The Taylor series expansion of a function $f(x)$ about $x = 0$ (Maclaurin series) is given by:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \dots$

For $f(x) = e^{-x^2}$, let's find the first four terms of the Taylor series expansion.

1. $f(x) = e^{-x^2}$

   At $x = 0$, $f(0) = e^{0} = 1$.

2. First derivative: $f'(x) = \frac{d}{dx}(e^{-x^2}) = -2x e^{-x^2}$ At $x = 0$, $f'(0) = 0$.

3. Second derivative: $f''(x) = \frac{d}{dx}(-2x e^{-x^2}) = -2 e^{-x^2} + 4x^2 e^{-x^2}$ At $x = 0$, $f''(0) = -2$.

4. Third derivative: $f^{(3)}(x) = \frac{d}{dx}(-2 e^{-x^2} + 4x^2 e^{-x^2}) = 12x e^{-x^2}$ At $x = 0$, $f^{(3)}(0) = 0$.

5. Fourth derivative: $f^{(4)}(x) = \frac{d}{dx}(12x e^{-x^2}) = 12 e^{-x^2} - 24x^2 e^{-x^2}$ At $x = 0$, $f^{(4)}(0) = 12$.

Now, substituting these values into the Taylor series formula, we get the first four terms:

$e^{-x^2} \approx 1 - \frac{2}{2!}x^2 + \frac{12}{4!}x^4$ $e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2}$

This is the Taylor series expansion of $e^{-x^2}$ up to four terms.

Would you like more details or have any questions?

Here are five related questions to explore further:

1. How is the radius of convergence of a Taylor series determined?
2. What is the general formula for the Taylor series expansion of any function?
3. How does the Taylor series for $e^{-x^2}$ compare to the series for $e^{x}$?
4. Can the Taylor series for $e^{-x^2}$ be used to approximate values of this function for small $x$?
5. How can we generalize this approach to find higher-order terms?

Tip: The more terms you use in a Taylor series, the better the approximation near the center point (here $x = 0$).