| Version 2 |
Version 1 |
| The Fejer kernel $F_n$ of order $n$ is defined as |
The Fejer kernel $F_n$ of order $n$ is defined as |
| $$F_n(t)=\frac{1}{n}\sum_{k=0}^{n-1}D_k(t),$$ |
$$F_n(t)=\frac{1}{n}\sum_{k=0}^{n-1}D_k(t),$$ |
| where $D_n$ is the Dirichlet kernel of order $n$. The Fejer kernel can be written as |
where $D_n$ is the Dirichlet kernel of order $n$. The Fejer kernel can be written as |
| \begin{equation}\label{eq:rep} |
\begin{equation}\label{eq:rep} |
| F_n(t)=\frac{1}{n}\left(\frac{\sin\frac{nt}{2}}{\sin\frac{t}{2}}\right)^2. |
F_n(t)=\frac{1}{n}\left(\frac{\sin\frac{nt}{2}}{\sin\frac{t}{2}}\right)^2. |
| \end{equation} |
\end{equation} |
| \textbf{Proof:} Since |
\textbf{Proof:} Since |
| $$D_n(t)=\frac{\sin\left(\left(n+\frac{1}{2}\right)t\right)}{\sin\frac{t}{2}}$$ |
$$D_n(t)=\frac{\sin\left(\left(n+\frac{1}{2}\right)t\right)}{\sin\frac{t}{2}}$$ |
| we have |
we have |
| $$\sin\frac{t}{2}D_n(t)=\sin\left(\left(n+\frac{1}{2}\right)t\right).$$ |
$$\sin\frac{t}{2}D_n(t)=\sin\left(\left(n+\frac{1}{2}\right)t\right).$$ |
| Therefore |
Therefore |
| \begin{align*} |
\begin{align*} |
|
n\sin^2\frac{t}{2}F_n(t)& =\sum_{k=0}^{n-1}\sin\left(\left(k+\frac{1}{2}\right)t\right)\sin\frac{t}{2}\\
|
n\sin^2\frac{t}{2}F_n(t)& =\sum_{k=0}^{n-1}\sin\left(\left(k+\frac{1}{2}\right)t\right)\sin\frac{x}{2}\\
|
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&=\frac{1}{2}\sum_{k=0}^{n-1}(\cos kt-\cos((k+1)t)\\
|
&=\frac{1}{2}\sum_{k=0}^{n-1}(\cos kx-\cos((k+1)x))\\
|
|
&=\frac{1}{2}(1-\cos nt)\\
|
&=\frac{1}{2}(1-\cos nx)\\
|
|
&=\sin^2\frac{nt}{2}.
|
&=\sin^2\frac{nx}{2}.
|
| \end{align*} |
\end{align*} |
| From this follows equation (\ref{eq:rep}). |
From this follows equation (\ref{eq:rep}). |
