I have the following $(2n-1)$th degree polynomial of $x$: $$y=f(x)=x(x^2+x+1)^{n-1}; n>0$$ The polynomial $f(x)$ is strictly increasing and has only one real root in $0$. Is that possible to analytically find the inverse of this polynomial for a given $n$?
I have suggested this polynomial because it is strictly increasing and has only one real root in $0$. I need these specifications. Do you know how to design a polynomial with these properties which its inverse can be calculated easily?
The polynomial is of degree $(2n-1)$; so, impossible to find its inverse as soon as $n \geq 3$.
However, for small enough values of $x$, we could expand as Taylor series to get $$y=x+(n-1) x^2+\frac{1}{2} (n-1) n x^3+\frac{1}{6} \left(n^3-7 n+6\right) x^4+\frac{1}{24} \left(n^4+2 n^3-25 n^2+46 n-24\right) x^5+\frac{1}{120} n \left(n^4+5 n^3-55 n^2+115 n-66\right) x^6+O\left(x^7\right)$$ and use series reversion to get $$x=y+(1-n) y^2+\frac{1}{2} \left(3 n^2-7 n+4\right) y^3-\frac{2}{3} \left(4 n^3-15 n^2+17 n-6\right) y^4+\frac{1}{24} \left(125 n^4-650 n^3+1135 n^2-826 n+216\right) y^5+\frac{1}{10} \left(-108 n^5+720 n^4-1725 n^3+1930 n^2-1027 n+210\right) y^6+O\left(y^7\right)$$