zeros of ces`aro sum approximations - Texas Tech Math

` ZEROS OF CESARO SUM APPROXIMATIONS ROGER W. BARNARD, KENT PEARCE∗ AND WILLIAM WHEELER† Abstract. P∞ n be analytic on D with positive Let D denote the open unit disk and let f (z) = n=0 an z monotone decreasing coefficients an . We answer several questions posed by J. Cima on the location of the zeros of polynomial approximants which he originally posed about outer functions. In particular, we show that the zeros of Ces` aro approximants to f are well-behaved in the following sense: (1) if an → 1, and aa0 ≤ amb , then ∂ D is the only accumulation set for the zeros of the Ces` aro sums of a n+1

m



P



P

∞ 1 n n f ; and (2) if f has a representation f (z) = ∞ n=0 g n+c z where g(x) = n=0 bn x 6≡ 0, bn ≥ 0, then we give sufficient conditions so that the convex hull of the zeros of the Ces` aro sums of f will contain D .

Key words. outer functions, polynomial approximations, zeros, Ces` aro sums, convex hulls AMS subject classifications. 30C15

1. Introduction. Let D denote the open unit disk. An outer function (see [8], p. 25) is a function that has the representation 1

f (z) = eiγ e 2π

R

2π eit +z 0 eit −z

log ψ(t)dt

for z ∈ D , where γ is a real number, ψ(t) ≥ 0, and log ψ(t) ∈ L1 . It is easy to show that if ψ ∈ Lp , then f ∈ H p . It is not obvious, from this definition, which functions are outer functions. One way of producing outer functions is to consider functions in the class H ∞ . It can be shown that an H ∞ function that is continuous and nonvanishing on the closed unit disk is an outer function (see [17], p. 105). Another way of generating outer functions is by considering Smirnov domains (see [13], p. 155). It also can be shown that if the conformal map f maps ∂ D onto a rectifiable starlike curve which bounds a Smirnov domain and if log f 0 (eit ) ∈ L1 , then f 0 is an outer function. A series of problems was recently posed by J. Cima [3] involving the location of the zeros of polynomial approximants to outer functions. The problems arose from discussions with numerical analysts who were investigating stability criteria for nonlinear systems of differential-difference equations. The characteristic equations for those systems typically involved fractions whose denominators included outer functions. One of Cima’s problems was to determine which polynomial approximants to outer functions inherit their non-vanishing property on D . Since outer functions are non-vanishing on D , it is desirable to have the approximants also non-vanishing on D . Taylor approximants, or partial sums, are natural choices one could think of selecting to approximate outer functions. It was shown in [2] that the Taylor approximating polynomials to outer functions in general can vanish on D , even if fairly restrictive geometric conditions are put on the outer functions. A second type of approximating polynomials that one could consider would be the function’s Ces` aro sums. Along this line, it was also shown in [2] that if f is a bounded convex function, then the Ces` aro sums of the outer function f 0 are non-vanishing on D . ∗ Department of Mathematics and Statistics, Texas Tech University, Lubbock, Texas 79409 ([email protected], [email protected]) † SAS Institute, Cary, NC 27513 ([email protected])

1

2

R.W. Barnard, K. Pearce, W. Wheeler

Cima’s problems about polynomial approximants to outer functions motivated our original interest in the zeros of Ces` aro approximants. However, the results which we have obtained do not require the source functions to be outer functions. The results extend to fairly general classes of functions with certain properties and can undoubtably extend to other approximants (See Section 7). Let M be the set of functions analytic in D with positive monotonically decreasing coefficients P∞in their series expansion. Let W be the set of functions of the form c > 0. g(x) = n=0 bn xn 6≡ 0, bn ≥ 0, g(x) < ∞ for x ∈ [0, 1c ], where   Let V be P∞ 1 n the set of functions that have the representation f (z) = n=0 g n+c z , where g ∈ W.PThen, Vn ⊂ M . We note that V contains various outer functions since ∞ z m , m ≥ 2, c ∈ (0, 1] and these f belong to f (z) = n=0 (n+c) m ∈ V with g(x) = x H ∞ and are continuous and non-vanishing on D . P∞ The Ces`aro sums (see [16], p. 142) of order γ ∈ N ∪ {0} of a series n=0 an z n can be defined as   n−k+γ n X n−k   ak z k , Cn(γ) (z) = n + γ k=0 n   a! a (1) . We will let Cn denote first order Ces` aro sums Cn . where = b b!(a − b)!

