Representing Functions as Power Series

Finding a Taylor series from derivatives can get messy fast. Try taking ten derivatives of $e^{-x^2}$! Luckily, there’s a shortcut: start from a series you already know, like $\dfrac{1}{1 - x}$ or $e^x$, and change it with algebra and calculus. This is how you get the series for $\ln(1 + x)$ and $\arctan x$, and how you can approximate integrals that have no elementary antiderivative.

Key ideas

Start from the known series

Everything on this page builds on the four Maclaurin series you’ve memorized:

$$\begin{aligned} e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} &&\text{all } x &\qquad \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} &&\text{all } x \\ \cos x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} &&\text{all } x &\qquad \frac{1}{1 - x} &= \sum_{n=0}^{\infty} x^n &&-1 \lt x \lt 1 \end{aligned}$$

The geometric series $1 + x + x^2 + \cdots$ converges to $\dfrac{1}{1 - x}$ only for $-1 \lt x \lt 1$. Any series built from it inherits that radius.

Substitution

Replace $x$ with an expression everywhere in a known series. The interval changes to match. For example, replacing $x$ with $-x^2$ in the geometric series:

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

valid when $|-x^2| \lt 1$, that is, $-1 \lt x \lt 1$.

To use the geometric series, first rewrite the function in the form $\dfrac{\text{first term}}{1 - \text{ratio}}$. For example, $\dfrac{1}{2 - x} = \dfrac{1/2}{1 - x/2}$.

Multiplying by x or a constant

Multiplying a series by $x^k$ or a constant multiplies every term. The interval doesn’t change. For example, $x\cos x = x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!} - \cdots$ for all $x$.

Differentiating and integrating term by term

Inside its interval of convergence, a power series can be differentiated or integrated one term at a time, just like a polynomial:

$$\frac{d}{dx}\sum c_n (x - a)^n = \sum n\,c_n (x - a)^{n-1}, \qquad \int \sum c_n (x - a)^n\,dx = C + \sum \frac{c_n (x - a)^{n+1}}{n + 1}$$

• The new series has the same radius of convergence.
• The endpoints may change: integrating can make a divergent endpoint converge, and differentiating can make a convergent endpoint diverge. Always recheck them.
• When you integrate, find $C$ by plugging in the centre.

Two important results

Integrating $\dfrac{1}{1 + t} = 1 - t + t^2 - \cdots$ from $0$ to $x$ gives

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

Integrating $\dfrac{1}{1 + t^2} = 1 - t^2 + t^4 - \cdots$ from $0$ to $x$ gives

$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n + 1}, \qquad -1 \le x \le 1$$

Both started from series that diverge at $x = \pm 1$, and both gained endpoints by integrating (Example 4).

Worked examples

Example 1: Substituting into the geometric series

Find the Maclaurin series and interval of convergence for $f(x) = \dfrac{1}{1 + 3x}$, and then for $g(x) = \dfrac{x}{1 + 3x}$.

Solution. Write $\dfrac{1}{1 + 3x} = \dfrac{1}{1 - (-3x)}$ and replace $x$ with $-3x$ in $\dfrac{1}{1 - x} = \sum x^n$:

$$f(x) = \sum_{n=0}^{\infty} (-3x)^n = 1 - 3x + 9x^2 - 27x^3 + \cdots$$

valid when $|-3x| \lt 1$, so $-\tfrac{1}{3} \lt x \lt \tfrac{1}{3}$. (At the endpoints the terms are $(\pm 1)^n$, which don’t approach $0$, so they diverge.)

Multiplying by $x$ doesn’t change the interval:

$$g(x) = \sum_{n=0}^{\infty} (-3)^n x^{n+1} = x - 3x^2 + 9x^3 - 27x^4 + \cdots, \qquad -\tfrac{1}{3} \lt x \lt \tfrac{1}{3}$$

Example 2: An integral with no elementary antiderivative

Find the Maclaurin series for $e^{-x^2}$. Then use the first three nonzero terms of a series for $\displaystyle\int_0^{1/2} e^{-x^2}\,dx$ to approximate the integral, and bound the error.

