# Error function etymology: Why the name?

I've recently been introduced to the error function:

$$\operatorname{erf}(z) =\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}dt$$

Naturally, I wondered about the origin of its name: The error of... what? I'm so used to uncertainty analysis in scientific experimentation that my mind went there first. It seems like the name sprung up in the context of statistical analysis. Here's what I found on Quora:

In the course of the 19th century the function from the theory of errors appeared in several contexts unrelated to probability, e.g. refraction and heat conduction. In 1871 J. W. Glaisher wrote that "Erf(x) may fairly claim at present to rank in importance next to the trigonometrical and logarithmic functions." Glaisher introduced the symbol Erf and the name error function for a particular form of the law as follows:

As it is necessary that the function should have a name, and as I do not know that any has been suggested, I propose to call it the Error-function ...

("On a Class of Definite Integrals, Philosophical Magazine, 42, 1871, p. 296)

But this quote sounds like Glaisher arbitrarily picked the name. He must have had some motivation in dubbing it error.

Anyway, what does error represent? Does this measure some deviation from the related Gaussian integral from $\int_0^\infty$?

EDIT The link provided by Dilip is useful, yet still did not explain the depth of this name to me. I am wholly unfamiliar with probability and error theory - where $erf(z)$ seems to have arisen - so excuse me if the answer should have been more obvious. (In fact, error theory sounds vaguely alien to me.) Is the function just commonplace in error theory, and for that reason named $erf$? This reasoning just seems a little, well, lackluster to me, but maybe it's true. Does $erf$ have a specific role that reflects its name?

• It's mostly pointless to worry too much about names. People choose names for all sorts of reasons. – Qiaochu Yuan May 1 '15 at 5:49

Sometimes (often) errors are modeled as being distributed normally, with probability distribution $$\varphi_\sigma(x)\,dx = \frac 1 {\sqrt{2\pi}} \exp\left( \frac{-1} 2 \left(\frac x \sigma\right)^2 \right) \, \frac{dx} \sigma$$ with expected value $0$ and standard deviation $\sigma$.
The cumulative probability distribution function is $$\Phi_\sigma(x) = \int_{-\infty}^x \varphi_\sigma(x)\,dx.$$ Up to a rescaling of $x$, this is the error function. The usual definition of the "error function" omits the factor of $1/2$, and thus the standard deviation of the distribution whose cumulative distribution function is the "error function" is not $1$. I am far from convinced that it ought to be rescaled in that way.