2

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 $a_i$?

Oleksandr Shchur
  • 153
  • 1
    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 $f^{-1}(x)=\sqrt{x}$. – asdf Jul 08 '19 at 09:00
  • https://math.stackexchange.com/questions/701061/is-there-a-way-of-approximating-a-polynomials-inverse – Nurator Jul 08 '19 at 09:26
  • @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 – Oleksandr Shchur Jul 08 '19 at 10:32
  • Can you specify perhaps what kind of approximation you want, e.g. by a polynomial, in what regime e.g. x large, x small? – Calvin Khor Jul 08 '19 at 10:37
  • Approximation in terms of any closed-form expression (as defined in https://en.wikipedia.org/wiki/Closed-form_expression) would be great. I am concerned with "small" values of $x$ (in the range (-10, 10)) – Oleksandr Shchur Jul 08 '19 at 10:53
  • Just to give another example where the inverse clearly exists: if f(x) = x^5 + c, f^(-1)(x) = (x - c)^(1/5). Maybe it's possible to identify the subset of polynomials (maybe stricter than just monotonic) where this kind of inverse can be found. Or at least find a test for whether whether a polynomial is in the set. – SteveWithamDuplicate Oct 12 '20 at 16:08

1 Answers1

1

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.

JJacquelin
  • 66,221
  • 3
  • 37
  • 87
  • 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. – Oleksandr Shchur Jul 08 '19 at 13:25
  • 1
    https://en.wikipedia.org/wiki/Quintic_function#Solvable_quintics summarizes quintics that are "solvable" with radicals. The category of monotonic quintics isn't mentioned as such. My amateur guess is your solution is in there but would take a bit of work to find, understand, and finally apply each time. Maybe one hint is that any solution to a monotonic quintic is the solution. A good first guess and Newton's method is the easiest path for me. – SteveWithamDuplicate Oct 12 '20 at 18:47