I have a polynomial of degree 5 $$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5$$ that is strictly increasing (the derivative is always greater than zero). I would like to (approximately) invert this polynomial, i.e. find $f^{-1}$ such that $f^{-1}(f(x)) = x$. I know that $f^{-1}$ is unique and well-defined on all of $\mathbb{R}$ because of monotonicty of $f(x)$.
One possible solution that I'm aware of is to compute the inverse $f^{-1}(y)$ by finding the real root of the polynomial $p(x) = f(x) - y$ using e.g. Newton's method. $$f(x) = y \iff f(x) - y = 0$$ Do there exists alternative non-iterative approaches for (approximately) solving this problem? Is it possible to approximate $f^{-1}$ given the coefficients $a_i$?
Given a function : $$y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5$$ such as $x$ and $y$ be real and the relationship be one-to-one, the inverse function $x(y)$ is the real root of the equation : $$A_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5=0\quad\text{where}\quad A_0=(a_0-y)$$ This is the quintic equation : http://mathworld.wolfram.com/QuinticEquation.html
In general (except for some particular values of the coefficients) the analytical solving for $x$ is not possible in terms of a finite number of elementary functions.
So, don't expect a non-iterative approche if you exclue the use of a convenient special function.
In the present case, the special functions involved are the Jacobi theta functions : http://mathworld.wolfram.com/JacobiThetaFunctions.html
This is an arduous analytical calculus. See the formal solution in http://mathworld.wolfram.com/QuinticEquation.html
On a practical viewpoint the use of numerical calculus is much simpler, but of course it doesn't satisfy your wish of non-iterative method.