Understanding Taylor series error function and Lagrange error bound

calculuserror functiontaylor expansion

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.

Best Answer

$R_n(x)$ 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 $|R_n(x)|$.

Related Question