Journal of Inequalities in Pure and
Applied Mathematics
AN INEQUALITY FOR CHEBYSHEV CONNECTION
COEFFICIENTS
volume 7, issue 2, article 67,
2006.
JAMES GUYKER
Department of Mathematics
SUNY College at Buffalo
1300 Elmwood Avenue
Buffalo, New York 14222-1095
USA.
EMail: guykerj@buffalostate.edu
Received 19 January, 2006;
accepted 11 March, 2006.
Communicated by: D. Stefanescu
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
021-06
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Abstract
Equivalent conditions are given for the nonnegativity of the coefficients of both
the Chebyshev expansions and inversions of the first n polynomials defined by
a certain recursion relation. Consequences include sufficient conditions for the
coefficients to be positive, bounds on the derivatives of the polynomials, and
rates of uniform convergence for the polynomial expansions of power series.
2000 Mathematics Subject Classification: 41A50, 42A05, 42A32.
Key words: Chebyshev coefficient, Ultraspherical polynomial, Polynomial expansion.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Chebyshev Expansion of Pn . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Bounds on Polynomial Derivatives . . . . . . . . . . . . . . . . . . . . . . 15
4
Polynomial Expansions of Power Series . . . . . . . . . . . . . . . . . . 19
References
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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1.
Introduction
In the classical theory of orthogonal polynomials a standard problem is to expand one set of orthogonal polynomials as a linear combination of another with
nonnegative coefficients (see [4], [6], [8], [12]). R. Askey [2] and R. Szwarc
[13] give general conditions ensuring nonnegativity in terms of underlying recurrence relations, while Askey [1] and W. F. Trench [14] determine when connection coefficients are positive. It is often desirable, especially in numerical
analysis (e.g., see [9], [10], [15]), to express orthogonal polynomials in terms
of Chebyshev polynomials Tn or cosine polynomials with nonnegative coefficients. In this note, we derive inequalities (2.2) on the connection coefficients
of certain Chebyshev expansions that imply positivity. Consequently, we obtain
bounds on polynomial derivatives, estimate uniform convergence of polynomial
expansions of power series, and illustrate the optimality of Chebyshev polynomials in these contexts.
For given real sequences αn , βn and nonzero γn we consider the sequence of
polynomials Pn defined by P−1 = 0, P0 = 1, and for n ≥ 0,
(1.1)
xPn = γn Pn+1 + βn Pn + αn Pn−1 .
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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For example the Jacobi polynomials
with
(α,β)
Pn
(α, β > −1) are defined by (1.1)
2(α + n)(β + n)
αn =
,
(α + β + 2n + 1)(α + β + 2n)
β 2 − α2
βn =
(α + β + 2n + 2)(α + β + 2n)
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and
γn =
2(n + 1)(α + β + n + 1)
.
(α + β + 2n + 1)(α + β + 2n + 2)
In [5] (see also [3]), Askey and G. Gasper characterize those Jacobi polynomials that are combinations of Chebyshev polynomials with nonnegative coefficients. Szwarc ([13, Corollary 1]) gives conditions on sequences of polynomials
satisfying (1.1) which imply that their Chebyshev connection coefficients are
nonnegative. Our results are based on this work and the following classical ex{α}
pansions of the normalized ultraspherical or Gegenbauer polynomials Pn :=
(α,α)
Pn
. We recall the factorial function (α)n := α(α + 1) · · · (α + n − 1) when
(α,α)
Pn
(1)
{− 12 }
n ≥ 1, and (α)0 := 1 for α 6= 0. Then Pn
(1.2)
An Inequality for Chebyshev
Connection Coefficients
Pn{α}
=
n
X
James Guyker
= Tn and
c(n, m)Tm
m=0
1
α 6= −
2
,
where c(n, m) = 0 if n − m is odd, and otherwise
(2 − δm0 ) α + 21 n−m α + 12 n+m n 2
2
c(n, m) =
n−m .
