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

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I suspect that 1 is just an excuse for stating the formula. Note that you must substitute $-x^2$ for $x$ For 2 you're basically right (but use the variable $t$!), however, you need to think about why you can interchange summation and integration. – t.b. Apr 23 '11 at 5:57
So in the value of sigma of (-1)^n*x^2n/n!, I should use t instead of x? Also, how can we integrate this, since this format I am not sure about: u-subs not nice. – mary Apr 23 '11 at 6:06
What better justification can you give other than the fact that we look at infinitely many rectangles, upon which we approximate the sigma summation with an integral? – mary Apr 23 '11 at 6:07
The formal computation starts with writing $\int \sum = \sum \int$ and then integrate term by term. This first equality is more subtle than you state in your second comment. This should be discussed justified somewhere after p.184 in Ross's book (google doesn't let me look at the relevant pages). I don't understand the second part of the first comment. Concerning the first part, note that you should write $F(x) = \int_{0}^x (\text{something depending on }t)\,dt$ (the integrand shouldn't depend on the boundaries of integration!). – t.b. Apr 23 '11 at 6:14
ALSO, in part b, we should still have a sigma in our answer, or should we get the taylor series expansion for it? – mary Apr 23 '11 at 6:15
up vote 1 down vote accepted

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.

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I only dealt with the formal part of the question because that's what seemed to cause problems here. I can expand on the (more interesting) theoretical part if you ask me to, but I'd appreciate if you had a closer look at the sections of Ross first and ask specific questions. – t.b. Apr 23 '11 at 7:25
can you please expand on the more theoretical part please? – mary Apr 27 '11 at 5:17
@user8917: Have a look a this question and the hints given by Eric Naslund. – t.b. Apr 28 '11 at 11:44

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