CC BY
7URAL MATHEMATICAL JOURNAL, Vol. 2, No. 1, 2016
ON AN EXTREMAL PROBLEM FOR POLYNOMIALS WITH FIXED MEAN VALUE1
Alexander G. Babenko
Krasovskii Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences; Institute of Mathematics and Computer Science, Ural Federal University,
Ekaterinburg, Russia, babenko@imm.uran.ru
Abstract: Let T+ be the set of nonnegative trigonometric polynomials r„ of degree n that are strictly positive at zero. For 0 < a < 2n/(n+2), we find the minimum of the mean value of polynomial (cos a — cos x)rn (x)/rn (0) over r„ G T+ on the period [—n, n).
Key words: Trigonometric polynomials, Extremal problem.
Let Tn be the space of trigonometric polynomials of degree n with real coefficients, and let T+ be the set of nonnegative polynomials from Tn that are strictly positive at zero. For a real a we define
1 r
Xn(a) = inf —-— Tn(x)(cos a - cosx) dx. (1)
rnGT+ 2nTn(0) J—n
In 1915, Fejer [4] (see also [2, vol. 2, Sec. 6, Problem 52]) proved the following statement.
n
Fejer's Theorem. Let the polynomial Tn(x) = a0 + ^ (av cos vx + bv sin vx) belong to the set
V=1
T+. Then
\Ja\ + b2 < 2a0 cos ■ (2)
This inequality turns into the equality for the polynomial
n + 2 \2// n
-x cos x — cos ■
tn(x) = ^ COS — / (COS x — COS —' (3)
This theorem is equivalent to the statement that
Xn{n/(n + 2)) =0. (4)
For 0 < a < n, put
/ . n + 1 . n + 3 . n + 3 . n + 1 \/ a Q(n+3)/2,a(x) ^ ^ sin 2 a sin 2 x - sin —— a sin 2 x) sin —, 0 < a < n, (5)
n + 3 n + 1
Q(n+3)/2,0(x) = aim Q(n+3)/2,a(x) = (n + 1) sin —— x - (n + 3) sin —— x.
In this paper we prove the following result.
xThe paper was originally published in a hard accessible collection of articles Approximation of Functions by Polynomials and Splines (The Ural Scientific Center of the Academy of Sciences of the USSR, Sverdlovsk, 1985), p. 15-22 (in Russian).
Theorem. Let n be a nonnegative integer and 0 < a < 2n/(n + 2). Then (1) takes the value
^sin 'n++3 a — sin a^ (1 — cos a) 2n
Xn(a) = (n + 3) sin n+1 a - (n + 1) sin n+3a' 0 <a < W+2' (6)
6
Xn(0) = lim Xn(a) =
a^oAnv 7 (n + 1)(n + 2)(n + 3)' and the infimum is attained for the polynomial
( ) ( Q(n+3)/2,a(x) \2 (7)
Tn,a(x) = 7-"- \ • f /o^ ' (7)
\(cos x — cos a)sm(x/2)/
where Q(n+3)/2,a is given by (5).
Note that Xn(a) > 0 for 0 < a < n/(n + 2) and %n(a) < 0 for n/(n + 2) < a < 2n/(n + 2). First we prove two auxiliary statements. Set a0 = n, ai = 2n/3, and for n > 2 let an be the first positive root of the equation
. n + 3 \/ n + 1 m + 1
sin — x) sin — x = Cn, C2m = —1, C2m-1 =--• (8)
2 2 m
It is easy to see that for r > 2 we have
a2r-2 = n/r, 2n/(2r + 1) <a2r-i <n/r• (9)
Lemma 1. If n is a nonnegative integer and 0 < a < an, then the function Q(n+3)/2,a defined by (5) has exactly [(n + 5)/2] zeros x0 = 0 < x1 = a < x2 < x3 < • • • < X[(n+3)/2] in the interval [0, n]. For each polynomial Tn+1 £ Tn+1 we have
where
1 fn ^ sin a — sin n±3 a , s
TT Tn+i(x) dx = -— . n+1--— 2 n+3 Tn(0)
2n J-n (n + 3) sin n++1 a — (n + 1) sin n+3 a
[(n+3)/2]
+ Z 9n+1 (xk){Tn+1(xk)+ Tn+1 ( xk)) , k=1
sin x
g2r-1(x) =
(10)
2r sin x — sin 2rx sin x
( ) = 2 r sin x — sin rx cos(r + 1)x g2r (x) = \ sin ra + sin(r + 1)a
x = n, (11)
x = n
4(r sin(r + 1)a + (r + 1) sin ra) Moreover, the numbers (n+2 — a)gn+1(x[(n+3)/2]), gn+1(xk), 1 < k < [(n + 1)/2], are nonnegative.
