$ f(x) = \ln(1+x)$
The previous part of this question required me to write down the remainder term for the taylor polynomial of order n.
My remainder term worked out to be:
$R_n(x) = (-1)^n \frac{x^{n+1}}{n+1}$
I checked this manually, and it seemed to be correct.
For this question I am quite sure I need to solve for $n$, such that $R_n(x) < 0.0002$.
I have:
$(-1)^n \frac{(0.3)^{n+1}}{n+1} < 0.0002 $
(I have $0.3$ for $x$, since $\ln(1+0.3) = \ln(1.3)$.) Am I on the right track here? I am a little stuck with the algebra. I have:
$ (-1)^n(0.3)^{(n+1)} < 0.0002(n+1) $
I have tried to take the natural log of both sides, but it seems like I can't isolate the $n$.
Any help would be greatly appreciated!