Formelsammlung Mathematik: Identitäten: Simpsonsche Extraktionsmethode

Ist

f(z)=\sum _{k\in \mathbb {Z} }a_{k}z^{k}

, so gilt

\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}}}

mit

\xi =e^{\frac {2\pi i}{n}}

.

Beweis

Aus $f(\xi ^{\ell }z)=\sum _{k\in \mathbb {Z} }a_{k}\,\xi ^{\ell k}\,z^{k}$ folgt ${\frac {1}{n}}\sum _{\ell =0}^{n-1}{\frac {f(\xi ^{\ell }z)}{\xi ^{\ell m}}}=\sum _{k\in \mathbb {Z} }a_{k}\,{\frac {1}{n}}\sum _{\ell =0}^{n-1}\xi ^{\ell (k-m)}\,z^{k}$ .

Und wegen ${\frac {1}{n}}\sum _{\ell =0}^{n-1}\xi ^{\ell (k-m)}=\left\{{\begin{matrix}1&{\text{falls}}&k\in [m]_{n}\\\\0&{\text{falls}}&k\notin [m]_{n}\end{matrix}}\right.$ ist dies $\sum _{k\in [m]_{n}}a_{k}z^{k}$ .

This article is issued from Wikibooks. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.