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Fix real numbers $a \ge 0$ and $b \gt 1$. For any integer $n\ge 1$, let $x_n$ be the unique positive solution of the equation

$$ \tag{1} 1 = \frac{1}{n^a x} + \frac{1}{n}\sum_{k=1}^\infty \frac{1}{1 + k^b x}. $$

Of course, $x_n$ depends on $a$ and $b$. It is easy to see (check (1) below) that $x_n \ge n^{-a}$. Less trivial is the fact that $$ \frac{x_n}{n^{-a}} - 1 = O(n^{-(1-a)}), \text{ if }a \in (0,1). \tag{0}. $$ Indeed, from (1), observe that

$$ x - n^{-a}= \frac{x}{n}\sum_{k=1}^\infty \frac{1}{1 + k^b x} = \frac{1}{n }\sum_{k=1}^\infty k^{-b} \frac{k^b x}{1 + k^b x} \le \frac{1}{n }\sum_{k=1}^\infty k^{-b} = O(1/n). $$

Question. For large $n$, what is good and asymptotically exact formula for $x_n$ ? That is, construct $\epsilon_n$ such that $x_n/n^{-a} - 1 = O(\epsilon_n)$, with $\epsilon_n$ decays as fast as possible.

Note. Fixed-point equations like (1) appear in random matrix theory.

A possibly sub-optimal solution

Claim. If $b \in (1,\infty)$ and $a \in [0,b)$, then for large $n$, it holds that $$ \frac{x_n}{n^{-a}} - 1=O(n^{- (1- a/b)}) = o(1). $$

Proof. Indeed, since the second term in the RHS of (1) is nonnegative, we deduce that $$ x_n \ge n^{-a}. \tag{2} $$

Now, observe that because the function $t \mapsto 1 / (1 + xt^b)$ is decreasing on $[0,\infty)$, one has $$ \sum_{k=1}^\infty \dfrac{1}{1 + k^b x} \le \int_0^\infty \dfrac{1}{1 + z^b x}\mathrm{d}z = cx^{-1/b}, $$

where $c:=\dfrac{\pi/b}{\sin(\pi/b)} \in (0,\infty)$, and the last step is thanks to this ME post https://math.stackexchange.com/a/247903/168758. Combining with (1) then gives $$ 1 = \frac{1}{n^a x_n} + O(\frac{1}{nx_n^{1/b}}). \tag{3} $$ Combining (2) and (3) gives $1 \le n^{-a}/x_n + O(n^{-(1-a/b)})$, and so $$ \frac{x_n}{n^{-a}} \le \frac{1}{1-O(n^{-(1-a/b)})} = 1+O(n^{-(1-a/b)}), \tag{4} $$

Combining with (1) gives the claimed result. $\quad\Box$

Points of improvement

I don't think I've made the best use of (3). Maybe it is possible to obtain a better $\epsilon_n$ via a more clever use of (3).

