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

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deltaquattro@gmail.com

Posts: 77
Registered: 7/21/06
Re: Terrible algebraic system
Posted: Mar 8, 2012 4:26 AM
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Il giorno mercoledì 7 marzo 2012 21:36:02 UTC+1, clicl...@freenet.de ha scritto:
> >
> > 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.


Hi Martin!

Excellent :)your result confirms my idea that probably the best way to deal with this beast, is to compute all roots and then find the only one which satisfies a>L>0, b>U>0, i.e., [1.1776225370286649805,1.1776225370286649805]. However, I need to implement this in a code, since I must solve the system automatically for arbitrary inputs. Unluckily, the rest of the code is in VBA :( so I have no access to complex arithmetics. I could try to implement a class for that...or rewrite all the rest of the code in some different language. Also, I'm not familiar with this WDK method. Anyway, I guess this is now mainly a numerical mathematics topic, so I will also ask for suggestions on sci.math.num-analysis. Thanks!

Sergio




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