Series representation of polynomial root

We consider the smooth functions $q:\Bbb R\to\Bbb R$, $f,F:\Bbb R^2\to\Bbb R$ given by $$ \begin{aligned} q(w) &= \frac {3w^3}{1-3w^3}\ , \\[3mm] f(q,x) &= (1+q)x^5-(2+3q)x^4+(1+3q)x^3-x^2+2x-1\\ &=(x-1)^3(x^2+x+1) +qx^3(x^2-3x+3)\\ &=(x-1)^3(x^2+x+1) +qx^2(x-1)^3 +qx^3 \\[3mm] F(w,x) &=f(q(w),x)\\ &=(x-1)^3(x^2+x+1) + \frac{3w^3}{1-3w^3}\;x^3(x^2-3x+3) \ , \end{aligned} $$ and note that $F(0,1)=f(0,1)=0$.

We apply the Implicit Function Theorem (IFT) for $f$, so that we obtain $x$ as a function of $q$ first. For this we note that the corresponding derivative $f'_q=x^3(x^2-3x+3)$ is not zero in $(1,0)$. Let $f=f(q)$ be this smooth solution with $f(0)=1$, $F(q,f(q))=0$.

We can not apply (IFT) to the function $F=F(w,x)$, but Kopal (1959) claims that $F$ admits an implicit function $g$ as a solution for $F(w,g(w))=0$, so that $g$ "factorizes" through the composition with $f$, explicitly $g(w)=f(q(w))$. Moreover, $g$ should have the described Taylor polynomial expansion around $0$ modulo $O(w^4)$.

It is then enough to verify this published solution. I will use CAS support, here sage, to investigate the situation. We will find all five solutions of the given problem in the power series ring of power series in $w$ (modulo some rather small power of $w$) over the field $F=\Bbb Q(u)$, $u$ being a primitive root of third order.

So let us consider $S$, the power series ring $\Bbb Q[[\; w\;]]$, and search for solutions $g=g(w)\in S$ for $F(w, g(w))=0$.

Because $F(0,x)=0$ has the roots (with multiplicities) $1,1,1,u,u^2$, we have five potential solutions $g$ starting respectively with these constant coefficients. We will search for (one, later three) solutions of the shape $g(w)=1+aw+O(w^2)$, and (later also) $u+aw+O(w^2)$, and $u^2+aw+O(w^2)$. Here, $a$ is an unknown.

We compute using sage in a power series ring, where we have also arranged for the existence of a "parameter" $a$, in the following setting

F.<u> = CyclotomicField(3)
S.<a> = PolynomialRing(F)
R.<w> = PowerSeriesRing(S, default_prec=20)

def F(q, x):
    return (1+q)*x^5   - (2+3*q)*x^4   + (1+3*q)*x^3   - x^2 + 2*x - 1

q = 3*w^3 / (1 - 3*w^3)

(with objects that should be mathematically clear) the following expansion of $F(w,\ 1+aw)$, taken modulo $O(w^5)$:

sage: F(q, 1 + a*w) + O(w^5)
(3*a^3 + 3)*w^3 + (3*a^4 + 6*a)*w^4 + O(w^5)

In case of a zero result in $R$, we have to arrange that the coefficient in $w^3$ vanishes, i.e. $a^3=-1$. There are three solutions, $-1,-u,-u^2$. And we will follow soon only the main path $a=-1$ (relevant for the OP) first. The choice of $a$ determines three brances $$ \begin{aligned} g_1(w) &= 1-w+O(w^2)\ ,\\ g_2(w) &= 1-uw+O(w^2)\ ,\\ g_3(w) &= 1-u^2w+O(w^2)\ . \end{aligned} $$ For $g_1$ we search then the next approximation of the shape $1-w+aw^2*O(w^3)$. The code

sage: F(q, 1 - w + a*w^2) + O(w^6)
(9*a - 3)*w^4 + (-9*a^2 - 6*a + 2)*w^5 + O(w^6)

leads to $(9a-3)=0$, so $a=1/3$, so $$ g_1(w) = 1-w+\frac 13w^2+O(w^3)\ . $$ The next coefficient is extracted from

sage: F(q, 1 - w + 1/3*w^2 + a*w^3) + O(w^7)
(9*a - 1)*w^5 + (-12*a + 70/9)*w^6 + O(w^7)

so we get $a=1/9$ and the better approximation: $$ g_1(w) = 1-w+\frac 13w^2+\frac 19w^3+O(w^4)\ . $$ We can go on in this manner:

sage: F(q, 1 - w + 1/3*w^2 + 1/9*w^3 + a*w^4) + O(w^8)
(9*a + 58/9)*w^6 + (-12*a - 9)*w^7 + O(w^8)

