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EDN: PCXYNS УДК 517.55
On Multiple Zeros of Entire Functions of Finite Order of Growth
Alexander M. Kytmanov* Olga V. Khodosf
Siberian Federal University Krasnoyarsk, Russian Federation
Received 10.07.2022, received in revised form 15.09.2022, accepted 20.10.2022 Abstract. The article is devoted to determination of the number of multiple zeros of entire functions of finite order of growth. Keywords: entire function, multiple zero.
Citation: A.M. Kytmanov, O.V. Khodos, On Multiple Zeros of Entire Functions of Finite Order of Growth, J. Sib. Fed. Univ. Math. Phys., 2023, 16(2), 239-244. EDN: PCXYNS.
Our work is devoted to the problem of multiple zeros of entire functions. For polynomials, this question is a classical problem, and its solution is included in algebra textbooks (see, for example, [1]).
Recall the statement. Consider a polynomial P(z) of degree n. Denote by Sj the power sums of the roots of a polynomial of degree j.
Theorem 1. In order for the polynomial P(z) to have multiple roots, it is necessary and sufficient that
n Si S2 Si S2 S3
D(P ) =
,2n-2
Sn — 1
Sn
S2
n2
0.
Sn — 1 Sn Sn+i
Here a0 is the leading coefficient of the polynomial P(z).
The determinant of D(P) is called the discriminant of the polynomial P(z). For entire functions, the question of multiple zeros needs to be clarified. An entire function may have no zeros at all, like, for example, the function ez, or an infinite number of zeros like sin z. Therefore, we have to consider various options here. 1. Let an entire function have the form
E<
k=0
f (z) =7 ya,k z-
f (0)= ao = 1.
(1)
The following statement is true ([2], corollary 1.4.1).
Theorem 2. In order for the function f (z) to be an entire function of finite order k0 that has no zeros, it is necessary and sufficient that the determinant
D(P)
2n 2
a1 a0 0. .0
2a2 a1 a0 . .0 = 0 for all h > h0, (2)
hau au-i au-2 . . ai
*akytmanov<8sfu-kras.ru https://orcid.org/0000-0002-7394-1480 [email protected] © Siberian Federal University. All rights reserved
0
0
where k0 is the minimal number with this property.
2. Consider an entire function of finite order of growth of the form (1). Find the order p of the function f. To do this, we apply the formula ([3], ch. 7)
M ln(l/|aw|) = 1 . n^w n ln n p'
If p is a fractional number, then the function f (z) is known to have an infinite number of zeros (see [3]). First we will assume that p is an integer.
Let us take a sequence of complex numbers so, s2, ... . It defines an infinite Hankel matrix
S =
s0 s1 s2 ...
S1 s2 s3
S2 S3 S4
...
(3)
The consecutive main minors of the matrix S are denoted by D0, D1, D2, ... .In addition, we set D-1 = 1.
If for every p G N there exists a minor of the matrix S of order p that is not equal to zero, then the matrix has infinite rank. If, starting from some p, all minors of larger orders are zero, then the matrix S has finite rank. The smallest such p is called the rank of the matrix. We recall a statement regarding matrices S of finite rank p ([4], ch. 16, Sec. 10).
Theorem 3. If an infinite Hankel matrix has finite rank p, then the minor Dp-1 = 0.
Thanks to the properties of entire functions, the power sums of ak
w i
ak = E ^ , k G N
n=1 an
are absolutely convergent series for k > p. Here, the zeros of the entire function f (z) are denoted by an. We will arrange them in ascending order of modules 0 < < la2l < ... < lanl .. .
The smallest such k is denoted by k0 and we denote sj = a2ko+j, j =0,1,.... Consider an infinite Hankel matrix S of the form (3). In the monograph ([2], Theorem 1.4.5) the following statement is proved
Theorem 4. In order for the function f to have a finite number of zeros, it is necessary and sufficient that the rank of the matrix S is finite, while the number of different zeros of f is equal to the rank of S.
