Understanding roots of complicated equations This is a very broad question apparently, but is there any systematic way to obtain (sort of) sharp estimates for roots of complicated equations? Or maybe bounds for the roots?
For example, consider the equation $$x = A - B\dfrac{e^x}{1+e^x}$$where $A,B$ are large positive integers with $B>A$. It is easy to see that this equation has a unique root, by observing that the function $f(x)=x - A + B\dfrac{e^x}{1+e^x}$ is increasing in $x$. But I want to get some pretty sharp estimates of where the root $x^*$ lies.
I have looked at several numerical examples and it seems that $x^*$ is very close to $\log(A/(B-A))$. But I cannot really prove it.
So I am asking, given a complicated equation, what is a good way to get a good estimate on where the root must lie?
 A: In fact, there is a formal solution to this equation. Rewrite it as
$$e^{-x}=-\frac{x-(A-B)}{x-A}$$ and the solution is given in terms of the generalized Lambert function (have a look at equation $(4)$). The bad news is that, from a practical point of view, it does not bring much.
What we could try is to use an high order iterative method which would give for example
$$x=-\frac{4 (2 A-B) \left(-B \left(-2 A^2+2 A B+(B+6) (B+12)\right)-96\right)}{(B+4)
   \left(B \left(-8 A^2+8 A B+B (B+36)+144\right)+192\right)}$$
Trying for $A=1234$ and $B=5678$, this would give as an estimate
$x=-1.28244$ while the solution is $x=-1.27997$. Pure luck ?
In this case, your guess is impressive since $-\log \left(\frac{2222}{617}\right)\sim -1.28129$ (close to mine - which is not !)
Trying two orders above (formula not typed - just too long), the guess is $x=-1.27994$.
A: Unfortunately, it doesn't exist a general method, because every equation can easily be slightly different from each other.
I show you what I'd do in this case,
First things first, I'd see the entire equation as a rational one. How do I do that? I simply found a common denominator. Then, solve a system of two equations (numerator and denominator). In order to obtain the right result, you must ask yourself: "which solution satisfies both equations at the same time?" Seeing this on a number line can help a lot.
Tell me if you have any doubt.
A: Consider the function $$x-\left(A-\frac{B e^x}{e^x+1}\right)$$
plugging $x=\log \left(\frac{A}{B-A}\right)$ we get
$$\log \left(\frac{A}{B-A}\right)+\frac{A B}{(B-A) \left(\frac{A}{B-A}+1\right)}-A=\log \left(\frac{A}{B-A}\right)$$
thus when $B\sim 2A$  we have $\log \left(\frac{A}{B-A}\right)\sim 0$ and $x^*=\log \left(\frac{A}{B-A}\right)$ is a good approximation of the solution.
If we take $A=1;\;B=10000$ the approximation gives $-9.2$ while the root is $-7.1$
For $A=60000;\;B=100000$ we get respectively $0.405465;\;0.405448$ with an error less than $10^{-4}$.
Hope this helps
