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NickH
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Minimal sum of squares
« on: Nov 15th, 2003, 4:37am » |
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If x, y, and z are real numbers, subject to the condition
xy + yz + zx = -1,
find the smallest value of
2x2 + 5y2 + 9z2.
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Icarus
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Re: Minimal sum of squares
« Reply #1 on: Nov 15th, 2003, 1:12pm » |
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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? 
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NickH
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Re: Minimal sum of squares
« Reply #2 on: Nov 15th, 2003, 1:38pm » |
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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?
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Icarus
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Re: Minimal sum of squares
« Reply #3 on: Nov 15th, 2003, 4:37pm » |
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on Nov 15th, 2003, 1:38pm, NickH wrote:| Are you sure 6 is not a minimum? |
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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.
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Eigenray
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Re: Minimal sum of squares
« Reply #4 on: Nov 16th, 2003, 3:30am » |
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Assuming there's an easy solution, and trying to find it, write
2x2 +5y2 + 9z2 = [x2 + y2 + (3z)2] + [x2 + (2y)2].
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SWF
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Re: Minimal sum of squares
« Reply #5 on: Nov 16th, 2003, 10:29am » |
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The 'easy' way (easy to follow, maybe not so simple to find):
(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
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Sir Col
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Re: Minimal sum of squares
« Reply #6 on: Nov 16th, 2003, 11:03am » |
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As you say, simple to follow, but what insight to find; kudos, SWF!
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Pcoughlin
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Re: Minimal sum of squares
« Reply #7 on: Nov 20th, 2003, 8:02am » |
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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
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towr
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Re: Minimal sum of squares
« Reply #8 on: Nov 20th, 2003, 8:15am » |
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(-1)2 = 1 
so 2(1)2 + 5(1)2 + 9(-1)2 = 2 + 5 + 9
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Sameer
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Re: Minimal sum of squares
« Reply #9 on: Jan 29th, 2004, 6:35am » |
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How about this... can someone find the flaw with this proof ? 
(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.
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towr
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Re: Minimal sum of squares
« Reply #10 on: Jan 29th, 2004, 10:09am » |
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on Jan 29th, 2004, 6:35am, Sameer wrote:| How about this... can someone find the flaw with this proof ? |
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::the four terms you start out with can't be 0 at the same time under the constraint xy+yz+zx =-1::
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Sameer
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Re: Minimal sum of squares
« Reply #11 on: Jan 29th, 2004, 10:35am » |
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on Jan 29th, 2004, 10:09am, towr wrote:
::the four terms you start out with can't be 0 at the same time under the constraint xy+yz+zx =-1::
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I know that ...but proof?
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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towr
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Re: Minimal sum of squares
« Reply #12 on: Jan 29th, 2004, 2:48pm » |
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on Jan 29th, 2004, 10:35am, Sameer wrote:
I know that ...but proof? |
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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..
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