
Re: Solving N1 equations in N unknowns with initial state?
Posted:
May 8, 2013 5:46 AM


On Tue, 7 May 2013, Don Del Grande wrote:
> I have a system of N1 equations in N unknowns. > If I start by assigning values to the N unknowns and then recalculate > the values recursively:
Huh? If you assign values to the n unknowns, then the n  1 equations are all true or one or more is false. > (a) Will they always reach a single solution (assuming there is a > solution for the equations in the first place), and > (b) If a single solution can be reached, is there a way to determine it > in terms of the initial values without having to go through the > recursion, and if so, what is it? > Example: > 3a  b  c  d = 9 > a + 3b  c  d = 3 > a  b + 3c  d = 3 > a  b  c + 3d = 9
That's not three equations in four unknowns; it's four equations in four unknowns.
> If you combine the first three equations, you get a + b + c  3d = 9, > which is 1 times the fourth equation, so it is effectively 3 equations > in 4 unknowns.
What do you mean by combining? I get a + b + c  3d = 9 by adding the first three equations. > If the values are initially set to a = 3, b = 1, c = 1, and d = 3, and > the equations repeated recalculated recursively, they eventually become > a = 9/4, b = 3/4, c = 3/4, and d = 9/4.
You can't set d. What d is, is determined by setting a,b and c. What's a recalculation?
> Can those values be determined in a nonrecursive method from the > equations and the initial values?
Given the three equations 3a  b  c  d = 9 a + 3b  c  d = 3 a  b + 3c  d = 3 that is 3a  b  c = 9 + d a + 3b  c = 3 + d a  b + 3c = 3 + d we have 2a + 2b  2c = 12 + 2d; a + b  c = 12 + d 2a + 2b + 4c = 2d; a + b + 2c = d 2b + 2c = 12 + 2d; b + c = 6 + d from which we should be able to calulate a,b,c in terms of d. So d is the only parameter that can be set and by setting d, a,b,c are sets.

