Finding terms of a Taylor series where $f(x)$ is a function with a power

Question

I've been stuck with this Taylor series problem for a while now.

We have that $$ f(x) = (1 + x^2)^{-2/3} $$ and it's centered at $0$.

So what I thought of doing was the $$ \frac{f^{n}(a)(x - a)^{n}}{n!} $$ method of finding up to the 4th derivative then making a conjecture of the summation but the derivatives of that function get pretty complicated and take a while to compute.

Is there a better way?

Thanks

Answer 1

Instead, find the Taylor series for the function $$f(x)=(1+x)^{-{2\over 3}}$$ and then just plug in $x^2$ to the result.

Answer 2

The series can be obtained using the binomial theorem: $$ \begin{align} &\left(1+x^2\right)^{-2/3}\\ &={\large\sum_{\normalsize k=0}^{\normalsize\infty}}\binom{-2/3}{k}x^{2k}\\ &=1+\tfrac1{1!}\left(-\tfrac23\right)x^2+\tfrac1{2!}\left(-\tfrac23\right)\left(-\tfrac53\right)x^4+\tfrac1{3!}\left(-\tfrac23\right)\left(-\tfrac53\right)\left(-\tfrac83\right)x^6+\dots\\[6pt] &=1-\frac23x^2+\frac59x^4-\frac{40}{81}x^6+\frac{110}{243}x^8-\frac{308}{729}x^{10}+\dots \end{align} $$