(2α + 1)n
2
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It follows that if α > − 12 and n − m is even, then c(n, m) > 0 and the sequence
hc(n, n)i is monotonically decreasing. Moreover, we have the inversions
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xn =
[ n2 ]
X
2 − δn,2k n
k=0
2n
k
{− 1 }
2
Pn−2k
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([·] is the greatest integer function) and
(1.3)
n
x =
n
X
m=0
d(n, m)Pm{α}
1
α 6= −
2
,
where d(n, m) = 0 if n − m is odd, and otherwise
(2α + 1 + 2m)(2α + 1)m n−m
+
1
n−m
2
n
2
d(n, m) =
1
n+1
m
2
α + 2 n+m+2
2
{α}
is positive and hd(n, n)i is decreasing. The polynomials Pn satisfy (1.1) for
sequences such that αn > 0 and βn = 0. Furthermore, α > − 21 if and only if
αn < 12 .
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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2.
The Chebyshev Expansion of Pn
The following characterizes those polynomials that satisfy (1.1) and enjoy similar expansions (1.2) and (1.3).
Theorem 2.1. Let Pn be defined by (1.1) for given sequences αn , βn and γn ,
and be normalized by Pn (1) = 1. With n fixed, we have that
(2.1)
Pk =
k
X
a(k, m)Tm
m=0
and
k
x =
k
X
b(k, m)Pm
(0 ≤ k ≤ n)
m=0
for nonnegative coefficients a(k, m) and b(k, m) such that the coefficients a(k, k)
and b(k, k) are monotonically decreasing if and only if 0 ≤ αi ≤ 12 and βi = 0
for i = 0, ..., n − 1. In this case, the following properties hold.
(i) a(k, m) = b(k, m) = 0 if k − m is odd.
(ii) If α1 > 0, then b(k, m) > 0 whenever k − m is even.
(iii) If n ≥ 3 and K is any subset of {0, ..., [ n−1
]}, then
2
X
X
(2.2)
a(n − 2, m − 2) ≤
a(n − 3, m − 1)
n−m
∈K 0
2
n−m
∈K ∗
2
where K 0 := {k ∈ K : k + 1 ∈ K} ∪ { n−2
} if n−2
is in K, and K 0 :=
2
2
{k ∈ K : k + 1 ∈ K} otherwise; and K ∗ := K 0 ∪ {k ∈ K : k − 1 ∈ K}.
(iv) If n ≥ 2 and αi <
is even.
1
2
for i = 1, ..., n − 1, then a(k, m) > 0 whenever k − m
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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(v) If αi < 12 for i = 1, ..., n − 3, then equality holds in (2.2) if and only if
either K is the whole set {0, ..., [ n−1
]} itself (i.e., (2.2) is just Pn−2 (1) =
2
Pn−3 (1) = 1) or K 0 and K ∗ are the empty sets.
Proof. Suppose that P0 , ..., Pn satisfy (1.1) and (2.1) with nonnegative coefficients such that Pk (1) = 1, and a(k, k) and b(k, k) are decreasing (0 ≤ k ≤ n).
It follows that a(0, 0) = a(1, 1) = 1 and
(2.3)
γk a(k + 1, j) = −αk a(k − 1, j) − βk a(k, j)
1
+ (a(k, j + 1) + (1 + δ1j )a(k, j − 1))
2
for k = 0, ..., n − 1 (We use the convention that entries of vectors or matrices
with any negative indices are zero; and a(i, j) = 0 if i < j. We also assume
α0 = 0.). Similarly, by substituting (2.1) into both sides of xk+1 = xxk we have
b(0, 0) = 1 and
(2.4)
b(k + 1, j) = γj−1 b(k, j − 1) + βj b(k, j) + αj+1 b(k, j + 1).
With j = k +1 we conclude γk , a(k, k) and b(k, k) are positive; and with j = k,
βk = a(k + 1, k) = b(k + 1, k) = 0 for k = 0, ..., n − 1. Similarly property (i)
follows by induction.
By the normalization we have γk + αk = 1. Thus by (2.3) and (2.4) with
j = k + 1, it follows that 12 ≤ γk ≤ 1, so 0 ≤ αk ≤ 12 (0 ≤ k ≤ n − 1), since
a(k, k) and b(k, k) are decreasing.
