MHB Direct solution for two unknowns in two equations

  • Thread starter Thread starter IanfromBristol
  • Start date Start date
  • Tags Tags
    Unknowns
AI Thread Summary
The discussion centers on solving two equations with two unknowns, specifically focusing on the equations involving coefficients C1, C2, C3, and known values Q1 and Q2. The original poster has attempted substitution and elimination methods but finds them leading to complex quartic expressions. They express a preference for iterative methods due to their stability and quick convergence, despite acknowledging the existence of a direct solution method for quartics. Another participant agrees that the iterative approach is simpler and more practical for this problem. The consensus leans towards using approximation methods for ease of computation.
IanfromBristol
Messages
2
Reaction score
0
I am writing a computer program to solve a physical problem which at some part involves the following two equations and two unknowns;

(1) \[ C_1(x^4-y^4) + C_2x = Q_1 \]
(2) \[ C_1(y^4-x^4) + C_3y = Q_2 \]

C1, C2 & C3 are coefficients which can readily be calculated and do not rely on knowing the values of X or Y. Q1 and Q2 are known values (effectively boundary conditions). I've tried substitution and elimination but end up with a much more complicated expression which looks to be a quartic equation. I know I can solve this iteratively and it does converge quickly and is stable. However, is there any method that can solve it directly and thus avoid the iterative approach?
 
Mathematics news on Phys.org
I see no way to avoid the quartic. There is a method to do this to get an exact answer (here) but I'd stick with approximation. It's much easier to work out.

-Dan
 
topsquark said:
I see no way to avoid the quartic. There is a method to do this to get an exact answer (here) but I'd stick with approximation. It's much easier to work out.

-Dan
Hi Dan, I took a look at that solution for the quartic and you're not wrong the iterative method (i.e. approximation) is far easier to work out!

Ian
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top