sage: F(q, 1 - w + 1/3*w^2 + 1/9*w^3 - 58/81*w^4 + a*w^5) + O(w^9)
(9*a - 11/27)*w^7 + (-12*a - 280/81)*w^8 + O(w^9)

sage: F(q, 1 - w + 1/3*w^2 + 1/9*w^3 - 58/81*w^4 + 11/243*w^5 + a*w^6) + O(w^10)
(9*a - 4)*w^8 + (-12*a + 8867/729)*w^9 + O(w^10)

sage: F(q, 1 - w + 1/3*w^2 + 1/9*w^3 - 58/81*w^4 + 11/243*w^5 + 4/9*w^6 + a*w^7) + O(w^11)
(9*a + 4979/729)*w^9 + (-12*a - 6523/2187)*w^10 + O(w^11)

Thus: $$ g_1(w) = 1-w+\frac 13w^2+\frac 19w^3 -\frac{58}{81}w^4 +\frac{11}{243}w^5 +\frac 4 9 w^6 -\frac{4979}{6561}w^7 +O(w^8)\ . $$ Because $F=F(w,x)$ depends "only on $w^3$", (i.e. factorizes through $w\to w^3$,) it is clear that $$ \begin{aligned} g_2(w) &= g(uw)\ ,\\ g_3(w) &= g(u^2w)\ , \end{aligned} $$ are also (exactly the other two) solutions in $1+O(w)$ (defined after passing to the cyclotomic field, or to $\Bbb C$).

Not needed indeed, but to have all solutions of the quintic in $x$, we search also for $g_4,g_5$.

sage: F(w, u + a*w) + O(w^2)
(-9*a - 4*u + 2)*w + O(w^2)

sage: F(w, u + (2-4*u)/9*w + a*w^2) + O(w^3)
(-9*a + 56/27*u - 98/27)*w^2 + O(w^3)

sage: F(w, u + (2-4*u)/9*w - (98-56*u)/27*w^2 + a*w^3) + O(w^4)
(-448/27*u + 784/27)*w^2 + (-9*a + 1316/81*u + 350/81)*w^3 + O(w^4)

sage: F(w, u + (2-4*u)/9*w - (98-56*u)/243*w^2 + a*w^3) + O(w^4)
(-9*a - 28/81*u + 1162/243)*w^3 + O(w^4)

sage: F(w, u + (2-4*u)/9*w - (98-56*u)/243*w^2 + (1162-84*u)/2187*w^3 + a*w^4) + O(w^5)
(-9*a - 7184/6561*u - 35482/6561)*w^4 + O(w^5)

From here we can extract $g_4(w)$, so that $g_4(w)=u+\frac{2-4u}9w+\dots$ , and for algebraic reasons, the Galois transformation $u\to u^2$ applied on $g_4$ delivers $g_5(w)=u^2+\dots$

A final remark. If we work algebraically as above, it is questionable if at each "step" we obtain a (linear) equation in $a$, that can be solved. (I.e. $a$ appears (linearly) with a non-zero coefficient.) This needs a more specialized study. But Kopal (1959) claims only a result modulo $O(w^4)$, not an analytic expansion (with controlled coefficients in $\Bbb Z$ localized in $3$).

Later EDIT: There are some questions in the comments, i am trying to comment on them. (Here, since my style is not fitting in the margin of the comment.) First of all, this is a wonderful question. Answering it in a target oriented manner does not give full explanation of the structure. So understanding the full setting has at least two points: (I) Which is the asymptotics of a solution $x=x(q)$ of the given quintic with $x(0)=1$? Is it a power series in $q$? (No.) A fractional power series maybe? (Yes. In $q^{1/3}$.) (II) Find it explicitly in terms of a given uniformizer. (In the OP it is $w$, a constant times $q^{1/3}+O(q^{2/3})$.) I did above only (II), checking the prescribed expansion in Kopal (1959). Some words on (I), and (II).

