Let's call $u_n$ the solution (it is unique) of $f_n(x)=x^5+nx-1=0$. I'd like to find an asymptotic expansion of $u_n$ of order two. I found $$u_n=\frac{1}{n}-\frac{1}{n^6}+o(\frac{1}{n^6})$$

The method I used:

  • We have $0 < u_n < 1$, hence $0 < n u_n =1-u_n^5 < 1$ and $\lim\limits_{n \to +\infty} u_n = 0$.
  • Therefore $\lim\limits_{n \to +\infty} u_n^5 = 0$ and $\lim\limits_{n \to +\infty} nu_n = 1$, which provides the first term of the development.
  • I followed with $v_n=u_n-\frac{1}{n}$ to get $(v_n+\frac{1}{n})^5+n v_n=0$. I proved that $0 < 1+ n v_n <1$ and followed to get the second term.

Somehow tedious...

I there a more systematic approach using calculus?

  • $\begingroup$ What about applying Newton's method till two consecutive approximations differ by less than the wanted error? Not very different from your fixed-point approach, anyway. $\endgroup$ – Jack D'Aurizio Sep 5 '15 at 16:12
  • $\begingroup$ This is mechanized in Maple. For example, see the three-terms asymptotics: $${n}^{-1}-{n}^{-6}+5\,{n}^{-11}+O \left( {n}^{-16} \right) . $$ $\endgroup$ – user64494 Sep 5 '15 at 16:19
  • $\begingroup$ @user64494. My question is in fact how is it mechanized! $\endgroup$ – mathcounterexamples.net Sep 5 '15 at 16:21
  • $\begingroup$ @ mathcounterexamples.net : Do you want to know how your calculator finds $\sqrt{29.78}?$ $\endgroup$ – user64494 Sep 5 '15 at 16:27
  • $\begingroup$ @user64494. Not precisely. I know it is usually based on finding an initial approximation and then using Newton's iterative method $u_{n+1}=\frac{1}{2}(u_n+\frac{29.78}{u_n})$. But not the details to be honest! $\endgroup$ – mathcounterexamples.net Sep 5 '15 at 16:31

Rearrange the equation like

$$ \frac{x}{1-x^5} = \frac{1}{n}. $$

The hypotheses of the Lagrange inversion theorem are satisfied, giving us the formula

$$ x = \sum_{k=1}^{\infty} \frac{1}{k!} \left\{ \left(\frac{d}{dx}\right)^{k-1} (1-x^5)^k \right\}_{x=0} n^{-k}, $$

valid as long as $|n|$ is large enough. The first terms of this series are

$$ x = n^{-1} - n^{-6} + 5n^{-11} - 35n^{-16} + 285n^{-21} - 2530n^{-26} + 23751 n^{-31} + \cdots. $$

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  • $\begingroup$ Thanks Antonio. Is there a proof of Lagrange inversion theorem in the book you mentionned in the other post: Bruijn's Asymptotic Methods in Analysis? $\endgroup$ – mathcounterexamples.net Sep 5 '15 at 17:29
  • $\begingroup$ No, it doesn't have a proof, only applications of the formula. The wikipedia page points to a book which does, though. $\endgroup$ – Antonio Vargas Sep 5 '15 at 17:31
  • $\begingroup$ I think the above expansion is found by CASs because the derivatives are huge. $\endgroup$ – user64494 Sep 5 '15 at 17:41
  • $\begingroup$ @user64494 yes, I used Mathematica to evaluate the derivatives. $\endgroup$ – Antonio Vargas Sep 5 '15 at 17:42

To compute the asymptotic expansion of $x$ by first principle, the proper way is to use the Lagrange Inversion formula as mentioned in another answer. I won't repeat it here. I'll like to point out this $x$ is closed related to something well studied.

Let $x = n^{1/4} y$, we have $$x^5 + nx - 1 = 0 \quad\iff\quad y^5 + y - n^{-5/4} = 0$$ The RHS has the form of a Bring radical,

$$\verb/BR(a)/ \stackrel{def}{=} \text{ root of } t^5 + t + a \text{ which is real when } a \text{ is real }$$

It is well known one can express the Bring radical in terms of Hypergeometric function.
If you look at Bring radical's wiki entry and the expansion there, you will find $$\begin{align} x &= n^{1/4}\verb/BR/(-n^{-5/4})\\ &= \frac{1}{n}{}_4F_3\left(\frac15,\frac25,\frac35,\frac45;\;\;\frac12, \frac34, \frac54;\;\;-\frac{5^5}{4^4 n^5}\right)\\ &= \sum_{k=0}^\infty \binom{5k}{k}\frac{(-1)^k}{(4k+1)n^{5k+1}} \end{align} $$ The expansion is not only asymptotic, it actually converges when $\displaystyle\;|n| > \frac{4^{4/5}}{5}$.

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