Mathematics

Session 1

Exponential Series & Logarithmic Series

Session Objectives

Session Objectives

• The number ‘e’

• Exponential Series

• Logarithmic Series

The number ‘e’

Let us consider the series

1 1 1

1 ...1 ! 2! 3!

The sum of this series is denoted by e.

1 1

e 1 ...1! 2!

To prove that

nlim 1e 1

n n

n 2 3n n 1 n n 1 n 21 1 1 11 1 n. ...

n n 2! n 3! n[Using binomial theorem]

The number ‘e’

n1 1 1 1 1 2

1 1 1 1 1 1 ...n 2! n 3! n n

When n will approach to

will approach to 0.

, 1 2 3, , , etc.,

n n n

nlim 1 1 11 1 1 ...

n n 2! 3!

nlim 1 1 1 11 1 ... e

n n 1! 2! 3!

n

r 0

lim 1 1 1 1 1e 1 1 ...

n n 1! 2! 3! r !

The number ‘e’ is an irrational number and its value lies between 2 and 3.

Exponential Series

2 3

x x x xe 1 ...

1! 2! 3!

is called exponential series.

We have

xn nxx lim lim1 1

e 1 1n nn n

1 1 2x x x x x

n n n1 x ...

2! 3!

Exponential Series

When n will approach to

will approach to 0.

, 1 2 3, , , etc.,

n n n

xnx nxlim lim1 1

1 1 en nn n

2 3x x

1 x ...2! 3!

2 3 r

x

r 0

x x xe 1 x ...

2! 3! r !

Some Results

2 3 r

x

r 0

x x x x1. e 1 ...

1! 2! 3! r !

2 3 r

rx

r 0

x x x x2. e 1 ... 1

1! 2! 3! r !

2 4x x x x

3. e e 2 1 ...2! 4!

x x 2 4 2r

r 0

e e x x x1 ...

2 2! 4! 2r !

Some Results

x x 3 5 2r 1

r 0

e e x x x4. x ...

2 3! 5! 2r 1 !

n

r 0

1 1 15. e 1 ...

1! 2! r !

r1

r 0

1 1 1 16. e 1 ... 1

1! 2! 3! r !

1

r 0

e e 1 1 17. 1 ...

2 2! 4! 2r !

Some Results

1

r 0

e e 1 1 17. 1 ...

2 3! 5! 2r 1 !

Some Important Deductions

n 0 n 0 n 0

1 1 1(i) e

n! n 1 ! n k !

n 1

1 1 1 1(ii) ... e 1

n! 1! 2! 3!

n 2

1 1 1(iii) ... e 2

n! 2! 3!

n 0

1 1 1 1(iv) ... e 1

1! 2! 3!n 1 !

n 0 n 1

1 1 1 1(vii) ... e 2

2! 3!n 2 ! n 1 !

Exponential Theorem

Let a > 0, then for all real values of x,

2

2xlog ax ee e

xa e 1 x log a log a ...

2!

General term of eax

2 3ax ax ax

e 1 ax ...2! 3!

r

r 1ax

T , r 0, 1, 2, ...r !

Logarithmic Series

Expansion of elog 1 x

If |x| < 1, then

2 3 4

ex x x

log 1 x x ...2 3 4

r

r 1

r 1

x1 ...(i)

r

Replacing x by –x,

2 3

ex x

log 1 x x ... (ii)2 3

r

r 1

x

r

Logarithmic Series

(i) – (ii)

e elog 1 x log 1 x

3 5x x2 x ... ... (iii)

3 5

2r 1

r 0

x2

2r 1

2r 1

e er 0

1 xlog 1 x log 1 x

2 2r 1

2r 1

er 0

1 1 x xlog

2 1 x 2r 1

Logarithmic Series

(i) (ii)

2 4x xlog 1 x log 1 x 2 ... ... (iv)

2 4

2r

r 1

1 xlog 1 x 1 x

2 2r

Putting x = 1 in (i), we get

1 1 1

log2 1 ... (v)2 3 4

Class Test

Class Exercise - 1

Find the sum of the series

3 3 32 3 4

1 ... .2! 3! 4!

Solution:

3 3 3 3

n 1

2 3 4 nS 1 ...

2! 3! 4! n!

3Let n A Bn Cn n 1 Dn n 1 n 2 ...(i)

3 2 3 2or n A Bn C n n D n 3n 2n

Comparing the coefficients of like powers of n fromboth sides, we getA = 0, B – C + 2D = 0, C – 3D = 0, D = 1

Solution contd..

A = 0, B = 1, C = 3, D = 1

From (i),

3n n 3n n 1 n n 1 n 2

n 1

n 3n n 1 n n 1 n 2S

n!

n 1

3n n 1 n n 1 n 2n

n! n! n!

n 1 n 2 n 3

1 3 1

n 1 ! n 2 ! n 3 !

= e + 3e + e = 5e

Class Exercise - 2

Find the sum the following series:

2 2

4 4 2 2

1x y x y x y x y x y

2!1

x y x y x y x y ... .3!

Solution:

The given series can be written as

3 32 2 4 4 2 21 1x y x y x y ...

2! 3!

2 2 4 4 6 61 1x y x y x y ...

2! 3!

Solution contd..

2 4 6 2 4 61 1 1 1x x x ... y y y ...

2! 3! 2! 3!

2 2x ye 1 e 1

2 3x x x x

e 1 ...1! 2! 3!

2 2x ye e

Class Exercise - 3

Find the coefficient of Xn in theexpansion of eex

.

Solution:

Let xe y

kxk kxxe y

k 0 k 0 k 0

ey ee e

k! k! k!

x 2x 3xe e e1 ...

