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volume 7, issue 2, article 67, 2006.

Received 19 January, 2006;

accepted 11 March, 2006.

Communicated by:D. Stefanescu

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Journal of Inequalities in Pure and Applied Mathematics

AN INEQUALITY FOR CHEBYSHEV CONNECTION COEFFICIENTS

JAMES GUYKER

Department of Mathematics SUNY College at Buffalo 1300 Elmwood Avenue Buffalo, New York 14222-1095 USA.

EMail:guykerj@buffalostate.edu

c

2000Victoria University ISSN (electronic): 1443-5756 021-06

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An Inequality for Chebyshev Connection Coefficients

James Guyker

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J. Ineq. Pure and Appl. Math. 7(2) Art. 67, 2006

http://jipam.vu.edu.au

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 expan- sion.

1. Introduction

In the classical theory of orthogonal polynomials a standard problem is to ex- pand 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 re- currence 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 sequencesα_n, β_nand nonzeroγ_nwe consider the sequence of polynomialsP_ndefined byP−1 = 0, P₀ = 1,and forn≥0,

(1.1) xP_n=γ_nP_n+1+β_nP_n+α_nPn−1.

For example the Jacobi polynomials Pn^(α,β) (α, β > −1) are defined by (1.1) with

α_n= 2(α+n)(β+n)

(α+β+ 2n+ 1)(α+β+ 2n),

β_n = β²−α²

(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 expansions of the normalized ultraspherical or Gegenbauer polynomials Pn^{α} :=

Pn^(α,α)

Pn^(α,α)(1).We recall the factorial function(α)_n:=α(α+ 1)· · ·(α+n−1)when n ≥1,and(α)₀ := 1forα6= 0. ThenP^{−

1 2}

n =T_nand

(1.2) P_n^{α} =

n

X

m=0

c(n, m)T_m

α6=−1 2

, wherec(n, m) = 0ifn−mis odd, and otherwise

c(n, m) =

(2−δm0) α+ ¹₂

n−m 2

α+ ¹₂

n+m 2

(2α+ 1)n

n

n−m 2

.

It follows that ifα >−¹₂ andn−mis even, thenc(n, m)>0and the sequence hc(n, n)iis monotonically decreasing. Moreover, we have the inversions

xⁿ = [ⁿ2] X

k=0

2−δ_n,2k 2ⁿ

n k

P^{−

1 2} n−2k

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([·]is the greatest integer function) and

(1.3) xⁿ=

n

X

m=0

d(n, m)P_m^{α}

α6=−1 2

, whered(n, m) = 0ifn−mis odd, and otherwise

d(n, m) =

(2α+ 1 + 2m)(2α+ 1)m n−m 2 + 1

n−m 2

2ⁿ⁺¹ α+ ¹₂

n+m+2 2

n m

is positive and hd(n, n)iis decreasing. The polynomialsPn^{α} satisfy (1.1) for sequences such that αn > 0andβn = 0. Furthermore, α > −¹₂ if and only if α_n< ¹₂.

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2. The Chebyshev Expansion of P

_n

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 byP_n(1) = 1. Withnfixed, we have that

(2.1) P_k=

k

X

m=0

a(k, m)T_m and x^k =

k

X

m=0

b(k, m)P_m (0≤k ≤n) for nonnegative coefficientsa(k, m)andb(k, m)such that the coefficientsa(k, k) andb(k, k)are monotonically decreasing if and only if0≤α_i ≤ ¹₂ andβ_i = 0 fori= 0, ..., n−1.In this case, the following properties hold.

(i) a(k, m) =b(k, m) = 0ifk−mis odd.

(ii) Ifα₁ >0, thenb(k, m)>0wheneverk−mis even.

