How do you use solve() to solve a complicated symbolic equation with one variable in terms of unknown constants?

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Kelly Catlin
Kelly Catlin on 20 Nov 2017
Commented: Walter Roberson on 23 Nov 2017
I have a a complicated symbolic equation that I would like to solve for x in terms of the letter-name constants. I received the following warning:
Warning: Unable to find explicit solution. For options, see help.
In solve (line 317)
solx =
Empty sym: 0-by-1
This is what I input:
syms x a b c d e f g
eqn = a*x - ((b^2)/2)*(a/b)*((1 - 4*exp(-2*(c-x)/b))^(-1)) - b*(a/b)*((1 - 4*exp(-2*(c-x)/b))^(-1))*(x+(((1/b)+d)^(-1))*log(1-(((b/(g*e))+(g/f))*((b/a)^(-0.5)))*((1-4*exp(-2*(c-x)/b))^(-0.5)))) - ((d*e)/2)*(((1/b)+((2*d)/b)+d^2)^(-1))*(1-2*exp((((1/b)+d)^(-1))*log(1-((b/(g*e))+(g/f))*((b/a)^(-0.5))*((1-4*exp(-2*(c-x)/b))^(-0.5)))*((1/b)+d)) + exp(2*(((1/b)+d)^(-1))*log(1 - ((b/g*e)+(g/f))*((b/a)^(-0.5))*((1-4*exp(-2*(c-x)/b))^(-0.5)))*((1/b)+d)) - (((1/b)+d)^(-1))*log(1 - ((b/(g*e))+(g/f))*((b/a)^(-0.5))*((1 - 4*exp(-2*(c-x)/b))^(-0.5)))*(1/b) - (2/b)*(((1/b)+d)^(-1)) + (2/b)*(((1/b)+d)^(-1))*exp((((1/b)+d)^(-1))*log(1 - ((b/(g*e))+(g/f))*((b/a)^(-0.5))*((1-4*exp(-2*(c-x)/b))^(-0.5)))) + (1/(2*b))*(((1/b)+d)^(-1)) - (1/(2*b))*(((1/b)+d)^(-1))*exp(2*(((1/b)+d)^(-1))*log(1-((b/(g*e))+(g/f))*((b/a)^(-0.5))*((1-4*exp(-2*(c-x)/b))^(-0.5)))*((1/b)+d)) == 0;
solx = solve(eqn, x)
Is there a way to proceed to a solution for x in terms of the letter-name constants? I have already used the simplify(eqn) command and several assume(x>0), assume(e>0), etc. parameters based on what I can assume from the physical constants, but to no avail.
  11 Comments
Walter Roberson
Walter Roberson on 23 Nov 2017
"Is there even a script for approximating with Newton's Method when constants are not explicitly defined?"
At that point what you are heading towards is doing a taylor series and solve()'ing what results.
In Maple the sequence would be
eqn3 := E+a*x-b+2*b*exp(-x*c)-b*exp(-2*x*c)-x*b/d-2*b*exp(-x*c)/(d*c)+2*b/(d*c)-b/((2*d)*c)+b*exp(-2*x*c)/((2*d)*c);
convert(taylor(eqn3, x = 0, 5), polynom);
solve(%);
simplify([allvalues](%),size) assuming nonnegative;
which would get you a wall of about 75 lines of unreadable closed-form solution. And since this is only degree 5 approximation, the approximation to the exponential is not going to be all that good, especially further away from x = 0. You cannot go for any higher degree because you would get a quintic or higher degree, for which there are no closed form solutions. Even degree 4 should only be treated as a vague approximation, perhaps a starting point for numeric solutions.

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