Approximations in physics classes, solved.

If you've taken a physics class, you may have encountered a frustrating situation like this one:

Instructor: Using small angle approximations, approximate $\tan x$ for small $x$.
Student: Okay.
$$\tan x=\frac{\sin x}{\cos x}\approx\frac x1=x.$$
I: Great, now approximate $\tan x-\sin x$.
S: Okay, based on earlier work,
$$\tan x-\sin x\approx x-x=0.$$
I: Nope, that's not a good approximation.
S: What? But you said $\tan x\approx x$ was good!
I: But you get zero, so it's a bad approximation. Instead, you should do something like
\begin{multline*}
\tan x-\sin x=\frac{\sin x(1-\cos x)}{\cos x}=\frac{\sin x(1-\cos^2x)}{\cos x(1+\cos x)}\\
=\frac{\sin^3x}{\cos x(1+\cos x)}
\approx\frac{x^3}{1\cdot(1+1)}=\frac{x^3}2.
\end{multline*}
S: What.

It was never clear to me in physics classes what makes an approximation good enough. It seemed like a hodgepodge of heuristics. At some point, though, I realized that what they are looking for is very simple: The first nonzero term of the Taylor series.

In each problem, there's always some quantity or difference that's "small," and the thing to be estimated (or sometimes its reciprocal) can be written as a Taylor series in terms of that small quantity. The first nonzero term of the Taylor series then provides a good estimate for whatever it is that you're trying to approximate.

The "first nonzero term of the Taylor series" approximations behave nicely. You can multiply quantities together and then approximate the product, or you can approximate them first and then multiply the approximations together, and you'll get the same answer. The same goes for compositions like $\ln(1+\sin x)$. And, the same goes for sums and differences, except when you would get zero. But, of course, the case where you do get zero is often the most interesting one, because then you compute the higher order effects.

In our example, that's exactly what happened. The Taylor series for tangent and sine are
\begin{align*}
\tan x&=x+\frac{x^3}3+\dotsb\\
\sin x&=x-\frac{x^3}6+\dotsb
\end{align*}
The first term of the difference is exactly $\frac{x^3}2$, and the earlier computation becomes an interesting example of how you can compute the third derivative of $\tan x-\sin x$ at zero knowing only the first derivative of $\sin x$ along with the identity $\sin^2x+\cos^2x=1$.