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CBSE · Class 11 · Mathematics

Logarithmic Series (Appendix 1: A.1.5)

2 concepts · 16 practice questions with answers · Free to read

The logarithmic series

From daily life

Calculators and computers need a way to find logarithms. One way is to add many small terms of an infinite series.

Simple definition

For |x| < 1: loge(1 + x) = x − x²/2 + x³/3 − x⁴/4 + … The result also holds for x = 1, giving loge 2 = 1 − 1/2 + 1/3 − 1/4 + … Replacing x by −x gives loge(1 − x) = −x − x²/2 − x³/3 − …

Formula: ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + …

Rule 1: Use the series — Put in a small x.

  • Easy: ln 1.1 (x = 0.1), three terms: 0.1 − 0.005 + 0.00033 = 0.09533 (true 0.09531).
  • Medium: ln 1.2 (x = 0.2), four terms: 0.18227 (true 0.18232).
  • Exam-type: ln(1 − x) = −x − x²/2 − x³/3 − …, all terms negative.

Rule 2: Slow at x = 1 — Many terms are needed.

  • Easy: ln 2 = 1 − 1/2 + 1/3 − 1/4 + … ; first four terms give 0.5833.
  • Medium: First four terms are far from ln 2 = 0.6931; the series is slow at x = 1.
  • Exam-type: Small |x| gives quick, accurate answers.
Remember

Signs alternate in ln(1 + x) and are all negative in ln(1 − x).

Common mistake

Using the series for x > 1.

Memory clue

Plus, minus, plus, minus; power over power.

Practice questions

Easy
  1. 1Write the first three terms of ln(1 + x).
  2. 2For which x is the series of ln(1 + x) valid?
  3. 3Use three terms to estimate ln 1.1.
Medium
  1. 4Find the coefficient of x³ in ln(1 + x).
  2. 5Write the first three terms of ln(1 − x).
  3. 6Use four terms to estimate ln 1.2.
Challenging
  1. 7Add the first four terms of 1 − 1/2 + 1/3 − 1/4 + … (rounded to 4 decimals).
  2. 8Write the first three terms of ln(1 + 2x).
Show answers (try them yourself first)
  1. 1x − x²/2 + x³/3
  2. 2−1 < x ≤ 1, and mainly |x| < 1
  3. 30.09533
  4. 41/3
  5. 5−x − x²/2 − x³/3
  6. 60.18227
  7. 70.5833
  8. 82x − 2x² + 8x³/3

Using the logarithmic series

From daily life

By combining two series we can make shortcuts for hard values and prove neat identities.

Simple definition

Subtracting the series of ln(1 − x) from ln(1 + x) gives ln[(1 + x)/(1 − x)] = 2(x + x³/3 + x⁵/5 + …). Also ln(1 + px + qx²) can be split into logs of factors when the quadratic factorises.

Formula: ln[(1+x)/(1−x)] = 2(x + x³/3 + x⁵/5 + …)

Rule 1: Find logs of numbers — Choose x so that (1 + x)/(1 − x) is the number.

  • Easy: ln 3: x = 1/2 gives (1.5)/(0.5) = 3; first three terms: 2(0.5 + 0.04167 + 0.00625) = 1.0958 (true 1.0986).
  • Medium: ln 1.5: x = 0.2; 2(0.2 + 0.00267 + 0.000064) = 0.40546 (true 0.40547).
  • Exam-type: ln(1 + x) + ln(1 − x) = ln(1 − x²) = −x² − x⁴/2 − …

Rule 2: Split a product — Factor first.

  • Easy: ln(1 + 5x + 6x²) = ln(1 + 2x) + ln(1 + 3x).
  • Medium: Coefficient of x² = −(2² + 3²)/2 = −13/2.
  • Exam-type: Sum of 1 − 1/2 + 1/3 − 1/4 + … equals ln 2 ≈ 0.693.
Remember

Use x so that (1 + x)/(1 − x) equals the number whose log you want.

Common mistake

Forgetting the factor 2 outside the bracket.

Memory clue

Twice the odd powers, over the odd numbers.

Practice questions

Easy
  1. 1Write the first two terms of ln(1 − x²).
  2. 2Write the first three terms of ln[(1 + x)/(1 − x)].
  3. 3Use x = 1/2 and three terms to estimate ln 3.
Medium
  1. 4Use x = 0.2 and three terms to estimate ln 1.5.
  2. 5Find the coefficient of x² in ln(1 + 3x).
  3. 6Find the coefficient of x² in ln(1 + 3x + 2x²).
Challenging
  1. 7Find the coefficient of x² in ln(1 + 5x + 6x²).
  2. 8Find the sum to infinity of 1 − 1/2 + 1/3 − 1/4 + … (to three decimals).
Show answers (try them yourself first)
  1. 1−x² − x⁴/2
  2. 22x + 2x³/3 + 2x⁵/5
  3. 31.0958
  4. 40.40546
  5. 5−9/2
  6. 6−5/2
  7. 7−13/2
  8. 80.693