Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them
We study the Yablonskii-Vorob'ev polynomials, which are special polynomials used to represent rational solutions of the second Painlevé equation. Divisibility properties of the coefficients of these polynomials, concerning powers of 4, are obtained and we prove that the nonzero roots of the Yab...
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irk-123456789-1465102019-02-10T01:25:17Z Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them Roffelsen, P. We study the Yablonskii-Vorob'ev polynomials, which are special polynomials used to represent rational solutions of the second Painlevé equation. Divisibility properties of the coefficients of these polynomials, concerning powers of 4, are obtained and we prove that the nonzero roots of the Yablonskii-Vorob'ev polynomials are irrational. Furthermore, relations between the roots of these polynomials for consecutive degree are found by considering power series expansions of rational solutions of the second Painlevé equation. 2010 Article Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them / P. Roffelsen // Symmetry, Integrability and Geometry: Methods and Applications. — 2010. — Т. 6. — Бібліогр.: 13 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 34M55 DOI:10.3842/SIGMA.2010.095 http://dspace.nbuv.gov.ua/handle/123456789/146510 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України |
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We study the Yablonskii-Vorob'ev polynomials, which are special polynomials used to represent rational solutions of the second Painlevé equation. Divisibility properties of the coefficients of these polynomials, concerning powers of 4, are obtained and we prove that the nonzero roots of the Yablonskii-Vorob'ev polynomials are irrational. Furthermore, relations between the roots of these polynomials for consecutive degree are found by considering power series expansions of rational solutions of the second Painlevé equation. |
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Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them |
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Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them |
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Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them |
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Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them |
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Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them |
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irrationality of the roots of the yablonskii-vorob'ev polynomials and relations between them |
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Irrationality of the Roots of the Yablonskii-Vorob'ev Polynomials and Relations between Them / P. Roffelsen // Symmetry, Integrability and Geometry: Methods and Applications. — 2010. — Т. 6. — Бібліогр.: 13 назв. — англ. |
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Symmetry, Integrability and Geometry: Methods and Applications |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 6 (2010), 095, 11 pages
Irrationality of the Roots of the Yablonskii–Vorob’ev
Polynomials and Relations between Them
Pieter ROFFELSEN
Radboud Universiteit Nijmegen, IMAPP, FNWI,
Heyendaalseweg 135, 6525 AJ Nijmegen, the Netherlands
E-mail: roffelse@science.ru.nl
Received November 13, 2010, in final form December 08, 2010; Published online December 14, 2010
doi:10.3842/SIGMA.2010.095
Abstract. We study the Yablonskii–Vorob’ev polynomials, which are special polynomials
used to represent rational solutions of the second Painlevé equation. Divisibility proper-
ties of the coefficients of these polynomials, concerning powers of 4, are obtained and we
prove that the nonzero roots of the Yablonskii–Vorob’ev polynomials are irrational. Fur-
thermore, relations between the roots of these polynomials for consecutive degree are found
by considering power series expansions of rational solutions of the second Painlevé equation.
Key words: second Painlevé equation; rational solutions; power series expansion; irrational
roots; Yablonskii–Vorob’ev polynomials
2010 Mathematics Subject Classification: 34M55
1 Introduction
In this paper we study the Yablonskii–Vorob’ev polynomials Qn, with a special interest in their
roots. These polynomials were derived by Yablonskii and Vorob’ev, while examining the hierar-
chy of rational solutions of the second Painlevé equation. The Yablonskii–Vorob’ev polynomials
are defined by the differential-difference equation
Qn+1Qn−1 = zQ2
n − 4(QnQ′′
n − (Q′
n)2), (1)
with Q0 = 1 and Q1 = z. From the recurrence relation, it is clear that the functions Qn are
rational, though it is far from obvious that they are polynomials, since in every iteration one
divides by Qn−1. The Yablonskii–Vorob’ev polynomials Qn are monic polynomials of degree
1
2n(n + 1), with integer coefficients. The first few are given in Table 1.
Yablonskii [1] and Vorob’ev [2] expressed the rational solutions of the second Painlevé equa-
tion,
PII(α) : w′′(z) = 2w(z)3 + zw(z) + α,
with complex parameter α, in terms of the Yablonskii–Vorob’ev polynomials, as summerized in
the following theorem:
Theorem 1. PII(α) has a rational solution iff α = n ∈ Z. For n ∈ Z the rational solution is
unique and if n ≥ 1, then it is equal to
wn =
Q′
n−1
Qn−1
− Q′
n
Qn
.
