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?

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

It is much simpler than what you've done:

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.

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  • $\begingroup$ 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. $\endgroup$ – Mads Jeppesen Apr 15 '18 at 15:54
  • $\begingroup$ You'll necver see it as a theorem: it's only the substitution method, and its justification results from the uniqueness of Taylor's polynomial. $\endgroup$ – Bernard Apr 15 '18 at 16:25
  • $\begingroup$ Hmmm okay, I guess that makes sense, but I am going to try and do it with the other method as that is the one I have learned and is expected to use $\endgroup$ – Mads Jeppesen Apr 15 '18 at 17:44
  • $\begingroup$ 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? $\endgroup$ – Bernard Apr 15 '18 at 17:53
  • $\begingroup$ I see what you are getting at, it is just that I am expected to solve it by finding the 3rd derivative of f(x) as that is the way that I have been taught to do it. I understand there are way quicker and easier methods to find the exact same thing. $\endgroup$ – Mads Jeppesen Apr 15 '18 at 18:25

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