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CBSE · Class 11 · Mathematics

Exponential Series (Appendix 1: A.1.4)

3 concepts · 24 practice questions with answers · Free to read

The number e

From daily life

Savings in a bank grow by interest on interest. A particular limit of this growth gives a special number, e, as important as π is for circles.

Simple definition

The sum 1 + 1/1! + 1/2! + 1/3! + 1/4! + … never ends, yet it settles to a fixed number denoted by e, about 2.71828. It always lies between 2 and 3. (Recall 0! = 1 and 1! = 1.)

Formula: e = 1 + 1/1! + 1/2! + 1/3! + …

Rule 1: Add term by term — Partial sums approach e.

  • Easy: First 3 terms: 1 + 1 + 1/2 = 2.5.
  • Medium: First 5 terms: 2.7083; first 6 terms: 2.7167.
  • Exam-type: First 10 terms: 2.718282; true e = 2.718282.

Rule 2: Why e is less than 3 — Compare with a G.P.

  • Easy: From the 3rd term on, 1/n! is less than 1/2^(n−1), because each new factor in n! is at least 3.
  • Medium: So the sum is below 1 + 1 + 1/2 + 1/4 + 1/8 + … = 1 + 2 = 3.
  • Exam-type: The sum is also more than 2, so 2 < e < 3.
Remember

Each term is the previous term divided by the next whole number.

Common mistake

Starting the series at 1/1! and forgetting the first term 1 (that is 1/0!).

Memory clue

Each term = previous ÷ next counting number.

Practice questions

Easy
  1. 1Add the first four terms of the e series.
  2. 2Add the first five terms of the e series.
  3. 3Find the values of 0! and 1!.
Medium
  1. 4Give e correct to two decimal places.
  2. 5Find 1/5!.
  3. 6Add the first six terms of the e series.
Challenging
  1. 7Between which two whole numbers does e lie?
  2. 8Why is 1/4! less than 1/2³?
Show answers (try them yourself first)
  1. 12.6667
  2. 22.7083
  3. 3Both are 1
  4. 42.72
  5. 51/120
  6. 62.7167
  7. 72 and 3
  8. 81/24 is less than 1/8; each extra factor of n! beyond 2 is at least 3, bigger than 2

The series for eˣ

From daily life

Replace 1 by a variable x in the same pattern, and we get a series that can stand for the whole function eˣ.

Simple definition

eˣ = 1 + x/1! + x²/2! + x³/3! + … It is valid for every real x. To expand e raised to an expression, put that expression in place of x.

Formula: e^x = 1 + x/1! + x²/2! + x³/3! + …

Rule 1: Substitute — Replace x by the given expression.

  • Easy: e^(2x) = 1 + 2x + 4x²/2 + 8x³/6 + … = 1 + 2x + 2x² + 4x³/3 + …
  • Medium: e^(−x) = 1 − x + x²/2 − x³/6 + …
  • Exam-type: e^(3x) = 1 + 3x + 9x²/2 + 9x³/2 + …

Rule 2: Coefficients — Coefficient of x^k in e^(cx) is c^k ÷ k!.

  • Easy: x² in e^(2x): 4/2 = 2.
  • Medium: x³ in e^(−x): −1/6.
  • Exam-type: x² in e^(3x+1) = e · e^(3x): e × 9/2 = 4.5e.
Remember

e^(cx) has coefficient c^k/k! for x^k.

Common mistake

Writing the coefficient as c/k! instead of c^k/k!.

Memory clue

Power of c over factorial.

Practice questions

Easy
  1. 1Find the coefficient of x² in eˣ.
  2. 2Find the coefficient of x² in e^(2x).
  3. 3Find the coefficient of x³ in e^(−x).
Medium
  1. 4Write the first four terms of e^(3x).
  2. 5Find the coefficient of x³ in e^(2x).
  3. 6Find the coefficient of x⁴ in e^(−2x).
Challenging
  1. 7Find the coefficient of x² in e^(3x+1).
  2. 8Find the coefficient of x in eˣ · e^(2x).
Show answers (try them yourself first)
  1. 11/2
  2. 22
  3. 3−1/6
  4. 41 + 3x + 9x²/2 + 9x³/2
  5. 54/3
  6. 62/3
  7. 79e/2
  8. 83 (since the product is e^(3x))

Finding values using the exponential series

From daily life

Calculators use series like this to give you eˣ. You can do the same by hand for small x.

Simple definition

Add the first few terms of eˣ. The more terms, the better the value; for small |x| few terms are enough.

Formula: e^x ≈ 1 + x + x²/2 + x³/6 + …

Rule 1: Small x — Few terms are enough.

  • Easy: e^0.1 ≈ 1 + 0.1 + 0.005 + 0.000167 = 1.10517 (true 1.10517).
  • Medium: e^(−0.2) with four terms: 1 − 0.2 + 0.02 − 0.00133 = 0.8187 (true 0.8187).
  • Exam-type: e^0.5 with five terms ≈ 1.6484 (true 1.6487).

Rule 2: Larger x needs more terms — Add terms until they become tiny.

  • Easy: e^(−1) with seven terms ≈ 0.368 (true 0.368).
  • Medium: e^1.5 with six terms ≈ 4.46 (true 4.48).
  • Exam-type: The series 1 + 2 + 4/2! + 8/3! + … equals e².
Remember

For negative x the signs alternate; add carefully.

Common mistake

Stopping after two terms when x is not tiny.

Memory clue

Add until the next term is tiny.

Practice questions

Easy
  1. 1Use four terms to estimate e^0.1.
  2. 2Use five terms to estimate e^0.5.
  3. 3Use seven terms to estimate e^(−1).
Medium
  1. 4Use six terms to estimate e^1.5, to two decimals.
  2. 5Use four terms to estimate e^(−0.2).
  3. 6Can eˣ ever be negative?
Challenging
  1. 7Find the sum of 1 + 2 + 4/2! + 8/3! + … (to two decimals).
  2. 8Find the sum of 1 − 1 + 1/2! − 1/3! + 1/4! − … (to three decimals).
Show answers (try them yourself first)
  1. 11.10517
  2. 21.6484
  3. 30.368
  4. 44.46
  5. 50.8187
  6. 6No, eˣ is always positive
  7. 77.39
  8. 80.368