On the remainder term of Gauss Radau quadratures for analytic functions

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1 Journal of Computational and Applied Mathematics ) On the remainder term of Gauss Radau quadratures for analytic functions Gradimir V. Milovanović a,1, Miodrag M. Spalević b,1, Miroslav S. Pranić c, a Department of Mathematics, Faculty of Electronic Engineering, University of Niš, P.O. Box 73, Niš, Serbia b Department of Mathematics and Informatics, Faculty of Science, University of Kragujevac, P.O. Box 60, Kragujevac, Serbia c Department of Mathematics and Informatics, Faculty of Science, University of Banja Luka, M. Stojanovića 2, Banja Luka, Bosnia and Herzegovina Received 25 September 2006; received in revised form 9 January 2007 Abstract For analytic functions the remainder term of Gauss Radau quadrature formulae can be represented as a contour integral with a complex kernel. We study the kernel on elliptic contours with foci at the points ±1 and a sum of semi-axes ϱ > 1 for the Chebyshev weight function of the second kind. Starting from explicit expressions of the corresponding kernels the location of their maximum modulus on ellipses is determined. The corresponding Gautschi s conjecture from [On the remainder term for analytic functions of Gauss Lobatto and Gauss Radau quadratures, Rocky Mountain J. Math ), ] is proved Elsevier B.V. All rights reserved. MSC: primary 41A55; secondary 65D30; 65D32 Keywords: Gauss Radau quadrature formula; Chebyshev weight function; Error bound; Remainder term for analytic functions; Contour integral representation 1. Introduction In this paper we prove a conjecture of Gautschi [1] for the Gauss Radau quadrature formula 1 1 wt)ft)dt = N λ ν fτ ν ) + R N f ) w, 1.1) ν=1 with respect to the Chebyshev weight function of the second kind wt) = w 2 t) = 1 t 2 and with a fixed node at 1 or 1). Let Γ be a simple closed curve in the complex plane surrounding the interval [ 1, 1] and D = int Γ its interior. If the integrand f is analytic in D and continuous on D, then the remainder term R N f ) w in 1.1) admits the contour integral Corresponding author. addresses: grade@elfak.ni.ac.yu G.V. Milovanović), spale@kg.ac.yu M.M. Spalević), pranic77m@yahoo.com M.S. Pranić). 1 The authors were supported in part by the Serbian Ministry of Science and Environmental Protection /$ - see front matter 2007 Elsevier B.V. All rights reserved. doi: /j.cam

2 282 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) representation R N f ) w = 1 2πi the kernel is given by K N z; w) = 1 ω N z; w) K N z; w)f z) dz, 1.2) Γ 1 1 wt)ω N t; w) z t and ω N z; w) = N ν=1 z τ ν ). It is clear that dt, z / [ 1, 1], K N z; w) = K N z; w). 1.3) The integral representation 1.2) leads to the error estimate R N f ) w lγ) ) ) max 2π K N z; w) max fz), 1.4) z Γ z Γ lγ) is the length of the contour Γ. In order to get estimate 1.4), one has to study the magnitude of K N z; w) on Γ. Note that the previous formulae hold for every interpolatory quadrature rule with mutually different nodes on [ 1, 1]. Many authors have used 1.4) to derive bounds of R N f ) w. Two choices of the contour Γ have been widely used: 1) a circle C r with a center at the origin and a radius r>1), i.e., C r ={z : z =r}, r>1, and 2) an ellipse E ϱ with foci at the points ±1 and a sum of semi-axes ϱ > 1, E ϱ ={z C z = 1 2 ϱeiθ + ϱ 1 e iθ ), 0 θ < 2π}. 1.5) When ϱ 1 the ellipse shrinks to the interval [ 1, 1], while with increasing ϱ it becomes more and more circle-like. The advantage of the elliptical contours, compared to the circular ones, is that such a choice needs the analyticity of f in a smaller region of the complex plane, especially when ϱ is near 1. Since the ellipse E ϱ has length le ϱ ) = 4ε 1 Eε), ε is the eccentricity of E ϱ, i.e., ε = 2/ϱ + ϱ 1 ), and π/2 Eε) = 1 ε 2 sin 2 θ dθ 0 is the complete elliptic integral of the second kind, estimate 1.4) reduces to ) R N f ) w 2Eε) 2 max K N z; w) f ϱ, ε =, 1.6) πε z E ϱ ϱ + ϱ 1 f ϱ = max z Eϱ fz). As we can see, the bound on the right-hand side in 1.6) is a function of ϱ,soitcanbe optimized with respect to ϱ > 1. This approach was discussed first for Gaussian quadrature rules, in particular with respect to the Chebyshev weight functions cf. [4,5]) w 1 t) = 1, w 2t) = 1 t 2, w 3 t) = 1 t t 1 t 1 t, w 4t) = 1 + t, and later has been extended to Bernstein Szegő weight functions [8] and some symmetric weight functions including especially the Gegenbauer weight functions [10], as well as to Gauss Lobatto and Gauss Radau cf. [1 3,6,9]), and to Gauss Turán cf. [7]) quadrature rules. In [1] Gautschi considered Gauss Radau and Gauss Lobatto quadrature rules relative to the four Chebyshev weight functions w i, i = 1, 2, 3, 4, and derived explicit expressions of the corresponding kernels K N z; w i ), i = 1, 2, 3, 4, in terms of the variable u = ϱe iθ ; they are the key for determining the maximum point of K N z; w i ), i = 1, 2, 3, 4, on Γ=E ϱ given by 1.5). Note that z=u+u 1 )/2. For Gauss Lobatto quadratures it was proved that K N z; w 1 ) attains

