A Singular Cosine Series

Consider the Fourier series of the Dirac delta ``function'' centered at $\theta=\theta_{1}$, which we denote by $\delta(\theta-\theta_{1})$. We may easily calculate its Fourier coefficients using the rules of manipulation of $\delta(\theta-\theta_{1})$, thus obtaining $\alpha_{k}=\cos(k\theta_{1})/\pi$ and $\beta_{k}=\sin(k\theta_{1})/\pi$ for all $k$. The series is therefore the complete Fourier series given by

$$\begin{eqnarray*} S_{\rm c} & = & \frac{1}{2\pi} + \frac{1}{\pi} \sum_{k=1... ...pi} + \frac{1}{\pi} \sum_{k=1}^{\infty} \cos(k\Delta\theta), \end{eqnarray*}$$

where $\Delta\theta=\theta-\theta_{1}$. Apart from the constant term this is in fact a DP cosine series on this new variable. Clearly, this series diverges at all points in the interval $[-\pi,\pi]$. Undaunted by this, we proceed to construct the FC series, with respect to the new variable $\Delta\theta$, which turns out to be

$$\bar{S}_{\rm c} = \frac{1}{\pi} \sum_{k=1}^{\infty} \sin(k\Delta\theta),$$

a series that is also divergent, this time almost everywhere. If we define $v=\exp(\mbox{\boldmath$\imath$}\theta)$ and $v_{1}=\exp(\mbox{\boldmath$\imath$}\theta_{1})$ the corresponding complex series $S_{v}$ is then given by

$$S_{v} = \frac{1}{2\pi} + \frac{1}{\pi} \sum_{k=1}^{\infty} \left(\frac{v}{v_{1}}\right)^{k},$$

where we included the $k=0$ term, and the corresponding complex power series $S_{z}$ is given by

$$S_{z} = \frac{1}{2\pi} + \frac{1}{\pi} \sum_{k=1}^{\infty} \left(\frac{z}{z_{1}}\right)^{k},$$

where $z=\rho v$ and $z_{1}=v_{1}$ is a point over the unit circle. The ratio test tells us that the disk of convergence of $S_{z}$ is the unit disk. This converges to a perfectly well-defined analytic function strictly inside the open unit disk. If we eliminate the constant term we get a series $S'_{z}$ which converges to an inner analytic function rotated by the angle $\theta_{1}$,

$$S'_{z} = \frac{1}{\pi} \sum_{k=1}^{\infty} \left(\frac{z}{z_{1}}\right)^{k}.$$

The dominant singularity is clearly at the point $z_{1}$, so we must use the factor $(z-z_{1})$ in the construction of the corresponding center series,

$$\begin{eqnarray*} C'_{z} & = & (z-z_{1})S'_{z} \\ & = & \frac{1}{\pi}\, ... ...z_{1}}\right)^{k} \right] \\ & = & -\, \frac{1}{\pi}\, z. \end{eqnarray*}$$

So we see that we get a remarkably simple result, since the center series can actually be added up exactly. We get therefore for the series $S'_{z}$

$$S'_{z} = -\, \frac{1}{\pi}\, \frac{z}{z-z_{1}},$$

and for the series $S_{z}$

$$S_{z} = \frac{1}{2\pi} - \frac{1}{\pi}\, \frac{z}{z-z_{1}}.$$

We may now take the real and imaginary parts of the $S_{v}$ series in order to obtain faster-converging representation of the original DP Fourier series and its FC series. The explanation of the reasons why this is a representation of the Dirac delta ``function'' requires taking limits to the unit circle carefully, and since they were given in the previous paper [1], they will not be repeated here. We have, for $z$ on the unit circle, so long as $\Delta\theta\neq 0$,

$$\begin{eqnarray*} \frac{z}{z-z_{1}} & = & \frac{z(z^{*}-z_{1}^{*})}{(z-z_{1})... ... \frac{1}{2}\, \frac{1+\cos(\Delta\theta)}{\sin(\Delta\theta)}, \end{eqnarray*}$$

and therefore

$$\begin{eqnarray*} S_{v} & = & \frac{1}{2\pi} - \frac{1}{2\pi} + \mbox{\bo... ...rac{1}{2\pi}\, \frac{1+\cos(\Delta\theta)}{\sin(\Delta\theta)}. \end{eqnarray*}$$

The original DP ``function'' is given by the real part, and therefore we get

$$f_{\rm c}(\theta) = 0,$$

which is the correct value for the Dirac delta ``function'' away from the singular point at $\Delta\theta=0$, and the corresponding FC function $f_{\rm s}(\theta)=\bar{f}_{\rm c}(\theta)$ is given by the imaginary part,

$$f_{\rm s}(\theta) = \frac{1}{2\pi}\, \frac{1+\cos(\Delta\theta)}{\sin(\Delta\theta)},$$

which is the same result we obtained in the previous paper [1].