Finding the real positive solution of $x^n+x-c^2;\ 2\lt n\in\mathbb{N},\,c\in\mathbb{R},$ comes from a practical problem I encountered.



> 
> **Question:**
> 
> 
> what can be said about exact solutions for specific values of $n$ and/or non-recursive functions $f(\xi,\eta)\mapsto x$ such that yields $\min\max\limits\_{n,c}|f(n,c)^n+f(n,c)-c^2|$.
> 
> 
> 


As the intended use of $f(n,c)$ is to obtain starting points for numeric root finders, that function should not be too complex and yield starting points that significantly reduce the number of iterations needed to achieve a given precision of the solution.



Real solutions of $x^n+x-c^2=0$
Hardly anything interesting can be said here for small $n$. So, let $n$ be large. Let $x=x\_{c,n}$ be the unique nonnegative zero of $f(x):=x^n+x-c^2$. We have to distinguish the following three cases.


*Case 1: $c^2<1$.* Then $x<c^2$ and hence
$$c^2-c^{2n}<x\_{c,n}<c^2,$$
so that $x\_{c,n}=c^2+o(1)$ exponentially fast (as $n\to\infty$).


*Case 2: $c^2=1$.* Let then $z\_{a,n}:=1-a\dfrac{\ln n}n$ for real $a>0$. Then for $x=z\_{a,n}$ we have
$$f(x)=\Big(1-a\dfrac{\ln n}n\big)^n-a\dfrac{\ln n}n
=n^{-a+o(1)}-a\dfrac{\ln n}n,$$
which is eventually (that is, for all large enough $n$) negative if $a>1$ and eventually positive if $a<1$. Therefore and because $f(x)$ is increasing in $x\ge0$,
$$x\_{c,n}=1-(1+o(1))\dfrac{\ln n}n.$$


*Case 3: $c^2>1$.* Let then $w\_{a,c,n}:=c^{2/n}e^{-a/n}$ for real $a>0$. Then for $x=w\_{a,c,n}$ we have
$$f(x)=c^{2/n}e^{-a/n}-c^2(1-e^{-a})\to1-c^2(1-e^{-a}), $$
which is $<0$ if $a>a\_c:=\ln\dfrac{c^2}{c^2-1}$ and $>0$ if $a<a\_c$. Therefore and because $f(x)$ is increasing in $x\ge0$,
$$x\_{c,n}=c^{2/n}e^{-(a\_c+o(1)/n}=1+\frac{\ln(c^2-1)+o(1)}n$$
(thanks to [Emil Jeřábek](https://www.wolframcloud.com/obj/ipinelis/Published/Untitled-6.nb) for the latter equality).




---


In a previous comment by the OP (now apparently deleted), it was indicated that the desired approximations to the root were intended to be fed to a recursive approximation method as initial approximations. [This Mathematica notebook](https://www.wolframcloud.com/obj/ipinelis/Published/Untitled-6.nb) shows that the approximations given above (with the $o(1)$'s omitted) work well in this role with the Newton method, even for $n$ as small as $5$ or $10$.


