Series

Lessons

Senior 6 · Mathematics

Series

4 lessons

Lesson 3 of 4

Power series, Maclaurin and Taylor series

25 min

How does a calculator know $\sin 0.5$ ?

A calculator's circuits can only do arithmetic, yet a calculator returns $\sin 0.5 = 0.479\,425\,5\ldots$ at once. One way to get such values from arithmetic alone is to write functions such as $\sin x$ , $e^{x}$ and $\ln(1 + x)$ as "polynomials with infinitely many terms". For instance, $0.5 - \frac{0.5^3}{6} + \frac{0.5^5}{120} = 0.479\,427$ already agrees with $\sin 0.5$ to five decimal places. This lesson shows where such series come from, and for which $x$ they are valid.

Definition

Power series in $x$ : a series $\sum_{r=0}^{\infty} a_r x^r = a_0 + a_1 x + a_2 x^2 + \cdots$ , where the $a_{r}$ are constants. More generally, a power series about $a$ is $\sum_{r=0}^{\infty} a_r (x - a)^r$ .

Interval of convergence: the set of $x$ for which the series converges. It is always an interval centred at $a$ , of half-width $R$ , the radius of convergence (possibly $R = 0$ or $R = +\infty$ ).

Example: the geometric series $1 + x + x^2 + \cdots = \frac{1}{1 - x}$ for $|x| \lt 1$ , so $R = 1$ .

Worked example 1: an interval of convergence

Find the interval of convergence of $\sum_{r=1}^{\infty}\frac{x^r}{r}$ .

Solution Apply the ratio test to the absolute values: $\left|\frac{x^{r+1}}{r + 1}\cdot\frac{r}{x^r}\right| = \frac{r}{r + 1}|x| \to |x|$ . So the series converges for $|x| \lt 1$ and diverges for $|x| \gt 1$ : $R = 1$ . The end points need separate tests. At $x = 1$ it is the harmonic series, which diverges. At $x = - 1$ it is $-1 + \frac{1}{2} - \frac{1}{3} + \cdots$ , which converges by Leibniz. The interval is $[- 1, 1)$ .

The power series $\sum_{r=1}^{\infty}\frac{x^r}{r}$ converges for $-1 \le x \lt 1$ and diverges elsewhere. The ratio test gives the open interval; each end point needs its own test.

Important

Maclaurin series. If a function can be written as a power series $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$ near 0, the coefficients are forced. Put $x = 0$ : $a_{0} = f (0)$ . Differentiate once and put $x = 0$ : $a_{1} = f^{'} (0)$ . Differentiate twice: $f^{''} (0) = 2 a_{2}$ . In general, differentiating $k$ times kills the terms before $a_{k} x^{k}$ , turns that term into $k!\,a_k$ , and leaves terms that still contain $x$ . So $a_k = \frac{f^{(k)}(0)}{k!}$ and

f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \cdots = \sum_{r=0}^{\infty}\frac{f^{(r)}(0)}{r!}x^r .

The sum of the first terms, up to $x^{n}$ , is the Maclaurin polynomial of degree $n$ .

Worked example 2: the series for $e^{x}$ and $\sin x$

Solution For $f (x) = e^{x}$ , every derivative is $e^{x}$ , so $f^{(r)}(0) = 1$ and

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{r=0}^{\infty}\frac{x^r}{r!} .

For $f(x) = \sin x$ , the derivatives cycle through $\cos x, -\sin x, -\cos x, \sin x$ , so at 0 they are $0, 1, 0, -1, 0, 1, \ldots$ . Only odd powers survive, with alternating signs:

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{r=0}^{\infty}\frac{(-1)^r x^{2r+1}}{(2r + 1)!} .

By the ratio test both converge for every real $x$ : for $e^{x}$ , $\left|\frac{x}{r + 1}\right| \to 0 \lt 1$ .

$\sin x$ (blue) and its Maclaurin polynomials $P_{1} = x$ (orange, dashed), $P_3 = x - \frac{x^3}{6}$ (green) and $P_5 = x - \frac{x^3}{6} + \frac{x^5}{120}$ (pink), plotted from computed points. Each extra term keeps the polynomial close to $\sin x$ over a wider interval; the full series equals $\sin x$ for every real $x$ .

The standard Maclaurin series

Function	Series	Valid for
$e^{x}$	$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$	all real $x$
$\sin x$	$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$	all real $x$
$\cos x$	$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$	all real $x$
$\ln(1 + x)$	$x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$	$-1 \lt x \le 1$
$(1 + x)^\alpha$	$1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!}x^3 + \cdots$	$\|x\| \lt 1$ (all $x$ if $\alpha \in \mathbb{N}$ )

The series for $\ln(1 + x)$ uses Lesson 2 of Unit 2: the $r$ th derivative of $\ln(1 + x)$ at 0 is $(-1)^{r-1}(r - 1)!$ , and dividing by $r!$ leaves $\frac{(-1)^{r-1}}{r}$ . At $x = 1$ it gives $\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \cdots$ , the alternating series of Lesson 2. The last row is the binomial series: when $\alpha$ is a natural number it stops, and it is the binomial expansion of S4 (Combinatorics).

Important

Euler's formula from the series (promised in Unit 1). Put $x = i\theta$ in the series for $e^{x}$ and use $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ :

\begin{aligned} e^{i\theta} &= 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots \\ &= \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \cdots\right) , \end{aligned}

and the two brackets are the series of $\cos\theta$ and $\sin\theta$ . So $e^{i\theta} = \cos\theta + i\sin\theta$ .

