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MathGroup Archive 2007

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Re: Re: Limit and Root Objects

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72688] Re: [mg72662] Re: Limit and Root Objects
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 15 Jan 2007 05:31:33 -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>

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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