Suppose not all the ln(pi)/ln(ri) are equal. |
Recall f(alpha) = q*alpha + tau(q). |
Differentiating, |
df/dalpha | = (dq/dalpha)*alpha + q + (dtau/dq)*(dq/dalpha) |
| =
(dq/dalpha)*(-dtau/dq) + q + (dtau/dq)*(dq/dalpha) = q |
|
Then d2f/dalpha2 = dq/dalpha. |
From alpha = -dtau/dq we see
dalpha/dq = -d2tau/dq2 < 0. |
Because dalpha/dq < 0, we have dq/dalpha < 0, and so
d2f/dalpha2 < 0. That is, the graph of f(alpha) is concave down. |