CC BY
5
Journal of Siberian Federal University. Mathematics & Physics 2024, 17(1), 18—26
EDN: BXZVAF УДК 517.538.5
Chebyshev Polynomials with Zeros outside the Open Arc Segment
Natalia N. Rybakova*
Siberian Federal University Krasnoyarsk, Russian Federation
Received 18.06.2023, received in revised form 04.09.2023, accepted 14.11.2023 Abstract. The problem of unitary polynomials of degree n with real coefficients least deviating from zero on an arbitrary fixed arc of a circle with a zero set outside an open segment of the same arc is considered. The description of the extremal polynomials of the solution of this problem is given and their norm depending on the degree of the polynomial and the arc length is obtained. Keywords: Chebyshev polynomials, polynomials least deviating from zero, zero set, polynomials with real coefficients.
Citation: N.N. Rybakova, Chebyshev Polynomials with Zeros outside the Open Arc Segment, J. Sib. Fed. Univ. Math. Phys., 2024, 17(1), 18-26. EDN: BXZVAF.
Let K be some compact complex plane, An be a class of unitary polynomials of degree n. For any polynomial Pn(z) from An norm on a compact K, we define the current: ||Pn||K = = max\Pn(z)\, with the symbol Z(Pn) denote the set of all its zeros.
The problem of finding the polynomials least deviating from zero by K with a zero set on some fixed subset of D c C is posed as follows: find the number En(K,D) = inf{IPnIK | Pn e An, Z(Pn) c D} and a polynomial Pn such that PIK = En(K, D), Z(P*) C D, which is called the extremal or Chebyshev polynomial. Finding such polynomials and other similar problems have been considered by many mathematicians (see, for example, [1-7]).
Chebyshev polynomials considered on the arc of a circle, without restriction on the location of zeros in the complex plane, were studied N. I. Akhiezer and many others (see, for example, [8]). For a narrower problem, with zeros on the arc of a circle, L. S. Maergoiz et al. [9] determined the extremal polynomials and their norm.
This article is devoted to finding the norm of Chebyshev polynomials with real coefficients and a zero set outside an open arc segment.
The author is grateful to his teacher L. S. Maergoiz for the task.
1. Two problems of finding extreme polynomials
Consider the arc ra = {z = eip e C : ^ a}, where 0 < a < n, interval Ga = = {z = cos a + iy : \y\ < sin a}, Ua — open arc segment ra.
We define two classes of polynomials: unitary polynomials with real coefficients with a zero set outside the open segment of the arc are
3n = {Pn e An : Z(Pn) C C \ Ua}
* nrybakova@sfu-kras.ru © Siberian Federal University. All rights reserved
and unitary polynomials with real coefficients with zero set on the boundary of the arc segment are
jn = {Pn e An : Z(Pn) C ra U Ga}
with norm
\\Pn\\ = max \Pn(z)\ = max |Pn(e^)|.
zEL a
Task A: Find the number En,a = infPn\\Pn\\ = minPn£^n \\Pn\\ and the polynomial Pn with the norm \\P,*\\ = En,a, which is called the extremal or Chebyshev polynomial.
Task B: Find the number Ena = inf ^ \\Pn\\ = minP ^ \\Pn\\ and the polynomial P* with the norm \\P,*\\ = Ena.
The main result of the work: Theorem 1. An extreme polynomial vacation abroad that can be seen
r
Tn(z) = slr) := n(z2 - 2akr)z + 1) for even n = 2r, k = l
Tn(z) = (z - 1)^nr) for odd n = 2r + 1, where a(n) = 1 - 2cos2 ^^ 1) sin2(a/2) k =1,...,r, \\Tn\\ =2sinn (a/2).
Remark. For n = 2 and cos a ^ 0, the solution of the problem is not unique: in addition to the polynomial T2, there is another extreme polynomial F2(z) = z2 — 2z cos a + cos a.
The polynomials from Theorem 1 and their norm were first written out in the article [9] in connection with solving the problem of finding extreme polynomials on the arc of a unit circle with a zero set coinciding with this arc.
The proof of the theorem relies on a number of lemmas.
Lemma 1. Tasks A and B have the same solution.
