wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> easy >> Minimal sum of squares
        
(Message started by: NickH on Nov 15th, 2003, 4:37am)
Title: Minimal sum of squares
Post by NickH on Nov 15th, 2003, 4:37am
If x, y, and z are real numbers, subject to the condition

xy + yz + zx = -1,

find the smallest value of

2x2 + 5y2 + 9z2.
Title: Re: Minimal sum of squares
Post by Icarus on Nov 15th, 2003, 1:12pm
A visit to Lagrange supplies 3 stationary values: 6, 5 - sqr(85) = -4.21954..., and 5 + sqr(85) = 14.21954....

It is clear though that the negative value is obtained for complex x, y, z. Plugging in some other values indicates as well that 6 is not a minimum, even when restricted to the real line. It looks like Lagrange was insufficiently constrained!

...Are you sure this is an "easy" problem? :)
Title: Re: Minimal sum of squares
Post by NickH on Nov 15th, 2003, 1:38pm
Well, it can be solved without Lagrange, and is perhaps "elementary" (in its special mathematical sense!) rather than "easy."

Are you sure 6 is not a minimum?   :)
Title: Re: Minimal sum of squares
Post by Icarus on Nov 15th, 2003, 4:37pm

on 11/15/03 at 13:38:10, NickH wrote:
Are you sure 6 is not a minimum?   :)


It appears I dropped a minus sign. So yes, 6 is the minimum.

While Lagrangian multipliers are not required, it is an easier approach than the more obvious "solve for z as a function of x and y, plug the result into the sum, differentiate with respect to x and to y, set both expressions to zero, then solve" approach.

Of course, some other approach may be easier yet.
Title: Re: Minimal sum of squares
Post by Eigenray on Nov 16th, 2003, 3:30am
Assuming there's an easy solution, and trying to find it, write
2x2 +5y2 + 9z2 = [hide][x2 + y2 + (3z)2] + [x2 + (2y)2][/hide].
Title: Re: Minimal sum of squares
Post by SWF on Nov 16th, 2003, 10:29am
The 'easy' way (easy to follow, maybe not so simple to find):[hide]
(x+y+3z)2 + (x+2y)2 [ge] 0
2x2 + 5y2 + 9z2 + 6(yz+xz+xy) [ge] 0
Since yz+xy+xy=-1,
2x2 + 5y2 + 9z2 [ge] 6

The minimum occurs when the original two squared terms are each zero, or
x=-6/[surd]21    y=3/[surd]21    z=1/[surd]21
[/hide]
Title: Re: Minimal sum of squares
Post by Sir Col on Nov 16th, 2003, 11:03am
As you say, simple to follow, but what insight to find; kudos, SWF!
Title: Re: Minimal sum of squares
Post by Pcoughlin on Nov 20th, 2003, 8:02am
It seems like this is smaller.  


IF x=1 and y=1 and z=-1

xy + yz + zx = -1  and

2x2 + 5y2 + 9z2 =

2(1)2 + 5(1)2 + 9(-1)2 =

-2
Title: Re: Minimal sum of squares
Post by towr on Nov 20th, 2003, 8:15am
(-1)2 = 1  ::)
so 2(1)2 + 5(1)2 + 9(-1)2 = 2 + 5 + 9
Title: Re: Minimal sum of squares
Post by Sameer on Jan 29th, 2004, 6:35am
How about this... can someone find the flaw with this proof ? ;D
(x+2y)^2+(y+2z)^2+(x+2z)^2+z^2 >= 0  

(x^2+4y^2+4xy) +(y^2+4z^2+4yz) +(x^2+4z^2+4zx) + z^2 >=0

2x^2+5y^2+9z^2+4(xy+yz+zx) >=0

2x^2+5y^2+9z^2-4>=0   ;since xy+yz+zx =-1

2x^2+5y^2+9z^2 >= 4  

Hence the minimal value is 4.
Title: Re: Minimal sum of squares
Post by towr on Jan 29th, 2004, 10:09am

on 01/29/04 at 06:35:30, Sameer wrote:
How about this... can someone find the flaw with this proof ? ;D
::[hide]the four terms you start out with can't be 0 at the same time under the constraint xy+yz+zx =-1[/hide]::
Title: Re: Minimal sum of squares
Post by Sameer on Jan 29th, 2004, 10:35am

on 01/29/04 at 10:09:51, towr wrote:
::[hide]the four terms you start out with can't be 0 at the same time under the constraint xy+yz+zx =-1[/hide]::

I know that ...but proof?
Title: Re: Minimal sum of squares
Post by towr on Jan 29th, 2004, 2:48pm

on 01/29/04 at 10:35:46, Sameer wrote:
I know that ...but proof?
Well, z2 can only be 0 if z=0, so x and y must be 0 as well, which means xy+xz+yz can't be -1 as it is allready 0..


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board