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Topic: An optimization problem
Replies: 19   Last Post: Sep 14, 2013 9:44 AM

 Messages: [ Previous | Next ]
 Peter Percival Posts: 2,623 Registered: 10/25/10
Re: An optimization problem
Posted: Sep 8, 2013 6:17 AM

quasi wrote:
> 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 real-valued 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.

Indeed so. What I meant when I said it was clear is that the proof is
very simple and can be be done in ones head.

For simplicity's sake let n = 2, x_1 = y and x_2 = z, then my claim is
that if

(1 + y^2 + z^2)/(1 + y + z)

has a minimum, then y = z there. Why is that clear? Because if it
weren't so then

(1 + y^2 + z^2)/(1 + y + z) =/= (1 + z^2 + y^2)/(1 + z + y).

--
Sorrow in all lands, and grievous omens.
Great anger in the dragon of the hills,
And silent now the earth's green oracles
That will not speak again of innocence.
David Sutton -- Geomancies

Date Subject Author
9/7/13 analyst41@hotmail.com
9/7/13 RGVickson@shaw.ca
9/7/13 quasi
9/7/13 RGVickson@shaw.ca
9/8/13 quasi
9/8/13 quasi
9/7/13 RGVickson@shaw.ca
9/7/13 William Elliot
9/8/13 quasi
9/8/13 Peter Percival
9/8/13 quasi
9/8/13 Peter Percival
9/8/13 quasi
9/8/13 Peter Percival
9/8/13 Timothy Murphy
9/9/13 AP
9/9/13 RGVickson@shaw.ca
9/11/13 analyst41@hotmail.com
9/12/13 RGVickson@shaw.ca
9/14/13 analyst41@hotmail.com