quasi
Posts:
12,021
Registered:
7/15/05


Re: An optimization problem
Posted:
Sep 8, 2013 6:17 AM


Peter Percival wrote: >quasi wrote: >> William Elliot wrote: >>> analyst41@hotmail.com wrote: >>> >>>> consider f(x1,x2,...xn) = {1 + sum(xi)^2)} / {1 +sum{x(i)} >>>> 0 <=xi <=1 for all i. >>>> >>>> The maximum function value of 1 occurs at either >>>> all x's = 0 or all x's = 1. >>>> Can an explicit formula be given for the minimum? >>> >>> Let g(x) = (1 + x^2)/(1 + x) and look for extreme values of g. >> >> The function g is not applicable. >> >> From the context, the OP intended >> >> f(x_1, ..., x_n) = (1 + sum((x_i)^2))/(1 + sum(x_i)) >> >> whereas you interpreted it (incorrectly) as >> >> f(x_1, ..., x_n) = (1 + (sum(x_i))^2))/(1 + sum(x_i)) >> >> Of the two possible interpretations above, only the first is >> consistent with the OP's claim that for all n, the maximum >> value of f is 1. > >I know nothing about such problems, but is it not clear that >_if_ there is a minimum,
Let X denote the standard unit cube in R^n.
Since f is a continuous realvalued function with domain X, and since X is compact, it follows that f achieves a minimum value on X.
>it occurs at some point where all the x_i's are equal?
No, that's not automatic  it requires proof.
I gave such a proof based on the CauchySchwarz inequality.
>Suppose that common value is x, then the minimum of > > (1 + nx^2)/(1 + nx) > >is sought. That occurs at x = (sqrt(1+n)1)/n, which (by a >happy accident, so to speak) is >= 0 and <= 1. > >Then argue that the minimum is global...
That was also already done.
quasi

