0

Show that the Fejer kernel $$F_N(x) = \sum_{-N}^{N} \left( 1 - \dfrac{|n|}{N}\right)e_n$$ where $e_n(x) = e^{2\pi i n x}$ is the trigonometric monomial, can be written as $$F_N(x) = \frac{e^{\pi i (N-1)x}\sin (\pi N x)}{\sin (\pi x)}$$ on the interval $(0,1)$.

Until so far I have researched, found and proven that $$F_N(x) = \frac{1}{N} \left|\sum_{i=0}^{N-1}e_n \right|^2$$ which is a gread leap for me, but not really for the whole proof. I also found a hint saying that I should use the geometric sums $$\sum_{i=0}^{N-1}e_n = \frac{e_N - e_0}{e_1 - e_0}$$ However, I do not know how that helps me. Any hints on how I can proceed would be greatly appreciated!

qarabala
  • 1,131

1 Answers1

1

The claim is wrong. As the result is a real function, there can be no non-trivial phase factor. The claimed sum formula could be for some intermediate result, for instance the geometric sum that you tried to compute next.

To advance to the next step, use that $$ e_k(x)-e_0(x)=e_{k/2}(x)(e_{k/2}(x)-e_{-k/2}(x))=2ie_{k/2}(x)\sin(k\pi x). $$

Lutz Lehmann
  • 131,652