In this case, we can write the function f as follows:
f (z) = e-Q(z) P(z), (4)
where Q(z) is a polynomial of degree p = p, and P(z) is a polynomial of some degree m
P (z) = J2 bk zk = 1 + b1 z + ... + bm zm
k=0
The number m is the number of roots of the function f(z) together with their multiplicities. To find the polynomial P(z), one needs to factorize the function f (z) (see Sec. 1.6.5 from [2]). Take the logarithm of both parts in the formula (4). Let
ln f (z) = ^ àk zk = ài z + ... + an zn + ...,
k=i
ln P (z) = ^2~hk zk = bi z + .. . + bn zn + ....
k=i
The coefficients of an can be found by the following formula (see [2], Lemma 1.2.1)
, _(-1)n-1
ài 1 0 . .0
2à2 ài 1 . .0
3à3 à2 ài . . .0 = 0 for all n > 1,
nan an-i an-2 . . ài
The coefficients bk are found from the theorem ( [2], Theorem 1.6.4, [6]).
Theorem 5. The formulas are valid
(m + p) am+p ... (m + p + 1) am+p+i ... (p + 1) ap+i
(m + p + 1) am+p+i ... (m + p + 2) am+p+2 ... (p + 2) àp+2
bk =
(2m + p - 1) a2m+p-i ... (2m + p) a2m+p ... (p + m) ap+m
(m + p) am+p (m + p + 1) am+p+i
(p + 1) àp+i (p + 2) àp+2
(2m + p - 1) a2m+p-i ... (p + m) ap+m
k = 1,... ,m. In the numerator k, the th column is replaced by the column
i-(m + p + 1) am+p+A -(m + p + 2) am+p+2
\ -(2m + p) a2m+p J
Here m this is the smallest k for which bk is different from zero.
Corollary 1. A function f (z) has multiple roots if and only if the polynomial P(z) has multiple roots.
Let us give an example. Consider the function
/2k 2k-2 \ f <z' = 1+2z+>:(«- V-2)! )
k=2
1 + 2z + z2
2z3 4z4 "3 3~
16z5
+ ... .
It is not difficult to calculate that the order of growth of this function is p = 1. By Lemma 1.2.1 of [2], the power sums of Sj with even numbers are 2, with odd numbers are 0. Therefore, the rank of the Hankel matrix is S
S
/2 0 2
0 2 0
2 0 2
... ... ...
...
...
is equal to 2.
n
n
k
z
From Theorem 5 and Lemma 1.2.1 from [2] we get
-1
02 =
03 =
2 1
2 2
2 1 0 2 2 1 212
-1,
04 =
-1 T
21 22 -2 1 2
00 10 1
16 2 i o
"T -3 12
05 =
2 1 0 2 2 1 -2 1 2 1
16 " 3
16
00 00 10 21
-16 -4 12
3 3 3 1 2
From here we find
bi = -
4 a 4 202 -2 -2
5 «5 3O3 0 0
3o3 202 0 -2
4O4 3O3 -2 0
b2 = -
3O3 4O4 0 -2
4O4 505 -2 0
3O3 202 0 -2
4O4 3O3 -2 0
-1.
The remaining bk is zero. Therefore, the polynomial P(z) is equal to
P(z) = 1 - z2.
It has two roots ±1 and has no multiple roots. Therefore, the function f (z) has no multiple roots.
3. Let an function f (z) of the form (1) have an infinite number of zeros, then the rank of the matrix S (3) is infinite. Multiple zeros can only have finite multiplicities. Therefore, if f (z) has an infinite number of zeros, then it has an infinite number of distinct zeros.
Multiple zeros are the common zeros of the function and its derivative, i.e., the zeros of the resultant. So the question is whether the function and its derivative have common zeros.
The approach to determining the resultant of two integer functions is considered in a number of papers [5-7], but for arbitrary entire functions of finite growth order it is not yet known how to find the common zeros of the function and its derivative.
Let an entire function f (z) have the order p. Due to the properties of entire functions, power sums ak
~ 1
TO
ak = J2 ak,
n=1 an
fee N,
are absolutely convergent series for k > p. Here, as before, an are zeros of the entire function f (z). We will arrange them in ascending order of modules 0 < |ai| < |a2| < ... < \a.n| .. . The smallest such k is denoted by k0. We assume that k0 is an integer.