The other rational solutions are given by w0 = 0 and for n ≥ 1, w−n = −wn.
mailto:roffelse@science.ru.nl
http://dx.doi.org/10.3842/SIGMA.2010.095
2 P. Roffelsen
Table 1.
Yablonskii–Vorob’ev polynomials
Q2 =4 + z3
Q3 =−80 + 20z3 + z6
Q4 = z(11200 + 60z6 + z9)
Q5 =−6272000− 3136000z3 + 78400z6 + 2800z9 + 140z12 + z15
Q6 =−38635520000 + 19317760000z3 + 1448832000z6 − 17248000z9 + 627200z12
+18480z15 + 280z18 + z21
Q7 = z(−3093932441600000− 49723914240000z6 − 828731904000z9 + 13039488000z12
+62092800z15 + 5174400z18 + 75600z21 + 504z24 + z27)
Q8 =−991048439693312000000− 743286329769984000000z3
+37164316488499200000z6 + 1769729356595200000z9 + 126696533483520000z12
+407736096768000z15 − 6629855232000z18 + 124309785600z21 + 2018016000z24
+32771200z27 + 240240z30 + 840z33 + z36
The rational solutions of PII can also be determined, using the Bäcklund transformations,
first given by Gambier [3], of the second Painlevé equation, by
wn+1 = −wn −
2n + 1
2w2
n + 2w′n + z
, w−n = −wn,
with “seed solution” w0 = 0; see also Lukashevich [4] and Noumi [5].
We note that the Yablonskii–Vorob’ev polynomials find many applications in physics. For
instance, solutions of the Korteweg–de Vries equation (Airault, McKean and Moser [6]) and the
Boussinesq equation (Clarkson [7]) can be expressed in terms of these polynomials. Clarkson
and Mansfield [8] studied the structure of the roots of the Yablonskii–Vorob’ev polynomials Qn
and observed that the roots, of each of these polynomials, form a highly regular triangular-like
pattern, for n ≤ 7, suggesting that they have interesting properties. This further motivates
studying the zeros of the Yablonskii–Vorob’ev polynomials.
In Section 2 the divisibility of the coefficients of the Yablonskii–Vorob’ev polynomials by
powers of 4 is examined. From the divisibility properties found, we conclude that nonzero
roots of the Yablonskii–Vorob’ev polynomials are irrational. In Section 3 we study power series
expansions of (functions related to) the rational solution wn of PII(n), around poles of wn. This
leads to relations between the roots of Qn−1 and Qn. These relations suggest deeper connections
between the zeros of Qn−1 and Qn. Similarly, we look at power series expansions of (functions
related to) the rational solution wn of PII(n) around 0, in Section 4. We obtain polynomial
expressions in n, with rational coefficients, for sums of fixed negative powers of the nonzero
roots of Qn.
2 Nonzero roots are irrational
The Yablonskii–Vorob’ev polynomials Qn are monic polynomials of degree 1
2n(n + 1), and
Taneda [9] proved:
• if n ≡ 1 (mod 3), then Qn
z ∈ Z[z3];
• if n 6≡ 1 (mod 3), then Qn ∈ Z[z3].
Therefore, we have
Qn = z
1
2
n(n+1) + an
1z
1
2
n(n+1)−3 + an
2z
1
2
n(n+1)−6 + · · ·+ an
[ 1
6
n(n+1)]z
1
2
n(n+1)−3[ 1
6
n(n+1)], (2)
for certain an
s ∈ Z, with convention an
0 = 1, where [·] denotes the floor function.
Roots of the Yablonskii–Vorob’ev Polynomials 3
Lemma 1. For every 0 ≤ m ≤
[
1
6n(n + 1)
]
, we have 4m | an
m.
Proof. We proceed by proving the following statement, by induction, for all M ∈ N:
For every 1 ≤ m ≤ M , for all n ∈ N, whenever m ≤
[
1
6n(n + 1)
]
, we have 4m | an
m, and
4M | an
M+1, 4M | an
M+2, . . . , 4M | an
[ 1
6
n(n+1)].
