## anonymous 4 years ago Unsure.... $f(x)=x^2tan^{-1}(x^3)$

1. anonymous

I'll post the remainder of the question in just a second...bear with me

2. anonymous

From the MacLaurin Table: $tan^{-1}(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}$

3. anonymous

This is what I've come up with so far: $x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+1}}{2n+1}$

4. anonymous

Hmmmm....could it be: $x^2\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}$?

5. anonymous

I meant: $\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}$

6. KingGeorge

I think both the first and third solutions you have are correct.

7. anonymous

I am inclined to agree

8. anonymous

Yes! So this would be correct? $\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+3}}{2n+1}$

9. KingGeorge

Looks good to me. If you wanted to, you could also say$(x^3)^{2n+3}=x^{6n+9}$

10. anonymous

ok Thanks!

11. KingGeorge

You're welcome.