General Mathematics Vol. 15, No. 1 (2007), 101–110
Development in series of orthogonal
polynomials with applications in
optimization1
Adrian Branga
Dedicated to Professor Ph.D. Alexandru Lupaş on his 65th anniversary
Abstract
Our aim in this paper is to find some developments in Chebyshev
series and using these to prove that min kQn kp = kTen kp , where
Ten (x) =
mial.
1
2n−1
en
Qn ∈Π
cos(n arccos x) is the n-th Chebyshev monic polyno-
2000 Mathematics Subject Classification: 33C45, 41A50, 42A05
1
Introduction
e n be the set of all real monic polinomials and Lpw [−1, 1],
Let Π
1 < p < ∞, the Lebesque linear space with the weight w(x) =
¸1/p
·1
R
p
.
|f (t)| w(t)dt
endowed with the norm kf kp =
−1
1
Received 30 December, 2006
Accepted for publication (in revised form) 12 February, 2007
101
√ 1
,
1−x2
102
Adrian Branga
Theorem 1.1 The extremal problem
(1)
has an unique solution.
e n,
kQn kp → min, Qn ∈ Π
Proof. It is known that the Lebesque normed linear space Lpw [−1, 1] is
strict convex for 1 < p < ∞.
Lemma 1.1 The extremal problem (1) is equivalent with the following problem:
¯p
Zπ ¯¯
n−1
¯
X
¯ 1
¯
(2)
ai · cos it¯ dt → min ,
¯ n−1 · cos nt +
¯2
¯
i=0
0
where a = (a0 , . . . , an−1 ) ∈ Rn .
e n , hence there is
Proof. {T0 , . . . , Tn } represents a basis in Π
a = (a0 , . . . , an−1 ) ∈ Rn so that:
Therefore
en (x) =
Q
1
2n−1
· Tn (x) +
n−1
X
i=0
ai · Ti (x) .
 π¯
¯p 1/p
Z ¯
n−1
¯
X
¯ 
en kp =  ¯¯ 1 · cos nt +
.
kQ
a
·
cos
it
¯ dt
i
¯ 2n−1
¯
i=0
0
(3)
Further we consider the function ϕ : Rn → R be the function
¯p
Zπ ¯¯
n−1
¯
X
1
¯
¯
ϕ(a0 , . . . , an−1 ) = ¯ n−1 · cos nt +
ai · cos it¯ dt.
¯2
¯
i=0
0
Hence
¯p
Zπ ¯¯
n−1
¯
X
∂ϕ
¯ 1
¯
ai · cos it¯ ·
(a0 , . . . , an−1 ) = p · ¯ n−1 · cos nt +
¯2
¯
∂ak
i=0
0
(4)
#
"
n−1
X
1
ai · cos it · cos kt dt , k = 0, 1, . . . , n − 1.
·sgn n−1 · cos nt +
2
i=0
Development in series of orthogonal polynomials with applications
103
∂ϕ ∗
(a ) = 0, k = 0, 1, . . . , n − 1, where
∂ak
a∗ = (0, . . . , 0) ∈ Rn , using Theorem 1.1 and Lemma 1.1 we deduce that
Ten (x) is the unique solution of the extremal problem (1).
Easily it obtains
If we shall prove that
∂ϕ ∗
p
(a ) = n−1 ·
∂ak
4
(5)
Zπ
| cos nt|p−1 · sgn[cos nt] · cos kt dt
0
and we denote:
(6)
Jn,k =
Zπ
| cos nt|p−1 · sgn[cos nt] · cos kt dt
0
where k = 0, 1, ..., n − 1.
2
Main results
Further for f, g ∈ C[0, 1] and α > −1 let consider the following inner
product
Z1
tα (1 − t)α
dt.
hf, giα = f (t)g(t)
B(α + 1, α + 1)
0
Lemma 2.1 (see [3]) If z ∈ [0, 1], −∞ < −4λ < min(0, 2α + 1), then
(7)
∞
B(λ + α + 1, α + 1) X (−1)k (−λ)k
(α) (α)
z =
·
· γk ϕk (z),
B(α + 1, α + 1)
(2α + λ + 2)k
k=0
λ
where
B(a, b) is the ”beta” function, a > −1, b > −1,
(8)
(c)k = c(c + 1) . . . (c + k − 1), k = 1, 2, . . . , (c)0 = 1, c ∈ R,

