4.3 Fractional and Negative Indices
AHL 1.10
- how to expand \((1 + x)^n\) when \(n\) is a negative integer or a fraction
- why such an expansion has infinitely many terms, and when it is valid
- how to expand \((a + bx)^n\) by first taking out a factor \(a^n\)
- how to combine expansions with products and partial fractions
- how to use the series for approximations
From a geometric series to the binomial series
In 2.4 you met the sum to infinity \(1 + r + r^2 + r^3 + \cdots = \dfrac{1}{1 - r}\) for \(|r| < 1\). Put \(r = -x\):
\[ (1 + x)^{-1} = \frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots, \qquad |x| < 1 . \]
Now try the binomial pattern from 4.1, written without factorials, with \(n = -1\):
\[ 1 + \frac{(-1)}{1}x + \frac{(-1)(-2)}{2!}x^2 + \frac{(-1)(-2)(-3)}{3!}x^3 + \cdots = 1 - x + x^2 - x^3 + \cdots \]
It gives the same series. In fact the pattern works for every rational power \(n\).
Extension of binomial theorem, \(n \in \mathbb{Q}\) Partly in the HL formula booklet
\[ (a + b)^n = a^n\left(1 + n\left(\frac{b}{a}\right) + \frac{n(n - 1)}{2!}\left(\frac{b}{a}\right)^2 + \cdots\right), \qquad \left|\frac{b}{a}\right| < 1 \]
In the HL booklet
With \(a = 1\) and \(b = x\) (the binomial series):
\[ (1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \cdots, \qquad |x| < 1 \]
- \(a\), \(b\): the two terms in the bracket (numbers or expressions), with \(a \neq 0\)
- \(n\): the power, any rational number (for example \(-2\) or \(\frac{1}{2}\))
- \(x\): the variable; the series is valid only for \(|x| < 1\)
The booklet does not state the condition \(\left|\dfrac{b}{a}\right| < 1\).
Two important differences from the positive-integer case:
- The series never stops. When \(n\) is a positive integer, the factor \((n - n)\) eventually appears and makes every later coefficient \(0\). For other \(n\) this never happens, so there are infinitely many terms.
- It is only valid when \(|x| < 1\). Outside this interval the series does not converge. You must always state the interval of validity.
Do not use \(\binom{n}{r}\) here: the factorial definition only works when \(n\) is a positive integer. Use the “falling” products \(n(n - 1)(n - 2)\cdots\) instead.
Example 1 Expand each of the following in ascending powers of \(x\) up to and including the term in \(x^3\). State the values of \(x\) for which each expansion is valid.
(a) \((1 + x)^{-3}\) \(\qquad\) (b) \((1 - 4x)^{\frac{1}{2}}\)
(a) \(n = -3\):
\[ 1 + (-3)x + \frac{(-3)(-4)}{2}x^2 + \frac{(-3)(-4)(-5)}{6}x^3 = 1 - 3x + 6x^2 - 10x^3 + \cdots, \quad |x| < 1 . \]
(b) \(n = \frac{1}{2}\), and “\(x\)” is \(-4x\):
\[ \begin{aligned} &1 + \tfrac{1}{2}(-4x) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}(-4x)^2 + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-4x)^3 \\ &= 1 - 2x - 2x^2 - 4x^3 + \cdots \end{aligned} \]
Valid when \(|-4x| < 1\), that is \(|x| < \dfrac{1}{4}\).
Expanding \((a + bx)^n\)
The series needs a 1 in front. So first take out \(a^n\):
\[ (a + bx)^n = a^n\left(1 + \frac{b}{a}x\right)^n, \qquad \text{valid for } \left|\frac{b}{a}x\right| < 1 . \]
Example 2 Expand, up to the term in \(x^3\), and state the interval of validity:
(a) \(\sqrt{4 + x}\) \(\qquad\) (b) \(\dfrac{1}{(2 + x)^2}\)
(a) \(\sqrt{4 + x} = (4 + x)^{\frac{1}{2}} = 4^{\frac{1}{2}}\left(1 + \frac{x}{4}\right)^{\frac{1}{2}} = 2\left(1 + \frac{x}{4}\right)^{\frac{1}{2}}\).
\[ 2\left(1 + \frac{1}{2}\cdot\frac{x}{4} - \frac{1}{8}\cdot\frac{x^2}{16} + \frac{1}{16}\cdot\frac{x^3}{64}\right) = 2 + \frac{x}{4} - \frac{x^2}{64} + \frac{x^3}{512} + \cdots \]
(the coefficients of \((1 + y)^{\frac{1}{2}}\) are \(1, \frac{1}{2}, -\frac{1}{8}, \frac{1}{16}\)). Valid for \(\left|\frac{x}{4}\right| < 1\), that is \(|x| < 4\).
