# Taylor polynomial and remainder

Observe the function $$f(x) = (1+x^2)^\frac{1}{3}\, ,for\ x\in\mathbb{R}$$

Find the Taylor polynomial $T_2$ for $f$ of degree 2 at the point $x_0 = 0$ and find a constant $C>0$ such that

$|f(x)-T_2(x)|\le C|x|^3 \, \forall x\in[-1,1]$

I have calculated the Taylor polynomial of 2 degree: $$T_2(x) = 1+0x+\frac{\frac{2}{3}}{2}x^2=1+\frac{1}{3}x^2$$ but I am not sure how to find the constant $C$.

I know that $|f(x)-T_2(x) = |(R_2f(x)|\le C|x|^3 = \frac {M_n}{(n+1)!}|x-x_0|^{n+1}$. Where $$M_n \ge \max{|f^{(n+1)}(t)|\ |t\in [x_0,x]}\; if \; x_0\le x$$ $$M_n \ge \max{|f^{(n+1)}(t)|\ |t\in [x,x_0]}\; if \; x\le x_0$$

This means that the constant $C = \frac {M_n}{(n+1)!}$. Therefore, I have to calculate the 3rd degree of $f(x)$ in order to find $M_n$. $$f^3(x)=\frac{10}{27}(1+x^2)^{-\frac{8}{3}} \cdot 8x^3 - \frac{2}{9}(1+x^2)^{-\frac{5}{3}} \cdot 8x - \frac{2}{9}(1+x^2)^{-\frac{5}{3}} \cdot 4x$$

This is where I get stuck, do I have to evaluate the points $1$ and $-1$ since $x\in [-1,1]$ or something else?

• Solve the same problem for $g(x)=(1+x)^{1/3}$ on $[0,1]$, and a Taylor of degree $1$. – user552631 Apr 15 '18 at 13:18

Start from Newton's expansion for $(1+u)^\alpha$, with $\alpha=\frac13$ at order $1$, together with Taylor-Lagrange formula: $$(1+u)^{\tfrac13}= 1+\frac13u-\frac19\frac1{(1+\xi)^{\tfrac53}}u^2\qquad\text{for some \;\xi\; between 0 and u.}$$ Then perform the substitution $u=x^2$: $$(1+x^2)^{\tfrac13}= \underbrace{1+\frac13x^2}_{T_2(x)}-\frac19\frac1{(1+\xi)^{\tfrac53}}x^4\qquad\text{for some \;\xi\; between 0 and x^2.}$$ Now, as $0< \xi< x^2\le 1$, we know that $$|f(x)-T_2(x)|\le\frac19x^4\le\frac19|x|^3, \quad\text{so }\; C=\frac19.$$ Furthermore we know the error will be negative.
• I have not, yet, been introduced to this method of finding the remainder, so if I have to use the definition of $$M_n \ge \max{|f^{(n+1)}(t)|\ |t\in [x_0,x]}\; if \; x_0\le x$$ which implies that I need the derivative of 3rd degree. – Mads Jeppesen Apr 15 '18 at 15:54
• Just a question: if you're asked to expand, say, $\mathrm e^{-x^2}$ at order $6$ and find a bound for the error term, do you really think you're expected to calculate the seven first derivatives of this composition? – Bernard Apr 15 '18 at 17:53