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In my notes, the definition of the Fejer kernel is $$ F_{n} = \sum_{j=-N}^{N} \left(1 - \frac{|j|}{N+1}\right) e^{ijt}. $$

But in most of the reference material I come across online, it is immediately defined as the average of the Dirichlet kernels

$$ F_{N} = \frac{1}{N+1} \left(D_{0} + \dots + D_{N}\right). $$

I've tried equating these two definitions by expanding $F_{n}$'s $e^{ijt}$ and using some trigonometry to get something looking like the $\sin$ representation of the Dirichlet kernel but it has not been going well.

Is there a simple way to prove that these two definitions are equivalent?

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  • $\begingroup$ Not sure what you're asking, you want a proof for $F_N(y)=\frac{1}{N+1} \bigg( \frac{\sin (\frac{N+1}{2})y)}{\sin (\frac{y}{2})} \bigg) ^2 \ ?$ $\endgroup$ – Itay4 Aug 4 '17 at 7:03
  • $\begingroup$ @Itay4 Sorry if I wasn't clear, I'm just trying to establish equivalence of the two definitions I outlined. $\endgroup$ – Eric Hansen Aug 4 '17 at 7:04
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$\begin{array}{rlll}F_N(y)~&=~ \frac{1}{N+1}\sum_{n=0}^ND_n(y)=\frac{1}{N+1}\sum_{n=0}^N\sum_{k=-n}^ne^{iky} \\ &=~ \frac{1}{N+1}\sum_{m=-N}^Ne^{imy}\#\{k \in [0,N]:m \in [-k,k]\} \\ &=~ \frac{1}{N+1}\sum_{m=-N}^Ne^{imy}(N+1-|m|)=\sum_{m=-N}^Ne^{imy}(1-\frac{|m|}{N+1}) \end{array}$

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  • $\begingroup$ Thank you! Would you mind explaining to me what your $\#$ notation is on the second line? $\endgroup$ – Eric Hansen Aug 4 '17 at 7:32
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    $\begingroup$ @EricHansen That's the counting of Dirichlet's kernels which is well explained in uniquesolution's answer. $\endgroup$ – Itay4 Aug 4 '17 at 7:34
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It is sufficient to compare the coefficients of $e^{ijx}$ in both expressions. What is the coefficient of $e^{ijx}$ in the second expression? Since $$D_n=\sum_{k=-n}^n e^{ikx}$$ the number of Dirichlet Kernels in the average that contain $e^{ijx}$ is clearly $N-j+1$, (because for $k<j$ the Dirichlet kernel $D_k$ does not contain $e^{ijx}$), and the coefficient is $1$ in each kernel, so all in all we have $\frac{1}{N+1}(N-j+1)=1-\frac{j}{N+1}$. Here we assumed $j>0$. Similarly, the number of times $e^{-ijx}$ appears in the second expression is also $N-j+1$, hence the expression $1-\frac{|j|}{N+1}$.

When I posted this I wasn't aware of Itay4's answer, which is essentially the same counting argument (although somewhat condensed).

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