Can this pattern of denominators be expanded?

Question

We can take a polynomial, $p(x)=\left(c_0 + c_1 x^1 + c_2 x^2\right)$, and get a repeating pattern of it by:

$$\frac{p(x)}{1-x}=c_0+(c_0+c_1)x+(c_0+c_1+c_2)x^2 + \dots + (c_0+c_1+c_2)x^m + \dots$$

...We can note that eventually the coefficients of the polynomial repeat. Also, another pattern appears by taking:

$$\frac{p(x)}{1-(x-x^2)}=c_0+$$ $$(c_0+c_1)x+$$ $$(c_1+c_2)x^2+$$ $$(-c_0+c_2)x^3+$$ $$(-c_0-c_1)x^4+$$ $$(-c_1-c_2)x^5+$$ $$(c_0-c_2)x^6+$$ $$(c_0+c_1)x^7+$$ $$(c_1+c_2)x^8+\dots$$

...We can note that the pattern of coefficients begins to repeat with $x^7$, which has the same coefficients as $x^1$.

I'm wondering if we can find more patterns that repeat in this manner, by simply modifying the denominator. So my question is, can we get more patterns like this?

IMPORTANT NOTE

I know that we can take the example above: $$\frac{p(x)}{1-(x-x^2)}$$ and change the $x$ powers to get a new pattern: $$\frac{p(x)}{1-(x^2-x^4)}$$

I'm wondering, instead, if we can expand the denominator, somehow, to get a new pattern, instead of changing the powers of $x$.

What I'm Really After

What is the general formula for $q(x)$ (or how can we find $q(x)$) such that

$$\frac{p(x)}{1-q(x)}$$

produces a pattern?

Answer 1

The coefficients of $\frac{p(x)}{q(x)}$ are periodic for all choices of $p$ (with degree less than that of $q$) if and only if the roots of $q$ are distinct roots of unity. If $q$ is required to have integer coefficients, then this is true if and only if $q$ is a product of distinct cyclotomic polynomials. This follows from taking the Taylor series expansions of partial fraction decompositions as Calvin Lin notes in the comments.

Answer 2

Note that the 'pattern' is basically just about the pattern of coefficients of the Maulaurin expansion of $\frac{1}{1-q(x) } $. Since, what we're doing, is just multiplying $p(x) = c_0 + c_1 x + c_2 x^2 $ to it.

As a more concrete statement, let's review the examples that you gave. We have

$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots $$

As such, we can easily conclude that $\frac{p(x) } {1-x} $ indeed has the form that you claim it should have.

Now, how about $\frac{ 1 } { 1-x+x^2}$? This is slightly harder to expand out directly (for example, some people might do it as $ 1 + (x-x^2) + (x-x^2)^2 + (x-x^2)^3 + \ldots$, but this is hard to get into the $\sum a_i x^i$ form that we really want. The reason why it works out so nicely, is because it expands out as (you can verify this by substitution $c_0 = 1, c_1 = 0 , c_2 = 0$):

$$1 + x - x^3 - x^4 + x^6 + x^7 + \ldots$$

In order for it to 'form a pattern', we want the coefficients to eventually repeat, or that $a_i = a_{k+i}$ for some period $k$, $\forall i \geq I$. Then, we can actually express

$$\sum a_i x^i = \sum_{i=0}^Ia_i x^i + \frac { \sum_{i=I+1}^{I+k} a_i x^i } {1-x^k} $$

Going back to your example, given that we want such a pattern, how do we determine the polynomial? You should be able to see that

$$1 + x - x^3 - x^4 + x^6 + x^7 + \ldots = \frac{ 1+x-x^3-x^4} { 1-x^6} = \frac{1}{1-x+x^2}.$$

And this is the reason why the polynomial that you chose works.

Now, how do we create more of such polynomials? That should be evident from the above.

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Note that Qiaochu's statement that "the coefficients are periodic if and only if the roots of $q$ are distinct roots of unity" follows directly from seeing that the denominator must be a factor of $1-x^k$.