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

$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$$

Integrate term by term:

$$\begin{aligned} \int_0^{1/2} e^{-x^2}\,dx &= \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \cdots\right]_0^{1/2} \\ &= \frac{1}{2} - \frac{1}{24} + \frac{1}{320} - \frac{1}{5376} + \cdots \end{aligned}$$

The first three terms give

$$\frac{1}{2} - \frac{1}{24} + \frac{1}{320} \approx 0.461458 \approx 0.461$$

This series alternates, and its terms decrease to $0$, so by the alternating series error bound, the error is at most $\dfrac{1}{5376} \approx 0.000186$. (Check: a calculator gives $0.461281$.)

Example 3: Differentiating the geometric series

Find a power series for $\dfrac{1}{(1 - x)^2}$ and its interval of convergence.

Solution. Notice that $\dfrac{d}{dx}\left(\dfrac{1}{1 - x}\right) = \dfrac{1}{(1 - x)^2}$. Differentiate $1 + x + x^2 + x^3 + \cdots$ term by term:

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

The radius stays $1$. At $x = \pm 1$, the terms $n(\pm 1)^{n-1}$ don’t approach $0$, so both endpoints diverge. The interval of convergence is $-1 \lt x \lt 1$.

Example 4: The series for ln(1 + x)

Find the Maclaurin series for $\ln(1 + x)$ and its interval of convergence.

Solution. Start from $\dfrac{1}{1 + t} = 1 - t + t^2 - t^3 + \cdots$ for $-1 \lt t \lt 1$, and integrate term by term:

$$\ln(1 + x) = \int_0^x \frac{1}{1 + t}\,dt = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$$

(Integrating from $0$ handles the constant: $\ln(1 + 0) = 0$, and the series is also $0$ at $x = 0$.)

The radius is still $1$. Now check the endpoints of the new series:

• At $x = 1$: $1 - \dfrac{1}{2} + \dfrac{1}{3} - \cdots$, the alternating harmonic series, converges (to $\ln 2$).
• At $x = -1$: $-1 - \dfrac{1}{2} - \dfrac{1}{3} - \cdots = -\sum \dfrac{1}{n}$, which diverges.

The interval of convergence is $-1 \lt x \le 1$. The original series for $\dfrac{1}{1 + x}$ diverged at $x = 1$; integrating gained that endpoint.

Common mistakes

Forgetting to change the interval after substituting. For $\dfrac{1}{1 - x^2/4}$, the condition is $\left|\dfrac{x^2}{4}\right| \lt 1$, so $|x| \lt 2$, not $|x| \lt 1$.

Not rewriting into geometric form first. $\dfrac{1}{2 - x}$ is not $\sum (x - 1)^n$ or $\sum (2 - x)^n$. Factor out the constant: $\dfrac{1}{2 - x} = \dfrac{1}{2} \cdot \dfrac{1}{1 - x/2} = \sum \dfrac{x^n}{2^{n+1}}$.

Forgetting the constant when integrating. Integrating term by term gives $C$ plus a series. Find $C$ by plugging in the centre, or write the integral from $0$ to $x$ as in Example 4.

Assuming the endpoints stay the same. Differentiation and integration keep the radius but can change what happens at the endpoints. Check them again in the new series.

Substituting in only some places. When you replace $x$ with $-x^2$, every $x$ becomes $-x^2$, including inside the powers: $(-x^2)^n = (-1)^n x^{2n}$.

Practice

1. (Warm-up) Find the Maclaurin series for $\dfrac{1}{1 + x}$ and its interval of convergence.

Solution

Replace $x$ with $-x$ in the geometric series:

$$\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots = \sum_{n=0}^{\infty} (-1)^n x^n, \qquad -1 \lt x \lt 1$$

(At $x = \pm 1$, the terms don’t approach $0$.)

2. (Warm-up) Find the first three nonzero terms and the general term of the Maclaurin series for $\sin(x^2)$.

Solution

Replace $x$ with $x^2$ in the sine series:

$$\sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n+1)!}$$

valid for all $x$.

3. (Warm-up) Find the first four nonzero terms and the general term of the Maclaurin series for $x e^x$.

