Taylor and Maclaurin Series

A Taylor polynomial stops at some degree $n$. If you never stop, you get a Taylor series: a power series that, for many functions, equals the function exactly on its interval of convergence. Four of these series come up so often that the AP exam expects you to know them by heart. Once you know them, you can build series for many other functions without taking a single derivative.

Key ideas

Taylor series and Maclaurin series

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’s called a Maclaurin series. Its partial sums are exactly the Taylor polynomials $P_0, P_1, P_2, \dots$

For the functions in this course, the Taylor series equals $f(x)$ at every $x$ in its interval of convergence. (For $e^x$, $\sin x$, and $\cos x$ you can prove this with the Lagrange error bound, which shrinks to $0$ as $n$ grows.)

The four series to memorize

Function	Series	General term	Converges for
$e^x$	$1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$	$\dfrac{x^n}{n!}$	all real $x$
$\sin x$	$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$	$\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$	all real $x$
$\cos x$	$1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$	$\dfrac{(-1)^n x^{2n}}{(2n)!}$	all real $x$
$\dfrac{1}{1 - x}$	$1 + x + x^2 + x^3 + \cdots$	$x^n$	$-1 \lt x \lt 1$

In each, the sum starts at $n = 0$. As always in calculus, $x$ is in radians for $\sin x$ and $\cos x$.

Patterns that help you remember

• $e^x$: every derivative is $e^x$, which is $1$ at $0$, so every coefficient is $\dfrac{1}{n!}$.
• $\sin x$: an odd function, so only odd powers, with alternating signs. Each power is divided by its own factorial.
• $\cos x$: an even function, so only even powers, with alternating signs.
• $\dfrac{1}{1 - x}$: a geometric series with first term $1$ and ratio $x$ (see infinite geometric series), so it only converges when $|x| \lt 1$.

A quick check: differentiate the $\sin x$ series term by term and you get the $\cos x$ series, just as $\dfrac{d}{dx}\sin x = \cos x$.

More terms, wider fit

Inside the interval of convergence, adding more terms makes the partial sums match the function over a wider and wider stretch. For $\sin x$, $\cos x$ and $e^x$ this keeps going for every $x$.

The partial sums of the cosine series of degree 4 and 10. The series converges to $\cos x$ for every $x$, so higher degrees fit over wider intervals.

Building a series from derivatives

To find a Taylor series directly:

1. Find several derivatives and evaluate them at $x = a$.
2. Look for a pattern in $f^{(n)}(a)$.
3. Write the first few terms, then the general term $\dfrac{f^{(n)}(a)}{n!}(x - a)^n$, simplified.

Often, though, it’s faster to start from one of the four known series; see representing functions as power series.

Worked examples

Example 1: Where the e to the x series comes from

Use derivatives to find the Maclaurin series for $e^x$, and write the general term. Then use it to write $\displaystyle\sum_{n=0}^{\infty} \frac{1}{n!}$ in closed form.

Solution. If $f(x) = e^x$, then $f^{(n)}(x) = e^x$ and $f^{(n)}(0) = 1$ for every $n$. So the coefficient of $x^n$ is $\dfrac{1}{n!}$:

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

At $x = 1$, this says $\displaystyle\sum_{n=0}^{\infty} \frac{1}{n!} = e^1 = e$.

Example 2: Where the sine series comes from

Use derivatives to find the Maclaurin series for $\sin x$.

Solution. The derivatives cycle with period $4$:

$n$	$0$	$1$	$2$	$3$	$4$	$5$
$f^{(n)}(x)$	$\sin x$	$\cos x$	$-\sin x$	$-\cos x$	$\sin x$	$\cos x$
$f^{(n)}(0)$	$0$	$1$	$0$	$-1$	$0$	$1$

The even-numbered terms vanish, and the odd ones alternate $1, -1, 1, \dots$:

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

Check the general term: $n = 0$ gives $x$, $n = 1$ gives $-\dfrac{x^3}{3!}$, $n = 2$ gives $\dfrac{x^5}{5!}$. ✓

Example 3: A series about x = 1

Find the Taylor series for $f(x) = \ln x$ about $x = 1$. Give the first four nonzero terms and the general term.