Observe that the case M = 0 is trivial. Now suppose the statement is true for M ∈ N. Then
there are bn
s ∈ Z, such that for every n ∈ N,
Qn = z
1
2
n(n+1) + 4bn
1z
1
2
n(n+1)−3 + 42bn
2z
1
2
n(n+1)−6 + · · ·+ 4Mbn
Mz
1
2
n(n+1)−3M + 4MPn,
where Pn ∈ Z[z] is zero or has degree less or equal to 1
2n(n+1)−3(M+1), and if m >
[
1
6n(n + 1)
]
,
then bn
m = 0.
To complete the induction, we need to show that for every n ∈ N, 4 | Pn. We prove this
by induction with respect to n. Observe that P0 = 0 and P1 = 0, therefore, indeed 4 | P0 and
4 | P1. Assume 4 | Pn−1 and 4 | Pn. Then 4MPn ≡ 0 (mod 4M+1), therefore, modulo 4M+1, we
have:
zmax(0,n(n+1)−3M+1) | zQ2
n, zmax(0,n(n+1)−3M+1) | 4QnQ′′
n,
zmax(0,n(n+1)−3M+1) | 4(Q′
n)2.
By the definition of Qn+1 (1),
Qn+1Qn−1 = zQ2
n − 4
(
QnQ′′
n − (Q′
n)2
)
,
so
zmax(0,n(n+1)−3M+1) | Qn+1Qn−1 (mod 4M+1). (3)
Let us consider Qn+1Qn−1. Since 4 | Pn−1, we have
4MPn−1 ≡ 0 (mod 4M+1),
therefore, modulo 4M+1,
Qn+1Qn−1 ≡ Qn+1z
1
2
n(n−1) + Qn+1
(
4bn−1
1 z
1
2
n(n−1)−3
+ 42bn−1
2 z
1
2
n(n−1)−6 + · · ·+ 4Mbn−1
M z
1
2
n(n−1)−3M
)
. (4)
Since
Qn+1 = z
1
2
(n+1)(n+2) + 4bn+1
1 z
1
2
(n+1)(n+2)−3
+ 42bn+1
2 z
1
2
(n+1)(n+2)−6 + · · ·+ 4Mbn+1
M z
1
2
(n+1)(n+2)−3M + 4MPn+1,
we have, modulo 4M+1,
zmax(0,n(n+1)−3M+1) | Qn+1
(
4bn−1
1 z
1
2
n(n−1)−3 + 42bn−1
2 z
1
2
n(n−1)−6
+ · · ·+ 4Mbn−1
M z
1
2
n(n−1)−3M
)
.
Hence, by (3) and (4),
zmax(0,n(n+1)−3M+1) | Qn+1z
1
2
n(n−1) (mod 4M+1),
4 P. Roffelsen
which implies
zmax(0, 1
2
(n+1)(n+2)−3M) | Qn+1 (mod 4M+1).
Since
Qn+1 = z
1
2
(n+1)(n+2) + 4bn+1
1 z
1
2
(n+1)(n+2)−3
+ 42bn+1
2 z
1
2
(n+1)(n+2)−6 + · · ·+ 4Mbn+1
M z
1
2
(n+1)(n+2)−3M + 4MPn+1,
we have, therefore, 4 | Pn+1. Hence, by induction, for all n ∈ N, 4 | Pn.
The lemma follows by induction on M . �
Let us denote the coefficient of the lowest degree term in Qn by
xn := an
[ 1
6
n(n+1)],
i.e. xn is the constant coefficient in Qn if n 6≡ 1 (mod 3), and xn is the coefficient of z in Qn if
n ≡ 1 (mod 3). Fukutani, Okamoto, and Umemura [10] proved that the roots of the Yablonskii–
Vorob’ev polynomials are simple, hence xn is nonzero. Let pn be the multiplicity of 2 in the
prime factorization of xn. As a consequence of Lemma 1, we obtain that pn ≥ 2
[
1
6n(n + 1)
]
.
We prove
pn =
[
1
3
n(n + 1)
]
.
Observe that xn = Qn(0) if n 6≡ 1 (mod 3), and xn = Q′
n(0) if n ≡ 1 (mod 3). Fuku-
tani, Okamoto, and Umemura [10] derived the following identity for the Yablonskii–Vorob’ev
polynomials:
Q′
n+1Qn−1 −Qn+1Q
′
n−1 = (2n + 1)Q2
n.
