I have the following problem: I want to find the value of $n$ (degree of the Taylor series) such that the taylor polynomial $Pn(x)$ approximates the function $f(x)$ with an error no greater than $10^{-t}$ in the interval $[a,b]$. So, i have this: $$\frac{\lvert f^{(n+1)}(\xi)\lvert}{(n+1)!} * \lvert(x-\bar{x})^{(n+1)}\lvert\le 10^{-t}$$

My problem is: How can i solve for n?, What can i do with the $n$ at $\lvert f^{(n+1)}(\xi)\lvert$ where $(n+1)$ is the derivative? The idea is to use the formula to calculate $n$ with any specific $f(x)$. Thank you.

  • $\begingroup$ Generally, just plug in a few values. Without more information on the specific problem, this is the best I advice I can give you. $\endgroup$ – Simply Beautiful Art Aug 25 '17 at 1:09
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    $\begingroup$ The best I think we can do is to first try to get an exact form or an approximation for $f^{(n)}(\xi)$. You can get some suggestions here and here. $\endgroup$ – Prasun Biswas Aug 25 '17 at 1:57

I tried to do something similar some months ago and this is what I got so far. Maybe it helps you. And no, I did not manage to solve this problem completely.

First thing, the function has to be analytic on that domain, so that the Taylor series converges to the function.

I will start with an observation related to the way you formulated the question. I will take $\bar x=0$ for convenience in writing the relations.

Let $T_n(x)$ be the Taylor polynomial of degree $n$ and $\epsilon$ the maximum value of the difference between $T_n(x)$ and $f(x)$ on a neighborhood around $\bar x = 0$. In other words, if you fix $\epsilon$, then $|T_n(x)-f(x)|<\epsilon$ only for $|x|<x_{\epsilon}$. So you can take values for $x$ only inside the interval $(-x_{\epsilon}, x_{\epsilon})$ so that $|T_n(x)-f(x)|<\epsilon$. There are 2 aspects here:

  1. If you take bigger values of $\epsilon$ but keep the degree of the Taylor polynomial fixed, the allowed interval $(-x_{\epsilon}, x_{\epsilon})$ for $x$ will get bigger.
  2. If you keep the same value for $\epsilon$ but take Taylor polynomials of greater degree, the interval $(-x_{\epsilon}, x_{\epsilon})$ will also get bigger.

Those result should be easy to check if you want to. What is important for your question is that you can approximate the function on an interval $(-x_{\epsilon}, x_{\epsilon})$, and that interval gets bigger as you take a Taylor polynomial of greater degree. So by incrementing the value of $n$ you 'cover' a bigger domain of what you considered as $[a,b]$. Now where is the problem with this. Based on observation 2. if you consider your Taylor polynomial centered at $\bar x = a$ and want to cover the interval $[a,b]$, you will actually need to cover the interval $[a-(b-a), b]$ which is double of what you initially needed. A different situation is if you start at the center of $[a,b]$ that is $\bar x = a+\frac{b-a}{2}$. In this scenario you will have to cover only $[a,b]$. So what should be clear from those 2 cases is that in the first one you will need a greater degree for the Taylor polynomial than in the second scenario, so your problem also depends on the variable $\bar x$. Thus, you have an equation with 3 variables ($\bar x$, $n$ and $\xi$), and trying to solve this won't get you very close to what you probably wanted. That is unless you impose some additional conditions.

Now, about the evaluation part, assuming that $\bar x = a+\frac{b-a}{2}$. As Prasun Biswas linked in his comment, in order to evaluate the reminder you'll have to be able to say something about $f^{n}(x)$. What I tried was to write my function as:

$$f(x) = \sum_{j=1}^{N} a_j sin(\omega _j x)$$

or using any other function that had a general formula for the $n$-th derivative. As an example I also considered writing the function as a discrete version of a Weierstrass transform.

In this form, you should be able to take each term $a_j sin(\omega _j x)$, plug it into the reminder formula, and see the contribution of each of them on the sum. At this point the $j$ term of the reminder should be something like this:

$$|a_j sin(\omega _j \xi)\frac{(x-\bar x)^{n+1}}{(n+1)!}\omega_j^{n+1}|$$

and your analysis should be focused on:

$$|\frac{(x-\bar x)^{n+1}}{(n+1)!}\omega_j^{n+1}|$$

I hope this helps you somehow.


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