Question

Find the power series for \[e^{-x^2/2}\]

Answer 1

Use the power series \(\displaystyle \large e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) and replace \(\large x\) with \(\large -\dfrac{x^2}{2}\).

Does this make sense?

Answer 2

Would I end up with \[\sum_{n=0}^{\inf}\frac{ {(\frac{ -x^2 }{ 2 }})^n }{ n! }\]

Answer 3

Yes, which you can rewrite as\(\large \displaystyle e^{-x^2/2} = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{2^n n!} \). :-)

Answer 4

ahh, thanks. And how would I approximate an integral to within 4 decimal places?

for example\[\frac{ 1 }{ \sqrt(2\pi) } \int\limits_{-1}^{1}e^{\frac{-x^2}{2}} dx = 0.68\]

Answer 5

Hmm...I'll have to think about that for a bit.