Conversely suppose that P0 , ..., Pn are normalized and given by (1.1) for
constants such that 0 ≤ αi ≤ 21 and βi = 0 (0 ≤ i ≤ n − 1). It follows that the
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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degree of Pk is k and thus Pk satisfies (2.1) for coefficients a(k, m) and b(k, m)
generated by (2.3) and (2.4) respectively. Since γk = 1 − αk > 0, the above
argument shows that property (i) holds and b(k, m) ≥ 0; and since 12 ≤ γk ≤ 1,
a(k, k) and b(k, k) are decreasing for k = 0, ..., n.
We will show that a(k, m) ≥ 0 simultaneously with inequality (2.2) by in1
duction on n. Now a(0, 0) = a(1, 1) = 1, a(2, 0) = 1−2α
, a(2, 2) = 2γ11 ,
2γ1
γ2 a(3, 1) = −α2 + 12 + 12 a(2, 0) and a(3, 3) = 4γ11 γ2 . Hence a(k, m) ≥ 0
when n is either 2 or 3. Also (2.2) is easily checked when n = 3.
Henceforth let n > 3 and suppose that a(i, i − 2k) ≥ 0 (0 ≤ i ≤ n − 1;
0 ≤ k ≤ [ 2i ]) and that (2.2) is true for all integers n0 such that 3 ≤ n0 < n. Let
j be given, 0 ≤ j ≤ [ n2 ], and let K be a subset of {0, ..., [ n−1
]}. We show that
2
a(n, n − 2j) ≥ 0 and that
X
X
a(n − 3, n − 2k − 1)
(2.5)
a(n − 2, n − 2k − 2) ≤
k∈K ∗
k∈K 0
James Guyker
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beginning with the latter.
We may assume that K is not the whole set {0, ..., [ n−1
]} since in the con2
trary case (2.5) reduces to Pn−2 (1) = 1 ≤ Pn−3 (1) = 1. Observe that K may
be written as a disjoint union of sets of the form {k, ..., k + m} such that there
is at least one integer between any two such sets. Since K 0 and K ∗ are then
corresponding disjoint unions of subsets of these sets, and both sides of (2.5)
may be separated into sums over these subsets accordingly, we may assume
K = {k, ..., k + m}.
Thus we wish to show
m
m
X
X
(2.6)
δk+i, n−2 a(n − 2, n − 2(k + i) − 2) ≤
a(n − 3, n − 2(k + i) − 1).
2
i=0
An Inequality for Chebyshev
Connection Coefficients
i=0
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By (2.3) we have
γn−3 a(n − 2, n − 2(k + i) − 2)
= −αn−3 a(n − 4, n − 2(k + i) − 2)
1
+ (a(n − 3, n − 2(k + i) − 1)
2
+ (1 + δ3,n−2(k+i) )a(n − 3, n − 2(k + i) − 3)),
where δ3,n−2(k+i) = 1 if and only if n is odd and k + i + 1 = n−1
= k+m
2
n−2
(∈ K). Moreover, a(n − 3, n − 2(k + i) − 3) = 0 if n is even and 2 = k + m.
Hence the left side of (2.6) may be combined as follows:
m
X
δk+i, n−2 a(n − 2, n − 2(k + i) − 2)
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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2
i=0
=−
αn−3
γn−3
m
X
Contents
δk+i, n−2 a(n − 4, n − 2(k + i) − 2)
2
i=0
1
m−1
X
γn−3
i=1
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1
+
a(n − 3, n − 2k − 1)
2γn−3
+ 1 + δk+m,[ n−1 ] a(n − 3, n − 2(k + m) − 1) .
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Therefore (2.6) becomes
m
X
αn−3
−
γn−3
(2.7)
−
δk+i, n−2 a(n − 4, n − 2(k + i) − 2)
2
i=0
m−1
X
!
a(n − 3, n − 2(k + i) − 1)
i=1
1 +
1 + δk+m,[ n−1 ] a(n − 3, n − 2(k + m) − 1)
2
2γn−3
1 − 2αn−3
≤
a(n − 3, n − 2k − 1) + a(n − 3, n − 2(k + m) − 1).
