Maths › Sequences and series › The general binomial expansion

The general binomial expansion

Let the exponent be negative or fractional and the binomial theorem keeps going, but the expansion never stops: an infinite series that only tells the truth inside a window of x-values. Learn the window and the series becomes a machine for square roots and reciprocals to any accuracy you like.

Year 12-13EDEXCEL 9MA0 4.1

Builds on The binomial expansion and Partial fractions.

IN THIS TOPIC

  • Expand (1 + x)n for negative and fractional n as far as a stated power.
  • Handle (a + bx)n by factoring out an, and state the validity window.
  • Combine expansions with partial fractions, and use them for numerical approximation.

WHAT YOU PROBABLY THINK

The expansion of (1 + x)−1 works for every x.

The series that never ends

For a positive whole n the binomial expansion stops after n + 1 terms because a coefficient eventually hits zero. For any other rational n nothing ever cancels, and the same formula, printed in the booklet, runs forever,

(1 + x)n = 1 + nx + n(n − 1)2!x2 + …IN THE FORMULAE BOOKLET

valid only for |x| < 1. Outside that window the terms grow instead of shrink and the series says nothing, which is the fate of the opening lie: at x = 1 the expansion of (1 + x)−1 alternates 1, 0, 1, 0 and never approaches the true value ½.

The curve of 1 over 1 plus x compared with its three-term expansion 1 minus x plus x squared: the two agree closely inside the validity window where the size of x is less than 1, and part company outside itvalid: |x| < 1series runs away1/(1 + x)
FIG. 11/(1 + x) against its three-term expansion. Inside |x| < 1 the two hug; outside, the series bends away and stops being about the curve at all.

WORKED EXAMPLE

A square root, expanded

Expand (1 − 2x)1/2 in ascending powers of x up to the x2 term, and state the range of validity.

Apply the formula with n = ½ and x replaced by −2x: 1 + ½(−2x) + [½ × (−½)/2](−2x)2.

The terms tidy to 1 − x − x2/2.

Validity needs |−2x| < 1, so |x| < ½.

The replacement goes in whole, sign and coefficient together, exactly as in the finite case, and the validity window shrinks to match.

Partial sums of the infinite expansion of 1 over 1 plus x at x equal to one half: the running totals 1, 0.5, 0.75, 0.625 close in on the true value two thirds from alternating sidespartial sums of 1 − x + x² − … at x = ½true value ⅔each extra term overshoots less
FIG. 2Why truncation works inside the window: the partial sums at x = ½ straddle the true value ⅔ and pinch onto it, each term correcting less than the last.

A leading coefficient, and old friends

When the bracket starts with a instead of 1, factor a out and expand what remains: (a + bx)n = an(1 + bx/a)n, valid for |bx/a| < 1. And partial fractions now pay their promised dividend, splitting an awkward rational function into linear brackets that each expand by this lesson's formula.

YOUR TURN

Factor out, then expand

Expand (4 + x)1/2 up to the x2 term, stating the validity, before opening the working.

Show the working

Factor: (4 + x)1/2 = 2(1 + x/4)1/2.

Expand: 2[1 + ½(x/4) + (½ × (−½)/2)(x/4)2] = 2 + x/4 − x2/64.

Validity: |x/4| < 1, so |x| < 4.

The factored 4 came out as 41/2 = 2 multiplying everything, and forgetting to raise it to the power is this topic's classic dropped mark.

TRY IT UNSEEN

Partial fractions feed the series

Using the decomposition (5x + 7)/((x + 1)(x + 2)) = 2/(x + 1) + 3/(x + 2), expand the original fraction up to the x2 term and state the validity.

Show the working

Each piece expands separately: 2(1 + x)−1 = 2 − 2x + 2x2 − …, and 3/(x + 2) = (3/2)(1 + x/2)−1 = 3/2 − 3x/4 + 3x2/8 − ….

Adding: 7/2 − 11x/4 + 19x2/8.

The windows are |x| < 1 and |x| < 2, and the expansion is only honest where both hold: |x| < 1, the tighter of the two.

A decimal check at x = 0.05 puts both sides at 3.368 to 4 significant figures. The combined validity is always the strictest window in play, because one divergent piece poisons the sum.

THE EXAM BIT

  • Quote the general formula, substitute the whole of bx sign included, and stop at the requested power.
  • Factor a out of (a + bx)n as an, and remember it multiplies every term of the expansion.
  • State the validity with every expansion, |bx/a| < 1 solved for x; the statement is a standing mark.
  • For combined expansions via partial fractions, expand each piece and report the tightest validity window.
  • For numerical estimates, say what x is, substitute into the truncation, and quote only the digits the truncation supports.

CHECK YOURSELF

Expand (1 + 3x)−2 in ascending powers up to x2, state the validity, and use the expansion to estimate 1.03−2.

Show a hint

n = −2, and the replacement is 3x; then x = 0.01.

Show the answer

1 + (−2)(3x) + [(−2)(−3)/2](3x)2 = 1 − 6x + 27x2, valid for |x| < ⅓.

At x = 0.01 the truncation gives 1 − 0.06 + 0.0027 = 0.9427.

The true value is 0.94260…, so the estimate holds to 4 significant figures, comfortably inside the window.

Rational exponents make the expansion infinite; |x| < 1, or |bx/a| < 1, is where it means anything.

Factor out the a, expand the rest, and always say where the series is valid.

CHECK YOUR PROGRESS

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  • Expand (1 + x)n for negative and fractional n as far as a stated power.
  • Handle (a + bx)n by factoring out an, and state the validity window.
  • Combine expansions with partial fractions, and use them for numerical approximation.

No animated video for this topic yet; these notes stand alone.