Taylor polynomial remainder theorem, why is $R_n^{n+1}(x) = f^{n+1}(x)$?

Asked Aug 13 '18 at 06:13

Active Aug 13 '18 at 06:13

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I am trying to grasp the concept of Taylor remainder.

Online lecture taught me that in equation

$R_n^{(n+1)}(x) = f^{(n+1)}(x) + T_n^{(n+1)}(x)$,

$T_n^{(n+1)}(x)$ becomes zero because in general nth polynomial when taken derivative n+1 times becomes zero. For example, $x^2$ when taken derivatives 3 times it becomes zero, 2x, 2, then 0.

So the lecture goes $R_n^{(n+1)}(x) = f^{(n+1)}(x)$.

But I am not getting it because why not $R_n^{(n+1)}$ and $f^{(n+1)}$ also do not become zero?

I don't quite get the idea of nth derivative of f function becoming error itself, though I am not sure if my interpretation is even right.

asked Aug 13 '18 at 06:13

강승태

What do $R_n^{(n+1)}(x)$ etc. denote? – Angina Seng Aug 13 '18 at 06:23
$R_n^{(n+1)}(x)$ denote an error function when taken (n+1)th derivative. Some people just write E instead of $R_n$ for Error. – 강승태 Aug 13 '18 at 06:25
$T_n^{(n+1)}(x)$ denotes taylor polynomial when taken (n+1)th derivative. – 강승태 Aug 13 '18 at 06:26

Taylor polynomial remainder theorem, why is $R_n^{n+1}(x) = f^{n+1}(x)$?

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