of Theorems. In some cases, the zerosP of the Ces` aro sums Cn of P∞2. Statement ∞ n n a z behave like the zeros of the partial sums S of a z . For example, n n n n=0 n=0 we know by Jentzsch’s Theorem (see [7], p. 352), that every point on the circle of P∞ n a z is a limit point of the set of zeros of {S }. Observing convergence of n n n=0 that Sn = (n + 1)Cn − nCn−1 , it was shown in [2] that Jentzsch’s Theorem can be extended to the first order Ces` aro sums Cn . So we know that some of the zeros of Cn will accumulate around the circle of convergence as n gets large. It is interesting to (γ) note that the limit set to the zeros of Sn and possibly Cn can contain much more than the circle of convergence; X. Qian and L. Rubel [14] have shown the following: Theorem A. If F is a finite set of points in D and K is any closed set outside D , then there is a power series with radius of convergence 1 such that the set of limit points to the zeros of its partial sums is F inside D and K ∪ ∂ D outside D . In some cases however, the circle of convergence is the complete limit set to zeros (γ) proof by S. of Sn and Cn . J. Cima and G. Csordas [4] have corrected and modified a P Izumi [9] to show that the maximum of the moduli of the zeros of Sn (z) = nk=0 ak z k approaches 1 as n → ∞, where {an } is a monotonically decreasing sequence such that an rm an+1 → 1 and limm→∞ am = 0, for any r ∈ (0, 1). We considerably extend the results of Cima and Csordas to Ces` aro sums of all orders in the following theorem. P∞ (γ) aro sums of order γ of f (z) = n=0 an z n , Theorem 2.1. Let Cn be the Ces` n → |z| < 1, where {an } is a positive monotonically decreasing sequence such that aan+1 a0 b 1, and am ≤ am for some a, b ∈ R . Then (γ) Cn (z) n!γ! n n→∞ (n+γ)! an z

lim

uniformly for |z| ≥ 1 + δ, δ > 0.

=

1 (1 − z1 )γ+1

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Zeros of Ces` aro Sum Approximations (γ)

Furthermore, let rn denote the maximum of the moduli of the zeros of Cn . Then, limn→∞ rn = 1. P∞ zk Cima considered the outer function f0 (z) = k=0 (k+1) 3 , which belongs to V and conjectured [3] that the convex hull of the zeros of the first order Ces` aro sums Cn would contain the disk D for all n ≥ 6. Various examples in M and V can be given to show that unless some additional conditions are imposed that will P this property 1 k not hold generally. E.g., consider g(x) = 1 + x/10 and f (x) = ∞ k=0 g( k+1 )z . Then, f ∈ V , but the first order Ces`aro sums Cn do not contain the disk D unless n ≥ 13. We verify Cima’s conjecture by finding conditions under which a fairly general class of functions in V satisfy Cima’s conjecture. The conditions which we find control the distributions of the zeros of the first order Ces` aro approximants Cn by controlling the moduli of the zeros of Cn and the distribution of the arguments of the zeros of Cn . We can obtain a lower bound to the moduli of the zeros of Cn by applying a generalization of the Enestr¨ om-Kakeya Theorem due to Anderson, Saff, and Varga [1]. Pn Theorem B. Let p(z) = k=0 ak z k , ak > 0, for 0 ≤ k ≤ n. Then, all the zeros of p are contained in the annulus ak ak ≤ |z| ≤ max . 0≤k 0. It can be shown (see Appendix) that   3π (n − k + 1) cos n+1 n−k − > 0, (k + c)j (k + c + 1)j for n ≥ 44, k = 0, . . . , n − 1, j ≥ 0, and also for n ≥ 14, k = 0, . . . , n − 1, j ≥ 1, c ∈ (0, 1].   Pn 1 k g Thus, the convex hull of the zeros of Cn (z) = k=0 n−k+1 n+1 k+c z will always contain the unit disk if n ≥ 44 - a result not true for the partial sums Sn (z) =  P Pn 1 k k g . (Consider, g(x) = 1 and f (x) = ∞ z k=0 k=0 x .) In addition, if g(0) = 0 k+c and c ∈ (0, 1], then the result is always true for n ≥ 14. 2. Examples of functions such that Theorem 2.2 holds for all n ≥ 6 are g(x) = bxm , where b > 0, m ≥ 7, c ∈ (0, 1].

