Binomial series for any index
For (1 + x)³ the expansion stops after four terms. But what about (1 + x)⁻¹ or the square root of (1 + x)? Their expansions never stop, and that is useful.
When the index m is a negative number or a fraction, (1 + x)m = 1 + m·x + [m(m − 1)/2!]·x² + [m(m − 1)(m − 2)/3!]·x³ + … The series has infinitely many terms and is valid only when |x| < 1, that is, −1 < x < 1.
Formula: (1 + x)^m = 1 + mx + m(m−1)/2! · x² + m(m−1)(m−2)/3! · x³ + …
Rule 1: Build the terms — Coefficient of x^k = m(m−1)…(m−k+1) ÷ k!.
- Easy: m = −1: coefficients 1, −1, 1, −1, …, so (1 + x)⁻¹ = 1 − x + x² − x³ + …
- Medium: m = 1/2: coefficients 1, 1/2, -1/8, 1/16; so √(1 + x) = 1 + x/2 − x²/8 + x³/16 − …
- Exam-type: m = −3: coefficient of x² is 6, of x³ is -10.
Rule 2: Check the condition — The series works only for |x| < 1.
- Easy: x = 0.1 in (1 + x)⁻¹: 1 − 0.1 + 0.01 − 0.001 ≈ 0.909, and 1/1.1 = 0.909… ✓.
- Medium: x = 3 in (1 + x)⁻¹: 1 − 3 + 9 − 27 + … grows without settling; it cannot equal 1/4.
- Exam-type: For a positive whole-number index the series stops, so no condition is needed.
Positive whole index: finite terms. Negative or fractional index: infinite terms and |x| < 1.
Using the infinite series for |x| ≥ 1.
Index not a whole number? Series never stops, keep |x| below 1.
Practice questions
- 1Write the first three terms of (1 + x)⁻¹.
- 2Write the first three terms of (1 + x)^(1/2).
- 3For which values of x is the series of (1 + x)⁻³ valid?
- 4Find the coefficient of x² in (1 + x)⁻².
- 5Find the coefficient of x³ in (1 + x)^(1/2).
- 6Find the coefficient of x² in (1 − x)⁻³.
- 7Find the coefficient of x³ in (1 + x)^(−1/2).
- 8Why can the series of (1 + x)⁻¹ not be used for x = 2?
Show answers (try them yourself first)
- 11 − x + x²
- 21 + x/2 − x²/8
- 3−1 < x < 1
- 43
- 51/16
- 66
- 7-5/16
- 8The terms 1, −2, 4, −8, … keep growing, so the series does not settle on a value
Expanding (a + b) to a general power
The same series can be used for numbers such as (4 + x)^(1/2) by first taking out the first term.
Write (a + b)m = am(1 + b/a)m and apply the series with x = b/a. The result is valid when |b| < |a|.
Formula: (a + b)^m = a^m (1 + b/a)^m, |b| < |a|
Rule 1: Take out the first term — Make the first term 1 inside the bracket.
- Easy: (2 + x)⁻¹ = (1/2)(1 + x/2)⁻¹ = 1/2 − x/4 + x²/8 − …, valid for |x| < 2.
- Medium: (4 + x)^(1/2) = 2(1 + x/4)^(1/2) = 2 + x/4 − x²/64 + …, valid for |x| < 4.
- Exam-type: (9 − x)^(−1/2) = (1/3)(1 − x/9)^(−1/2) = 1/3 + x/54 + x²/648 + …, valid for |x| < 9.
Rule 2: Four special cases — Valid for |x| < 1.
- Easy: (1 − x)⁻¹ = 1 + x + x² + x³ + …
- Medium: (1 + x)⁻² = 1 − 2x + 3x² − 4x³ + …
- Exam-type: (1 − x)⁻² = 1 + 2x + 3x² + 4x³ + …
Always check |b| < |a| before expanding.
Writing (a + b)^m = a^m + b^m. It is not.
Pull out the big term first.
Practice questions
- 1Write (2 + x)⁻¹ in the form (1/2)(1 + …) and give the first three terms.
- 2For which x is the expansion of (3 + x)⁻¹ valid?
- 3Write the first four terms of (1 − x)⁻¹.
- 4Find the first three terms of (4 + x)^(1/2).
- 5Find the first three terms of (1 + 2x)⁻².
- 6Find the first three terms of (9 − x)^(−1/2).
- 7Find the coefficient of x² in (1 − 3x)^(−1/2).
- 8Write the first four terms of (1 + x)⁻² and of (1 − x)⁻².
Show answers (try them yourself first)
- 11/2 − x/4 + x²/8
- 2|x| < 3
- 31 + x + x² + x³
- 42 + x/4 − x²/64
- 51 − 4x + 12x²
- 61/3 + x/54 + x²/648
- 727/8
- 81 − 2x + 3x² − 4x³ and 1 + 2x + 3x² + 4x³
Approximate values using the series
To estimate the square root of 1.02 without a calculator, treat 1.02 as 1 + 0.02 and use the first few terms.
When x is small, the first two or three terms of the series already give a good value. The smaller |x| is, the better the approximation.
Formula: √(1 + x) ≈ 1 + x/2 − x²/8
Rule 1: Small x — Use the first three terms.
- Easy: √1.02 ≈ 1 + 0.01 − 0.00005 = 1.00995 (true 1.00995).
- Medium: 1/1.04 ≈ 1 − 0.04 + 0.0016 = 0.9616 (true 0.9615).
- Exam-type: (1.1)⁻² ≈ 1 − 0.2 + 0.03 = 0.83 (true 0.826); x = 0.1 is larger, so the error is bigger.
Rule 2: Cube roots, negative x — Same method.
- Easy: ∛1.06 ≈ 1 + 0.02 − 0.0004 = 1.0196 (true 1.01961).
- Medium: √0.98 = √(1 − 0.02) ≈ 1 − 0.01 − 0.00005 = 0.98995.
- Exam-type: √99 = 10·√(1 − 0.01) ≈ 10(1 − 0.005 − 0.0000125) = 9.949875.
Write the number as 1 + x with a small x.
Using the series when x is not small, like √3 as 1 + 2 …
Small x, few terms, good answer.
Practice questions
- 1Use three terms to estimate √1.02.
- 2Use three terms to estimate 1/1.04.
- 3Use three terms to estimate √1.1.
- 4Use three terms of (1 + x)⁻² to estimate 1/(1.1)².
- 5Use three terms to estimate ∛1.06.
- 6Use three terms to estimate √0.98.
- 7Use three terms of (1 + x)^(−1/2) to estimate 1/√1.04.
- 8Estimate √99 by writing it as 10√(1 − 0.01) and using three terms.
Show answers (try them yourself first)
- 11.00995
- 20.9616
- 31.04875
- 40.83
- 51.0196
- 60.98995
- 70.9806
- 89.949875