Positive real solution to $x^n=x+1$

Question

I am interested in finding the positive real solution to $x^{n}=x+1$, where $n<0, n>1$. For example, if $n = 2$, then $x$ is the golden ratio: $\frac{1+\sqrt{5}}{2} \approx 1.618$. As $n$ approaches $\infty$, $y$, the solution to the equality, approaches $1$.

To find the solution, I fixed $y$. Then $$y^n=y+1$$ Solving for $n$ yields $$n = \log_{y}(y+1)$$

We now have a solution only in terms of $n$ and $y$. However, I would like to find $y$ as a function of $n$, not the other way around.

If this is not possible, as I suspect, I would like to find an approximation. For my attempt to find an approximation, I examined $$n = \log_{s}(s+1)$$ with $s$ being $1-\frac{1}{y_1}$. I graphed this and got something that asymptotically looked like a line, with a gap from $n = 0$ to $n = 1$, of the form $y_1=kn+1$, where $k \approx -1.4427$. This then means that $$\frac{1}{1-y}=kn+1 \rightarrow y \approx -\frac{1}{kn+1}+1$$ . This matches up with the graph. In the graph, $n$ is on the $x$-axis.

My questions:

Is there a closed form for the inverse of $n = \log_y(y+1)$?

What is the exact value of $k$?

What is a better approximation, perhaps of the form $y=c_0+\frac{1}{c_1n+d_1}+ \frac{1}{c_2n^2+d_2} +...$?

Edit:

Letting $$t = 1-\frac{1}{y_1-1+\frac{1}{kn+1}}$$ I graphed $n=\log_t(t+1)$ This was asymptotically equal to $$y_1=kn+2$$ However, I'm not sure how to use this in order to get a better estimate.

Edit 2:

I appreciate the answers so far. However, using a simple sum of inverse powers will not work near $x = 0$. At $x = 0$, the approximations of the form $y=a_0+a_1x^{-1}+a_2x^{-2}+...$ will diverge to $-\infty$. In fact, successive approximations get worse for $-1<x<0$. On the other hand, $$y = -\frac{1}{kn+1}+1$$ approaches $0$ as $n$ approaches $0$.

Answer 1

Let us consider the equivalent equation $x = (x+1)^{1/n}$; then, since we are interested in the behavior as $n \to \pm\infty$ (and since $\frac{1}{n}$ already appears in the equation), let us set $\epsilon := \frac{1}{n}$.

Now, consider the function $F(\epsilon, x) := x - (x+1)^\epsilon$. We have that $F$ is differentiable in a neighborhood of $(0, 1)$; $F(0, 1) = 0$; and $\frac{\partial F}{\partial x}(0, 1) = 1 \ne 0$. Therefore, by the Implicit Function Theorem, there is a differentiable function $f(\epsilon)$ on a neighborhood of $\epsilon = 0$ such that $f(0) = 1$ and $F(\epsilon, f(\epsilon)) = 0$. This function $f(\epsilon)$ is exactly the solution to the equation that we were looking for as a function of $\epsilon$.

Furtherfore, $f'(0) = -\frac{\partial F/\partial \epsilon(0, 1)}{\partial F/\partial x(0, 1)} = \ln 2$. It follows that $f(\epsilon) = 1 + \epsilon \ln 2 + o(\epsilon) = 1 + \frac{\ln 2}{n} + o(\frac{1}{n})$.

If we want to get more accurate approximations, then observe that since $F$ is $C^\infty$ near $(0, 1)$, then $f \in C^\infty$ in a neighborhood of 0 also, so Taylor series approximations will give the desired refinements. The higher derivatives of $f$ at 0, in turn, can be found by a standard implicit differentiation of the equation $f(\epsilon) - (f(\epsilon) + 1)^\epsilon = 0$.

Answer 2

I doubt there is a closed form, but for a slightly better approximation, including Daniel Schepler's $\ln(2) \approx 0.6931471806 \approx \frac{1}{1.4427}$, you might try something like $$y \approx 1+\log_e(2) n^{-1}+0.5868n^{-2}+ 0.529 n^{-3}+ 0.49 n^{-4}$$ giving the following values - not brilliant for small $n$ as you would benefit from more terms and more precise coefficients but not bad for larger $n$

    n    y suggested     y^n              y^n - (1+y)
    2   1.5900235903    2.5281750176    -0.0618485726
    3   1.3218910355    2.3098669885    -0.0120240470
    4   1.2201414826    2.2163623827    -0.0037791000
    5   1.1671174361    2.1655727910    -0.0015446451
    6   1.1346516906    2.1339065049    -0.0007451857
    7   1.1127428917    2.1123398903    -0.0004030014
    8   1.0969649796    2.0967280452    -0.0002369344
    9   1.0850611332    2.0849126584    -0.0001484747
   10   1.0757607181    2.0756628744    -0.0000978437
   20   1.0361935465    2.0361870122    -0.0000065343
   50   1.0141019740    2.0141017411    -0.0000002330
  100   1.0069906857    2.0069906570    -0.0000000287
  200   1.0034804723    2.0034804668    -0.0000000056
  500   1.0013886458    2.0013886448    -0.0000000010
 1000   1.0006937345    2.0006937342    -0.0000000003
 2000   1.0003467204    2.0003467202    -0.0000000001
 5000   1.0001386529    2.0001386529    -0.0000000000
10000   1.0000693206    2.0000693206    -0.0000000000

Answer 3

The root is, for large $n$, $1+\frac{\ln(2)}{n}+O(\frac1{n^2}) $.