Using this identity at 0, we obtain
xn+1xn−1 =
{
(2n + 1)x2
n if n ≡ 0 (mod 3),
−(2n + 1)x2
n if n ≡ 2 (mod 3).
By evaluating equation (1) at 0,
xn+1xn−1 = 4x2
n, if n ≡ 1 (mod 3).
Therefore, we have the following recursion for (xn)n:
x0 = 1, x1 = 1 and
xn+1xn−1 =
(2n + 1)x2
n if n ≡ 0 (mod 3),
4x2
n if n ≡ 1 (mod 3),
−(2n + 1)x2
n if n ≡ 2 (mod 3).
So, we obtain the following recursion for (pn)n:
p0 = 0, p1 = 0 and
pn+1 =
{
2pn − pn−1 if n 6≡ 1 (mod 3),
2 + 2pn − pn−1 if n ≡ 1 (mod 3).
Using this recursion, the formula pn =
[
1
3n(n + 1)
]
, can be proven directly, by induction.
Roots of the Yablonskii–Vorob’ev Polynomials 5
Remark 1. Kaneko and Ochiai [11] found an explicit expression for the coefficients xn. But
deriving the formula pn =
[
1
3n(n + 1)
]
directly from this expression seems to be a difficult task.
Theorem 2. The nonzero roots of the Yablonskii–Vorob’ev polynomials are irrational.
Proof. Let n 6≡ 1 (mod 3). Suppose x is a rational root of Qn. Since Qn ∈ Z[z] is monic, by
Gauss’s lemma, x ∈ Z. By Lemma 1,
Qn ≡ z
1
2
n(n+1) (mod 4),
so x is even. Let y := x
2 , then, by equation (2),
0 = (2y)
1
2
n(n+1) + an
1 (2y)
1
2
n(n+1)−3 + an
2 (2y)
1
2
n(n+1)−6 + · · ·+ an
1
6
n(n+1)−1
(2y)3 + an
1
6
n(n+1)
.
By Lemma 1, for every m ≤ 1
6n(n + 1), we have 4m | an
m. Hence
2
1
2
n(n+1) | (2y)
1
2
n(n+1), 2
1
2
n(n+1)−1 | an
1 (2y)
1
2
n(n+1)−3,
2
1
2
n(n+1)−2 | an
2 (2y)
1
2
n(n+1)−6, . . . , 2
1
2
n(n+1)− 1
6
n(n+1)+1 | a 1
6
n(n+1)−1(2y)3.
So
2
1
3
n(n+1)+1 | an
1
6
n(n+1)
= xn,
which implies
pn ≥
1
3
n(n + 1) + 1.
But pn = 1
3n(n + 1), a contradiction, hence roots of Qn are irrational.
If n ≡ 1 (mod 3), we can apply the same reasoning to Qn
z , and show that roots of Qn
z are
irrational. Therefore, nonzero roots of Qn are irrational. �
This result raises the question whether the Yablonskii–Vorob’ev polynomials, excluding the
trivial factor z in case n ≡ 1 (mod 3), are irreducible in Q[z]. Kametaka [12] showed that for
n ≤ 23, the Yablonskii–Vorob’ev polynomials Qn are indeed irreducible.
3 Relations between roots of the Yablonskii–Vorob’ev
polynomials
By Theorem 1, for n ≥ 1, the unique rational solution of PII(n) is given by
wn =
Q′
n−1
Qn−1
− Q′
n
Qn
.
Fukutani, Okamoto, and Umemura [10] proved that the roots of the Yablonskii–Vorob’ev poly-
nomials are simple, hence
wn =
1
2
n(n−1)∑
k=1
1
z − zn−1,k
−
1
2
n(n+1)∑
k=1
1
z − zn,k
, (5)
where the zm,k are the roots of Qm. From equation (5) and the fact that wn is the rational
solution of PII(n), we obtain relations between the zeros of Qn−1 and Qn.