1! 2! 3!

n nn

n 0 n 0 n 0

2x 3x1 x 1 11 ...

1! n! 2! n! 3! n!

Solution contd..

Coefficient of

n nxn e 1 1 1 2 1 3x in e ...

1! n! 2! n! 3! n!

n n1 1 2 3...

n! 1! 2! 3!

Class Exercise - 4

Sum the series

2 2 31 2 1 2 2 1 2 2 21 ... .

2! 3! 4!

Solution:

We have

2 2 31 2 1 2 2 1 2 2 21 ...

2! 3! 4!

2 n 1

n 1

1 2 2 ... 2

n!

n n

n 1 n 1

2 1 1 2 1.

2 1 n! n!

n

n 1 n 1

2 1

n! n!

Solution contd..

2 32 2 2 1 1 1... ...

1! 2! 3! 1! 2! 3!

2e 1 e 1

2e e

Class Exercise - 5

Sum the series

4 11 22 37 56

... .1! 2! 3! 4! 5!

Solution:

We have 4 11 22 37 56

...1! 2! 3! 4! 5!

Now we will find the nth term of the numerator

Sn = 4 + 11 + 22 + 37 + 56 + ... + tn – 1 + tn

Sn = 4 + 11 + 22 + 37 + ... + tn – 1 + tn

– – – – – – – –_____________________________________

Subtracting, 0 = 4 + 7 + 11 + 15 + 19 + ... (tn – tn – 1) – tn

Solution contd..

tn = 4 + [7 + 11 + 15 + 19 + ... + t0(n – 1) terms]

n 14 2 7 n 2 4

2

n 14 4n 6

2

= 2n2 + n + 1

The given series =

2

n 1

2n n 1

n!

2

n 1 n 1 n 1

n n 12

n! n! n!

Solution contd..

n 1 n 1 n 1

n 1 12

n 1 ! n 1 ! n!

n 1 n 1 n 1

n 1 1 1 12

n 1 ! n 1 ! n!

n 1 n 1 n 1 n 1

1 1 1 12 2

n 2 ! n 1 ! n 1 ! n!

= 2e + 3e + (e – 1)

= 6e – 1

Class Exercise - 6

Find the sum of infinite series

5 7 9

... .1.2.3 3.4.5 5.6.7

Solution:

Let Tr be the nth term of the infinite series.

n

2n 3T , n 1, 2, 3, ...

2n 1 2n 2n 1

2n 3 A B C...(i)

2n 1 2n 2n 1 2n 1 2n 2n 1

Solution contd..

2n 3 A 2n 2n 1 B 2n 1 2n 1

C 2n 1 2n ...(ii)

Comparing the coefficients of n2, n andconstant term from both sides of theequation (ii), we get4A + 4B + 4C = 0, 2A – 2C = 2 and –B = 3

Solving the above equations, we getA = 2, B = –3, C = 1

Solution contd..

From (i)

n

2 3 1T

2n 1 2n 2n 1

Sum of the given series

nn 1

T

n 1

2 3 1

2n 1 2n 2n 1

n 1

2 2 1 1

2n 1 2n 2n 2n 1

n 1 n 1

1 1 1 12

2n 1 2n 2n 2n 1

Solution contd..

1 1 1 1 1 1 1 1 1 12 1 ... ...

2 3 4 5 6 2 3 4 5 6

e e2 log 2 log 2 1

e3log 2 1

Class Exercise - 7

If

find t.

2 3

2 3

x 1 1 x 1 1 x 1t . . ...,

x 1 2 3x 1 x 1

Solution:

We have

2 3

2 3

x 1 1 x 1 1 x 1t . . ...

x 1 2 3x 1 x 1

2 3x 1 x 1 x

...x 1 2 x 1 3 x 1

2 3

1 1 1 1 1. . ...

x 1 2 3x 1 x 1

Solution contd..

x 1log 1 log 1

x 1 x 1

1 xlog log

x 1 x 1

x x 1log logx

x 1 1

Class Exercise - 8

Find the sum of the series

2 3 41 1 2 1 3 1

....2 5 3 5 4 5

Solution:

We have the series

2 3 41 1 2 1 3 1

...2 5 3 5 4 5

n

n 2

n 1 1.

n 5

n

n 2

1 11

n 5

n n

n 2 n 2

1 1 1.

5 n 5

Solution contd..

2 3 2 31 1 1 1 1 1

... . . ...5 5 2 5 3 5

11 125 log 1

1 5 515

1 5 4 1

log25 4 5 5

1 4log

4 5

Class Exercise - 9

Show that

e e e

2 4 6

2 4 6

log x 2h 2log x h log x

h 1 h 1 h. . ... .

2 3x h x h x h

Solution:

RHS =

2 4

e e 2 4

h 1 h2log x h log x .

2x h x h

6

6

1 h. ...

3 x h

Solution contd..

22

e e e 2

hlog x h log x log 1

x h

22

e e e 2

x 2hxlog x h log x log

x h

2 2

e 2

x h x 2hxlog

x x h

elog x 2h LHS

Class Exercise - 10

If

prove that

2x y 2x y and x 1 and y 1,

3 5 3 5y y x xy ... 2 x ... .

3 5 3 5

Solution:

We have 2x y 2x y 2y x 1 2x

2x 1 1

2x yBy componendo and dividendo

2

2

x 1 2x 1 y

1 yx 1 2x

Solution contd..

2

2

1 x 1 y

1 y1 x

2

e e1 y 1 x

log log1 y 1 x

e e

1 y 1 xlog 2log

1 y 1 x

3 5 3 5y y x x2 y ... 2 x ... (Pr oved)

3 5 3 5

Thank you