(I) Find a dependency $x=x(q)$ explicitly or at least say something about its "asymptotics". This is beautifully done in the answer of Simply Beautiful Art, please consider giving full cretit to his answer, and he explains exactly the first step(s) to be done. He is writing $x=1+y$, so $y$ is now around zero, and (equivalently written) gets the Taylor expansion of the function $f$ from the given equation $f(q,x)=0$, i.e. $f(q, 1+y)=0$ around $(q,y)=(0,0)$. Let us do it here also: $$ \begin{aligned} f(q.1+y) &=f(q,x)_{\text{in }x=1+y} \\ &=\Big(\ (x-1)^3(x^2+x+1) +qx^2(x-1)^3 +qx^3\ \Big)_{\text{in }x=1+y} \\ &\equiv y^3(1+1+1+2y+y+O(y^2)) \\ &\qquad+qy^3(1+2y+O(y^2)) \\ &\qquad\qquad+q(1+3y+O(y^2)) \\ &\equiv 3y^3+q+\dots\ . \end{aligned} $$ In the last line we have omitted terms in "higher order", where the "order" is still open, we may want to give different weights to $q$ and $y$. At any rate, for any such degree "order", the first components are $3y^3$ and $q$. To solve for $y$ depending on (a fractional power series expansion in) $q$, we have to arrange for the asymptotic $q\sim 3y^3$, i.e. $y\sim (q/3)^{1/3}$. In such a situation i would substitute $q=3z^3$, but for some reason, Kopal (1959) considers rather $q=3w^3/(1-3w^3)$. At any rate, this amounts to the same final information on the asymptotic of the solution around $q=0$. To obtain explicitly the expansion in terms of $z$, we can do the same as in this answer...

F.<u> = CyclotomicField(3)
S.<a> = PolynomialRing(F)
R.<z> = PowerSeriesRing(S, default_prec=20)

def f(q, x):
    return (1+q)*x^5   - (2+3*q)*x^4   + (1+3*q)*x^3   - x^2 + 2*x - 1

q = 3*z^3

and then

sage: f(q, 1+a*z) + O(z^5)
(3*a^3 + 3)*z^3 + (3*a^4 + 6*a)*z^4 + O(z^5)

sage: # so a = -1 above to cancel the term in z^3

sage: f(q, 1 -z + a*z^2) + O(z^6)
(9*a - 3)*z^4 + (-9*a^2 - 6*a + 2)*z^5 + O(z^6)

sage: # so a = 1/3 above to cancel the term in z^4

sage: f(q, 1 -z + 1/3*z^2 + a*z^3) + O(z^7)
(9*a - 1)*z^5 + (-12*a - 11/9)*z^6 + O(z^7)

sage: # so a = 1/9 above to cancel the term in z^5

sage: f(q, 1 -z + 1/3*z^2 + 1/9*z^3 + a*z^4) + O(z^8)
(9*a - 23/9)*z^6 + (-12*a + 9)*z^7 + O(z^8)

sage: # so a = 23/81 above to cancel the term in z^6

sage: f(q, 1 -z + 1/3*z^2 + 1/9*z^3 + 23/81*z^4 + a*z^5) + O(z^9)
(9*a + 151/27)*z^7 + (-12*a - 685/81)*z^8 + O(z^9)

sage: # so a = -151/243 above to cancel the term in z^7, and so on

and we get in this manner the asymptotic solution $$ 1-z+\frac 13z^2+\frac 19z^3+\frac {23}{81}z^4-\frac{151}{243}z^5+\dots $$ i.e. explicitly in terms of $q$ $$ 1 -\frac 1{\sqrt [3]3}q^{1/3} +\frac 1{3\sqrt [3]3}q^{2/3} +\frac 1{27}q +\frac {23}{243\sqrt [3]3}q^{4/3} -\frac{151}{729\sqrt [3]9}q^{5/3} +\dots\ . $$ (II) But even if we have the information above, the expansion in terms of $q$, the question wants the expansion in terms of $w$, so we have to redo the computation, or substitute the rather unnatural relation between $q$ and $w$... Instead, i did it directly in my answer, translating $f(q,x)=0$ as an equation $F(w,x)=0$ for which we have the information from Kopal (1959) that a power series expansion can be found. I really wanted to get his coefficients (and some more, supposing that they have some inner rule, but there is no such rule).


Let $x=y+1$ to get

$$(1+q)y^5+(3+2q)y^4+(3+q)y^3+q(y+1)^2=0$$

Note then we are considering $y\to0$ as $q\to0$, where $(y+1)^2\to1$ and $y^5$ and $y^4$ tend to zero faster than $y^3$, with coefficient tending to $3$. Thus, the dominating term is given by

$$3y^3\sim-q\implies y^3\sim-\frac q3\implies y\sim-\sqrt[3]{\frac q3}$$

which agrees with the first term of your expansion. It is notable that the writing this in terms of $q/(1+q)$ is a conformal mapping likely used to increase the interval of $q$ for which the series converges, since as $q\to-1$ at least one root will grow unboundedly as we lose a degree of the polynomial.

From there it is simply a matter of letting $w^3$ be defined as it is, rewriting everything in terms of $w$ using $q=3w^3/(1-3w^3)$, and expanding $y$ as a series.

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Polynomials

Sequences And Series

Roots

Quintic Equations

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