0
0
0
We will introduce, as in the previous section, power sums sj infinite Hankel matrix S of the form (3). Its rank is infinite. Consider its submatrices of the order m:
a2k°+j, j = 1
and an
S 2
We introduce finite power sums
a.
2k°+j
and matrices
S m Sm
Consider an infinite matrix
\sm sm+1 sm+2
so s1 s2. . . sm
s1 s2 s3. - - Sm+1
s2 S3 S4 . - - Sm+2
1 an n=1 n
m Sm+1
m Sm+2
A
S2m+1)
Y.—, k e N,
f 4 r\k
s
m m m
m+1 m
m+2
S2m+1/
/ ^ 1 k o 1 k° a3° ...
1 1 1
ak°+i a k ° + 1 ak°+1
1 1 1
ak°+2 v... ak°+'2 ak3°+3 ...
(5)
(6)
(7)
Then we have
S = A-A'
where A' is the transpose of the matrix A. If the function f has multiple zeros, then the matrix A has the same columns.
Denote by Am the main submatrix of the matrix A of order m. Then Sm = Am ■ A'm. If the function f (z) has multiple zeros, then det Am = 0, starting from some m. Since the matrix Am is a Vandermonde matrix up to a nonzero multiplier, the opposite is also true: if its determinant is 0, at least two of its columns coincide. Thus, the next statement is true.
Lemma 1. In order for the function f (z) to have multiple zeros, it is necessary and sufficient that det Am = 0 starting from some m.
Since Sm = Am ■ Am, the following statement is true
Proposition 1. In order for the function f (z) to have multiple zeros, it is necessary and sufficient that det Smm = 0 starting from some m.
In order to find det Sm, we first need to factorize the function f (see point 2). Suppose that after factorization, the function f (z) takes the form
' w = g(1 - j-
m a=
k
m
m
s
j
m
s
m
Thus, the function f (z) is a function of genus zero (or an entire function of the first order of growth of minimal type ([3], Chapter 7). In this case, the series
E-
a, j=0 j
absolutely converges. Then the coefficients ak of the function f (z) take the form
ak = 5>l)fc---•
k=0
The work was supported by the Krasnoyarsk Mathematical Center, funded by the Ministry of Education and Science of the Russian Federation (Agreement 075-02-2022-876).
References
[1] A.Kurosh, Higher algebra, Moscow, Mir, 1972.
[2] A.M.Kytmanov, Algebraic and transcendental systems of equation, Krasnoyarsk, SFU, 2019 (in Russian).
[3] A.Markouchevitch, Theory of Functions of a Complex Variable, Vol. 2, Chelsea Publishing Series, 1977.
[4] F.Gantmacher, The Theory of Matrices, Chelsea Publishing Series, 1959.
[5] N.G.Chebotarev, Collected works, Vol. 2, Moscow-Leningrad, AN SSSR, 1949 (in Russian).
[6] A.M.Kytmanov, Ya.M.Naprienko, One approach to finding the resultant of two entire function, Complex variables and elliptic equations, 62(2017), 269-286.
DOI: 10.1080/17476933.2016.1218855
[7] A.M.Kytmanov, O.V.Khodos, An Approach to the Determination of the Resultant of Two Entire Functions, Russian Mathematics, 62(2018), 42-51.
DOI: 10.3103/S1066369X18040059
[8] M.G.Krein, M.A.Naimark, The Method of Symmetric and Hermitian Forms in the Theory of the Separation of the Roots of Algebraic Equation, Linear and Multilinear Algebra, 10(1981), no. 4, 265-308. DOI: 10.1080/03081088108817420
[9] E.I.Jury, Inners and stability of dynamic system, Wiley, New York-London-Sydney-Toronto, 1974.
Кратные нули целых функций конечного порядка роста
Александр М. Кытманов Ольга В. Ходос
Сибирский федеральный университет Красноярск, Российская Федерация
Аннотация. Статья посвящена определению числа кратных нулей целой функции конечного порядка роста.
Ключевые слова: целая функция, кратный нуль.
