Functions as Power Series — AP® Calculus BC Notes

Four Maclaurin series are worth memorizing: $\frac{1}{1 - x}$ , $e^x$ , $\sin x$ and $\cos x$ . Almost every other series you need can be built from them without computing a single higher derivative. This topic is about the four moves that do the building, and about what they do to the interval of convergence.

Move 1: substitute

A known series holds for any input that meets its condition. Replace $x$ by an expression, and replace it in the condition too.

Worked example 1. Find a power series for $\frac{1}{1 + 4x^2}$ and its interval of convergence.

Start with $\frac{1}{1 - u}$ $= 1 + u + u^2 + \cdots$ for $|u| < 1$ , and substitute $u = -4x^2$ :

$\frac{1}{1 + 4x^2} = 1 - 4x^2 + 16x^4 - 64x^6 + \cdots = \sum_{n=0}^{\infty} (-4)^n x^{2n}.$

Condition: $|-4x^2| < 1$ , so $x^2 < \frac{1}{4}$ , that is $|x| < \frac{1}{2}$ . At $x = \pm\frac{1}{2}$ every term has size 1, so both endpoints diverge: the interval is $\left(-\frac{1}{2}, \frac{1}{2}\right)$ .

Substitution works the same way in the other series: $e^{-x^2}$ $= \sum \frac{(-x^2)^n}{n!}$ $= 1 - x^2$ $+ \frac{x^4}{2!} - \cdots$ for every $x$ , and $\cos(3x)$ $= 1 - \frac{9x^2}{2!}$ $+ \frac{81x^4}{4!} - \cdots$

Move 2: multiply by a power of x

Multiplying each term by $x^k$ shifts every power up by $k$ and does not change where the series converges. For example, $x\cos x$ $= x - \frac{x^3}{2!} + \frac{x^5}{4!} - \cdots$ for every $x$ .

Move 3: differentiate term by term

Inside its interval, a power series can be differentiated one term at a time, and the result is the derivative of its sum:

$\frac{d}{dx}\sum_{n=0}^{\infty} c_n x^n = \sum_{n=1}^{\infty} n c_n x^{n-1}.$

Differentiating $\frac{1}{1 - x} = \sum x^n$ gives $\frac{1}{(1 - x)^2}$ $= \sum_{n=1}^{\infty} n x^{n-1}$ $= 1$ $+ 2x$ $+ 3x^2$ $+ \cdots$ for $|x| < 1$ . Each term is differentiated with respect to $x$ ; the index $n$ is a fixed number in each term.

Move 4: integrate term by term

Figure 1. The geometric series matches

\frac{1}{1 - x}

only on

(-1, 1)

. LevlPrep original diagram.

Worked example 2. Find the Maclaurin series for $\ln(1 + x)$ and its interval of convergence.

Substitute $u = -t$ : $\frac{1}{1 + t}$ $= 1 - t + t^2 - t^3 + \cdots$ for $|t| < 1$ .

Integrate from 0 to $x$ , term by term: $\ln(1 + x)$ $= \int_0^x \frac{1}{1 + t}\,dt$ $= x - \frac{x^2}{2}$ $+ \frac{x^3}{3} - \frac{x^4}{4}$ $+ \cdots$ $= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$ .

The radius stays 1. Endpoints: at $x = 1$ the series is the alternating harmonic series, which converges; at $x = -1$ every term is $-\frac{1}{n}$ , which diverges. The interval is $(-1, 1]$ , one endpoint more than the series we started from.

Integrating from 0 fixes the constant: the integral is 0 at $x = 0$ , matching $\ln 1 = 0$ . In the same way, integrating $\frac{1}{1 + t^2}$ $= 1 - t^2 + t^4 - \cdots$ gives $\arctan x$ $= x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ for $|x| \le 1$ .

What happens to the interval

Move	Radius	Endpoints
substitute $u = g(x)$	from the new condition	test again
multiply by $x^k$	unchanged	unchanged
differentiate term by term	unchanged	may lose an endpoint: test again
integrate term by term	unchanged	may gain an endpoint: test again

The figure is a reminder of why the interval matters: outside $(-1, 1)$ the partial sums of $1 + x + x^2 + \cdots$ swing or climb away, while $\frac{1}{1 - x}$ stays finite. A series stands for its function only on its interval of convergence.

Using a built series

Once you have a series, you can read off derivatives at the center ( $f^{(n)}(0) = n! \cdot c_n$ ), evaluate sums such as $\sum \frac{n}{2^n}$ $= \frac{1}{2}\sum n\left(\frac{1}{2}\right)^{n-1}$ $= \frac{1}{2} \cdot \frac{1}{(1 - 1/2)^2}$ $= 2$ , and approximate integrals with no elementary antiderivative, such as $\int_0^x e^{-t^2}\,dt$ $= x - \frac{x^3}{3}$ $+ \frac{x^5}{10} - \cdots$ . When the resulting values alternate with sizes decreasing to 0, the alternating series error bound says how accurate the estimate is.