Taylor Expansion at x=0 of 1/(1-x^4)

Question

What is the Taylor expansion of $\frac{1}{1-x^4}$ at $x=0$? I was told it is $1 + x^4 + x^8$ + ... but the function's derivatives at $0$ are $0$ so shouldn't the Taylor Series at $0$ just be 1?

Answer 1

Using \begin{align} \frac{1}{1-x} = \sum^\infty_{n=0}x^n \end{align} we see that \begin{align} \frac{1}{1-x^4} = \sum^\infty_{n=0}x^{4n}. \end{align}

Moreover, observe that \begin{align} f^{(4)}(x) = -24 \frac{(1 + 65 x^4 + 155 x^8 + 35 x^{12})}{(-1 + x^4)^5} \end{align} which means \begin{align} f^{(4)}(0)= 24 = 4!. \end{align} Likewise, we can show that other than derivatives which are multiples of 4, the other terms are zeros.