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

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quasi

Posts: 10,188
Registered: 7/15/05
Re: An optimization problem
Posted: Sep 8, 2013 6:17 AM
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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.

I gave such a proof based on the Cauchy-Schwarz 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



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