An algebraic equation; $ x^n -x =a $

Question

Let $n\ge 2$ be an integer and $a$ be a real number such that $|a|\le (n-1)n^{-\frac n{n-1}}$. Can we find the real solutions of the algebraic equation $$ x^n -x =a. $$

Answer 1

First, if $n$ is odd, then $f(x) = x^n-x$ is unbounded above and below, and so $f(x)=a$ has a solution for any $a$.

If $n$ is even, $f(x)$ is unbounded above and is convex, since $f''(x) = n(n-1)x^{n-2}\geq 0.$ Any local minimum of $f$ is therefore its global minimum.

Step 1: Find the global minimum of $f$ by solving $f'(x) =0.$

Step 2: What is the value of $f$ at this minimum?

Step 3: What can you conclude about the range of $f(x)?$

In neither case will you in general be able to solve for $x$ analytically.

Answer 2

By Lagrange inversion theorem, whenever $|a|\le(n-1)n^{n/(1-n)}$, one solution may be given as

$$x_0=\sum_{k=0}^\infty\binom{nk}k\frac{(-a)^{(n-1)k+1}}{(n-1)k+1}$$

One may rewrite this in terms of a generalized hypergeometric function by letting

$$c_k=\binom{nk}k\frac1{(n-1)k+1}\left(\frac{(n-1)^{n-1}}{n^n}\right)^k$$

And seeing that

$$\frac{c_{k+1}}{c_k}=\frac{\left(k+\frac1n\right)\left(k+\frac2n\right)\dots\left(k+\frac nn\right)}{\left(k+\frac2{n-1}\right)\left(k+\frac3{n-1}\right)\dots\left(k+\frac n{n-1}\right)(k+1)}$$

And by letting

$$t=\left(\frac a{1-n}\right)^{n-1}n^n$$

we get the generalized hypergeometric function:

$$x_0=-a\sum_{k=0}^\infty c_kt^k=-a~_nF_{n-1}\left(\frac1n,\frac2n,\dots,\frac nn;\frac2{n-1},\frac3{n-1},\dots,\frac n{n-1};t\right)$$