2γn−3
Let us suppose first that k + m 6= [ n−1
]. Then (2.7) may be rearranged as
2
follows:
− αn−3
(2.8)
m−1
X
i=0
−
≤
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a(n − 3, n − 2(k + i) − 1)
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a(n − 4, n − 2(k + i) − 2)
m−1
X
An Inequality for Chebyshev
Connection Coefficients
1
− αn−3 (a(n − 3, n − 2k − 1) + a(n − 3, n − 2(k + m) − 1)),
2
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0
which is clearly true if αn−3 = 0 so we assume αn−3 6= 0. With n = n − 1,
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replacing i by i + 1 in the second sum, we seek to show
(2.9)
m−2
X
0
m−1
X
0
a(n − 2, n − 2(k + i) − 2) ≤
i=0
a(n0 − 3, n0 − 2(k + i) − 1).
i=0
n0 −1
0
Since k + m < [ n−1
],
we
have
K
=
k,
...,
k
+
m
−
1}
⊆
{0,
...,
[
]
.
2
2
Moreover, (K 0 )0 = {k, ..., k + m − 2} where the second prime is with respect
to n0 . Hence (2.9) follows from the induction hypothesis
X
X
a(n0 − 2, n0 − 2k − 2) ≤
a(n0 − 3, n − 2k − 1).
(2.10)
k∈(K 0 )0
k∈(K 0 )∗
James Guyker
Finally suppose that k + m = [ n−1
] so that (2.7) becomes
2
(2.11)
− αn−3
m
X
δk+i, n−2 a(n − 4, n − 2(k + i) − 2)
2
i=0
−
m
X
An Inequality for Chebyshev
Connection Coefficients
!
a(n − 3, n − 2(k + i) − 1)
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i=1
≤
1
− αn−3 a(n − 3, n − 2k − 1).
2
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As above we may assume αn−3 6= 0 and wish to show
(2.12)
m−1
X
i=0
0
0
a(n −2, n −2(k+i)−2) ≤
m
X
δk+i, n−2 a(n0 −3, n0 −2(k+i)−1).
2
i=0
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If n is even then n0 is odd and (K 0 )0 (with respect to n0 ) is {k, ..., k + m − 1}.
Thus (2.12) is a consequence of the induction hypothesis as above. On the other
hand, if n is odd, then (K 0 )0 = {k, ..., k + m − 1} and (2.12) again reduces to
the hypothesis.
Q
1
Next we verify a(n, n−2j) ≥ 0. Since a(n, n) = nj=2 2γn−j+1
> 0 suppose
further that j ≥ 1. Let
n−1
K0 := k : k integer, 0 ≤ k ≤
, k 6= j − 1, j .
2
n−2
,
2
If n is even and j =
define K1 := K0 ; otherwise let
n−2
K1 := k : k integer, 0 ≤ k ≤
, k 6= j − 1 .
2
[ n−1
]
2
[ n−2
].
2
Note P
that a(n − 2, n − 2k − 2) = 0 if k =
>
Furthermore, the
sum
k∈K
/ 0 a(n − 2, n − 2k − 2) is zero unless j + 1 is in K0 (in which
k+1∈K0
case the sum is a(n − 2, n − 2j − 2)). Solving the equations Pn−2 (1) = 1 and
Pn−1 (1) = 1 for a(n − 2, n − 2j) and a(n − 1, n − 2j + 1) + a(n − 1, n − 2j − 1)
respectively, and substituting them into γn−1 a(n, n − 2j) as given by (2.3), we
have the following identities.
γn−1 a(n, n − 2j)
X
1
1
1 − 2αn−2
= − αn−1 +
αn−1 −
1−
a(n − 2, n − 2k − 2)
2
2
2γn−2
k∈K1
αn−2 X
a(n − 3, n − 2k − 1)
+ αn−1 δn,2j+2 a(n − 2, 0) +
2γn−2 k∈K
0
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Connection Coefficients
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1
−
2γn−2
1 − 2αn−2
αn−2 +
2
X
a(n − 2, n − 2k − 2)
k∈K00
1
+ δ1,n−2j a(n − 1, n − 2j − 1)
2
X
1
=
a(n − 2, n − 2k − 2)
− αn−1
2
k∈K
/ 1
X
1
1
+
− αn−2
a(n − 2, n − 2k − 2)
2γn−2 2
k∈K
1
0
k∈K
/ 0
1
+ αn−1 δn,2j+2 a(n − 2, 0) + δ1,n−2j a(n − 1, n − 2j − 1)
2


X
X
αn−2 
+
a(n − 3, n − 2k − 1) −
a(n − 2, n − 2k − 2) .
