1. Mathematics

2. Session 1 Exponential Series & Logarithmic Series

3. Session Objectives

4. T he number ‘e’ E xponential Series L ogarithmic Series

5. The number ‘e’ Let us consider the series The sum of this series is denoted by e. To prove that [Using binomial theorem]

6. The number ‘e’ When n will approach to will approach to 0. The number ‘e’ is an irrational number and its value lies between 2 and 3.

7. Exponential Series is called exponential series. We have

8. Exponential Series When n will approach to will approach to 0.

9. Some Results

12. Some Important Deductions

13. Exponential Theorem Let a > 0, then for all real values of x, General term of e ax

14. Logarithmic Series Expansion of If |x| < 1, then Replacing x by –x,

15. Logarithmic Series (i) – (ii)

16. Logarithmic Series Putting x = 1 in (i), we get

17. Class Test

18. Class Exercise - 1 Find the sum of the series Solution: Comparing the coefficients of like powers of n from both sides, we get A = 0, B – C + 2D = 0, C – 3D = 0, D = 1

19. Solution contd.. A = 0, B = 1, C = 3, D = 1 From (i), = e + 3e + e = 5e

20. Class Exercise - 2 Find the sum the following series: Solution: The given series can be written as

21. Solution contd..

22. Class Exercise - 3 Find the coefficient of X n in the expansion of e ex. Solution: Let

23. Solution contd.. Coefficient of

24. Class Exercise - 4 Sum the series Solution: We have

25. Solution contd..

26. Class Exercise - 5 Sum the series Solution: We have Now we will find the nth term of the numerator S n = 4 + 11 + 22 + 37 + 56 +... + t n – 1 + t n S n = 4 + 11 + 22 + 37 +... + t n – 1 + t n – – – – _____________________________________ Subtracting, 0 = 4 + 7 + 11 + 15 + 19 +... (t n – t n – 1 ) – t n

27. Solution contd.. t n = 4 + [7 + 11 + 15 + 19 +... + t 0 (n – 1) terms] = 2n 2 + n + 1 The given series =

28. Solution contd.. = 2e + 3e + (e – 1) = 6e – 1

29. Class Exercise - 6 Find the sum of infinite series Solution: Let T r be the nth term of the infinite series.

30. Solution contd.. Comparing the coefficients of n 2, n and constant term from both sides of the equation (ii), we get 4A + 4B + 4C = 0, 2A – 2C = 2 and –B = 3 Solving the above equations, we get A = 2, B = –3, C = 1

31. Solution contd.. Sum of the given series

32. Solution contd..

33. Class Exercise - 7 If find t. Solution: We have

34. Solution contd..

35. Class Exercise - 8 Find the sum of the series Solution: We have the series

36. Solution contd..

37. Class Exercise - 9 Show that Solution: RHS =

38. Solution contd..

39. Class Exercise - 10 If prove that Solution: We have By componendo and dividendo

40. Solution contd..

41. Thank you