Fractal Graph?!

Calculus Level 3

Try graphing the inequality

lim ⁡ n → ∞ ( ∑ i = 0 n digit ( x , i − n ) ⋅ digit ( y , i − n ) ) < 1 \displaystyle {\lim_{n\to\infty}\left(\sum\limits_{i=0}^n{\text{digit}(x, i-n)\cdot{}\text{digit}(y, i-n)}\right)<1}

where digit ( k , i ) { \text{digit}(k, i)} is the i'th digit of a right of the decimal point in base 2. (i.e. digit ( k , i ) = ⌊ k 2 i ⌋ m o d 2 {\text{digit}(k, i)=\lfloor{\frac{k}{2^i}\rfloor}\mod 2}

Notice anything? What is the fractal dimension of this shape (in the range, x ∈ [ 0 , 1 ) x\in[0, 1) and y ∈ [ 0 , 1 ) y\in[0, 1) ?

log₂(3) about 1.3058 π

Chase Marangu by chase marangu , 21, USA

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1 solution

Chase Marangu
Mar 1, 2020

on the interval x ∈ [ 0 , 1 ) ∩ y ∈ [ 0 , 1 ) x\in[0, 1)\cap y\in[0, 1) the limiting point set is a linear-transformed sierpinski triangle. Scaling it up by 2x side length increases it's "fractal mass" by 3x (as there are 3x as many sierpinski triangles) Therefore the answer is l o g 2 ( 3 ) log_2{(3)} . Analogously scaling up a square by 2x side length gives 4x as many squares of original side length so l o g 2 ( 4 ) = 2 log_2{(4)}=2 and for cube l o g 2 ( 8 ) = 3 log_2{(8)}=3

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