Hi,  

I have a complex-valued equation that turns out to be a cubic when
properly arranged.   However, I'm having difficulty finding reasonably
simple closed form symbolic solutions for the roots (i.e.,
non-numerical solutions).   The equation can be expressed as follows:

  f1*(e1-x)/(e1+2*x) + f2*(e2-x)/(e2+2*x) + (1-f1-f2)*(e3-x)/(e3+2*x)=0

Solve for x.   

If it matters, f1 and f2 are real-valued between 0 and 1.  e1,e2,e3
are  complex-valued, both real and imaginary components are positive.

I expect a concise expression for each of the three roots (x) without
the use of excessive substitution.  By excessive substitution I mean
you shouldn't have more than 3 levels of nested substitution (i.e.,
a==b-3, b==c+1, c==e1+2).  The goal is to provide a symbolic solution
for a paper I'm writing.  You will receive acknowledgement (if you
desire) in addition to the pay.

If the equation is unclear, you can see a prettier version at:
http://rain.aos.wisc.edu/~jbenjam/pretty.pdf

Regards, Benjamin Johnson

---

Clarification of Question by snowkid-ga on 04 Aug 2005 12:14 PDT

Concise implies a form that could easily fit into a few lines of
equation.  The more compact the better.  Answer is subject to
verification (I have the numerical solutions).

---

Request for Question Clarification by mathtalk-ga on 05 Aug 2005 09:27 PDT

Hi, snowkid-ga:

I'm interested as to what else is known beforehand about the parameters:

  - Is f1 + f2 < 1 ?

  - Can any of the parameters be zero?

Note that although you can "convert" the rational equation to a cubic
polynomial equation, this risks introducing "artifact" roots.  In
general roots of the rational equation will be roots of the
polynomial, but the converse need not be true.

To illustrate this formally, consider the case where f1 = f2 = 1/3 and
each of the e's is zero.  The rational equation has no solution, but
"simplifying" to a cubic polynomial yields x^3 = 0.

Even without this complication, your equation with its five parameters
is not "simpler" than the general cubic.  So any explicit solution
will have at least as much content as a "closed form" solution for the
cubic equation.  Your request for "a concise expression" reminds me of
Einstein's dictum that we should make things as simple as possible,
but no simpler.


regards, mathtalk-ga

---

Clarification of Question by snowkid-ga on 05 Aug 2005 17:43 PDT

>>  - Is f1 + f2 < 1 ?

 0.0 <= f1+f2 <= 1.0 

although the case where f1+f2 == 0 doesn't have much physical meaning
in my model, it's mathematically possible.

e1,e2,e3 and x are complex-valued and have positive real and imaginary
components. (I don't think this will help simplify the problem, but
you may know more about this than I do.)

Thanks for taking a shot at it.  I realize that the answer may be as
simple as the cubic form of the equation (which itself is several
lines), but with appropriate substitutions, it should be possible to
create something that can reasonably fit into a journal article/paper.

---

Clarification of Question by snowkid-ga on 08 Aug 2005 05:53 PDT

Mathtalk, 

I'm already satisfied with tibcol's answer, so please don't continue
unless you can come up with a much more simplified or elegant
solution.   FWIW, I've already spotted the typographical errors in
tibcol's answer and corrected those.

Something that I didn't mention in the original question is that the
equation describes a method for obtaining an "effective" dielectric
permittivity (x) for a particle composed of three components
(e1,e2,e3) terms (for example a hailstone composed of ice, water, and
air).  In general this can be extended to N terms.  In physics this is
sometimes called the Bruggeman method or Effective Medium Theory.

sum_(i=1:N)  f_i (e_i - x)/(e_i + 2*x) = 0

where sum_(i=1:N) f_i = 1.

It's very difficult (for me) to find a general expression for the average
dielectric permittivity "x" in the general case, I'm not even sure it's
possible!  The motivation for doing so is programming convenience, so
it's not critical.   However, I do believe that a general form exists
to solve this kind of problem based on similar types of problems.