Definition

Taylor series of $f$ about $x = a$ : the same idea centred at $a$ ,

\begin{aligned} f(x) &= \sum_{r=0}^{\infty}\frac{f^{(r)}(a)}{r!}(x - a)^r \\ &= f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots \end{aligned}

A Maclaurin series is the Taylor series about $a = 0$ . Choose $a$ where $f$ and its derivatives are easy to evaluate and close to the $x$ you need.

Example: about $a = 1$ , $\ln x = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \cdots$ for $0 \lt x \le 2$ .

Worked example 3: $\sqrt{4.1}$ from a Taylor series

Solution Take $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ about $a = 4$ : $f (4) = 2$ , $f'(x) = \frac{1}{2}x^{-\frac{1}{2}}$ , so $f'(4) = \frac{1}{4}$ ; $f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}$ , so $f''(4) = -\frac{1}{32}$ . Hence

\sqrt{x} \approx 2 + \frac{x - 4}{4} - \frac{(x - 4)^2}{64},

\sqrt{4.1} \approx 2 + 0.025 - 0.000\,156 = 2.024\,844 .

A calculator gives $2.024\,846$ : three terms are already within $0.000\,002$ of it.

New series from old

Inside its interval of convergence, a power series may be substituted into, multiplied, and differentiated term by term. This is usually much quicker than differentiating again and again.

Substitution: $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$ , for all $x$ .
Multiplication: $e^x\sin x = \left(1 + x + \frac{x^2}{2} + \cdots\right)\left(x - \frac{x^3}{6} + \cdots\right) = x + x^2 + \frac{x^3}{3} + \cdots$ .
Differentiation: differentiating the series of $\sin x$ term by term gives $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ , the series of $\cos x$ , as it should.

Brook Taylor (1685–1731), in a small copy of a portrait by Louis Goupy, about 1720, National Portrait Gallery, London. He published the general series in 1715; the open book in his hands is on linear perspective, another subject he wrote on. Painting: public domain, via Wikimedia Commons.

Colin Maclaurin (1698–1746), from an engraving by Edward Harding published in 1798. Maclaurin used the special case about 0 in his Treatise of Fluxions (1742), so it carries his name. Engraving: public domain (Smithsonian Institution Libraries), via Wikimedia Commons.

Warning

A series is only valid inside its interval of convergence. Putting $x = 2$ into $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ does not give $\ln 3$ : the terms $\frac{2^r}{r}$ grow without limit, and the series diverges. Always state the interval with the series.

Tip

Learn the five standard series with their intervals; build everything else from them by substitution, multiplication or differentiation. In an exam, "find the first three non-zero terms" nearly always wants this method rather than repeated differentiation.

3Blue1Brown: Taylor series (Chapter 11, Essence of calculus). Watch how each new term is chosen to match one more derivative at the centre point, and the pictures of polynomials hugging cos x.

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Practice

State the interval of validity with every series.

By repeated differentiation, find the first four non-zero terms of the Maclaurin series of $\cos x$ .
Write the first four terms of the Maclaurin series of $e^{2x}$ .
Find the series of $\ln(1 - x)$ , and hence show that $\ln\frac{1 + x}{1 - x} = 2\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots\right)$ for $|x| \lt 1$ .
Use the binomial series to expand $(1 + x)^{-2}$ as far as $x^{3}$ .
Find the interval of convergence of $\sum_{r=1}^{\infty}\frac{x^r}{r\,3^r}$ .
Find the Taylor series of $e^{x}$ about $x = 1$ , as far as the term in $(x - 1)^{3}$ .
Find the Maclaurin series of $e^x\sin x$ as far as $x^{3}$ , and check the coefficient of $x^{2}$ by differentiation.
Differentiate the series of $e^{x}$ term by term. What do you notice, and why is it expected?

Show answersHide answers

Derivatives at 0: $1, 0, - 1, 0, 1, 0, - 1$ . So $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots$ , all $x$ .
$1 + 2x + 2x^2 + \frac{4}{3}x^3 + \cdots$ , all $x$ .
Replace $x$ by $- x$ : $\ln(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots$ for $-1 \le x \lt 1$ . Subtracting it from $\ln(1 + x)$ , the even powers cancel and the odd ones double.
$\alpha = -2$ : $1 - 2x + \frac{(-2)(-3)}{2}x^2 + \frac{(-2)(-3)(-4)}{6}x^3 = 1 - 2x + 3x^2 - 4x^3 + \cdots$ , for $|x| \lt 1$ .
Ratio $\frac{r}{r + 1}\cdot\frac{|x|}{3} \to \frac{|x|}{3}$ , so $|x| \lt 3$ . At $x = 3$ : harmonic, diverges. At $x = - 3$ : $\sum\frac{(-1)^r}{r}$ , converges. Interval $[- 3, 3)$ .
All derivatives at 1 equal $e$ : $e\left[1 + (x - 1) + \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{6}\right] + \cdots$ , all $x$ .
$x + x^2 + \frac{x^3}{3} + \cdots$ . Check: $f''(x) = 2e^x\cos x$ , so $f^{''} (0) = 2$ and the coefficient is $\frac{2}{2!} = 1$ ✓.
The derivative of $\frac{x^r}{r!}$ is $\frac{x^{r-1}}{(r - 1)!}$ , so the series reproduces itself: it confirms $\frac{d}{dx}e^x = e^x$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the graphs (plotted from computed points), the worked examples, the practice items and the quiz questions.
Images: Brook Taylor, after Louis Goupy, National Portrait Gallery, London, public domain; Colin Maclaurin, engraving by Edward Harding (1798), Smithsonian Institution Libraries, public domain; both via Wikimedia Commons.
Video: Taylor series | Chapter 11, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1How does a calculator know sin⁡0.5 \sin 0.5 sin0.5?

2The standard Maclaurin series

3New series from old

How does a calculator know $\sin 0.5$ ?

The standard Maclaurin series

New series from old