Proof. If the polynomial Pn(z) belongs to the class j3n, but does not belong to the class j3n, then there is at least one root of it that does not belong to ra U Ga.
Let b0 e ra U Ga be such a root of the polynomial Pn(z). Consider the polynomial Pn(z) such that Z(Pn) = Z(Pn)/b0 U b0 and \\Pn\\ < \\Pn\\. There are three possible cases:
1. \bo\ = \r • eiY\ > 1 h Re(b0) > cos a; 60 = eiY e ra.
2. Re(b0) < cos a h Im(b0) ^ sin a; b0 = cos a + iIm(b0) e Ga.
3. Re(b0) < cos a h Im(b0) > sin a; b0 = cos a + i sin ae ra. □
2. The case of an even number of roots on G
a
Consider the following result. Lemma 2. For polynomials
L2(z) = (z — cos a — iy)(z — cos a + iy),
where 0 ^ y < sin a, and
P2(z) = (z — e^ )(z — e-i^),
where ^ is determined from the equality cos ^ = (sin2 a — y2)/2 + cos a, the inequality \L2(z)\ ^ \P2(z)\ is valid for all z e ra, equality is achieved only at points 1 and e±ia.
Proof. Denote \L2(eip)\2 := q(cosy>), \P2(eip)\2 := p(cosy>). Note that p(cosa) = q(cosa) and p(1) = q(1), by this \L2(z)\ = \P2(z)\ for z e {1; ena}.
The inequality p(t) — q(t) < 0 holds for all t from the interval (cos a; 1), because cos a and 1 are the roots of a square trinomial
p(t) — q(t) = 4(sin2 a — y2)(t2 — t(cos a + 1) + cos a),
and the multiplier sin2 a — y2 > 0 according to the lemma, all this indicates that the control is unmanageable \L2(z)\ > \P2(z)\ for everything z e ra \ {1; ena}. □
Let's consider a few lemmas-consequences.
Lemma 3. For the norm of polynomials of the form
L2(z) = (z — cos a — iy)(z — cos a + iy),
where 0 ^ y < sin a, the inequality is valid \L2(z)\ ^ 2 sin2 a/2, and the equality holds only for polynomials F2 (z) = z2 — 2z cos a + cos a and only in the case of 0 ^ a ^ n/2.
Proof. The norm of the polynomial P2 described in Lemma 2 is equal to 2 sin2 a/2, if and only if
it coincides with the polynomial T2 (see Theorem 1), and in other cases is greater than this value.
(2)
We define the polynomial L2 of Lemma 2 when P2 = T2. Equating cos y and a,k , described in
1 + cos a 2 2
Lemma 2 and Theorem 1, respectively, we get cos y = -2-, hence y = cos a — cos a ^ 0,
y2 is non-negative only for 0 ^ a ^ . The desired polynomial is — L2(z) = z2 — 2z cos a + cos a. It is not difficult to check that its norm is
2 sin2 a/2. □
Lemma 4. For any polynomial Pn e 3n, where n > 2, having an even number of roots on Ga (taking into account multiplicity) and at least two roots, the inequality ||Pn|| > 2sinn a/2 is valid.
Proof. If we consider any polynomial Pn e 3n, where n ^ 2, having an even number of roots on Ga (taking into account multiplicity), then the inequality ||Pn|| > 2sinn a/2 is valid for it, because if we replace each pair of complex-conjugate roots from Ga of the polynomial Pn e 3n with a pair of roots from ra described in Lemma 2, we get the polynomial Pn(z).
Because if Z(Pn) e ra, then 0y > 2sinn a/2 (see [9]). By lemma 2 \Pn(z)\ < \Pn(z)\ for all z e ra. Consequently, HPnH > HPnH > 2sinn a/2.
If Pn = Tn, what exists means £ such that £ =1, £ = e±ia and \Tn(£)\ = 2sinn a/2 (the paper [10] describes the properties of such polynomials, in particular, such as: there is n +1 points on ra such that at these points the module of the polynomial Tn is equal to its norm). This is possible because n > 2. Therefore, the inequality
UPnU > \Pn(£)\ > \Pn(£)\ = \Tn(£)\ = 2sinn a/2
is valid.
