Appendix: Reduction of Domain Intervals

The fact that we discuss the properties of real functions only within the interval $[-\pi,\pi]$ is not a limitation from the physics point of view. It is easy to show that all situations in physics application can be reduced to this interval. Let us consider a physical variable $x$ within the closed interval $[a,b]$, and let $g(x)$ be a real function describing some physical quantity within that interval. The size of the interval does not matter. For example, if $x$ is a measure of length, the interval could be the size of a small resonant cavity or it could be the size of the galaxy. In any case it is still a closed interval. The same is true if $x$ represents an energy, since it is always true that only limited amounts of energy are available for any given experiment or realistic physical situation. So long as $x\in[a,b]$, regardless of the magnitude or physical nature of the dimensionfull physical variable $x$, we can rescale it to fit within $[-\pi,\pi]$. It is a simple question of making a change of variables, by defining the new dimensionless variable

$$\theta = 2\pi\,\frac{x-a}{b-a}-\pi,$$

so that $\theta\in[-\pi,\pi]$. The inverse transformation is given by

$$x = \frac{b-a}{2\pi}\,\theta + \frac{b+a}{2}.$$

The function $g(x)$ can now be transformed into a function $f(\theta)$,

$$g(x) = f(\theta),$$

where $f(\theta)$ describes the same physics as $g(x)$. Let us now consider the transformation of the first-order low-pass filter from one system of variables to the other. Given a value of $\theta$, if we vary it by $\epsilon$, we have a corresponding variation of $x$, which we will call $\varepsilon$. It then follows that we have a relation between $\varepsilon$ and $\epsilon$,

$$\begin{eqnarray*} x\pm\varepsilon & = & \frac{b-a}{2\pi}\,(\theta\pm\epsilon)... ...Rightarrow \\ \varepsilon & = & \frac{b-a}{2\pi}\,\epsilon. \end{eqnarray*}$$

The limit $\epsilon \to 0$ is now seen to be clearly equivalent to the limit $\varepsilon\to 0$, and the definition of the filtered function $f_{\epsilon}(\theta)$ has a counterpart $g_{\varepsilon}(x)$,

$$\begin{eqnarray*} f_{\epsilon}(\theta) & = & \frac{1}{2\epsilon} \int_{\thet... ...2\varepsilon} \int_{x-\varepsilon}^{x+\varepsilon}dx'\, g(x'). \end{eqnarray*}$$

It follows that if $f(\theta)$ is a combed function and we thus have that

$$f(\theta) = \lim_{\epsilon\to 0} f_{\epsilon}(\theta),$$

then we also have that

$$g(x) = \lim_{\varepsilon\to 0} g_{\varepsilon}(x),$$

so that $g(x)$ is also a similarly combed function. We may thus conclude that it suffices to consider and discuss only the set of combed functions within $[-\pi,\pi]$.