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Zeros of Ces` aro Sum Approximations

4. Proof of Theorem 2.1. Proof of Theorem 2.1. Fix δ > 0 and let |z| ≥ 1 + δ. Then, n ∞ (n + γ)! X C (γ) (z) X (n − k + γ)!n! (k + γ)! 1 n k −k = − z − a z n!γ! k n γ!k! (1 − z1 )γ+1 n!γ!an z n k=0 (n − k)!(n + γ)! (n+γ)! an z k=0 n ∞ X (n − k + γ)! a X (k + γ)! −k k k−n z z − = (n − k)!γ! an γ!k! k=0 k=0 n ∞ X (k + γ)! a X (k + γ)! −k n−k −k z z − = γ!k! an γ!k! k=0 k=0 n   ∞ X (k + γ)! a X (k + γ)! n−k z −k − 1 z −k − = γ!k! an γ!k! k=0 k=n+1 m X (k + γ)! an−k 1 ≤ − 1 (1 + δ)k γ!k! an k=0 n X (k + γ)! an−k 1 + − 1 γ!k! an (1 + δ)k (4.1)

+

k=m+1 ∞ X k=n+1

1 (k + γ)! γ!k! (1 + δ)k

Now, we will estimate each sum above. Let  > 0,  < 2δ . P (k+γ)! 1 First, consider the sum ∞ k=n+1 γ!k! (1+δ)k . By applying the ratio test, we see that the series converges for δ > 0. Thus, there exists an integer N1 such that for all n ≥ N1 , ∞ X 1 (k + γ)!  < . γ!k! (1 + δ)k 3

k=n+1

Pn 1 an−k − 1 Second, consider the sum k=m+1 (k+γ)! (1+δ)k . Since γ!k! an exists an integer N2 ≥ N1 such that for all n ≥ N2 , (4.2)

0< n

Let Km = max

am−1 a0 a1 am , am , . . . , am

an−1 < 1 + . an

o =

a0 am .

Claim: For n ≥ m ≥ N2 , 0
0, β0 <  such that 0 < (1 + β0 )N3 − 1
0. Then, for any integer N1 there exist infinitely many integers n greater than N1 such that, (4.5)

1 < 1 + δ < rn .

γ+1  . Then, by the above there is an integer N2 Choose  > 0 such that  < 1+δ 2+δ such that for all n ≥ N2 , C (γ) (z) 1 n − n!γ! γ+1 −  n a z 1 − z1 (n+γ)! n γ+1  1+δ −  > 0. ≥ 2+δ (γ) Thus, Cn (z) > 0 for n ≥ N2 , |z| ≥ 1 + δ. But this contradicts (4.5). Therefore, 1 ≤ lim inf rn ≤ lim sup rn ≤ 1, n→∞

which implies limn→∞ rn = 1.

n→∞

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R.W. Barnard, K. Pearce, W. Wheeler

5. Statements and Proofs of Auxiliary Lemmas. We need the following aro lemmas to prove Theorem  2.2.  In these lemmas, Cn denotes the first order Ces` P∞ 1 n sums of f (z) = n=0 g n+c z , where g ∈ W .   Lemma 5.1. Cn (z) z n+1 − 1 6= iλ for λ > 0, z = Reiθ , θ ∈ (0, π), R is sufficiently large. P n k Lemma 5.2. If k=1  (n − k + 1)x sin(ky) > 0 for x > 0, y ∈ R , then Pn 1 k k=1 (n − k + 1)g k+c x sin(ky) > 0 for g ∈ W, x > 0. Pn

x sin(w )

j p (x), where wj = (2j+1)π n+1 , |1−xeiwj |4 j 2 n+1 ) + 2(n + 1)(1 − x cos(wj )). pj (x) = (x − 1)(n + 2 + x n+1 − 1) 6= iλ, λ > 0 for z = xeiwj , where x > 0, wj = Lemma 5.4. Cn (z)(z (2j+1)π n+1 n+1 , j = 0, 1, . . . , b 2 c − 1, n ≥ 2.   Pn k 1 (−x) > 0 for g ∈ W, x ≥ 0. Lemma 5.5. For n even, k=0 (n − k + 1)g k+c

Lemma 5.3.

