# Given the function $f(x) = (1+x^2)^{1/3}$ find the Taylor polynomial and constant $C > 0$ such that $|f(x) - T_2(x)| \leq C|x|^3$

Given the function $$f(x) = (1+x^2)^{1/3}$$ I have to find the Taylor polynomial for f of order two centered at $$x_0 = 0$$.

I know that I can use the binomial series to find that

$$T_2(x) = \sum_{n= 0}^1 \binom{1/3}{n}x^{2n} = 1 + \frac{1}{3} x^2$$ but I am actually not sure why I have to use $$n = 1$$ instead of $$n = 2$$ in the sum when I have to find the Taylor polynomial for f of order $$2$$, indicating that $$n = 2$$. Can you explain why?

Now I have to find a constant $$C > 0$$ such that $$|f(x) - T_2(x)| \leq C|x|^3 \ \text{for all} \ x \in [-1,1]$$ By definition I know that $$f(x) = T_2(x) - (R_nf)(x)$$ and that $$|(R_nf)(x)| \leq \frac{M_n}{(n+1)!}|x-x_0|^{n+1}$$ where $$M_n \geq \max \{|f^{(n+1}(t)| \ : t \in [x_0,x] \}$$ which gives me $$|f(x)-T_2(x)| = |(R_nf)(x)| \leq C |x|^3 = \frac{M_n}{(n+1)!}|x-x_0|^{n+1}$$ Thus for $$n = 2$$ and $$x_0 = 0$$ we must have $$C|x|^3 = \frac{M_n}{3!} |x-0|^{2+1} \Leftrightarrow C = \frac{M_n}{3!}$$ which means that I have to find $$f^{(3)} = M_n$$ and divide it by $$3!$$ if I am not mistaken. But this does not make sense to me really unless I evaluate in $$x_0 = 0$$ but then it doesn't give the right answer which I know is $$1/9$$. I have tried to read Taylor polynomial and remainder but I just don't understand it. These definitions I have written are the only definitions which are used in my book so I think I have to solve it this way. And is there an easy way to calculate $$f^{(3)}$$ using the binomial series instead of calculating it manually? Can you help me?

Consider the Taylor series of $$(1+u)^{1/3}$$: $$(1+u)^{1/3}=1+\tfrac13 u-\tfrac19 u^2+\tfrac 5{81}u^3-\dotsm,$$ which converges for $$|u|<1$$. By the substitution $$u=x^2$$, you obtain the Taylor series of $$(1+x^2)^{1/3}=1+\tfrac13 x^2-\tfrac19 x^4+\tfrac 5{81}x^6-\dotsm$$ You can easily check it is an alternating series, with decreasing general term, so it converges for $$|x|<1$$, and by the uniqueness of Taylor polynomials, it expansion at order $$2$$ is indeed $$(1+x^2)^{1/3}=1+\tfrac13 x^2+R_2(x).$$
As to the estimation of the remainder, you don't have to calculate the derivatives explicitly: as we have an alternating series, by Leibniz' theorem, we know that $$R_2(x)$$ is nonpositive, and $$|R_2(x)|\le\tfrac19 x^4.$$