If the polynomials Pn(z) and Tn(z) are not identically equal, then the inequality holds for the norm of the polynomial
UPn|| > UPnU > !Tn! = 2sinn a/2. □
3. Properties of Chebyshev polynomials on a segment
To solve our problem, we need several well-known facts about polynomials with real coefficients, including weight functions, on the segment [—1; 1] of the real axis. Let's write down a special case of the theorem [11, Chapter II, page 66].
k
Lemma 5. Let s(x) be a weight function and a family of polynomials Qk (x) = (x — rj), then
j=1
for an extremal polynomial Qk(x), such that
lls(x) • Qk (x)ll[-1;1] = min ||s(x) • Q k(x)||[ —1;1], Qk(x)
the statement is true: there are k + 1 points Xj ,j = 1,... ,k +1, such that \s(Xj) • Qk (Xj )| = ||s(x) • Qk (x)||— i;i], j = 1,...,k + 1.
1 n
Properties of Chebyshev polynomials tn(x) = 0n_ 1 cos(n arccos x) = EI (x — Yj) on the
2n j=i 1 n
segment [—1; 1] can be viewed in [12], in particular, ||tn|| —1;1] = _ 1, Yj =0 if Vj, j =
2n j=1
n— 1
= 1,... ,n — 1 are roots of the derivative of the Chebyshev polynomial, then ^ nj = 0.
j=1
Let —1 = no, 1 = nn, the Chebyshev polynomial deviates most from zero on the segment [—1; 1] at the points nj, j = 0,... ,n, and if n is odd, then tn(1) = tn(nn—2) = ... = tn (n1) = 2^—1
and tn( — 1) = tn(n2) = ... = tn (nn—1) = — 2—1; if n is even, then tn (m) = tn(ns) = ... =
tn(nn—1) = — 2—1 and tn( —1) = tn(n2) = ... = tn(1) = 2^11.
Lemma 6. If the norm of the polynomial is f (x) on the segment [—1; 1] is not greater than the norm of the polynomial tn(x), then there are at least n points pj (taking into account multiplicity) such that
— 1 < P1 < n1 < P2 < n2 < ... < nn —1 < Pn < 1,
n
tn(Pj) — f (Pj) = 0, j = 1,..., n, at the same time , the inequality —1 Pj ^ 1 is fulfilled.
j=1
Lemma 7. The unitary polynomial least deviating from zero on the segment [a, b] has a norm ,'b — a\n 1
equal to
&)
2n—1'
4. The case of an odd number of roots on Ga
In the case of an odd number of roots on Ga, to solve problem B it is enough to consider polynomials with no multiple roots (see lemma 2 and the proof of property 4 of [10] is easily transferred to this case too)
k
Pn(z) = (z — cos a)]J(z — eiLpj )(z — e—ilfi>), (1)
j=1
where \ ^ a,j = 1,... ,k для n = 2k + 1, k G N and
к
Pn(z) = (z - cos a)(z - 1) - eiV3 )(z - e~i<Pi), (2)
j=i
for n = 2k + 2, k G N. It is easy to check that the polynomial (2) is not extremal at k = 0. Lemma 8. If cos a = 0, n ^ 2, then for all polynomials of the form
n — 1
Pn(z) = (z - cos a)]](z - ei^), (3)
j=i
where \ ^ a, j = 1,... ,n - 1, in particular, for (1) and (2), the inequality \\Pn У > 2sinn a/2.
Proof. For the norm of the polynomial (3) the inequality is true \\Pn(z)\\ ^ minzera \z - cosa\ ■ \\Tn—1(z)\, where Tn—1(z) is an extremal polynomial (see Theorem 1) of degree n - 1.
\\Pn(z)\\ > min \z - cos a\ ■ 2 ■ sinn—1(a/2) > 2sinn a/2. □
zera
Find the moduli of the polynomials (1) and (2) if a = п/2, for z = eiLp G Га:
к
\Pn(eip)\ = + cos2 a - 2cos a cos ф ■ ^ \2(cos фj - cos ф)\,
j=1
к
\Pn(eip)\ = v7! + cos2 a - 2 cos a cos ф ■у/ 2 - 2 cos ф ■ ^ \2(cos фj - ео8ф)\ .
j=i
Applying the transformation х(ф) = sign(cos a) f2 fsin - Л, we obtain \Pn(eiV)\ =
sin a/2
= 2(sina/2)n ■ 2(n—2)/2\ln(x)\, where for polynomials (1) and (2) respectively
к
in (x)^v/axTi n (x - vj) (4)
j=i
and
к
ln(x) = Vax + 1 ■ V(1 + sign(cos a)x) ■ JJ(x - Vj), (5)
j=i
a = \ cos a\, Vj = x^j), j = 1,... ,r. If the norm of polynomials is defined on Га, then the norm of functions ln (x) is on the segment [— 1; 1] of the real axis.
