Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32023] Re: equations
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Wed, 19 Dec 2001 04:28:59 -0500 (EST)
  • References: <3AEECC45-F384-11D5-A700-00039311C1CC@tuins.ac.jp>
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote:
> 
> The point I was trying to make was somewhat different, I think. Of
> course any polynomial equation with symbolic coefficients can be
> regarded as an equation over some function field in characteristic zero,
> and since the function field has an algebraic closure we know that there
> will be n-roots, where n is the degree of the equation. We can also
> express the roots in terms of parametrized Root objects, in the manner
> indicated. But these Root objects are of course "multivalued functions"
> with branch points depending on the parameters. I think what people mean
> when they want a "formula" is an expression that will always give ,in
> some algorithmic way, a root of the original equation for any value of
> the parameters.

The solution above will do that. The reason is that a given solution may
be expressed in terms of a particular parametrised Root[] function. When
you plug in values for parameters all instances of that function
consistently become the same algebraic number. The fact that these
functions are in some sense multivalued and may not behave nicely around
subvarieties in parameter space where crossings occur is not terribly
relevant. It really is like the situation where we contrast the pair
set1=(a,-a} with set2={Sqrt[a^2],-Sqrt[a^2]}. For any value of a the
sets are identical but for some values the first elements agree (as do
the second) while for other values of a we have equality between first
element of set1 and second element of set2, and vice versa. We agree (I
trust) that set1 is somehow "nicer" than set2. But in fact either might
be used to represent solutions to x^2==a^2. In more complicated examples
I doubt it is possible to give a solution in terms of a "nice" set, that
is, one where the individual solutions are well behaved everywhere in
parameter space. Hence solution sets are akin to set2. But all they same
they are valid.


> However, for each choice of the parameters it will be
> necessary to decide which branches to take, and I this can't be
> specified until the parameters are assigned values. So while we can say
> that roots exist, and we can refer to them as "the roots" I don't think
> we can write a computer program that will automatically choose the
> correct branches to give us a solution for very choice of parameters.
> (Or can we?)

As per above, yes. This follows from the fact that solutions can be
represented in terms of a single parametrized Root[] function. For
example, try

polys = {x^3+a*x+7+y, y^2+b*x*y+11*x-3};
soln = Solve[polys==0, {x,y}];

Each solution is expressed in terms of a single Root[] function

Root[46 + 11*#1 + 14*a*#1 - 7*b*#1 + a^2*#1^2 - a*b*#1^2 + 14*#1^3 + 
   2*a*#1^4 - b*#1^4 + #1^6 & , j]

where j is in the set {1,2,3,4,5,6}.


> The point of my comments is that a formula in this context
> must mean "an algorithm which can be implemented on a computer", and
> since the issue of choosing the correct branch is basically topological
> and not algebraic I do not think such an algorithm can exist at present.
> However I may be wrong here, so I would like to be corrected on this
> point.
> 
> (That would then raise another issue, why the solutions returned by
> eliminate do not seem to work for certain values of the parameters).

I'm not entirely certain what you mean here. If you eliminate variables
you get partial solutions that "work" but you will be unable to
propagate to full solutions involving the eliminated variables. If you
have an example where this statement does not hold then I'd like to see
it.


> Andrzej Kozlowski
> Toyama International University
> JAPAN
> http://platon.c.u-tokyo.ac.jp/andrzej/


Daniel Lichtblau
Wolfram Research


  • Prev by Date: Re: equations
  • Next by Date: Urgent Plot problem
  • Previous by thread: Re: equations
  • Next by thread: Re: equations