6 P. Roffelsen
Theorem 3. For 1 ≤ j ≤ 1
2n(n− 1):
1
2
n(n−1)∑
k=1, k 6=j
1
zn−1,j − zn−1,k
−
1
2
n(n+1)∑
k=1
1
zn−1,j − zn,k
= 0,
1
2
n(n−1)∑
k=1, k 6=j
1
(zn−1,j − zn−1,k)2
−
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)2
=
zn−1,j
6
,
1
2
n(n−1)∑
k=1, k 6=j
1
(zn−1,j − zn−1,k)3
−
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)3
= −n + 1
4
,
1
2
n(n−1)∑
k=1, k 6=j
1
(zn−1,j − zn−1,k)5
−
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)5
= zn−1,j
(
n + 1
24
− 1
36
)
.
For 1 ≤ j ≤ 1
2n(n + 1):
1
2
n(n−1)∑
k=1
1
zn,j − zn−1,k
−
1
2
n(n+1)∑
k=1, k 6=j
1
zn,j − zn,k
= 0,
1
2
n(n−1)∑
k=1
1
(zn,j − zn−1,k)2
−
1
2
n(n+1)∑
k=1, k 6=j
1
(zn,j − zn,k)2
= −zn,j
6
,
1
2
n(n−1)∑
k=1
1
(zn,j − zn−1,k)3
−
1
2
n(n+1)∑
k=1, k 6=j
1
(zn,j − zn,k)3
= −n− 1
4
,
1
2
n(n−1)∑
k=1
1
(zn,j − zn−1,k)5
−
1
2
n(n+1)∑
k=1, k 6=j
1
(zn,j − zn,k)5
= zn,j
(
n− 1
24
+
1
36
)
.
Proof. Let 1≤j≤ 1
2n(n−1) and define ω := zn−1,j and u := wn− 1
z−ω . Since gcd(Qn−1, Qn) = 1,
see Fukutani, Okamoto, and Umemura [10], equation (5) shows that u is holomorphic in a neigh-
bourhood of ω. Hence u has a power series expansion, say
∞∑
m=0
am(z − ω)m,
which converges in an open disc centered at ω.
Since wn is a solution of PII(n), u satisfies
(z − ω)2u′′ = 6u + 6(z − ω)u2 + 2(z − ω)2u3 + (n + 1)(z − ω)2 + ω(z − ω)
+ (z − ω)3u + ω(z − ω)2u.
Hence we have the following identity in an open disc centered at ω:
∞∑
m=2
(m− 1)mam(z − ω)m = 6
∞∑
m=0
am(z − ω)m + 6(z − ω)
( ∞∑
m=0
am(z − ω)m
)2
+ 2(z − ω)2
( ∞∑
m=0
am(z − ω)m
)3
+ (n + 1)(z − ω)2 + ω(z − ω)
Roots of the Yablonskii–Vorob’ev Polynomials 7
+ (z − ω)3
∞∑
m=0
am(z − ω)m + ω(z − ω)2
∞∑
m=0
am(z − ω)m.
By considering coefficients of (z − ω)n, n = 0, 1, 2, 4, it is easy to deduce that a0 = 0, a1 = −ω
6 ,
a2 = −n+1
4 and a4 = ω
(
n+1
24 − 1
36
)
. Note that a3 does not follow from considering coefficients
of (z − ω)3.
By Taylor’s theorem and equation (5),
am =
u(m)(zn−1,j)
m!
= (−1)m
1
2
n(n−1)∑
k=1, k 6=j
1
(zn−1,j − zn−1,k)m+1
−
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)m+1
.
The first half of the theorem follows, the second half is proved analogously. �
Note that countably many nontrivial relations can be found between the am in the above
proof, by considering the coefficient of (z − ω)n, for n ∈ N.
In Kudryashov and Demina [13] similar relations for the roots of Qn are obtained using
the Korteweg–de Vries equation. In particular, the following results are presented in [13] for
1 ≤ j ≤ 1
2n(n + 1):
1
2
n(n+1)∑
k=1, k 6=j
1
(zn,j − zn,k)2
= −zn,j
12
,
1
2
n(n+1)∑
k=1, k 6=j
1
(zn,j − zn,k)3
= 0,
1
2
n(n+1)∑
k=1, k 6=j
1
(zn,j − zn,k)5
= −zn,j
144
.