(iii) Ifn ≥3andK is any subset of{0, ...,[ⁿ⁻¹₂ ]}, then

(2.2) X

n−m 2 ∈K⁰

a(n−2, m−2)≤ X

n−m 2 ∈K^∗

a(n−3, m−1)

whereK⁰ := {k ∈ K : k+ 1 ∈ K} ∪ {ⁿ⁻²₂ }if ⁿ⁻²₂ is in K, andK⁰ :=

{k∈K :k+ 1∈K}otherwise; andK^∗ :=K⁰∪ {k∈K :k−1∈K}.

(iv) Ifn≥2andα_i < ¹₂ fori= 1, ..., n−1, thena(k, m)>0wheneverk−m is even.

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(v) If α_i < ¹₂ fori = 1, ..., n−3, then equality holds in (2.2) if and only if eitherK is the whole set {0, ...,[ⁿ⁻¹₂ ]}itself (i.e., (2.2) is just Pn−2(1) = Pn−3(1) = 1) orK⁰andK^∗ are the empty sets.

Proof. Suppose that P₀, ..., P_n satisfy (1.1) and (2.1) with nonnegative coefficients such thatPk(1) = 1,anda(k, k)andb(k, k)are decreasing (0≤k ≤n).

It follows thata(0,0) =a(1,1) = 1and

(2.3) γ_ka(k+ 1, j) =−α_ka(k−1, j)−β_ka(k, j) +1

2(a(k, j+ 1) + (1 +δ_1j)a(k, j−1)) 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 ofx^k+1 =xx^kwe have b(0,0) = 1and

(2.4) b(k+ 1, j) =γj−1b(k, j−1) +β_jb(k, j) +α_j+1b(k, j+ 1).

Withj =k+1we concludeγ_k, a(k, k)andb(k, k)are positive; and withj =k, β_k =a(k+ 1, k) =b(k+ 1, k) = 0fork = 0, ..., n−1. Similarly property (i) follows by induction.

By the normalization we haveγ_k +α_k = 1. Thus by (2.3) and (2.4) with j = k+ 1, it follows that ¹₂ ≤ γ_k ≤ 1,so0 ≤α_k ≤ ¹₂ (0 ≤k ≤n−1), since a(k, k)andb(k, k)are decreasing.

Conversely suppose that P₀, ..., P_n are normalized and given by (1.1) for constants such that0≤α_i ≤ ¹₂ andβ_i = 0(0≤i≤n−1). It follows that the

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degree ofP_kiskand thusP_ksatisfies (2.1) for coefficientsa(k, m)andb(k, m) generated by (2.3) and (2.4) respectively. Since γk = 1−αk > 0, the above argument shows that property (i) holds andb(k, m)≥0; and since ¹₂ ≤γ_k≤1, a(k, k)andb(k, k)are decreasing fork = 0, ..., n.

We will show thata(k, m) ≥ 0simultaneously with inequality (2.2) by induction on n. Now a(0,0) = a(1,1) = 1, a(2,0) = ^1−2α_2γ ¹

1 , a(2,2) = _2γ¹

1, γ2a(3,1) = −α2+ ¹₂

+ ¹₂a(2,0) and a(3,3) = _4γ¹

1γ2. Hence a(k, m) ≥ 0 whennis either 2 or 3. Also (2.2) is easily checked whenn= 3.

Henceforth let n > 3and suppose that a(i, i−2k) ≥ 0 (0 ≤ i ≤ n −1;

0 ≤k ≤ [₂ⁱ]) and that (2.2) is true for all integersn⁰ such that3≤ n⁰ < n. Let j be given,0 ≤ j ≤ [ⁿ₂], and let K be a subset of{0, ...,[ⁿ⁻¹₂ ]}. We show that a(n, n−2j)≥0and that

(2.5) X

k∈K⁰

a(n−2, n−2k−2)≤ X

k∈K^∗

a(n−3, n−2k−1) beginning with the latter.

We may assume thatK is not the whole set{0, ...,[ⁿ⁻¹₂ ]} since in the con- trary case (2.5) reduces toPn−2(1) = 1 ≤ Pn−3(1) = 1.Observe thatK 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 K⁰ 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 (2.6)

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−2, n−2(k+i)−2)≤

m

X

i=0

a(n−3, n−2(k+i)−1).

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By (2.3) we have

γn−3a(n−2, n−2(k+i)−2)

=−αn−3a(n−4, n−2(k+i)−2) + 1

2(a(n−3, n−2(k+i)−1)

+ (1 +δ3,n−2(k+i))a(n−3, n−2(k+i)−3)), where δ_3,n−2(k+i) = 1 if and only if n is odd and k +i+ 1 = ⁿ⁻¹₂ = k +m (∈K). Moreover,a(n−3, n−2(k+i)−3) = 0ifnis even and ⁿ⁻²₂ =k+m.

Hence the left side of (2.6) may be combined as follows:

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−2, n−2(k+i)−2)

=−αn−3

γn−3 m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−4, n−2(k+i)−2) + 1

γn−3 m−1

X

i=1

a(n−3, n−2(k+i)−1)

+ 1

2γ_n−3

a(n−3, n−2k−1) +

1 +δ_k+m,[ⁿ⁻¹

2 ]

a(n−3, n−2(k+m)−1)

.

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Therefore (2.6) becomes

(2.7) −αn−3

γn−3 m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−4, n−2(k+i)−2)

−

m−1

X

i=1

a(n−3, n−2(k+i)−1)

!

+ 1

2γ_n−3

1 +δ_k+m,[ⁿ⁻¹

2 ]

a(n−3, n−2(k+m)−1)

≤ 1−2αn−3

2γn−3

a(n−3, n−2k−1) +a(n−3, n−2(k+m)−1).

Let us suppose first that k +m 6= [ⁿ⁻¹₂ ]. Then (2.7) may be rearranged as follows:

(2.8) −αn−3 m−1

X

i=0

a(n−4, n−2(k+i)−2)

−

m−1

X

i=1

a(n−3, n−2(k+i)−1)

!

≤ 1

2 −αn−3

(a(n−3, n−2k−1) +a(n−3, n−2(k+m)−1)), which is clearly true if αn−3 = 0 so we assume αn−3 6= 0. With n⁰ = n −1,

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replacingibyi+ 1in the second sum, we seek to show (2.9)

m−2

X

i=0

a(n⁰−2, n⁰−2(k+i)−2)≤

m−1

X

i=0

a(n⁰−3, n⁰−2(k+i)−1).

Since k + m < [ⁿ⁻¹₂ ], we have K⁰ =

k, ..., k+m−1} ⊆ {0, ...,[ⁿ⁰₂⁻¹] . Moreover,(K⁰)⁰ ={k, ..., k+m−2}where the second prime is with respect ton⁰. Hence (2.9) follows from the induction hypothesis

(2.10) X

k∈(K⁰)⁰

a(n⁰−2, n⁰−2k−2)≤ X

k∈(K⁰)^∗

a(n⁰−3, n−2k−1).

Finally suppose thatk+m= [ⁿ⁻¹₂ ]so that (2.7) becomes (2.11) −αn−3

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−4, n−2(k+i)−2)

−

m

X

i=1

a(n−3, n−2(k+i)−1)

!

≤ 1

2−αn−3

a(n−3, n−2k−1).

As above we may assumeα_n−3 6= 0and wish to show (2.12)

m−1

X

i=0

a(n⁰−2, n⁰−2(k+i)−2)≤

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n⁰−3, n⁰−2(k+i)−1).

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Ifn is even thenn⁰ is odd and(K⁰)⁰ (with respect ton⁰) is{k, ..., k+m−1}.

Thus (2.12) is a consequence of the induction hypothesis as above. On the other hand, if nis odd, then(K⁰)⁰ ={k, ..., k+m−1}and (2.12) again reduces to the hypothesis.

