Rational function, sequences, polynomials and roots of unity

Question

Let $$f(x) = \sum_{n\geq 0} a_n x^n = \frac{P(x)}{(1-x)^d}$$ be a rational function.

(a) Prove: There is a polynomial $P_2(x)$ so $$\sum\limits_{n\geq 0} a_{2_n} x^n = \frac{P_2(x)}{(1-x)^d}$$

(b) Let $r \geq 1 \in \mathbb{N}$. Show that an polynomial $P_r$ exists so that $$\sum_{n\geq 0} a_{rn} x^n = \frac{P_r(x)}{(1-x)^d}$$

Hint: Use the $r$th roots of unity which are defined by $\exp\left(\large \frac{2\pi ik}{r}\right), 0 \leq k \leq r - 1$.

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(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 $\large \frac{P(x)}{Q(x)}$. 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!

Answer 1

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

Answer 2

Here's what you need to know about roots of unity. Let $\zeta=e^{2\pi i/r}$. Then the $r$th 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?