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Find a power series for F, such that $F'(x)=e^{-x^2}$.

Don't understand how to come up with the solution

kiwifruit
  • 707
  • Do you know the power series for $e^x$, and if so, do you know how to get the power series for $e^{-x^2}$ from it? – Scott H. Apr 15 '14 at 05:10

3 Answers3

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We have

$$F'(x)=e^{-x^2}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{n!}$$ with radius of convergence $+\infty$ hence by integrating term by term we find:

$$\int_0^xF'(t)dt=F(x)-F(0)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)n!}$$

2

Recall that the power series for $e^x$ is $$ \sum_{n=0} ^\infty { \frac{x^n}{n!} } . $$ Thus, the power series for $e^{-x^2}$ is $$ \sum_{n=0} ^\infty { \frac{(-x^2)^n}{n!} } = \sum_{n=0} ^\infty { \frac{(-1)^n x^{2n}}{n!} } . $$ Integrating term by term yields $$ F(x) = \int F'(x) = \int \sum_{n=0} ^\infty { \frac{(-1)^n x^{2n}}{n!} } = \sum_{n=0} ^\infty { \frac{(-1)^n x^{2n + 1}}{(2n+1)n!} } . $$

izœc
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The power series for $F'(X)$ is $$F'(x) = \sum_0^\infty \frac{(-x)^{2n}}{n!} = \sum_0^\infty \frac{(-1)^nx^{2n}}{n!}$$ We now integrate each side and get: $$ \begin{align} F(X) &= \sum_0^\infty \frac{(-1)^n (x)^{2n+1}}{(2n+1)(n!)} \end{align}$$

druckermanly
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