How Accurate Is the Maclaurin Series for \( f(x) = x^2e^{-2x^2} \)?

[RadiationX]

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

 

[RadiationX]

does anyone have a review for me?

 

[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.

 

[dextercioby]

Are u sure...?

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

Daniel.

 

[RadiationX]

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

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

my oginal post of part (b) was incorrect. now this is correct