What you have to do is just expand or simplify the rational equation into its
ax^3 +bx^2 +cx +d = 0 form.
Then use the Cubic Formula to solve for x.

-------------
For less confusion, let:
K = f1`
L = f2
M = (1 -f1 -f2)
u = e1
v = e2
w = e3

--------------
Target is "ax^3 +bx^2 +cx +d = 0"  ------***

Doing it on paper here, I found:

a = -4(K+L+M)  ----negative 4(K+L+M).

b = 2K(2u -v -w) +2L(2v -u -w) +2M(2w -u -v)

c = K(2uv +2uw -vw) +L(2uv +2vw -uw) +M(2uw +vw -uv)

d = (k+L+M)(uvw)

------------
Then, you can use Cardano's version of the Cubic formula:

x = {q +[q^2 +(r-p^2)^3]^(1/2)}^(1/3) +{q -[q^2 +(r-p^2)^3]^(1/2)}^(1/3) +p

That is:
x = cubrt{q +sqrt[q^2 +(r-p^2)^3]} +cubrt{q -sqrt[q^2 +(r-p^2)^3]} +p

where 

sqrt is squareroot
cubrt is cuberoot

p = -b/(3a),   q = p3 + (bc-3ad)/(6a^2),   r = c/(3a)

---
Or, you search the Internet for the Cubic Formula.

Correction:

q = p^3 +(bc-3ad)/(6a^2)

It may be convenient to treat the equation with f3 = 1 - f1 - f2
(and/or to avoid subscripts altogether, as ticbol-ga suggests).  After
all the equation:

   f1(e1 - x)      f2(e2 - x)      f3(e3 - x)
  -----------  +  -----------  +  ----------- = 0
   (e1 + 2x)       (e2 + 2x)       (e3 + 2x)

with f1,f2,f3 nonnegative (but not all zero) is no more general than:

   f1 + f2 + f3 = 1

because the latter circumstance can always be contrived by dividing
both sides by this sum.  However for consistency and convenience we
will assume this normalization of the f's.

The terms of the form (e - x)/(e + 2x) can be considered as "Moebius"
or linear fractional transformations of the complex variable x.  As
such they have the interesting geometric interpretation in the complex
plane as maps that preserve circles (with lines as a "special case" of
circles).

Rewriting the terms:

   (e - x)/(e + 2x)  =  1 - 3x/(e + 2x)

leads to an alternative form in z = 1/x that may be attractive:

      f1           f2           f3         1
   --------  +  --------  +  --------  =  ---
   e1*z + 2     e2*z + 2     e3*z + 2      3

which depends on the condition f1 + f2 + f3 = 1.

The "degenerate" cases in which the "degree" is essentially less than
cubic, that can still arise, given snowkid-ga's Clarification above,
are as follows:

   -  One (or two) of the f's are zero (degree drops accordingly).

   -  Two (or all) of the e's are equal.

Of course the linear case (two f's are zero, or all e's are equal) can
be solved by inspection, and the quadratic cases (two nonzero f's, or
two e's equal) are easy to write symbolically.

regards, mathtalk-ga

I'll have to run it through a computer algebra package to check my
arithmetic, but the monic form of the cubic I get, assuming the
restrictions discussed above to avoid degenerate cases, is:

  x^3 + (1/2)[e_1*(1 - 3f_1) + e_2*(1 - 3f_2) + e_3*(1 - 3f_3)]*x^2
  
  - (1/4)[e_1*e_2*(2 - 3f_3) + e_1*e_3(2 - 3f_2) + e_2*e_3*(2 - 3f_1)]*x
  
  - (e_1*e_2*e_3)/4  =  0

where f_3 = 1 - f_1 - f_2.

I'd previously discussed how to use Cardano's forumla here:

[Q: Need general solution to cubic equation]
http://answers.google.com/answers/threadview?id=433886

and again in this follow-up thread about a software implementation:

[Q: for mathtalk-ga only: Need more help with cubic equations]
http://answers.google.com/answers/threadview?id=441538

But I need to spend time with the computer algebra package checking my
work before I submit an Answer!

regards, mathtalk-ga