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Topic: Terrible algebraic system
Replies: 18   Last Post: Mar 8, 2012 4:26 AM

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clicliclic@freenet.de

Posts: 1,245
Registered: 4/26/08
Re: Terrible algebraic system
Posted: Mar 7, 2012 3:36 PM
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deltaquattro@gmail.com schrieb:
>
> Il giorno venerdì 24 febbraio 2012 17:21:59 UTC+1, Waldek Hebisch ha
> scritto:

> > > Hmmm! Tried following this approach, i.e. solving f(b)=0 with
> > > Netwon's and then plugging b in a=g(b), but it doesn't work. The
> > > solution doesn't satisfy the initial system. Try for example
> > >
> > > L=-1
> > > U=1
> > > p=0.0013498980316301
> > >
> > > I get
> > >
> > > a=-1
> > > b=1
> > >
> > > which clearly is wrong, i.e., a and b don't solve the initial
> > > system, even though f(b)=f(1)=0 and a=g(b)=g(1)=-1. I rewrite the
> > > initial system here:
> > >
> > > ((a-L)^2/a)*(2/(a+b))-p=0
> > > ((b-U)^2/b)*(2/(a+b))-p=0
> > >

> >
> > In first message you wrote:
> >

> > > with a>L>0, b>U>0.
> >
> > Now you put L < 0, which violates this condition.
> >

> > > Am I doing something wrong, or is FriCAS/Derive solution wrong?
> >
> > FriCAS solution is "generic" in the sense that:
> >
> > 1) All solutions of your system are contained in FriCAS
> > solution
> > 2) It solves your system if there is no division by zero when
> > plugging FriCAS solution into your system
> > 3) Unless there are some special relation between L and U
> > there will be no division by zero when plugging FriCAS
> > solution into your system.
> >
> > In case of your system special relations are: L = 0, U = 0,
> > L + U = 0. If no of the above relations hold then you
> > need not worry about division by zero (in particular
> > no worry when U>0 and L>0). When one of relations holds,
> > you need to decide if you want to you your original
> > system (that is throw out (a, b) which cause division
> > by zero), or you want to simplify it and solve new,
> > simpler system.
> >
> > BTW: Since a = -1, b = 1 is the only real solution that
> > FriCAS found, this means that your orignal system even
> > after simplification has no real solutions.
> >

>
> Ok, tried again: I am using the formulation where a>L>0, b>U>0. Please
> forget about the other!
>
> L = 1
> U = 1
> p = 0.022750132
>
> With Newton's method, I solve
>
> 2 * b ^ 4 * (p - 1) * (p - 2) + 2 * b ^ 3 * (L * p * (p - 2) - U * (p
> ^ 2 - 8 * p + 8)) + b ^ 2 * (L ^ 2 * p ^ 2 + 8 * L * U * p + U ^ 2 *
> (p ^ 2 - 16 * p + 24)) - 4 * b * U ^ 2 * (L * p + 2 * U * (2 - p)) + 2
> * U ^ 4 * (2 - p) = 0
>
> for b, starting from b=U. I get b = 0.859373503 (!), which is not
> bigger than U. I plug this result in
>
> a = (2 * b ^ 3 * (p - 1) * (p - 2) + 2 * b ^ 2 * (L * p * (p - 2) - U
> * (p ^ 2 - 8 * p + 8)) + b * (L ^ 2 * p ^ 2 + 8 * L * U * p + 4 * U ^
> 2 * (5 - 3 * p)) - 4 * U ^ 2 * (L * p + U * (2 - p))) / (U ^ 2 * p *
> (p - 2))
>
> getting a = 1.163638391, which is not smaller than L. Trying to plug
> these solutions back in the original system,
>
> ((a-L)^2/a)*(2/(a+b))-p=0
> ((b-U)^2/b)*(2/(a+b))-p=0
>
> I get residuals
>
> 2.34718E-12
> 1.9082E-16
>
> which show that the system is indeed solved with excellent accuracy!
> However, my constraints
>
> a>L
> b>U
>
> are violated. Any idea how to "enforce" these constraints in the
> problem solution? My guess: I should find all 4 solutions of the
> equation for b by Newton's method, plug all of them one at a time in
> the equation and discard all but the ones which satisfy constraints.
> How can I find all the solutions with Newton's method? I usually only
> use it to find a single zero. Can you help me? Thanks,
>


Here are 20-digit solutions for the cases you mentioned:

[L:=-1,U:=1,p:=0.0013498980316301]

[[-1,1],[-1.0006754048796738625+0.00067586120557976235658*#i,1.0~
006754048796738625+0.00067586120557976235658*#i],[-1.00067540487~
96738625-0.00067586120557976235658*#i,1.0006754048796738625-0.00~
067586120557976235658*#i]]

[L:=1,U:=1,p:=0.022750132]

[[1.1776225370286649805,1.1776225370286649805],[0.85937350255160~
052169,1.1636383912592832729],[1.1636383912592832729,0.859373502~
55160052169],[0.86893696171357480864,0.86893696171357480864]]

Weierstrass-Durand-Kerner iteration is related to Newton's method and
produces all zeros of a polynomial in parallel. Complex arithmetic is
needed for this since roots may be complex even for real polynomials.

Martin.



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