2γn−2 k∈K
0
0
k∈K0
Each of the terms in the last expression is nonnegative, the final one a result of
the induction assumption on (2.2). Therefore a(n, n − 2j) ≥ 0.
Property (i) has already been shown so for (ii), assume α1 > 0. Since
b(0, 0) = 1 and b(k + 1, 0) = α1 γ0 b(k − 1, 0) + α1 α2 b(k − 1, 2), we have that
b(k + 1, 0) > 0 for k + 1 even. But also b(k + 1, 0) = α1 b(k, 1) so b(k, 1) > 0
for odd k. Hence (ii) is now straightforward from (2.4).
For property (iv), suppose αi < 21 (i = 1, ..., n − 1). By the induction argument above, the first term of the last identity for γn−1 a(n, n − 2j) is positive
since k = j − 1 ∈
/ K1 . Since the other terms are nonnegative, a(n, n − 2j) > 0.
Next let αi < 12 (i = 1, ..., n − 3) and suppose that equality holds in (2.2)
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Connection Coefficients
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for some subset K of {0, ..., [ n−1
]}. As above, it follows that equality holds in
2
(2.2) over each subset of K of the form {k, ..., k + m}. Hence assume K =
then K 0 and K ∗ are empty. If K = { n−2
},
{k, ..., k+m}. If m = 0 and k 6= n−2
2
2
1
then by equality in (2.11) we have −αn−3 a(n − 4, 0) = 2 − αn−3 a(n − 3, 1)
which is impossible since the left side is nonpositive and the right side is positive
by property (iv).
Therefore, let m ≥ 1. If k + m 6= [ n−1
] then equality holds in (2.8), and by
2
(2.9) and (2.10) we have

− αn−3 

X
k∈(K 0 )∗
=
a(n0 − 3, n0 − 2k − 1) −
X
a(n0 − 2, n0 − 2k − 2)
An Inequality for Chebyshev
Connection Coefficients
James Guyker
k∈(K 0 )0
1
− αn−3 (a(n − 3, n − 2k − 1) + a(n − 3, n − 2(k + m) − 1)).
2
This is impossible since both sides must be zero which implies a(n − 3, n −
2(k + m) − 1) = 0.
Finally, if k +m = [ n−1
] then equality holds in (2.11) and a similar argument
2
shows a(n − 3, n − 2k − 1) = 0 which by property (iv) implies k = 0 and thus
K must be the whole set {0, ..., [ n−1
]}.
2
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3.
Bounds on Polynomial Derivatives
It is well known [9] that Tn is bounded by one and for 1 ≤ k ≤ n
2
2
2
2
2
(k) Tn (x) ≤ Tn(k) (1) = n (n − 1)(n − 4) · · · (n − (k − 1) )
1 · 3 · 5 · · · (2k − 1)
(k) (k)
when −1 ≤ x ≤ 1; and if 1 ≤ k < n and Tn (x) = Tn (1), then x = ±1.
More generally, it follows from a result of R.J. Duffin and A.C. Schaeffer ([7],
[9, Thm. 2.24]) that if Pn is anypolynomial
of degree n that is bounded by one
(k) (k)
in [−1, 1] and 1 ≤ k < n, then Pn (x) ≤ Tn (1) with equality holding only
when Pn = ±Tn and x = ±1. For the polynomials in Theorem 2.1, we may be
more precise.
Corollary 3.1. Let P0 , ..., Pn be defined by (1.1) with 0 ≤ αi ≤
γi = 1 − αi for i = 0, ..., n − 1. Then
1
,
2
βi = 0, and
(a) |Pn (x)| ≤ 1 = Pn (1) = Tn (1); and if n ≥ 1, |Pn (x)| = 1, and αi <
i = 1, ..., n − 1, then x = ±1.
1
2
for
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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(b) If 1 ≤ k ≤ n, then
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(k) Pn (x) ≤ Pn(k) (1) ≤ Tn(k) (1)
for all x in [−1, 1]. Moreover in this case:
(k) (k)
(i) If k < n and Pn (x) = Pn (1), then x = ±1.
