4
$\begingroup$

I am in high school and find math interesting so lately I have been trying to learn as much about it as I can. I recently began studying Taylor Series as it pertains to the research areas that I am currently following and have two questions. I am trying to conceptualize Lagrange's error bound and the remainder function.

I have read that:

$f(x)=f(c)+f'(c)(x-c)+...+\frac{f^{(n)}(c)}{n!}(x-c)^{n}+R_n(x)$

where $R_n(x)$ is the remainder given by:

$\frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.

From what I have read, I understand that the remainder gives the difference between a given Taylor polynomial and its original function, $f(x)$. I have also read about La Grange's error bound which differs from the remainder.

My first question is, if Lagrange's error bound gives the maximum "remainder" for a given $x$, how does it differ from the remainder function $R_n(x)$? Are they equivalent? If a function can be found from its Taylor polynomial and Remainder term or any combination vice-versa, what is the purpose of Lagrange's error bound?

My second question is, how is the error term function, $R_n(x)$ derived? I understand that it can be found from the difference between a function and its Taylor polynomial but I am not sure where it originates from in general form.

Any help with conceptualizing this is greatly appreciated.

$\endgroup$
4
$\begingroup$

$R_n(x)$ is defined to be the difference between the function and its $n$th order Taylor polynomial (not series).

It is an important consequence of the Mean Value Theorem that there exists $z$ between $x$ and $c$ such that the remainder can be written as $R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!} (x-c)^{n+1}$. However in general we do not know more about the exact value of $z$ (or $f^{(n+1)}(z)$) beyond the fact that it lies between $x$ and $c$.

Since we often cannot compute the remainder explicitly, we would like to bound it. For example, it would be nice to say $|R_n(x)| \le 0.01$; this would tell us that the value of the $n$th order Taylor polynomial at $x$ is within $0.01$ of the actual function value at $x$. Error bounds use assumptions on $f$ to bound $|R_n(x)|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.