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MathGroup Archive 2001

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Re: equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32036] Re: equations
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Wed, 19 Dec 2001 04:29:28 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

I am sure you must be right. I guessed I was confused because I have 
never considered carefully how root isolation and their numbering as 
Root[f,1], ....Root[f,n] work. Of course we know that :

In[2]:=
Solve[a + b*x^2 + c*x^3 + d*x^4 + e*x^5 == 0, x]

Out[2]=
{{x -> Root[a + b*#1^2 + c*#1^3 + d*#1^4 + e*#1^5 & , 1]},
   {x -> Root[a + b*#1^2 + c*#1^3 + d*#1^4 + e*#1^5 & , 2]},
   {x -> Root[a + b*#1^2 + c*#1^3 + d*#1^4 + e*#1^5 & , 3]},
   {x -> Root[a + b*#1^2 + c*#1^3 + d*#1^4 + e*#1^5 & , 4]},
   {x -> Root[a + b*#1^2 + c*#1^3 + d*#1^4 + e*#1^5 & , 5]}}

means no more than that there are 5 roots of a fifth degree equation. At 
this point the ordering of the roots is purely formal. It's only when 
you substitute values for the parameters than Mathematica isolates the 
roots and the ordering acquires a meaning (so one can think of these 
"solutions" as (topologically badly behaved) functions of the parameters 
that return roots).  I was not sure that this however would be remain 
true when you have a complex system of equations and solutions involve 
root objects with coefficients that themselves involve root objects and 
so on. It does not seem quite obvious to me that a root of a system of 
equations can be consistently represented as something like Root[f, 5] 
where f again involves other root objects, in such a way that this 
returns a root of the system for all values of the parameters (it's the 
choice of numbering that worried me).  Presumably there is some theorem 
here, perhaps a rather trivial one. On the other hand, it now seems that 
nothing very sophisticated is going on in such a case, a suitable 
Groebner basis is found which eliminates variables in turn and then the 
roots are simply numbered purely formally, just as in the above 
example.  Still, I am puzzled by Fred's claim that the solutions 
obtained by means of ELiminate in th eexample that started this 
discussion do not work for some values of the parameters (I may be still 
misunderstanding him). Since I think they must be solutions (by 
elimination theory) I assumed that was something to do with the above 
discussion. Of course it may be due to something quite different, for 
example problems with precision in numerical computations.

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


On Tuesday, December 18, 2001, at 11:40  PM, Daniel Lichtblau wrote:

> 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.
>
>
>
>
> Daniel Lichtblau
> Wolfram Research
>



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