<s>[INST] Maclaurin series of $\tan(x+x^2)$ to order 3 I know that the maclaurin series of $\tan(x)$ is $\tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+...$, then shouldn't be $\tan(x+x^2)=(x+x^2)+\frac{1}{3}(x+x^2)^3+\frac{2}{15}(x+x^2)^5+...$? 
Mathematica actually gives me $\tan(x+x^2)=x+x^2+\frac{x^3}{3}+o(x^4)$ to order 3.
 [/INST] Expanding the monomial terms gives Mathematica's result – and you can stop at $(x+x^2)^3$:
$$(x+x^2)+\frac13(x+x^2)^3=x+x^2+\frac13(x^3\color{lightgrey}{+3x^4+3x^5+x^6})=x+x^2+\frac{x^3}3$$
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