 1, k = 0
(α)
,
γk =
2k + 2α + 1 (2α + 2)k

·
, k = 1, 2, . . .
k + 2α + 1
k!
104
Adrian Branga
(α)
(9)
(α,α)
ϕk (z) = Rk
(2z − 1) is the ultraspherical polynomials,
with the condition that the series converges uniformly on [0, 1].
In addition
(
0, k 6= j
(α)
(α)
(10)
hϕk , ϕj iα =
.
1
(α) , k = j
γk
Lemma 2.2 If z ∈ [0, 1], λ > 0 then
(11)
¡λ/2¢
∞
· k!
Γ(λ + 1) X
¡λ k
¢·
¢ · γk · Tk∗ (z),
z = λ ¡λ
2 Γ 2 + 1 k=0 Γ 2 + k + 1
λ
2
where Tk∗ (z) is the k-th Chebyshev polynomial on [0, 1] and
(
1, k = 0
,
γk =
2, k = 1, 2, . . .
with the condition that the series converges uniformly on [0, 1].
In addition
(
0, k 6= j
∗
∗
(12)
.
hTk , Tj i− 1 =
1
2
,k = j
γk
Proof. In (7) we consider λ := λ2 , α := − 21 and using (8)–(10) we deduce
(11) and (12).
Lemma 2.3 If λ > 0 then
¡ ¢
Z1
πΓ(α + 1) · λ/2
j!
λ
dz
∗
¢
¡λ j
¢ , j = 0, 1, . . . .
= λ ¡λ
(13) z 2 · Tj (z) p
2 Γ 2 +1 ·Γ 2 +1+j
z(1 − z)
0
Proof. From (11), (12) we find
E
D λ
z 2 , Tj∗ (z)
− 12
¡λ/2¢
j!
Γ(λ + 1)
¢ · ¡λ j
¢
= λ ¡λ
2 Γ 2 +1 Γ 2 +j+1
and using the definition of inner product we obtain (13).
Development in series of orthogonal polynomials with applications
Further for f, g ∈ C[−1, 1] we use the following inner product
hf, gi =
Z1
−1
dt
.
f (t)g(t) √
1 − t2
We consider the following development in Chebyshev series
λ
(14)
|z| =
∞
X
k=0
ck (λ) · Tk (z), z ∈ [−1, 1], λ > 0,
where
­
® 1
1
ck (λ) = γk · |z|λ , Tk (z) = γk ·
π
π
(15)
Z1
|z|λ Tk (z) √
−1
Substituting z = −t in (15) and using the relation
Tk (−t) = (−1)k Tk (t),
we find
(16)
c2j+1 (λ) = 0, k = 0, 1, . . . ,
hence
|z|λ =
(17)
where
∞
X
k=0
2
c2k (λ) = γk ·
π
(18)
c2k (λ) · T2k (z),
Z1
0
dz
z λ T2k (z) √
.
1 − z2
It is known that
Tmn (z) = Tm (Tn (z)), m, n ∈ N,
hence
(19)
T2k (z) = Tk (T2 (z)) = Tk (2z 2 − 1) .
dz
.
1 − z2
105
106
Adrian Branga
Using (19) in (18) and substituting z 2 = t we obtained
1
c2k (λ) = γk ·
π
(20)
Z1
0
λ
dt
.
t 2 Tk∗ (t) p
t(1 − t)
From equalities (13), (17) and (20) we conclude with
Lemma 2.4 If λ > 0, z ∈ [−1, 1] we have the following development in
Chebyshev series
∞
X
λ
(21)
c2k (λ) · T2k (z),
