Consider solving the following system for unknowns A1, A2, B, D, G:
w = 2 B^2 + D^2;
z = A1 B + A2 B + G D;
y = 2 A1 A2 + G^2;
x = A1^2 + A2^2 + G^2;
a2 A1 + a2 A2 + a G = x2 B + x4 B + x3 D;
Where w, z, y, x, a2, x2, x4, x3 are known quantities.
Any ideas ?
***** EDIT: This problem is solved. See comment below. *******
by using some substitution and understanding of equation you can solve this problem.
1st equation represent ellipse so we can easily find value of (B and D), and so more......!!!
there are some specific method to solve the system of non linear equation so you can use them.
Konrad,
Do you need to solve this particular system or just any non-linear one?
Are you interested in analytical form of solution(s), i.e. A_1 = f1(w, z, x, ...), A_2=f2(..), and so on, or rather numerical values of A_1, A_2 etc. are required, even if only approximate?
As far as I know, there is no general scheme to solve non-linear system analytically, even in case of only polynomial expressions,as in your original post.
When numerical approach is sufficient then you have to remember that, in general case, you will have to start with some guessed set of approximate values of your unknowns. The Newton-Raphson procedure may be fine, if convergent. Unfortunately, it may give you only one solution for a given starting point. There will be no way to prove that there are no solutions, should it be the case.
If you know the domain where your unknowns may reside then interval arithmetic maybe the solution (if you haven't heard about interval arithmetic, ask Wikipedia:
http://en.wikipedia.org/wiki/Interval_arithmetic ).
This method will give you all (real or even complex) roots located in a starting region or a proof there are none, no matter that it is purely numerical approach.
Hi Marek, i need to solve this particular system. It would be best to find an analytical solution, but I would settle for an EASY method to find numerical approximations. I need a simple method to solve this system, as I will need to solve it many times for many values of w, z, y, x, a2, x2, x4, x3. I tried finding a closed formula for some special cases like D=0 or B = 0 and had success, but I'm interested in formulas for a solution in general case. Perhaps I can solve this system as an optimization problem using Lagrange multipliers like the one below ?
min A1^2 + A2^2 + B^2 + G^2 + D^2
with constraints:
w = 2 B^2 + D^2;
z = A1 B + A2 B + G D;
y = 2 A1 A2 + G^2;
x = A1^2 + A2^2 + G^2;
a2 A1 + a2 A2 + a G = x2 B + x4 B + x3 D;
Where w, z, y, x, a2, x2, x4, x3 are known quantities.
Hi Marek, I think this equation is vague, can you bring it to the table of solution by shedding some light on it?
@Konrad: Maybe the extra condition:
min A_1^2 + A_2^2 + B^2 + G^2 + D^2
has something to do with your original problem (never presented earlier), but it is NOT a part of the equation system you have shown. It is taken out of the hat. Your extra condition, after some reformulation, may nevertheless have sense, when your problem is suspected to have no true solutions and you are going to "solve" it as well as possible. But read on, maybe you'll find some interesting ideas below.
@Ogunniyi Olasunkanmi:
Suppose you know that x is somewhere in interval [a,b], that is a
Thank you Marek and Juan for your detailed answers.
However, I wonder why no one tried to actually solve it using just plain algebra and common sense ? I figured that for this little system it is too time consuming for me to code all this interval arithmetic or even Newton-Rapson. "It's just a few equations, that suggest "completion of square" method, why do I need to do all this to get to the solution ?". So today, i followed Jaipratap Chauhan's suggestion and after a few cups of coffee and some twiddling with these equations I FINALLY managed to find an analytical solution using the substitution:
T = A1 + A2, t = x + y, s = x2 + x4.
Here, the only new variable is T. The other two, t and s are just shorthand constants.
The key observation was that adding together
y = 2 A1 A2 + G^2;
x = A1^2 + A2^2 + G^2;
enables us to write everywere T instead od A1+A2, reducing this system to four equations with four variables, which can then be solved for T,B,G and D. When you subtract these two equations you get A1 - A2 = +-sqrt(x-y), which together with all the rest gives you the final solutions of the original system.
Regards.
Great! It's a very good trick for this particular case. Did you think about its generalization? Maybe only to multivariate polynomial systems of degree not exceeding 2? Such solution would be probably applicable to many different real world problems.
In fact, I was thinking to create new "unknowns" like that:
x1 = x; x2 = y; x3 = z;
x4 = xy; x5 = xz; x6 = yz; x7 = x^2; x8 = y^2; x9 = z^2, and so on.
The system written in terms of those new unknowns is linear --> easy, when the number of unknowns is equal to the number of equations. But this need not to be case ...
@Joseph Uphoff: A bit cryptic as for me. How about many solutions/no solution case?
I'm glad you solved your system with a clever substitution. But you should be aware that systems of multivariate polynomials can often be solved with the Dixon resultant. Your system would be quite easy for that method. Google "Dixon Resultant."
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