Exponential and Logarithmic Series PDF
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This document covers exponential and logarithmic series, including properties of the number 'e', and various related formulas. It presents the concepts in a clear, step-by-step manner, useful for students studying advanced mathematics.
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284 E3 Exponential Series 60 284 Exponential and Logarithmic Series 7.1 Definition (The number e). n ID 1 The limiting value of 1 when n tends to infinity is denoted by e n n U 1 1 1 1 1 i.e., e = e lim 1 1 ......... = 2.71 (Nearly) n n 1! 2! 3! 4 ! 7.2 Proper...
284 E3 Exponential Series 60 284 Exponential and Logarithmic Series 7.1 Definition (The number e). n ID 1 The limiting value of 1 when n tends to infinity is denoted by e n n U 1 1 1 1 1 i.e., e = e lim 1 1 ......... = 2.71 (Nearly) n n 1! 2! 3! 4 ! 7.2 Properties of e. 1 1 n 1 for n 2 ) n! 2 (2) The value of e correct to 10 places of decimals is 2.7182818284 D YG (1) e lies between 2.7 and 2.8. i.e., 2.7 < e < 2.8 (since (3) e is an irrational (incommensurable) number (4) e is the base of natural logarithm (Napier logarithm) i.e. ln x log e x and log 10 e is known as Napierian constant. log 10 e 0.43429448 , ln x 2.303 log 10 x 1 since ln x log 10 x. log e 10 and log e 10 2.30258509 log 10 e U 7.3 Exponential Series. ST For x R , e x 1 x x2 x3 xr xn ...... ...... or e x n! 1! 2! 3! r! n 0 The above series known as exponential series and e x is called exponential function. Exponential function is also denoted by exp. i.e. exp A e A ; exp x e x 7.4 Exponential Function ax, where a > 0. x a x e loge a e x loge a ax e x....(i) , where log e a. We have, e x 1 Replacing x by x in this series, e x 1 x 1! 2x2 2! x x2 x3 xr ........ ........ 1! 2! 3! r! 3x3 3! ..... rxr r! ..... Exponential and Logarithmic Series 285 Hence from (i), a x 1 log e a (log e a)2 2 (log e a)r x r x x ....... ........ 1! 2! r! 7.5 Some Important Results from Exponential Series. We have the exponential series xn n! …..(i) n 0 (2) Replacing x by –x in (i), we obtain e x 1 (3) Putting x 1 in (i) and (ii), we get, e 1 (1)n n! n 0 e x e x x3 x5 x ....... 2 3! 5! e x e x x2 x4 x6 1 ....... 2 2! 4! 6! x 2 n 1 n 0 xn n! …..(ii) x 2n n 0 (2n)! (2n 1)! n 0 e e 1 1 1 1 1 ....... 2 2! 4 ! n 0 (2n)! e e 1 1 1 1 1 ....... 2 3! 5! n 0 (2n 1)! , D YG (5) (1)n (4) From (i) and (ii), we obtain 1 1 1 1 ...... 1! 2! 3! n 0 n! ID 1 1 1 ...... 1! 2! 3! x x2 x3 ...... 1! 2! 3! U e 1 1 60 x x2 ........ 1! 2! E3 (1) e x 1 Note : e 1 11! 21! 31! 41! ......... r1! e2 r 1 1 1 1 ......... 2! 3! 4 ! r! 1 r2 7.6 Some Standard results. (1) 1 n 0 1 n 0 1 n! 2! 