2
$\begingroup$

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}) $

$\endgroup$
3
  • 2
    $\begingroup$ Not sure what your question is here. It's true that evaluating the estimate by hand here would be very tedious, but that's just because you're going to third order and have some fractions with large integers involved. $\endgroup$
    – Ian
    Apr 19, 2020 at 20:07
  • 2
    $\begingroup$ That being said there are some careless mistakes in your error calculation. First off note that the original thing multiplies by $2$, so if you are estimating $R_3$ after factoring out the $2^{12}$ then you need to multiply what you get by $2$ at the end. Second you didn't raise to the fourth power at the end. Finally the $M$ should be the maximum absolute fourth derivative, which does not occur at $\xi=0$ but instead at $\xi=3/2^7$. $\endgroup$
    – Ian
    Apr 19, 2020 at 20:11
  • $\begingroup$ @Ian I guess my question is whether my go at the problem is correct or not. I feel like those numbers are quite stupid for a textbook problem and I am afraid that I made some error . $\endgroup$
    – mr.pink
    Apr 19, 2020 at 20:16

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Here $x=-3/2^7<0$ so the series doesn't alternate. $\endgroup$
    – Ian
    Apr 19, 2020 at 20:31
  • $\begingroup$ That being said, in the case of $(1+x)^a$ with $x>0,0<a<1$, the Leibniz estimate is literally the same as the Lagrange-type Taylor estimate, because the max of $|f^{(n+1)}|$ is at the point of expansion. $\endgroup$
    – Ian
    Apr 19, 2020 at 20:34
  • $\begingroup$ Sorry, it see I skimmed through the question. I'll remove the answer. The Leibniz estimate is faster to compute. $\endgroup$
    – Bernard
    Apr 19, 2020 at 20:36
  • $\begingroup$ oh I see sou you suggest expanding the $\mathcal{O}(x^4)$ and finding the error like that ? $\endgroup$
    – mr.pink
    Apr 19, 2020 at 20:46
  • $\begingroup$ No, not exactly. I advised (erroneously) to use the Leibniz' bound for alternating series. But it can't work since in your case, it is not an alternating series. But if it workes (incase $x>0$, this rule yields instantly the constant of $O(x^4)$. $\endgroup$
    – Bernard
    Apr 19, 2020 at 20:52
0
$\begingroup$

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 $L(4000)$.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks for contributing an answer. But how different is this approach? $\endgroup$ Apr 19, 2020 at 20:13
  • $\begingroup$ I thought it just for ease of calculation. They are exactly the same approach. This is a very standard question, and that is one of the most common approaches. $\endgroup$ Apr 19, 2020 at 20:20
  • $\begingroup$ could you explain what does it mean to "write the taylor approximation centered at a of third degree as a function of x" - as I can't figure out the meaning of this sentence $\endgroup$
    – Noa Even
    May 3, 2020 at 15:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .