P4.3 The binomial series

What is the binomial series for (1 + x)ⁿ?

In P2 the binomial expansion of \((1 + x)^n\) stopped after \(n + 1\) terms. When \(n\) is negative or a fraction, the same coefficient pattern never stops: you get an infinite series that equals the function only for small enough \(x\). Three skills follow from that one formula: writing out the first few terms correctly, stating where they are valid, and putting them to use.

It is valid for \(|x| < 1\) when \(n\) is not a positive integer. When \(n\) is a positive integer, the factor \((n - n) = 0\) appears and every later term is zero, so you get the finite P2 expansion, valid for all \(x\).

Why the series is infinite

For \(n = 3\) the coefficients are 1, 3, \(\frac{3\times2}{2} = 3\), \(\frac{3\times2\times1}{6} = 1\) and then \(\frac{3\times2\times1\times0}{24} = 0\): the expansion stops, giving \((1 + x)^3 = 1 + 3x + 3x^2 + x^3\). For \(n = \frac12\) the factors on top are \(\frac12, -\frac12, -\frac32, -\frac52, \ldots\) and none of them is ever 0, so the terms never stop. The \({}^nC_r\) key on a calculator only works for positive whole \(n\); for rational \(n\), build each coefficient by hand from the pattern.

A reliable layout is to write each term as "coefficient pattern × (whole term)":

Here \(t\) is whatever replaces \(x\) inside \((1 + t)^n\), such as \(4x\), \(-3x\) or \(-\frac{x}{4}\), always with its sign and in a bracket.

When is the expansion valid?

The simplest case shows why. \(\frac{1}{1-x} = (1 - x)^{-1} = 1 + x + x^2 + x^3 + \cdots\) is a geometric series with ratio \(x\), which converges only when \(|x| < 1\). Inside the range, adding more terms gets closer to the true value: at \(x = \frac12\), \(1 + \frac12 + \frac14 + \frac18 = \frac{15}{8}\), approaching 2. Outside it, the terms grow and the sums run away. When \(n\) is a positive integer there is no restriction: the expansion is finite and true for all \(x\).

How do you expand (a + bx)ⁿ?

The series in the formulae list needs a 1 in front. If the bracket starts with any other number, factor it out first.

Products of a polynomial and a series

Expand the binomial part first, up to the power you need. Multiply by each term of the polynomial and collect only the terms up to the required power: terms beyond it are not needed and are not accurate anyway. The range of validity is that of the binomial part; multiplying by a polynomial does not change it.

How do you find unknown constants and coefficients?

Write the general expansion with the unknowns left in: \((1 + kx)^n = 1 + nkx + \frac{n(n-1)}{2}k^2x^2 + \cdots\). Equate coefficients with the given ones: each matching power of \(x\) gives one equation. Two unknowns need two equations. When one equation contains \(nk\) and another \(n(n-1)k^2\), square the first and divide to eliminate \(k\). To find a single coefficient, you only need that one term: the coefficient of \(x^r\) in \((1 + t)^n\) uses \(r\) factors on top and \(r!\) underneath.

How do you use the series to approximate?

Choose a value of \(x\) that turns the bracket into the number you want, and that lies inside the range of validity; the smaller \(|x|\), the better. Substitute into the truncated series: with \(|x|\) small each term is much smaller than the one before, so the first few terms give an accurate value. To find a surd such as \(\sqrt6\), find a number close to 1 whose square root is a simple multiple of \(\sqrt6\), e.g. \(0.96 = \frac{96}{100} = \frac{16 \times 6}{100}\). Judge accuracy with

How do partial fractions help you expand?

A rational function such as \(\frac{1 + 8x}{(1 - 2x)(2 + x)}\) has no single binomial form, but each partial fraction does: \(\frac{A}{1 - 2x} = A(1 - 2x)^{-1}\). Split into partial fractions, expand each one with the binomial series, then add the coefficients of each power. A repeated factor gives a term like \(\frac{C}{(1 + 2x)^2} = C(1 + 2x)^{-2}\): still one binomial expansion. Each piece has its own range of validity; the sum is valid only where all of them are valid, so take the smallest range.

Exam tips for P4.3

Common misconceptions

How this topic connects

P2 Binomial expansion and series

The coefficient pattern is the same as \({}^nC_r\), and the geometric series and its sum to infinity explain why \(\frac{1}{1-x}\) needs \(|x| < 1\).

P4.1 Partial fractions and P4.6 Integration

Partial fractions turn a rational function into pieces you can expand (this page) or integrate (to logarithms). Ideas such as \(\sqrt{1 + x} \approx 1 + \frac{x}{2}\) for small \(x\) sit alongside the small-angle approximations of trigonometry, and are a first look at the power series of later courses.

IAL Physics

Binomial approximations simplify formulas when one quantity is much smaller than another, for example \((1 + h/R)^{-2} \approx 1 - 2h/R\) for gravitational field strength a small height \(h\) above a planet of radius \(R\).

Key formulae

Check your understanding

Specification checklist

• (P4 4.1) I can expand \((1 + x)^n\) for any rational \(n\) as far as a required power, building each coefficient from \(\frac{n(n-1)\cdots(n-r+1)}{r!}\).
• (P4 4.1) I can state that the series is valid for \(|x| < 1\), and give the range of validity for expansions such as \((1 + kx)^n\).
• (P4 4.1) I can expand \((a + bx)^n\) by writing it as \(a^n\left(1 + \frac{bx}{a}\right)^n\), and state that it is valid for \(\left|\frac{bx}{a}\right| < 1\).
• (P4 4.1) I can multiply a binomial series by a polynomial and find unknown constants or a particular coefficient from given information.
• (P4 4.1) I can use a truncated expansion with a suitable small value of \(x\) to approximate a root or a value, and comment on its accuracy.
• (P4 4.1) I can split a rational function into partial fractions, expand each part, combine them and state the range of validity of the result.