How to find the Maclaurin series for $e^{-x^2}$ I don't know how to get $$1-x^2+\frac{x^4}{2!}-\cdots.$$
I think it is too complex, if not impossible, to just use the definition of Maclaurin series.

Using the definition:
consider the situation that all terms containing $x$ equal to zero,
$f(0)=1$, $\,\,f'(0)=0$, $\,\,f''(0)=-2$, $\,\,f'''(0)=0$, $\,\,f^{(4)}(0)=4$, $\,\,f^{(5)}(0)=0$, $\,\,f^{(6)}(0)=16$
so I get the result: $\sum_{n=0}^∞(-2)^n\frac{x^{2n}}{(2n)!}$
Why...
 A: Take the power serie of $e^Y$
$$e^{Y} = \sum_{n=0}^{\infty} \frac{Y^n}{n!}$$
Now replace $Y$ by $-x^2$, and you get
$$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$$
Then you have
$$e^{-x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}$$
A: why can't you substitute in $$e^u = 1 + \frac u{1!} +  \frac {u^2}{2!} +  \frac {u^3}{3!} +\cdots$$ for $u = -x^2$ and get $$e^{-x^2} = 1 - \frac {x^2}{1!} +  \frac {x^4}{2!} -  \frac {x^6}{3!} +\cdots$$
A: Just to add to the other methods here (which are much more efficient and much less tedious) you can also use the Maclaurin formula, which involves differentiating $f(x)$ as many times as needed, finding the values of these derivatives at $x=0$, and then dividing each term by the correct factorial and multiplying by the correct power of $x$.
$$f(x) = e^{-x^2}\implies f(0)=1$$
$$f'(x) = -2xe^{-x^2}\implies f'(0)=0$$
$$f''(x) = -2(1\times e^{-x^2}+(-2x)\times x e^{-x^2})=-2e^{-x^2}(1-2x^2)\implies f''(0)=-2$$
$$f'''(x) = -2(-4xe^{-x^2}+(1-2x^2)(-2xe^{-x^2}))\implies f'''(0)=0$$
etc. The tricky part is doing the differentiation of this function (and I may well have made a mistake as I don't have pen and paper on me!) but I'll leave you to practise your differentiation with the fourth derivative - I strongly recommend it if, as you say, you keep making mistakes doing it this way. Because there will always be an $e^{-x^2}$ factor, it's worth factorising your answer each time as then you only have to apply the product rule once each time.
The Maclaurin series will then be:
$$f(x) = f(0) + {f'(0)x\over1!}+ {f''(0)x^2\over2!}+ {f'''(0)x\over3!}+\cdots$$
$$=1+0+{-2x^2\over2!}+0+\cdots$$
$$=1-x^2+\cdots$$
