapes are sharply truncated in the upper right because the coefficients are all too positive, and cause divergence at the sharp boundary.

Now the unanswered question is why does the remainder of the images look so close to the M-set?  The orbits all wander around in 3 dimensions and still give the same results as the M-set. My conjecture is that a projection of the 3-D orbits onto the 2-D planes are the same as the M-set orbits. Otherwise it is too coincidental. To prove that is going to be an interesting exercise.

An even harder problem is why does C11 have cross-sections that look like the M-set? See
http://www.insidetheoutbox.net/recmaps/recmap-index.php?genre=c11

[Steve]

Schlega:
You were interested in idempotent operations (v² = v). I didn't get the connection till now, that a projection in linear algebra is an idempotent operator.  I succeeded in finding a projection to map the C3 space to a plane that contains the Mandelbrot recursions. It is a -45 degree projection (perpendicular to the i plane) of the j plane onto the real/i plane.  The M-set will lie on the real/i plane, and the j plane disappears.

Or mathematically the projection is defined as follows:
Let a point in C3 be x = 1(a) + i(b) + j(c)
Define projection P(x) = 1(a-c) + i(b) + j(0)
(The real component has the j component subtracted from it because of the -45 degree projection)
(The j component is set to zero because the cross section we are looking at is at j=0)
It is obvious that P(P(x)) = P(x) and is idempotent.

Squaring any C3 point, x, is given by
x² = 1(a² +2bc) + i(c²+2ab) + j(b²+2ac)
(using the relations i² = j and j³ = 1)

But since we are looking at a j=0 cross section, c is zero,
x² = 1(a²) + i(2ab) + j(b²)
Now apply the projection operator:
P(x²) = 1(a²-b²) + i(2ab)
This is identical to the square of the complex number and thus leads to the M-set.

However the 45 degree projection causes an elongation. This can be seen in the slices of C3 - the circular bulbs become ellipses.

[Schlega]

But since we are looking at a j=0 cross section, c is zero,
x² = 1(a²) + i(2ab) + j(b²)
Now apply the projection operator:
P(x²) = 1(a²-b²) + i(2ab)
This is identical to the square of the complex number and thus leads to the M-set.

I'm not sure if this is the full explanation just yet. x² is no longer in the c = 0 plane, so continuing to the next iteration would no longer project to the right place.

As for C11 and the other fractals on your site, if you are willing to share the code you use to generate them I'd love to look at some 3D slices. (Of course, if you would prefer to do that yourself I'll just wait to see your results)

[Steve]

Yes, actually I proved it for the first iteration. Other iterations would have to be proved by induction: if the n^th projection of the orbit of C3 is equal to the n^th M-set orbit, then the n+1 projection is also equal etc.  I don't know how to do it yet.  It seems like you would have to equate n^th order polynomials rather than iterations, and that sounds messy, but there might be a simpler way.

I can give you the inner loop computations for C11, but while there are only 20 floating point ops for each iteration of C3, there are 164 Flops for C11.  Rather than continue that on the "Hello, I'm new Here" board it should be on a math or theory board. I will post whatever groups you want. I presume you want only the inner loops, because my software is all in Java, and I don't know what you need.  Let me know.

[Schlega]

Thank you. The groups I'd like to look at first are C11 and A4b3p. The inner loops are all I need, but I would also like to look at how you generate them automatically. I'm not much of a programmer, but I try to learn.

You're right about this needing its own thread, so I'll stop replying in this topic now.