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A cubic of the form $$x^3+1-x/b=0$$ has has three real roots

Using the Lagrange inversion theorem one of the roots is given by

$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$

How do you find the second one? I cannot find any info online. I am looking for a simialr series solution. I suspect is is of the form

$$x = \sum_{k=0}^\infty \binom{3k+2}{k} \frac{b^{3k+2} }{(3)k+2} $$

Semiclassical
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CarP24
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    In the limit $b\to 0$, the equation (which is equivalent to $bx^3+b-x=0$) the equation collapses to $x=0$. What this really amounts to is that two of the three roots escape to infinity as $b\to 0$ and only the $x=0$ root remains. In particular, this must include the second real root. But your proposed second solution would go to 0 as well as $b\to 0$, so I don't think this can be correct. I suggest instead seeking series solutions in powers of $1/b$, as the roots converge to cubic roots of $-1$ in the $b\to \infty$ limit. – Semiclassical Aug 14 '21 at 23:11
  • Actually, one other point to note: The cubic only has three real roots if $b<2^{2/3}/3$. That renders my comment about "powers of $1/b$" rather unhelpful and it should be disregarded. – Semiclassical Aug 14 '21 at 23:37

4 Answers4

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The related OEIS sequences A206300 and A224884 help to answer your question. Define $$ a_n := \frac{2^{2n-1}}{n!}\frac{\Gamma(3n/2-1/2)}{\Gamma(n/2+1/2)}. \tag{1} $$ Now define $$ y_2 := \sum_{n=0}^\infty -a_n z^{3n-1},\quad y_3 := \sum_{n=0}^\infty\, (-1)^na_n z^{3n-1},\quad y_1 := -y_2-y_3. \tag{2} $$ These are the three roots of the cubic $$ y^3 - y/z^2 + 4 = 0. \tag{3}$$

The power series expansions are $$ y_1 = 4z^2 + 64z^8 + 3072z^{14} + 196608z^{20} +\dots, \tag{4} $$ $$ y_2 = z^{-1} - 2z^2 - 6z^5 - 32z^8 - 210z^{11} - \dots, \tag{5}$$ $$ y_3 = -z^{-1} - 2z^2 + 6z^5 - 32z^8 +210z^{11} - \dots. \tag{6}$$ The roots of the cubic $$ x^3 + 1 - x/b = 0 \tag{7} $$ are the roots of the previous cubic divided by $\,2^{2/3}\,$ and where $\,z = b^{1/2}/2^{2/3}.$

The power series expansions are $$ x_1 = b + b^4 + 3b^7 + 12b^{10} + 55b^{13} + \dots, \tag{8} $$ $$x_2 = b^{-1/2}-\frac12 b - \frac38 b^{5/2} - \frac12 b^4+\dots, \tag{9}$$ $$x_3 =-b^{-1/2}-\frac12 b+\frac38 b^{5/2}-\frac12 b^4+\dots. \tag{10}$$

Note that the coefficient of $\,x^{3n+1}\,$ in $\,x_1\,$ is OEIS sequence A001764.

Note that $\,x_1 = b(1+x_1^3).\,$ Use it iteratively to generate power series truncations of $\,x_1.$

Note that $$ (x-x_1)(x-x_2)(x-x_3) = x^3-x/b+1, \tag{11}$$ which implies $$ x_1+x_2+x_3 = 0,\quad x_1 x_2 x_3 = -1. \tag{12}$$ Thus, given a value for $\,x_1\,$ the other two roots $\,x_2,x_3\,$ satisfy a quadratic.

Oscar Lanzi
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Somos
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Use the fact that the remaining two roots must solve a quadratic equation. If $r_1$ is your known root, then the other two roots must have a sum equal to $(1/b)-r_1$ and a produxt equal to $-1/r_1$ (why)? So you have a quadratic equation for the remaining two roots:

$x^2-[(1/b)-r_1]x-1/r_1=0$

Solve by the quadratic formula, substitute for $r_1$ and (tediously) collapse it into a series.

Oscar Lanzi
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Too long for a comment.

Looking for truncated series expansions, you can directly solve for the roots solving the cubic using the trigonometric method.

The solutions are given by $$x_k=\frac{2 \cos \left(\frac{2 \pi }{3}k-\frac{1}{3} \cos ^{-1}\left(-\frac{3\sqrt{3}}{2} b^{3/2}\right)\right)}{\sqrt{b}\sqrt{3}}\qquad \text{with} \qquad k=0,1,2$$

We have $$x_1=b+b^4+3 b^7+12 b^{10}+55 b^{13}+273 b^{16}+1428 b^{19}+7752 b^{22}+O\left(b^{25}\right)$$ and the coefficients corrrespond to sequence $A001764$ in $OEIS$ that is to say $$x_1=\sum_{n=0}^\infty \frac{ \binom{3 n}{n}}{2 n+1}b^{3 n+1}$$ which is the one you obtained using Lagrange inversion theorem.

Similarly, $$x_0=\frac{1}{\sqrt{b}}-\frac{b}{2}-\frac{3 b^{5/2}}{8}-\frac{b^4}{2}-\frac{105 b^{11/2}}{128}-\frac{3 b^7}{2}-\frac{3003 b^{17/2}}{1024}-6 b^{10}+O\left(b^{23/2}\right)$$ $$x_2=-\frac{1}{\sqrt{b}}-\frac{b}{2}+\frac{3 b^{5/2}}{8}-\frac{b^4}{2}+\frac{105 b^{11/2}}{128}-\frac{3 b^7}{2}+\frac{3003 b^{17/2}}{1024}-6 b^{10}+O\left(b^{23/2}\right)$$ already given by @Somos.

You also have $x_0+x_1+x_2=0$ and $x_0x_1x_2=-1$ also as mentioned by @Somos.

Claude Leibovici
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$x^3 - \frac {x}{b} =-1$

Let $x = 2\frac {\sqrt {3b}}{3b}\cos\theta$

$\frac {8\sqrt {3b}}{9b^2} \cos^3\theta - \frac {2\sqrt{3b}}{3b^2}\cos\theta = 1\\ \frac {2\sqrt{3b}}{9b^2}(4\cos^3\theta - 3\cos\theta) = 1\\ \cos3\theta = \frac {3b\sqrt {3b}}{2}\\ \theta = \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right), \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right) + \frac {2\pi}{3}, \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right) + \frac {4\pi}{3} \\ x = 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)\right), 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)+\frac {2\pi}{3}\right), 2\frac {\sqrt {3b}}{3b}\cos\left(\frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right)+\frac {4\pi}{3}\right)$

user317176
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