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MathGroup Archive 2007

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RE: Re: Re: Re: Limit and Root Objects

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72692] RE: [mg72668] Re: [mg72644] Re: [mg72620] Re: Limit and Root Objects
  • From: "David Park" <djmp at earthlink.net>
  • Date: Mon, 15 Jan 2007 05:47:43 -0500 (EST)

Paul Abbot asked: "Again, ignoring root ordering, why
isn't it possible for all these roots to maintain their identity and so
be continuous functions of the parameter? And wouldn't such continuity
be nicer than enforcing root ordering?"

(I'm not certain if I exactly understand Paul's question. Does he mean that
each individual root function, in the set of root functions, must be
continuous, or does he allow global reindexing on the entire set to achieve
continuity?)

So, what exactly does your mathematical theorem say? Does it say that given
a one-parameter finite order polynomial it is not possible to continuously
reindex the set of root solutions so that the reindexed roots will vary
continuously in the complex plane as a function of the parameter? Or that
you can't do this when the polynomial contains a complex coefficient? If it
says that, I don't believe it.

It certainly is possible to arrange things so that the roots vary
continuously, and it can even be useful.

It reminds me of a Groucho Marx quip, something about: "Are you going to
listen to me or are you going to go by what you see in front of your eyes?"

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/



From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl]

I don't understand what you mean by "continuous enough" and why you
think and animation can "disprove" a mathematical proof?

Please read the argument:

http://forums.wolfram.com/mathgroup/archive/2006/Dec/msg00496.html

It is very simple. Remember the point is that you cannot define such
a function that will be continuous over the entire 6 dimensional (3
complex dimensions) space of polynomials of degree 3 with complex
roots (we actually normally remove the discriminant from the space).
You are using a smaller subspace ( you have just one complex
parameter) and over a smaller subspace naturaly it is possible to
have a continuous root.
If you look carefully at Adam's argument you will be easily able to
see what must go wrong over the entire space of complex cubics.

This in fact is a good illustration of what a double edged weapon
graphics and animations are in studying mathematics: they can just as
easily mislead your intuition and lead you to wrong conclusions as to
right ones. Which is one reason why I think one should never rely too
much on such tools when teaching mathematics. Proofs are proofs and
no number of "convincing animations" and "experimental mathematics"
can replace them.

Andrzej Kozlowski,

Department of Mathematics, Informatics and Mechanics
Warsaw University, Poland


On 14 Jan 2007, at 01:54, David Park wrote:

