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. |