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(k)
(k)
(ii) If Pn (1) = Tn (1), then Pn = Tn . In particular, if n ≥ 2 and αi < 12
(k)
(k)
(i = 1, ..., n − 1), then Pn (1) < Tn (1).
P
Proof. By Theorem 2.1, Pn = nm=0 a(n, m)Tm where a(n, m) ≥ 0. Therefore
n
X
|Pn (x)| ≤
a(n, m) |Tm (x)| ≤ Pn (1) = 1.
m=0
Suppose that n ≥ 1 and |Pn (x)| = 1. If n = 1, then x = ±1, so assume
n ≥ 2 and αi < 21 (i = 1, ..., n − 1). Then Pn (x) = ±Pn (1) and hence
a(n, m)(1 ± Tm (x)) = 0 for all m. Since T0 = 1, T1 = x and T2 = 2x2 − 1,
property (iv) of Theorem 2.1 implies
that
D
E x = ±1.
(k)
Next assume k ≥ 1. Since Tm (1)
is increasing,
m
n
(k) X
(k)
Pn (x) = a(n, m)Tm (x)
≤
≤
m=k
n
X
a(n, m) Tm(k) (x)
m=k
n
X
a(n, m)Tm(k) (1) = Pn(k) (1)
James Guyker
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m=k
≤ Tn(k) (1)
An Inequality for Chebyshev
Connection Coefficients
n
X
a(n, m) ≤ Tn(k) (1).
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(k) (k)
Assume 1 ≤ k < n and Pn (x) = Pn (1). Then
0=
n
X
a(n, m) Tm(k) (1) − Tm(k) (x) ,
m=k
(k) where each term is nonnegative. Since a(n, n) > 0 we have that Tn (x) =
(k)
Tn (1) so x = ±1.
P
(k)
(k)
Finally suppose that k ≥ 1 and Pn (1) = Tn (1). Then nm=k a(n, m) =
(k)
(k)
1 so a(n, m) = 0 for m < k. Also a(n, m)Tm (1) = a(n, m)Tn (1) so
a(n, m) = 0 for m = k, ..., n−1. Therefore Pn = a(n, n)Tn and thus a(n, n) =
1.
However the previous case is impossible if n ≥ 2 and αi < 21 (i = 1, ..., n −
1) since by property (iv) of Theorem 2.1 we would have a(n, 0) > 0 or a(n, 1) >
0.
(k)
Remark 1. For fixed k the sequence Pn (1) of bounds is increasing. In fact
(k)
by (1.1), Pn (1) may be generated recursively as follows: Initially we have
(1)
(k)
(k−1)
(k)
k
P1 (1) = 1, Pk (1) = 1−αkk−1 Pk−1 (1) (k ≥ 2) and set ekk := k+α
P (1).
1−αk k
Then for n ≥ 1,
(k)
Pn+1 (1) = Pn(k) (1) + enk ≥ Pn(k) (1) ≥ 0,
where the differences enk are defined by
enk :=
αn
k
en−1,k +
P (k−1) (1).
1 − αn
1 − αn n
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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{α}
Ultraspherical polynomials y = Pn with α ≥ − 12 satisfy the differential
equation
(1 − x2 )y 00 − 2(α + 1)xy 0 + n(n + 2α + 1)y = 0
(k)
and thus a closed form for Pn (1) is possible in this case since
Pn(k) (1) =
n(n + 2α + 1) − (k − 1)(2α + 1) − (k − 1)2 (k−1)
Pn
(1).
2(α + k)
This extends known Chebyshev and Legendre identities ([9, p. 33], [11, p.
251]).
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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4.
