Sequence for generating function $f(x)=\frac{1}{1+x^4}$

Question

The question says: Find the sequence $(a_k)$ such that the function $\frac{1}{1+x^4}$ is the ordinary generating function of $(a_k)$, that is

$\frac{1}{1+x^4}=\sum_{n=0}^{\infty}a_n.x^n$.

I have tried as follows:

$\frac{1}{1+x^4}=\frac{1}{1-(-x^4)}=\sum_{n=0}^{\infty}(-1)^nx^{4n}$.

But I can't express $x^n$ because it exists in 4n form. how can I express it in $x^n$ form?

Answer 1

What if you say $$\frac{1}{1+x^4} = \sum_{n=0}^{\infty}a_nx^n$$ if $$a_n = \begin{cases} (-1)^{n/4},\text{ if }n\mod 4 =0, \\ 0,\text{ otherwise.} \end{cases}$$

Answer 2

If you really need a closed formula for $a_n$ ... $$ \frac{1}{1+x^4} = \sum_{n=0}^{\infty} \frac{(-1)^{n(n+1)(n+2)(n+3)/8}-(-1)^{(n+1)(n+2)(n+3)(n+4)/8}}{2} \;x^n $$

Answer 3

$$\sum_{n=0}^{\infty} (-1)^{n} x^{4n}$$

$$\sum_{n=0, n \mod 4 \equiv 0}^{\infty} (-1)^{n/4} x^n$$

$$=\sum_{n=0}^{\infty} (-1)^{n/4} \frac{w_1^n+w_2^n+w_3^n+w_4^n}{4} x^n$$

Where $w_1,w_2,w_3,w_4$ are the $4$ distinct fourth roots of unity.