# Find a constant $C>0$ such that $R_nf(x) \leq C|x|^3$

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

$$T_2(x) = \sum_{n= 0}^1 \binom{1/3}{n}x^{2n} = 1 + \frac{1}{3} x^2$$

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 $$|f(x)-T_2(x)| = |(R_nf)(x)| = \frac{f^{(3)}(t)}{3!}|x|^3 \leq \max_{t \in [-1,1]} |f^{(3)}(t)\frac{|x|^3}{3!}$$ which means I have to find the third derivative of f(x) which I have found as $$f'''(x) = -\frac{8x\left(-x^2+9\right)}{27\left(x^2+1\right)^{\frac{8}{3}}}$$ and then find an upper bound for $$\max_{t \in [-1,1]} |f^{(3)}(t) \frac{|x|^3}{3!}$$ but I am not sure how to? I just picked some randoms number in the interval and found that x around $$0.5$$ was close. But is there an exact way? I guess I could just draw it but this is an earlier exam question so I would not have my computer with me.

I am asked to do it this way so I am not allowed to use that this is an alternating series.

Can you help me?

You need a upper bound on $$|f^{(3)}(t)|=\frac{|8t(9-t^2)|}{|27(t^2+1)^{\frac{8}{3}}|}$$ for $$t\in [-1,1]$$, so you find an upper bound for the numerator and a lower bound for the denominator :
$$|f^{(3)}(t)|=\frac{|8t(9-t^2)|}{|27(t^2+1)^{\frac{8}{3}}|} \leq \frac{|8 \times 1 \times 9|}{|27\times 1^{\frac{8}{3}}|}= \frac{8}{3}$$
• Hi Ewan. That makes sense. I guess you have used the triangle inequality? I understand the lower bound for the denominator as $t = 0$ gives the lower bound but in the numerator it seems like you have used both $t = 0$ and $t=1$. Is this allowed? Thanks for your help. – Mathias Apr 2 at 10:36
• The numerator is a product fo three terms, $8 \times |t| \times |9-t^2|$. Each term has its own upper bound, and may correspond to different values of $t$ – Ewan Delanoy Apr 2 at 10:43
• Just to be sure does this imply that $C = M_n / 3! = (8/3)/6 = 4/9?$ Thanks in advance. – Mathias Apr 2 at 11:35
• $\frac{\frac{8}{3}}{3!}=\frac{4}{9}$ is indeed an upper bound though it is not optimal. The optimal constant is $4-\sqrt{2}$ but you can't obtain it using only Taylor expansions. – Ewan Delanoy Apr 2 at 11:45