Taylor polynomial of $e^{\frac{x^2}{2}}$ at $x=0$

Question

How can I write the taylor polynomial of $e^{\frac{x^2}{2}}$ at $x=0$?

When I use brute force I get $\sum_{k=0}^{\infty} \frac{(x^2)^k}{2^k k!}$

but when I do it by myself all the derivatives are zero.

Can anyone explain?

Answer 1

Let $f : x \mapsto e^{x^2/2}$.

First, $f(0)=1$ so the first coefficient is $a_0=1$.

Differentiate one time $$ f'\left(x\right)=xe^{x^2/2} $$ that value $0$ when $x=0$ so $a_1=0$. Again ; $$ f''\left(x\right)=\left(xe^{x^2/2} \right)'=e^{x^2/2}+x\left(xe^{x^2/2}\right) $$ that equals $1$ when $x=0$ so the coefficient $a_2=1/2$ as expected. You can keep doing it to find the first terms of the polynomial.

Answer 2

The Taylor's expansion for $y\to0$ is

$$e^y=1+y+\frac{y^2}{2!}+...+\frac{y^k}{k!}+o(x^k)$$

then for $\frac{x^2}{2}\to 0$ we can simply replace and obtain

$$e^\frac{x^2}{2}=1+\frac{x^2}{2}+\frac{x^4}{8}+...+\frac{x^{2k}}{2^k\cdot k!}+o(x^{2k})$$