Re: How to solve for L?
- From: "Ioannis" <morpheus@xxxxxxxxxxxx>
- Date: Sat, 25 Nov 2006 20:47:38 +0200
"bluelabel" <bluelabel.invalid@xxxxxxxx> wrote in message
news:45687baf$0$7643$4fafbaef@xxxxxxxxxxxxxxxxxxxxxx
hello, I have the following system of 3 equations:
2x
1 + L(--------------- + 1) = 0
1+(x^2+y^2-2)
2y
-1 + L(--------------- - 1) = 0
1+(x^2+y^2-2)
arctg (x^2 + y^2 -2) = 2-x+y
The only thing I know for sure is that solving for L leads to the value
L=-1/3, giving (x,y)=(1,-1). But despite my many attempts I did not find a
way to determine L.
Could you show me how to do that?
Thank you.
PS.
The equations system arises from the problem of determining the critical
points of the function f(x,y) = x-y with the condition arctg(x^2+y^2-2) =
2-x+y, applying Lagrange's multipliers method.
If you add the first two equations, you get:
L(2*x+2*y)/(1+(x^2+y^2-2))=0, =>
L = 0 or y = -x.
L = 0 is obviously rejected since then equation (1) would become, 1 = 0
Therefore, y = -x (4)
Substituting y = -x on the third equation (x^2 + y^2 - 2) = tan(2 - x + y),
you get:
2*(x^2-1) = tan(2*(1-x)) (5)
(5) has the obvious solution x=1, which leads to the rest (via (4) and either
the first or second) as:
y = -1 and L = -1/3.
However (5) seems to have infinitely many solutions, inside the intervals
(k*Pi/2,(k+1)*Pi/2), k in Z.
For example, there is a solution for x in (Pi/2, Pi), as:
x ~= 1.882446503, which leads to
y ~= -1.882446503 and
L ~= -0.6178589355
Obviously (5) cannot be solved analytically so it seems to me the system has
infinitely many solutions.
--
Ioannis
-------
The best way to predict reality, is to know exactly what you DON'T want.
.
- Follow-Ups:
- Re: How to solve for L?
- From: bluelabel
- Re: How to solve for L?
- References:
- How to solve for L?
- From: bluelabel
- How to solve for L?
- Prev by Date: connectivity
- Next by Date: Re: Cantor Confusion
- Previous by thread: How to solve for L?
- Next by thread: Re: How to solve for L?
- Index(es):
Relevant Pages
|
