Multifractals

Guaranteeing f(alpha_max) = f(alpha_min) = 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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