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For all $n \geq 2$, let's consider $P_n = X^n-nX+1$. Using Rolle theorem we can then show that $P_n$ has a single root $u_n$ in $]0,1[$.

I would like to get an asymptotic expansion with $4$ terms of $u_n$.

First I need to find the limit of $u_n$ so we have : $u_n^n -nu_n+1 = 0$ thus : $l^n/n - l = -1/n$ so $l = 0$ if we assume that $u_n$ has a limit.

Yet from now I don't really know how to proceed.

ShBh
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ZingZong
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  • @PeterForeman: this is confirmed by $\dfrac1{n^n}-n\dfrac1n+1\approx0$. –  Jun 16 '20 at 18:26
  • What you mean by expansion in 4 terms? $X=\frac{1}{n}+O\left(\frac{1}{n^{n+1}}\right)$, you want a solution up to $O(n^4)$ or up to the first four non null terms? – ECL Jun 16 '20 at 18:41

3 Answers3

4

If this can help:

A first approximation is $x_n=\dfrac1n$. A better approximation can be found in the form $\dfrac{1+t}n$. We write

$$\left(\frac{1+t}n\right)^n-n\left(\frac{1+t}n\right)+1=0,$$

$$\left(\frac{1+t}n\right)^n\approx\frac{1+nt}{n^n}=t,$$

$$t\approx \frac1{n^n-n},$$

and

$$x_n\approx\frac{1+\dfrac1{n^n-n}}n.$$

E.g., $x_5\approx0.2000641025641$ and $x_5^5-5x_5+1\approx3.28\cdot10^{-10}$

That process can be continued.

Another method is to write the Newton's iterates, starting from the initial approximation. Accuracy is very high, but the expressions quickly grow.

4

Hopefully I didnt do any mistake:

Note that

$$P_n(\frac{1}{n}+\frac{1}{n^{n+1}})= (\frac{1}{n}+\frac{1}{n^{n+1}})^n - \frac{1}{n^n}>0$$

Now, let $\alpha >1$. We have $$P_n(\frac{1}{n}+\frac{\alpha}{n^{n+1}}) <0 \Leftrightarrow \\ \frac{1}{n}+\frac{\alpha}{n^{n+1}} < \frac{\sqrt[n]{\alpha}}{n} \Leftrightarrow \\ n(\sqrt[n]{\alpha} -1) \geq \frac{\alpha}{n^{n-1}}$$

Now, $$\lim_n n(\sqrt[n]{\alpha} -1) =\lim_n \frac{\alpha^\frac{1}{n}-\alpha^0}{\frac{1}{n}-1}=\ln(\alpha) >0$$ and $$\lim_n \frac{\alpha}{n^{n-1}}=0$$

This shows that for all $\alpha >1$ there exists some $N$ so that, for all $n>N$ we have $$P_n(\frac{1}{n}+\frac{\alpha}{n^{n+1}}) <0 $$

It follows that asymptotically, we have $$\frac{1}{n}+\frac{1}{n^{n+1}} < u_n < \frac{1}{n}+\frac{\alpha}{n^{n+1}} \qquad \forall \alpha >1$$

[i.e. For each $\alpha >1$, there exists a $N$ such that the above holds for all $n>N$]

N. S.
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3

Here's an idea to get a complete asymptotics (obviously not in fixed powers of $n$, too).

For a fixed $n>1$, the solution $w=w_n(z)$ of $w=1+zw^n$ has a known power series $$w_n(z)=\sum_{k=0}^\infty\binom{nk}{k}\frac{z^k}{(n-1)k+1}$$ (a way to get it is basically Lagrange's inversion theorem). Thus, if $v_n=nu_n$ for our $u_n$, then $$v_n=1+n^{-n}(v_n)^n\implies v_n=w_n(n^{-n})\implies u_n=n^{-1}w_n(n^{-n}).$$

metamorphy
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