Let $f(x) = x^n-x-1$. $f(0) = -1, f(1) = -1, f(2) = 2^n-3 \gt 0$ for $n \ge 2$.

$f'(x) =nx^{n-1} -1 \ge n-1$ for $x \ge 1$. Therefore only one real root with $x > 1$. Let $x_0$ be that root.

Since $(1+\frac1{n})^{n+1} \gt e$, $f(1+\frac1{n-1}) =(1+\frac1{n-1})^n-(1+\frac1{n-1})-1 \gt e-2-\frac1{n-1} \gt 0$ for $n-1 \gt \frac1{.7}$ or $n \ge 3$.

Therefore $1 \lt x_0 \lt 1+\frac1{n-1} $.

Let's look for a root of the form $1+\frac{c}{n} $.

$\begin{array}\\ f(1+\dfrac{c}{n}) &=(1+\dfrac{c}{n})^n-(1+\dfrac{c}{n})-1\\ &=e^{n\ln(1+c/n)}-2-\dfrac{c}{n}\\ &=e^{n(c/n-c^2/(2n^2)+O(1/n^3)))}-2-\dfrac{c}{n}\\ &=e^{c-c^2/(2n)+O(1/n^2)}-2-\dfrac{c}{n}\\ &=e^{c}e^{-c^2/(2n)+O(1/n^2)}-2-\dfrac{c}{n}\\ &=e^{c}(1-c^2/(2n)+O(1/n^2))-2-\dfrac{c}{n}\\ &=e^{c}-e^{c}(c^2/(2n)+O(1/n^2))-2-\dfrac{c}{n}\\ &=e^{c}-2-\frac{c}{n}(1+\frac{ce^{c}}{2})+O(1/n^2)\\ \end{array} $

Therefore, if $e^{c} = 2$, or $c = \ln(2) $, $f(1+\dfrac{c}{n}) =-\frac{\ln(2)}{n}(1+\ln(2))+O(\frac1{n^2}) $ and, for any other $c$, $f(1+\dfrac{c}{n}) =e^c-2 +O(\frac1{n}) $.

Answer 4

Another possible way to get an approximation.

Let us build a $[1,k]$ Padé approximant of $x^n-x-1$ around $x=1$ which would write $$x^n-x-1\sim\frac {a_0^{(k)}+a_1^{(k)}(x-1)} {1+\sum_{p=1}^k b_p^{(k)}(x-1)^k}$$ and the approximate solution would be $$x_{(k)}=1-\frac{a_0^{(k) }} {a_1^{(k) } }$$ Using for example $k=6$, this would give $$x_{(6)}=1+\frac{32781 n^5-133728 n^4+209727 n^3-154728 n^2+51156 n-5040 } {47293 n^6-232966 n^5+463695 n^4-471470 n^3+252512 n^2-64224 n+5040 }$$ which, expanded as a series for infinitely large values of $n$ would give $$x_{(6)}=1+\frac{32781}{47293 n}+O\left(\frac{1}{n^2}\right)$$ and $\frac{32781}{47293}=0.693147$ which is, for six significant figures, $\log(2)$.

Building the same table as in Henry's answer $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 2 & 1.61904761905 & 1.61803398875 \\ 3 & 1.32472324723 & 1.32471795727 \\ 4 & 1.22074383382 & 1.22074408461 \\ 5 & 1.16730393752 & 1.16730397826 \\ 6 & 1.13472418394 & 1.13472413840 \\ 7 & 1.11277574322 & 1.11277568428 \\ 8 & 1.09698161049 & 1.09698155780 \\ 9 & 1.08507028837 & 1.08507024549 \\ 10 & 1.07576609989 & 1.07576606609 \\ 20 & 1.03619371886 & 1.03619371713 \\ 50 & 1.01410197382 & 1.01410197638 \\ 100 & 1.00699068416 & 1.00699068585 \\ 200 & 1.00348047141 & 1.00348047235 \\ 500 & 1.00138864541 & 1.00138864580 \\ 1000 & 1.00069373431 & 1.00069373451 \\ 2000 & 1.00034672026 & 1.00034672036 \\ 5000 & 1.00013865287 & 1.00013865291 \\ 10000 & 1.00006932057 & 1.00006932057 \end{array} \right)$$