Equivalent of $\sin(x)/x = \operatorname{sinc}(x)$ for $(\cos(x)-1)/x$

Question

In a work calculation I have come across an expression that involves $\sin(x)/x$ which is undefined for $x=0$, even though for my problem, in the $x=0$ case, $\sin(x)/x$ should naturally be $1$ from the context of the problem.

This is handily fixed by substituting $\sin(x)/x = \operatorname{sinc}(x)$, which is defined as $\operatorname{sinc}(x) = \sin(x)/x$ for all nonzero $x$, and $\operatorname{sinc}(0) = 1$.

However, my work calculation also has an expression $(\cos(x)-1)/x$, which has an analogous problem; for $x=0$, it should simplify to $0$ (which is the limit for $x\to0$) but it is, of course, undefined.

As in the $\operatorname{sinc}$ case, my goal is to obtain an expression that's defined everywhere, with no inconvenient 'special case' for $x=0$.

Is there, perhaps, an obvious way to write $(\cos(x)-1)/x$ in terms of $\operatorname{sinc}(x)$ that I am missing? Or, alternatively, is there an analogous function defined for this case?

Answer 1

$$1-\cos x = 2 \sin^2\left(\frac{x}{2}\right)$$ $$\frac{\cos x-1}{x} = \frac{-2\sin^2\left(\frac{x}{2}\right)}{x} = \frac{-2\operatorname{sinc}^2\left(\frac{x}{2}\right)}{x}\left(\frac{x}{2}\right)^2= -\frac{x}{2}\operatorname{sinc}^2\left(\frac{x}{2}\right)$$

Answer 2

you may want to have a look at $\frac{1-\cos{x}}{x^2}$. the limit when $x \to 0$ is $\frac 1 2$.

and they share many properties, see my question

essentially, they're function series, $\frac 1{1!} - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + ...$ for the sinc case, and $\frac 1{2!} - \frac{x^2}{4!} + \frac{x^4}{6!} - \frac{x^6}{8!} + ...$ for the cosine case.