> This looks continuous enough to me, with and without complex
> coefficients.
>
> Needs["Graphics`Animation`"]
> Needs["Graphics`Colors`"]
>
> frame[a_] :=
>   Module[{roots = rootset[a], newroots, rootpermutations,
> distances, pick,
>       locations},
>     If[memoryrootset === Null, newroots = roots,
>       rootpermutations = Permutations[roots];
>       distances =
>         Plus @@ Abs[Part[roots, #] - memoryrootset] & /@
> basepermutations;
>       pick = Part[Position[distances, Min[distances], 1], 1, 1];
>       newroots = Part[roots, Part[basepermutations, pick]]];
>     memoryrootset = newroots;
>     locations = {Re[#], Im[#]} & /@ newroots;
>
>     Show[Graphics[
>         {LightCoral, AbsolutePointSize[15],
>           Point /@ locations,
>           Black,
>           MapThread[Text[#1, #2] &, {Range[5], locations}],
>
>           Text[SequenceForm["a = ",
>               NumberForm[a, {3, 2}, NumberPadding -> {" ", "0"}]],
>             Scaled[{0.1, 0.95}], {-1, 0}]}],
>
>       AspectRatio -> Automatic,
>       TextStyle -> {FontFamily -> "Courier", FontSize -> 12,
>           FontWeight -> "Bold"},
>       Frame -> True,
>       PlotRange -> {{-3, 3}, {-3, 3}},
>       PlotLabel -> SequenceForm["Continuous Roots of" , displaypoly],
>       ImageSize -> 400]
>     ]
>
> Case 1
>
> polynomial = x^5 - a x - 1 == 0;
> displaypoly = polynomial /. a -> HoldForm[a];
> rootset[a_] = x /. Solve[polynomial, x]
>
> memoryrootset = Null;
> basepermutations = Permutations[Range[5]];
> Animate[frame[a], {a, -6, 10, 0.25}]
> SelectionMove[EvaluationNotebook[], All, GeneratedCell]
> FrontEndTokenExecute["OpenCloseGroup"]; Pause[0.5];
> FrontEndExecute[{FrontEnd`SelectionAnimate[200,
> AnimationDisplayTime -> 0.1,
>       AnimationDirection -> ForwardBackward]}]
>
> Case 2
>
> polynomial = x^5 + (1 + I) x^4 - a x - 1 == 0;
> displaypoly = polynomial /. a -> HoldForm[a];
> rootset[a_] = x /. Solve[polynomial, x];
>
> memoryrootset = Null;
> basepermutations = Permutations[Range[5]];
> Animate[frame[a], {a, -6, 10, 0.5}]
> SelectionMove[EvaluationNotebook[], All, GeneratedCell]
> FrontEndTokenExecute["OpenCloseGroup"]; Pause[0.5];
> FrontEndExecute[{FrontEnd`SelectionAnimate[200,
> AnimationDisplayTime -> 0.1,
>       AnimationDirection -> ForwardBackward]}]
>
> But this is a combinatoric algorithm and if there are too many
> roots it
> might begin to get costly. But it is perfectly practical for these
> examples.
>
> David Park
> djmp at earthlink.net
> http://home.earthlink.net/~djmp/
>
> From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl]
>
> On 12 Jan 2007, at 11:05, Paul Abbott wrote:
>
>> In article <em8jfr$pfv$1 at smc.vnet.net>,
>>  Andrzej Kozlowski <andrzej at akikoz.net> wrote:
>>
>>> What you describe, including the fact that the numbering or roots
>>> changes is inevitable and none of it is not a bug. There cannot
>>> exist
>>> an ordering of complex roots that does not suffer from this problem.
>>> What happens is this.
>>> Real root objects are ordered in the natural way. A cubic can have
>>> either three real roots or one real root and two conjugate complex
>>> ones. Let's assume we have the latter situation. Then the real root
>>> will be counted as being earlier then the complex ones. Now suppose
>>> you start changing the coefficients continuously. The roots will
>>> start "moving in the complex plane", with the real root remaining on
>>> the real line the two complex roots always remaining conjugate
>>> (symmetric with respect to the real axis). Eventually they may
>>> collide and form a double real root. If this double real root is now
>>> smaller then the the "original real root" (actually than the root to
>>> which the original real root moved due the the changing of the
>>> parameter), there will be a jump in the ordering; the former root
>>> number 1 becoming number 3.
>>> This is completely unavoidable, not any kind of bug, and I am not
>>> complaining about it. It takes only elementary  topology of
>>> configuration spaces to prove that this must always be so.
>>
>> But is there a continuous root numbering if the roots are not
>> ordered?
>>
>> What I mean is that if you compute the roots of a polynomial, which
>> is a
>> function of a parameter, then if you assign a number to each root,
>> can
>> you follow that root continuously as the parameter changes? Two
>> examples
>> are presented below.
>>
>> Here is some code to animate numbered roots using the standard root
>> ordering, displaying the root numbering:
>>
>>  rootplot[r_] := Table[ListPlot[
>>    Transpose[{Re[x /. r[a]], Im[x /. r[a]]}],
>>    PlotStyle -> AbsolutePointSize[10],
>>    PlotRange -> {{-3, 3}, {-3, 3}},
>>    AspectRatio -> Automatic,
>>    PlotLabel -> StringJoin["a=", ToString[PaddedForm[Chop[a], {2,
>> 1}]]],
>>    Epilog -> {GrayLevel[1],
>>     MapIndexed[Text[#2[[1]], {Re[#1], Im[#1]}] & , x /. r[a]]}],
>>       {a, -6, 10, 0.5}]
>>
>> First, we have a polynomial with real coefficients:
>>
>>   r1[a_] = Solve[x^5 - a x - 1 == 0, x]
>>
>> Animating the trajectories of the roots using
>>
>>   rootplot[r1]
>>
>> we observe that, as you mention above, when the complex conjugate
>> roots
>> 2 and 3 coalesce, they become real roots 1 and 2 and root 1 becomes
>> root
>> 3. But, ignoring root ordering, why isn't it possible for these
>> roots to
>> maintain their identity (I realise that at coelescence, there is an
>> arbitrariness)?
>>
>> Second, we have a polynomial with a complex coefficient:
>>
>>   r2[a_] = Solve[x^5 + (1+I) x^4 - a x - 1 == 0, x]
>>
>> Animating the trajectories of the roots using
>>
>>   rootplot[r2]
>>
>> we observe that, even though the trajectories of the roots are
>> continuous, the numbering switches:
>>
>>   2 -> 3 -> 4
>>   5 -> 4 -> 3
>>   3 -> 4 -> 5
>>   4 -> 3 -> 2
>>
>> and only root 1 remains invariant. Again, ignoring root ordering, why
>> isn't it possible for all these roots to maintain their identity
>> and so
>> be continuous functions of the parameter? And wouldn't such
>> continuity
>> be nicer than enforcing root ordering?
>>
>> Cheers,
>> Paul
>>
>> _____________________________________________________________________
>> _
>> _
>> Paul Abbott                                      Phone:  61 8 6488
>> 2734
>> School of Physics, M013                            Fax: +61 8 6488
>> 1014
>> The University of Western Australia         (CRICOS Provider No
>> 00126G)
>> AUSTRALIA                               http://physics.uwa.edu.au/
>> ~paul
>
>
> 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. For a proof see my
> and Adam Strzebonski's posts in the same thread. Adam Strzebonski
> gave a very elementary proof of the fact that a continuous root
> cannot be defined on the space of complex polynomials of degree d. I
> quoted a more powerful but not elementary theorem of Vassiliev, which
> describes the minimum number of open sets that are needed to cover
> the space of complex polynomials of degree d, so that there is a
> continuous root defined on each open set. In fact, Vassiliev gives
> the exact number only in the case when d is prime, in which case d
> open sets are needed. For example, for polynomials of degree 3 at
> least 3 sets are needed . If it were possible to define a  continuous
> root, then of course only one set would suffice. In the case when d
> is not prime no simple formula seems to be known, but it is easy to
> prove that that the number is >1, (e.g. by means of Adam
> Strzebonski's proof).
>
> Andrzej Kozlowski
>
>



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