(b) \((2 + x)^{-2} = 2^{-2}\left(1 + \frac{x}{2}\right)^{-2} = \frac{1}{4}\left(1 + \frac{x}{2}\right)^{-2}\). The coefficients of \((1 + y)^{-2}\) are \(1, -2, 3, -4\):
\[ \frac{1}{4}\left(1 - x + \frac{3x^2}{4} - \frac{x^3}{2}\right) = \frac{1}{4} - \frac{x}{4} + \frac{3x^2}{16} - \frac{x^3}{8} + \cdots, \quad |x| < 2 . \]
Combining series
Example 3 Let \(f(x) = \dfrac{3 - 3x}{(1 + x)(1 - 2x)}\).
(a) Express \(f(x)\) in partial fractions.
(b) Hence expand \(f(x)\) in ascending powers of \(x\) up to the term in \(x^3\), and state the interval of validity.
(a) \(3 - 3x \equiv A(1 - 2x) + B(1 + x)\). \(x = -1\): \(6 = 3A\), \(A = 2\). \(x = \frac{1}{2}\): \(\frac{3}{2} = \frac{3}{2}B\), \(B = 1\).
\[ f(x) = \frac{2}{1 + x} + \frac{1}{1 - 2x} \]
(b)
\[ \begin{aligned} 2(1 + x)^{-1} &= 2 - 2x + 2x^2 - 2x^3 + \cdots \qquad (|x| < 1) \\ (1 - 2x)^{-1} &= 1 + 2x + 4x^2 + 8x^3 + \cdots \qquad \left(|x| < \tfrac{1}{2}\right) \end{aligned} \]
Adding: \(f(x) = 3 + 6x^2 + 6x^3 + \cdots\)
Both series must be valid, so we need the smaller interval: \(|x| < \dfrac{1}{2}\).
Example 4 Find, in simplest form, the coefficient of \(x^r\) in the expansion of \(\dfrac{1}{(1 - 3x)^2}\), where \(|x| < \dfrac{1}{3}\).
The coefficients of \((1 - y)^{-2}\): the term in \(y^r\) is
\[ \frac{(-2)(-3)\cdots(-2 - r + 1)}{r!}(-y)^r = \frac{(-1)^r(r + 1)!}{r!}(-1)^r y^r = (r + 1)y^r . \]
So \((1 - y)^{-2} = 1 + 2y + 3y^2 + 4y^3 + \cdots\) With \(y = 3x\), the coefficient of \(x^r\) is \((r + 1)3^{r}\).
Example 5 (a) Expand \(\sqrt{\dfrac{1 + x}{1 - x}}\) in ascending powers of \(x\) up to the term in \(x^2\).
(b) State the interval of values of \(x\) for which the expansion is valid.
(a) Write it as a product: \((1 + x)^{\frac{1}{2}}(1 - x)^{-\frac{1}{2}}\).
\[ (1 + x)^{\frac{1}{2}} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots, \qquad (1 - x)^{-\frac{1}{2}} = 1 + \frac{x}{2} + \frac{3x^2}{8} + \cdots \]
Multiply, keeping terms up to \(x^2\):
\[ 1 + \frac{x}{2} + \frac{3x^2}{8} + \frac{x}{2} + \frac{x^2}{4} - \frac{x^2}{8} = 1 + x + \frac{x^2}{2} + \cdots \]
(b) Both series need \(|x| < 1\), so the expansion is valid for \(-1 < x < 1\).
Approximations
Choose \(x\) so that the expression matches the number you want, and \(x\) is inside the interval of validity. The smaller \(x\) is, the better the approximation.
Example 6 Use the first three terms of the expansion of \((1 + x)^{\frac{1}{2}}\) to find an approximation for \(\sqrt{10}\), giving your answer as a fraction.
\(\sqrt{10} = \sqrt{9 \times \frac{10}{9}} = 3\left(1 + \frac{1}{9}\right)^{\frac{1}{2}}\). Use \(x = \frac{1}{9}\), which is small and satisfies \(|x| < 1\).
\[ 3\left(1 + \frac{1}{2}\cdot\frac{1}{9} - \frac{1}{8}\cdot\frac{1}{81}\right) = 3\left(1 + \frac{1}{18} - \frac{1}{648}\right) = 3 \times \frac{683}{648} = \frac{683}{216} . \]
\(\frac{683}{216} = 3.16204\ldots\), while \(\sqrt{10} = 3.16228\ldots\) — correct to 3 decimal places.
(Writing \(\sqrt{10} = (1 + 9)^{\frac{1}{2}}\) with \(x = 9\) would be wrong: \(x = 9\) is outside \(|x| < 1\).)
