# Finding a Maclaurin Series

1. May 22, 2005

I need some peer review of this quesion. Let $$f(x)=x^2e^{-2x^2}$$

(a) find the maclaurin series of f

(b) find the 100th degree maclaurin polynomial of f.

for part a i have:$$\sum_{n=0}^{\infty}\frac{(-2)^nx^{2n+2}}{n!}$$

and for b:$$\sum_{n=0}^{99}\frac{(-2)^nx^{2n+2}}{n!}$$

Last edited: May 22, 2005
2. May 23, 2005

does anyone have a review for me?

3. May 23, 2005

### Galileo

It looks good. I'm not sure what you mean by the 100th degree MacLaurin polynomial, but if you mean the expansion up to the 100th power of x you have too many terms, since it goes up to the 2(99)+2=200th power of x.

4. May 23, 2005

### dextercioby

Are u sure...?

What is $$f^{\mbox{iv}}(x)$$...?And then,of course,$$f^{\mbox{iv}}(x)$$...?

Daniel.

5. May 26, 2005

This is the 100th degree maclaurin polynomial let $$100=2n+2$$ so that $$n=49$$ now we have this:$$\sum_{n=0}^{49}\frac{(-2)^nx^{2n+2}}{n!}$$