Maclaurin Power Series for 1/(4x^2+1) and Integration of e^-x^2

[fsm]

I was hoping someone could check my work:

Find the maclaurin power series for the function:

a. f(x)=1/(4x^2+1)

b. f(x)= $$\int$$ e^-x^2 dx

For a I got $$(-1)^n*2nx^n$$. For b I don't know where to start.

 

[fsm]

For b I can get the approx to be

1-x^2+(x^4/2!)-(x^6/3!)+(x^8/4!)

So i get (x^-2n)/n!

 

[courtrigrad]

1 (a) $$f(x) = \frac{1}{1+4x^{2}}$$

This is equaled to $$\sum_{n=0}^{\infty} (-4)^{n}x^{2n}$$(b) $$e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$$

Thus $$e^{-x^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{n!}$$ and

$$\int \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{n!} = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{(2n+1)n!}$$

 

[fsm]

Thanks for the help. I just got b and was coming back to post my answer.