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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.

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    Wow, I suppose I take that back. The fact that $\exp(2)$ is optimal seems interesting... – 2017-02-17
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    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. – 2017-02-17
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    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 $f(n)=1+\log n/n + o(\log n/n)$. – 2017-02-18
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    @ChristianRemling Pretty close to what I was thinking. Again checking, that holds much better, as far as I can tell. – 2017-02-18
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    @SimplyBeautifulArt I was able to develop the first two terms in the asymptotic expansion and have posted as a solution. Hope you enjoy. ;-)) – 2017-02-18
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    I made an important edit – 2017-02-18
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    Sorry for the mistake – 2017-02-18
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    I wonder about a closed form for f(n) ... Like an integral or such – 2017-02-18
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    Very similar to [this question](http://math.stackexchange.com/q/485408/5531). – 2017-02-18
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    making edits that totally change a possible answer to a questions are really crappy... – 2017-02-18

2 Answers 2

0

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 $n \to \infty$.

4

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!

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    I am sorry to refer to this old post, but: How do you get from (1) to (2)? I do not see this immadiately. – 2018-01-22
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    Exponentiate both sides of $(1)$ and multiply by $n-1$. – 2018-01-22
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    @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? – 2018-01-22
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    Note that we are analyzing the asymptotic behavior as $n\to \infty$. – 2018-01-22
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    @MarkViola Yes, indeed. But for me its not immediately clear that the argument of $W$, i.e. $(n-1)e^{O(n\varepsilon^2(n))}$, is "large" as $n\to\infty$. – 2018-01-23
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    Sorry, I cannot follow you! Am I right that you suppose $\varepsilon(n)$ to be any zero sequence? If yes, it is not true in general, I think, that $ne^{O(n\varepsilon^2(n)}$ is large for $n\to\infty$. In particular, I do not see why $f(n)\in O(n\varepsilon^2(n)$ is non-negative for $n\to\infty$ in general. – 2018-01-23
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    @MarkViola Thats of course right. Nonetheless, the statement that the expression $n\cdot e^{O(n\varepsilon^2(n))}$ is large for $n\to\infty$ is not true in general. For example, consider $n\cdot e^{-\sqrt{n}}$, then this tends to $0$ as $n\to\infty$ but $-\sqrt{n}\in O(n\varepsilon^2(n))$ for suitable $\varepsilon(n)$, say $\varepsilon(n)=n^{-1/4}$. – 2018-01-24
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    @MarkViola First of all thank you for your patience. Then, maybe I should describe my problem more formally. So, I take **any** $f(n)\in O(n\cdot \varepsilon^2(n))$, where this is meant as $n\to\infty$ and $\varepsilon(n)$ is a zero sequence. This means, by definition, that there exist some positive real number $M$ and some $n_0$ such that $\lvert f(n)\rvert\leq M\lvert n\cdot \varepsilon^2(n)\rvert$ for all $n\geq n_0$. Now, could you please explain me, how to see formally, that $n\cdot e^{f(n)}$ is large as $n\to\infty$? Maybe this formal explanation could solve my still existing confusion. – 2018-01-24
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    @MarkViola But this is only for the special choice $f(n)=n\varepsilon^2(n)$ in my notation. – 2018-01-24
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    @MarkViola You could have $\varepsilon(n)=n^{-1/4}$, $f(n)=-\sqrt{n}$. This also satisfies $f(n)\in O(n\varepsilon^2(n))$ since $\lvert f(n)\rvert=\lvert\sqrt{n}\rvert\leq M \lvert n\varepsilon^2(n)\rvert=M\lvert\sqrt{n}\rvert$, but $n e^{f(n)}=n e^{-\sqrt{n}}\to 0$. – 2018-01-24
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    @mathfemi You are correct. But note from $(1)$ that $n\epsilon^2 \to 0$. Hence, $e^{O(n\epsilon2)}\to 1$. – 2018-01-24
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    Why does (1) imply that $n\varpsilon^2(n)\to 0$? – 2018-01-24
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    Note that if $n\varepsilon^2 \to C$, then for large $n$, $\varepsilon =O( n^{-1/2})$. But then $n\varepsilon = O(n^{1/2})$ and $(1)$ cannot hold. – 2018-01-24
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    Again, I cannot follow you. Why does $n\varepsilon(n)^2\to C$ (I guess you mean some constant $C\neq 0$) imply that $\varepsilon(n)\in O(n^{-1/2})$ as $n\to\infty$? – 2018-01-25
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    @mathfemi In Equation $(1)$, as $n\to \infty$, $\varepsilon \to 0$ and hence, $\log(\varepsilon) \to -\infty$. We must have, therefore, that $n\varepsilon \to \infty$. Now, if $n\varepsilon^2$ also approaches $\infty$ or approaches a finite number, say $C$, then for large enough $n$, $\varepsilon >\frac12 C n^{-1/2}$. But then for $n$ sufficiently large, $n\varepsilon >\frac12 C n^{1/2}$. But then, Equation $(1)$ cannot hold. – 2018-01-25
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    Do not see how to get from (2) to (3) since $a=e^xx$ exacly if $x=W(a)$, but you have to different expressions in the exponents, namely $(n-1)\epsilon(n)$ and $O(n\epsilon^2(n))$. – 2018-09-11
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    @Rhjg If you read the comments, I believe that your question will be answered therein. If not, please let me know. – 2018-09-11
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    @Rhjg I'll have a look. It's been quite a while since I posted this answer. – 2018-09-11
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    @MarkViola It was my fault! Its clear now – 2018-09-12
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    @Rhjg Pleased to hear. We simply use the definition of $W$. – 2018-09-12