Multifractals

Dimension of the Second-Highest Section

Because p₁ = 0.1 and p₂ = p₃ = p₄ = 0.3, the regions containing the largest number of points are those whose addresses contain no 1s. That is, it is the fractal determined by T₂, T₃, and T₄, a gasket.

The regions with the second-largest number of points are those whose addresses contain exactly one 1.

In the square with address 1, this is exactly the gasket consisting of all points with addresses 1a₂a₃a₄..., where the a_i take on all values except 1.

In the squares with addresses 21, 31, and 41, we find the gaskets consisting of those points with addresses 21a₃a₄a₅..., 31a₃a₄a₅..., and 41a₃a₄a₅....

Continuing, we find

a gasket of side length 1/2 in the square with address 1,	three gaskets of side length 1/4 in the squares with addresses 21, 31, and 41,
nine gaskets of side length 1/8 in the squares with addresses 221, 231, 241, 321, 331, 341, 421, 431, and 441,	27 gaskets of side length 1/16 in the squares with addresses 2221, 2231, 2241, 2321, 2331, 2341, 2421, 2431, 2441, 3221, 3231, 3241, 3321, 3331, 3341, 3421, 3431, 3441, 4221, 4231, 4241, 4321, 4331, 4341, 4421, 4431, and 4441.

Unlike our earlier examples of iterative constructions, here successive generations both remove parts of previously added squares, and add new squares.

To find the dimension of this set, we focus on the gaskets and determine the squares that will stay empty as consequence of the gasket structure.

From the number of empty squares, we can determine the number of occupied squares, hence the dimension.

The calculation is simple, but involves several steps, so is presented on another page.

Return to Dimension of the Second-Highest Section.