Remainders of Fourier Series

We will now derive certain expressions for the partial sums and for the corresponding remainders of the Fourier series. In order to do this, let $f(\theta)$ be a zero-average integrable real function defined on $[-\pi,\pi]$ and let the real numbers $\alpha_{0}=0$, $\alpha_{k}$ and $\beta_{k}$, for $k\in\{1,2,3,\ldots,\infty\}$, be its Fourier coefficients. We then define the complex coefficients $c_{0}=0$ and $c_{k}$ shown in Equation (5), and thus construct the corresponding proper inner analytic function $w(z)$ within the open unit disk, using the power series $S(z)$ given in Equation (6), which, as was shown in [#!CAoRFI!#], always converges for $\vert z\vert<1$. Considering that $c_{0}\equiv 0$, the partial sums of the first $N$ terms of this series are given by

$$S_{N}(z) = \sum_{k=0}^{N} c_{k}z^{k},$$ (51)

where $N\in\{1,2,3,\ldots,\infty\}$, a complex sequence for each value of $z$ which, for $\vert z\vert<1$, we already know to converge to $w(z)$ in the $N\to\infty$ limit. Note however that, since $S_{N}(z)$ is a polynomial of order $N$ and therefore an analytic function over the whole complex plane, this expression itself can be consistently considered for all finite $N$ and all $z$, and in particular for $z$ on the unit circle, where $\vert z\vert=1$. One can also define the corresponding remainders of the complex power series, in the usual way, as

$$R_{N}(z) = w(z) - S_{N}(z).$$ (52)

We will now prove the following theorem.

Theorem 5   : Given an arbitrary zero-average integrable real function $f(\theta)$ on the unit circle and the corresponding Fourier-conjugate real function $g(\theta)$, if $S_{N}^{F}(\theta)=\Re[S_{N}(1,\theta)]$ is the $N^{\rm th}$ partial sum of the Fourier series of $f(\theta)$, and if $R_{N}^{F}(\theta)=\Re[R_{N}(1,\theta)]$ is the corresponding remainder of that Fourier series, then we have that this partial sum and this remainder are given by the following integrals:

$$S_{N}^{F}(\theta) = \frac{1}{2\pi}\, \int_{-\pi}^{\pi}d\theta_{1}\, \frac { \sin\!\le... ...0em}{2ex}(N+1/2)\Delta\theta\right] } { \sin(\Delta\theta/2) }\, f(\theta_{1}),$$

$$R_{N}^{F}(\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...0em}{2ex}(N+1/2)\Delta\theta\right] } { \sin(\Delta\theta/2) }\, g(\theta_{1}),$$ (53)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$.

Note that the integral in the expression of the partial sum is the known Dirichlet integral, while the one in the expression of the reminder is similar but not identical to it.

Proof 5.1   :

In order to prove this theorem, let us consider the complex partial sums $S_{N}(z)$ as given in Equation (51). In addition to this, the complex coefficients $c_{k}$ may be written as integrals involving $w(z)$, with the use of the Cauchy integral formulas,

$$c_{k} = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz\, \frac{w(z)}{z^{k+1}},$$ (54)

for $k\in\{0,1,2,3,\ldots,\infty\}$, where $C$ can be taken as a circle centered at the origin, with radius $\rho\leq 1$. The reason why we may include the case $\rho=1$ here is that, as was shown in [#!CAoRFI!#], as a function of $\rho$ the expression above for $c_{k}$ is not only constant within the open unit disk, but also continuous from within at the unit circle. In this way the coefficients $c_{k}$ may be written back in terms of the inner analytic function $w(z)$. If we substitute this expression for $c_{k}$ back in the partial sums of the complex power series shown in Equation (51) we get

$$S_{N}(z) = \sum_{k=0}^{N} z^{k}\, \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{k+1}}$$

$$= \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}} \sum_{k=0}^{N} \left( \frac{z}{z_{1}} \right)^{k},$$ (55)

where $z$ can have any value, but where we must have $\vert z_{1}\vert\leq 1$. The sum is now a finite geometric progression, so that we have its value in closed form,

$$S_{N}(z) = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}}\, \frac {1-(z/z_{1})^{N+1}} {1-(z/z_{1})}$$

$$= \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(... ...ox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)}.$$ (56)

