telegrafyx

2022-06-26

Let

$f(x)=\sum _{n\ge 0}{a}_{n}{x}^{n}=\frac{P(x)}{(1-x{)}^{d}}$

be a rational function.

(a) Prove: There is a polynomial ${P}_{2}(x)$ so

$\sum _{n\ge 0}{a}_{{2}_{n}}{x}^{n}=\frac{{P}_{2}(x)}{(1-x{)}^{d}}$

(b) Let $r\ge 1\in \mathbb{N}$. Show that an polynomial ${P}_{r}$ exists so that

$\sum _{n\ge 0}{a}_{rn}{x}^{n}=\frac{{P}_{r}(x)}{(1-x{)}^{d}}$

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 $f(x)$ is a rational function, it can be defined as a fraction of two polynomials $\frac{P(x)}{Q(x)}$. But that is unfortunately all I know about this.

Could you please help me going on?

(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.

Could you please help me a bit?

Thanks in advance!

$f(x)=\sum _{n\ge 0}{a}_{n}{x}^{n}=\frac{P(x)}{(1-x{)}^{d}}$

be a rational function.

(a) Prove: There is a polynomial ${P}_{2}(x)$ so

$\sum _{n\ge 0}{a}_{{2}_{n}}{x}^{n}=\frac{{P}_{2}(x)}{(1-x{)}^{d}}$

(b) Let $r\ge 1\in \mathbb{N}$. Show that an polynomial ${P}_{r}$ exists so that

$\sum _{n\ge 0}{a}_{rn}{x}^{n}=\frac{{P}_{r}(x)}{(1-x{)}^{d}}$

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 $f(x)$ is a rational function, it can be defined as a fraction of two polynomials $\frac{P(x)}{Q(x)}$. But that is unfortunately all I know about this.

Could you please help me going on?

(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.

Could you please help me a bit?

Thanks in advance!

candelo6a

Beginner2022-06-27Added 24 answers

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: $1+{\zeta}^{m}+{\zeta}^{2m}+\cdots +{\zeta}^{(r-1)m}$. 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?

OK?

glycleWogry

Beginner2022-06-28Added 8 answers

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

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