From these relations and Theorem 3, we obtain the following corollary:
Corollary 1. For 1 ≤ j ≤ 1
2n(n− 1):
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)2
= −zn−1,j
4
,
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)3
=
n + 1
4
,
1
2
n(n+1)∑
k=1
1
(zn−1,j − zn,k)5
= −zn−1,j
(
n + 1
24
− 1
48
)
.
For 1 ≤ j ≤ 1
2n(n + 1):
1
2
n(n−1)∑
k=1
1
(zn,j − zn−1,k)2
= −zn,j
4
,
1
2
n(n−1)∑
k=1
1
(zn,j − zn−1,k)3
= −n− 1
4
,
1
2
n(n−1)∑
k=1
1
(zn,j − zn−1,k)5
= zn,j
(
n− 1
24
+
1
48
)
.
In Theorem 3, we have obtained 4 times 1
2n(n− 1) plus 4 times 1
2n(n+1) equations satisfied
by the 1
2n(n+1) roots of Qn, suggesting that these equations can be used to determine the roots
of the polynomials Qn recursively. If so, then these equations may be of use to derive properties
of the roots of the Yablonskii–Vorob’ev polynomials. We shall not pursue this issue further here.
8 P. Roffelsen
4 Sums of negative powers of roots
In Section 2, the rational solutions wn of PII(n) were studied around roots of the Yablonskii–
Vorob’ev polynomials. In this section, we consider wn at 0.
Let n ≡ 0 (mod 3), then 0 is not a root of Qn−1 or Qn. Therefore, by equation (5), wn is
holomorphic in a neighbourhood of 0. So wn has a power series expansion, say
∞∑
m=0
amzm,
which converges on an open disc centered at 0.
By Taylor’s theorem and equation (5), we have
am = −
1
2
n(n−1)∑
k=1
1
zm+1
n−1,k
−
1
2
n(n+1)∑
k=1
1
zm+1
n,k
.
Let ω := e
2πi
3 . Since n ≡ 0 (mod 3), Qn ∈ Z[z3]. Therefore, the roots of Qn are invariant under
multiplication by ω. Hence
1
2
n(n+1)∑
k=1
1
zm+1
n,k
=
1
2
n(n+1)∑
k=1
1
(ωzn,k)m+1
=
1
ωm+1
1
2
n(n+1)∑
k=1
1
zm+1
n,k
,
therefore, if m 6≡ 2 (mod 3),
1
2
n(n+1)∑
k=1
1
zm+1
n,k
= 0. (6)
By the same reason, if m 6≡ 2 (mod 3),
1
2
n(n−1)∑
k=1
1
zm+1
n−1,k
= 0.
So am = 0, if m 6≡ 2 (mod 3), and in an open disc centered at 0,
wn(z) =
∞∑
m=0
a3m+2z
3m+2.
Since wn is a solution of PII(n), we have the following identity in an open disc centered at 0:
∞∑
m=0
(3m + 1)(3m + 2)a3m+2z
3m = 2
( ∞∑
m=0
a3m+2z
3m+2
)3
+
∞∑
m=0
a3m+2z
3m+3 + n.
Comparing coefficients gives a2 = 1
2n, a5 = 1
40n and a8 = 1
2240n + 1
224n3. We have obtained the
following relations for n ≡ 0 (mod 3):
1
2
n(n−1)∑
k=1
1
z3
n−1,k
−
1
2
n(n+1)∑
k=1
1
z3
n,k
= −n
2
,
1
2
n(n−1)∑
k=1
1
z6
n−1,k
−
1
2
n(n+1)∑
k=1
1
z6
n,k
= − n
40
,
Roots of the Yablonskii–Vorob’ev Polynomials 9
1
2
n(n−1)∑
k=1
1
z9
n−1,k
−
1
2
n(n+1)∑
k=1
1
z9
n,k
= − 1
2240
n− 1
224
n3.
If n ≡ 1 (mod 3), then u := wn + 1
z is holomorphic at 0 and satisfies
z2u′′ = 6u− 6zu2 + 2z2u3 + z3u + (n− 1)z2.
