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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.

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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 $$|R_2(x)|\le\tfrac19 x^4.$$

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  • $\begingroup$ What kind of uniqueness did you refer to? You can truncate the Taylor series to get the Taylor polynomial every time, because the series is just the limit of the polynomial. $\endgroup$ – Botond Mar 26 at 18:11

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