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 $a_i$?

  • $\begingroup$ An issue is that the inverse function need not be a polynomial - for example if you take $f(x)=x^2$, then its inverse is $f^{-1}(x)=\sqrt{x}$. $\endgroup$ – asdf Jul 8 at 9:00
  • $\begingroup$ math.stackexchange.com/questions/701061/… $\endgroup$ – Nurator Jul 8 at 9:26
  • $\begingroup$ @Nurator, thank you, I am aware of that question. However, the answers there mostly talk about (1) existence of the inverse (which definitely exists in my case) and (2) iterative methods for finding the inverse $\endgroup$ – Alexander Shchur Jul 8 at 10:32
  • $\begingroup$ Can you specify perhaps what kind of approximation you want, e.g. by a polynomial, in what regime e.g. x large, x small? $\endgroup$ – Calvin Khor Jul 8 at 10:37
  • $\begingroup$ Approximation in terms of any closed-form expression (as defined in en.wikipedia.org/wiki/Closed-form_expression) would be great. I am concerned with "small" values of $x$ (in the range (-10, 10)) $\endgroup$ – Alexander Shchur Jul 8 at 10:53

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.

  • $\begingroup$ My hope was that even though quintic equations don't have closed-form roots in general, it would still be possible to find them in a relatively simple way due to monotonicity, (and hence uniqueness of the real root) in my special case. However, if that's not possible, then my question is answered I guess. $\endgroup$ – Alexander Shchur Jul 8 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.