The binomial theorem expands $(1 + x)^{k}$ when $k$ is a whole number, and the expansion stops. Newton noticed that the same pattern works for any exponent, like $k = \frac{1}{2}$ or $k = -1$, except that the series never stops. It is a known series you can build from Taylor coefficients.
Why it works
For $f(x) = (1 + x)^{k}$: $f'(x) = k(1 + x)^{k-1}$, $f''(x) = k(k - 1)(1 + x)^{k-2}$, and in general $f^{(n)}(0) = k(k - 1)\cdots(k - n + 1)$. Dividing by $n!$ gives the coefficient
$$\binom{k}{n} = \frac{k(k - 1)\cdots(k - n + 1)}{n!}.$$
When $k$ is a whole number, a factor $(k - k)$ eventually appears and every later coefficient is 0: the familiar finite expansion. Otherwise the series goes on forever, and the ratio test gives radius 1.
Key idea
$$\begin{aligned} (1 + x)^{k} &= \sum_{n=0}^{\infty}\binom{k}{n}x^{n} \\ &= 1 + kx + \frac{k(k - 1)}{2!}x^{2} \\ &\quad + \frac{k(k - 1)(k - 2)}{3!}x^{3} + \cdots \end{aligned}$$ for $|x| < 1$.
Worked example: A square root
Find the first four terms of the series for $\sqrt{1 + x}$.
- $k = \frac{1}{2}$: $\binom{1/2}{1} = \frac{1}{2}$, $\binom{1/2}{2} = \frac{\frac{1}{2}\cdot(-\frac{1}{2})}{2} = -\frac{1}{8}$, $\binom{1/2}{3} = \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{6} = \frac{1}{16}$.
- $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{8} + \frac{x^{3}}{16} - \cdots$ for $|x| < 1$.
- Estimate: $$\begin{aligned} &\sqrt{1.1} \\ &\approx 1 + 0.05 - 0.00125 + 0.0000625 \\ &\approx 1.04881 \end{aligned}$$ (true value $1.048809\ldots$).
Answer: $1 + \frac{x}{2} - \frac{x^{2}}{8} + \frac{x^{3}}{16}$
Worked example: Substituting
Find the first three nonzero terms for $\frac{1}{\sqrt{1 - x^{2}}}$.
- $k = -\frac{1}{2}$, $u = -x^{2}$: $\binom{-1/2}{1} = -\frac{1}{2}$ and $\binom{-1/2}{2} = \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} = \frac{3}{8}$.
- $$\begin{aligned} (1 + u)^{-1/2} &= 1 - \frac{u}{2} + \frac{3u^{2}}{8} - \cdots \\ &= 1 + \frac{x^{2}}{2} + \frac{3x^{4}}{8} + \cdots. \end{aligned}$$
- Integrating this term by term gives the series for $\arcsin x$.
Answer: $1 + \frac{x^{2}}{2} + \frac{3x^{4}}{8}$
Common mistake
Using the series on $(a + x)^{k}$ without factoring out $a$. The formula needs the form $(1 + u)^{k}$. For $\sqrt{4 + x}$, write $2\left(1 + \frac{x}{4}\right)^{1/2} = 2 + \frac{x}{4} - \frac{x^{2}}{64} + \cdots$, which converges for $|x| < 4$, not $|x| < 1$. Check: at $x = 0$ the series must give $f(0) = 2$.
Worked example: A single coefficient
Find the coefficient of $x^{3}$ in $(1 + x)^{1/3}$.
- $$\begin{aligned} \binom{1/3}{3} &= \frac{\frac{1}{3}\left(\frac{1}{3} - 1\right)\left(\frac{1}{3} - 2\right)}{3!} \\ &= \frac{\frac{1}{3}\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{5}{3}\right)}{6}. \end{aligned}$$
- $= \frac{10/27}{6} = \frac{5}{81}$.
Answer: $\frac{5}{81}$