Math Core

If Taylor polynomials get better as the degree goes up, what happens if you never stop? You get a Taylor series: a power series that, for the most important functions in calculus, equals the function exactly on its interval of convergence. A handful of these series are so useful that the AP exam expects you to know them by heart.

From polynomials to series

Definition

Taylor series

If $f$ has derivatives of all orders at $x = a$ , the Taylor series for $f$ about $x = a$ is

\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots.

When $a = 0$ , it is called the Maclaurin series for $f$ .

The $n$ th partial sum of the Taylor series is just the Taylor polynomial $P_n(x)$ . For the functions in this lesson, the Lagrange remainder $R_n(x)$ goes to $0$ as $n \to \infty$ for every $x$ in the interval of convergence, so the series really does equal the function there.

cos x with Maclaurin polynomials of degree 2, 4 and 8. Adding terms extends the good fit farther from 0; the full series equals cos x for every x.Open in grapher →

Does the series equal the function?

Having a Taylor series isn't quite the same as having the function. The series is built only from derivatives at a single point, and there are two separate questions to ask:

Where does the series converge? Answer this with the ratio test, as in the previous lesson.
Where it converges, does it converge to $f(x)$ ? Answer this by showing the Lagrange remainder goes to $0$ .

For $e^x$ , for instance, the remainder after $n$ terms is at most $\dfrac{M|x|^{n+1}}{(n+1)!}$ , where $M$ is the largest value of $e^t$ between $0$ and $x$ . Factorials outgrow powers, so this goes to $0$ for every $x$ , and the series equals $e^x$ everywhere. The same argument, with $M = 1$ , works for $\sin x$ and $\cos x$ . For $\frac{1}{1-x}$ you already know the answer from geometric series. On the AP exam, you can take for granted that the standard series below equal their functions on their intervals of convergence.

The series to memorize

Essential Maclaurin series

\begin{aligned} e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots && \text{all } x \\ \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots && \text{all } x \\ \cos x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots && \text{all } x \\ \frac{1}{1 - x} &= \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots && -1 < x < 1 \end{aligned}

Some patterns make these easier to remember. The series for $\sin x$ has only odd powers, matching the fact that $\sin$ is an odd function; $\cos x$ has only even powers because $\cos$ is even. Both alternate in sign. The series for $e^x$ has every power, all positive. And $\frac{1}{1-x}$ is the geometric series.

Two more series, which you'll derive in the next lesson, round out the list:

\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \;\; (-1 < x \le 1), \qquad \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \;\; (-1 \le x \le 1).

Building series by substitution

You rarely need to compute derivatives from scratch. Instead, substitute into a known series. To get the Maclaurin series for $e^{3x}$ , replace $x$ by $3x$ in the series for $e^x$ :

e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \frac{(3x)^3}{3!} + \cdots = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \cdots.

You can also multiply a series by a power of $x$ : $\; x\cos x = x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!} - \cdots$ .

Common mistake

When you substitute, put the whole replacement in parentheses and raise it to the power. $(3x)^2 = 9x^2$ , not $3x^2$ . And $(-x^2)^n = (-1)^n x^{2n}$ : the sign and the exponent both change.

Recognizing a series

Running the process backward is a common exam question: identify the function whose series you are looking at, then evaluate it. For example,

\sum_{n=0}^{\infty} \frac{(-1)^n \pi^{2n+1}}{(2n+1)!\,2^{2n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^n (\pi/2)^{2n+1}}{(2n+1)!} = \sin\frac{\pi}{2} = 1.

Worked examples

Worked example: Substituting a power

Find the first four nonzero terms of the Maclaurin series for $e^{-x^2}$ .

Solution. Replace $x$ with $-x^2$ in the series for $e^x$ :

e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots.

Worked example: A Taylor series not centered at 0

Find the Taylor series for $f(x) = e^x$ about $x = 2$ .

Solution. Every derivative is $e^x$ , so $f^{(n)}(2) = e^2$ for all $n$ :

e^x = \sum_{n=0}^{\infty} \frac{e^2}{n!}(x - 2)^n = e^2 + e^2(x - 2) + \frac{e^2}{2}(x-2)^2 + \cdots.

Worked example: Summing a series by recognition

Find the exact sum of $\displaystyle\sum_{n=0}^{\infty} \frac{3^n}{n!}$ and of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}$ .

Solution. The first matches $\sum \frac{x^n}{n!}$ with $x = 3$ , so it equals $e^3$ . The second matches the cosine series with $x = 1$ (since $1^{2n} = 1$ ), so it equals $\cos 1$ .

Worked example: A coefficient deep in the series

Find the coefficient of $x^{10}$ in the Maclaurin series for $\cos(x^2)$ .

Solution. Substitute $x^2$ into the cosine series: $\cos(x^2) = \sum \dfrac{(-1)^n x^{4n}}{(2n)!}$ . The power $x^{10}$ would need $4n = 10$ , which has no whole-number solution. So the coefficient is $0$ . (Only powers $x^0, x^4, x^8, x^{12}, \dots$ appear.)

Tip

Since the Taylor coefficient of $(x-a)^n$ is $\dfrac{f^{(n)}(a)}{n!}$ , a series gives you every derivative at the center for free: $f^{(n)}(a) = n! \cdot c_n$ . For example, from $e^{-x^2}$ above, the $x^6$ coefficient is $-\frac16$ , so the sixth derivative at $0$ is $6!\cdot\left(-\frac16\right) = -120$ .