Does there exist a set of fractals whose sum defines a differentiable field?
I'm not sure what you mean, but I think I can shed some light on it.
Generally, when you want to answer the question: "Can two ugly things sum to a nice thing?", you can usually answer in the affirmative by picking any ugly thing, then subtracting it from a nice thing. If this subtraction yields another ugly thing, then you have two ugly things that add to a nice thing.
Nice way to explain, really nice ... you should write a book Hurkyl
nice explanation Hurkyl...
Is it possible that a superposition of fractal fields yields a linear field? I hope this is sufficiently ugly, Hurkyl (please forgive my mathematical illiteracy).
The problem is, those terms just don't go together. I really have no idea what you are trying to say.
A fractal pattern is assumed discontinuous and nonlinear. I was wondering whether the "superposition" (sum of values for every correspondent point) over two or more such fractals could generate a continuous, linear pattern.
Consider a fractal of dimension 1.2 projected onto a fractal of dimension 1.8; might this mapping ever represent a differentiable geometry of dimension 3, or even of dimension 2?
Again, you're using words in an alien way. "represent a differential geometry"?
The simple answer is of course you can sum two fractal curves of non-integer dimension and get something nice:
embed your favourite fractal curve y=f(x), and 1-f(x) in the plane, then the sum will be 1just as hurkyl said.
Is there a simple proof that fractals are not differentiable?
... or are fractals differentiable by fractional derivatives?
Separate names with a comma.