Finding Taylor coefficients from derivatives is slow. Once you know a handful of Maclaurin series, you can build new ones by substituting, multiplying by powers of $x$, adding, and differentiating or integrating. That is how series are used in practice.
Why it works
A power series that converges to $f$ on an interval is unique: if two power series agree on an interval around the center, they have the same coefficients (they must, since $c_n = \frac{f^{(n)}(a)}{n!}$ either way). So any honest way of producing a power series for a function, such as substituting $x^{2}$ into the series for $e^{x}$, gives the Taylor series.
| Function | Maclaurin series | Converges for |
|---|---|---|
| $\frac{1}{1 - x}$ | $1 + x + x^{2} + x^{3} + \cdots$ | $|x| < 1$ |
| $e^{x}$ | $1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots$ | all $x$ |
| $\sin x$ | $x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots$ | all $x$ |
| $\cos x$ | $1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots$ | all $x$ |
| $\ln(1 + x)$ | $x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots$ | $-1 < x \le 1$ |
| $\arctan x$ | $x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \cdots$ | $-1 \le x \le 1$ |
Key idea
Build new series from known ones: substitute ($e^{-x^{2}}$ from $e^{u}$ with $u = -x^{2}$), multiply by a power of $x$, add or subtract, differentiate or integrate. The interval of convergence follows from the substitution.
Worked example: Substitute
Find the first three nonzero terms of the Maclaurin series of $e^{-x^{2}}$.
- $e^{u} = 1 + u + \frac{u^{2}}{2!} + \cdots$ with $u = -x^{2}$.
- $e^{-x^{2}} = 1 - x^{2} + \frac{x^{4}}{2} - \cdots$, valid for all $x$.
Answer: $1 - x^{2} + \frac{x^{4}}{2}$
Worked example: Multiply
Find the Maclaurin series of $x\cos(2x)$.
- $$\begin{aligned} \cos(2x) &= 1 - \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} - \cdots \\ &= 1 - 2x^{2} + \frac{2}{3}x^{4} - \cdots. \end{aligned}$$
- Multiply by $x$: $x - 2x^{3} + \frac{2}{3}x^{5} - \cdots$.
Answer: $x - 2x^{3} + \frac{2}{3}x^{5} - \cdots$
Worked example: Recognize a sum
Find $\sum_{n=0}^{\infty}\frac{(-1)^{n}\pi^{2n}}{(2n)!}$.
- This is $\cos x = \sum\frac{(-1)^{n}x^{2n}}{(2n)!}$ with $x = \pi$.
- So the sum is $\cos\pi = -1$.
Answer: $-1$
Common mistake
Substituting only into the first term, or forgetting to raise the coefficient. In $e^{2x}$, the term $\frac{u^{3}}{3!}$ becomes $\frac{(2x)^{3}}{6} = \frac{8x^{3}}{6}$, not $\frac{2x^{3}}{6}$. Write the substitution in parentheses, $(2x)^{n}$, and expand each term. A quick check: the coefficient of $x$ must be $f'(0)$, which for $e^{2x}$ is 2.