Multifractals

Unique value for β(q)

To see that for each q the equation (p₁^q)(r₁^β(q)) + ... + (p_N^q)(r_N^β(q)) = 1 determines a unique value of β(q), for each q define a function g(β) by

g(β) = (p₁^q)(r₁^β) + ... + (p_N^q)(r_N^β)

Certainly, g(β) is a continuous function.

Recall that 0 < r_i < 1 and 0 < p_i < 1 for all i. Then

as β → ∞, each r_i^β → 0, and so g(β) → 0,

as β → -∞, each r_i^β → ∞, and so g(β) → ∞, and

finally, dg/dβ = (p₁^q)(r₁^βln(r₁)) + ... + (p_N^q)(r_N^βln(r_N)) < 0 because each ln(r_i) < 0.

Combining these three observations, we see the graph of g(β) must look something like this

That is, g(β) decreases from large positive values to nearly 0, so g(β) = 1 for exactly one value of β, and this is β(q) for the given value of q.

If all the r_i take on a common value, r, the equation

(p₁^q)(r₁^β(q)) + ... + (p_N^q)(r_N^β(q)) = 1

can be solved for β(q):

β(q) = -ln(p₁^q + ... + p_N^q)/ln(r).

Return to Multifractals from IFS.