FM5002-HW5-2.22.12

# FM5002-HW5-2.22.12 - Financial Mathem atics 5002 Hom ework...

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Financial Mathematics 5002 : Homework 5 (0042, 0043) Due on 22 Februrary 2012 Scot Adams Solutions 0042 − 1. a.Compute I 2 I 5 P 4 I cos L 2 z R D z. b.Compute I 5 P 4 I 4 cos L 2 z R D c.Compute I 2 I 4 cos L 2 z R D d.Show that the complexlimit lim h R 0 cos (2(3+2 I +h)) M cos (2(3+2 I )) h exists and is equal to M 2 sin (2(3+2 I )). a. I 2 I 5 P 4 I cos L 2 z R D z E I 2 I 5 P 4 I e 2 iz Me M 2 iz 2 D z E 1 2 I 2 I 5 P 4 I e 2 iz D z M I 2 I 5 P 4 I e M 2 iz D z E L 5 P 2 I R 2 I 0 1 e 2 I L 2 IP t L 5 P 2 I RR D t M I 0 1 e M 2 I L 2 t L 5 P 2 I RR D t E L 5 P 2 I R 2 I 0 1 e M 4 M L 4 M 10 I R t D t M I 0 1 e 4 P L 4 M I R t D t E L 5 P 2 I R 2 e M 8 P I M 4 M L 4 M 10 I R M e 8 M I 4 L 4 M 10 I R E I 4 L Me M 8 P I P e M 4 8 M I P e 4 R . b. I 5 P 4 I 4 cos L 2 z R D z E 1 2 I 5 P 4 I 4 e 2 iz D z M I 5 P 4 I 4 e M 2 iz D z E L M 1 M 4 I R 2 I 0 1 e 2 I L 5 P 4 t L M 1 M 4 I RR D t M I 0 1 e M 2 I L 5 P 4 t L M 1 M 4 I RR D t E L M 1 M 4 I R 2 I 0 1 e L M 8 P I R P L 8 M 2 I R t D t M I 0 1 e M L M 8 P I R M L 8 M 2 I R t D t E L M 1 M 4 I R 2 e 8 I M 8 P I 8 M 2 I M e M 8 I 8 M I M L 8 M 2 I R E I 4 L Me 8 I P e M 8 P I M 8 I P e 8 M I R . c. I 2 I 4 cos L 2 z R D z E I 2 I 5 P 4 I cos L 2 z R D z P I 5 P 4 I 4 cos L 2 z R D z E I 4 L e M 4 P e 4 M 8 I 8 I R . d.Observe that this is just the derivative of cos L 2 L 3 P 2 I P z RR . We check that the purely real and purely complexlimits exist and are equal : lim h R 0 cos L 2 L 3 P 2 I P h RR M cos L 2 L 3 P 2 I RR h E lim h R 0 L e L M 4 P 6 I R P 2 I h P e L 4 M 6 I R M 2 I h R M L e M 4 P 6 I P e 4 M 6 I R 2 h E

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lim h R 0 2 I E L M 4 P 6 I R P 2 I h M 2 I E L 4 M 6 I R M 2 I h 2 e I E L M 4 P 6 I R M I E L 4 M 6 I R eM 2 Sin l 6 P 4 I r , by L’ Hopital’ s rule. On the other hand, lim h R 0 cos L 2 L 3 P 2 I P h I RR M cos L 2 L 3 P 2 I RR h I e lim h R 0 L E L M 4 P 6 I R M 2 h P E L 4 M 6 I R P 2 h R M L E M 4 P 6 I P E 4 M 6 I R 2 h I e lim h R 0 M 2 E L M 4 P 6 I R M 2 h P 2 E L 4 M 6 I R P 2 h 2 I eI E L M 4 P 6 I R MI E L 4 M 6 I R 2 Sin l 6 P 4 I r . Thus, the limit exists and is equal to M 2 Sin l 6 P 4 I r . 0042 − 2. a.Compute I 2 I 5 P 4 I L z R 4 D z. b.Compute I 5 P 4 I 4 L z R 4 D c.Compute I 2 I 4 L z R 4 D d.Show that the complexlimit lim h R 0 L 3 P 2 I P h R 4 M L 3 P 2 I R 4 h does not exist. a. I 2 I 5 P 4 I L z R 4 D z e L 5 P 2 I R I 0 1 L 2 I P t L 5 P 2 I R R 4 D t e L 5 P 2 I R I 0 1 L 4 P 8 t P 29 t 2 R 2 D t e L 5 P 2 I R I 0 1 L 16 P 64 t P 296 t 2 P 464 t 3 P 841 t 4 R D t e 6463 3 P 12926 I 15 . b. I 5 P 4 I 4 L z R 4 D z e L M 1 M 4 I R I 0 1 L 5 P 4 I P t L M 1 M 4 I R R 4 D t e L M 1 M 4 I R I 0 1 L 41 M 42 t P 17 t 2 R 2 D t e L M 1 M 4 I R I 0 1 L 1681 M 3444 t P 3158 t 2 M 1428 t 3 P 289 t 4 R D t 10687 15 M 42748 I 15 . c. I 2 I 4 L z R 4 D z e L 4 M 2 I R I 0 1 L 2 I P t L 4 M 2 I R R 4 D t e L 4 M 2 I R I 0 1 L 4 M 8 t P 20 t 2 R 2 D t e L 4 M 2 I R I 0 1 L 16 M 64 t P 224 t 2 M 320 t 3 P 400 t 4 R D t e 704 3 M 352 I 3 N I 2 I 5 P 4 I L z R 4 D z P I 5 P 4 I 4 L z R 4 D z e 21628 15 M 29822 I 15 d.lim h R 0 3 P 2 I P h 4 M 3 P 2 I 4 h e lim h R 0 156 h P 62 h 2 P 12 h 3 P h 4 h e 156. On the other hand, lim h R 0 3 P 2 I P h I 4 M 3 P 2 I 4
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