Use a known power series expansion to find the power series
representation of the integral function $g(x) =\int_0^x e^{-t^2}dt$
centered at $a=0$
My approach
Note that $g'(x) = e^{-x^2}$. We also know the Maclaurin Series for
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.Then, $$g'(x)=e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$$
$$g(x) = \int \bigg[\sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}\bigg]
dx$$ $$g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int x^{2n}dx $$
$$g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)} +K,
\text{ where $K$ is unknown}$$Consider $x=0$, we figured out that $g(0) = 0 +K = 0$, hence $K=0$. Hence
$$g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)}$$
Is this right? I cannot differentiate between power series and Maclaurin series or stuffs like that. Can anybody clarify on this too?
Best Answer
It is correct, although I would have integrated each term between $0$ and $x$ instead of the indefinite integral.
Power series are series of the form $\sum_{n=0}^\infty a_n(x-x_0)^n$. A Taylor series is a series where $a_n=f^{(n)}(x_0)/n!$ and $f$ is a $C^\infty$ on a neighborhood of $x_0$. If $x_0=0$, it is called a Maclaurin series.