Polynomial Expansions of Power Series
A standard application of the theory of orthogonal polynomials is the least
squares or uniform approximation of functions by partial sums of generalized
Fourier expansions in terms of orthogonal polynomials, especially Chebyshev
polynomials. The coefficients of the expansion are given by an inner product
used in generating the polynomials. In our case, we may define Fourier coefficients forP
expansions of power series in terms of the polynomials that satisfy
ai xi be a convergent power series
(1.1): Let
Pon (−1, 1), and for every n let
Pn be a polynomial of degree n. Then xn = nm=0 b(n,P
m)Pm for some numbers b(n, m); and we define the Fourier coefficient cj of ai xi with respect to
the sequence hPn i by
X
cj :=
ai b(i, j)
i
Pn
whenever this sum converges. Note that cnj :=
P i=0 ai b(i, j) is then the jth
coefficient in the expansion of the partial sum ni=0 ai xi : since b(i, j) = 0 for
j > i,
n
n
n
n
X
X
X
X
i
ai x =
ai
b(i, j)Pj =
cnj Pj .
i=0
i=0
j=0
j=0
We have the following estimate where kf k denotes the uniform norm
max{|f (x)| : −1 ≤ x ≤ 1}. The optimal property of Chebyshev expansions
extends a result of T.J. Rivlin and M.W. Wilson ([10], [9, Thm. 3.17]).
Corollary 4.1. Let hPn i be given by (1.1) with
P hαn i, hβn i and hγn i nonnegative,
and suppose that Pn (1) = 1 for all n. If
ai converges absolutely, then the
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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coefficient cj of
P
ai xi with respect to hPn i exists for every j and
n
X
X
ai b(i, j) ≤
|ai | .
cj −
i>n
i=0
Moreover, if
P
i−j even
iai converges absolutely, then
n
X
X
X
i
ai x −
cj Pj (x) ≤
(|ai | + |iai |)
j=0
i>n
1
2
for all x in [−1, 1]. In this case, if αi ≤ and βi = 0 for iP
= 0, ..., n − 1, and if
ai ≥ 0 for all i and dk is the kth Chebyshev coefficient of ai xi , then
n
n
X
X
X
X
(4.1)
ai x i −
c j Pj ≥ ai x i −
dk Tk .
j=0
James Guyker
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k=0
InP
addition, if αi < 21 (i = 1, ..., n − 1), then equality holds in (4.1) if and only
if ai xi is a polynomial of degree at most n.
Proof. Asume that hαn i, hβn i and hγn i are nonnegative,
Pnand Pn (1) = 1 for all
n
n. By (1.1) the degree of Pn is n for all n. Thus x = m=0 b(n, m)Pm where
b(n, m)
Pn≥ 0 by (2.4), and b(n, m) ≤ 1 by the normalization since we have
n
1 = m=0 b(n, P
m).
P
Suppose that
|ai | converges. Then i ai b(i, j) converges absolutely by
the comparison test so cj exists and
X
X
ai b(i, j) ≤
|ai | .
i>n
i>n
An Inequality for Chebyshev
Connection Coefficients
i−j even
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P
P
Next assume |iai | < ∞. Then |ai | < ∞ so cj exists for every j. Thus by
Corollary 3.1 we have
n
n
X
X
X
X
i
i
ai x −
cj Pj (x) ≤ ai x + (cj − cnj )Pj (x)
j=0
i>n
≤
X
|ai | +
i>n
j=0
n
X
|cj − cnj | ,
j=0
An Inequality for Chebyshev
Connection Coefficients
where
n X
n
X
X 1
X
n+1X
|iai | ≤
|iai | .
|cj − cnj | =
ai b(i, j) ≤
i
n + 1 i>n
j=0
j=0 i>n
j=0 i≥n+1
James Guyker
Suppose further that αi ≤ 12 , βi = 0 (i = 0, ..., n − 1) and ai is nonnegative for
Pj
all i. Then cj ≥ 0 for all j and P
by Theorem
2.1,
P
=
j
k=0 a(j, k)Tk where
P
a(j, k) ≥ 0. Since we also have ai xi =
cj Pj uniformly on [−1, 1] in this
case, it follows that
n
X
X
X
X
i
ai x −
c j Pj = c j Pj =
cj
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n
X
j=0
j>n
j>n
and similarly
n
X
X
X
dk Tk =
dk .
ai x i −
k=0
k>n
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But
X
dk Tk =
X
ai x i
=
X
c j Pj
=
X
cj
j
∞
X
a(j, k)Tk
k=0
!
