A Regular Sine Series with Even $k$

Consider the Fourier series of the two-cycle unit-amplitude sawtooth wave. As is well known it is given by the sine series

$$S_{\rm s} = -\, \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{1}{2j}\, \sin[(2j)\theta].$$

The corresponding FC series is then

$$\bar{S}_{\rm s} = -\, \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{1}{2j}\, \cos[(2j)\theta],$$

the complex $S_{v}$ series is given by

$$S_{v} = -\, \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{1}{2j}\, v^{2j},$$

and the complex power series $S_{z}$ is given by

$$S_{z} = -\, \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{1}{2j}\, z^{2j}.$$

The ratio test tells us that the disk of convergence of $S_{z}$ is the unit disk. If we consider the inner analytic function $w(z)$ within this disk we observe that $w(0)=0$, as expected. We have for the function and its derivative

$$\begin{eqnarray*} w(z) & = & -\, \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{1... ...}{dz} & = & -\, \frac{4}{\pi} \sum_{j=1}^{\infty} z^{2j-1}. \end{eqnarray*}$$

This is a geometrical series and therefore we may write the derivative in closed form,

$$\begin{eqnarray*} \frac{dw(z)}{dz} & = & -\, \frac{4}{\pi}\, \frac{z}{1-z^{... ...rac{2}{\pi}\, \left( \frac{1}{1+z} - \frac{1}{1-z} \right). \end{eqnarray*}$$

This function has two simple poles at the points where the two-cycle sawtooth wave is discontinuous. These are hard singularities of degree $1$. It is now very simple to integrate in order to obtain the inner analytic function $w(z)$ in closed form, remembering that we should have $w(0)=0$,

$$\begin{eqnarray*} w(z) & = & \frac{2}{\pi}\, \left[ \ln(1+z) + \ln(1-z) \right] \\ & = & \frac{2}{\pi}\, \ln\!\left(1-z^{2}\right). \end{eqnarray*}$$

This function has logarithmic singularities at the points where the two-cycle sawtooth wave is discontinuous. These are borderline hard singularities. If we now rationalize the argument of the logarithm in order to write it in the form

$$1-z^{2} = R\,{\rm e}^{\mbox{\boldmath$\imath$}\alpha},$$

then the logarithm is given by

$$\ln\!\left(1-z^{2}\right) = \ln(R)+\mbox{\boldmath$\imath$}\alpha,$$

which allows us to identify the real and imaginary parts of $w(z)$. We may write for the argument

$$\begin{eqnarray*} R\,{\rm e}^{\mbox{\boldmath$\imath$}\alpha} & = & 1 - \rh... ...ox{\boldmath$\imath$} \left[ \rho^{2} \sin(2\theta) \right]. \end{eqnarray*}$$

From this it follows, after some algebra, that we have

$$\begin{eqnarray*} R(\rho,\theta) & = & \sqrt{1-2\rho^{2}\cos(2\theta)+\rho^{4... ...ft[\frac{\rho^{2}\sin(2\theta)}{1-\rho^{2}\cos(2\theta)}\right]. \end{eqnarray*}$$

The part which is of interest now is the imaginary one, which is related to the series $S_{\rm s}$,

$$\Im[w(z)] = -\, \frac{2}{\pi}\, \arctan\! \left[\frac{\rho^{2}\sin(2\theta)}{1-\rho^{2}\cos(2\theta)}\right].$$

The $\rho\to 1$ limit of this quantity is the function $f_{\rm s}(\theta)$. If we consider the cases $\theta=0$ and $\theta=\pm\pi$, we immediately get zero, because the argument of the arc tangent is zero, and therefore we get $\alpha=0$. Therefore we have

$$\begin{eqnarray*} f_{\rm s}(0) & = & \lim_{\rho\to 1} \Im[w(\rho,0)] \\ &... ...i) & = & \lim_{\rho\to 1} \Im[w(\rho,\pm\pi)] \\ & = & 0. \end{eqnarray*}$$

These are the correct values for the Fourier-series representation of the two-cycle sawtooth wave at these points. For $\theta\neq 0$ and $\theta\neq\pm\pi$ we get

$$\begin{eqnarray*} f_{\rm s}(\theta) & = & \lim_{\rho\to 1} \Im[w(z)] \\ &... ...}{\pi} \arctan\!\left[\frac{\cos(\theta)}{\sin(\theta)}\right]. \end{eqnarray*}$$

In order to solve this equation we first write it as

$$\begin{eqnarray*} f_{\rm s}(\theta) & = & \frac{2}{\pi}\, \alpha(\theta), \... ...ghtarrow \\ \tan(\alpha) & = & -\, \frac{1}{\tan(\theta)}. \end{eqnarray*}$$

Keeping in mind that we must have $-\pi\leq\theta\leq\pi$ and $-\pi/2\leq\alpha\leq\pi/2$, and paying careful attention to the signs of the sine and cosine of either angle, one concludes that the solution is given by

$$% \renewedcommand{arraystretch}{2.4} \begin{array}{rclcc... ...pi}{\displaystyle 2}} & \mbox{for} & \theta<0, \end{array}$$

which means that for $f_{\rm s}(\theta)$ we have

$$% \renewedcommand{arraystretch}{2.4} \begin{array}{rclcc... ...playstyle \pi}} + 1 & \mbox{for} & \theta<0. \end{array}$$

These are the correct values for the two-cycle unit-amplitude sawtooth wave. The FC function is related to the real part

$$\Re[w(z)] = \frac{1}{\pi}\, \ln\!\left[1-2\rho^{2}\cos(2\theta)+\rho^{4}\right].$$

In the $\rho\to 1$ limit this function is singular at $\theta=0$ and $\theta=\pm\pi$. Away from these logarithmic singularities we may write

$$\begin{eqnarray*} \bar{f}_{\rm s}(\theta) & = & \lim_{\rho\to 1} \Re[w(z)] ... ...\ & = & \frac{1}{\pi}\, \ln\!\left[4\sin^{2}(\theta)\right]. \end{eqnarray*}$$

We see therefore that the FC function to the two-cycle sawtooth wave is a function with logarithmic singularities at the two points where the original function is discontinuous. The derivative of this conjugate function is easily calculated, and turns out to be

$$\frac{d\bar{f}_{\rm s}(\theta)}{d\theta} = \frac{2}{\pi}\, \frac{\cos(\theta)}{\sin(\theta)}.$$

This is a quite a simple function, with two simple poles at $\theta=0$ and $\theta=\pm\pi$. Note that the logarithmic singularities of the FC function $\bar{f}_{\rm s}(\theta)$ are integrable ones, as one would expect since the Fourier coefficients are finite.

Note that we get a free set of integration formulas out of this effort. Since the coefficients $a_{k}$ can be written as integrals involving $\bar{f}_{\rm s}(\theta)$, we have at once that for even $k=2j$

$$\int_{-\pi}^{\pi}d\theta\, \ln\!\left[4\sin^{2}(\theta)\right] \cos(k\theta) = -\, \frac{4\pi}{k},$$

for $k>0$, while for odd $k=2j+1$

$$\int_{-\pi}^{\pi}d\theta\, \ln\!\left[4\sin^{2}(\theta)\right] \cos(k\theta) = 0.$$

Note that $\bar{f}_{\rm s}(\theta)$ is an even function of $\theta$, so that these integrals are not zero by parity arguments, and therefore we also have

$$\int_{0}^{\pi}d\theta\, \ln\!\left[4\sin^{2}(\theta)\right] \cos(k\theta) = 0,$$

for $k=2j+1$.