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Re: Re: Limit and Root Objects

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  • Subject: [mg72681] Re: [mg72662] Re: Limit and Root Objects
  • From: Andrzej Kozlowski <akoz at>
  • Date: Mon, 15 Jan 2007 05:03:22 -0500 (EST)
  • References: <em606c$2o1$> <> <em8jfr$pfv$> <> <eoa71a$mp6$> <> <>

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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