Re: Maximise polynomial degree N


marmitage@xxxxxxxxx wrote:
Thanks for taking a look at it - I've solved for the restricted cases
when x=a=b or when a or b = 1 or 0. The problem is that in the more
general case the differentials are polynomials of degree > 4 and so
have no general solution in radicals... is this correct in this case?

Secondly the effect on how n is set optimally seems to be fairly
weak... the properties on n* only hold seem to hold when everything is
close to optimal

The one "solvable" relation that pops out for an interior
critical point is from setting df/da = 0, where:

f(a,b) = 1 + b + (a-b) b^n + (b-1-a) b^n a^(N-n)

in which case we obtain:

df/da = b^n * [ 1 + (N-n) (b-1-a) a^(N-n-1) - a^(N-n) ]

Because b > 0 in the interior of the region of interest, we
have:

0 = 1 - (N-n) (1-b) a^(N-n-1) - (N-n+1) a^(N-n)

This is a first-degree polynomial equation for b, and thus:

b = 1 - [1 - (N-n+1) a^(N-n)]/[(N-n) a^(N-n-1)]

This relationship can be exploited in a couple of ways.
One is to obtain conditions on a so that a <= b < 1.
The other is to substitute for b in f(a,b) and therefore
obtain a single parameter optimization (for fixed n)
over some reduced range of values for a determined
from a <= b < 1.

I take it from your description of numerical experiments
that interior maximums do occur. That's contrary to
what my intuition was, but all the more interesting in
view of that.


regards, chip

.

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