Solution verification: finding a Maclaurin series for $f(x)$, interval of convergence, and $f^{(100)}(0)$

Question

I have to find Maclaurin series for function $f(x) = e^{(x^2+1)}\cdot x$, it's interval of convergence and $f^{(100)}(0)$. I know that $e^x$ = $\sum_{i=0}^{\infty} {x^n\over n!}$. Placing $x^2$ instead of x we got $\sum_{i=0}^{\infty} {x^{2n}\over n!}$. Multiplying that by $e\cdot x$ we finally get $e^{(x^2+1)}\cdot x$ and Maclaurin series looks as follows : $\sum_{i=0}^{\infty} ({x^{2n+1}\over n!}\cdot e)$. What would be interval of convergence and $f^{(100)}(0)$?

Answer 1

Your function $f$ is an odd function. So, $f'$ is even, $f''$ is odd, $f^{(3)}$ is even, and so on… In particular, $f^{(100)}$ is an odd function, and therefore $f^{(100)}(0)=0$.