3 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) its maximum on E ϱ on the real axis cf. [1, Theorem 4.1]). For w 2, w 3 and w 4 only empirical results and conjectures on the location of the maximum point on E ϱ were presented. These conjectures have been proved by Schira see [9]). For Gauss Radau quadratures with a fixed node at 1, Gautschi proved that the corresponding kernel K N z; w) for Chebyshev weight functions w = w 1 and w = w 4 attains its maximum modulus on E ϱ on the negative real axis cf. [1, Theorems 4.4 and 4.5]). For the remaining cases w = w 2 and w = w 3 only empirical results and conjectures on the location of the maximum point on E ϱ for the corresponding kernels are presented in [1]. They are also mentioned in [9], but without the proof. In this paper we derive an analytic proof of the conjecture for w = w 2. Thus, we are concerned with the Gauss Radau quadrature rule 1.1) with respect to the weight function w = w 2 with N = nodes and a fixed node at 1 or 1. Because of symmetry, it is enough to consider only one case, e.g., when τ 1 = 1. It is well known that the node polynomial in this case can be expressed as ω n+1 z; w) = 1 + z)π n z; w R ), π n ; w R ) denotes the monic polynomial of degree n orthogonal with respect to the weight function w R t) = 1 + t)w 2 t) = 1 + t) 3/2 1 t) 1/2. 2. The maximum of the kernel for Chebyshev weight function of the second kind Gautschi [1, Eq. 3.16)] derived the explicit representation of the kernel on E ϱ K n+1 z; w 2 ) = π u u 1 )u 1 + n + 2)/)) u n+1 u n+2 u n+2) + n + 2)/))u n+1 u n+1) ), 2.1) z = u + u 1 )/2 and u = ϱe iθ. Using 2.1) we can determine the modulus of the kernel on E ϱ. It is easy to prove u u 1 = 2 a 2 cos 2θ, 2.2) u 1 + n n + 2 d sin 2 θ ϱ 2, 2.3) a j = a j ϱ) = 1 2 ϱj + ϱ j ), j N, ϱ > 1, 2.4) and d = dϱ) = 1 n ϱ ) n + 2 ϱ ) Further, un+2 u n+2) + n + 2 u n+1 u n+1)) 2 [ = ϱ n+2 ϱ n+2)) cosn + 2)θ + n + 2 ϱ n+1 ϱ n+1)) ] 2 cos)θ [ + ϱ n+2 + ϱ n+2)) sinn + 2)θ + n + 2 ϱ n+1 + ϱ n+1)) ] 2 sin)θ =[ϱ 2n+2) + ϱ 2n+2) cos 2 n + 2)θ] ) n [ + ϱ 2n+1) + ϱ 2n+1) cos 2 )θ] n = 4 n + 2 Bθ), ) [ ϱ 2n+3 + ϱ 2n+3)) cos θ ϱ + ϱ 1) ] cos2n + 3)θ

