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Let $u_0 = 1$ and $u_{n+1} = \frac{u_n}{1+u_n^2}$ for all $n \in \mathbb{N}$.

I can show that $u_n \sim \frac{1}{\sqrt{2n}}$, but I would like one more term in the asymptotic development, something like $u_n = \frac{1}{\sqrt{2n}}+\frac{\alpha}{n\sqrt{n}} + o\bigl(\frac{1}{n^{3/2}}\bigr)$.

Here is the outline of my proof of $u_n \sim \frac{1}{\sqrt{2n}}$:

  • $(u_n)$ is decreasing, and bounded from below by $0$, hence converges.
  • The limit $\ell$ satisifies $\ell = \frac{\ell}{1+\ell^2}$, hence $\ell = 0$.
  • A computation gives $v_n = u_{n+1}^{-2} - u_n^{-2} \to 2$.
  • Using Cesàro lemma, $\frac{1}{n} \sum_{k=0}^{n-1} v_k \to 2$.
  • Hence $\frac{1}{n} u_n^{-2} \to 2$.
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Let $a_n=\dfrac1{u_n^2}$. Then $$ a_{n+1}-a_n = \left(\frac{1+u_n^2}{u_n}\right)^2 - \frac1{u_n^2} = 2+u_n^2 = 2+\frac1{a_n}. $$ From this and $a_2=4$ we can find a successive sequence of estimates for $n\ge2$: \begin{align*} a_n &\ge 2n; \\ a_n &\le 2n + \sum_{k=2}^{n-1}\frac1{2k} < 2n+\frac12\log n; \\ a_n &\ge 2n + \sum_{k=2}^{n-1}\frac1{2k+\frac12\log k} > 2n+\int_2^n \frac{dx}{2x+\frac12\log x} \\ &> 2n+\int_2^n \frac{dx}{2x} - \int_2^n \frac{\frac12\log x}{2x(2x+\frac12\log x)}dx = 2n+\frac12\log n - \mathcal{O}(1). \end{align*}

Then $$ u_n = a_n^{-1/2} = \frac1{\sqrt{2n}} \left(1+\frac{\log n}{4n}+\mathcal{O}\bigg(\frac1n\bigg)\right)^{-1/2} \\ = \frac1{\sqrt{2n}} \left(1-\frac12 \cdot \frac{\log n}{4n} + \mathcal{O}\bigg(\frac1n\bigg) \right) = \frac1{\sqrt{2n}} - \frac1{8\sqrt2}\cdot \frac{\log n}{n^{3/2}} + \mathcal{O}\left(\frac1{n^{3/2}}\right). $$

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  • $\begingroup$ Ooops, I started the sequence with $u_1=1$. But it changes only the constant in the error term. $\endgroup$ – G. Kós Dec 6 '14 at 18:47
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    $\begingroup$ I wonder if this method can be iterated to yield abitrary deep asymptotic expansions. $\endgroup$ – Ewan Delanoy Dec 9 '14 at 19:00
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Note: This answer does not contain a proof, but I'm fairly sure it's right. Posted in the hope that perhaps the answer will guide you towards a proof thereof.

Looking at the graph you can figure out that in fact the next correction looks to be $n^{-3/2} \log n$ dominating the next $n^{-3/2}$ term. Given this, one can do a plausible self-consistency check using the recurrence relation to find that the answer.

Specifically, suppose

$$u_n \sim \frac \alpha {\sqrt{2n}} + \frac {\beta \log n}{n^{3/2}}+\text{a sufficiently nice smaller series in particular }\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$$

Then $$u_{n+1}=\frac{u_n}{1+u_n^2}$$ implies $$u_{n+1}-\frac{u_n}{1+u_n^2} \sim \frac 1 2 (\alpha - 2 \alpha^3) \frac 1 {n^{3/2}}+\cdots $$ so that $\alpha=0,-1/\sqrt 2,1/\sqrt 2$ are the options. You have checked that $\alpha=1/\sqrt{2}$.

Substituting this back in gives $$u_{n+1}-\frac{u_n}{1+u_n^2} \sim \left(-\frac 1 {8\sqrt 2} - \beta\right) \frac 1 {n^{5/2}}+\cdots $$ and hence it seems the only consistent result has $$\boxed{\displaystyle u_{n} \sim \frac{1}{\sqrt{2n}}-\frac{\log n}{8\sqrt 2 n^{3/2}}}$$

Numerically, I find the error $$\epsilon = \frac{u_n - \frac{1}{\sqrt{2n}}}{-\frac{\log n}{8\sqrt 2 n^{3/2}}} - 1\approx \frac{3.4}{\log n} \to 0$$ suggesting that indeed this is correct with the next term in the series being the expected $n^{-3/2}$ term.

However, the coefficient of the $n^{-3/2}$ term cannot be determined by this self-consistency procedure, reflecting the fact that one can check this actually depends on the initial datum $u_0$.

Left to do: (Not that I think I'll come back to this myself, but others can!)

  • Rigorously show the above logarithmic correction.
  • Figure out the dependence of the next term on the initial data.

Mathematica code for my verification:

min = 2;
max = 5000000;
dat = RecurrenceTable[{u[n + 1] == u[n]/(N[1, 50] + u[n]^2), u[0] == 1}, u, {n, min, max}];
errordat = 
  Table[{M, (dat[[M - min + 1]] - 
        1/Sqrt[2 M])/(-Log[M]/(8 Sqrt[2] M^(3/2))) - 1}, {M, min, max, 100000}];
nextconst = ((dat[[max - min + 1]] - 
       1/Sqrt[2 max])/(-Log[max]/(8 Sqrt[2] max^(3/2))) - 1)*Log[max]
Show[ListPlot[errordat, AxesOrigin -> {0, 0}, PlotStyle -> PointSize[Large]], Plot[nextconst/Log[x], {x, min, max}, PlotStyle -> Red, PlotRange -> All]]

Output: Estimate of $3.446...$ for next constant, and a plot of the $\epsilon$ against $n$ with a fitted curve based on a logarithmic next correction: enter image description here Obviously the decay to 0 here is slow, which is to be expected given the predicted logarithmic correction. Feel free to go to larger ranges.

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  • $\begingroup$ Wouldn't it be better to plot the difference between the asymptotic (1st term) and numerical results and see if the result brings about the correction term? Showing that the error goes to zero is not good enough. $\endgroup$ – Ron Gordon Dec 6 '14 at 19:08
  • $\begingroup$ I'm not entirely sure what you mean. The requirement that $\epsilon \to 0$ is precisely what it means for the boxed expression to be the first two terms of an asymptotic series, by definition. $\endgroup$ – Sharkos Dec 7 '14 at 11:58

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