telegrafyx

2022-06-26

Let
$f\left(x\right)=\sum _{n\ge 0}{a}_{n}{x}^{n}=\frac{P\left(x\right)}{\left(1-x{\right)}^{d}}$
be a rational function.
(a) Prove: There is a polynomial ${P}_{2}\left(x\right)$ so
$\sum _{n\ge 0}{a}_{{2}_{n}}{x}^{n}=\frac{{P}_{2}\left(x\right)}{\left(1-x{\right)}^{d}}$
(b) Let $r\ge 1\in \mathbb{N}$. Show that an polynomial ${P}_{r}$ exists so that
$\sum _{n\ge 0}{a}_{rn}{x}^{n}=\frac{{P}_{r}\left(x\right)}{\left(1-x{\right)}^{d}}$
Hint: Use the rth roots of unity which are defined by $\mathrm{exp}\left(\frac{2\pi ik}{r}\right),0\le k\le r-1$
(a) I don't know what this d is about (and no one else did). Might be an absolute term.
As $f\left(x\right)$ is a rational function, it can be defined as a fraction of two polynomials $\frac{P\left(x\right)}{Q\left(x\right)}$. But that is unfortunately all I know about this.
(b) I don't know how the $r$th roots of unity (and therefore numbers $x$ for which applies: ${x}^{r}=1$) can help me solving this? I don't find any approach.

candelo6a

Here's what you need to know about roots of unity. Let $\zeta ={e}^{2\pi i/r}$. Then the rth roots of unity are the numbers $1,\zeta ,{\zeta }^{2},\dots ,{\zeta }^{r-1}$. Let m be some integer, and raise all these numbers to the power m, and add them: $1+{\zeta }^{m}+{\zeta }^{2m}+\cdots +{\zeta }^{\left(r-1\right)m}$. That's the sum of a geometric progression. If $m$ is a multiple of $r$ then each term in the sum is 1 so the sum is $r$. If $m$ is not a multiple of $r$ then you should check that the formula for the sum of a geometric progression tells you that the sum is zero.
OK?

glycleWogry

For (a), try comparing $f\left(-x\right)$ and $f\left(x\right)$. This should give you some idea how to use the hint in given in (b).