# extracting every ${n}^{\mathrm{th}}$ term of a series

Roots of unity^{} can be used to extract every ${n}^{\mathrm{th}}$ term of a series. This method is due to Simpson [1759].

Theorem. Let $\omega ={e}^{2\pi i/k}$ be a primitive ${k}^{\mathrm{th}}$ root of unity. If $f(x)={\sum}_{j=0}^{\mathrm{\infty}}{a}_{j}{x}^{j}$ and $n\not\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modk)$, then

$$\sum _{j=0}^{\mathrm{\infty}}{a}_{kj+n}{x}^{kj+n}=\frac{1}{k}\sum _{j=0}^{k-1}{\omega}^{-jn}f({\omega}^{j}x)$$ |

Proof. This is a consequence of the fact that ${\sum}_{j=0}^{k-1}{\omega}^{jm}=0$ for $m\not\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modk)$.

Consider the term involving ${x}^{r}$ on the right-hand side. It is

$$\frac{1}{k}\sum _{j=0}^{k-1}{\omega}^{-jn}{a}_{r}{\omega}^{jr}{x}^{r}=\frac{1}{k}{a}_{r}{x}^{r}\sum _{j=0}^{k-1}{\omega}^{j(r-n)}$$ |

If $r\not\equiv n\phantom{\rule{veryverythickmathspace}{0ex}}(modk)$, the sum is zero. So the term involving ${x}^{r}$ is zero unless $r\equiv n\phantom{\rule{veryverythickmathspace}{0ex}}(modk)$, in which case it is ${a}_{r}{x}^{r}$ since each element of the sum is $1$.

Note that this method is a generalization^{} of the commonly known trick for extracting alternate terms of a series:

$$\frac{1}{2}(f(x)-f(-x))$$ |

produces the odd terms of $f$.

Title | extracting every ${n}^{\mathrm{th}}$ term of a series |
---|---|

Canonical name | ExtractingEveryNmathrmthTermOfASeries |

Date of creation | 2013-03-22 16:23:34 |

Last modified on | 2013-03-22 16:23:34 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 10 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 11-00 |