# 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].

Let $\omega=e^{2\pi i/k}$ be a primitive $k^{\mathrm{th}}$ root of unity. If $f(x)=\sum_{j=0}^{\infty}a_{j}x^{j}$ and $n\not\equiv 0\pmod{k}$, then

 $\sum_{j=0}^{\infty}a_{kj+n}x^{kj+n}=\frac{1}{k}\sum_{j=0}^{k-1}\omega^{-jn}f(% \omega^{j}x)$

Proof. This is a consequence of the fact that $\sum_{j=0}^{k-1}\omega^{jm}=0$ for $m\not\equiv 0\pmod{k}$.

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 $r\not\equiv n\pmod{k}$, the sum is zero. So the term involving $x^{r}$ is zero unless $r\equiv n\pmod{k}$, 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 ExtractingEveryNmathrmthTermOfASeries 2013-03-22 16:23:34 2013-03-22 16:23:34 rm50 (10146) rm50 (10146) 10 rm50 (10146) Theorem msc 11-00