SIAM J. NUMER. ANAL. Vol. 27, No. 1, pp. 219-224, February 1990 1990 Society for Industrial and Applied Mathematics 015 A NOTE ON THE CONTOUR INTEGRAL REPRESENTATION OF THE REMAINDER TERM FOR A GAUSS-CHEBYSHEV QUADRATURE RULE* WALTER GAUTSCHI, E. TYCHOPOULOS, AND R. S. VARGA Abstract. It is shown that the kernel K,,(z), n (even) -> 2, in the contour integral representation of the remainder term of the n-point Gauss formula for the Chebyshev weight function of the second kind, as z varies on the ellipse,={z: e-i, 0<O<27r},= p> 1, assumes its largest modulus on the imaginary axis if p >= p,,+, where z=pei+p- P,,+t is the root of a certain algebraic equation. If < p <P,,+l, the maximum is attained near the imaginary axis within an angular distance less than /(2n + 2). The bounds {P,,+l} decrease monotonically to 1. Key words. Gauss-Chebyshev quadrature, remainder term for analytic functions, kernel of contour integral representation AMS(MOS) subject classification. 65D32 1. We are dealing here with the remainder term R.(f) of the Gaussian quadrature rule for the Chebyshev weight function of the second kind, (1.1) f(t)(1- t) / dt= A<fl>f(r")) + R.(f), --1 u=l where )=cos (ur/(n+ l)), A)= r sin2 (vcr/(n+ l))/(n+ v= l,2, n. We assume that f is analytic inside of, and continuous on, an ellipse (1.2) o={z" z= 1/2(u + u-1), u p ei, O<- O <-- 2.a } with foci at z + 1 and with the sum of the semi-axes equal to p, t9 > 1. The remainder R,,(f) of (1.1) has the form (cf. [1]) (1.3) R, (f) 27r---./p so that f K,,(z)f(z) dz, IR.(f)l<=,, l(o) max If(z)l max IK.(z)l, where l(o) denotes the length of o. Since f(z) and eo are assumed known, the first two terms on the right side of (1.3 ) can be calculated, and our interest then is in determining where on o the kernel K.(z) assumes its maximum modulus. In view of K.() K.(z) and K(-)=-K.(z), the modulus of K. is symmetric with respect to both coordinate axes" (1.4) Thus, consideration may be restricted to the first quarter of, i.e., to the interval 0- < O -<_ 7r/2 in (1.2). * Received by the editors July 25, 1988; accepted for publication January 3, 1989. Department of Computer Sciences, Purdue University, West Lafayette, Indiana 47907. The work of this author was supported in part by National Science Foundation grant CCR-8704404. Department of Mathematics, National Technical University, 157 73, Athens, Greece. Institute for Computational Mathematics, Kent State University, Kent, Ohio 44242. The work of this author was supported by the Air Force Office of Scientific Research. 219
220 W. GAUTSCHI, E. TYCHOPOULOS, AND R. S. VARGA It is known that, when n is odd, the maximum of [K,(z) on p is attained on the imaginary axis 1, Thm. 5.2]. It is remarked in [ 1 that when n is even, the maximum "... is attained slightly off the imaginary axis." The purpose of this note is to amplify this statement and make it more precise. Defining (1.5) a=a(p)=1/2(p+p-), j=1,2,3,..., p>l, we will prove, in fact, the following theorem. THEOREM 1. For each positive integer n with n >-2, let p, > 1 be the unique root of (1.6) a,(p)_ 1 (p> 1). a,(p) n Then, if n >- 2 is even, we have (1.7) max [K,(z)[ K, ( (p _p-l)) p- if p>=pn+, \z / i.e., the maximum of [Kn(z)[ on gep, when p >-p+l, is attained on the imaginary axis. If l<p<p+, then the maximum in (1.7) is attained at some z=z*= 1/2(p ei*+ e-i*) p with (n/(n + l))tr/2 < O*< r/2. In Table 1 we display p, for n =2(1)40 to 10 decimal places. Since a(p)/a,(p)<2p/p"=2p -(- for p> 1, putting p=p, we obtain from (1.6) that (1.8) nl/(n-1) < Pn < (2n)1 - o. The next theorem establishes monotonic- This shows, in particular, that p. 1 as n ity of the p. and sharpens the bounds in (1.8). THEOREM 2. The roots p. > 1 of (1.6) satisfy (1.9) p. > p.