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

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quasi

Posts: 10,257
Registered: 7/15/05
Re: An optimization problem
Posted: Sep 8, 2013 6:40 AM
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Peter Percival wrote:
>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).


No, your appeal to symmetry doesn't support your claim unless
you know that there is a _unique_ point in R^2 for which the
function achieves a minimum.

If all you know is that the function f is symmetric in the
variables y and z and that f achieves a global minimum value
somewhere in the domain, that's not enough to conclude that
the global minimum value can be achieved at some point for
which y = z.

quasi



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