Lemma 9. If all roots of the derivative of the function ln(x), (for (4) and (5)) n ^ 2 belong to the segment [-1; 1], then the inequality
\\ln\\ = max \ln(x)\ >ri (6)
rn^1 2 2
is true for the norm of the function.
Proof. If all roots of the derivative of the function ln(x), n > 2 and a = 0 belong to the segment [—1; 1], then the norm of the function ||ln|| = max|x|^1 \ln(x)\ on the segment [—1; 1] coincides with the norm of the function on the segment [—1; 1] with the norm of the function on the segment
— I ;1
a
1 /1 + 1/ a\ n 1
. If rn(x) = ll (x) — 2n—T, then by Lemma 7 |lrnHI[ — 1/a;1] > a • (-• .
/1 + 1/a \ n
Notice that 0 < a < 1. The function m(a) = a • ( -2-j in the interval (0; 1) is decreasing
for n > 2. Thus m(1) = 1, hence, ||rn|| — 1/a;1] > _ 1. The lemma is proved. □
Lemma 10. For the norm of the function (4) for which Y1 vj < 2, the inequality (6) holds.
k1
j <
j=1
_ k
Proof. Consider the family of functions ln(x) = %/ax + 1 • n (x — Vj), \vj\ < 1, j = 1,..., k, for
j=1
which the inequality ||\ln|| < n-2 is satisfied. If rn(x) = ln(x) — 0n_ 1, then by assumption 1 2 2 2" ||rn(x)|[ —1;1] = 2^ .
Consider the Chebyshev polynomial on the segment [—1; 1]
1n
tn(x) = 2^ cos(n arccos x) = ]^[(x — Yj).
j=1
The roots of the derivative of the Chebyshev polynomial are nj(j = 1,... ,n — 1). Since, by assumption, ||\rn|| = ||tn||, then Lemma 6 is valid. Thus, the identity
n
tn(x) — rn(x) = (1 — a) H(x — Pj)
j=1
is valid for Pj, j = 1,... ,n are described in Lemma 6:
n k 1 n
n(x — Yj) — (ax + 1) n(x — Vj)2 + = (1 — a) n(x — Pj)
j=1 j=1 j=1
Let us equate the coefficients of the polynomials at xn—1 of the right and left parts of the previous equality:
n k n
Yj— I —2aJ2vj + 11 = —(1 — a)J2 Pj.
j=1 v j=1 / j=1
Using the inequality of Lemma 7 and properties of Chebyshev polynomials, we obtain
k
— 1 + a < —2a^2/vj + 1 < 1 — a
j=1
nan
1 2 — a
v. <-. (7)
2 ^ j 2a V '
j=1
Hence, if for the coefficients Vj,j = 1,... ,k of the function ln(x) = %/ax + 1 • n (x _ Vj) the
j=i
inequality (7) is not satisfied, then for the norm of this function the inequality (6) holds for the norm of this function. □
The following two lemmas are proved in the same way.
k
Lemma 11. For the norm of the function (5) at cos a > 0; for which Y1 vj &
j=i
inequality (6) holds.
l; -
a J
k
Lemma 12. For the norm of the function (5) at cos a < 0 for which vj &
j=i
inequality (6) holds.
0;
l — a
the
, the
Lemma 13. For a function ln(x) (4) for which there exists k +1 points Xj € [—1; 1] such that \ln(Xj)| = ||ln||, j = 1,... ,k + 1 and having the largest deviation from zero on the segment [—1; 1] at the roots of the derivative located between the points Vj,j = 1,... ,k and at the ends of the segment, the inequality (6) holds.
