QUADRATIC CONVERGENCE OF THE TANH-SINH ...

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8 JONATHAN M. BORWEIN AND LINGYUN YEOur goal is to estimate the approximation error||E h,N || = ||Êh,N ||which will be computed by repeated use of Lemma 3.2 and (3.8).4. Evaluation of the error norm ||Êh,N ||Let ̂σ h,N := ̂T h − ̂T h,N . Then (3.8) implies that(4.1)||Êh,N || 2 = ||Êh + ̂σ h,N || 2= Êh,(z)Êh,(w) ̂K(z, w) +} {{ }Êh,(z)̂σ h,N,(w) ̂K(z, w)} {{ }e 1+ ̂σ h,N,(z) Ê h,(w) ̂K(z, w) + ̂σ h,N,(z)̂σ h,N,(w) ̂K(z, w)} {{ } } {{ }e 3e 2e 4.In (4.1), the error is divided into four parts denoted by e 1 , e 2 , e 3 , and e 4 . In thefollowing three subsections, we will evaluate these quantities term by term.4.1. Evaluation of e 2 and e 3 . The second and third terms of (4.1) are actuallyequal. Indeed:(4.2)e 3 = ̂σ h,N,(z) Ê h,(w) ̂K(z, w) = Ê h,(z)̂σ h,N,(w) ̂K(z, w)⎛⎞= Êh,(z) ⎝h ∑ ̂K(z, nh) ⎠= Êh,(z)= h ∑|n|>N⎛|n|>N⎝h ∑|n|>NÊ h F nh⎞F nh (z) ⎠whereF nh := P nh,(w) ̂K(z, w)(4.3)=π 2 cosh(z) cosh(nh)4 cosh ( π2 sinh(z) − π 2 sinh(nh)) cosh ( π2 sinh(z)) cosh ( π2 sinh(nh)).To calculate (4.2), first we compute ÎF nhÎF nh ===∫ ∞−∞∫ ∞−∞P nh,(w) ̂K(z, w)dzπ 2 cosh(z) cosh(nh)4 cosh ( π2 sinh(z) − π 2 sinh(nh)) cosh ( π2 sinh(z)) cosh ( πdz2sinh(nh))cosh(z)π 2 ∫cosh(nh) ∞4 cosh ( π2 sinh(nh)) −∞cosh ( π2 sinh(z) − π 2 sinh(nh)) cosh ( πdz.2sinh(z))
QUADRATIC CONVERGENCE OF THE TANH-SINH QUADRATURE RULE 9Using the change of variable u := π 2 sinh(x), and letting a := π 2sinh(nh), we get(4.4)(4.5)Applying the identitywe see thatÎF nh = π cosh(nh)2 cosh(a)(4.4) === π cosh(nh)cosh(a)∫ ∞−∞∫ ∞−∞1cosh(u − a) cosh(u) du1cosh(a) + cosh(2u − a) du.sinh(a)cosh(a) + cosh(b) = 11 + e b−a − 11 + e b+a ,∫π cosh(nh) ∞()1cosh(a) sinh(a) −∞ 1 + e 2u−2a − 11 + e 2u duπ cosh(nh)cosh(a) sinh(a) · aπ 2 sinh(2nh)(4.6)=2 sinh (π sinh(nh)) = ÎF nh.∞XBy a similar manipulation, we get ̂T n F nh =π 2 h cosh(nh)(4.7)sinh(π sinh(nh)) 1 + e − 1π sinh(rh)−π sinh(nh) 1 + e sinh(rh)cosh(rh).πTherefore(4.8) h ∑Ê h F nh = ∑×|n|>N{ ∞∑r=−∞r=−∞1|n|>Nπ 2 cosh(nh)sinh (π sinh(nh))()}11 + e − 1h cosh(rh) − sinh(nh) .π sinh(rh)−π sinh(nh) 1 + e π sinh(rh)Consider the summation∞∑()1S n :=1 + e − 1h cosh(rh),π sinh(rh)−π sinh(nh) 1 + e π sinh(rh)r=−∞and note that for h > 0∣S n = ∣e π sinh(nh) − 1∣= sinh(π/2 sinh(nh))≤sinh(π/2 sinh(nh))∞∑r=−∞= h sinh(π/2 sinh(nh))π/2 sinh(rh)e h cosh(rh)e π sinh(nh) + e π sinh(rh) 2 cosh (π/2 sinh(rh))∞∑r=−∞∞∑r=−∞{2 e π/2 sinh(nh) π/2 sinh(rh)e h cosh(rh)e π sinh(nh) + e π sinh(rh) cosh (π/2 sinh(rh))h cosh(rh)cosh(π/2 sinh(rh))1 + 2∞∑r=1cosh(rh)cosh(π/2 sinh(rh))}.
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