Trying a solution of the form Ck ~ Aka, (7.101) where A is an arbitrary constant and a is to be determined, gives 1 ka. 2k (k + 1)" - 1- (7.102) Now (*+r - (1+)" ~ k° (1+). = ka (7.103) Comparison of equation (7.103) with the right side of equation (7.102) shows that 1 a = (7.104) 2' and the dominant behavior of Ck is A Ck = (7.105) Higher-order correction may be obtained by using the representation A Ck = · dg, (7.106) where A2 A1 di -1+ k A3 k3 (7.107) Substitution of equation (7.106) into equation (7.99) gives 1 7 3 dr+1 1+ 1- k 2k 24k2 16k3 (7.108) 1 1 +.. dz. 12k2 12k3 Note that nok- term appears in the k dependent function on the right side of equation (7.108). Let us now calculate the coefficient A1 in equation (7.107). To do this, we substitute di = 1+ ek k (7.109)
Trying a solution of the form Ck ~ Aka, (7.101) where A is an arbitrary constant and a is to be determined, gives 1 ka. 2k (k + 1)" - 1- (7.102) Now (*+r - (1+)" ~ k° (1+). = ka (7.103) Comparison of equation (7.103) with the right side of equation (7.102) shows that 1 a = (7.104) 2' and the dominant behavior of Ck is A Ck = (7.105) Higher-order correction may be obtained by using the representation A Ck = · dg, (7.106) where A2 A1 di -1+ k A3 k3 (7.107) Substitution of equation (7.106) into equation (7.99) gives 1 7 3 dr+1 1+ 1- k 2k 24k2 16k3 (7.108) 1 1 +.. dz. 12k2 12k3 Note that nok- term appears in the k dependent function on the right side of equation (7.108). Let us now calculate the coefficient A1 in equation (7.107). To do this, we substitute di = 1+ ek k (7.109)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.3: Algebraic Expressions
Problem 32E
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