Multifractals

Visual interpretations of f(α) curves

Addresses 2 and 4 appear to have about the same number of points. Addresses 1 and 3 have about the same number of points, and fewer than in addresses 2 and 4. So we think

p₂ = p₄ > p₁ = p₃

Because r₁ = r₂ = r₃ = r₄, the minimum probability corresponds to the maximum α, and the maximum probability corresponds to the minimum α.

Then α_min = log(p₂)/log(1/2) and α_max = log(p₁)/log(1/2)

The maximum probability occurs when just the transformations T₂ and T₄ are applied. These generate the line between corners 2 and 4, so this line fills in most densely, which we observe. Then f(α_min) is the dimension of the line between corners 2 and 4, that is, f(α_min) = 1. We see this on the left endpoint of the f(α) curve.

The minimum probability occurs when just the transformations T₁ and T₃ are applied. These generate the line between corners 1 and 3, so this line fills in least densely, which we observe. Then f(α_max) is the dimension of the line between corners 1 and 3, that is, f(α_max) = 1. We see this on the right endpoint of the f(α) curve.

The four transformations T₁, T₂, T₃, and T₄ generate the filled-in unit square, a 2-dimensional shape. consequently max(f(α)) = 2. We see this as the highest point of the f(α) curve.

Here we see less diffrence in density between squares 1 and 2 than we saw in the previous example, so here p₁ and p₂ are closer than they are in the previous example, and consequently α_min and α_max are closer together here than in the previous example.

Return to visual interpretations of f(α) curves.


Addresses 2 and 4 appear to have about the same number of points. Addresses 1 and 3 have about the same number of points, and fewer than in addresses 2 and 4. So we think
p₂ = p₄ > p₁ = p₃
Because r₁ = r₂ = r₃ = r₄, the minimum probability corresponds to the maximum α, and the maximum probability corresponds to the minimum α.
Then α_min = log(p₂)/log(1/2) and α_max = log(p₁)/log(1/2)
The maximum probability occurs when just the transformations T₂ and T₄ are applied. These generate the line between corners 2 and 4, so this line fills in most densely, which we observe. Then f(α_min) is the dimension of the line between corners 2 and 4, that is, f(α_min) = 1. We see this on the left endpoint of the f(α) curve.
The minimum probability occurs when just the transformations T₁ and T₃ are applied. These generate the line between corners 1 and 3, so this line fills in least densely, which we observe. Then f(α_max) is the dimension of the line between corners 1 and 3, that is, f(α_max) = 1. We see this on the right endpoint of the f(α) curve.
The four transformations T₁, T₂, T₃, and T₄ generate the filled-in unit square, a 2-dimensional shape. consequently max(f(α)) = 2. We see this as the highest point of the f(α) curve.
Here we see less diffrence in density between squares 1 and 2 than we saw in the previous example, so here p₁ and p₂ are closer than they are in the previous example, and consequently α_min and α_max are closer together here than in the previous example.