A-Level Further Mathematics – Chapter 8: Series and Maclaurin Expansions Further Pure
Contents
Infinite series lie at the heart of analysis and have practical importance throughout applied mathematics, physics, and engineering. In this chapter we study two distinct but complementary topics. The method of differences gives us a powerful technique for evaluating finite sums in closed form. Maclaurin expansions allow us to represent functions as infinite power series, enabling approximations, limit calculations, and the evaluation of integrals that resist standard techniques.
8.1 Method of Differences
The Method of Differences (Telescoping)
If we can write the general term of a series as $f(r) = g(r) - g(r+1)$ for some function $g$, then the partial sum telescopes:
$$\sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} \bigl[g(r) - g(r+1)\bigr] = g(1) - g(n+1)$$
The most common way to find $g$ is to express $f(r)$ using partial fractions.
Example 8.1.1 — A simple telescoping sum
Show that $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$, and hence find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)}$.
Partial fractions: $\frac{1}{r(r+1)} = \frac{A}{r}+\frac{B}{r+1}$. Setting $r=0$ gives $A=1$; setting $r=-1$ gives $B=-1$. So $\frac{1}{r(r+1)} = \frac{1}{r}-\frac{1}{r+1}$. ✓
Telescoping: $\sum_{r=1}^{n}\Bigl(\frac{1}{r}-\frac{1}{r+1}\Bigr) = \frac{1}{1}-\frac{1}{n+1} = \frac{n}{n+1}$.
Example 8.1.2 — Sum involving a product of three consecutive integers
Use the identity $\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\Bigl[\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}\Bigr]$ to find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}$.
$\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\Bigl[\frac{1}{1\cdot2}-\frac{1}{(n+1)(n+2)}\Bigr] = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$.
Example 8.1.3 — Difference involving a factorial-like product
Show that $r(r+1) - (r-1)r = 2r$, and hence find $\displaystyle\sum_{r=1}^{n} r$ using the method of differences.
$r(r+1)-(r-1)r = r[(r+1)-(r-1)] = 2r$. So $r = \tfrac{1}{2}[r(r+1)-(r-1)r]$.
$\sum_{r=1}^{n}r = \frac{1}{2}[n(n+1)-0] = \frac{n(n+1)}{2}$. ✓
Example 8.1.4 — Finding the sum of $\sum r(r+2)$ by splitting
Show that $r(r+2) = \tfrac{1}{3}\bigl[r(r+1)(r+2)-(r-1)r(r+1)\bigr]$ and use this to find $\displaystyle\sum_{r=1}^{n}r(r+2)$.
$(r-1)r(r+1) = r(r^2-1)$ and $r(r+1)(r+2) = r(r^2+3r+2)$. The difference is $r[(r^2+3r+2)-(r^2-1)] = r(3r+3) = 3r(r+1)$, so $r(r+1) = \tfrac{1}{3}[r(r+1)(r+2)-(r-1)r(r+1)]$... adjusting the approach: note $r(r+2) = r(r+1)+r$ so
$\sum_{r=1}^{n}r(r+2) = \sum r(r+1) + \sum r = \frac{n(n+1)(n+2)}{3} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6}[2(n+2)+3] = \frac{n(n+1)(2n+5)}{6}$.
Example 8.1.5 — Partial fractions with repeated roots
Find $\displaystyle\sum_{r=1}^{n}\frac{2r+1}{r^2(r+1)^2}$.
Notice $\frac{2r+1}{r^2(r+1)^2} = \frac{1}{r^2}-\frac{1}{(r+1)^2}$ (verified by expanding). Hence the sum telescopes to $\frac{1}{1}-\frac{1}{(n+1)^2} = \frac{(n+1)^2-1}{(n+1)^2} = \frac{n(n+2)}{(n+1)^2}$.
Exam Tip
In method-of-differences questions, the examiners typically ask you first to "show that" the partial-fraction decomposition holds — always verify this result before using it in the sum. Write out the first few and last few terms explicitly to make the telescoping clear.
