The subject of our current study is a Comb $P(x)$ of rectangular distributions: $$ P(x) = \sum_{L=-\infty}^{+\infty} p(x-L.\Delta) \, \Delta $$ $$ \MET p(x) = \left\{ \begin{array}{lll} 0 & \mbox{for} & x \le -\half \sigma \\ 1/\sigma & \mbox{for} & -\half\sigma \le x \le +\half\sigma \\ 0 & \mbox{for} & +\half \sigma \le x \end{array} \right. $$ The geometry of this is a rectangle with width $\sigma$ and height $1/\sigma$, resulting in an area $1$, thus establishing that the function $p(x)$ is normed. The spread $\sigma > 0$ of a rectangular distribution, despite of its name, is not exactly a standard deviation, as is clear from the following: $$ \int_{-\infty}^{+\infty} x^2 p(x) \, dx = \int_{-\half \sigma}^{+\half \sigma} x^2/\sigma \, dx = \left[ \frac{x^3}{3\sigma} \right]_{-\half \sigma}^{+\half \sigma} = \frac{\sigma^2}{8\times 3}\times 2 $$ $$ \hieruit \mbox{standard deviation} = \sigma /(2\sqrt{3}) $$ $P(x)$ is developed into the standard Fourier series for combs of hat functions: $$ P(x) = 1 + 2 \times \sum_{k = 1}^{\infty} A(k \omega) \, cos(k \omega x) \MET A(y) = \int_{-\infty}^{+\infty} p(x) \cos(x y) \, dx $$ By hand, because I'm not too lazy: $$ A(y) = \int_{-\half\sigma}^{+\half\sigma} 1/\sigma \cos(x y) \, dx = \left[ \frac{\sin(y x)}{y \sigma}\right]_{-\half\sigma}^{+\half\sigma} = \frac{\sin(\half y \sigma)}{\half y \sigma} = \sinc(\half y \sigma) \hieruit $$ $$ P(x) = 1 + 2 \times \sum_{k=1}^\infty \frac{\sin(\half k\omega\sigma)}{\half k\omega\sigma} \cos(k\omega x) \MET \omega = \frac{2\pi}{\Delta} $$ The Fourier analysis of a comb of Rectangles is thus similar to the Fourier analysis of a comb of Triangles. For $\sigma = m\Delta$, with $m > 0$ integer, the outcome is simply $P(x) = 1$, without any wiggles. On the other hand, especially for $\sigma = (m + \half) \Delta$, it's much harder to get rid of those wiggles. Furthermore, proceeding as usual is virtually impossible because of the terms with $1/k$ , together with the fact that the harmonic series is known to be divergent: $$ \sum_{k=1}^\infty\frac{1}{k} = \infty $$ So it seems that the only possibility left is our Improved Error Analysis: $$ \sum_{k=1}^\infty \left[ \frac{\sin(\half k\omega\sigma)}{\half k\omega\sigma} \right]^2 < (\epsilon/2)^2 $$ It is seen that the left hand side of this expression is exactly the same as the left hand side of the normal error analysis for triangular distributions. Therefore we can immediately conclude that: $$ \sigma > \frac{\Delta}{\sqrt{3/2}\,\epsilon} $$