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Let's consider the real roots of the following equation in x: $$x^5+\epsilon x =1.$$ Here $\epsilon\ge0$ is a real parameter. When $\epsilon=0$ the only root is $x=1.$ For a general $\epsilon>0$ we know that there is no closed form solution, so this is the perfect candidate for a series expansion: $$x(\epsilon)=1+a_1\epsilon+a_2\epsilon^2+a_3\epsilon^3+\dots$$ If we work out the coefficients $$\mathbf{a}=(-\frac{1}{5},-\frac{1}{25},-\frac{1}{125},0,-\frac{21}{15625},\dots)$$ we come to the conclusion that $$a_{4+5k}=0,\text{ for }k=0,1,2,\dots$$ My question is why does this happen every five indices? My guess is that it has to do with the order of the equation. However when I do the actual computation it's not at all obvious and it seems to happen by chance! Terrible!

All ideas/comments welcome

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  • $\begingroup$ can you supply more terms of the series? So as to see the numerators? $\endgroup$
    – KonKan
    Sep 19, 2016 at 7:29
  • $\begingroup$ Nothing fancy,$-\frac{5381601456 \epsilon ^{20}}{59604644775390625}+\frac{137307401 \epsilon ^{18}}{476837158203125}+\frac{79893138 \epsilon ^{17}}{95367431640625}+\frac{28881767 \epsilon ^{16}}{19073486328125}+\frac{6491331 \epsilon ^{15}}{3814697265625}-\frac{175398 \epsilon ^{13}}{30517578125}-\frac{105672 \epsilon ^{12}}{6103515625}-\frac{39767 \epsilon ^{11}}{1220703125}-\frac{9367 \epsilon ^{10}}{244140625}+\frac{286 \epsilon ^8}{1953125}+\frac{187 \epsilon ^7}{390625}+\frac{78 \epsilon ^6}{78125}+\frac{21 \epsilon ^5}{15625}-\frac{\epsilon ^3}{125}-\frac{\epsilon ^2}{25}$ $\endgroup$
    – Georgy
    Sep 19, 2016 at 7:39
  • $\begingroup$ And we know for a fact that we can't sum the series otherwise we could have a closed form solution of a quintic equation. $\endgroup$
    – Georgy
    Sep 19, 2016 at 7:45
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    $\begingroup$ what about other orders for the equation: $x^3+\epsilon x =1$, $x^4+\epsilon x =1$, $x^7+\epsilon x =1$ ... etc. Have you checked for a similar phenomenon? $\endgroup$
    – KonKan
    Sep 19, 2016 at 7:49

2 Answers 2

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Let us do the same for $$x^n+\epsilon x =1$$ and use $$x=1+\sum_{i=1}^{2n} a_i\epsilon^i$$ Now, using the binomial theorem or Taylor expansion around $\epsilon=0$, by identification, we can find that $$a_1=-\frac 1n$$ $$a_2=-\frac{(n-3)}{2 n^2}$$ $$a_3=-\frac{(n-4) (n-2)}{3 n^3}$$ $$a_4=-\frac{(n-5) (2 n-5) (3 n-5)}{24 n^4}$$ $$a_5=-\frac{(n-6) (n-3) (n-2) (2 n-3)}{10 n^5}$$ $$a_6=-\frac{(n-7) (2 n-7) (3 n-7) (4 n-7) (5 n-7)}{720 n^6}$$ $$a_7=-\frac{(n-8) (n-4) (n-2) (3 n-8) (3 n-4) (5 n-8)}{315 n^7}$$ $$a_8=-\frac{(n-9) (n-3) (2 n-9) (2 n-3) (4 n-9) (5 n-9) (7 n-9)}{4480 n^8}$$ $$a_9=-\frac{(n-10) (n-5) (n-2) (2 n-5) (3 n-10) (3 n-5) (4 n-5) (7 n-10)}{4536 n^9}$$In this, you can notice that $a_k$ has, as a factor, $(n-k-1)$ and this explains that.

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  • $\begingroup$ this was nice! +1 $\endgroup$
    – KonKan
    Sep 19, 2016 at 7:54
  • $\begingroup$ @KonKan. Thank you ! May I confess that my passion for Taylor series started ... 60 years ago ? $\endgroup$ Sep 19, 2016 at 7:56
  • $\begingroup$ Ok! I feel a little better now! I have to admit that at first I was ... shocked when I saw you already had posted the solution, while I was still thinking ... which strategy to choose ! $\endgroup$
    – KonKan
    Sep 19, 2016 at 8:02
  • $\begingroup$ We can verify the start of the periodic appearance of zeros for $n=2, \ 3,\ 4,\ 5$. $\endgroup$ Sep 19, 2016 at 15:32
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The solution of $x^5+\varepsilon x=1$ can be computed through Bring radical.
If we set $x=\varepsilon^{1/4} z$, by Lagrange inversion theorem we have: $$ x(\varepsilon) = \varepsilon^{1/4}\,\text{BR}(\varepsilon^{-5/4})=\color{red}{-\sum_{k\geq 0}\binom{5k}{k}\frac{(-1)^k\varepsilon^{-(5k+1)}}{4k+1}} $$

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