# Asymptotic to a sequence of algebraic numbers.

Let $f(n)$ be the largest real solution of

$$x^n - x^{n-1} = 1$$

As $n$ grows to positive infinity we get the asymptotic :

$$f(n) = 1 + \frac{\ln(n)}{n} + \frac{\exp(2)}{n^2} + ...$$

Where the value $\exp(2)$ is optimal !

( and $...$ means smaller term(s) )

Notice $f(2)$ is the golden mean.

How to show this asymptotic ?

Edit

Corrected the formula.

• Wow, I suppose I take that back. The fact that $\exp(2)$ is optimal seems interesting... – Simply Beautiful Art Feb 17 '17 at 23:25
• I checked up to $n=10^{10}$, and for $n>11900$, $f(n)>1+\frac{\exp(2)}n$, and it seems to me that there is no constant $c$ such that $f(n)=1+\frac cn+\dots$, as after $n=2\times10^9$, or so, $f(n)=1+\frac{2.5\exp(2)}n+\dots$ looks more optimal. – Simply Beautiful Art Feb 17 '17 at 23:34
• This is incorrect: if $x=1+b/n$, then the LHS behaves like $e^bb/n$ for large $n$. I believe the correct asymptotics is $f(n)=1+\log n/n + o(\log n/n)$. – user138530 Feb 18 '17 at 0:19
• @ChristianRemling Pretty close to what I was thinking. Again checking, that holds much better, as far as I can tell. – Simply Beautiful Art Feb 18 '17 at 0:26
• making edits that totally change a possible answer to a questions are really crappy... – tired Feb 18 '17 at 11:07

A rough outline.

The first few terms in the asymptotic are

$$f(n) = 1 + \frac{W(n)}{n} + \frac{W(n)^2}{2n^2} + \cdots,$$

where $W$ is the Lambert-W function, so the stated asymptotic is incorrect.

First show that, with $x = 1 + \frac{W(n)}{n} + \frac{z}{n}$, where $z = O(1)$, we have

$$x^n - x^{n-1} - 1 \to e^z - 1$$

uniformly as $n \to \infty$. Conclude by Hurwitz's theorem that

$$f(n) = 1 + \frac{W(n)}{n} + \frac{\epsilon_n}{n}$$

with $\epsilon_n \to 0$ as $n \to \infty$. Substitute this into the equation

$$f(n)^n - f(n)^{n-1}-1 = 0$$

and apply asymptotic simplifications to conclude that $\epsilon_n \sim W(n)^2/(2n)$ as $n \to \infty$.

Let $f(n)=1+\epsilon(n)$. Then, $f^n(n)-f^{n-1}(n)=1$ becomes

$$(n-1)\log(1+\epsilon(n))+\log(\epsilon(n))=0$$

As $n\to \infty$, $\epsilon(n)\to 0$. Hence, we have

$$(n-1)\epsilon(n)+O(n\epsilon^2(n))+\log(\epsilon(n))=0 \tag 1$$

We can write $(1)$ equivalently as

$$(n-1)\epsilon(n)e^{(n-1)\epsilon(n)}=(n-1)e^{O(n\epsilon^2(n))}\tag 2$$

which using Lambert's W function is given by

$$\epsilon(n)=\frac{1}{n-1}W\left((n-1)e^{O(n\epsilon^2(n))}\right)\tag 3$$

Using the first term in the large argument asymptotic expansion of $W$ yields

\begin{align} \epsilon(n)&\sim \frac{1}{n-1}\log((n-1)e^{O(n\epsilon^2(n))})\\\\ &\sim\frac{\log(n-1)}{n-1}\\\\ &\sim\frac{\log(n)}{n} \end{align}

Hence, we find that the first two terms in the expansion of $f(n)$ for large $n$ is given by

$$\bbox[5px,border:2px solid #C0A000]{f(n)\sim 1+\frac{\log(n)}{n}}$$

And we are done!

• I am sorry to refer to this old post, but: How do you get from (1) to (2)? I do not see this immadiately. – mathfemi Jan 22 '18 at 11:05
• Exponentiate both sides of $(1)$ and multiply by $n-1$. – Mark Viola Jan 22 '18 at 16:38
• @MarkViola Another question: You are using the large Argument Expansion of W. Could you please explain why the Argument of W, which here is $(n-1)e^{O(n\varepsilon^2(n))}$, is "large" here? – mathfemi Jan 22 '18 at 19:47
• Note that we are analyzing the asymptotic behavior as $n\to \infty$. – Mark Viola Jan 22 '18 at 22:02
• @Rhjg I'll have a look. It's been quite a while since I posted this answer. – Mark Viola Sep 11 '18 at 14:40