Power Series and Sequences and Series of Functions 
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*Observe that $e^{-x^2} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{n!} x^{2n}$ for $x$ an element of the reals. 

*Express $F(x) = \int_{0}^x e^{-t^2}\,dt$ as a power series


For part 1, I don't understand the purpose, are we just suppose to look at the power series for $e^x$ and substitute $x$ with $x^2$ to get the result to be proven? It seems rather trivial, and I am not understanding why we can do that. It makes sense, intuitively.
For part 2, all I got to do is to put in $\sum_{n = 0}^{\infty} \frac{(-1)^n}{n!} x^{2n}$ and integrate this with respect to $dx$, right? Also, when they say express, what else can I do for this.
BTW, this is 26.4 in Ross Analysis Elementary Calculus.
 A: For the first part use the fact that $e^{x} = \sum_{n=0}^{\infty} \frac{1}{n!} x^{n}$ and replace $x$ by $-x^2$ in this formula. This yields $$e^{-x^2} = \sum_{n=0}^{\infty} \frac{1}{n!} (-x^2)^n = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^{2n}.$$
This is the Taylor series expansion for $e^{-x^2}$ around $0$. To see this, you need to use the chain rule and the Taylor formula: Write $g(x) = f(-x^2)$ and derive a formula for $g^{(n)}(x)$ in terms of $f^{(n)}(x)$.
The second part can be done purely formally first:
$$F(x) = \int_{0}^{x} e^{-t^2}\,dt = \int_{0}^{x} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} t^{2n}\,dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{x} t^{2n}\,dt$$
and, as you computed correctly, this last expression evaluates to
$$F(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)} x^{2n+1}$$
which solves the exercise. However, we need to justify the third equation sign above, the one equating $\int \sum = \sum \int$. As I said, this is a bit subtle (and certainly the main part of the exercise!). The key here is uniform convergence of the power series of $e^{-t^2}$ on the interval $[0,x]$, and this should be discussed in detail in §25 or §26 of Ross's book.
Added. You can also go backwards and test that differentiating the power series $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)} x^{2n+1}$ term by term yields the power series for $e^{-x^2}$. Again, this involves interchanging two operators $\frac{d}{dx} \sum = \sum \frac{d}{dx}$ formally. The formal computation yields what you want, but it must be justified rigorously, and again this involves uniform convergence of the power series on compact intervals and somewhat subtle analysis.
