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 > >

**References**:**Re: Limit and Root Objects***From:*Paul Abbott <paul@physics.uwa.edu.au>

**Re: Limit and Root Objects***From:*John Doty <jpd@whispertel.LoseTheH.net>

**Re: a problem with integrate**

**Re: Limit and Root Objects**

**Re: Limit and Root Objects**

**Re: Re: Limit and Root Objects**