how to solve an equation of the form $\displaystyle \frac{x^n}{1-x}=c$ I >was wondering how can I solve an equation of the form: $\displaystyle \frac{x^n}{1-x}$$=c$ (because I face this form a lot) for large values of $n$ like:
$$\displaystyle \frac{x^5}{1-x}=9.984$$ 
and even when $n=50$. I can't do long division especially when $n$ is too large, this takes time and effort. Is there any other idea? Thank you.
 A: Your equation is,
$$\frac{x^n}{1-x} = c$$
which is equivalent to solving the polynomial of degree $n$, $x^n + cx - c = 0$. According to the Abel-Ruffini theorem, there is no algebraic solution (a solution in radicals) for $n \geq 5$. Thus, one must resort to numerical methods to obtain an estimate of the solution.
A simple method which is also easily implemented in virtually any programming language is the Newton-Raphson method. Taking $f(x) = x^n + cx - c$, if $x_0$ is a guess for $f(x_0)= 0$, then we can get a better root $x_1$ through,
$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$$
and we can continue iterating in this way to obtain a sufficiently accurate value, though there are some caveats and the success of the method depending on $f(x)$ may be highly dependent on the initial choice of $x_0$.
Another method you could look into is Halley's method which is virtually just as easily implemented and has its own boons compared to the Newton-Raphson method. Both are part of a larger class known as Householder methods.
A: There is a nice series for a solution in negative powers of $c$, which should converge when $c$ is large:
$$ \eqalign{x &= 1 - \frac{1}{c} + \frac{n}{c^2} - \frac{n(3n-1)}{2c^3} + \frac{n(4n-1)(4n-2)}{6c^4} + \ldots\cr
&= \sum_{k=0}^\infty \frac{1}{k! (-c)^k} \prod_{j=0}^{k-2} (kn-j)\cr}$$
A: If we think about the quantities in
$$\frac{x^n}{1-x} = c$$
we can interpret it that the fractional relation between an event with probability $x$ happening independently $n$ times and the complementary with probability $1-x$ happens once. 
You can therefore probably solve it with one of the many probabilistic method or simulation.

Another option is to rewrite $x^n-c(1-x)=0$ and then use for example use a Companion matrix power iteration. 
I wonder if there could be a relation between this method and a probabilistic method with Markov chain transition matrices... I can update if I find something.
A: There isn't a simple formula or a method that will algebraically solve you such equations. The only way to handle them is with approximation methods from Numerical Analysis, which theory and execution can be translated to computer algorithm codes. In some short words, recursive numerical methods exists that generally through a form of a sequence will converge to the solution of any equation, always with or without an error, depending on the equation and the method you chose.
