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Suppose I have a fifth degree polynomial: $f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ and that it does not factor nicely or have any nice roots.

Is there a way to approximate the inverse function $f^{-1}$? Whether by hand or by computer aid.

I know that one can use Newton's Method and approximate roots, but I don't think that helps the problem at hand.

I was not able to find any information on this after a lot of searching of google and stackexchange.

Joseph DiNatale
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You can approximate the inverse function as long as you can properly define what "inverse" means. Suppose that, for any given $y$, you can establish some rule that allows you to choose one specific $x$ such that $y=f(x)$. This gives you a definition of the inverse function $x = g(y) = f^{-1}(y)$. I'm not sure that it's easy (or even possible) to make this inverse function continuous, even in the simple case of quintic polynomials. It's possible on intervals where $f$ is monotone, though, which might be enough for your purposes.

Now we need a method of approximating $g$. You can calculate a few specific values of $g(y)$, using some root-finding method (like Newton's method), and then interpolate these values with a polynomial. If $g$ is continuous on the interval of interest, this should work well both in theory and in practice. Use Chebyshev nodes in your interpolation to avoid nasty wiggliness.

To do the root-finding and construct the interpolating polynomial, I highly recommend the chebfun system.

bubba
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  • It's not possible to choose the inverse continuously. If we had such a continuous $f^{-1}$ the image would be connected and contain the complement of a bounded subset of $\mathbb{R}$ (since $f$ is eventually monotone) and so $f^{-1}$ would be surjective. But this is a contradiction since $f^{-1}$ can't map onto two points on which $f$ takes the same value. – Kevin Carlson Mar 06 '14 at 01:28
  • @Kevin -- Thanks. I don't fully follow your argument because my topology skills are rusty from decades of non-use. But surely there are cases where a quintic polynomial is strictly monotone, and therefore does have a global inverse?? Anyway, as I said, existence of a "local" inverse on some interval might be enough to solve the OP's problem. But that's for him (or her) to say. – bubba Mar 06 '14 at 01:42
  • Yes, of course a monotone quintic has a global inverse. I'm just using the intermediate value theorem-if $f^{-1}$ hits $a$ and $f(b)=f(a)$ for $b>a$ then $f^{-1}$ can't hit $b$, but if it were continuous it would since it hits arbitrarily large $c$. It's true that a local inverse could be plenty useful. I'd never heard of Chebyshev nodes, looks cool. – Kevin Carlson Mar 06 '14 at 01:45
  • If you like Chebyshev nodes, then you might like chebfun: http://www2.maths.ox.ac.uk/chebfun/ – bubba Mar 06 '14 at 01:48
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This may be more of a computational science question/answer, but from a practical standpoint it's absolutely possible to approximate the inverse of a function on some domain where it is one-to-one.

Here's one way to do it: say you need your inverse to be accurate on $x_s < x < x_e$. Then, parameterize $f^{-1}$ however you see fit, and minimize squared errors on inversion efficacy:

\begin{equation} \int_{x_s}^{x_e} (f^{-1}(f(x)) - x)^2 dx \end{equation}

In a practical/computational setting, $f^{-1}$ will be defined by some parameters. For example, you could have the inverse be another quintic, with different coefficients: $f^{-1}(x) = b_0 + b_1x + b_2x^2 + b_3x^3 + b_4x^4 + b_5x^5$. Then the integral would evaluate to something messy but exact, and you could solve for the inverse coefficients by setting the gradient of the integral to zero (linear problem).

nonbasketless
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There's no reason $f^{-1}$ should exist: $f$ is surjective but needn't be one-to-one. Newton's method is a very good way to compute a local inverse, though-applying it to $f-r$ gives you a point at which $f$ equals $r$, i.e. one possible choice for "$f^{-1}(r)$", though there might be up to four other possibilities. None of this is special to quintics or even to polynomials, of course.

Kevin Carlson
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  • Kevin Carlson - Thanks, I guess it would make sense that f need not exist since it is not bijective. But suppose I only need to calculate $f^{-1}$ for a few values in the domain? Is there a way of computing those? – Joseph DiNatale Mar 06 '14 at 01:07
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    I'm unclear on how your question isn't answered by "Yes-use Newton's method." – Kevin Carlson Mar 06 '14 at 01:08
  • @KevinCarlson Theorem 1.1, McMullen : There is no generally convergent purely iterative algorithm for computing the roots of a polynomial of degree greater than $3$ - Newton's method falls into that category. – Balarka Sen Apr 25 '14 at 09:27
  • Thanks for the comment. McMullen's general convergence requires that I can pick the point to initiate Newton's method on as a rational function of the coefficients of my polynomial, but it's quite elementary that this is impossible, because this rational function would have to avoid points at which $f'=0$. But there is a neighborhood around every root of every polynomial (or continuously differentiable function) on which Newton's method converges, which make it sufficient for many applications. That said, Lagrange inversion is much more powerful. – Kevin Carlson Apr 25 '14 at 14:04