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

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quasi

Posts: 10,398
Registered: 7/15/05
Re: An optimization problem
Posted: Sep 8, 2013 3:13 AM
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>quasi wrote:
>>analyst41 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?

>
>Yes.
>
>The minimum value is
>
> ((2n)*((n+1)-sqrt(n^2+n)))/sqrt(n^2+n)
>
>which occurs when all x's are equal to
>
> (sqrt(n^2+n) - n)/n


As Ray Vickson noted, the values above are wrong.

I was rushing out so had no time to check the values but,
while my values were wrong, my strategy was correct.

Here's a corrected version, complete with reasoning ...

Fix a positive integer n.

Let X = {x = (x_1,...,x_n) in R^n | 0 <= x_i <= 1 for all i}.

For x in X, let

sum(x_i) = x_1 + ... + x_n

sum((x_i)^2) = (x_1)^2 + ... + (x_n)^2

f(x) = (1 + sum((x_i)^2))/(1 + sum(x_i))

For s in [0,n] let

X_s = {x in X | sum(x_i) = s}

For x in X_s, the Cauchy-Schwarz inequality yields

s^2 <= n*sum((x_i)^2)

with equality iff x_i = s/n for all i.

Thus,

sum((x_i)^2) >= (s^2)/n

=> 1 + sum((x_i)^2) >= 1 + (s^2)/n

=> (1 + sum((x_i)^2))/(1 + s) >= (1 + (s^2)/n)/(1 + s)

=> f(x) >= (1 + (s^2)/n)/(1 + s)

with equality iff x_i = s/n for all i.

Thus, on X_s, f has minimum value (1 + (s^2)/n)/(1 + s)
which occurs uniquely at x = (s/n,...,s/n).

Hence, to find the minimum value of f on X, it suffices to
find the minimum value of

v(s) = (1 + (s^2)/n)/(1 + s)

for s in [0,n].

At the endpoints, v(0) = v(n) = 1.

For s in (0,n),

s/n < 1

=> (s^2)/n < s

=> 1 + (s^2)/n < 1 + s

=> v(s) < 1

Thus, since v(0) = v(n) = 1, the minimum value of v(s) for s
in the closed interval [0,n] occurs at a value of s in the open
interval (0,n) such that v'(s) = 0.

For s in (0,n), v'(s) = 0 occurs uniquely at

a = -1 + sqrt(n+1)

Thus, the minimum value of v on X is

v(a) = (2*((n+1)-sqrt(n+1)))/(n*sqrt(n+1))

which occurs uniquely at the point

x = (x_1, ..., x_n)

where

x_i = a/n = (-1 + sqrt(n+1))/n

for all i.

quasi



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