There are now two relevant cases to be considered here, the case in which $\vert z\vert<\vert z_{1}\vert$ and the case in which $\vert z\vert>\vert z_{1}\vert$. In the first case, since the explicit simple pole of the integrand at the position $z_{1}=z$ lies within the integration contour, we have in the first term the Cauchy integral formula for $w(z)$, and therefore we get

$$S_{N}(z) = w(z) - \frac{z^{N+1}}{2\pi\mbox{\boldmath$\i... ...$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)}.$$ (57)

This is the equation that allows us to write an explicit expression for the remainder of the complex power series within the open unit disk, thus making it easier to discuss its convergence there. In the other case, in which $\vert z\vert>\vert z_{1}\vert$, the explicit simple pole of the integrand at the position $z$ lies outside of the integration contour, and therefore by the Cauchy-Goursat theorem we just have zero in the first term, so that we get

$$S_{N}(z) = -\, \frac{z^{N+1}}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)}.$$ (58)

This provides us, therefore, with an explicit expression for the partial sums, but not for the remainder. The only other possible case is that in which $\vert z\vert=\vert z_{1}\vert$, in which both $z_{1}$ and $z$ are over the circle $C$ of radius $\rho_{1}$, and therefore so is the explicit simple pole of the integrand at the position $z_{1}=z$. In this case, just as we did before in Section 3, we may slightly deform the integration contour $C$ in order to have it pass on one side or the other of the simple pole of the integrand at $z_{1}=z$. If we use a deformed contour $C_{\ominus}$ that excludes the pole from its interior, then we have, instead of Equation (57),

$$S_{N}(z) = 0 - \frac{z^{N+1}}{2\pi\mbox{\boldmath$\imat... ..._{C_{\ominus}}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)},$$ (59)

while if we use a deformed contour $C_{\oplus}$ that includes the pole in its interior, then we have, just as in Equation (57),

$$S_{N}(z) = w(z) - \frac{z^{N+1}}{2\pi\mbox{\boldmath$\i... ...t_{C_{\oplus}}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)}.$$ (60)

Once more, since by the Sokhotskii-Plemelj theorem [#!sokhplem!#] the Cauchy principal value of the integral over $C$ is the arithmetic average of these two integrals, taking the average of Equations (59) and (60) we obtain the expression

$$S_{N}(z) = \frac{w(z)}{2}\, - \frac{z^{N+1}}{2\pi\mbox{... ...}\!\!\oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)},$$ (61)

where both $z_{1}$ and $z$ are now on the circle $C$ of radius $\rho_{1}$. Since we have that the corresponding remainder of the series is defined as given in Equation (52), we get a corresponding expression for the remainder, in terms of the same integral,

$$R_{N}(z) = \frac{w(z)}{2}\, + \frac{z^{N+1}}{2\pi\mbox{... ...}\!\!\oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)},$$ (62)

where both $z_{1}$ and $z$ are on the circle $C$ of radius $\rho_{1}$. We have therefore the pair of equations

$$S_{N}(z) = \frac{w(z)}{2}\, - I_{N}(z),$$

$$R_{N}(z) = \frac{w(z)}{2}\, + I_{N}(z),$$ (63)

and we must now write the integral $I_{N}(z)$ explicitly in terms of $\rho_{1}$, $\theta_{1}$ and $\theta$,

$$I_{N}(z) = \frac{z^{N+1}}{2\pi\mbox{\boldmath$\imath$}}\, \mbox{\rm PV}\!\!\oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}^{N+1}(z_{1}-z)}$$