+ for all n >- 2. Moreover, if A. := (2n) /", and if I., n >-2, is the unique positive root (by Descartes" rule of signs) of (1.10) M.(/x) 0, M,,(/x) :=/x"+l- n(/x2+ 1), then (1.11) h. < p. </z. for all n >-_2. It is easily seen that the bounds in (1.11) are sharper than those in (1.8), except when n 2, in which case the lower bounds are both equal to 2. TABLE The roots p,, > of (1.6). n p,, n p,, n p,, n p,, 11 1.3290434092 21 1.1956793660 31 1.1427199553 2 2.2966302629 12 1.3068931058 22 1.1884619640 32 1.1390810161 3 1.9318516526 13 1.2878200461 23 1.1818092074 33 1.1356405646 4 1.7390838834 14 1.2712053026 24 1.1756552136 34 1.1323822718 5 1.6180339887 15 1.2565878778 25 1.1699441267 35 1.1292915806 6 1.5341771340 16 1.2436169389 26 1.1646282627 36 1.1263554696 7 1.4722691130 17 1.2320204906 27 1.1596666536 37 1.1235622539 8 1.4244774799 18 1.2215842188 28 1.1550238943 38 1.1209014162 9 1.3863414780 19 1.2121368378 29 1.1506692205 39 1.1183634632 10 1.3551231521 20 1.2035397132 30 1.1465757653 40 1.1159398028
REMAINDER TERM FOR GAUSS-CHEBYSHEV QUADRATURE 221 The proofs of Theorems 1 and 2 will be given in 3 and 4, respectively. Section 2 contains some auxiliary results. 2. LEMMA 1. For each positive integer n, set sin trr (2.1) q.(o ) := 0<=o--<_ 1, sin ((1 cr)tr/(n + 1)) where o,(1):=lim,lo,(o ). Then o,(o ) increases monotonically from o,(o)=o to o, (1)= n + 1 as tr varies from zero to 1. Proof. Since sin crr sin [(1-tr)Tr], we can write o.(tr) as sin[(n+l)u] (1-tr)Tr o,(tr) where =: so that 0 _-< u _-< r/(n + 1). Furthermore, sin [(n + 1)u] sin u sin u n + 1 U.(x) (cos u=: x), where U,(x) is the Chebyshev polynomial (of the second kind) of degree n. It is well known that U,(x) is increasing from U,(cos(Tr/(n+l)))=O to U,(1)=n+l as x increases from cos (r/(n + 1)) to 1 (hence tr increases from zero to 1), from which the assertions of Lemma 1 follow. LEMMA 2. Let o, be as in Lemma 1, and set (2.2) 0, (o ) := cos o Tr + (n + 1)o (tr) cos 7r 0<tr<l. n+l Then d/,(tr) increases monotonically from,(0)= 1 to $,(1)=(n+1)2-1 as tr varies from zero to 1. Proof The limit values follow directly from the limit values of o. in Lemma 1. Differentiating (2.2), we get 4,;(o)-Trsintrr+q.(tr)sin( 1-7r ) +(n+ 1)o (r) cos (ln_tr) n+l 1 sin o er + r sin o-- + (n + 1 q (o ) cos 1 r =(n+ 1)q;(o-) cos which is positive by Lemma 1. [3 (3.1) (3.2) 3. + 1 Proof of Theorem 1. From 1, eq. (5.9)] we have (r_,p.+,lk.(z)l)2 a2(p)- cos 20 02n+2(p) COS 2(n+ 1)0 z 1/2(p e + p-1 e-,o) e L" By (1.4), it,suffices to consider 0 -< 0 <-7r/2. Denote.(O) a2- cos 20 a2.+2-- COS 2(n + 1)0
222 W. GAUTSCHI, E. TYCHOPOULOS, AND R. S. VARGA where a2, a2n+2 are as defined in (1.5). By symmetry, Let n 7/- (3.4) O,.--- n+12 Since cos 20 -> cos 20, for 0.<= O _-< O,, we have a 2 cos 2 (3.5).(o)_-<.(o.), 0-<o-<o., a,+2-1 where the equality on the right follows from cos 2(n + 1)O, cos ntr 1, since n is even. Differentiating (3.2) gives (3.6) [az,,+-cos2(n+l)o]tc,(o)+t,,(o) 2(n+l)sin2(n+l)O=2sinZO, from which it follows that (azn+-- 1)teen(On) 2 sin 20,, hence (3.7) Letting maxo_<_o=<=/2 t,(o)= t,(o*), we conclude from (3.5) and (3.7) that (3.8) O, < O* _-<-. 2 Differentiating (3.6) once more, and then setting O (3.9) 4 (a2n+2+ 1)to =(n+ 1) 2 a+ 1 azn+2 0, 7r/2, gives 1 (n+ 1) since a2+ 1 2a 2a,+. From the definition of p, (cf. (1.6)) and from the fact that a(p)/a,,+(p) for p > 1 decreases monotonically, we get from (3.9) that (3 10) "(/)0 iffpp.