Let $ f: \mathbb{R} \rightarrow \mathbb{R}, f(x):=\cos (x) \mathrm{e}^{x} . $ Find a minimal $ N \in \mathbb{N} $ such that for the $ N $ -th Taylor-polynomial in $0$ applies:
$\left|f(x)-T_{N} f(x, 0)\right| \leq 10^{-6} \quad\left(x \in\left[-\frac{1}{10}, \frac{1}{10}\right]\right)$
Also specify the $ N $ -th Taylor polynomial in $0$. Hint: You can use $\cos(x)\geq \frac{1}{2}(x \in\left[-\frac{1}{10}, \frac{1}{10}\right])$.
Can someone explain me how to solve this?
For all positive integers $n$ we have $$ f(x) = \sum_{k=0}^n \frac{(1-i)^k+(1+i)^k}{(2k)!}x^k + R_n(x), $$ where $$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} $$ for some $\xi$ between $0$ and $x$. Taking $x=\frac1{10}$ and $\xi=0$, we have $$ R_n(x) = \frac{1}{2} \left((1-i)^{n+1}+(1+i)^{n+1}\right)\frac{(1/10)^n}{(n+1)!}. $$ Now, since $|1-i|^{n+1} = |1+i|^{n+1}=2^{(n+1)/2}$, it follows that $$ |R_n(x)| \leqslant \frac{2^{(n+1)/2}}{10^n\cdot (n+1)!}. $$ For $n=4$ we have $$ |R_4(x)| \leqslant \frac{1}{150000 \sqrt{2}}\approx 4.714045\cdot 10^{-6}, $$ which is greater than $10^{-6}$, and for $n=5$ we have $$ |R_5(x)| \leqslant \frac{1}{9000000} = \frac19\cdot 10^{-6} < 10^{-6}. $$ Hence the minimal $n$ such that the remainder is less than $10^{-6}$ is $n=5$.