Find the Taylor polynomial for a function $f(x) = (1+x^2)^{1/3}$

Question

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

Answer 1

sorry your evaluation is wrong ist is NOT 9^(5/3) but just 9 so f''(0)=2/3 you have much less work, wenn you set x=0 in your first equation. so you have not to evaluate the derivative in (...)'*x

Answer 2

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 $$\bigl(1+x^2\bigr)^{1/3}=\underbrace{1+\tfrac 13x^2}_{T_2(x)}+o\bigl(x^2\bigr).$$