$$= \frac{\rho_{1}^{N+1}\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$... ...{1}} - \rho_{1} \,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\theta} \right) }$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...,\theta_{1})} {1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}},$$ (64)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$. Once again we must rationalize the integrand, and using once more the result shown in Equation (22) we get

$${ I_{N}(\rho_{1},\theta) }$$

$$= \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ... e}^{\mbox{\boldmath\scriptsize$\imath$}\Delta\theta/2}} {\sin(\Delta\theta/2)}$$

$$= \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...\mbox{\boldmath\scriptsize$\imath$}(N+1/2)\Delta\theta}} {\sin(\Delta\theta/2)}$$

$$= \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...ox{\boldmath$\imath$} \sin(N_{1}\Delta\theta) \right] } {\sin(\Delta\theta/2)},$$ (65)

where $\Delta\theta=\theta_{1}-\theta$ and $N_{1}=N+1/2$, with $N\in\{1,2,3,\ldots,\infty\}$. Expanding the numerator in the integrand of this integral we have

$${ \left[ \rule{0em}{2ex} v(\rho_{1},\theta_{1}) - \mbox{\boldmath... ...{1}\Delta\theta) - \mbox{\boldmath$\imath$} \sin(N_{1}\Delta\theta) \right] }$$

$$= - \left[ \rule{0em}{2ex} \sin(N_{1}\Delta\theta) u(\rho_{1},\theta_{1}) - \cos(N_{1}\Delta\theta) v(\rho_{1},\theta_{1}) \right] +$$

$$- \mbox{\boldmath$\imath$} \left[ \rule{0em}{2ex} \sin(N_{1}\Delt... ...(\rho_{1},\theta_{1}) + \cos(N_{1}\Delta\theta) u(\rho_{1},\theta_{1}) \right],$$ (66)

and therefore we are left with the following expression for our integral,

$${ I_{N}(\rho_{1},\theta) }$$

$$= -\, \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1... ...1}) - \cos(N_{1}\Delta\theta) v(\rho_{1},\theta_{1}) } {\sin(\Delta\theta/2)} +$$

$$-\, \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int... ...{1}) + \cos(N_{1}\Delta\theta) u(\rho_{1},\theta_{1}) } {\sin(\Delta\theta/2)},$$ (67)

where $\Delta\theta=\theta_{1}-\theta$ and $N_{1}=N+1/2$, with $N\in\{1,2,3,\ldots,\infty\}$. Once again we note that, since by construction the real and imaginary parts $u(\rho_{1},\theta)$ and $v(\rho_{1},\theta)$ of $w(\rho_{1},\theta)$ for $\rho_{1}=1$ are integrable real functions on the unit circle, and since there are no other dependencies on $\rho_{1}$ in this equation, we may now take the $\rho_{1}\to 1_{(-)}$ limit of this expression, in which the principal value acquires its usual real meaning on the unit circle, thus obtaining

$${ I_{N}(1,\theta) }$$

$$= -\, \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1... ...{0em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } {\sin(\Delta\theta/2)} +$$

$$-\, \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int... ...e{0em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } {\sin(\Delta\theta/2)},$$ (68)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$, in terms of which we now have the pair of equations at the unit circle

$$S_{N}(1,\theta) = \frac{w(1,\theta)}{2}\, - I_{N}(1,\theta),$$

$$R_{N}(1,\theta) = \frac{w(1,\theta)}{2}\, + I_{N}(1,\theta).$$ (69)