8.2 Standard Power Series
Definition — Maclaurin Series
If $f$ is infinitely differentiable at $x = 0$, its Maclaurin series is:
$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$
The series converges to $f(x)$ within its radius of convergence. Outside this radius the partial sums diverge.
Theorem 8.1 — Standard Maclaurin Series
The following series are valid for the stated ranges:
| Function | Maclaurin Series | Validity |
|---|---|---|
| $e^x$ | $1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = \displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$ | All $x$ |
| $\sin x$ | $x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ | All $x$ |
| $\cos x$ | $1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}$ | All $x$ |
| $\ln(1+x)$ | $x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots = \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$ | $-1 < x \le 1$ |
| $(1+x)^n$ | $1 + nx + \dfrac{n(n-1)}{2!}x^2 + \cdots$ | $|x|<1$ (if $n\notin\mathbb{N}$) |
Example 8.2.1 — Using the series for $e^x$
Write down the first four non-zero terms of the Maclaurin series for $e^{-2x}$ and state its range of validity.
Replace $x$ with $-2x$ in the series for $e^x$: $e^{-2x} = 1 + (-2x) + \frac{(-2x)^2}{2!} + \frac{(-2x)^3}{3!} + \cdots = 1 - 2x + 2x^2 - \frac{4x^3}{3} + \cdots$
Valid for all $x$.
Example 8.2.2 — Series for $\ln(1-x)$
By replacing $x$ with $-x$ in the series for $\ln(1+x)$, write down the Maclaurin series for $\ln(1-x)$, and hence find the series for $\ln\!\left(\frac{1+x}{1-x}\right)$.
$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots$, valid for $-1 \le x < 1$.
$\ln\!\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots = 2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$, valid for $|x|<1$.
8.3 Deriving Maclaurin Series from Scratch
When a function is not directly obtainable by substitution into a standard series, we return to the definition: differentiate repeatedly, evaluate at $x = 0$, and build the series term by term.
Example 8.3.1 — Maclaurin series for $\tan x$ (first three non-zero terms)
Let $f(x) = \tan x$. We need $f(0), f'(0), f''(0), f'''(0), f^{(4)}(0), f^{(5)}(0)$.
$f(0) = 0$. $f'(x) = \sec^2 x$, so $f'(0) = 1$.
$f''(x) = 2\sec^2 x \tan x$, so $f''(0) = 0$.
$f'''(x) = 2\sec^4 x + 4\sec^2 x\tan^2 x = 2\sec^2 x(\sec^2 x + 2\tan^2 x)$, so $f'''(0) = 2$.
$f^{(4)}(0) = 0$ (by parity — $\tan x$ is odd so only odd powers appear). $f^{(5)}(0) = 16$.
$$\tan x = x + \frac{2}{3!}x^3 + \frac{16}{5!}x^5 + \cdots = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$$
Valid for $|x| < \pi/2$.
Example 8.3.2 — Maclaurin series for $\arctan x$
Since $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$, we can use the geometric series:
$$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots \quad (|x|<1)$$
Integrating term by term with $\arctan 0 = 0$:
$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}$$
Valid for $|x| \le 1$. Setting $x = 1$ gives the famous result $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$ (Leibniz formula).
Example 8.3.3 — Maclaurin series for $\ln(\cos x)$
Let $g(x) = \ln(\cos x)$. Then $g'(x) = -\tan x$.
Using the series from Example 8.3.1: $g'(x) = -\tan x = -x - \frac{x^3}{3} - \frac{2x^5}{15} - \cdots$
Integrating with $g(0) = \ln 1 = 0$:
$$\ln(\cos x) = -\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45} - \cdots$$
Example 8.3.4 — Maclaurin series for $(1+x)^{1/2}$
Using the binomial expansion with $n = \tfrac{1}{2}$:
$\sqrt{1+x} = 1 + \tfrac{1}{2}x + \frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}x^2 + \frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}x^3 + \cdots$
$= 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + \cdots$, valid for $|x| \le 1$.
