Multifractals

Unique value for tau(q)

To see that for each q the equation (p1q)(r1tau(q)) + ... + (pNq)(rNtau(q)) = 1 determines a unique value of tau(q), for each q define a function g(tau) by
g(tau) = (p1q)(r1tau) + ... + (pNq)(rNtau)
Certainly, g(tau) is a continuous function.
Recall that 0 < ri < 1 and 0 < pi < 1 for all i.
    As tau -> infinity, each ritau -> 0, and so g(tau) -> 0.
    As tau -> -infinity, each ritau -> infinity, and so g(tau) -> infinity.
    Finally, dg/dtau = (p1q)(r1tauln(r1)) + ... + (pNq)(rNtauln(rN)) < 0
    because each ln(ri) < 0.
Combining these three observations, we see the graph of g(tau) must look something like this
That is, g(tau) decreases from large positive values to nearly 0, so g(tau) = 1 for exactly one value of tau, and this is tau(q) for the given value of q.
 
If all the ri take on a common value, r, the equation
(p1q)(r1tau(q)) + ... + (pNq)(rNtau(q)) = 1
can be solved for tau(q):
tau(q) = -ln(p1q + ... + pNq)/ln(r).