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Re: Re: Limit and Root Objects
*To*: mathgroup at smc.vnet.net
*Subject*: [mg72681] Re: [mg72662] Re: Limit and Root Objects
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 15 Jan 2007 05:03:22 -0500 (EST)
*References*: <em606c$2o1$1@smc.vnet.net> <4586C045.2020805@metrohm.ch> <em8jfr$pfv$1@smc.vnet.net> <200701121105.GAA02818@smc.vnet.net> <eoa71a$mp6$1@smc.vnet.net> <200701140934.EAA07476@smc.vnet.net> <44E0EF18-2C95-4BA3-85CA-26AED2AB629E@mimuw.edu.pl>
One correction: the roots of a quadratic below should have been (of
course) 1/2(-b+Sqrt[b^2-4c]) and 1/2(-b+I Sqrt[4c-b^2]), by the usual
high school formula.
Andrzej Kozlowski
On 14 Jan 2007, at 16:04, Andrzej Kozlowski wrote:
>
> On 14 Jan 2007, at 10:34, John Doty wrote:
>
>> Andrzej Kozlowski wrote:
>>
>>> In the cases of polynomials with real coefficients it is indeed
>>> possible to define a continuous root. It is certianly not
>>> possible to
>>> do so for polynomials with complex coefficients.
>>
>> Given the importance of polynomials with real coefficients in many
>> application areas (control theory, for example), I think it could be
>> very handy to have a continuous root function available as a tool,
>> even
>> if it doesn't generalize to the complex coefficient case.
>>
>
> The problem is that the fact that a continuous root exists does not
> mean that there is a nice and convenient formula that can be used
> to implement it. I don't know of any and I rather doubt that there
> is one.
> To see why consider just the case of quadratics x^2 + b x + c ==0.
> The set of real quadratics can be identified with the real plane
> R^2, by taking a quadratic to the pair (b,c). When we remove the
> discriminant, that is the parabola b^2==4 c, we get two open
> regions, consisting of the points where b^2-4c >0 and b^-4c<0. Over
> each region we can define a continuous square root in two ways,
> e.g. we can take -b+Sqrt[b^2-4c] when b^2>4c and -b+I Sqrt[4c-b^2]
> where b^2<4c. This gives us the required continuous root, but of
> course we had to make some arbitrary choices. In the complex case
> this will not work because the function Sqrt[z] cannot be defined
> continuously over the whole complex plane.
>
> I think I can prove that a continuous root always exists in the
> space of real polynomials but withe proof gives no idea of any
> formula. Essentially, the one shows that the space obtained by
> removing the discriminant from the space of all real polynomials of
> degree d consists of [d/2]+1 contractible components. This implies
> that a continuous root exists on each component but tells one
> nothing about any "formula" for it. I have never seen any such
> formula (but then I have never been interested in this question
> before). Anyway, when the degree of the polynomials is larger then
> 4 you won't even have any radical expressions to work with, so you
> will have to use something like Root objects and patch them in a
> continuous way over the whole space of real polynomials using
> Piecewise. There may also exist such formulas expressed in terms of
> inverse trigonometric functions, but the problem with them is that
> they are very hard to manipulate algebraically. There is nothing
> like RootReduce and in general Simplify and FullSimplify will not
> be of much use. So even if one could find a formula of this kind it
> would probably be of pretty limited use.
>
> Andrzej Kozlowski
>
>
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