Logarithmic terms in a system of equations

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Main Question or Discussion Point

(I hope this is not a double posting) I want to solve this system of equations, containing logarithmic terms:

##7\ln(a/b)+A = 7\ln(d/e)+D = 7\ln(g/h)+G##
##7\ln(a/c)+B = 7\ln(d/f)+E = 7\ln(g/i)+H##
##7\ln(b/c)+C = 7\ln(e/f)+F = 7\ln(h/i)+I##
##a\phi_1+d\phi_2+g\phi_3=X##
##b\phi_1+e\phi_2+h\phi_3=Y##
##c\phi_1+f\phi_2+i\phi_3=Z##
[itex]a+b+c=1[/itex]
[itex]d+e+f=1[/itex]
[itex]g+h+i=1[/itex]

where the uppercase and [itex]\phi[/itex] coefficients are known and [itex]a,b,c...i[/itex] are the unknown coefficients.

The following are also true, but might not be important since the values are known:
[itex]\phi_1+\phi_2+\phi_3=1[/itex]
[itex]X+Y+Z=1[/itex]

My strategy so far:

I introduce the unknowns [itex]n_1,n_2, n_3[/itex] to link the equations as:

##7\ln(a)-7\ln(b)-n_1 = -A##
##7\ln(a)-7\ln(c)-n_2 = -B##
##7\ln(b)-7\ln(c)-n_3 = -C##
##7\ln(d)-7\ln(e)-n_1=-D##
##7\ln(d)-7\ln(f)-n_2=-E##
##7\ln(e)-7\ln(f)-n_3=-F##
##7\ln(g)-7\ln(h)-n_1=-G##
##7\ln(g)-7\ln(i)-n_2=-H##
##7\ln(h)-7\ln(i)-n_3=-I##

The corresponding system of equations maybe looks like this:

[tex]
\begin{bmatrix}
7L & -7L & & & & & & & & -1 & & \\
7L & & -7L & & & & & & & & -1 & \\
& 7L & -7L & & & & & & & & & -1\\
& & & 7L & -7L & & & & & -1 & & \\
& & & 7L & & -7L & & & & & -1 & \\
& & & & 7L & -7L & & & & & & -1\\
& & & & & & 7L & -7L & & -1 & & \\
& & & & & & 7L & & -7L & & -1 & \\
& & & & & & & 7L & -7L & & & -1\\
\phi_1 & & & \phi_2 & & & \phi_3 & & & & & \\
& \phi_1 & & & \phi_2 & & & \phi_3 & & & & \\
& & \phi_1 & & & \phi_2 & & & \phi_3 & & & \\
1 & 1 & 1 & & & & & & & & & \\
& & & 1 & 1 & 1 & & & & & & \\
& & & & & & 1 & 1 & 1 & & & \\
\end{bmatrix}

\begin{bmatrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h\\
i\\
n_1\\
n_2\\
n_3
\end{bmatrix}
=
\begin{bmatrix}
-A\\
-B\\
-C\\
-D\\
-E\\
-F\\
-G\\
-H\\
-I\\
X\\
Y\\
Z\\
1\\
1\\
1
\end{bmatrix}
[/tex]
where the "L" terms indicate that there is actually a natural log of the variable.

My main problem is: What do I do with the logarithmic terms (the 7L terms indicate 7*ln(unknown))??? I want something that can use computer solvers so I need to build a system kind of like I've done, but I'm not sure how to do it even if I am on the right track. Do I need to decompose the system into a logarithmic part and a linear part first? or what?
 
Last edited:

Answers and Replies

  • #2
mathman
Science Advisor
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439
I don't know if it helps much, but you could divide the first 3 equations by 7 and take exponential of each term such as: ##7 ln(a/b)+A \to (a/b)e^{A/7}##.
 
  • #3
Stephen Tashi
Science Advisor
7,208
1,326
I want something that can use computer solvers
If you mean computer solvers for systems of simultaneous linear equations, I think you are out of luck. What other types of solvers would you consider?

A teacher once told me "Every problem can be phrased as an optimization problem". For example, we can use an equation ##f(a,b,c) = g(a,b,c)## to define a term in a penality function given by ##(f(a,b,c) - g(a,b,c))^2##. We can state the problem of solving simultaneous equations as a different problem involving minimizing the penalty function created from some equations subject to constraints given by other equations. There are all sorts of computer programs that solve optimization problems.
 
  • #4
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418
The system is overdetermined, because there are nine unknowns and twelve equations.

The two equations in the third line are redundant and can be discarded, as they can be derived from those in the first two, although they reveal a relationship between some of the constants, viz, that:
B-C-A = E-D-F = H-G-I

So we have ten equations in nine unknowns.

What about the following strategy?

Temporarily treat three of the unknowns as knowns ('known unknowns'), so that you have six 'unknown unknowns', and solve the system comprising the last six equations, which is linear and hence easy to work with. That will give you formulas for each of the six unknown unknowns in terms of the constants and the three known unknowns. Substituting those formulas into the four equations in the first two lines gives us four equations in the three known unknowns. That's a much smaller system that may be more amenable to numerical approaches. The search space is only three-dimensional rather than nine-dimensional. The system is still overdetermined, so either there will be no solutions or one of the equations will be redundant.

The trick would be to select our three 'known unknowns' in a way that doesn't make any of the last six equations redundant by not containing any unknowns. For instance choosing a, b, c would make the 3rd last equation redundant. Choosing a, d, g makes the 6th last equation redundant. Perhaps try choosing a, e, i.
 
  • #5
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0
Could you solve this system with Newton's method?
 

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