Re: Help (polynomial)


<john_ramsden@xxxxxxxxxxxxxx> ha scritto nel messaggio
news:1117123480.240174.154430@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
>
> big.ass wrote:
>>
>> > JR wrote
>> >
>> > Denoting p(x) by (x^2 - 2u.x + u^2.v).(x^2 + 2u.x + u^2.w) + 1,
>> > your condition m^2 + n^2 < 2k is equivalent to 0 < v + w.
>> >
>> > But denoting p(x) by (x^2 - 2a.x + a^2.b).(x^2 + 2a.x + a^2.c),
>> > the condition for two real and two non-real roots is equivalent
>> > to (1 - b).(1 - c) < 0.
>>
>> A bit careful here: I can't assume that p(x) can be written
>> in that form because it is just what have to prove.
>
> I shouldn't have used x as a place-holder in my polynomials,
> because your x is my x - u. Using t or something instead of x
> would have been less misleading.
>
> But that aside, and barring slips on my part, my conditions
> above allow an alternative but equivalent formulation of the
> problem. My first polynomial product can be expressed as:
>
> ((x-u)^2 + u^2.(v-1)).((x-u)^2 + 4u.(x-u) + u^2.(w+3)) + 1
>
> So your k, m, n are related to my u, v, w by:
>
> k, m + n, m.n = u^2.(v-1), - 4u, u^2.(w + 3)
>
> and your condition, i.e. (m + n)^2 - 2mn < 2k, becomes:
>
> 16.u^2 - 2.u^2.(w + 3) < 2u^2.(v - 1)
>
> which, assuming u != 0, is equivalent to:
>
> 6 < v + w
>
> Drat - see what I mean about slips? But the principle
> is unchanged.
>
> The possible (?) advantage of my notation is that there
> is no power of x^3 in the expanded polynomial, which
> thus allows the product of quadratics to take the form
>
> (x^2 - 2a.x + ..).(x^2 + 2a.x + ..)


I see no advantage. It seems too laborious too.
Terms in x^3 are still there. If you make a substitution (with Mathematica
for example) you get:
1 - ((4 u + u^2*(3 + w)x - x^2))(u^2*(-1 + v) + x^2)
that even symplified leads to a complete polynomial of degree 4 in wich the
terms in x^3 are
3 u^2 x^3 - u^2 w x^3.

and they won't go away.


>


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