Exam Tip
When deriving series by repeated differentiation, always check whether $f(x)$ is an odd or even function. Odd functions ($f(-x)=-f(x)$) have only odd-power terms; even functions have only even-power terms. This halves the work and provides a useful check.
8.4 Operations on Series
Once we know the Maclaurin series for individual functions, we can build series for combinations by algebraic manipulation: multiplication, composition, addition, and subtraction.
Example 8.4.1 — Multiplying series: $e^x \sin x$
Find the series for $e^x\sin x$ up to and including $x^4$.
$e^x = 1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\tfrac{x^4}{24}+\cdots$
$\sin x = x - \tfrac{x^3}{6}+\cdots$
Multiply term by term, collecting by degree:
$x^1$: $1\cdot x = x$
$x^2$: $x\cdot x = x^2$
$x^3$: $\tfrac{x^2}{2}\cdot x + 1\cdot(-\tfrac{x^3}{6}) = \tfrac{x^3}{2}-\tfrac{x^3}{6} = \tfrac{x^3}{3}$
$x^4$: $\tfrac{x^3}{6}\cdot x + x\cdot(-\tfrac{x^3}{6}) = \tfrac{x^4}{6}-\tfrac{x^4}{6} = 0$
$$e^x\sin x = x + x^2 + \frac{x^3}{3} + 0 \cdot x^4 + \cdots$$
Example 8.4.2 — Composing series: $e^{\sin x}$
Find the Maclaurin series for $e^{\sin x}$ up to $x^3$.
Let $u = \sin x = x - \tfrac{x^3}{6} + \cdots$. Then $e^u = 1 + u + \tfrac{u^2}{2} + \tfrac{u^3}{6} + \cdots$
$u = x - \tfrac{x^3}{6}$, $u^2 = x^2 + O(x^4)$, $u^3 = x^3 + O(x^5)$.
$$e^{\sin x} = 1 + \bigl(x-\tfrac{x^3}{6}\bigr) + \tfrac{x^2}{2} + \tfrac{x^3}{6} + \cdots = 1 + x + \frac{x^2}{2} + \frac{x^3 - x^3}{6} + \cdots = 1 + x + \frac{x^2}{2} + 0\cdot x^3 + \cdots$$
More carefully: $-\tfrac{x^3}{6}+\tfrac{x^3}{6} = 0$, so the $x^3$ coefficient is $0$.
Example 8.4.3 — Finding a limit using series (no L'Hôpital)
Find $\displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}$.
$e^x - 1 - x = \tfrac{x^2}{2} + \tfrac{x^3}{6} + \cdots$
$\frac{e^x-1-x}{x^2} = \frac{1}{2} + \frac{x}{6} + \cdots \xrightarrow{x\to 0} \boxed{\frac{1}{2}}$.
Example 8.4.4 — Limit with both sin and $\ln$
Find $\displaystyle\lim_{x\to 0}\frac{\ln(1+x) - \sin x}{x^2}$.
$\ln(1+x) = x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots$, $\sin x = x - \tfrac{x^3}{6}+\cdots$.
Numerator: $(x-\tfrac{x^2}{2}+\cdots)-(x-\cdots) = -\tfrac{x^2}{2} + O(x^3)$.
$\frac{-x^2/2}{x^2} \to -\frac{1}{2}$.
Example 8.4.5 — Product $\cos x \cdot \ln(1+x)$
Find the first three non-zero terms in the Maclaurin series for $\cos x \cdot \ln(1+x)$.
$\cos x = 1 - \tfrac{x^2}{2}+\cdots$, $\ln(1+x) = x - \tfrac{x^2}{2}+\tfrac{x^3}{3}-\cdots$.
Product up to $x^3$:
$1\cdot(x-\tfrac{x^2}{2}+\tfrac{x^3}{3}) + (-\tfrac{x^2}{2})\cdot x + \cdots = x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \tfrac{x^3}{2} + \cdots = x - \frac{x^2}{2} - \frac{x^3}{6} + \cdots$
8.5 Applications of Series
Maclaurin series have several important practical applications: approximating function values, evaluating integrals that have no elementary closed form, and obtaining refined small-angle approximations.
