$$f(x)=(x+1)e^{x^2}$$
Find the Taylor expansion up to order $4$ around $x_0=1$, and find an approximation of $f(\frac 1 2)$ with error less than $10^{-3}$.
I can take several derivatives and do it by hand, but I wonder if there is a way to use the well known expansion of $e^x$, which is valid everywhere but is centered in $0$.
Thanks!
Of course there is: first you have to rewrite all as a polynomial in $x-1$: \begin{align}f(x)&=(x+1)e^{x^2}\\&=((x-1)+2)e^{(x-1)^2}e^{2(x-1)}\cdot e^1\\ &=((x-1)+2)\sum\frac{(x-1)^{2n}}{n!}\cdot\sum\frac{2(x-1)^m}{m!}\cdot e\\ &=\sum a_n(x-1)^n\end{align} And now find the $a_n$... but honestly this might take more time than calculating the derivatives...
The other way (as Hamza pointed out) is to write \begin{align}f(x)&=(x+1)e^{x^2}=(x+1)\cdot\sum_n\frac{x^{2n}}{n!}\\ &=((x-1)+2)\cdot\sum_n\frac{((x-1)+1)^{2n}}{n!}\\ &=((x-1)+2)\cdot\sum_n\frac{\sum_k {2n\choose k}(x-1)^k}{n!}\end{align} Then change the order of summation to get again something of the form $\sum a_n(x-1)^n$