Next we verifya(n, n−2j)≥0. Sincea(n, n) =Qn j=2

1

2γn−j+1 >0suppose further thatj ≥1. Let

K₀ :=

k :kinteger,0≤k ≤

n−1 2

, k 6=j −1, j

. Ifnis even andj = ⁿ⁻²₂ , defineK1 :=K0; otherwise let

K1 :=

k :kinteger,0≤k ≤

n−2 2

, k 6=j −1

.

Note that a(n−2, n −2k −2) = 0 ifk = [ⁿ⁻¹₂ ] > [ⁿ⁻²₂ ]. Furthermore, the sum P

k /∈K0 k+1∈K0

a(n −2, n −2k − 2) is zero unless j + 1 is in K₀ (in which case the sum isa(n−2, n−2j −2)). Solving the equationsPn−2(1) = 1and Pn−1(1) = 1fora(n−2, n−2j)anda(n−1, n−2j+ 1) +a(n−1, n−2j−1) respectively, and substituting them intoγn−1a(n, n−2j)as given by (2.3), we have the following identities.

γ_n−1a(n, n−2j)

= 1

2−αn−1+ X

k∈K1

αn−1− 1 2

1− 1−2αn−2

2γn−2

a(n−2, n−2k−2) +α_n−1δ_n,2j+2a(n−2,0) + αn−2

2γn−2

X

k∈K0

a(n−3, n−2k−1)

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− 1 2γn−2

αn−2+ 1−2αn−2

2

X

k∈K₀⁰

a(n−2, n−2k−2) +1

2δ1,n−2ja(n−1, n−2j −1)

= 1

2−α_n−1

X

k /∈K1

a(n−2, n−2k−2)

+ 1

2γn−2

1

2−αn−2

X

k∈K1 k /∈K0 0

a(n−2, n−2k−2)

+αn−1δ_n,2j+2a(n−2,0) + 1

2δ1,n−2ja(n−1, n−2j−1) + αn−2

2γ_n−2



 X

k∈K0

a(n−3, n−2k−1)− X

k∈K₀⁰

a(n−2, n−2k−2)



. Each of the terms in the last expression is nonnegative, the final one a result of the induction assumption on (2.2). Thereforea(n, n−2j)≥0.

Property (i) has already been shown so for (ii), assume α₁ > 0. Since b(0,0) = 1andb(k+ 1,0) =α₁γ₀b(k−1,0) +α₁α₂b(k−1,2),we have that b(k+ 1,0)>0fork+ 1even. But alsob(k+ 1,0) = α₁b(k,1)sob(k,1)>0 for oddk. Hence (ii) is now straightforward from (2.4).

For property (iv), supposeα_i < ¹₂ (i = 1, ..., n−1). By the induction argument above, the first term of the last identity forγn−1a(n, n−2j)is positive sincek =j−1∈/K₁. Since the other terms are nonnegative,a(n, n−2j)>0.

Next letαi < ¹₂ (i = 1, ..., n−3)and suppose that equality holds in (2.2)

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for some subset K of{0, ...,[ⁿ⁻¹₂ ]}. As above, it follows that equality holds in (2.2) over each subset of K of the form {k, ..., k +m}. Hence assume K = {k, ..., k+m}. Ifm = 0andk 6= ⁿ⁻²₂ thenK⁰andK^∗are empty. IfK ={ⁿ⁻²₂ }, then by equality in (2.11) we have−αn−3a(n−4,0) = ¹₂ −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, letm ≥1. Ifk+m 6= [ⁿ⁻¹₂ ]then equality holds in (2.8), and by (2.9) and (2.10) we have

−αn−3



 X

k∈(K⁰)^∗

a(n⁰−3, n⁰−2k−1)− X

k∈(K⁰)⁰

a(n⁰ −2, n⁰−2k−2)





= 1

2 −αn−3

(a(n−3, n−2k−1) +a(n−3, n−2(k+m)−1)).

This is impossible since both sides must be zero which implies a(n−3, n− 2(k+m)−1) = 0.