Example 8.5.1 — Approximating $e^{0.1}$
Use the first five terms of the series for $e^x$ to estimate $e^{0.1}$, and bound the error.
$e^{0.1} \approx 1 + 0.1 + \tfrac{0.01}{2} + \tfrac{0.001}{6} + \tfrac{0.0001}{24} = 1 + 0.1 + 0.005 + 0.0001\overline{6} + 0.000004\overline{16} \approx 1.10517$.
The true value is $1.10517091...$, so the error is less than $2\times10^{-7}$. The error is bounded by the next term $\tfrac{(0.1)^5}{5!} \approx 8.3\times10^{-8}$.
Example 8.5.2 — Estimating $\int_0^1 e^{-x^2}dx$
The function $e^{-x^2}$ has no elementary antiderivative (it is related to the error function). Use the series to estimate $\int_0^{0.5}e^{-x^2}dx$ to 5 decimal places.
$e^{-x^2} = 1 - x^2 + \tfrac{x^4}{2} - \tfrac{x^6}{6} + \cdots$
$\int_0^{0.5}e^{-x^2}dx \approx \Bigl[x - \tfrac{x^3}{3}+\tfrac{x^5}{10}-\tfrac{x^7}{42}\Bigr]_0^{0.5}$
$= 0.5 - \tfrac{0.125}{3}+\tfrac{0.03125}{10}-\tfrac{0.0078125}{42} \approx 0.5 - 0.04167+0.003125-0.000186 \approx 0.46127$.
Example 8.5.3 — Higher-order small-angle approximations
For small $\theta$ (in radians), the standard approximation $\sin\theta\approx\theta$ has error $O(\theta^3)$. A better approximation is $\sin\theta \approx \theta - \tfrac{\theta^3}{6}$. Show that this gives a relative error less than $0.01\%$ for $|\theta| < 0.1$ rad ($\approx 5.7°$).
The next term is $\tfrac{\theta^5}{120}$. At $\theta=0.1$: $\tfrac{(0.1)^5}{120} = \tfrac{10^{-5}}{120} \approx 8.3\times10^{-8}$. Compare with $\sin(0.1)\approx0.09983$: relative error $\approx 8.3\times10^{-7}$, well below $0.01\%$.
Example 8.5.4 — Evaluating $\int_0^1\frac{\sin x}{x}dx$
Since $\frac{\sin x}{x} = 1 - \tfrac{x^2}{3!}+\tfrac{x^4}{5!}-\cdots$ (the sinc function), integrate term by term:
$\int_0^1\frac{\sin x}{x}dx = \Bigl[x-\tfrac{x^3}{18}+\tfrac{x^5}{600}-\tfrac{x^7}{35280}\Bigr]_0^1 = 1-\tfrac{1}{18}+\tfrac{1}{600}-\tfrac{1}{35280}+\cdots \approx 0.94608$.
Exam Tip
When you are asked to "use the series to find an approximation", show enough terms that the final answer is clearly correct to the required accuracy. State explicitly which term you are using as the "error bound" term, and verify that it is small enough. The alternating series estimation theorem guarantees that the error is less than the absolute value of the first neglected term, provided the terms are decreasing in modulus.
Practice Problems
Problem 1
Show that $\dfrac{4}{(2r-1)(2r+3)} = \dfrac{1}{2r-1}-\dfrac{1}{2r+3}$, and hence find $\displaystyle\sum_{r=1}^{n}\frac{4}{(2r-1)(2r+3)}$.
Show solution
Telescoping (terms are $\tfrac{1}{1}-\tfrac{1}{5}+\tfrac{1}{3}-\tfrac{1}{7}+\cdots$). The groups of non-cancelling terms at start and end give:
$\sum = \frac{1}{1}+\frac{1}{3}-\frac{1}{2n+1}-\frac{1}{2n+3}$.
Problem 2
Find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)}$ by first expressing the general term in partial fractions.
Show solution
The non-cancelling terms are $\frac{1}{2}\Bigl(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\Bigr) = \frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}$.
Problem 3
Write down the Maclaurin series for $\cos(2x)$ up to and including the $x^6$ term, and state the range of validity.
Show solution
$\cos(2x) = 1 - \frac{(2x)^2}{2}+\frac{(2x)^4}{24}-\frac{(2x)^6}{720}+\cdots = 1-2x^2+\frac{2x^4}{3}-\frac{4x^6}{45}+\cdots$
Valid for all $x$.
Problem 4
Find the first four non-zero terms in the Maclaurin series for $\arcsin x$ by differentiating appropriately. State the range of validity.
Show solution
$(1-x^2)^{-1/2} = 1+\frac{x^2}{2}+\frac{3x^4}{8}+\frac{5x^6}{16}+\cdots$
Integrate with $\arcsin 0 = 0$:
$\arcsin x = x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\cdots$, valid for $|x|\le1$.
Problem 5
Find the Maclaurin series for $e^x\cos x$ up to and including $x^4$.
Show solution
Coefficient of $x^0$: $1$. Of $x^1$: $1$. Of $x^2$: $\frac{1}{2}-\frac{1}{2}=0$. Of $x^3$: $\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$. Of $x^4$: $\frac{1}{24}-\frac{1}{4}+\frac{1}{24}=-\frac{1}{6}$.
$e^x\cos x = 1+x-\frac{x^3}{3}-\frac{x^4}{6}+\cdots$
Problem 6
Use the Maclaurin series to find $\displaystyle\lim_{x\to 0}\frac{\cos x - 1 + \frac{1}{2}x^2}{x^4}$.
Show solution
Numerator: $\cos x - 1 + \frac{x^2}{2} = \frac{x^4}{24}-\cdots$
$\frac{x^4/24}{x^4} \to \frac{1}{24}$.
Problem 7
Evaluate $\displaystyle\int_0^{0.2}\ln(1+x^2)\,dx$ using the first three terms of the appropriate Maclaurin series. Give your answer to 4 significant figures.
Show solution
$\int_0^{0.2}\approx\Bigl[\frac{x^3}{3}-\frac{x^5}{10}+\frac{x^7}{21}\Bigr]_0^{0.2}$
$=\frac{0.008}{3}-\frac{0.00032}{10}+\frac{0.0000128}{21}\approx0.002667-0.000032+0.0000006\approx0.002636$.
Problem 8
The Maclaurin series for $f(x) = (1+x)\ln(1+x)$ begins $x+\frac{x^2}{2}-\frac{x^2}{2}+\cdots$. Find the first four non-zero terms by multiplying the series for $\ln(1+x)$ by $(1+x)$.
Show solution
$(1+x)\ln(1+x) = \ln(1+x)+x\ln(1+x)$
$= (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots) + (x^2-\frac{x^3}{2}+\frac{x^4}{3}-\cdots)$
$= x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{12} - \cdots$
Problem 9
Find the sum to infinity of $\displaystyle\sum_{r=2}^{\infty}\frac{1}{r^2-1}$.
Show solution
Telescoping partial sum (non-cancelling terms at start): $\frac{1}{2}\Bigl(1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\Bigr)$.
As $n\to\infty$: $\frac{1}{2}\cdot\frac{3}{2}=\frac{3}{4}$.
Problem 10
By writing $\sin^2 x = \frac{1}{2}(1-\cos 2x)$, find the Maclaurin series for $\sin^2 x$ up to and including $x^6$, and verify the result by squaring the series for $\sin x$ directly.
Show solution
Method 2: $\sin x = x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$ Squaring: $x^2-\frac{x^4}{3}+(terms)$… the $x^6$ coefficient from $2(x)(\frac{x^5}{120})+(-\frac{x^3}{6})^2 = \frac{x^6}{60}+\frac{x^6}{36}=\frac{3x^6+5x^6}{180}=\frac{8x^6}{180}=\frac{2x^6}{45}$ ✓.