Using now the infinite collection of identities in Equation (50) for the case $k=N$, which allow us to write $w(1,\theta)/2$ in terms of integrals similar to those in $I_{N}(1,\theta)$,

$${ \frac{w(1,\theta)}{2} }$$

$$= \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$+ \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int_{... ...0em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } { \sin(\Delta\theta/2) },$$ (70)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$, we may write for the complex partial sums

$${ S_{N}(1,\theta) }$$

$$= \frac{w(1,\theta)}{2}\, - I_{N}(1,\theta)$$

$$= \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$+ \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int_{... ...em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$+ \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\... ...{0em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } {\sin(\Delta\theta/2)} +$$

$$+ \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int_{... ...e{0em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } {\sin(\Delta\theta/2)}.$$ (71)

As one can see in this equation, all the terms involving $\cos\!\left[\rule{0em}{2ex}(N+1/2)\Delta\theta\right]$ cancel off, and therefore we are left with

$$S_{N}(1,\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$+ \frac{\mbox{\boldmath$\imath$}}{2\pi}\, \mbox{\rm PV}\!\!\int_{... ...0em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } { \sin(\Delta\theta/2) },$$ (72)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$. We now observe that, since by hypothesis $f(\theta)$ is integrable on the unit circle, the Cauchy principal value refers only to the possible explicit non-integrable singularity of the integrands, due to the zero of the denominators at $\theta_{1}=\theta$. However, since the numerators of the integrands are also zero at that point, the integrands are not really divergent at all at that point, so that from this point on we may drop the principal value. We have therefore our final results for the real partial sums, for both $u(1,\theta)$ and $v(1,\theta)$,

$$S_{N}^{F,u}(\theta) = \Re[S_{N}(1,\theta)]$$

$$= \frac{1}{2\pi}\, \int_{-\pi}^{\pi}d\theta_{1}\, \frac { \sin\!\le... ...m}{2ex}(N+1/2)\Delta\theta\right] } { \sin(\Delta\theta/2) }\, u(1,\theta_{1}),$$

$$S_{N}^{F,v}(\theta) = \Im[S_{N}(1,\theta)]$$

$$= \frac{1}{2\pi}\, \int_{-\pi}^{\pi}d\theta_{1}\, \frac { \sin\!\le... ...m}{2ex}(N+1/2)\Delta\theta\right] } { \sin(\Delta\theta/2) }\, v(1,\theta_{1}),$$ (73)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$. These are the well-known results for the partial sums, in terms of Dirichlet integrals [#!FSchurchill!#]. Note that the two equations above have exactly the same form, which is to be expected, since the result holds for all zero-average integrable real functions, including of course both $u(1,\theta_{1})$ and $v(1,\theta_{1})$. Therefore, given an arbitrary zero-average integrable real function $f(\theta)$ on the unit circle, we have that the partial sums of its Fourier series are given by

$$S_{N}^{F}(\theta) = \frac{1}{2\pi}\, \int_{-\pi}^{\pi}d\... ...\rule{0em}{2ex}(\theta_{1}-\theta)/2\right]}\, f(\theta_{1}),$$ (74)

where $N\in\{1,2,3,\ldots,\infty\}$. Note that, although this result is already very well known, we have showed here that it does follow from our complex-analytic structure. This completes the proof of the first part of Theorem 5.

Once more, it is interesting to observe that this relation can be interpreted as a linear integral operator acting on the space of zero-average integrable real functions defined on the unit circle, this time resulting in the $N^{\rm th}$ partial sum of the Fourier series of a zero-average integrable real function, a partial sum which is itself a zero-average integrable real function. The integration kernel of this integral operator ${\cal D}_{\rm s}[N,f(\theta)]$ depends only on $N$ and on the difference $\theta-\theta_{1}$, and is given by

$$K_{{\cal D}_{\rm s}}(N,\theta-\theta_{1}) = \frac{1}{2\pi... ...]} {\sin\!\left[\rule{0em}{2ex}(\theta_{1}-\theta)/2\right]},$$ (75)

where $N\in\{1,2,3,\ldots,\infty\}$, so that the action of the operator on $f(\theta)$ can be written as

$${\cal D}_{\rm s}[N,f(\theta)] = \int_{-\pi}^{\pi}d\theta_{1}\, K_{{\cal D}_{\rm s}}(N,\theta-\theta_{1}) f(\theta_{1}).$$ (76)