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \((4 + x)^{\frac{1}{2}} = 4\left(1 + \frac{x}{4}\right)^{\frac{1}{2}}\) | \(4^{\frac{1}{2}} = 2\) in front | The factor taken out is \(a^n\), not \(a\). |
| Forgetting the interval of validity | always state it | The series is false outside it. |
| \(|4x| < 1 \Rightarrow |x| < 4\) | \(|x| < \frac{1}{4}\) | Divide by 4. |
| Using \(\binom{-3}{2}\) with factorials | \(\frac{(-3)(-4)}{2!}\) | Factorials of negative numbers do not exist. |
| Using the wider interval for a sum of two series | the narrower one | Both series must converge. |
Exercises
1 Expand \((1 + x)^{-2}\) up to the term in \(x^3\), and state the interval of validity.
\(1 - 2x + 3x^2 - 4x^3 + \cdots\), \(\;|x| < 1\).
2 Expand \(\sqrt{1 - x}\) up to the term in \(x^3\).
\(1 + \frac{1}{2}(-x) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}x^2 + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}(-x)^3 = 1 - \dfrac{x}{2} - \dfrac{x^2}{8} - \dfrac{x^3}{16} + \cdots\), \(\;|x| < 1\).
3 Expand \(\dfrac{1}{\sqrt{1 + 2x}}\) up to the term in \(x^3\), and state the interval of validity.
\((1 + 2x)^{-\frac{1}{2}} = 1 - x + \dfrac{3}{2}x^2 - \dfrac{5}{2}x^3 + \cdots\), \(\;|x| < \dfrac{1}{2}\).
4 Expand \((8 + 3x)^{\frac{2}{3}}\) up to the term in \(x^3\), and state the interval of validity.
\(8^{\frac{2}{3}}\left(1 + \frac{3x}{8}\right)^{\frac{2}{3}} = 4\left(1 + \frac{x}{4} - \frac{x^2}{64} + \frac{x^3}{384}\right) = 4 + x - \dfrac{x^2}{16} + \dfrac{x^3}{96} + \cdots\)
Valid for \(\left|\frac{3x}{8}\right| < 1\), that is \(|x| < \dfrac{8}{3}\).
5 Expand \(\dfrac{1}{3 - x}\) up to the term in \(x^2\), and state the interval of validity.
\(\dfrac{1}{3}\left(1 - \frac{x}{3}\right)^{-1} = \dfrac{1}{3}\left(1 + \dfrac{x}{3} + \dfrac{x^2}{9}\right) = \dfrac{1}{3} + \dfrac{x}{9} + \dfrac{x^2}{27} + \cdots\), \(\;|x| < 3\).
6 Find the coefficient of \(x^4\) in the expansion of \((1 - 3x)^{-2}\).
From Example 4, the coefficient of \(x^r\) is \((r + 1)3^r\). For \(r = 4\): \(5 \times 81 = 405\).
7 Expand \(\dfrac{\sqrt{1 + x}}{1 - x}\) up to the term in \(x^2\).
\(\left(1 + \frac{x}{2} - \frac{x^2}{8}\right)\left(1 + x + x^2\right) = 1 + x + x^2 + \frac{x}{2} + \frac{x^2}{2} - \frac{x^2}{8} = 1 + \dfrac{3x}{2} + \dfrac{11x^2}{8} + \cdots\), \(\;|x| < 1\).
8 (a) Express \(\dfrac{x + 5}{(1 - x)(2 + x)}\) in partial fractions.
(b) Hence expand it up to the term in \(x^2\), and state the interval of validity.
(a) \(x + 5 \equiv A(2 + x) + B(1 - x)\): \(x = 1\): \(A = 2\); \(x = -2\): \(B = 1\). \(\;\dfrac{2}{1 - x} + \dfrac{1}{2 + x}\).
(b) \(2(1 + x + x^2) + \dfrac{1}{2}\left(1 - \dfrac{x}{2} + \dfrac{x^2}{4}\right) = \dfrac{5}{2} + \dfrac{7x}{4} + \dfrac{17x^2}{8} + \cdots\)
Valid for \(|x| < 1\) (the first needs \(|x| < 1\), the second \(|x| < 2\)).
9 Use the first three terms of \((1 + x)^{\frac{1}{2}}\) to estimate \(\sqrt{1.04}\).
\(x = 0.04\): \(\;1 + 0.02 - \dfrac{0.0016}{8} = 1.0198\). (True value \(1.019\,80\ldots\))
10 Explain why the expansion \(\dfrac{1}{1 + x} = 1 - x + x^2 - \cdots\) cannot be used to evaluate \(\dfrac{1}{3}\) by putting \(x = 2\).
\(x = 2\) is outside the interval of validity \(|x| < 1\). The series \(1 - 2 + 4 - 8 + \cdots\) does not converge, so it cannot equal \(\frac{1}{3}\).