1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $\displaystyle\;|n| > \frac{4^{4/5}}{5}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.