Proof. First we consider the case when n = 0 and 0 < a < n. The function Q3/2,a(x) = 2sin(x/2) (cos x — cos a) has two zeros x0 = 0, x1 = a in the interval [0,n]. We have
1 i'n /x , cos a . T1(a) + t1(—a) ^
— T1(x) dx =-- T1(0) + -, T1 £ T1,
2n J-n cos a — 1 2(1 — cos a)
since this formula is valid for the polynomials 1, sin x, cos x and thus the lemma follows for n = 0.
Now let n = 1 and 0 < a < 2n/3. Then the function Q2,a(x) = 4 cos 0, sin x(cos x — cos a) has three zeros x0 = 0, x1 = a, x2 = n in the interval [0, n]. The quadrature formula
1 i'n /x , 1 — 2 cos a . t2 (a) + t2(—a) 1 + 2cos a , , , . ^
2-J. n T2<x)dx = 4(1—cosa T2(0) + 2(1—J a) + sor^oa <T2«+T2<—• T*£ T2'
holds, for it holds for the polynomials sin x, sin2x, (1 + cos x)(cos a — cos x), 1 — cos2x, (1 — cos x)(cos a — cos x) which generate the space T2. This proves the lemma for n = 1.
Next we consider the case of an odd n = 2r — 1, r > 2, and 0 < a < a2r-1. The function (5) can be written in the form Qr+1a(x) = (cos x — cos a) sin xSr-1,a(x)/ sin(a/2), where
sinra sin(r + 1)x — sin(r + 1)a sinrx
Sr-1,a(x) = T n : (12)
(cos x — cos a) sin x
is a cosine polynomial of degree r — 1. To study the zeros of the polynomial Sr-1,a, we write it in
i r- n t \ (f (x) — f (a)) sin ra sin rx .. . sin(r + 1)x TTT,
the form Sr-1 a(x) =-----, where f (x) =-. When x runs over the
sin x(cos x — cos a) sin rx
intervals (0, n/r), ((r — 1)n/r,n), and (kn/r, (k + 1)n/r), 1 < k < r — 2, then the values of f run continuously over the intervals ((r + 1)/r, —to), (+to, —(r + 1)/r), and (+to, —to), respectively. Thus, taking into account the definition (8) of a2r-1, we see that for each a in the interval (0, a2r-1) the polynomial Sr-1,a has exactly r — 1 zeros x2 <x3 < ... <xr in the interval (a,n). Moreover, these zeros are all simple since Sr-1,a has degree r — 1. It is known [3, p. 403, formulae 30, 31, 33] that
n . ( 1,m>v,m + v = 2k — 1;
1 ' sin mx cos vx , , , s
— -dx = < 0, m > v, m + v = 2k; (13)
n Jo sin x I „ ^
JU ^ 0, m < v.
It follows that for the polynomial (12) we have
1 r
— Sr-1a(x)cos vx(1 + cos x)(cos a — cos x) dx = sin(r + 1)a — sin ra, v = 0,1,...,r — 1. n Jo
Consequently, for each cosine polynomial Cr-1 of degree r — 1 we have 1 fn
— Sr-1,a(x)Cr-1(x)(1 + cosx)(cos a — cosx) dx = (sin(r + 1)a — sinra)Cr-1(0). (14) n Jo
Thus, the polynomial Sr-1,a is orthogonal with the weight (cosx — cosa)(1 — cosx)(1 + cosx) to all cosine polynomials of degree r — 2.
We will need the following known result (e.g., [1, pp. 162, 163]). Let the weight u(x) and the points a1,... ,am in the interval [0,n] be given. A quadrature formula of the form
,, n m v
/ C2v+m-1(x)u(x) dx = ^2 A£C2v+m-1(a£) + ^ BkC2v+m-1(xk) ■J0 1=1 k=1
which is exact for cosine polynomials of degree 2v + m — 1 exists if and only if there exists a cosine polynomial Sv of degree v which is orthogonal to all cosine polynomials of degree v — 1 with the weight u(x)(cosx — cosa1)... (cosx — cos am). The zeros of the polynomial Sv coincide with the nodes x1, x2, . . . , xv ; they should be all distinct and differ from the fixed nodes a1,..., am.
By this result, there exist numbers e0,... ,er+1 such that for each cosine polynomial C2r of degree 2r we have
1 f n r+1
- C2r (x) dx = y £k C2r (xk), (15)
nJo k=0
where x2,x3, ...,xr are the zeros of the polynomial Sr-1,a in the interval (a, n), x0 = 0, x1 = a, xr+1 = n.