4 284 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) Bθ) = Bϱ, θ) = n + 2 a2 n+2 cos2 n + 2)θ) + n + 2 a2 n+1 cos2 )θ) + a 2n+3 cos θ a 1 cos2n + 3)θ. 2.6) In this way we get [ ] K n+1 z; w 2 ) = π 2 a 2 cos 2θ)d sin 2 1/2 θ/2)) ϱ n+3/2. 2.7) Bθ) For the location of the maximum point of K n+1 z; w 2 ) on E ϱ the following conjectures are presented in [1, p. 224]:if n 3, the maximum is attained at z = 2 1 ϱ + ϱ 1 ) on every ellipse E ϱ, ϱ > 1; if n 4 there exist parameters ϱ 0 = ϱ 0 n) and ϱ 0 =ϱ 0 n), ϱ 0 > ϱ 0, such that the maximum is attained at z= 1 2 ϱ+ϱ 1 ) when ϱ 1, ϱ 0 ) ϱ 0, + ). Otherwise, the maximum point moves on the ellipse E ϱ from some close to the imaginary axis to the negative real axis as ϱ increases. With increasing n the parameters ϱ 0 converge to one rather rapidly cf. [1, Table 4.2]). Hence, the part of the conjecture when n 4 and ϱ 1, ϱ 0 ) is less important for practical use because the corresponding maximum tends to infinity for increasing n and ϱ 1, ϱ 0 ). Therefore, the error bound 1.4) is rather poor in this case. Theorem 2.1. For each integer n 4 there exists ϱ 0 = ϱ 0 n) such that the kernel of the )-point Gauss Radau formula with fixed node at 1) for the Chebyshev weight function of the second kind attains its maximum modulus on E ϱ on the negative real axis when ϱ > ϱ 0. For n = 1, 2, 3, the maximum of the kernel K n+1 z; w 2 ) attains its maximum on the negative real axis on every ellipse E ϱ ϱ > 1). The maximum is given by max K n+1 z; w 2 ) = z E ϱ K n ϱ + ϱ 1 ); w 2 ) = π ϱ n+1 ϱ ϱ 1 )n + 2)/)) ϱ 1 ) ϱ n+2 ϱ n+2) n + 2)/))ϱ n+1 ϱ n+1) ). Proof. First we show that the denominator of the previous fraction, i.e., the function Vϱ,n)= ϱ n+2 ϱ n+2) n + 2 ϱn+1 ϱ n+1) ) is always positive ϱ > 1, n N). If n is fixed Vϱ) := Vϱ,n)), we get V ρ) = n + 2)ϱ 1)ϱ n ϱ n+3) )>0 and V1) = 0. We consider four separate cases: n = 1, n = 2, n = 3, and n 4. Case n = 1: Using 2.1) we get K 2 z; w 2 ) = πu 2 1)3u + 2) u2u 6 + 3u 5 3u 2) = πϱ 1 u + 1)/3u + 2) 2u 3 + u 2 + u + 2. Now, we prove that both of the expressions in the denominator for fixed ϱ and θ [0, π] attain their minimum modulus when θ = π. The first expression can be written as u + 1 3u + 2 = u + 2 ), from which it is evident that it attains its minimum modulus when θ = π 1/3u + 2) < 1 and attains maximum when θ = π).

5 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) For the second expression we can get the following equality: 2u 3 + u 2 + u = 32x 3 + 8ϱ 2 + 1)x 2 + 4ϱ 4 22ϱ 2 + 4)x + 4ϱ 6 3ϱ 4 3ϱ 2 + 4), x = Ru) = ϱ cos θ, x [ ϱ, ϱ]. We denote the right-hand side of the last equality as F ϱ x) = Fx).Nowit suffices to prove Fx) F ϱ) 0, for x [ ϱ, ϱ] and ϱ > 1: Fx) F ϱ) = 32x 3 + ϱ 3 ) + 8ϱ 2 + 1)x 2 ϱ 2 ) + 4ϱ 4 22ϱ 2 + 4)x + ϱ) = 2x + ϱ)[16x 2 + 4ϱ 2 16ϱ + 4)x + 2ϱ 4 4ϱ 3 + 5ϱ 2 4ϱ + 2)]. For the discriminant of the polynomial in brackets D there holds D = 16ϱ 1) 2 7ϱ 2 + 6ϱ + 7) 0, which completes the proof of this case. Case n = 2: Using 2.1) we get ) K 3 z; w 2 ) = π u u 1 ) u u 3 u 4 u i.e., = π u u 3 u 3) = πu2 1)4u + 3) u3u 8 + 4u 7 4u 3) 4u + 3 3u 6 + 4u 5 + 3u 4 + 4u 3 + 3u 2 + 4u + 3, K 3 z; w 2 ) = πϱ 1 gu), 2.8) gu) = 3u6 + 4u 5 + 3u 4 + 4u 3 + 3u 2 + 4u u + 3 By 2.8) and 1.3) it suffices to prove that gu) attains its minimum when θ = π u = ϱ), i.e., gu)gu) g ϱ)g ϱ) 0, θ [0, π]. After some calculation we get gu)gu) g ϱ)g ϱ) = y = cos θ and Pϱ,y)= 144ϱ 4 3 4ϱ) 2 y 5 4ϱ 2 y + 1) 3 4ϱ) 2 4u Pϱ,y), + ϱ ϱ ϱ ϱ ϱ + 864)y 4 + ϱ 2 576ϱ ϱ ϱ ϱ ϱ ϱ + 324)y 3 + ϱ384ϱ ϱ ϱ 6 196ϱ 5 192ϱ 4 504ϱ ϱ 2 900ϱ + 216)y ϱ ϱ ϱ 8 600ϱ ϱ 6 432ϱ ϱ 4 768ϱ ϱ 2 432ϱ + 81)y + 42ϱ ϱ ϱ 9 337ϱ ϱ 7 432ϱ ϱ 5 418ϱ ϱ 3 288ϱ ϱ 81).

