The binomial series flashcards

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Binomial series (rational n)
(1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + …, an infinite series when n is not a positive integer.
Range of validity
The set of x for which the infinite series adds up to the function; for (1 + t)ⁿ it is |t| < 1.
Truncated expansion
The first few terms of the series, used as a polynomial approximation to the function for small x.
Ascending powers of x
Terms written in the order constant, x, x², x³, … — the order the binomial series produces them.
Percentage error
|approximation − true value| ÷ true value × 100%.
Coefficient of xʳ in (1 + x)ⁿ
n(n − 1)(n − 2)…(n − r + 1)/r!: r factors on top, starting at n and going down by 1.
Binomial series for (1 + x)ⁿ, n rational
1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + …, valid for |x| < 1. (Yes)
(a + bx)ⁿ in the (1 + t)ⁿ form
aⁿ(1 + bx/a)ⁿ, valid for |bx/a| < 1, i.e. |x| < |a/b|. (No — derive it)
(1 − x)⁻¹
1 + x + x² + x³ + …, valid for |x| < 1. (No — derive it)
(1 + x)⁻¹
1 − x + x² − x³ + …, valid for |x| < 1. (No — derive it)
(1 + x)⁻²
1 − 2x + 3x² − 4x³ + …, valid for |x| < 1. (No — derive it)
(1 + x)^½
1 + x/2 − x²/8 + x³/16 − …, valid for |x| < 1. (No — derive it)
Validity of (1 + kx)ⁿ
|kx| < 1, i.e. |x| < 1/|k|. (No — derive it)
Partial sums of 1/(1 − x) plotted against the curve
Inside −1 < x < 1 they close in on the curve as terms are added; outside, they move further away.
A curve and its quadratic approximation from the binomial series
They touch and agree closely near x = 0, then separate towards the edges of the range of validity.
Where the range of validity of a rational function's series ends
At the vertical asymptote nearest to the origin, e.g. x = ½ for (1 + 8x)/((1 − 2x)(2 + x)).
Size of successive terms when x is small (e.g. x = 0.02)
Each term is far smaller than the one before (1, 0.02, 0.0002, …), so a few terms give high accuracy.
Why the series is finite for positive integer n
The factor (n − n) = 0 appears in every coefficient from x^(n+1) on, so all later terms vanish.
Why (1 − x)⁻¹ needs |x| < 1
Its series is geometric with ratio x, which converges only when |x| < 1.
Why take aⁿ out of (a + bx)ⁿ
The series is for (1 + t)ⁿ, so the constant inside the bracket must be 1.
Validity when adding several expansions
The sum is valid only where every part is valid: take the narrowest range.
How to choose x for an approximation
Small, inside the range of validity, and turning the bracket into a number whose root is a simple multiple of the one you want.
Why partial fractions help with series
Each partial fraction A/(1 + kx) or C/(1 + kx)² is a single binomial expansion; add them term by term.
Expand √(1 + 4x) up to x³.
1 + 2x − 2x² + 4x³, valid for |x| < ¼.
Expand (1 − 3x)⁻² up to x³ and state the validity.
1 + 6x + 27x² + 108x³, valid for |x| < ⅓.
First three terms of 1/√(4 − x).
½(1 − x/4)^(−½) = ½ + x/16 + 3x²/256, valid for |x| < 4.
(1 + kx)ⁿ = 1 − 2x + 6x² + …. Find n and k.
nk = −2 and n(n − 1)k² = 12 ⇒ (n − 1)/n = 3 ⇒ n = −½, k = 4.
Estimate √6 from √(1 − 2x) ≈ 1 − x − ½x² − ½x³.
x = 0.02: √0.96 = 2√6/5 ≈ 0.979796, so √6 ≈ 2.44949.
Expand (1 + 8x)/((1 − 2x)(2 + x)) up to x².
= 2/(1 − 2x) − 3/(2 + x) = ½ + 19x/4 + 61x²/8, valid for |x| < ½.
Coefficient of x² in 1/√(1 − x).
(−½)(−3/2)/2 × (−x)² ⇒ 3/8.
Expand (3 − x)/√(1 + 2x) up to x².
(3 − x)(1 − x + 3x²/2) = 3 − 4x + 11x²/2, valid for |x| < ½.
(1 + 3x)⁻¹: the x² term
(−1)(−2)/2 × (3x)² = +9x², not 3x² — bracket the whole term.
Taking 4 out of (4 + x)^½
4^½ = 2 goes outside: 2(1 + x/4)^½, not 4(1 + x/4)^½.
Validity of (1 − 3x)⁻²
|3x| < 1 ⇒ |x| < ⅓, not |x| < 1.
Using ⁿCᵣ on a calculator for n = −2 or ½
It does not apply; build each coefficient from n(n − 1)…(n − r + 1)/r!.
Quick check on a rational-function series
Put x = 0: the constant term must equal f(0).
Substituting a value outside the range
The series no longer equals the function — e.g. x = 1 in √(1 + 4x) gives 5, not √5.

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