Solution. Find derivatives and look for the pattern:

$$\begin{aligned} f'(x) &= x^{-1} && f'(1) = 1 \\ f''(x) &= -x^{-2} && f''(1) = -1 \\ f'''(x) &= 2x^{-3} && f'''(1) = 2 \\ f^{(4)}(x) &= -6x^{-4} && f^{(4)}(1) = -6 \end{aligned}$$

The pattern is $f^{(n)}(1) = (-1)^{n+1}(n - 1)!$ for $n \ge 1$, and $f(1) = \ln 1 = 0$. The coefficient of $(x - 1)^n$ is

$$\frac{(-1)^{n+1}(n - 1)!}{n!} = \frac{(-1)^{n+1}}{n}$$

So

$$\ln x = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x - 1)^n}{n}$$

Example 4: From a formula for the derivatives

A function $f$ has derivatives of all orders, with $f^{(n)}(0) = \dfrac{n + 1}{2^n}$ for $n = 0, 1, 2, \dots$

• (a) Write the first four terms and the general term of the Maclaurin series for $f$.
• (b) Find the radius of convergence.

Solution. (a) The values are $f(0) = 1$, $f'(0) = 1$, $f''(0) = \dfrac{3}{4}$, $f'''(0) = \dfrac{1}{2}$. Dividing by $n!$:

$$f(x) = 1 + x + \frac{3/4}{2!}x^2 + \frac{1/2}{3!}x^3 + \cdots = 1 + x + \frac{3}{8}x^2 + \frac{1}{12}x^3 + \cdots$$

The general term is $\dfrac{(n + 1)\,x^n}{2^n\, n!}$.

(b) Ratio test:

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n + 2)|x|^{n+1}}{2^{n+1}(n+1)!} \cdot \frac{2^n\, n!}{(n + 1)|x|^n} = \frac{n + 2}{n + 1} \cdot \frac{|x|}{2(n + 1)}$$

This goes to $0$ for every $x$, so the series converges for all real $x$: $R = \infty$.

Common mistakes

Mixing up the sine and cosine series. Sine has odd powers and starts with $x$ (because $\sin 0 = 0$); cosine has even powers and starts with $1$ (because $\cos 0 = 1$). Check by plugging in $x = 0$.

Dividing by the wrong factorial. In the sine series, $x^5$ is divided by $5!$, not $3!$ or $(2n)!$. The factorial always matches the power: $x^k$ goes with $k!$.

Using the 1/(1 − x) series outside its interval. The series $1 + x + x^2 + \cdots$ only equals $\dfrac{1}{1 - x}$ for $-1 \lt x \lt 1$. At $x = 2$ the function equals $-1$, but the series diverges.

Writing a general term that doesn’t match the first terms. Always test your general term with $n = 0, 1, 2$ to see that it reproduces the terms you wrote out, as in Example 2.

Forgetting radians. The series for $\sin x$ and $\cos x$ are only true in radians. $\sin 30$ means $30$ radians, not $30^\circ$.

Practice

1. (Warm-up) Write the first four nonzero terms and the general term of the Maclaurin series for $\cos x$.

Solution

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$

2. (Warm-up) Find the sum of each series by recognizing a known Maclaurin series.

• (a) $1 + 2 + \dfrac{2^2}{2!} + \dfrac{2^3}{3!} + \cdots$
• (b) $1 - \dfrac{\pi^2}{2!} + \dfrac{\pi^4}{4!} - \dfrac{\pi^6}{6!} + \cdots$

Solution

(a) This is the $e^x$ series with $x = 2$, so the sum is $e^2$.

(b) This is the $\cos x$ series with $x = \pi$, so the sum is $\cos \pi = -1$.

3. (Warm-up) What is the sum of $1 + 0.3 + 0.3^2 + 0.3^3 + \cdots$? What happens to $1 + 2 + 2^2 + 2^3 + \cdots$, even though $\dfrac{1}{1 - 2} = -1$?

Solution

The first is the $\dfrac{1}{1 - x}$ series with $x = 0.3$, which is inside $-1 \lt x \lt 1$, so the sum is $\dfrac{1}{0.7} = \dfrac{10}{7}$.

The second has $x = 2$, outside the interval of convergence. The partial sums grow without bound, so the series diverges. The formula $\dfrac{1}{1 - x}$ doesn’t apply there.