6 286 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) Now it suffices to prove Pϱ,y) 0. If we take ϱ = x + 1 and y = t 1, x>0, t [0, 2], wehave Px + 1,t 1) = 42x i=0 b i t)x i, b 10 t) = 624t + 29), b 9 t) = 2416t 2 + 3t + 28), b 8 t) = 576t t 2 887t , b 7 t) = 1536t t t t , b 6 t) = 21152t t t t t ), b 5 t) = 42592t t t t t ), b 4 t) = 44644t t t t t ), b 3 t) = 44176t t t t t + 539), b 2 t) = 41944t t t t t + 49), b 1 t) = 48t6t 2 14t + 7) 2, b 0 t) = 4t6t 2 14t + 7) 2. By computing their zeros, it can be seen that all functions b i t), i = 0, 1,...,10, are nonnegative when t [0, 2]. Case n = 3: We prove this case in the same way as the previous one. From 2.1) we get ) K 4 z; w 2 ) = π u u 1 ) u u 4 u 5 u u 4 u 4) = πu2 1)5u + 4) u4u u 9 5u 4) = π 5u + 4 u 4u 8 + 5u 7 + 4u 6 + 5u 5 + 4u 4 + 5u 3 + 4u 2 + 5u + 4, i.e., K 4 z; w 2 ) = πϱ 1 hu), 2.9) hu) = 4u8 + 5u 7 + 4u 6 + 5u 5 + 4u 4 + 5u 3 + 4u 2 + 5u u + 4 By 2.9) and 1.3) it suffices to prove that hu) attains its minimum when θ = π u = ϱ), i.e., hu)hu) h ϱ)h ϱ) 0, θ [0, π]. After some calculation we get 4ϱ 2 y + 1) hu)hu) h ϱ)h ϱ) = 4 5ϱ) 2 5u + 4 Rϱ,y). 2 After denoting ϱ = x + 1 and y = t 1, x>0, t [0, 2], wehave Rx + 1,t 1) = 90x i=0 c i t)x i,

7 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) c 14 t) = 508t + 11), c 13 t) = 2050t t + 81), c 12 t) = 1600t t t , c 11 t) = 41000t 4 840t t t ), c 10 t) = 23200t 5 400t t t t ), c 9 t) = 20800t t t t t t ), c 8 t) = 25600t t t t t t t , c 7 t) = t t t t t t t ), c 6 t) = t t t t t t t , c 5 t) = t t t t t t t ), c 4 t) = t t t t t t t ), c 3 t) = t t t t t t t + 675), c 2 t) = t t t t t t t + 225), c 1 t) = 64t 16t t 2 54t + 15) 2, c 0 t) = 4t 16t t 2 54t + 15) 2. By computing their zeros, it can be seen that all functions c i t), i = 0, 1,...,14, are nonnegative when t [0, 2], which completes the proof of this case. Case n 4: By 2.7) it is enough to prove a 2 cos 2θ)d sin 2 θ/2)) Bθ) a 2 1)d 1), Bπ) for sufficiently large ϱ ϱ > ϱ 0 ) and θ [0, π). The last inequality is reduced to [ [a 2 1) + 2 sin 2 θ] d 1) + cos 2 θ 2 a 2 1)d 1) ] Bπ) { n + 2 [a2 n+2 1) + sin2 n + 2)θ] + n + 2 [a2 n+1 1) + sin2 )θ] +a 2n+3 2 cos 2 θ ) 2 1 a 1 2 cos 2 2n + 3) θ )} 2 1, and further, using the value of Bπ) from 2.6) on the right-hand side, a 2 1)d 1)Bπ) + [a 2 1)cos 2 θ2 + d sin 2 θ ) ] 2 sin 2 θ Bπ) 2 [ a 2 1)d 1)Bπ) + a 2 1)d 1) 2a 2n+3 cos 2 θ 2 2a 1 cos 2 2n + 3) θ 2 + n + 2 sin2 n + 2)θ + n + 2 ] sin2 )θ.

