Htykuut

Let $f:(-infty, 1) to mathbb{R}$ be a function, $f(x)=e^x+ln(1-x)$.

Find $n in mathbb{N}$ such that the error when approximating
$e^{1.1}+ln(1.1)$ by their Taylor polynomial $T_{n,f,0}(x)$, with $n < 0.0001$.

I'm really confused and I have no idea where to start.

I know that $f(x)=T_{n,f,0}(x)+R_{n}(x)$ where $R_{n}(x)$ is the reminder of the Taylor polynomial.

$e^{1.1}+ln(1.1)=(1-frac{x^3}{6}-frac{5 x^4}{24}-frac{23 x^5}{120}-...)-n$ Where $n$ is the given error?

Sorry if this is so confusing but I'm really lost!

Your $f(x)$ does not match the function you are asked to form the Taylor series of because it has $ln (1-x)$ and the $e^x$ term is $0.1$, not $1.1$ when $x=0.1$. Your Taylor series is not correct-it should have an $x$ on the left, and the Taylor series of $e^{1+x}+ln(1+x)$ is not what you have shown-at $x=0$ it should be $e$, not $1$.

Once you get the right Taylor series, $n$ is the number of terms you need to keep to get the error at $x=0.1$ to be less than $0.0001$. You can look up the remainder term of the Taylor series, which will be $0.1^{n+1}$ times the $n+1^{st}$ derivative of the function. You can bound the $n+1^{st}$ derivative in the interval $[0,0.1]$ and evaluate the upper bound for the error this gives you.