Note that, for v = 1, 2, . . . , r, the zeros of the polynomial
, . sin rxv sin(r + 1)x — sin(r + 1)xv sin rx
Sr-1,xv (x) = -7-^--(16)
(cos x — cos xv) sin x
coincide with the zeros of the polynomial (cos x — cos a)Sr-1, a(x)/(cos x cos x~v ). Thus,
Sr- 1,Xv
(x) = Av (cos x — cos a)Sr-1;Q,(x)/(cos x — cos xv), (17)
where Av is a constant that does not depend on x.
It is not difficult to check that, for v = 1,2, • • • ,r, the polynomial (16) satisfies the equations
, . Dr (x — xv) — Dr (x + xv) sin kxv sin kx Sr-1,xv (x) =-TT-.-= 2V-:-, (18)
2 sin x ' sin x
k=1
where
is the Dirichlet kernel. Using (18), we obtain
sin x
Dr (x) = 1 + 2V cos kx = , 2, „ ^ k=1 sin(x/2)
1 fn
— Sr-1 Xv (x)(sin x)2 dx = sin xv, v = 1,(19) n Jo
Using (19), (17) and (18), one can calculate the following coefficients of the quadrature formula
(15): r
= 1/(2 V(sin kxv)2) =—-.sin^ ( +1) , 1 < v < r (20)
/ V ^ ) r sin xv — sin rxv cos(r + 1)xv
k=1
By (14), we have
1 r
— Sr-1 a(x)(1+cos x)(cos a — cos x) dx = sin(r + 1)a — sin ra• n Jo
Using (15) and (12), we obtain from here that
sin ra — sin(r + 1)a (21)
2((r + 1) sin ra — r sin(r + 1)a)
By (13) and (12) we conclude that 1 fn
— Sr-1 a(x)(1 — cos x)(cos x — cos a) dx = (—1)r( sin ra + sin(r + 1)aV (22)
n Jo
Formulae (22), (15) and (12) imply
sin ra + sin(r + 1)a
£r+1 = —^---:--—r • (23)
2 ((r + 1) sin ra + r sin(r + 1)aj
It is easy to check that
2n \
— a)er > 0 (24)
2r + 1 "J 'r ~
for 0 < a < a2r-1. The statement of the lemma for n = 2r — 1, r > 2, now follows from (20), (21), (23) and (24).
Finally, let us consider the case when n = 2r — 2, r > 2, and 0 < a <n/r. Function (5) can be written in the form Q(2r+1)/2,a(x) = sin(x/2)(cosx — cosa)6r-1 a(x)/sin(a/2), where
2T-1 a sin 2r+ix — sin ^T+la sin ^T-x (^>(x) — a)) sin a sin x
sin a sin ±iLTTLx — sin ±lttl a sin■x
©r_ 1 a(x) = 2 ~ ^ 2 — 2 ~ — 2 - -- 2 -; (25)
r 'a (cos x — cos a)sin(x/2) (cos x — cos a)sin(x/2) '
here, p(x) = (sin ^i^/sin 2r—1x. When x runs over the intervals (0, 2n/(2r — 1)), (2(r— 1)n/(2r— 1), n) and (2kn/(2r — 1), 2(k + 1)n/(2r — 1)), 1 < k < r — 2, then the values of the function p run continuously over the intervals ((2r + 1)/(2r — 1), —to), (+to, —1) and (+to, —to), respectively. Thus, for 0 < a <n/r the polynomial 0r-1,a has exactly r — 1 simple zeros x2 <x3 < ... <xr in the interval (a, n). With the help of (13) and (25), repeating the arguments used in the proof of formula (14), we see that
1 fn ( 2r + 1 2r — 1 \
Qr-1,a(x)Cr-1(x)(cos a — cosx) dx = ( sin—2— a — sin—2— ^Cr-1 (0) (26) Jo V 2 2 /
I r , ^ 2
for all cosine polynomials Cr-1 of degree r — 1. Thus, the polynomial 0r-1,a is orthogonal to all cosine polynomials of degree r — 2 with the weight (1 — cos x)(cos x — cos a). It follows that there exist numbers 5o,51,..., 5r such that the quadrature formula
1 i n r
- C2r-1(x) dx = V 5k C2r-1(xk), (27)
n l0 z—'
o k=o
where x2,x3,... ,xr are the zeros of the polynomial 0r-1,a in the interval (a,n), x0 = 0, x1 = a, is exact for all cosine polynomials C2r-1 of degree 2r — 1. Note that, for v = 1, 2, . . . , r, the polynomial
2r— 1 • 2r+1 • 2r+1 • 2r — 1 , „ sin 2r—1 xv sin 2L+1 x — sin 2L+1 xv sin 2r—1 x , ,
®r-1,xv (x) =--2-) . ( .2;-^ (28)
(cos x — cos xv )sin(x/2)
satisfies the equation
&r-1,Xv(x) = Bv(cosx — cosa)6r-1;Q,(x)/(cosx — cosxv), 1 < v < r, (29)
where Bv is a constant that does not depend on x.
