# On an Optimal Quadrature Formula in a Hilbert Space of Periodic Functions
## 5 Optimal Coefficients of the Quadrature Formula (4)  
In the present section, we prove Theorem 2. In order to prove the theorem, it is enough to solve the systems (38) and (39) with respect to coefficients \(C_{k},k=1,2,\ldots,N\). For a solution of this system, we search in the form  
\[\hat{C}_{k}=C(\omega,N,m)\exp\biggl{(}\mathrm{i}\omega\frac{2\pi k}{N}\biggr{)}, \tag{40}\]  
where \(C(\omega,N,m)\) is an unknown function. We are directly convinced that \(\hat{C}_{k}\) values, defined by Formula (40), \(k=1,2,\ldots,N\), satisfy Equality (39). Substituting (40) into (38), we obtain  
\[\frac{-1}{\omega^{2m}}\exp\biggl{(}\frac{2\pi\mathrm{i}\omega k^{\prime}}{N} \biggr{)}+C(\omega,N,m)\sum_{k=1}^{N}\exp\biggl{(}\frac{\mathrm{i}\omega 2\pi k}{N}\biggr{)}\sum_{\beta \
eq 0}\frac{\exp\biggl{(}\frac{2\pi\mathrm{i}\beta(k-k^{\prime})}{N} \biggr{)}}{2\pi\beta^{2m}}=0. \tag{41}\]  
We introduce the following notation:  
\[z=C(\omega,N,m)\sum_{\beta\
eq 0}\frac{\exp\biggl{(}\frac{-2\pi\mathrm{i}\beta k ^{\prime}}{N}\biggr{)}}{2\pi\beta^{2m}}\sum_{k=1}^{N}\exp\biggl{(}\frac{2\pi \mathrm{i}k}{N}(\omega+\beta)\biggr{)}. \tag{42}\]  
It is clear that  
\[\sum_{k=1}^{N}\exp\biggl{(}\frac{2\pi\mathrm{i}k}{N}(\omega+\beta)\biggr{)}= \left\{ \begin{array}{ll}0,&\text{ if }\frac{\omega+\beta}{N}\text{ is not integer,}\\ N,&\text{ if }\frac{\omega+\beta}{N}\text{ is integer.}\end{array} \right.\]  
Therefore, denoting \(t=\frac{\omega+\beta}{N}\) and then \(\beta=tN-\omega\), we rewrite (42) in the form  
\[z = \frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{-2\pi\mathrm{i}(tN -\omega)k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}} \tag{43}\] \[= \frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{2\pi\mathrm{i} \omega k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}}.\]  
From (41)-(43), it follows that  
\[\frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{2\pi\mathrm{i} \omega k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}}-\frac{1}{\omega^{2m}}\exp\biggl{(}\frac{2\pi\mathrm{i}k^{\prime}\omega }{N}\biggr{)}=0,\] \[k^{\prime}=1,2,\ldots,N.\]  
Hence, we have  
\[\frac{1}{\omega^{2m}}\exp\biggl{(}\frac{2\pi\mathrm{i}\omega k^{ \prime}}{N}\biggr{)}\biggl{[}\frac{N}{2\pi}C(\omega,N,m)\sum_{t=-\infty}^{ \infty}\frac{1}{(1-\frac{tN}{\omega})^{2m}}-1\biggr{]}=0,\] \[k^{\prime}=1,2,\ldots,N.\]  
From the uniqueness of the solution for the systems (38) and (39), we obtain  
\[C(\omega,N,m) = \frac{2\pi}{N}\Biggl{(}\sum_{t=-\infty}^{\infty}\frac{1}{(1- \frac{tN}{\omega})^{2m}}\Biggr{)}^{-1}.\]Hence, we directly obtain  
\[C(\omega,N,m) = \frac{2\pi}{N}\bigg{(}\frac{N}{\omega}\bigg{)}^{2m}\bigg{(}\sum_{t =-\infty}^{\infty}\frac{1}{(t-\frac{\omega}{N})^{2m}}\bigg{)}^{-1}. \tag{44}\]  
We now refer to calculations of  
\[f(z) = \bigg{(}\sum_{t=-\infty}^{\infty}\frac{1}{(t-z)^{2m}}\bigg{)}^{-1}. \tag{45}\]  
As is known [11],  
\[\sum_{t=-\infty}^{\infty}\frac{1}{(z-t)^{2}} = \bigg{(}\frac{\pi}{\sin(\pi z)}\bigg{)}^{2}. \tag{46}\]  
Taking the derivative of order \(2m-2\) with respect to \(z\) from both sides of (46) and taking (45) into account, we obtain  
\[= \frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{2\pi\mathrm{i} \omega k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}}.\]0  
Now we calculate \(\frac{d^{2m-2}}{dz^{2m-2}}\Big{(}\frac{\pi}{\sin\left(\pi z\right)}\Big{)}^{2}\). For this, we use [PERSON]'s known formula  
\[= \frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{2\pi\mathrm{i} \omega k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}}.\]1  
Denoting \(\lambda:=\exp(2\pi{\rm i}z)\), we have  
\[= \frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{2\pi\mathrm{i} \omega k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}}.\]2  
\[= \frac{N}{2\pi}C(\omega,N,m)\exp\biggl{(}\frac{2\pi\mathrm{i} \omega k^{\prime}}{N}\biggr{)}\sum_{t=-\infty}^{\infty}\frac{1}{(tN-\omega) ^{2m}}.\]3  
where \(D=\lambda\frac{d}{d\lambda}\), and \(D^{2m-2}=\lambda\frac{d}{d\lambda}D^{2m-3}\).