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

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Waldek Hebisch

Posts: 228
Registered: 12/8/04
Re: Terrible algebraic system
Posted: Feb 24, 2012 11:21 AM
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deltaquattro@gmail.com wrote:
> 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.

--
Waldek Hebisch
hebisch@math.uni.wroc.pl



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