what is the Maclaurin Series of this function?

Question

Can anyone explain to me how to find the Maclaurin series of: $$f(x)=(x^2+1)e^{\frac{-x^2}{4}}$$ and why does it converge for every x? thanks,

Answer 1

You should know the basic formula $$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots\ .$$ Substitute $t=-x^2/4$ to get $$e^{-x^2/4}=1-\frac{x^2}{4}+\frac{x^4}{4^22!}-\frac{x^6}{4^33!}+\cdots\ .$$ This converges for all $x$ by the ratio test, or because $e^t$ converges for all $t$. Then multiply by $1+x^2$, which is a polynomial, so no convergence problems, to get $$(1+x^2)e^{-x^2/4}=1+x^2-\frac{x^2}{4}-\frac{x^4}{4}+\frac{x^4}{4^22!}+\frac{x^6}{4^22!}-\frac{x^6}{4^33!}-\frac{x^8}{4^33!}+\cdots\ .$$ You can now simplify by collecting terms, and you should not find it too hard to obtain a general formula for the coefficient of $x^{2k}$.

Answer 2

Hint

As suggested by Mhenni Benghorbal, use the Maclaurin series of $e^{−t}$, replace $t$ by $-x^2/4$, multiply by $(x^2+1)$ and expand.

I am sure that you can take from here and get your result.