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?
Thanks in advance.
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 $$|R_2(x)|\le\tfrac19 x^4.$$