Considering that its kernel is given by a Dirichlet integral, one might call this the Dirichlet operator, so that the $N^{\rm th}$ partial sum of the Fourier series of $f(\theta)$ is given by the action of this operator on the zero-average integrable real function $f(\theta)$,

$$S_{N}^{F}(\theta) = {\cal D}_{\rm s}[N,f(\theta)],$$ (77)

where $N\in\{1,2,3,\ldots,\infty\}$. Note that ${\cal D}_{\rm s}[N,f(\theta)]$ constitutes in fact a whole collection of linear integral operators acting on the space of zero-average integrable real functions.

Proof 5.2   :

Using once more the very same elements that were used above for the complex partial sums, we may also write corresponding results for the complex remainders,

$${ R_{N}(1,\theta) }$$

$$= \frac{w(1,\theta)}{2}\, + I_{N}(1,\theta)$$

$$= \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$+ \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int_{... ...em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$-\, \frac{1}{4\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1... ...{0em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } {\sin(\Delta\theta/2)} +$$

$$-\, \frac{\mbox{\boldmath$\imath$}}{4\pi}\, \mbox{\rm PV}\!\!\int... ...e{0em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } {\sin(\Delta\theta/2)}.$$ (78)

As one can see in this equation, this time all the terms involving $\sin\!\left[\rule{0em}{2ex}(N+1/2)\Delta\theta\right]$ chancel off, and therefore we are left with

$$R_{N}(1,\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...em}{2ex}(N+1/2)\Delta\theta\right] v(1,\theta_{1}) } { \sin(\Delta\theta/2) } +$$

$$-\, \frac{\mbox{\boldmath$\imath$}}{2\pi}\, \mbox{\rm PV}\!\!\int... ...0em}{2ex}(N+1/2)\Delta\theta\right] u(1,\theta_{1}) } { \sin(\Delta\theta/2) },$$ (79)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$. We have therefore our results for the real remainders, for both $u(1,\theta)$ and $v(1,\theta)$,

$$R_{N}^{F,u}(\theta) = \Re[R_{N}(1,\theta)]$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...m}{2ex}(N+1/2)\Delta\theta\right] } { \sin(\Delta\theta/2) }\, v(1,\theta_{1}),$$

$$R_{N}^{F,v}(\theta) = \Im[R_{N}(1,\theta)]$$

$$= -\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1... ...m}{2ex}(N+1/2)\Delta\theta\right] } { \sin(\Delta\theta/2) }\, u(1,\theta_{1}),$$ (80)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$. Recalling that $f(\theta)=u(1,\theta)$ and that $g(\theta)=v(1,\theta)$ almost everywhere over the unit circle, this completes the proof of Theorem 5.

We believe that these are new results, written in terms of integrals which are similar to the Dirichlet integrals, but not identical to them. Note that the remainder of the series of $f(\theta)$ is given as an integral involving its Fourier-conjugate function $g(\theta)$, and vice versa. Therefore, we conclude that the convergence condition of the Fourier series of a given real function does not depend directly on that function, but only indirectly, through the properties of its Fourier-conjugate real function.

Since we know that these two real functions are related by the compact Hilbert transform, we may write these equations as

$$R_{N}^{F,u}(\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...ta\theta\right] } { \sin(\Delta\theta/2) }\, {\cal H}_{\rm c}[u(1,\theta_{1})],$$

$$R_{N}^{F,v}(\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...ta\theta\right] } { \sin(\Delta\theta/2) }\, {\cal H}_{\rm c}[v(1,\theta_{1})],$$ (81)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$. Note that the two results are now identical in form. Therefore, given an arbitrary zero-average integrable real function $f(\theta)$ on the unit circle, we have our final result for the remainder of its Fourier series,

$$R_{N}^{F}(\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\i... ...{ \sin(\Delta\theta/2) }\, {\cal H}_{\rm c}[f(\theta_{1})],$$ (82)

where $\Delta\theta=\theta_{1}-\theta$ and $N\in\{1,2,3,\ldots,\infty\}$.