3! 4! ......... e 2 1 1 (2) 1 1 (4) ST (7) (2n)! 1 2! 4! 6! ......... 1 1 1 1 n 0 (8) (9) e x 1 2 3 (6) 1 1 1 1 1 1 (n 1)! 2! 3! 4! ......... e 2 1 1 1 1 n 1 e e 2 1 1 1 1 e e ......... ( 2 n 1 )! 1 ! 3 ! 5 ! 2 n 1 1 (n 1)! 1! 2! 3! ......... e 1 1 1 1 1 ......... e 2 2! 3! 4 ! n 0 (n 2)! 1 n 0 n2 (5) n! 1! 2! 3! .......... e 1 n 1 n 0 U (3) n! (n 1)! (n k )! e 1 1 (2n 2)! 1 n 1 (2n 1)! 1 n 0 n x x x x ......... ......... 1! 2! 3! n! Tn 1 General term in the expansion of e x xn 1 and coefficient of x n in e x n! n! 286 Exponential and Logarithmic Series 2 x3 xn x x ........ (1)n ........ n! 1! 2! 3! Tn 1 General term in the expansion of e x (1)n (11) e ax 1 (ax) (ax) 2 (ax)3 (ax)n ......... ......... 1! 2! 3! n! Tn 1 General term in the expansion of e ax n e n 0 n! n n 1 n! (13) n3 5e n 0 n! (15) n2 n 1 n! n4 15 e n 0 n! ID 2 4 6 8 ......... 1! 3! 5! 7! (a) 1 / e (b) e (c) 2e n4 n 1 n! [JMI CET 2000] (d) 3e 2 4 6 8 (1 1) (1 3) (1 5) (1 7) ......... ........ 1! 3! 5! 7! 1! 3! 5! 7! U Solution: (b) n2 2e n 0 n! n3 n 1 n! Example: 1 (14) (ax)n an and coefficient of x n in e ax n! n! E3 (12) (1)n xn and coefficient of x n in e x n! n! 60 (10) e x 1 1 1 1 1 1 1 1 e e 1 e e 1 ........ 1 ........ e 2! 4 ! 6! 2 2 1! 3! 5! 7! 2 4 6 ......... 3! 5 ! 7 ! D YG Example: 2 2002] (a) e Solution: (d) Here Tn (2n 1) 1 1 1 S (2n 1)! (2n)! (2n 1)! e e 1 2 1 1 U 3 3 Solution: (d) Example: 5 n 1 1 1 1 1 1 1 ....... ....... 2! 4! 6! 3! 5! 7! [MNR 1976; MP PET 1997] (c) 4 e (d) 5 e 3 1 2 3 n ........ ........ 1! 2! 3! n! n3 Sn n! 4 n 1 (2)n n! We have 1 n (d) 1/e 1 1 e 1 e n! 5e n3 n 1 The coefficient of x n in the expansion of (a) T (b) 3 e 3 ST S Here Tn Example: 4 23 3 3 4 3 ......... = 2! 3! 4 ! 1 (a) 2e Solution: (d) (c) e 2 (b) 2 e e e S 2 Example: 3 [MNR 1979; MP PET 1995, e 7x e x e 3x (b) e 7x e x e 3x 4 n 1 2n n! e 4 x e 2 x n 0 is (c) 4 n 1 (2)n 1 n! (d) 4 n (2)n n! e 7x e x 4 n (2)n (4 x )n (2 x )n coefficient of x n in 3x n! n! n! e n 0 12 123 1234 ........ 2! 3! 4! [Roorkee 1999; MP PET 2003] Exponential and Logarithmic Series 287 Solution: (d) Tn (b) 3 e n n(n 1) 1 (n 1) 1 n 1 n! Sn 2 (n 1)! 2.n! Tn n 1 1 2 n 1 1 (n 2)! 2 (n 1)! (c) e/2 (n 1)! 2 e 1 e n 1 (d) 3e/2 2 2 1 1 (n 1)! 2 (n 2)! (n 1)! 3e 2 Logarithmic Series 7.7 Logarithmic Series. 60 (a) e Expansion of log e (1 x ); if | x | 1, then log e (1 x ) x E3 An expansion for log e (1 x ) as a series of powers of x which is valid only when, | x | 1 , x2 x3 x4 ........ 2 3 4 7.8 Some Important Results from the Logarithmic Series. or log e (1 x ) x x2 x3 x4 ........ 2 3 4 x 2 x4 (i) log e (1 x ) log e (1 x ) log e (1 x 2 ) 2 ........ , (1 x 1) 4 2 D YG (2) x2 x3 x4 ........ 