Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $

Question

Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.

And I want to find the series representation of this function. I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fractions $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{2(1+x)}+\frac{1}{2(1+x^{2})}$.

I know the series representations of these functions $(1+x)^{n}$ and $\frac{1}{1-x}$, and I guess I should use one of them. But the result looks weird. Can you help me, please, may be I can't see some easy way to solve it.

Answer 1

$$f(x)=\frac{1}{1+x+x^2+x^3}=\frac{1-x}{1-x^4}=\sum_{n\geq 0}\chi(n)\,x^n $$ where $\chi(n)$ equals $1$ if $n\equiv 0\pmod{4}$, $-1$ if $n\equiv 1\pmod{4}$ and zero otherwise.

Answer 2

There are two possibilities here: either we use the binomial theorem on $(1+x(1+x+x^2))^{-1}$, which looks awful, or we use a trick: multiplying and dividing by $1-x$, we have $$ \frac{1-x}{(1-x)(1+x+x^2+x^3)} = \frac{1-x}{1-x^4}, $$ because the middle terms all cancel out. Now, we can expand the denominator, $$ \frac{1-x}{1-x^4} = (1-x)\sum_{k=0}^{\infty} x^{4k}, $$ and it's pretty easy to multiply through and get the final answer from here.

Answer 3

Note $$\frac{1}{1 + x + \cdots + x^{n-1}} = \frac{1-x}{1-x^n}.$$ Since we know $$\frac{1}{1-z} = 1 + z + z^2 + \cdots = \sum_{k=0}^\infty z^k, \quad |z| < 1,$$ we have $$\frac{1-x}{1-x^n} = (1-x)\sum_{k=0}^\infty (x^n)^k = (1-x)(1+x^n + x^{2n} + x^{3n} + \cdots).$$ We conclude that the desired series expansion is $$1 - x + x^n - x^{n+1} + x^{2n} - x^{2n+1} + x^{3n} - x^{3n+1} + \cdots,$$ for integers $n > 1$. Your case is $n = 4$.

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It is also worth pointing out that because of the restriction $|z| < 1$ for the geometric series expansion we used above, the given series expansion does not converge even when the original function is well-defined. This is addressed by performing the series expansion about $z = \infty$ rather than $z = 0$. We do this by writing $$\frac{1}{1-z} = \frac{z^{-1}}{z^{-1} - 1} = -z^{-1} \sum_{k=0}^\infty z^{-k} = -\sum_{k=1}^\infty z^{-k}, \quad |z| > 1,$$ hence $$\frac{1-x}{1-x^n} = -(1-x) \sum_{k=1}^\infty (x^n)^{-k} = x^{-n+1} - x^{-n} + x^{-2n+1} - x^{-2n} + x^{-3n+1} - x^{-3n} + \cdots, \quad |x| > 1.$$