8 288 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) Table 1 n ϱ 0 n ϱ x Fig. 1. The typical graph of Gϱ). After dividing this by cos 2 θ/2) we have Since cos 2 2n + 3)θ/2) 2 a 2 1)d 1)a 2n+3 2a 2 1)d 1)a 1 cos 2 θ/2) [a 2 1) + 8 d sin 2 θ/2))sin 2 θ/2)]bπ) [ + a 2 1)d 1) n + 2 sin2 n + 2)θ + n + 2 ] sin2 )θ 0. cos 2 2n + 3)θ/2) max θ [0,π] cos 2 = 2n + 3) 2, θ/2) C = Cϱ) := max 8 d sin 2 θ ) θ [0,π] 2 sin 2 θ { 8d 1) 2 = 2d 2 { min θ [0,π] n + 2 sin2 n + 2)θ + n + 2 } sin2 )θ = 0, if d 2, if d<2, 2.10)

9 G.V. Milovanović et al. / Journal of Computational and Applied Mathematics ) we conclude that the left-hand side of the last inequality is greater than or equal to Gϱ) = G n ϱ), Gϱ) = 2a 2 1)d 1)a 2n+3 2a 2 1)d 1)a 1 2n + 3) 2 a C)Bπ). Using 2.4), 2.5) and 2.10) we get Gϱ) = 1 8 n + 2 ) ϱ 2n+6 + n + 2 2n+5 i= 2n+6) l i n)ϱ i, ϱ ϱ d, 2.11) ϱ d is the unique zero greater than 1 of the equation dϱ)=2. When ϱ 1, ϱ d ), Gϱ) is less than the function from right-hand side of 2.11). Since Gϱ) is continuous when ϱ > 1 and lim ϱ + Gϱ) =+, it follows that Gϱ)>0, for each ϱ > ϱ 0, ϱ 0 is the largest zero of Gϱ). The values of ϱ 0 for some values of n are displayed in Table 1. All values of ϱ 0, except when n = 4, are optimal in a sense that for ϱ < ϱ 0 and ϱ > ϱ 0 ) the point z = 1 2 ϱ + ϱ 1 ) is not a maximum point. The optimal value of ϱ 0 when n = 4 is The values for n = 5,...,10 coincide with the values ϱ n given in [1, Table 4.2]. A typical graph of Gϱ) is displayed in Fig. 1. Here n = 4. References [1] W. Gautschi, On the remainder term for analytic functions of Gauss Lobatto and Gauss Radau quadratures, Rocky Mountain J. Math ) [2] W. Gautschi, S. Li, The remainder term for analytic functions of Gauss Radau and Gauss Lobatto quadrature rules with multiple end points, J. Comput. Appl. Math ) [3] W. Gautschi, S. Li, Gauss Radau and Gauss Lobatto quadratures with double end points, J. Comput. Appl. Math ) [4] W. Gautschi, E. Tychopoulos, R.S. Varga, A note on the contour integral representation of the remainder term for a Gauss Chebyshev quadrature rule, SIAM J. Numer. Anal ) [5] W. Gautschi, R.S. Varga, Error bounds for Gaussian quadrature of analytic functions, SIAM J. Numer. Anal ) [6] D.B. Hunter, G. Nikolov, On the error term of symmetric Gauss Lobatto quadrature formulae for analytic functions, Math. Comp ) [7] G.V. Milovanović, M.M. Spalević, Error bounds for Gauss Turán quadrature formulae of analytic functions, Math. Comp ) [8] F. Peherstorfer, On the remainder of Gaussian quadrature formulas for Bernstein Szegő weight functions, Math. Comp ) [9] T. Schira, The remainder term for analytic functions of Gauss Lobatto quadratures, J. Comput. Appl. Math ) [10] T. Schira, The remainder term for analytic functions of symmetric Gaussian quadratures, Math. Comp )