2 3 4 U log e (1 x ) x ID (1) Replacing x by x in the logarithmic series, we get x3 x5 x3 x5 1 x (ii) log e (1 x ) log e (1 x ) 2 x ........ ........ or log e 2x 3 3 5 5 1 x (3) The series expansion of log e (1 x ) may fail to be valid if |x| is not less than 1. It can be proved that the logarithmic series is valid for x=1. Putting x=1 in the logarithmic series. We get, log e 2 1 1 1 1 1 1 1 1 1 ........ ........ 2 3 4 5 6 1.2 3.4 5.6 U (4) When x 1 , the logarithmic series does not have a sum. This is in conformity with the fact that log(1-1) is not a finite quantity. ST 7.9 Difference between the Exponential and Logarithmic Series. (1) In the exponential series e x 1 x x2 x3 .......... all the terms carry positive signs 1! 2! 3! whereas in the logarithmic series log e (1 x ) x x2 x3 x4 ........ the terms are alternatively 2 3 4 positive and negative. (2) In the exponential series the denominator of the terms involve factorial of natural numbers. But in the logarithmic series the terms do not contain factorials. (3) The exponential series is valid for all the values of x. The logarithmic series is valid when |x|< 1. 288 Exponential and Logarithmic Series (0.5)2 (0.5)3 (0.5)4 ........ 2 3 4 1 (b) log 10 2 3 (a) log e 2 Solution: (a) We know that, x (0.5)2 (0.5)3 (0.5)4 3 ........ log e (1 0.5) log e 2 3 4 2 1 1 1 1 ........ 1.2 2.3 3.4 4.5 [Roorkee 1992; MP PET 1999; AIEEE 2003] 4 (a) log e e (b) log e We know that, log e 2 e 4 (c) log e 4 1 1 1 ....... 1.2 3.4 5.6 (d) log e 2 ID Solution: (a) 1 (d) log e 2 (c) log e (n!) x2 x3 x4 ........ log e (1 x ) 2 3 4 Putting x = 0.5, we get, 0.5 Example: 7 [MP PET 1995] 60 0.5 E3 Example: 6 1 1 1 Also log e 2 1 ....... 2.3 4.5 6.7 ………(i) ………(ii) Example: 8 1 1 1 1 1 1 1 1 4 ........ log e 4 log e e log e ........ 1.2 2.3 3.4 4.5 1.2 2.3 3.4 4.5 e D YG 2 log e 2 1 U 1 1 1 1 By adding (i) and (ii), we get, 2 log e 2 1 ........ 1.2 2.3 3.4 4.5 The coefficient of x n in the expansion of log e (1 3 x 2 x 2 ) is 2n 1 (a) (1)n n Solution: (b) (b) (1)n 1 n [2 1] n 2n 1 n (1)n 1 xn n (1)n 1 n 1 (2 x )n n U n 1 1 2n (1)n 1 x n n n n 1 ST 2 n 1 (1)n 1 (2 n 1) So coefficient of x n (1)n 1 n n 1 2n n x (1)n 1 n n 1 (1) n (1)n 2 ........ 2 The equation x log x (2 x ) 25 holds for (a) x 6 (b) x 3 [MP PET 1992] (c) x 3 Given equation x log x (2 x ) 25 (2 x )2 25 hold for x 3 Example: 10 x6 x9 ........ , then x If y x 3 2 3 (a) 1 ey 3 (d) x 7 2 Solution: (c) Solution: (c) (d) None of these We have, log e (1 3 x 2 x 2 ) log e (1 x ) log e (1 2 x ) Example: 9 (c) (b) 1 ey 3 [MNR 1975] 1 (c) (1 e y ) 3 1 (x 3 )2 (x 3 )3 y x 3 ........ log e (1 x 3 ) e y 1 x 3 x (1 e y ) 3 2 3 (d) (1 e y )3