k k=1 (n − k + 1)x

sin(wj k) =

  Proof of Lemma 5.1. Expanding Cn (z) z n+1 − 1 and letting z = Reiθ gives     n n   X X 1 1 n−k+1 n−k+1 g g z k z n+1 − zk Cn (z) z n+1 − 1 = n+1 k+c n+1 k+c k=0 k=0   n h i X 1 n−k+1 g = Rk Rn+1 ei(k−n−1)θ − eikθ . n+1 k+c k=0    n+1 i(k−n−1)θ  Pn 1 k e − eikθ . We will show that Re(S(θ)) > R Let S(θ) = k=0 n−k+1 n+1 g k+c R 0. Now, R is sufficiently large. This will imply that Cn (z) z 1

R

Re(S(θ)) = 2n+1

    n X 1 n−k+1 cos(kθ) g Rk−n cos((k − n − 1)θ) − . n+1 k+c Rn+1

k=0

Notice that as R → ∞, when k = n, 1 g n+1



1 R2n+1 Re(S(θ))

→ 0. The dominant term of

1 R2n+1 Re(S(θ))

is,

     g 1  n+c 1 1 1 cos(nθ) √ − n+1 cos(−θ) − ≥ n+c Rn+1 (n + 1) 2 R >0

for θ ∈ (0, π4 ], and R sufficiently large. Now considering the imaginary part we have     n X 1 n−k+1 sin(kθ) 1 k−n g Im(S(θ)) = R sin((k − n − 1)θ) − . R2n+1 n+1 k+c Rn+1 k=0

1 R2n+1 Im(S(θ))

is, when k = n,   1      g n+c 1 sin(nθ) sin(nθ) 1 g sin(−θ) − n+1 = − sin(θ) + n+1 n+1 n+c R (n + 1) R 0. Then, multiplying by xc−1 > 0 and integrating gives n X

(n − k + 1)

k=1

xk+c sin(ky) > 0. k+c

Repeating this process we obtain n X

(n − k + 1)

k=1

xk+c sin(ky) > 0 (k + c)m

for x > 0, m ∈ N . Now, multiplying by am x−c ≥ 0 and summing over m gives n ∞ X X

(n − k + 1)xk

m=0 k=1

am sin(ky) > 0. (k + c)m

P∞ Note that the double sum is positive since am ≥ 0 and g(x) = m=0 am xm 6≡ 0. Since g(x) < ∞ for x ∈ [0, 1c ], we can interchange the summations and apply the definition of g to obtain n X

 (n − k + 1)g

k=1

1 k+c

 xk sin(ky) > 0.

Proof of Lemma 5.3. Since sin(0) = 0 and sin θ = Im eiθ we have " # n n n X X X
k (n − k + 1)xk sin(wj k) = Im (n + 1) kxk eiwj k . xeiwj − (5.1) k=1

k=0

By applying the identity (5.1) becomes "

Pm k=0

yk =

1−y m+1 1−y

k=0

and simplifying, the right-hand side of

# n
d X 1 − xn+1 eiwj (n+1) k xeiwj −x Im (n + 1) 1 − xeiwj dx k=0   (n + 1)(1 − xeiwj ) − x(1 + xn+1 )eiwj (5.2) . = Im (1 − xeiwj )2  
ba gives us Multiplying out the numerator in (5.2) and using Im ab = Im |b| 2  Im " = Im

(5.3)

(n + 1)(1 − xeiwj ) − x(1 + xn+1 )eiwj (1 − xeiwj )2 1 |1 −

4 xeiwj |





n + 1 − x(n + 2 + xn+1 )eiwj − 2(n + 1)xe−iwj

+2x2 (n + 2 + xn+1 ) +(n + 1)x2 e−2iwj − x3 (n + 2 + xn+1 )e−iwj



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R.W. Barnard, K. Pearce, W. Wheeler

Now, taking the imaginary part and applying the trigonometric identity sin(2y) = 2 sin(y) cos(y), (5.3) becomes 

1 |1 −

4 xeiwj |

−x(n + 2 + xn+1 ) sin(wj ) + 2(n + 1)x sin(wj )

−(n + 1)x2 sin(2wj ) + x3 (n + 2 + xn+1 ) sin(wj ) x sin(wj )  2 (x − 1)(n + 2 + xn+1 ) + 2(n + 1)(1 − x cos(wj )) . = 4 iw |1 − xe j | Proof of Lemma 5.4. For z = xeiwj , Cn

(z)(z n+1

  n X 1 n−k+1 g − 1) = xk eiwj k xn+1 e−i(n+1)wj n+1 k+c k=0   n X 1 n−k+1 g − xk eiwj k n+1 k+c k=0   n X 1 n−k+1 g =− xk (xn+1 + 1)eiwj k n+1 k+c k=0