Moreover, the polynomial (28) can be rewritten in the form
^ , , v^ ( 2k — 1 2k — 1 w x ,,
@r-1,xv (x) = 2^ ^ sin—2— xv sin—2— xj! sin ^. (30)
k=1
This implies the equation
1 fn / x \ 2 x
n J @r-1,xv (x^ sin 2 J dx = sin -2, 1 < v < r. (31)
Formulae (31), (27), (29) and (30) yield
2k — 1 2 2 sin xv
5v = 1/ 2> sin-xJ =-v-, 1 < v < r. (32)
2 2r sin xv — sin 2rxv
k=1
By (26) we obtain
1 I &r-1,Xv(x)(cos a — cos x) dx = sin 2r + 1 a — sin 2r 2 1 a. (33)
Jo 22
n J o
Using (33), (27) and (25), we get
( 2r — 1 2r + 1 \/( 2r — 1 2r + 1 \
5o = ^ sin —2— a — sin —2— ajj y(2r + 1) sin —^— a — (2r — 1) sin —^— a J. (34)
The statement of the lemma for n = 2r — 2, r > 2, now follows from (32) and (34). This completes the proof of the lemma. □
Lemma 2. Let n be a nonnegative integer, 0 < a < an if n is even and 0 < a < an if n is odd. For each polynomial Tn £ Tn we have
1 fn s (sin nt3 a — sin m a) (1 - cos a) , s
— Tn(X)(cos a — cos x) dx = -— 3) . ra+1-(- 1) . ra+3 Tn (0)
2n J-n (n + 3) sin a — (n + 1) sin n++3a
[(n+1)/2]
+ E 9n+i (xk )(cos a — cos xk)(Tn(xk) + Tn(—Xk)), k=1
(35)
where x\ < x2 < ■■■ < X[(n+1)/2] are the zeros of the polynomial (7) in the interval (a,n], and the numbers gn+1(xk), k = 1, 2,..., [(n + 1)/2], are defined by equations (11). Moreover, the coefficients gn+1(xk)(cos a — cosxk), k = 1,2,..., [(n — 1)/2], are nonnegative, as well as the
number (n+2 — a)g«+i(x[(n+i)/2])(cosa — cosx[(n+i)/2]).
Proof. For 0 < a < an, the statement is a straightforward consequence of Lemma 1. Let Tn be an arbitrary polynomial of degree n, then the right-hand side of (35) and the coefficients of this quadrature formula tend uniformly to the claimed (bounded) values as a ^ 0, and the statement of the lemma follows for a = 0. The case of a = an with even n can be proved in a similar way. As for the case of odd n, note that for an odd n > 3 we have gn+1(x[(n+1)/2])(cos a — cos x[(n+1)/2]) = g(-n)(cosa + 1) ^ as a an, while gn+1(x[(n_1)/2])(cosa — cosx[(n-1)/2]) ^ as a an. □
Proof of the theorem. The statement of the theorem follows from the fact that for each nonnegative polynomial Tn and each number a in the interval [0,2n/(n + 2)] we have, by Lemma 2, the inequality
1 fn ,w , (sin a — sin n++1 a)(1 — cos a) , s
— Tn(x)(cos a — cosx) dx >—-2-n-2—-to- Tn(0).
2n J_n n J J ~ (n + 3) sin ^a — (n + 1) sin ^a y>
This inequality turns into the equality for the polynomial Tnaa. This proves the theorem. □
Acknowledgments
The author is grateful to Professor V. V. Arestov for the statement of the problem as well as to Doctor E. E. Berdysheva for the excellent translation of the paper into English.
REFERENCES
1. Krylov V.I. Approximate calculation of integrals. Fizmatgiz, Moscow. 1959. [Russian]
2. Polya, G., Szego, G. Problems and theorems in analysis. Springer, Berlin. Vol. 2. 1998.
3. Prudnikov, A. P., Brychkov, Yu. A., Marichev, O.I. Integrals and series. Nauka, Moscow. 1981. [Russian]
4. Fejer L. Über trigonometrische Polynome. Gesammelte Arbeit, Bd. 1, Akad. Kiado, Budapest. 1970. P. 842-872.
