I'm currently going through "An Introduction to Mechanics" by Kleppner and Kolenkow. On page 38 of the text, he glosses over the small-angle approximation of $sin(x)$ and $cos(x)$.

Specifically, he uses the Taylor series of $sin(x)$ and $cos(x)$ to show that $sin(x) \approx x$ and $cos(x) \approx 1 - \frac{1}{2}x^2$.

He ends this section by remarking that "These expressions, which are sometimes called the small angle approximations, are valid up to terms of order $x^3$, denoted by $O(x^3)$".

I don't understand what being valid up to terms of order $x^3$ means. Could someone please explain?

Thanks in advance.


1 Answer 1


Generally speaking, if a polynomial approximation $p (x)$ is accurate to $O (x^n)$, then

$$f (x) = p (x) + \varepsilon (x)$$

where $\varepsilon (x)$ approaches $0$ more rapidly than $x^n$; i.e., $\lim_{x \to 0} \frac{\varepsilon (x)}{x^n} = 0$.

For example, in the "small-angle" linear approximation for $f (x) = \sin (x)$, we have $p (x) = x$. Using Taylor expansion, we have an explicit expression for $\varepsilon (x)$, which is

$$\varepsilon (x) = -\frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \dots$$

The linear approximation $p$ is accurate to $O (x^2)$, because

$$\lim_{x \to 0} \frac{\varepsilon (x)}{x^2} = \lim_{x \to 0} -\frac{1}{3!} x + \frac{1}{5!} x^3 - \dots = 0$$

But $p$ is not accurate to $O (x^3)$, since

$$\lim_{x \to 0} \frac{\varepsilon (x)}{x^3} = -\frac{1}{6} \ne 0$$

In a similar fashion, the quadratic-accurate polynomial approximation to $f (x) = \cos x$ is $p (x) = 1 - \frac{1}{2} x^2$. It turns out that this approximation is accurate to $O (x^3)$, since

$$\lim_{x \to 0} \frac{\varepsilon (x)}{x^3} = \lim_{x \to 0} \frac{\frac{1}{4!} x^4 - \frac{1}{6!} x^6 + \dots}{x^3} = \lim_{x \to 0} \frac{1}{4!} x - \frac{1}{6!} x^3 + \dots = 0$$


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