Can this equation be written in terms of x?

Question

I've been solving this equation, and I managed to boil the question down to: $x^n +1/x=2x^{n-1}$

Is it possible to write this equation so that it appears as $x=$ some function of n?

Answer 1

You are looking for the zero's of function$$f(x)=x^n +\frac 1x-2x^{n-1}$$ It is simple to show that, excluding the trivial $x=1$, there is a second root which is closer and closer to $x=2$.

We can obtain approximations considering instead that we look for the zero of function $$g(x)=\log \left(2 x^n-x^{n+1}\right)$$ Expanding as a Taylor series around $x=2$, we have $$g(x)=n\log(2)+\log(2-x)-\frac n2 (2-x)+O\left((x-2)^2\right)$$ Neglecting the higher order terms, we have the explicit solution $$\color{blue}{x_0=2+\frac 2 n W\left(-2^{-(n+1)}\, n\right)}$$ where $W(.)$ is Lambert function.

Trying this first approximation, some results $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 3 & 1.841435700 & 1.839286757 \\ 4 & 1.927789323 & 1.927561975 \\ 5 & 1.965975508 & 1.965948249 \\ 6 & 1.983586353 & 1.983582844 \\ 7 & 1.991964665 & 1.991964197 \\ 8 & 1.996031243 & 1.996031180 \\ 9 & 1.998029479 & 1.998029470 \\ 10 & 1.999018634 & 1.999018634 \end{array} \right)$$

If you want to improve these results, make one single iteration of Newton, Halley or Householder methods. The formulae are totally analyical (a bit too long to be printed here). This will give you the $x_1$'s according to the selected method.

Some results $$\left( \begin{array}{ccccc} n & \text{Newton} & \text{Halley}& \text{Householder}& \text{solution} \\ 3 & 1.83930665042 & 1.83928676586 & 1.83928675546 & 1.83928675521 \\ 4 & 1.92756239793 & 1.92756197539 & 1.92756197548 & 1.92756197548 \\ 5 & 1.96594824862 & 1.96594823664 & 1.96594823665 & 1.96594823665 \\ 6 & 1.98358284382 & 1.98358284342 & 1.98358284342 & 1.98358284342 \\ 7 & 1.99196419662 & 1.99196419661 & 1.99196419661 & 1.99196419662 \\ 8 & 1.99603117974 & 1.99603117974 & 1.99603117974 & 1.99603117974 \\ 9 & 1.99802947026 & 1.99802947026 & 1.99802947026 & 1.99802947026 \\ 10 & 1.99901863271 & 1.99901863271 & 1.99901863271 & 1.99901863271 \end{array} \right)$$

Answer 2

I prefer to make a separate second answer since it is based on a different approach.

Consider that $$x f(x)=1 +2^{n+1}\sum_{k=0}^\infty 2^{-k} \Bigg[\binom{n+1}{k}-\binom{n}{k}\Bigg](x-2)^k$$ Truncate to some order and perform a series reversion. Skipping the intermediate steps, we should end with $$x=2+t+n\sum_{k=2}^p (-1)^{k+1}\, \frac{P_{k}(n)} {a_k}\, t^k \quad \text{where} \quad t=-2^{-n}$$ where the first $a_k$ $$\{2,8,24,384,240,46080,40320,2064384,\cdots\}$$ correspond to sequence $A050971$ in $OEIS$.

The first polynomials are $$\left( \begin{array}{cc} k & P_k(n) \\ 2 & 1 \\ 3 & (3 n+1) \\ 4 & (2 n+1) (4 n+1) \\ 5 & (5 n+1) (5 n+2) (5 n+3) \\ 6 & (2 n+1) (3 n+1) (3 n+2) (6 n+1) \\ 7 & (7 n+1) (7 n+2) (7 n+3) (7 n+4) (7 n+5) \\ 8 & (2 n+1) (4 n+1) (4 n+3) (8 n+1) (8 n+3) (8 n+5) \end{array} \right)$$ where some interesting patterns appear for odd and even values of $k$.

As a function of $p$, are calculated the solutions for $n=3,4,5$. $$\left( \begin{array}{cccc} p & x_{(3)} & x_{(4)} & x_{(5)} \\ 2 & 1.851562500 & 1.929687500 & 1.966308594 \\ 3 & 1.844238281 & 1.928100586 & 1.966003418 \\ 4 & 1.841461182 & 1.927711487 & 1.965957522 \\ 5 & 1.840293884 & 1.927605927 & 1.965949895 \\ 6 & 1.839770675 & 1.927575417 & 1.965948545 \\ 7 & 1.839525625 & 1.927566207 & 1.965948296 \\ 8 & 1.839407126 & 1.927563338 & 1.965948248 \\ \cdots & \cdots & \cdots & \cdots \\ \infty & 1.839286755 & 1.927561975 & 1.965948237 \end{array} \right)$$

Edit

It is interesting to compare the two series in both answers $$2+\frac 2 n W\left(\frac{n }{2}t\right)=2+t-\frac{n }{2}t^2+\frac{3 n^2 }{8}t^3-\frac{n^3 }{3}t^4+\cdots$$ while here we have $$2+t-\frac{n }{2}t^2+\frac{3 n^2 }{8} \left(1+\frac{1}{3 n}\right)t^3-\frac{n^3 }{3}\left(1+\frac{3}{4 n}+\frac{1}{8 n^2}\right)t^4\cdots$$

Answer 3

If you multiply by $x$ you get $$x^{n+1}-2x^n=-1$$ or if you prefer $$x^n(x-2)=-1$$ Since this product has to be negative, you can't take any roots (unless you settle to some complex value) or logarithm. In any case there's no way you can possibly write $x$ in terms of $n$.

The only thing you can get is: $$x=\Big(\frac{1}{2-x}\Big)^{\frac{1}{n}}$$ but it is of no use unless you are doing iterations of some sort.