# HOW TO FIND EXPANSION OF LOGARITHMIC FUNCTIONS

In this section, you will learn how to find expansion of logarithmic functions.

## Logarithmic Functions

The series Σ n = 1 to ∞ (−1)n+1 xn /n is called a logarithmic series. This series converges for all values of x satisfying |x| < 1. This series converges when x = 1 also.

For all values of x satisfying |x| < 1, the sum of the series is log(1 + x). Thus

log(1 + x) = x − (x2/2) + (x3/3) − (x4/4) + · · ·

for all values of x satisfying |x| < 1.

By taking −x in place of x we get

log(1 − x) = −x − x2/2 − x3/3 − x4/4 − ·· ·

for all values of x satisfying |x| < 1.

Now log [ (1+x)/(1−x)] = log(1+x) − log(1 − x).

Using this we get

log [ (1 + x)/(1 − x)]  = 2 [x + x3/3  + x5/5 +............]

Now log [ (1-x)/(1+x)] = log(1-x) − log(1+x).

Using this we get

log [ (1 - x)/(1 + x)]  = -2 [x + x3/3  + x5/5 +............]

## Expansion of Logarithmic Functions - Practice Questions

Question 1 :

Write the first 4 terms of the logarithmic series

log (1 + 4x)

Solution :

log(1 + x) = x − (x2/2) + (x3/3) − (x4/4) + · · ·

log(1 + 4x)  =  4x − [(4x)2/2] + [(4x)3/3] − [(4x)4/4] +..........

=  4x − (16x2/2) + (64x3/3) − (256x4/4) +............

=  4x − 8x2 + (64x3/3) − 64x4 +............

Required condition is |x| < 1/4

Question 2 :

Write the first 4 terms of the logarithmic series

log(1 − 2x)

Solution :

log(1 − x) = −x − x2/2 − x3/3 − x4/4 − ·· ·

log(1 - 2x)  =  -2x − [(2x)2/2] - [(2x)3/3] − [(2x)4/4] +..........

=  -2x − (4x2/2) - (8x3/3) − (16x4/4) +............

=  -2x − 2x2 + (8x3/3) − 4x4 +............

Required condition is |x| < 1/2

Question 3 :

Write the first 4 terms of the logarithmic series

log [(1+3x)/(1−3x)]

Solution :

log [ (1 + x)/(1 − x)]  = 2 [x + x3/3  + x5/5 +............]

log [ (1 + x)/(1 − x)]

= 2 [3x + (3x)3/3  + (3x)5/5 + (3x)7/7............]

= 2 [3x + (27x3/3)  + (243x5/5) + (2187x7/7)+............]

Required condition is |x| < 1/3

Question 4 :

Write the first 4 terms of the logarithmic series

log [(1-2x)/(1+2x)]

Solution :

log [ (1 - x)/(1 + x)]  = -2 [x + x3/3  + x5/5 +............]

log [(1-2x)/(1+2x)]

= -2 [2x + (2x)3/3 + (2x)5/5  + (2x)7/7 +............]

= -2 [2x + (8x3/3)  + (32x5/5) + (128x7/7)+............]

Required condition is |x| < 1/2

Question 5 :

If y = x + x2/2 + x3/3 + x4/4 + · · · , then show that x = y − y2/2! + y3/3! − y4/4! + · · · .

Solution :

y = x + x2/2 + x3/3 + x4/4 + · · ·

Let us take negative signs on both sides.

-y = -(x + x2/2 + x3/3 + x4/4 + · · · )

-y  =  -x - x2/2 - x3/3 - x4/4 - · · ·

-y  =  log (1 - x)

e-y  =  1 - x

x =  1 - e-y

x = 1 - [1 - y/1! + y2/2! - y3/3! + ............]

x  =  1 - 1 + y/1! - y2/2! + y3/3! - ............

x = y/1! - y2/2! + y3/3! - y4/3 + ............

Hence proved.

Question 6 :

If p − q is small compared to either p or q, then show that Solution :  Hence proved.

From this, we have to find the value of 8th root of (15/16)

p = 15, q = 16 and n = 8

=  [15 (8 + 1) + 16 (8 - 1)]/[15 (8 - 1) + 16 (8 + 1)]

=  [15 (9)+16(7)]/[15 (7) + 16 (9)]

=  [135 + 112]/[105 + 144]

=  247/249

=  0.9919

Question 7 :

Find the coefficient of x4 in the expansion of (3−4x+x2)/e2x .

Solution :

ex  =  1 + x/1! + x2/2! + x3/3! + ..............

(3−4x+x2)/e2x  =  (3−4x+x2)(1/e2x)

Expansion for ex :

1 + x/1! + x2/2! + x3/3! + x4/4! ..............

Expansion for e2x :

1 - 2x/1! + (-2x)2/2! + (-2x)3/3! + (-2x)4/4! ..............

1 - 2x/1! + 4x2/2 - 8x3/6 + 16x4/24 ..............

=  (3−4x+x2) (1 - 2x + 2x2 - 4x3/3 + 2x4/3 ..............)

Coefficient of x4

=  2x4 + (16x4/3) + 2x4

=  (2 + 16/3 + 2) x4

=  (4 + 16/3)x4

=  28x4/3

=  (28/3)x4

Question 8 :

Find the value

Solution :   Formula for

log [(1 + x)/(1 - x)]  =  2 [x + x3/3 + x5/5 +..........]

[x + x3/3 + x5/5 +..........]  =  (1/2)log [(1 + x)/(1 - x)]

=  (3/2) log(1 + (1/3)/(1-(1/3)) + (1/2)log(1 + (1/9)/(1-(1/9))

=  (3/2) log(4/2) + (1/2)log(10/9)/(8/9)

=  (3/2) log 2 + (1/2) log (5/4)

=  (1/2) log 23 + (1/2) log (5/4)

=  (1/2) [log 8 + log (5/4)]

=  (1/2) log 10 After having gone through the stuff given above, we hope that the students would have understood how to find expansion of logarithmic functions

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