Solution

Multiply the $e^x$ series by $x$:

$$x e^x = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}$$

4. (Core) Find a power series for $\dfrac{1}{2 - x}$ and its interval of convergence.

Solution

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

Valid when $\left|\dfrac{x}{2}\right| \lt 1$: $-2 \lt x \lt 2$ (the endpoints give terms of constant size $\tfrac{1}{2}$, so they diverge).

5. (Core) Integrate a geometric series to find the Maclaurin series for $\ln(1 - x)$, and find its interval of convergence.

Solution

$\dfrac{d}{dx}\ln(1 - x) = -\dfrac{1}{1 - x}$, so

$$\ln(1 - x) = -\int_0^x \frac{1}{1 - t}\,dt = -\int_0^x \left(1 + t + t^2 + \cdots\right)dt = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots = -\sum_{n=1}^{\infty} \frac{x^n}{n}$$

The radius is $1$. At $x = 1$: $-\sum \dfrac{1}{n}$ diverges. At $x = -1$: $-\sum \dfrac{(-1)^n}{n}$ converges by the alternating series test. The interval is $-1 \le x \lt 1$.

6. (Core) Let $f(x) = \displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n^2}$, which converges for $-1 \le x \le 1$. Find a power series for $f'(x)$ and its interval of convergence.

Solution

Differentiate term by term:

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

The radius is still $1$. At $x = 1$: $\sum \dfrac{1}{n}$ diverges. At $x = -1$: $\sum \dfrac{(-1)^{n-1}}{n}$ converges by the alternating series test. The interval is $-1 \le x \lt 1$: differentiating lost the endpoint $x = 1$.

7. (Core) Use the first two nonzero terms of a series to approximate $\displaystyle\int_0^1 \sin(x^2)\,dx$, and bound the error.

Solution

From Question 2, $\sin(x^2) = x^2 - \dfrac{x^6}{6} + \dfrac{x^{10}}{120} - \cdots$, so

$$\int_0^1 \sin(x^2)\,dx = \left[\frac{x^3}{3} - \frac{x^7}{42} + \frac{x^{11}}{1320} - \cdots\right]_0^1 = \frac{1}{3} - \frac{1}{42} + \frac{1}{1320} - \cdots$$

Two terms: $\dfrac{1}{3} - \dfrac{1}{42} = \dfrac{13}{42} \approx 0.310$.

The series alternates with terms decreasing to $0$, so the error is at most $\dfrac{1}{1320} \approx 0.00076$. (Since that first omitted term is positive, $\dfrac{13}{42}$ is an underestimate; a calculator gives $0.310268$.)

8. (Challenge) Use the series from Example 3 to find the exact sum of $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots$

Solution

Multiply $\dfrac{1}{(1 - x)^2} = \displaystyle\sum_{n=1}^{\infty} n x^{n-1}$ by $x$:

$$\frac{x}{(1 - x)^2} = \sum_{n=1}^{\infty} n x^n, \qquad -1 \lt x \lt 1$$

At $x = \tfrac{1}{2}$, which is inside the interval:

$$\sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{1/2}{(1/2)^2} = 2$$

9. (Challenge) Let $f(x) = \arctan\left(\dfrac{x}{2}\right)$.

• (a) Find the first three nonzero terms and the general term of the Maclaurin series for $f$.
• (b) Find the interval of convergence.
• (c) Find $f^{(5)}(0)$.

Solution

(a) Replace $x$ with $\dfrac{x}{2}$ in the arctan series:

$$f(x) = \frac{x}{2} - \frac{(x/2)^3}{3} + \frac{(x/2)^5}{5} - \cdots = \frac{x}{2} - \frac{x^3}{24} + \frac{x^5}{160} - \cdots$$

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

(b) The arctan series converges for $-1 \le x \le 1$, so this converges for $-1 \le \dfrac{x}{2} \le 1$, that is, $-2 \le x \le 2$.

(c) The coefficient of $x^5$ is $\dfrac{1}{160} = \dfrac{f^{(5)}(0)}{5!}$, so

$$f^{(5)}(0) = \frac{120}{160} = \frac{3}{4}$$