SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 489 39. By Example 7, tan 31 { = " S q=0 (31) q { 2q+1 2q +1 for |{| ? 1. In particular, for { = 1 I 3 , we have 6 = tan 31 1 I 3 = " S q=0 (31) q 1@ I 3 2q+1 2q +1 = " S q=0 (31) q 1 3 q 1 I 3 1 2q +1 , so = 6 I 3 " S q=0 (31) q (2q + 1)3 q =2 I 3 " S q=0 (31) q (2q + 1)3 q . 11.10 Taylor and Maclaurin Series 1. Using Theorem 5 with " S q=0 eq({ 3 5) q , eq = i (q) (d) q! , so e8 = i (8) (5) 8! . 3. Since i (q) (0) = (q + 1)!, Equation 7 gives the Maclaurin series " S q=0 i (q) (0) q! { q = " S q=0 (q + 1)! q! { q = " S q=0 (q + 1){ q . Applying the Ratio Test with dq =(q + 1){ q gives us lim q<" d q+1 d q = lim q<" (q + 2){ q+1 (q + 1){ q = |{| lim q<" q +2 q +1 = |{|· 1= |{|. For convergence, we must have |{| ? 1, so the radius of convergence U =1. 5. q i (q) ({) i (q) (0) 0 (1 3 {) 32 1 1 2(1 3 {) 33 2 2 6(1 3 {) 34 6 3 24(1 3 {) 35 24 4 120(1 3 {) 36 120 . . . . . . . . . (1 3 {) 32 = i (0) + i 0 (0){ + i 00 (0) 2! { 2 + i 000 (0) 3! { 3 + i (4) (0) 4! { 4 + ··· =1+2{ + 6 2 { 2 + 24 6 { 3 + 120 24 { 4 + ··· =1+2{ +3{ 2 +4{ 3 +5{ 4 + ··· = " S q=0 (q + 1){ q lim q<" d q+1 d q = lim q<" (q + 2){ q+1 (q + 1){ q = |{| lim q<" q +2 q +1 = |{| (1) = |{| ? 1 for convergence, so U =1. 7. q i (q) ({) i (q) (0) 0 sin { 0 1 cos { 2 3 2 sin { 0 3 3 3 cos { 3 3 4 4 sin { 0 5 5 cos { 5 . . . . . . . . . sin { = i (0) + i 0 (0){ + i 00 (0) 2! { 2 + i 000 (0) 3! { 3 + i (4) (0) 4! { 4 + i (5) (0) 5! { 5 + ··· =0+ { +0 3 3 3! { 3 +0+ 5 5! { 5 + ··· = { 3 3 3! { 3 + 5 5! { 5 3 7 7! { 7 + ··· = " S q=0 (31) q 2q+1 (2q + 1)! { 2q+1 lim q<" d q+1 d q = lim q<" 2q+3 { 2q+3 (2q + 3)! · (2q + 1)! 2q+1 { 2q+1 = lim q<" 2 { 2 (2q + 3)(2q + 2) =0 ? 1 for all {, so U = ". TX.10 F.
