<s>[INST] Find the Taylor polynomial for a function $f(x) = (1+x^2)^{1/3}$ I have to find the Taylor polynomial $T_2(x)$for a function $f$ of grade $2$ around $x_0 = 0$ given by
$$
f(x) = (1+x^2)^{\frac{1}{3}} \ \text{for} \ x \in \mathbb{R}
$$
I am very new to Taylor polynomials so I think what I have done is correct but I am just unsure whether or not they should give something 'pretty'. I know that they are approximations so they probably shouldn't but are you able to tell me whether or not my calculations are correct as I need this to be correct to be able to do some other questions which relies on me having the Taylor polynomial correct.
I have done the following:
\begin{align*}
   T_2(x) & = \sum_{n=0}^2 \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n = f(x_0) + f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 \\
   & = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2
\end{align*}
Thus
$$
f(0) = (1+0^2)^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1
$$
and
$$
f'(0) = \frac{2}{3} \cdot \frac{1}{(1+0^2)^{\frac{2}{3}}} \cdot 0 = 0
$$
and lastly
\begin{align*}
  f''(x) & = (f'(x))' = \bigg( \frac{2}{3} \cdot \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x \bigg)' = \frac{2}{3} \cdot \bigg(\frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot x \bigg)' \\
  & = \frac{2}{3} \cdot \Bigg( \bigg(\frac{1}{(1+x^2)^{\frac{2}{3}}} \bigg)' \cdot x + \frac{1}{(1+x^2)^{\frac{2}{3}}}  \cdot x' \Bigg) \\
  & = \frac{2}{3} \cdot \Bigg(   - \frac{1}{((1+x^2)^\frac{2}{3})^2} \cdot \frac{2}{3} \cdot (1+x^2)^{\frac{2}{3}-1} \cdot 2x  \cdot x +  \frac{1}{(1+x^2)^{\frac{2}{3}}} \cdot 1  \Bigg)\\ 
  & = \frac{2}{3} \cdot \Bigg(  - \frac{1}{(1+x^2)^\frac{4}{3}}  \cdot \frac{4}{3} \cdot \frac{1}{(1+x^2)^\frac{1}{3}} \cdot x^2 +  \frac{1}{(1+x^2)^{\frac{2}{3}}} \Bigg) \\
  & =  \frac{2}{3} \cdot \Bigg( - \frac{4x^2}{3(1+x^2)^\frac{5}{3}} + \frac{1}{(1+x^2)^{\frac{2}{3}}} \Bigg) = \frac{2}{3} \cdot \Bigg (- \frac{4x^2}{3(1+x^2)^\frac{5}{3}} + \frac{3(1+x^2)}{3(1+x^2)^{\frac{5}{3}}} \Bigg) \\
  & = \frac{2}{3} \cdot \Bigg ( \frac{x^2+3}{3(1+x^2)^{\frac{5}{3}}} \Bigg) = \frac{2(x^2+3)}{9(1+x^2)^{\frac{5}{3}}}
\end{align*}
which if we evaluate in $x_0 = 0$ gives
$$
f''(0) = \frac{6}{9^{\frac{5}{3}}}
$$
Thus
$$
T_2(x) = 1 + \frac{6}{2! \cdot 9^{\frac{5}{3}}}\cdot x^2
$$
I know that this is a lot of calculations but I hope some of you still wants to help me.
Thank you very much.
Regards
Mathias
 [/INST] The simplest consists in using the standard binomial expansion at order $\mathbf 1$:
$$(1+u)^{1/3}=1+\tfrac 13u+o(u) $$
and substituting $u=x^2$, whence
$$\bigl(1+x^2\bigr)^{1/3}=\underbrace{1+\tfrac 13x^2}_{T_2(x)}+o\bigl(x^2\bigr).$$
</s>