|z| =
k=0
where
(22)
¡ ¢
γk · Γ(λ + 1) · λ/2
· k!
¢
¡ λk
¢ , k = 0, 1, . . . .
¡λ
c2k (λ) = λ
2 ·Γ 2 +1 ·Γ 2 +1+k
The previous result allow us to obtain:
Theorem 2.1 If 1 < p < ∞ the following equality holds
(23)
(24)
where
p−1
|Tn (x)|
= a0 (p) +
| cos nt|p−1 = a0 (p) +
∞
X
j=1
aj (p) · T2jn (x), x ∈ [−1, 1] ,
∞
X
j=1
aj (p) · cos(2jnt), t ∈ [0.π] ,
¡ p−1 ¢
Γ(p) · j2 · j!
Γ(p)
¢ , j = 1, 2, . . .
(25) a0 (p) = p−1 2 ¡ p+1 ¢ , aj (p) = p−2 ¡ p+1 ¢ ¡ p+1
2 Γ
2 Γ 2 Γ 2 +j
2
Proof. In (21) we consider λ := p − 1, 1 < p < ∞, z := Tn (x), x ∈ [−1, 1]
and we find (23). If we consider in (23) t := arccos x, t ∈ [0, π] we deduce
(24).
Further we consider the following situation:
1. Suppose that
n is even, n = 2s, s ∈ N ∗ , and k is odd, k = 2m + 1, m = 0, s − 1,
or n is odd, n = 2s + 1, s ∈ N , and k is even, k = 2m, m = 0, s.
Substituting in (6) t = π − x we deduce
Development in series of orthogonal polynomials with applications
107
Theorem 2.2 If n = 2s, s ∈ N ∗ , and k = 2m + 1, m = 0, s − 1, or
n = 2s + 1, s ∈ N , and k = 2m, m = 0, s we have
(26)
J2s,2m+1 = 0 and J2s+1,2m = 0.
2. Suppose that n is even, n = 2s, s ∈ N ∗ , and k is even, k = 2m,
m = 0, s − 1.
, i = 1, 2s, be the zeros of cos(2st) on [0, π] and t0 = 0,
Let ti = (2i−1)π
4s
t2s+1 = π.
It is known that
s
(27)
cos(2st) > 0, for t ∈ ∪ (t2i , t2i+1 )
i=0
and
s
(28)
cos(2st) < 0, for t ∈ ∪ (t2i−1 , t2i ) .
i=1
Therefore
−
+
+ J2s,2m
J2s,2m = J2s,2m
(29)
where
(30)
+
J2s,2m
tZ
2i+1
s
X
(cos 2st)p−1 · cos 2mt dt,
=
i=0 t
2i
(31)
−
J2s,2m
=−
Zt2i
s
X
i=1 t
2i−1
From (24) it follows that
(− cos 2st)p−1 · cos 2mt dt.
108
Adrian Branga
Lemma 2.5 If m = 0, 1, . . . , s − 1, s ∈ N ∗ , then
+
J2s,2m
tZ
2i+1
s
X
cos(2mt)dt+
= a0 (p) ·
i=0 t
2i
(32)
+
∞
X
j=1
tZ
2i+1
s
X
aj (p)
cos(4jst) · cos(2mt)dt ,
i=0 t
2i
−
J2s,2m
= −a0 (p) ·
Zt2i
s
X
cos(2mt)dt−
i=1 t
2i−1
(33)
−
∞
X
j=1
aj (p) ·
Zt2i
s
X
i=1 t
cos(4jst) · cos(2mt)dt .
2i−1
After an easily computation we obtain the results.
Lemma 2.6 If i = 0, . . . , s, m = 0, . . . , s − 1, s ∈ N ∗ , then
tZ
2i+1
(34)
t2i
cos(2mt)dt =









