A plane self-similar set of dimension e-1

Find an IFS whose invariant set has dimension $e-1$. Generate an image of your set.


Note: There are lots of potential answers to this question so everyone should feel free to make a contribution.

Comments

  • Our dimension is [imath] e-1 [/imath]. We will choose an arbitrary large value [imath] r_0=0.99 [/imath] and solve the equation:

    [dmath] 0.99^{e-1} + r^{e-1}=1 [/dmath]

    Which gives us [imath] r \approx 0.0937498 [/imath] after plugging it into Mathematica.

    We then plug these two [imath] r [/imath] values into our IFS visualizer, under spiral, and get the following:

    r0 = 0.99;
    r1 = 0.0937498;
    t = 15*degree;
    [
    rotate(t).compose(scale(r0)),
    scale(r1,[1,0])
    ]

  • @theoldernoah Close! But not quite. :/

    Remember - for the dimension theorems to be applicable, we need the open set condition to be satisfied. From an intuitive perspective, this means that the pieces shouldn't overlap.

    Now, here's your picture with an outline super-imposed:

    Here's a similar picture but, instead of the outline, we have the image of the outline under one application of the IFS:

    We see the overlap pretty clearly.

    You'll need to fiddle with the rotation parameter in the IFS Visualizer. I'd recommend a larger value of $r_0$ as well.

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