I want to find an analytical solution $x$ as a function of parameters $(e,u,r,t)\in\mathbb{R}^4$ that satisfies the following condition: $$x+r\frac{e}{e+t+x^2u}+\frac{xr^2\frac{e}{e+t+x^2u}}{1+r\frac{e}{e+t+x^2u}}=e,$$ which after rearranging yields $$x\left(1+r\frac{e}{e+t+x^2u}\right)+r\frac{e}{e+t+x^2u}\left(1+r\frac{e}{e+t+x^2u}\right)+xr^2\frac{e}{e+t+x^2u}=e\left(1+r\frac{e}{e+t+x^2u}\right).$$ Multiplying by the denominators (i.e., multiplying by $(e+t+x^2u)^2)$, this yields a 5th degree polynomial in $x$. The good news is that a 5th degree polynomial always has at least one solution in the real numbers. The bad news is that there may not be an explicit expression for the solution(s), and even if there were, one could not really work with them (too long/complicated).

Given this problem:

  1. Is there a simpler/better way to find solutions? Some trick you can think of? Since many terms appear repeatedly, maybe there is some potential for simplification.

  2. If not, is there a nice way to approximate the condition, say by a 2nd degree Taylor approximation, and then solve it? My problem with the Taylor approximation is that the term has to be approximated at some point $x^*$, which ideally is the solution, but since I want to make the approximation to find a solution in the first place, this seems difficult. So at which point would I approximate?

All I am able to do right now is find solutions for special cases. For example, $e\to 0$ leads to $x=0$. But these are really just special cases.

Ideally, I would like to have a solution (or solutions) that can be expressed as some explicit function of the parameters $(e,u,r,t)$, i.e., some $x^*=f(e,u,r,t)$. The $f$ should be simple enough so that one can see how the parameters $(e,u,r,t)$ change the solution (e.g., if $e$ increases, then the solution $x^*$ decreases). To achieve this, an approximation is fine.

Any help is very much appreciated! Thanks!

  • 1
    $\begingroup$ Yes, you have a polynomial of degree $5$ to solve. For generic values of the parameters, it is irreducible. For example, with $e=r=t=u=1$, the polynomial is ${x}^{5}-{x}^{4}+6\,{x}^{3}-4\,{x}^{2}+8\,x-3$, which has Galois group $S_5$ and thus is not solvable by radicals. $\endgroup$ Aug 14, 2015 at 15:51
  • $\begingroup$ Series solutions may be available, though. Are there any parameters that can be considered "small" or "large" for this purpose? $\endgroup$ Aug 14, 2015 at 15:52
  • $\begingroup$ Thanks for clarifying the 5th degree polynomail issue, I suspected as much! Typically $|r|$ will not be small, but it is still interesting to see if assuming small $|r|$ allows to obtain workable solutions. $\endgroup$
    – Nameless
    Aug 14, 2015 at 16:12
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    $\begingroup$ @Nameless: Incidentally, your quintic is irreducible but solvable in radicals for an infinite number of rational coefficients $c_i$. These $c_i$ are special cases that depend on solving the simple $x^2+5y^2=z^2$. Not really useful for your question, but just an interesting factoid. :) $\endgroup$ Aug 14, 2015 at 17:09

1 Answer 1


For example, if $r, t, u$ are small, so that we write $r = R\epsilon$, $t = T\epsilon$, $u =U\epsilon$, there is a series solution that starts $$ x = e-R\epsilon-{\frac {R \left( {e}^{2}R-U{e}^{2}-T \right) }{e}}{ \epsilon}^{2}++{\frac {R \left( RU{e}^{4}-{U}^{2}{e}^{4}+{R}^{2}{e}^{3} +{R}^{2}{e}^{2}+{e}^{2}RT-2\,RU{e}^{2}-2\,TU{e}^{2}-{T}^{2} \right) }{ {e}^{2}}}{\epsilon}^{3} +\ldots $$

  • $\begingroup$ Can you elaborate a bit on the steps you took or give a source where the approach is explained? It looks as if you assumed $r,t,u$ are small, which allows you to write a Taylor series for the entire term, but then how do you solve this series for $x$? I may be wrong though..;) $\endgroup$
    – Nameless
    Aug 14, 2015 at 16:10
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    $\begingroup$ Actually I used Maple, but basically the method is: write $x = a_0 + a_1 \epsilon + a_2 \epsilon^2 + \ldots$, substitute in to the polynomial, expand, collect terms in each power of $\epsilon$, and solve one by one. $\endgroup$ Aug 14, 2015 at 16:50
  • $\begingroup$ I see! You approximate $x$ with a series of $\epsilon$s, substitute it in the polynomial, and then match coefficients. Interesting! The $\epsilon$s need to be small so that higher order terms are not as important. But I don't quite understand why that requires $r,t,u$ to be small and not $e$? $\endgroup$
    – Nameless
    Aug 14, 2015 at 17:57
  • $\begingroup$ You could try playing the same game with any nonempty set of parameters, as long as you can find a solution when those parameters are $0$. For no very good reason I happened to pick $\{r,t,u\}$. $\endgroup$ Aug 14, 2015 at 22:58
  • $\begingroup$ For example, a solution in powers of $e$: $$ x = \left(1 - \frac{r}{t}\right) e + \dfrac{r(r^2-rt+1)}{t^2} e^2 + \ldots $$ $\endgroup$ Aug 14, 2015 at 23:08

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