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

The series Σ n = 1 to ∞ (−1)^{n+1} x^{n }/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 − (x^{2}/2) + (x^{3}/3) − (x^{4}/4) + · · ·

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

By taking −x in place of x we get

log(1 − x) = −x − x^{2}/2 − x^{3}/3 − x^{4}/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 + x^{3}/3 + x^{5}/5 +............]

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

Using this we get

log [ (1 - x)/(1 + x)] = -2 [x + x^{3}/3 + x^{5}/5 +............]

**Question 1 :**

Write the first 4 terms of the logarithmic series

log (1 + 4x)

**Solution :**

log(1 + x) = x − (x^{2}/2) + (x^{3}/3) − (x^{4}/4) + · · ·

Instead of x, apply 4x

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

= 4x − (16x^{2}/2) + (64x^{3}/3) − (256x^{4}/4) +............

= 4x − 8x^{2} + (64x^{3}/3) − 64x^{4} +............

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 − x^{2}/2 − x^{3}/3 − x^{4}/4 − ·· ·

Instead of x, apply 2x

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

= -2x − (4x^{2}/2) - (8x^{3}/3) − (16x^{4}/4) +............

= -2x − 2x^{2} + (8x^{3}/3) − 4x^{4} +............

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 + x^{3}/3 + x^{5}/5 +............]

Instead of x, apply 3x

log [ (1 + x)/(1 − x)]

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

= 2 [3x + (27x^{3}/3) + (243x^{5}/5) + (2187x^{7}/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 + x^{3}/3 + x^{5}/5 +............]

Instead of x, apply 2x

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

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

= -2 [2x + (8x^{3}/3) + (32x^{5}/5) + (128x^{7}/7)+............]

Required condition is |x| < 1/2

**Question 5 :**

If y = x + x^{2}/2 + x^{3}/3 + x^{4}/4 + · · · , then show that x = y − y^{2}/2! + y^{3}/3! − y^{4}/4! + · · · .

**Solution :**

y = x + x^{2}/2 + x^{3}/3 + x^{4}/4 + · · ·

Let us take negative signs on both sides.

-y = -(x + x^{2}/2 + x^{3}/3 + x^{4}/4 + · · · )

-y = -x - x^{2}/2 - x^{3}/3 - x^{4}/4 - · · ·

-y = log (1 - x)

e^{-y} = 1 - x

x = 1 - e^{-y}

x = 1 - [1 - y/1! + y^{2}/2! - y^{3}/3! + ............]

x = 1 - 1 + y/1! - y^{2}/2! + y^{3}/3! - ............

x = y/1! - y^{2}/2! + y^{3}/3! - y^{4}/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 x^{4} in the expansion of (3−4x+x^{2})/e^{2x} .

**Solution :**

**e ^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + ..............**

(3−4x+x^{2})/e^{2x }= (3−4x+x^{2})(1/e^{2x)}

Expansion for e^{x} :

**1 + x/1! + x ^{2}/2! + x^{3}/3! +**

Expansion for e^{2x} :

**1 - 2x/1! + (-2x) ^{2}/2! + (-2x)^{3}/3! +**

**1 - 2x/1! + 4x ^{2}/2 - 8x^{3}/6 +**

= (3−4x+x^{2}) (**1 - 2x + 2x ^{2} - 4x^{3}/3 +**

**Coefficient of x ^{4}**

= 2x^{4} + (16x^{4}/3) + 2x^{4}

= (2 + 16/3 + 2) x^{4}

= (4 + 16/3)x^{4}

= 28x^{4}/3

= (28/3)x^{4}

**Question 8 :**

Find the value

**Solution :**

Formula for

log [(1 + x)/(1 - x)] = 2 [x + x^{3}/3 + x^{5}/5 +..........]

[x + x^{3}/3 + x^{5}/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 2^{3} + (1/2) log (5/4)

= (1/2) [log 8 + log (5/4)]

= (1/2) log 10

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