Taylor series of $e^{-z^2}$ around $z_0=0$

Question

I am trying to find the Taylor series of $e^{-z^2}$ around $z_0$=0

I found the general formula for the $n^{th}$ derivative: $f^{(n)}(z)=(-2z)^ne^{-z^2}$

To find the Taylor series, I need to plug in $z_0=0$. However, this will lead to $f^{(n)}(z)=0$, so the Taylor series will be equal to 0.

What goes wrong here?

Answer 1

It is much simpler to proceed by substitution: setting $t=-z^2$, you obtain instantly $$\mathrm e^{-z^2}=1-z^2+\frac{z^4}{2}-\frac{z^6}{6}+\dots+(-1)^n\frac{z^{2n}}{n!}+\dotsm$$

Answer 2

Hint

It equals to zero only for odd $n$. You can also easily find the Taylor expansion of $e^z$ around $z=0$ by using $${d\over dz}e^z=e^z$$and then substitute $z\to -z^2$

Answer 3

$$(P(z)e^{-z^2})'=(P'(z)-2zP(z))e^{-z^2}$$ allows you to compute all derivatives by recurrence. We have

$$P_0(z)=1$$ $$P_1(z)=-2z$$ $$P_2(z)=-2+4z^2$$ $$P_3(z)=12z-8z^3$$ $$P_3(z)=12-48z^2+16z^4$$ $$P_4(z)=-120z+160z^3-32z^5$$ $$\cdots$$

Unfortunately, the pattern is not immediately apparent, though there is one: https://en.wikipedia.org/wiki/Hermite_polynomials#Explicit_expression.

As said by others, it is much easier to consider the development of $e^{-z}$ and substitute $z^2$ for $z$.

Answer 4

$$ e^{x} = 1 + x + \dfrac {x^{2}}{2!} +\dfrac {x^{3}}{3!} + \cdots = \sum_{n=1}^{\infty} \; \dfrac{x^{n-1}}{(n -1)!} $$

Now replace the value of $ x $ by $ -z^{2} $ in the above Taylor series representation:

$$ \boxed { e^{-z^{2}} = 1 - z^{2} + \dfrac {z^{4}}{2!} - \dfrac {z^{6}}{3!} + \cdots = \sum_{n=1}^{\infty} \; (-1)^{n -1} \; \dfrac{z^{2n-2}}{(n -1)!}} $$