The Orthogonality Relations

The two most central elements of the structure of Fourier theory are the set of orthogonality relations and the completeness relation. Let us then show that these also follow from the structure of complex analysis. We start with the orthogonality relations, which we already gave in Equation (1) of the introduction. Of course the integrals involved are simple ones, and can be calculated by elementary means. Our objective here, however, is not to just calculate them but to show that they are a consequence of the analytic structure of the complex plane. We can do this by simply considering the Cauchy integral formulas for the coefficients $a_{k'}$ of the Taylor series of a simple power $z^{k}$, with $k\geq 0$,

$$\begin{eqnarray*} a_{k'} & = & \frac{1}{k!}\, \frac{d^{k'}z^{k}}{dz^{k'}}(0)... ...mbox{\boldmath$\imath$}} \oint_{C}dz\, \frac{z^{k}}{z^{k'+1}}, \end{eqnarray*}$$

where $C$ is a circle or radius $\rho$ centered at the origin, with $0<\rho\leq 1$. On the one hand, if $k'>k$ we get $a_{k'}=0$ due to the multiple differentiation of the power function, which is differentiated more times than the power itself. On the other hand, if $k'<k$ we get $a_{k'}=0$ when we calculate the derivatives and apply the result at zero, since in this case there is always at least one factor of $z$ left, or alternatively due to the Cauchy-Goursat theorem, because in this case the integrand is analytic and thus the integral is zero. If $k'=k$, however, we get $a_{k'}=1$, which we can get either directly from the result of the differentiation, or from the fact that in this case the integral is given by

$$\oint_{C}dz\, \frac{1}{z} = 2\pi\mbox{\boldmath$\imath$},$$

as one can easily verify, either directly or by the residues theorem. In any case we get the result

$$\frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz\, \frac{z^{k}}{z^{k'+1}} = \delta_{kk'}.$$

We now write the integral explicitly on the circle of radius $\rho$, using $\theta$ as the integration variable, with $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$ and thus with $dz=\mbox{\boldmath$\imath$}zd\theta$,

$$\begin{eqnarray*} \delta_{kk'} & = & \frac{1}{2\pi\mbox{\boldmath$\imath$}} ... ...\pi}d\theta\, \cos(k\theta) \sin(k'\theta) \right] \right\}. \end{eqnarray*}$$

One can see that the integrals in the imaginary part are zero due to parity arguments. In fact, these constitute some of the orthogonality relations, those including sines and cosines. We are left with

$$\delta_{kk'} = \frac{\rho^{k-k'}}{2\pi} \left[ \int_{-\... ...{-\pi}^{\pi}d\theta\, \sin(k\theta) \sin(k'\theta) \right].$$ (4)

For $k=0=k'$ the second term vanishes, and the equation becomes a simple identity, which is in fact one of the other orthogonality relations. If, on the other hand, we have $k+k'>0$, we now consider the integral, on the same circuit,

$$\oint_{C}dz\, z^{k}z^{k'-1} = 0,$$

which is zero due to the Cauchy-Goursat theorem, since the integrand is analytic within the circle $C$ for $k+k'\geq 1$. Writing the integral explicitly on the circle we get

$$\begin{eqnarray*} 0 & = & \int_{-\pi}^{\pi}d\theta\, \mbox{\boldmath$\imath$... ...\pi}d\theta\, \cos(k\theta) \sin(k'\theta) \right] \right\}. \end{eqnarray*}$$

Once again the integrals in the imaginary part are zero by parity arguments, and thus we are left with

$$\int_{-\pi}^{\pi}d\theta\, \cos(k\theta) \cos(k'\theta) = \int_{-\pi}^{\pi}d\theta\, \sin(k\theta) \sin(k'\theta),$$

since $\rho\neq 0$. If we now go back to our previous expression in Equation (4) we see that the two integrals that appear there are equal to each other, so we may write that

$$\begin{eqnarray*} \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(k\theta) \co... ...ta) & = & \rho^{k'-k} \delta_{kk'} \\ & = & \delta_{kk'}, \end{eqnarray*}$$

where the factor involving $\rho$ is irrelevant since the right-hand sides are only different from zero if $k=k'$. We get therefore the complete set of orthogonality relations

$$\begin{eqnarray*} \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(k\theta) \co... ..._{-\pi}^{\pi}d\theta\, \cos(k\theta) \sin(k'\theta) & = & 0, \end{eqnarray*}$$

for $k\geq 1$ and $k'\geq 1$, which are the relevant values for DP Fourier series. It is interesting to note that the orthogonality relations are valid on the circle of radius $\rho$ with $0<\rho\leq 1$, without the need to actually take the $\rho\to 1$ limit. We thus get a bit more than we bargained for in this case, for it would have been sufficient to establish these relation only on the unit circle. Note that this is different from what happened during the calculation of the coefficients of the Fourier series. However, in both cases the results come from the Cauchy integral formulas and the Cauchy-Goursat theorem, and in either case the same real integral appears, defining the usual scalar product in the space of real functions on the periodic interval.