Finally, ifk+m= [ⁿ⁻¹₂ ]then equality holds in (2.11) and a similar argument showsa(n−3, n−2k−1) = 0which by property (iv) impliesk= 0and thus K must be the whole set{0, ...,[ⁿ⁻¹₂ ]}.

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3. Bounds on Polynomial Derivatives

It is well known [9] thatT_nis bounded by one and for1≤k ≤n T_n^(k)(x)

≤T_n^(k)(1) = n²(n²−1)(n² −4)· · ·(n²−(k−1)²) 1·3·5· · ·(2k−1)

when−1 ≤ x ≤ 1; and if1 ≤ k < n and

Tn^(k)(x)

= Tn^(k)(1), thenx = ±1.

More generally, it follows from a result of R.J. Duffin and A.C. Schaeffer ([7], [9, Thm. 2.24]) that ifP_nis any polynomial of degreenthat is bounded by one in[−1,1]and1≤k < n, then

Pn^(k)(x)

≤Tn^(k)(1)with equality holding only whenP_n =±T_nandx=±1. For the polynomials in Theorem2.1, we may be more precise.

Corollary 3.1. Let P₀, ..., P_nbe defined by (1.1) with0≤α_i ≤ ¹₂, β_i = 0,and γ_i = 1−α_i fori= 0, ..., n−1.Then

(a) |P_n(x)| ≤1 =P_n(1) =T_n(1); and ifn ≥1,|P_n(x)|= 1,andα_i < ¹₂ for i= 1, ..., n−1, thenx=±1.

(b) If1≤k ≤n,then

P_n^(k)(x)

≤P_n^(k)(1) ≤T_n^(k)(1) for allxin[−1,1]. Moreover in this case:

(i) Ifk < nand

Pn^(k)(x)

=Pn^(k)(1),thenx=±1.

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(ii) IfPn^(k)(1) =Tn^(k)(1), thenP_n=T_n. In particular, ifn≥2andα_i < ¹₂ (i= 1, ..., n−1), thenPn^(k)(1)< Tn^(k)(1).

Proof. By Theorem 2.1, Pn = Pn

m=0a(n, m)Tm wherea(n, m) ≥ 0. There- fore

|P_n(x)| ≤

n

X

m=0

a(n, m)|T_m(x)| ≤P_n(1) = 1.

Suppose that n ≥ 1 and |P_n(x)| = 1. If n = 1, then x = ±1, so assume n ≥ 2 and α_i < ¹₂ (i = 1, ..., n − 1). Then P_n(x) = ±P_n(1) and hence a(n, m)(1±Tm(x)) = 0for allm. SinceT0 = 1, T1 = xandT2 = 2x² −1, property (iv) of Theorem2.1implies thatx=±1.

Next assumek ≥1. SinceD

Tm^(k)(1)E

mis increasing, P_n^(k)(x)

=

n

X

m=k

a(n, m)T_m^(k)(x)

≤

n

X

m=k

a(n, m)

T_m^(k)(x)

≤

n

X

m=k

a(n, m)T_m^(k)(1) =P_n^(k)(1)

≤T_n^(k)(1)

n

X

m=k

a(n, m)≤T_n^(k)(1).

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Assume1≤k < nand

Pn^(k)(x)

=Pn^(k)(1). Then 0 =

n

X

m=k

a(n, m) T_m^(k)(1)−

T_m^(k)(x)

,

where each term is nonnegative. Since a(n, n) > 0we have that

Tn^(k)(x) = Tn^(k)(1)sox=±1.

Finally suppose thatk ≥ 1andPn^(k)(1) = Tn^(k)(1). ThenPn

m=ka(n, m) = 1 so a(n, m) = 0 for m < k. Also a(n, m)Tm^(k)(1) = a(n, m)Tn^(k)(1) so a(n, m) = 0form=k, ..., n−1.ThereforeP_n =a(n, n)T_nand thusa(n, n) = 1.

