Approximate the value using Taylor series

Question

I have to approximate the value of $\sqrt[12]{4000}$ by using tTaylor series. We were told that using the third order will be enough.I tried it but I am really feeling like I got something wrong, because I wouldnt be able to compute it without using my calculator and I just dont feel like that this is right. Could somebody please explain where I am making my mistake ? The numbers should be all ok, checked that twice with wolfram. Thanks, here is my go at it :

The term can be rewritten as : $\sqrt[12]{4000}=\sqrt[12]{2^{12}-96}=2\left(1-\frac{3}{2^7}\right)^{\frac{1}{12}}$ so I want to find the Taylor series for function $\sqrt[12]{1-x}$ which is :

$ 1-\frac{x}{12}-\frac{11 x^{2}}{288}-\frac{253 x^{3}}{10368} $ Now evaluating this at $x=3/2^7$ and multiplying by 2 I get the evaluation of : 1.99605.... which is an OK estimate.

Now I have to compute the error:

$ R_{k}(x)=\frac{f^{(k+1)}\left(\xi\right)}{(k+1) !}(x-a)^{k+1} $

where $\xi \in |0,x|$. Now for my case $k=3$ and therefore :

$ R_{3}(x)=\frac{M}{(4) !}(x)^{4}$.

Now $M$ is nothing else but the fourth derivative of $\sqrt[12]{1-\xi}$ which I can compute as :$ -\frac{8855}{20736(1-\xi)^{47 / 12}}$ Now I am trying to find the upper bound for the error, so I use $\xi=0$ for which $M$ is maximal. Evaluating this for said $\xi$ I get $M \approx -0.427$

Then plugging that back in I evaluate the error as :

$ R_3(x)=\frac{-0.427}{4!}(\frac{3}{2^7}) $

Answer 1

I would recommend trying for the function $x^{1/12}$ instead of $(1-x)^{1/12}$. Then take the center as close as possible to 4000. So, consider $2^{12}$ which is 4096. Let $a=2^{12}$. Write the taylor approximation centered at $a$ of third degree as a function of x, call this function $L(x)$. Finally, put $x=4000$ to get your answer $L(4000)$.

Answer 2

You have an error in your Taylor series: as the Newton expansion of $(1+x)^\alpha$ is $$(1+x)^\alpha=1+\alpha x+\alpha(\alpha -1)\,\frac{x^2}{2!}+\alpha(\alpha -1)(\alpha-2)\,\frac{x^3}{3!}+\dotsm,$$ if $\,0<\alpha<1$, this is an alternating series if $x>0$, so you can apply Leibniz'rule for the error, namely it is bounded by the first omitted term, and it has the same sign.

Therefore, if you expand at order $3$, you know the error will be negative and, in absolute value, less than $$\frac{\alpha(\alpha -1)(\alpha-2)(\alpha-3)}{4!}\,x^4.$$ Edit:

Unfortunately, this doesn't work here since we have a negative value for $x$, so the series is no more alternating.