Once again, it is interesting to observe that this relation can be interpreted as a linear integral operator acting on the space of integrable zero-average real functions defined on the unit circle. The operator ${\cal D}_{\rm c}[N,g(\theta)]$, acting on the compact Hilbert transform $g(\theta)$ of such a function, results in the $N^{\rm th}$ remainder of the Fourier series of the original function $f(\theta)$, a remainder which, if it exists at all, is itself a zero-average integrable real function. The integration kernel of the integral operator depends only on $N$ and on the difference $\theta-\theta_{1}$, and is given by

$$K_{{\cal D}_{\rm c}}(N,\theta-\theta_{1}) = \frac{1}{2\pi... ...]} {\sin\!\left[\rule{0em}{2ex}(\theta_{1}-\theta)/2\right]},$$ (83)

where $N\in\{1,2,3,\ldots,\infty\}$, so that the action of the operator on an arbitrarily given zero-average integrable real function $g(\theta)$ can be written as

$${\cal D}_{\rm c}[N,g(\theta)] = \mbox{\rm PV}\!\!\int_{-\... ...}\, K_{{\cal D}_{\rm c}}(N,\theta-\theta_{1}) g(\theta_{1}).$$ (84)

This new operator, which we might refer to as the conjugate Dirichlet operator, is similar to the Dirichlet operator, and is such that the $N^{\rm th}$ remainder of the Fourier series of the real function $f(\theta)$ is given by the action of this operator on the Fourier-conjugate function $g(\theta)$ of the real function $f(\theta)$,

$$R_{N}^{F}(\theta) = {\cal D}_{\rm c}[N,g(\theta)],$$ (85)

where $N\in\{1,2,3,\ldots,\infty\}$. Note once more that ${\cal D}_{\rm c}[N,g(\theta)]$ constitutes in fact a whole collection of linear integral operators acting on the space of zero-average integrable real functions. Note also that, since by hypothesis $f(\theta)$ and $g(\theta)$ are integrable on the unit circle, the Cauchy principal value refers only to the explicit non-integrable singularity of the integration kernel at $\theta_{1}=\theta$. In this operator notation we have therefore that the remainder of the Fourier series of an arbitrarily given zero-average integrable real function $f(\theta)$ is given by the composition of ${\cal D}_{\rm c}[N,g(\theta)]$ with ${\cal H}_{\rm c}[f(\theta)]$,

$$R_{N}^{F}(\theta) = {\cal D}_{\rm c}\!\left[\rule{0em}{2ex}N,{\cal H}_{\rm c}[f(\theta)]\right].$$ (86)

Note that, according to the inverse of the relation shown in Equation (42) we may write as well that

$$R_{N}^{F}(\theta) = {\cal H}_{\rm c}^{-1}\!\left[\rule{0em}{2ex}{\cal D}_{\rm c}[N,f(\theta)]\right],$$ (87)

which constitutes an equivalent way to express the remainder of the Fourier series of $f(\theta)$ in terms of the function itself.

Finally note that, given the results obtained here for the partial sums and remainders of the Fourier series, the expressions in the infinite collection of identities shown in Equation (50) have now, in fact, the very simple interpretation that was alluded to there, since we now see that they can in fact be written as

$$f(\theta) = S_{N}^{F,f}(\theta) + R_{N}^{F,f}(\theta),$$

$$g(\theta) = S_{N}^{F,g}(\theta) + R_{N}^{F,g}(\theta),$$ (88)

where $N\in\{1,2,3,\ldots,\infty\}$, a fact which greatly clarifies the nature of that infinite collections of identities.