A simpler way to define Mandelbrot sets?

[Tglad]

In fact this might not be a new way for a lot of people, but anyway...

Many-valued algebra (http://www.springerlink.com/content/lq53075471u24w50/) is the simple idea that you can consider a set of numbers to be a single many-valued quantity. For example the solution to x^2 = 4 is the quantity {-2, 2}.
Anyway, this leads to the definition for a Julia set:

A Julia set is the solution to z^2 + c = z.
It is the many-valued quantity that satisfies this equation.

This definition avoids the technical details of iterating infinitely many times, in fact it doesn't imply how you should arrive at the value at all, just states its basic property.
Just like sqrt(3) doesn't mention whether you iterate or not to find the result, it really is just a rearrangement of- the solution to x^2 = 3.

To make a similar definition for the Mandelbrot set requires knowing about a feature of many-valued quantities, you can have two sorts of equality. Normal equality {-1,1} = {1,-1} and (what I call) identity {-1,1}=={-1,1}. Identity is used to operate on the same indices of each many-value quantity. For example, if z = {-1,1}, then z + z = {-2,0,2}, that is how many-valued quantities sum. But if two zs in that equation are identical then z + z = {-2,2}. If we indicate identical quantities by suffixing a number, this makes it clearer. z1+z1 = {-2,2}, but z1+z1+z = {-3,-1,1,3}.
With this in mind you can define the Mandelbrot set as:

The Mandelbrot set is the solution to z1^2 + z1 = z
The quantity such that applying z^2+z to each element returns the quantity z.

[Schlega]

I might be misunderstanding you, but I don't think this definition works. For example, if z is the entire complex plane, the equation will be satisfied. Other solutions are the nonnegative reals and {0}. I think the set of all points with periodic orbits would also work.

To fix it, I would say let z_i be a bounded set satisfying z_i1²+z_i1=z_i, where i is a member of an arbitrary indexing set. The mandelbrot set would be the union of all z_i.

[Tglad]

I think you have to exclude the full complex plane as an answer (or alternately you could consider there to be two solutions, the full plane and the answer of interest).
Taking a real number analogy, if you make the equation x = 2*(x - 1), the obvious answer is x = 2.
Using many-valued algebra there are two answers to this, the value 2 and the set of all real numbers {R}. Since the 'full' set is a valid answer to most equations I think it can be considered the non-interesting solution and put aside by default in most situations.
Perhaps a clearer way to define it is to have a class of many-valued quantities which are bounded subsets of complex plane, call them B. We then require a solution within this class of numbers.
The other point you bring up is whether subsets of the mandelbrot/julia are valid solutions. By analogy, if we ask what is the solution to cos(x) = 1, then you could say 0, or {0, 2pi} or {0, 4pi, 78pi}. I think that the non-arbitrary answer is to give the full set of results- x = {n*2pi for all integer n}. The question has not constrained which answer it should be so the full set of answers is sensible. Equally, if I ask you to draw abs(x)<2 on the real axis I would expect you to draw a line between -2 and 2, without leaving arbitrary gaps.

I realised after posting that the formula I gave was wrong for the mandelbrot since mandelbrot iterations go outside the mandelbrot (but the julia set equation is correct).
Perhaps better is, the set of points c such that

z^2 + c = z   (for some many-value quantity z in B)
c element of z

[Nahee_Enterprises]

I realised after posting that the formula I gave was wrong for the mandelbrot
since mandelbrot iterations go outside the mandelbrot... 

Interesting!!!  A whole week goes by since the posting and you did not bother to Modify it to correct the wrong formula.            

[Timeroot]

I think the formal definition of the Julia Set is pretty much just that. Except it says to exclude all fixed points. It's incorrect to state that the nonnegative reals satisfy this, because if c=i it will immediately map to something else. Also, 0 is almost never a solution, because 0^2+c=0 if and only if c=0.

I think the correct way to state it would be, "The Julia Set J_c is the set of all solutions to the equation z^2+c=z such that for no z1 and value of n does (z1^2+c)ⁿ=z1." (Here the ⁿ denotes iteration.)

No, that's actually not correct, because a Julia set also includes unstable fixed points... but this is still "almost correct", in the sense that only a finite (I think?) number of the points are included. And even if there are infinite points excluded, they have a smaller cardinality than the Julia set.

Of course, there are some Julia sets for other functions which, under the formal definition, fill the entire complex plane. Our computers don't treat them that way, though, so we get the meaningless pretty pictures 

[Tglad]

Nahee, I don't respond to rudely worded comments. Improve your attitude please.

Schlega, also do take this stuff with a pinch of salt (I'm no expert you see!). I'm thinking out loud really, its sort of an interesting problem to try and find the simplest definition for the mandelbrot, given that it is so unique.

Timeroot, I guess a different way to explain a julia set which is probably the very easiest, is that a julia set is a shape which when squared results in the same shape.

[Timeroot]

Except that could also include any number of basins of attraction of the Fatou Set. The formal Julia set is one set satisfying that. The filled Julia Set is another. So is the complex plane. So is everything but the Julia Set...  Math is evil! Muahaha!

[Schlega]

I'm not an expert, either (I didn't even realize that iterations go outside the mandelbrot set). It is an interesting problem, though.

I've been playing with mandelbrot polynomials, and I think I found a definition that doesn't involve iterations, but is computationally impractical. It's the set of all z such that does not diverge to infinity.

Nevermind. That doesn't work either.