0
$\begingroup$

To show the remainder of the taylor polynomial for $e^x$ about $x=0$, goes to $0$, is all I need to do suggest there is some $c \in (0,x)$ such that $r_n(x)=\frac{e^cx^n}{n!}$, where $r_n$ is the remainder term of the taylor polynomial, then note that this approaches $0$ as $n \rightarrow \infty$?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is essentially correct but observe that $c$ can depend on $n$. But for any fixed $x$ you can bound $e^c$ by $\max\{1,e^x\}$ and then take the limit of $|r_n(x)|$ as $n\to\infty$.

$\endgroup$
2
  • $\begingroup$ If I know the remainder term goes to zero for any function can I bound the term $f^{(n)}(c)$ by $\text{max}\{1,f(x)\}$? $\endgroup$ Feb 10, 2021 at 3:59
  • $\begingroup$ No. This works here because $f^{n}(x)=e^x$ And this function is increasing. So for any $x>0$ you have $f^n(c)=e^c\le e^x$ and for $x<0$ $f^n(c)<e^0=1$. The bounds depend on the particular function and its derivatives. $\endgroup$
    – GReyes
    Feb 10, 2021 at 6:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .