Multifractals

Guaranteeing f(alphamax) = f(alphamin) = 0.

Suppose all the ln(pi)/ln(ri) are distinct.
We have seen that if i is not m, then limq -> infinityhi(q) = 0. A similar argument shows that if i is not M, then limq -> -infinityhi(q) = 0.
Then as q -> infinity,
p1qr1tau(q) + ... + pNqrNtau(q) = 1
becomes
pmqrmtau(q) = 1.
Taking ln of both sides and solving for tau(q), we see that as q -> infinity,
tau(q) -> -(ln(pm)/ln(rm))*q
That is, the q -> infinity asymptote of tau(q) is the line tau = -(ln(pm)/ln(rm))*q, which passes through the origin.
As q -> infinity the tangent line approaches the asymptote,
alpha = -dtau/dq approaches alphamin,
and f(alphamin) is the tau-value at which the asymptote intersects the tau-axis.
Consequently, f(alphamin) = 0.
 
By a similar argument we see f(alphamax) = 0.

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