=
X X
k
An Inequality for Chebyshev
Connection Coefficients
cj a(j, k) Tk
j≥k
James Guyker
since
X
X
X
X
cj a(j, k) ≤
cj ≤
cj Pj (1) =
ai < ∞.
j≥k
j≥k
SinceP
the coefficients in a uniformly convergent Chebyshev expansion are unique,
dk = j≥k cj a(j, k). Therefore
X
dk =
X
cj
j>n
k>n
=
X
X
=
j>n
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k>n
cj
j
X
j>n
X
a(j, k)
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cj −
a(j, k) −
k=0
n X
X
k=0 j>n
n
X
!
a(j, k)
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cj a(j, k) ≤
X
j>n
cj .
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Finally, assume that αi < 12 (i = 1, ..., n − 1). By property (iv) of Theorem 2.1,
a(j, 0) + a(j, 1) > 0 for j P
> n so ifP
equality holds in the last inequality then
cj = 0 for all j > n. Thus ai xi =
cj Pj is a polynomial of degree at most
n.
P
Remark 2. If hPn i is defined as in Corollary 4.1 and, more generally, (iai )2
converges, then by the Schwarz inequality it follows that
! 12
!1
X
X
X1 2
|ai | ≤
(iai )2
i2
i>n
i>n
i>n
P
so |ai | < ∞ and cj exists for every j. Moreover in this case we have
n
X
X
ai x i −
cj Pj (x)
j=0
n X
X
X
i
≤
ai x +
ai b(i, j)
i>n
j=0 i>n
! 21
!1
! 12
!1
n
X
X x i 2 2 X
X
X b(i, j) 2 2
≤
+
(iai )2
(iai )2
i
i
i>n
j=0
i>n
i>n
i>n
!1
! 12
! 12
X1 2 X
X
n
+
2
≤ (n + 2)
(iai )2
≤ √
(iai )2
2
i
n
i>n
i>n
i>n
where the last inequality follows from the proof of the integral test.
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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References
[1] R. ASKEY, Orthogonal expansions with positive coefficients, Proc. Amer.
Math. Soc., 16 (1965), 1191–1194.
[2] R. ASKEY, Orthogonal expansions with positive coefficients. II, SIAM J.
Math. Anal., 2 (1971), 340–346.
[3] R. ASKEY, Jacobi polynomial expansions with positive coefficients and
imbeddings of projective spaces, Bull. Amer. Math. Soc., 74 (1968), 301–
304.
An Inequality for Chebyshev
Connection Coefficients
[4] R. ASKEY, Orthogonal Polynomials and Special Functions, Regional
Conference Series in Applied Mathematics, 21, SIAM, Philadelphia, PA,
1975.
James Guyker
[5] R. ASKEY AND G. GASPER, Jacobi polynomial expansions of Jacobi
polynomials with non-negative coefficients, Proc. Camb. Phil. Soc., 70
(1971), 243–255.
Contents
[6] H. BATEMAN, Higher Transcendental Functions, V. 2, Bateman
Manuscript Project, McGraw-Hill, New York, 1953.
[7] R.J. DUFFIN AND A.C. SCHAEFFER, A refinement of an inequality of
the brothers Markoff, Trans. Amer. Math. Soc., 50 (1941), 517–528.
[8] E.D. RAINVILLE, Special Functions, Macmillan, New York, 1960.
[9] T.J. RIVLIN, The Chebyshev Polynomials, Wiley, New York, 1974.
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[10] T.J. RIVLIN AND M.W. WILSON, An optimal property of Chebyshev
expansions, J. Approx. Theory, 2 (1969), 312–317.
[11] G. SANSONE, Orthogonal Functions, Interscience, New York, 1959.
[12] G. SZEGÖ, Orthogonal Polynomials, Amer. Math. Soc. Colloq. Publ., 23,
Amer. Math. Soc., Providence, RI, 1939.
[13] R. SZWARC, Connection coefficients of orthogonal polynomials, Canad.
Math. Bull., 35 (1992), 548–556.
[14] W.F. TRENCH, Proof of a conjecture of Askey on orthogonal expansions
with positive coefficients, Bull. Amer. Math. Soc., 81 (1975), 954–956.
[15] M.W. WILSON, Nonnegative expansions of polynomials, Proc. Amer.
Math. Soc., 24 (1970), 100–102.
An Inequality for Chebyshev
Connection Coefficients
James Guyker
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