Free Pearson Edexcel International A-level Mathematics · P4

General binomial series and approximations

What you’ll learn

  • Expand a rational power about x = 0.
  • State and combine validity intervals.
  • Use a truncated expansion for an approximation.

Understand the idea

The binomial series extends the finite expansion to negative and fractional powers. Unless the exponent is a non-negative integer, the series usually continues indefinitely and is valid for an inner term of magnitude less than one. The coefficient pattern uses falling products of the exponent divided by a factorial.

First factor out the constant so the bracket begins with 11. In (a+bx)n(a+bx)^n, write an(1+(b/a)x)na^n(1+(b/a)x)^n when this real-valued form is appropriate. Expand to the requested power and keep enough terms when multiplying two expansions.

For a product or a sum of expansions, the final valid interval must satisfy every contributing restriction. An approximation is generally better when the substituted inner term is small, because omitted higher powers then shrink quickly. Do not label a truncated expression as an exact equality.

Key formulae

(1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\frac{n(n-1)(n-2)}{3!}u^3+\cdots

∣u∣<1for the general binomial series|u|<1\quad\text{for the general binomial series}

(a+bx)n=an(1+bax)n(a+bx)^n=a^n\left(1+\frac ba x\right)^n

A truncated binomial series is local rather than exact

Conditions. Compare √(1 + u) with the second-order binomial approximation 1 + u/2 − u²/8 over −0.5 ≤ u ≤ 0.5, inside the general series range |u| < 1.

u / dimensionless; Function value / dimensionless.

Both agree at u = 0 and closely nearby. At u = 0.5 the approximation is 1.21875 versus √1.5 ≈ 1.224745. More visible agreement on this scale does not make the truncated expression an identity.

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Worked example

Expand (1+2x)−1(1+2x)^{-1} through the term in x3x^3 and state the interval of validity.

  1. Use the exponent n=−1n=-1 and inner term u=2xu=2x.
  2. The coefficients give 1−(2x)+(2x)2−(2x)31-(2x)+(2x)^2-(2x)^3.
  3. The condition is ∣2x∣<1|2x|<1, so ∣x∣<1/2|x|<1/2.

Answer: 1−2x+4x2−8x3+⋯1-2x+4x^2-8x^3+\cdots, valid for ∣x∣<1/2|x|<1/2.

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