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Can fractals sum to a linear function?

  1. Oct 31, 2004 #1
    Does there exist a set of fractals whose sum defines a differentiable field?
  2. jcsd
  3. Oct 31, 2004 #2


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    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.
  4. Oct 31, 2004 #3
    Nice way to explain, really nice ... you should write a book Hurkyl
  5. Oct 31, 2004 #4
    nice explanation Hurkyl...

  6. Oct 31, 2004 #5
    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).
  7. Oct 31, 2004 #6


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    The problem is, those terms just don't go together. I really have no idea what you are trying to say.
  8. Oct 31, 2004 #7

    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?
  9. Nov 1, 2004 #8

    matt grime

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    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.
  10. Nov 1, 2004 #9
    Is there a simple proof that fractals are not differentiable?
  11. Nov 1, 2004 #10
    ... or are fractals differentiable by fractional derivatives?
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