Proof. Let the conditions of the lemma be satisfied. This is possible if and only if, at the ends of the segment and in the roots of the derivative belonging to the segment [—1; 1], the modulus of the function coincides with its norm. Find some positive number L such that ||ln(x)||[-i,i] =
L -. Consider the transformation rn(x) = in(x) — tt~\ , where n = 2k + 1. For the
22fc-1 ' ^^^^ 'nVA/ "nVA/ 2n-1 :
function bn(x) = tn(x) — rn(x)/L Lemma 7 holds, bn(pj) = 0, where pj, j = 1,...,n, are
described in Lemma 6. Note that pn = 1, the remaining arrangement of zeros of the polynomial
k k
bn is possible if Vj < V2j. Indeed, let vk > n2k, (2k = n — 1), then the function bn on
j=i j=i
the segment [n2k; 1] has at least two roots (if the root is at n2k, it has multiplicity two). Hence, the total zeros of the function bn are more than n, then bn = 0. This is impossible because
of the difference between the polynomials tn(x) and rn(x)/L. Similarly consider the inequality
k
vk-1 > n2k-2, etc. The equality n2j = 1/2 follows easily from the properties of the Chebyshev
j=i
polynomial. The inequality (6) follows from Lemma 10. □
The proof of the next two lemmas differs slightly from the proof of the previous lemma.
Lemma 14. For a function ln(x) = %/ax + 1 • \J(1 — x) • 0,k=i(x — Vj), a = 0, for which there exist k +1 points Xj € [—1; 1] such that \ln(Xj)\ = ||ln||, j = 1,... ,k +1 and having the largest deviation from zero on the interval [—1; 1] in the roots of the derivative located on the interval [—1; 1] and at the point -1, the inequality (6) holds.
Lemma 15. For a function ln(x) = %/ax + 1 • \J(x + 1) • Y\!j=1(x — Vj), a = 0, for which there exist k +1 points Xj € [—1; 1] such that \ln(Xj)\ = ||ln||, j = 1,... ,k +1 and having the largest deviation from zero on the interval [—1; 1] in the roots of the derivative located on the interval [—1; 1] and at the point 1, the inequality (6) holds.
Lemma 16. Let n ^ 2. The norms of the polynomials (1) and (2) are greater than 2sinn a/2.
Proof. Let a = 0. If P* is an extremal polynomial among (1) and (2), then a function l*t(x) of the form (4) or (5), respectively, associated with the extremal polynomial. The function ln(x) falls under the conditions of Lemma 5: a/ax + 1 an^ax + 1 • \J(1 + sign(cos a)x) are weight
functions. Hence, there are k + 1 points Xj e [—1; 1] such that \n(Xj)\ = ||ln||, j = 1,... ,k + 1 (see Lemma 5). The derivative of the function ln(x) has k roots. There are two possible cases: either all roots of the derivative function ln(x) lie on the segment [—1; 1], or one root does not belong to the segment [—1; 1], and in both the first and second cases k — 1 roots are located between the points vj,j = 1,...,k. Moreover, in the second case, the function ln(x) has the largest deviation from zero on the segment [—1; 1] at the roots of the derivative located between the points Vj,j = 1,... ,k and at one or two ends of the segment, depending on the structure of this function.
All possible variants have been considered above. If a = 0, then Lemma 8 is valid. □
Theorem 1 is proved.
The main result was reported at the conference [13].
This work is supported by the Krasnoyarsk Mathematical Center and financed by the Ministry of Science and Higher Education of the Russian Federation (Agreement no. 075-02-2023-936).
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Многочлены Чебышева с нулями вне открытого сегмента дуги
Наталья Н. Рыбакова
Сибирский федеральный университет Красноярск, Российская Федерация
Аннотация. Рассмотрена задача об унитарных многочленах степени п с вещественными коэффициентами, наименее уклоняющихся от нуля на произвольной фиксированной дуге окружности, с нулевым множеством вне открытого сегмента той же самой дуги. Дано описание экстремальных многочленов решения этой задачи и получена их норма, зависящая от степени полинома и длины дуги.
Ключевые слова: многочлены Чебышева, многочлены, наименее уклоняющиеся от нуля, нулевое множество, многочлены с вещественными коэффициентами.
