Zurück zu Identitäten
Aus f ( ξ ℓ z ) = ∑ k ∈ Z a k ξ ℓ k z k {\displaystyle f(\xi ^{\ell }z)=\sum _{k\in \mathbb {Z} }a_{k}\,\xi ^{\ell k}\,z^{k}} folgt 1 n ∑ ℓ = 0 n − 1 f ( ξ ℓ z ) ξ ℓ m = ∑ k ∈ Z a k 1 n ∑ ℓ = 0 n − 1 ξ ℓ ( k − m ) z k {\displaystyle {\frac {1}{n}}\sum _{\ell =0}^{n-1}{\frac {f(\xi ^{\ell }z)}{\xi ^{\ell m}}}=\sum _{k\in \mathbb {Z} }a_{k}\,{\frac {1}{n}}\sum _{\ell =0}^{n-1}\xi ^{\ell (k-m)}\,z^{k}} . Und wegen 1 n ∑ ℓ = 0 n − 1 ξ ℓ ( k − m ) = { 1 falls k ∈ [ m ] n 0 falls k ∉ [ m ] n {\displaystyle {\frac {1}{n}}\sum _{\ell =0}^{n-1}\xi ^{\ell (k-m)}=\left\{{\begin{matrix}1&{\text{falls}}&k\in [m]_{n}\\\\0&{\text{falls}}&k\notin [m]_{n}\end{matrix}}\right.} ist dies ∑ k ∈ [ m ] n a k z k {\displaystyle \sum _{k\in [m]_{n}}a_{k}z^{k}} .
![{\displaystyle \sum _{k\in [m]_{n}}a_{k}z^{k}={\frac {1}{n}}\,\sum _{\ell =0}^{n-1}{\frac {f(\xi ^{\ell }z)}{\xi ^{\ell m}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe833123ed3685cc18014d13f81667a512e29fe1)
![{\displaystyle {\frac {1}{n}}\sum _{\ell =0}^{n-1}\xi ^{\ell (k-m)}=\left\{{\begin{matrix}1&{\text{falls}}&k\in [m]_{n}\\\\0&{\text{falls}}&k\notin [m]_{n}\end{matrix}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c70aceea98ffdf41b6afe9138304f4af00633e82)
![{\displaystyle \sum _{k\in [m]_{n}}a_{k}z^{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a5da811e5b642f4cd344161601c387ed65b5b43)