However the previous case is impossible ifn≥2andα_i < ¹₂ (i= 1, ..., n− 1) since by property (iv) of Theorem2.1we would havea(n,0)>0ora(n,1)>

0.

Remark 1. For fixed k the sequence Pn^(k)(1) of bounds is increasing. In fact by (1.1), Pn^(k)(1) may be generated recursively as follows: Initially we have P₁⁽¹⁾(1) = 1, P_k^(k)(1) = _1−α^k

k−1P_k−1^(k−1)(1) (k ≥ 2)and sete_kk := ^k+α_1−α^k

kP_k^(k)(1).

Then forn ≥1,

P_n+1^(k)(1) =P_n^(k)(1) +e_nk ≥P_n^(k)(1) ≥0, where the differencese_nk are defined by

e_nk := α_n

1−α_nen−1,k+ k

1−α_nP_n^(k−1)(1).

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Ultraspherical polynomials y = Pn^{α} with α ≥ −¹₂ satisfy the differential equation

(1−x²)y⁰⁰−2(α+ 1)xy⁰+n(n+ 2α+ 1)y = 0 and thus a closed form forPn^(k)(1)is possible in this case since

P_n^(k)(1) = n(n+ 2α+ 1)−(k−1)(2α+ 1)−(k−1)²

2(α+k) P_n^(k−1)(1).

This extends known Chebyshev and Legendre identities ([9, p. 33], [11, p.

251]).

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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 for expansions of power series in terms of the polynomials that satisfy (1.1): LetP

a_ixⁱ be a convergent power series on (−1,1), and for everynlet P_nbe a polynomial of degreen. Thenxⁿ =Pn

m=0b(n, m)P_m for some num- bersb(n, m); and we define the Fourier coefficientc_j ofP

a_ixⁱ with respect to the sequencehP_niby

c_j :=X

i

a_ib(i, j) whenever this sum converges. Note that c_nj := Pn

i=0a_ib(i, j) is then the jth coefficient in the expansion of the partial sumPn

i=0a_ixⁱ: sinceb(i, j) = 0for j > i,

n

X

i=0

a_ixⁱ =

n

X

i=0

a_i

n

X

j=0

b(i, j)P_j =

n

X

j=0

c_njP_j.

We have the following estimate where kfk 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. LethP_nibe given by (1.1) withhα_ni,hβ_niandhγ_ninonnegative, and suppose that P_n(1) = 1 for alln. IfP

a_i converges absolutely, then the

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coefficientc_j ofP

a_ixⁱ with respect tohP_niexists for everyj and

c_j−

n

X

i=0

a_ib(i, j)

≤ X

i>n i−jeven

|a_i|. Moreover, ifP

iai converges absolutely, then

Xa_ixⁱ−

n

X

j=0

c_jP_j(x)

≤X

i>n

(|a_i|+|ia_i|)

for allxin[−1,1]. In this case, ifα_i ≤ ¹₂ andβ_i = 0fori= 0, ..., n−1, and if a_i ≥0for alliandd_kis thekth Chebyshev coefficient ofP

a_ixⁱ, then (4.1)

Xa_ixⁱ−

n

X

j=0

c_jP_j

≥

Xa_ixⁱ−

n

X

k=0

d_kT_k .

In addition, ifα_i < ¹₂ (i= 1, ..., n−1), then equality holds in (4.1) if and only ifP

a_ixⁱ is a polynomial of degree at mostn.

Proof. Asume that hα_ni, hβ_niandhγ_niare nonnegative, andP_n(1) = 1for all n. By (1.1) the degree ofP_nisnfor alln. Thusxⁿ =Pn

m=0b(n, m)P_m where b(n, m) ≥ 0 by (2.4), and b(n, m) ≤ 1 by the normalization since we have 1ⁿ =Pn

m=0b(n, m).

Suppose thatP

|a_i| converges. Then P

ia_ib(i, j) converges absolutely by the comparison test soc_j exists and

X

i>n

aib(i, j)

≤ X

i>n i−jeven

|ai|.

