Taylor error approximation

Question

So a Taylor polynomial is given by the Taylor formula, but how do I approximate the error?

I see on wikipedia:

$$R_k = \frac{f^{(k+1)}(s)}{(k+1)!} (x-a)^{k+1}$$

Do I just pick any $s$ between $x$ and $a$? Example if my interval is $[1,4]$ I could pick for example 2 or 3...Or even 3,3?

Also do I withdraw it from my polynomial like with linear approximations or do I just put it on the back..? Perhaps you could give me a simple example on error approximation.

Linear error approximation:

$$f(x)-L(x)$$

Answer 1

That is Lagrange's form of the remainder. What you know is that $a \le s \le x$ (or the other way around, if $x < a$). To get a bound on the error, pick the value of $s$ that maximizes $f^{(k + 1)}(x)$ in that range. That might be hard to do, in which case get some upper bound, as in the end you want a (more or less crude) bound of a (hopefully small) quantity, not an exact value.

Answer 2

Do I just pick any $s$ between $x$ and $a$?

It won't give you much, the order of magnitude of the error will stay the same. Typically when you wish to reduce the error, you expand the series for further elements. The order of magnitude of error is basically the $k$ of the $(x - a)^{k + 1}$, when you enlarge the $k$ the error reduces.

Also do I withdraw it from my polynomial like with linear approximations

Linear approximation is private case of Taylor expansion with $R_1$

Answer 3

You estimate the error by setting:

$|R_k|=|\frac {f^{k+1}(s)}{(k+1)!}(x-a)^{k+1}|\leq \sup_{s\in(a,x)}|f^{k+1}(s)||\frac {(x-a)^{k+1}}{(k+1)!}| $

We want to consider the size of our error $R_k $, lets call our function that we are trying to approximate $f$, then:

$ |f-approx|=|R_k | = |\frac {f^{k+1}(s)}{(k+1)!}(x-a)^{k+1}|\leq \sup_{s\in(a,x)}|f^{k+1}(s)||\frac {(x-a)^{k+1}}{(k+1)!}| $ This allows us to find an upper bound for the error, and this is what we work with as we cannot just "pick" a value of s.