≡ T (x). Taking the imaginary part of T gives us Im(T (x)) = −

  n X 1 n−k+1 g xk (xn+1 + 1) sin(wj k). n+1 k+c k=0

n+1 It suffices to show that Im(T (x)) <  0 for  x > 0, j = 0, 1, . . . , b 2 c − 1. Im(T (x)) < 0 Pn 1 xk sin(wj k) > 0. By Lemma 5.2, it suffices if and only if k=1 (n − k + 1)g k+c Pn (n − k + 1)xk sin(wj k) > 0 for x > 0. First, we will show that to Pnshow that k=1 k k=1 (n − k + 1)x sin(wj k) > 0 for x ≥ 1, n ≥ 2. By Lemma 5.3, n X

(n − k + 1)xk sin(wj k) =

k=1

x sin(wj )

4 pj (x),

|1 − xeiwj |

where pj (x) = (x2 − 1)(n + 2 + xn+1 ) + 2(n + 1)(1 − x cos(wj )). Since sin(wj ) > 0 for j = 0, 1, . . . , b n+1 2 c − 1, we need to show that pj (x) > 0 for x > 0. First, for x = 1,   n+1 0≤j≤ − 1, n ≥ 2. pj (1) > 0, 2 For x > 1, pj (x) > (x2 − 1)(n + 2 + xn+1 ) + 2(n + 1)(1 − x) > (x − 1) [2(n + 3) − 2(n + 1)] > 0.

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Zeros of Ces` aro Sum Approximations

For x ∈ (0, 1), first consider cos(wj ) ≤ 0. Then,

n−1 4

≤ j ≤ b n+1 2 c − 1. So that wj ∈

π

2,π




and

pj (x) ≥ (x2 − 1)(n + 2 + xn+1 ) + 2(n + 1) > n − xn+1 > 0, n ≥ 2. For x ∈ (0, 1) and 0 ≤ j ≤ cos(wj ). So that

n−1 4

− 1, we have that wj ∈ (0, π2 ) and cos(w0 ) ≥

−2(n + 1)x cos(w0 ) ≤ −2(n + 1)x cos(wj ) which implies p0 (x) ≤ pj (x). |1−xeiw0 |4 Pn k Now, p0 (x) = x sin(w0 ) k=1 (n − k + 1)x sin(w0 k) > 0 for x > 0, since sin(w0 k) > 0
for k = 1, . . . , n. Therefore, pj (x) > 0 for x ∈ (0, 1), wj ∈ 0, π2 . Proof of Lemma 5.5. It is sufficient to show that n X

(5.4)

(n − k + 1)(−x)k > 0

k=0

for x > 0, because (5.4) implies n X

(n − k + 1)y k (−x)k > 0

k=0

for x > 0, y > 0, which in turn implies, adapting the argument in the proof of Lemma 5.2, the conclusion of the lemma. However, we can write n X

1 + 2x + n + nx + (−x) x2 (1 + x)2 n

k

(n − k + 1)(−x) =

k=0

which implies the positivity of (5.4). 6. Proofs of Theorem 2.2 and Corollary 1. Proof of Theorem 2.2. We first apply Rouch´e’s Theorem to show that Cn has exactly one zero in each of the n < arg(z) < (2j+3)π sectors: (2j+1)π n+1 n+1 , j = 0, 1, . . . , b 2 c − 1, n ≥ 5. n+1 − 1) and let Γ be the following contour: Γ is the line segment Let h(z) = i(z z = xeiwj , 0 ≤ x ≤ R, followed by the arc z = Reiθ , wj ≤ θ ≤ wj+1 , followed by the line segment z = xeiwj+1 , 0 ≤ x ≤ R, where wj = (2j+1)π n+1 . The zeros of h 2πk

are z = ei n+1 , k = 0, 1, . . . , n. Then, h has exactly one zero in each of the sectors (2j+1)π (2j+3)π n < arg(z) < (2j+3)π < arg(z) < − (2j+1)π n+1 n+1 , − n+1 n+1 , j = 0, 1, . . . , b 2 c − 1, n ≥ 5. Now, it is clear that the following statements are equivalent |Cn + h| < |Cn | + |h| Cn h 6= λ ≥ 0.

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R.W. Barnard, K. Pearce, W. Wheeler

  Hence, we need to show that Cn (z) i(z n+1 − 1) 6= λ on Γ, where λ ≥ 0, or equiva  lently, Cn (z) z n+1 − 1 6= iλ, λ ≥ 0. Fix n ≥ 5. Choose R > 1 so that the zeros of   Cn are contained in |z| < R and such that Cn (z) z n+1 − 1 6= iλ, λ > 0, by Lemma   5.1. Then, Cn (z) z n+1 − 1 6= iλ, for λ ≥ 0, z = Reiθ , θ ∈ (0, π). Now we need to show for n ≥ 5 that for z = xeiwj , 0 ≤ x ≤ R, wj = j = 0, 1, . . . , b n+1 2 c − 1,   λ ≥ 0, Cn (z) z n+1 − 1 6= iλ

(2j+1)π n+1 ,

and in the case that n is even, that for z = −x, 0 ≤ x ≤ R,   λ ≥ 0. Cn (z) z n+1 − 1 6= iλ For x = 0,     1 6= iλ, Cn (0) 0n+1 − 1 = −g c

λ ≥ 0.

For x > 0, we have by Lemma 5.4 that   Cn (z) z n+1 − 1 6= iλ   for λ > 0, z = xeiwj . For λ = 0, suppose that Cn (z0 ) z0n+1 − 1 = 0, for some z0 = x0 eiwj , x0 > 0. Then, Cn (z0 ) = 0 or z0n+1 − 1 = 0. But, z n+1 − 1 6= 0 on the rays z = xeiwj , x ≥ 0. If Cn (z0 ) = 0, then Im (Cn (z0 )) = 0. But for z = xeiwj ,   n X 1 n−k+1 g xk sin(wj k) > 0 Im (Cn (z)) = n+1 k+c k=0

for x > 0, by the proof of Lemma 5.4.  n+1 − 1 6= iλ for λ ≥ 0, z = xeiwj , 0 ≤ x ≤ R, wj = Thus, Cn (z) z

(2j+1)π n+1 ,

j = 0, 1, . . . , b n+1 2 c − 1, n ≥ 5. In the case that n is even, Lemma 5.5 implies that Cn (z) 6= 0 for z = −x, x ≥ 0. Hence,   Cn (z) z n+1 − 1 6= iλ for λ ≥ 0, z = −x, x ≥ 0. This gives us that Cn has b n2 c zeros in the upper half plane, and b n2 c zeros in the lower half plane, since Cn has real coefficients. This also tells us that the maximum angle between two “consecutive” zeros in the upper half plane or two “consecutive” 4π . Also, the maximum angle between two “consecuzeros in the lower half plane is n+1 tive” zeros where one zero is in the upper half plane and the other zero is in the lower 6π . half plane is n+1 To show that the convex hull of the zeros of Cn contains the unit disk, consider the points rj eiαj , rj > 1, j = 1, . . . , n that are distributed around the unit circle

13

Zeros of Ces` aro Sum Approximations

in the sectors

(2j+1)π n+1

< arg(z)
1 −  q1

n−c 2 x2 2



 =

n+c+2 2



2 cos

3π n+1



 −

n−c 2 .

n+c 2

The minimum

2 .

for x > 0, we have

n−c 2



  (n + c + 2)2 1 9π 2 (n + c)2 1− − 2 4 2 (n + 1) 4   2 2 1 9π 9π >0 > 4(n + 1) 1 − − 2 (n + 1)2 2 >

for n ≥ 14. Now, we will show that qj < qj+1 for j ≥ 0. Simplifying the difference qj+1 − qj gives     3π j (k + c) − (n − k)(k + c − 1) . qj+1 − qj > (k + c + 1) (n − k + 1) cos n+1 Now, let   3π (k + c) − (n − k)(k + c − 1) l(k) = (n − k + 1) cos n+1          3π 3π 3π k2 + (n + 1) cos − c cos −n+c−1 k = 1 − cos n+1 n+1 n+1   3π c − n(c − 1). +(n + 1) cos n+1 Now l0 (k) = 0, implies k =

n+1−c . 2



 3π (n + c − 1)2 . − l n+1 4  
3π 2 2 > 0 if and only if (n + c + 1) cos Thus, l n+1−c 2 n+1 − (n + c − 1) > 0. Now, 

n+1−c 2



Hence,

(n + c + 1)2 cos

for n ≥ 12, c ∈ (0, 1].



(n + c + 1)2 cos = 4

3π n+1



2 n+2 n+1  2 2 14 9π ≥ 4n − >0 2 13

− (n + c − 1)2 > 4n −

9π 2 2



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R.W. Barnard, K. Pearce, W. Wheeler REFERENCES

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