By considering the power series expansion of u = wn + 1
z around 0, the following relations are
found:
1
2
n(n−1)∑
k=1
1
z3
n−1,k
−
1
2
n(n+1)∑
k=1, zn,k 6=0
1
z3
n,k
=
1
4
(n− 1),
1
2
n(n−1)∑
k=1
1
z6
n−1,k
−
1
2
n(n+1)∑
k=1, zn,k 6=0
1
z6
n,k
=
1
56
(n− 1) +
3
112
(n− 1)2,
1
2
n(n−1)∑
k=1
1
z9
n−1,k
−
1
2
n(n+1)∑
k=1, zn,k 6=0
1
z9
n,k
=
1
2800
(n− 1) +
9
5600
(n− 1)2 +
1
448
(n− 1)3.
If n ≡ 2 (mod 3), then u := wn − 1
z is holomorphic at 0 and satisfies
z2u′′ = 6u− 6zu2 + 2z2u3 + z3u + (n + 1)z2.
By considering the power series expansion of u = wn − 1
z around 0, the following relations are
found:
1
2
n(n−1)∑
k=1, zn−1,k 6=0
1
z3
n−1,k
−
1
2
n(n+1)∑
k=1
1
z3
n,k
=
1
4
(n + 1),
1
2
n(n−1)∑
k=1, zn−1,k 6=0
1
z6
n−1,k
−
1
2
n(n+1)∑
k=1
1
z6
n,k
=
1
56
(n + 1)− 3
112
(n + 1)2,
1
2
n(n−1)∑
k=1, zn−1,k 6=0
1
z9
n−1,k
−
1
2
n(n+1)∑
k=1
1
z9
n,k
=
1
2800
(n + 1)− 9
5600
(n + 1)2 +
1
448
(n + 1)3.
Remark 2. Considering higher order coefficients, we see that for every threefold m ≥ 3,
polynomial expressions in n, with rational coefficients, depending on n (mod 3), exist for
1
2
n(n−1)∑
k=1, zn−1,k 6=0
1
zm
n−1,k
−
1
2
n(n+1)∑
k=1, zn,k 6=0
1
zm
n,k
.
As a corollary of these relations, by induction, we obtain:
1
2
n(n+1)∑
k=1, zn,k 6=0
1
z3
n,k
=
n
4
if n ≡ 0 (mod 3),
0 if n ≡ 1 (mod 3),
−n + 1
4
if n ≡ 2 (mod 3),
10 P. Roffelsen
1
2
n(n+1)∑
k=1, zn,k 6=0
1
z6
n,k
=
1
40
n2 +
1
80
n if n ≡ 0 (mod 3),
− 1
560
n2 − 1
560
n +
1
280
if n ≡ 1 (mod 3),
1
40
n2 +
3
80
n +
1
80
if n ≡ 2 (mod 3),
1
2
n(n+1)∑
k=1, zn,k 6=0
1
z9
n,k
=
n + 7n2 + 10n3
4480
if n ≡ 0 (mod 3),
2− n− n2
22400
if n ≡ 1 (mod 3),
−20− 85n− 115n2 − 50n3
22400
if n ≡ 2 (mod 3).
By Remark 2, for every threefold m ≥ 3, polynomial expressions in n, with rational coefficients,
depending on n (mod 3), exist for
1
2
n(n+1)∑
k=1, zn,k 6=0
1
zm
n,k
.
If m 6≡ 0 (mod 3), see equation (6), then
1
2
n(n+1)∑
k=1, zn,k 6=0
1
zm
n,k
= 0.
So, for all n, m ∈ N,
1
2
n(n+1)∑
k=1, zn,k 6=0
1
zm
n,k
∈ Q,
even though the nonzero roots of the Yablonskii–Vorob’ev polynomials are irrational.
Acknowledgements
I wish to thank Erik Koelink for his enlightening discussions and introducing me to the world
of the Painlevé equations. I am also grateful to Peter Clarkson for his interest and useful links
to the literature.
References
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Roots of the Yablonskii–Vorob’ev Polynomials 11
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http://dx.doi.org/10.1142/S0219530508001250
http://dx.doi.org/10.1088/0951-7715/16/3/201
http://dx.doi.org/10.2969/jmsj/1191418760
http://arxiv.org/abs/math.QA/0205178
http://dx.doi.org/10.1007/BF03167047
http://dx.doi.org/10.1016/j.physleta.2007.03.081
http://arxiv.org/abs/nlin.SI/0610058
1 Introduction
2 Nonzero roots are irrational
3 Relations between roots of the Yablonskii-Vorob'ev polynomials
4 Sums of negative powers of roots
References
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