4. (Core) Find the Taylor series for $e^x$ about $x = 3$. Give the first three terms and the general term.

Solution

Every derivative of $e^x$ is $e^x$, so $f^{(n)}(3) = e^3$ for every $n$:

$$e^x = e^3 + e^3(x - 3) + \frac{e^3}{2!}(x - 3)^2 + \cdots = \sum_{n=0}^{\infty} \frac{e^3 (x - 3)^n}{n!}$$

5. (Core) Find the Taylor series for $\cos x$ about $x = \dfrac{\pi}{2}$. Give the first three nonzero terms and the general term.

Solution

At $x = \dfrac{\pi}{2}$, the derivatives $\cos x, -\sin x, -\cos x, \sin x, \cos x, -\sin x, \dots$ have values $0, -1, 0, 1, 0, -1, \dots$ So only odd powers appear:

$$\cos x = -\left(x - \frac{\pi}{2}\right) + \frac{1}{3!}\left(x - \frac{\pi}{2}\right)^3 - \frac{1}{5!}\left(x - \frac{\pi}{2}\right)^5 + \cdots$$$$= \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!}\left(x - \frac{\pi}{2}\right)^{2n+1}$$

6. (Core) A function $f$ has $f^{(n)}(0) = (-1)^n 2^n$ for all $n \ge 0$. Write the first four terms and the general term of its Maclaurin series. Which familiar function is $f$?

Solution

$$f(x) = 1 - 2x + \frac{4}{2!}x^2 - \frac{8}{3!}x^3 + \cdots = 1 - 2x + 2x^2 - \frac{4}{3}x^3 + \cdots$$

The general term is $\dfrac{(-1)^n 2^n x^n}{n!} = \dfrac{(-2x)^n}{n!}$. That’s the $e^x$ series with $x$ replaced by $-2x$, so $f(x) = e^{-2x}$.

7. (Core) Find the Taylor series for $f(x) = \dfrac{1}{x}$ about $x = 2$, and its interval of convergence.

Solution

$f^{(n)}(x) = (-1)^n n!\, x^{-(n+1)}$, so $f^{(n)}(2) = \dfrac{(-1)^n n!}{2^{n+1}}$. Dividing by $n!$:

$$\frac{1}{x} = \frac{1}{2} - \frac{x - 2}{4} + \frac{(x - 2)^2}{8} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n (x - 2)^n}{2^{n+1}}$$

This is geometric with ratio $-\dfrac{x - 2}{2}$, so it converges when $\dfrac{|x - 2|}{2} \lt 1$, that is, $0 \lt x \lt 4$. At $x = 0$ and $x = 4$ the terms have constant size $\dfrac{1}{2}$, so they don’t approach $0$ and both endpoints diverge. The interval of convergence is $0 \lt x \lt 4$.

8. (Challenge) Use the Maclaurin series for $e^x$ to find $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.

Solution

$$e^x - 1 - x = \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$$$\frac{e^x - 1 - x}{x^2} = \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} + \cdots$$

As $x \to 0$, every term after the first goes to $0$, so

$$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{1}{2}$$

(L’Hôpital’s rule twice gives the same answer.)

9. (Challenge) The Maclaurin series for a function $g$ is $\displaystyle\sum_{n=0}^{\infty} \frac{(n + 1)x^n}{3^n}$.

• (a) Find $g^{(4)}(0)$.
• (b) Find the interval of convergence.

Solution

(a) The coefficient of $x^4$ is $\dfrac{5}{3^4} = \dfrac{5}{81}$, and it equals $\dfrac{g^{(4)}(0)}{4!}$. So

$$g^{(4)}(0) = 4! \cdot \frac{5}{81} = \frac{120}{81} = \frac{40}{27}$$

(b)

$$L = \lim_{n \to \infty} \frac{n + 2}{n + 1} \cdot \frac{|x|}{3} = \frac{|x|}{3}$$

so $|x| \lt 3$. At $x = \pm 3$ the terms are $(n + 1)$ and $(-1)^n (n + 1)$, which don’t approach $0$, so both endpoints diverge. The interval of convergence is $-3 \lt x \lt 3$.