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anonymous
 3 years ago
Unsure....
\[f(x)=x^2tan^{1}(x^3)\]
anonymous
 3 years ago
Unsure.... \[f(x)=x^2tan^{1}(x^3)\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll post the remainder of the question in just a second...bear with me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From the MacLaurin Table: \[tan^{1}(x)=\sum_{n=0}^{\infty}(1)^n\frac{x^{2n+1}}{2n+1}=x\frac{x^3}{3}+\frac{x^5}{5}\frac{x^7}{7}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is what I've come up with so far: \[x^2\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+1}}{2n+1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmmm....could it be: \[x^2\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+3}}{2n+1}\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I meant: \[\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+3}}{2n+1}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2I think both the first and third solutions you have are correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am inclined to agree

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes! So this would be correct? \[\sum_{n=0}^{\infty}(1)^n\frac{(x^3)^{2n+3}}{2n+1}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2Looks good to me. If you wanted to, you could also say\[(x^3)^{2n+3}=x^{6n+9}\]
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