π
, i = 0, s, m = 0
4s
π
, i = 1, s − 1, m = 0
2s
1
· sin mπ
, i = 0, s, m = 1, s − 1
2m
2s
mπ
1
· sin 2s · cos 2imπ
, i = 1, s − 1, m
m
s
= 1, s − 1
Lemma 2.7 If i = 0, . . . , s, m = 0, . . . , s − 1, s ∈ N ∗ , j = 1, 2, . . ., then
tZ
2i+1
cos(4jst) · cos(2mt)dt =
t2i


0,
i
=
0,
s,
m
=0
(35) 

j+1

 (−1) · m
mπ
· sin
, i = 0, s, m = 1, s − 1
2
2
2
=
2(4j s − m )
2s



(−1)j+1 m
mπ
2imπ


· sin
· cos
, i = 1, s − 1, m = 1, s − 1.
2
2
2
4j s − m
2s
s
Development in series of orthogonal polynomials with applications
109
Lemma 2.8 If i = 1, . . . , s, m = 0, . . . , s − 1, s ∈ N ∗ , then
(
Zt2i
(36) cos(2mt)dt =
t2i−1
π
,
2s
1
·
m
i = 1, s, m = 0
sin mπ
· cos (2i−1)mπ
, i = 1, s, m = 1, s − 1.
2s
s
Lemma 2.9 If i = 1, . . . , s, m = 0, . . . , s − 1, s ∈ N ∗ , j = 1, 2, . . ., then
Zt2i
cos(4jst) · cos(2mt)dt =
t2i−1
(37) 
 0, i = 1, s, m = 0
=
(2i − 1)mπ
(−1)j+1 · m
mπ

· cos
, i = 1, s, m = 1, s − 1.
· sin
2
2
2
4j s − m
2s
s
Lemma 2.10 If m = 0, 1, . . . , s − 1, s ∈ N ∗ , then
(38)
s
X
i=1
s
X
(2i − 1)mπ
2imπ
= 0 and
= 0.
cos
cos
s
s
i=1
Taking into account the equalities (34)–(38) from Lemma 2.5 it follows
that.
Theorem 2.3 If m = 0, 1, . . . , s − 1, s ∈ N ∗ , then
( π
· a0 (p), m = 0
+
2
J2s,2m
=
(39)
,
0, m = 1, s − 1
(40)
−
J2s,2m
=
(
π
· a0 (p), m = 0
2
.
0, m = 1, s − 1
−
Using the results from the Theorem 2.3 we conclude with
Theorem 2.4 If m = 0, 1, . . . , s − 1, s ∈ N ∗ , then
(41)
J2s,2m = 0 .
110
Adrian Branga
3. Suppose that n is odd, n = 2s + 1, s ∈ N ∗ , and k is odd, k = 2m + 1,
m = 0, 1, . . . , s − 1.
Likewise that in section 2 it follows that
Theorem 2.5 If m = 0, 1, . . . , s − 1, s ∈ N ∗ , then
(42)
J2s+1,2m+1 = 0 .
From Theorem 2.2, 2.4, 2.5 and relation (5) we conclude with
Theorem 2.6 Ten (x) is the unique solution of the extremal problem (1).
References
[1] Bernstein S., Sur les polynomes orthogonaux relatifs à un segment fini,
Journ. de Math., tome IX, 1930, p.127–137.
[2] Ghizzetti A., Ossicini A., Quadratere formulae, Birkhäuser Verlag,
Basel, 1970.
[3] Lupaş A., The Approximation by Means of Some Linear Positive Operators, Approximation Theory, Proc. International Dortmund Meeting,
IDoMat 95, Akademic Verlag, Vol.86, p.201–230.
[4] Szegö G., Orthogonal Polynomials, Amer. Math. Soc., Colloquium
Publ., vol.23, Fourth edition, 1978.
Department of Mathematics,
Faculty of Sciences,
University ”Lucian Blaga” of Sibiu,
Dr. Ion Ratiu 5-7, Sibiu, 550012, Romania
E-mail address: adrian branga@yahoo.com