Re: On limitations of the Mautsch Principle (was Re: ineqality)

On 2 Dec 2005 02:31:58 +0100, Thomas Mautsch <mautsch@xxxxxxxxxxxx>
wrote:

>How shocking! ;-)
>After two internet-free nights, I return to this newsgroup
>to see my name associated to a principle (thanks, Dave!)
>which becomes increasingly useless
>with every new message in this threat:
> :-(
>In news:<9buqo1t7kiq0stepi05gp4tet16p2pac74@xxxxxxx> schrieb quasi:
>> On Wed, 30 Nov 2005 09:11:15 GMT, "Dirk Van de moortel"
>><dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>>>"quasi" <quasi@xxxxxxxx> wrote:
>[ ... ]
>>>> >However, to defend the principle as it is, how about this ...
>[ ... ]
>>>> Conjecture:
>>>>
>>>> Let f(x_1, x_2, ..., x_n) be a
>[ homogeneous ]
>>>> symmetric polynomial with real coefficients such that
>>>> for all x_1, x_2, ..., x_n>=0,
>>>>
>>>> f(x_1, x_2,..., x_n) >= 0 with equality iff x_1 = x_2 = ... = x_n.
>>>>
>>>> Then in the expansion of the polynomial
>>>>
>>>> f(y_1, y1+y_2, ..., y_1+y_2+...+y_n)
>>>>
>>>> there are no negative coefficients.
>>>
>>>Counter examples:
>>>Dave Rusin's
>>> ( x^2 + y^2 )^2 - 3 x y (2 x^2 - 3 x y + 2 y^2)
>>> x y ( x^2 + y^2 )^2 - 3 x^2 y^2 (2 x^2 - 3 x y + 2 y^2)
>>>and Thomas Mautsch's
>>> (x+y-z)^2 (y+z-x)^2 (z+x-y)^2
>>
>> The above examples are _not_ counterexamples to my conjecture.
>>
>> A key requirement in my conjecture is that equality hold iff all
>> variables are equal. Neither of the above examples satisfies that
>> requirement.
>
>Once you have a counterexample like Dave Rusin's from above,
>you can construct very many counterexamples with special properties:
>
>Take for example Dave's example
> P(x,y) = ( x^2 + 3 - 3 x y)^2,
>and add to it a constant epsilon
>times an arbitrary symmetric polynomial Q(x,y)
>of the same homogeneity as P(x,y)
>that satisfies
> Q(x,y) > 0 for all x,y>=0, except (0,0).
>Take e.g.
> Q(x,y) = x^6 + y^6.
>
>Then for all small enough epsilon>0,
>the resulting polynomial
> P(x,y) + epsilon Q(x,y)
>will also be a counter example to my original conjecture,
>but with the additional property that it is only zero
>at (x,y)=(0,0), i.e. for x=y.
>
>Then this polynomial, and also the products
>
> (x-y)^(2k) * ( P(x,y) + epsilon Q(x,y) )
>
>will be counterexamples to your conjecture.

I don't believe it -- let's see it explicitly.

To break my conjecture you have to find a polynomial
f(x_1,x_2,..,x_n), where n>=2, with real coefficients, such that:

(1) f is symmetric and homogeneous

(2) for all x_1,x_2...,x_n>=0, f(x_1,x_2,..,x_n)>=0
with equality iff x_1=x_2...=x_n

(3) in the expansion of f(y_1,y_1+y_2,...,y_1+y_2+...+y_n) there is at
least one term with a negative coefficient.

I have a proof of my conjecture in sketch form, so I won't believe
it's broken until I see an actual counterexample.

quasi
.

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