dohmatob
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  • what is your index $k$ doing? – dezdichado Nov 13 '22 at 23:50
  • The typo has been fixed. – dohmatob Nov 13 '22 at 23:51
  • Where is $a$ in $(1)$? What is $\lambda$? How did you get that equality with $c$ after $(2)$? The last equation should be $(4)$. – Gary Nov 14 '22 at 10:24
  • The typos have been fixed. Thanks. – dohmatob Nov 14 '22 at 11:45
  • How did you get $$ \sum\limits_{k = 0}^\infty {\frac{1}{{1 + k^b x}}} = \frac{{\pi /b}}{{\sin (\pi /b)}}x^{ - 1/b} ,? $$ – Gary Nov 14 '22 at 12:16
  • Sorry, it shouldn't be = but $\le$. This is from comparing series and integral. – dohmatob Nov 14 '22 at 12:37
  • what is the context of this question? – Conrad Nov 14 '22 at 19:49
  • @dohmatob You had already $y_n = n^{-a}$. You want an approximation of $\epsilon_n$ (for example, $\epsilon = d \cdot n^{-\gamma}$)? Or perhaps a bound for $\epsilon_n$ is sufficient? – NN2 Nov 14 '22 at 20:35
  • Yes. That's the idea. I already have $x_n/y_n = 1 + \epsilon_n$ with $y_n=n^{-a}$ and $\epsilon_n = O(n^{-\min(1,a/n)})$. The question is whether this can be improved (i.e replace $n^{-\min(1,a/b)}$ with something the decays faster). – dohmatob Nov 14 '22 at 21:53
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    @dohmatob We must have $a<b$, if not, for $n$ sufficently large, $1 \le \frac{x_n}{n^{-a}}<\frac{1}{1+\beta n^{-1}}$ with $0<\beta <1$. This implies a contradiction. So, you can replace $\mathcal{O}(n^{-\min(1,a/b)})$ by $\mathcal{O}(n^{-a/b})$ – NN2 Nov 14 '22 at 22:34
  • @NN2 I agree with you. But, then this means I've probably made an error somewhere in my arguments, before (4) :/ – dohmatob Nov 14 '22 at 22:47
  • ok, i think I found the culprit. I wrote $\sum_{k=1}^\infty f(k) \le \int_1^\infty f(t)dt$ instead of $\sum_{k=1}^\infty f(k) \le \int_0^\infty f(t)dt$. This mistake introduced the $1/n$ in $1+1/n$. Fixing... – dohmatob Nov 14 '22 at 23:04
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    When going to $(3)$, you forgot the $1/n$ in front of the sum in $(1)$. – Gary Nov 15 '22 at 00:18
  • Indeed, and this induced yet another bug in the final result. My corrected bound is $x_n/n^{-a} - 1 = O(n^{-(1-a/b) })$, valid for $b \in (1,\infty)$ and $a \in [0,b)$. – dohmatob Nov 15 '22 at 00:36
  • Since posting this question yesterday, you have made (thus far) 23 edits. Please do not do this---it is disruptive, as it continually bumps your question to the top of the front page. Decide what you are going to ask, write that post, and leave it alone. If you are trying to nail the formatting, please use the Sandbox on meta. – Xander Henderson Nov 15 '22 at 13:14

1 Answers1

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I think that the idea of using the continuous case is a way to go for possible approximations. Multiplying by $x$ to remove the asymptote, we need to find the zero of function $$f(x)= x-\frac{\pi \csc \left(\frac{\pi }{b}\right)}{b\, n}x^{1-\frac{1}{b}}-n^{-a}$$ Let

$$x=\frac y{n^a}\qquad \qquad \alpha=\frac{\pi}b \csc \left(\frac{\pi }{b}\right)\,n^{\frac{a}{b}-1}\qquad \qquad \beta=1-\frac 1b$$
$$f(y)=y-\alpha \,y^\beta-1$$ The first derivative cancels at $$y_*=(\alpha \beta )^{\frac{1}{1-\beta }}$$ and then a first estimate using series $$y_0=y_*+\sqrt{-2 \frac{f(y_*)}{f''(y_*)}}$$ and the first iterate of Newton method $$y_1=y_0-\frac{f(y_0)}{f'(y_0)}$$ is totally explicit.

Trying with $(a=\pi,b=e,n=11)$ $$y_*=1.49611\quad \quad y_0= 5.39684 \quad \quad y_1=7.86537 \quad \implies \quad x_1=0.00420812$$ while the solution is $y=7.62709$ that is to say $x=0.00408063$.

Trying with the summation, the solution is $x=0.00369922$

Edit

Probably easier would be to let $$y=z^{\frac{1}{\beta }}\qquad \qquad \gamma=\frac{b}{b-1} >1$$ $$f(z)=z^\gamma-\alpha z-1$$

Claude Leibovici
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  • Thanks for the contribution (upvoted). As stated in the question, i'm really only interested in the asymptotics (for large $n$), not in the actual values of the root. – dohmatob Nov 18 '22 at 13:06
  • @dohmatob. The asymptotics is "almost" given by $y_1$. I just recover the results of $10^6$ random runs and compared $y_1$ and the exact result (for the summation) : $R^2=0.988$. – Claude Leibovici Nov 19 '22 at 02:48