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Next assumeP

|ia_i| <∞.ThenP

|a_i|<∞soc_j exists for everyj. Thus by Corollary3.1we have

Xa_ixⁱ−

n

X

j=0

c_jP_j(x)

≤

X

i>n

a_ixⁱ

+

n

X

j=0

(c_j−c_nj)P_j(x)

≤X

i>n

|a_i|+

n

X

j=0

|c_j −c_nj|, where

n

X

j=0

|c_j−c_nj|=

n

X

j=0

X

i>n

a_ib(i, j)

≤

n

X

j=0

X

i≥n+1

1

i |ia_i| ≤ n+ 1 n+ 1

X

i>n

|ia_i|.

Suppose further thatα_i ≤ ¹₂,β_i = 0(i = 0, ..., n−1) anda_i is nonnegative for all i. Thenc_j ≥ 0 for allj and by Theorem 2.1, P_j = Pj

k=0a(j, k)T_k where a(j, k) ≥0.Since we also haveP

a_ixⁱ = P

c_jP_j uniformly on[−1,1]in this case, it follows that

Xa_ixⁱ−

n

X

j=0

c_jP_j

=

X

j>n

c_jP_j

=X

j>n

c_j and similarly

Xa_ixⁱ−

n

X

k=0

d_kT_k

=X

k>n

d_k.

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But

Xd_kT_k=X a_ixⁱ

=X

c_jP_j

=X

j

c_j

∞

X

k=0

a(j, k)T_k

=X

k

X

j≥k

cja(j, k)

! Tk

since

X

j≥k

cja(j, k)

≤X

j≥k

cj ≤X

cjPj(1) =X

ai <∞.

Since the coefficients in a uniformly convergent Chebyshev expansion are unique, d_k=P

j≥kc_ja(j, k).Therefore X

k>n

dk =X

j>n

cj

X

k>n

a(j, k)

=X

j>n

c_j

j

X

k=0

a(j, k)−

n

X

k=0

a(j, k)

!

=X

j>n

cj−

n

X

k=0

X

j>n

cja(j, k)≤X

j>n

cj.

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Finally, assume thatα_i < ¹₂ (i= 1, ..., n−1). By property (iv) of Theorem2.1, a(j,0) +a(j,1) > 0 forj > n so if equality holds in the last inequality then c_j = 0for allj > n. ThusP

a_ixⁱ =P

c_jP_j is a polynomial of degree at most n.

Remark 2. IfhP_niis defined as in Corollary4.1and, more generally,P (ia_i)² converges, then by the Schwarz inequality it follows that

X

i>n

|a_i| ≤ X

i>n

(ia_i)²

!¹₂ X

i>n

1 i²

!¹₂

soP

|a_i|<∞andc_j exists for everyj. Moreover in this case we have

Xaixⁱ−

n

X

j=0

cjPj(x)

≤

X

i>n

a_ixⁱ

+

n

X

j=0

X

i>n

a_ib(i, j)

≤ X

i>n

(ia_i)²

!¹₂ X

i>n

xⁱ i

2!¹₂ +

n

X

j=0

X

i>n

(ia_i)²

!¹₂ X

i>n

b(i, j) i

2!¹₂

≤(n+ 2) X

i>n

1 i²

!¹₂ X

i>n

(ia_i)²

!¹₂

≤ n+ 2

√n

X

i>n

(ia_i)²

!¹₂

where the last inequality follows from the proof of the integral test.

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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.

[4] R. ASKEY, Orthogonal Polynomials and Special Functions, Regional Conference Series in Applied Mathematics, 21, SIAM, Philadelphia, PA, 1975.

[5] R. ASKEY AND G. GASPER, Jacobi polynomial expansions of Jacobi polynomials with non-negative coefficients, Proc. Camb. Phil. Soc., 70 (1971), 243–255.

[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.