+. If 1 <p <p,+l, i.e., t, (Tr/2) > 0, it is clear from the second relation in (3.3) that O* < r/2 in (3.8), proving the second statement of the theorem. If p => p,+, i.e., tc, (Tr/2)=<0, we now show that (3.11) tc,(a)>0 for O,,<a<- (Opn+l). 2 We introduce the variable r by n A- o- 7r (3.12) O--- 0<o-< 1. n+l 2 Using (n+tr)/(n+ 1)= 1-(1-o-)/(n+ 1), we can rewrite (3.6) in the form a2n +2 cos tr.rr ]:z (3.13) K.(O) a2.+2- (n + 1)a2qn (r) q. (r), 2sin ((1 r)tr/(n + 1)) with qn(tr) and q.(tr) as defined in Lemmas 1 and 2, respectively. By the assumption P => P.+I, which implies a.+ => (n + 1)al, hence 2 a.+z 2an+l l>=2(n+l)a-l=(n+l)2(a:z+l)-l, al 1,
REMAINDER TERM FOR GAUSS-CHEBYSHEV QUADRATURE 223 and using Lemmas 1 and 2, we find that the right-hand side of (3.13) is larger than or equal to (n+ 1)2(a + 1)- 1 (n + 1)azq, (o-) q, (or) >(n+l)(a+l)-l-(n+l)az-[(n+l)2-1]=o, Therefore, (O) > 0 for O, < O < 7r/2, showing that 0"= 7r/2 in (3.8). 4. We precede the proof of Theorem 2 with the following lemma. LEMMA 3. With A, and Ix, as defined in Theorem 2, there holds (4.1) A. >/x,+ for all n >= 2. 0<r<l. Proof Since M,(/x), n=>2, in (1.10) has a unique positive zero /x,, and since M,(+c) +c, it is evident that M,+l(/X)> 0 implies /x>/x,+. It suffices, therefore, to show that (4.2) M,+I(A,) > 0 for n-> 2. This is clearly true when n 2, since A. 2 2 and M3(2) 1. We may thus assume that n=>3. We have - M,+l(&,)= A. A-(n+ 1)(2+ 1)=A 2n-(n+ 1)(A+ 1) -(n + l) + (n -1)& 2 I, n log 2n, there follows M,+(A,) -(n + 1)+(n 1) e 2 (n + 1)+(n 1)[1 + 21, + e 2,, -(1 + 21,)] When we write A, e,,, 2[-1 +(n- 1)1,] + (n- 1)[e,. (1 + 21,)]. Here, the expression in the last bracket is clearly positive, and an elementary calculation shows that -l+(n-1)l.=-l+(1-n-) log2n>o if n>=3. Proof of Theorem 2. To establish (1.9), it suffices to prove the inequalities (1.11), since combining them with the inequality in Lemma 3 immediately gives p, > A, > /x,+ > p,+ for all n _-> 2. Now (1.6) is equivalent to (4.3) L,(p) := pzn rip,+,_ np,- + 1 =0. Clearly, L, (by Descartes rule of signs) has at most two positive zeros. Since L,(0) 1, L,(1) 2-2n <0, and L,(+)= +, there are exactly two positive zeros, one in (0, 1) and the other in (1, c). (Because p and p- occur symmetrically in (1.6), one zero is the reciprocal of the other.) The larger of the two, as in Theorem 1, is denoted by We have L,(p) p2,+ pzn-z_pzn-2_ np,-,(p2+ 1)+ 1 p"-( o + ) p"- no"- ( p + ) +, so that the equation in (4.3), after division by p,-.(p2+ 1), can be written in the form n-1 n-1 /9 1 (4.4) P 2 n-1 p2 n. Since L,(p,) 0, dropping the third term on the left of (4.4), we arrive at n--1 Pn n+l 2 p, ------<n or p, -n(p,+l)<o.
224 W. GAUTSCHI, E. TYCHOPOULOS, AND R. S. VARGA In terms of the function M. in (1.10), this says M.(p.) <0, and hence implies the right inequality in (1.11). To prove the left inequality of (1.11), we will show that (4.5) L.(A.) < 0 for all n >- 2. We now express L.(p) from (4.3) as L, (p) 0" 2no + 1 + 2no no p + Since I (2n) / e,,, =0 (O -2n)+l+no 2- p+ l n -1 log 2n, this gives L,(I,) 1 + 4n(1 cosh 1,) 1 2nl-4n[cosh 1, 1 -sl]. The expression in brackets, when expanded in Taylor s series, involves only positive terms and hence is positive, while 1-2nl=l-2(log2n)<O for all n_->2. This establishes (4.5) and completes the proof of Theorem 2. REFERENCE W. GAUTSCHI AND R. S. VA,GA, Error bounds for Gaussian quadrature of analytic functions, SIAM J. Numer. Anal., 20 (1983), pp. 1170-1186.