QUADRATIC CONVERGENCE OF THE TANH-SINH ...

14 JONATHAN M. BORWEIN AND LINGYUN YEX(4.23)+ h2 π 2cosh(nh) cosh(rh)2 cosh π sinh(nh) − π sinh(rh)cosh π sinh(nh)cosh sinh(rh).π r,−n>N 2 2 2 2XThen, because of (4.11), equation (4.23) is equal toh 2 π 2cosh(nh) cosh(rh)X sinh(rh)2 cosh π sinh(nh) + π sinh(rh)cosh π sinh(nh)cosh π r,n>N 2 2 2 2≤ h2 π 2!2cosh(nh)(4.24)= Oe − 2eN−1π2 cosh(π/2 sinh(nh))n>NAlso, (4.22) can be written ash 2 π 2 ∑ cosh 2 (nh)2 cosh 2 (π/2 sinh(nh))n>N+ 2 ∑cosh(nh) cosh(rh)cosh ( π2 sinh(nh) − π 2 sinh(rh)) cosh ( π2 sinh(nh)) cosh ( π2 sinh(rh)).n>r>NThe first term above is less than(4.25) O) (e − 2eN−1πand the second term is less than the first. Equation (4.24) and (4.25) together give( )(4.26) e 4 = ̂σ h,N,(z)̂σ h,N,(w) ̂K(z, w) = O e − 2eN−1π .5. The main resultsCombining (4.12), (4.20) and (4.26) we getTheorem 5.1. (Tanh-sinh convergence.) (a) For f ∈ H 2 and ψ(x) = tanh( π 2 sinh(x)),the error bound can be evaluated as:(||Êh,N || 2 = ||E h,N || 2 = O e −A/h) )+ O(C(h)e − eN−1πwhere the order constant is independent of both N and h. Here as before∞∑ cosh(rh)(5.1) C(h) := 1 + 2cosh(π/2 sinh(rh)) ≤ 1 + 4πhr=1(b) This method exhibits quadratic convergence as we let N → ∞ and h → 0 + ,while keeping Nh a constant.Proof. (a) We estimate||Êh,N || 2 = ||E h,N || 2 = e 1 + e 2 + e 3 + e 4(= O e −A/h) + O(= O e −A/h) + O(C(h)e − eN−1π(C(h)e − eN−1π)+ O).) (e − 2eN−1πTo obtain the estimate for C(h) observe thathC(h) ≤ h + 2∫ ∞0cosh(t)cosh(π/2 sinh(t)) dt = h + 4 π∫ ∞0( π)sech2 x dx,