<s>[INST] Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$ 
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$

My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x)))=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{4}(x), r_{4}(x)=o(x)$$
That is why fifth expression of Taylor's is $\frac{2}{15}x^5$ for me. However Mathematica say that: $$\sin \tan x=x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^5)$$ Where have I a mistake?
 [/INST] Taylor's expansions up to the same order can be composed. So you have to consider the Taylor polynomial of degree $5$ for $\sin x$, and substitute $x$  with the degree $5$ Taylor polynomial for $\tan x$, truncating the successive powers of the latter polynomial at degree $5$.
As a start-up, here is the computation for degree $3$:
$$\bigl(x+\tfrac13x^3 +\tfrac2{15}x^5\bigr)^2=x^2+\tfrac19x^4+o(x^5),$$
\begin{align}
&\text{so that}\hspace{6em} &\bigl(x+\tfrac13x^3 +\tfrac2{15}x^5\bigr)^3&=\bigl(x^2+\tfrac19x^4+o(x^5)\bigr)\bigl(x+\tfrac13x^3 +\tfrac2{15}x^5\bigr)&\hspace{6em} \\
&&&=x^3+\tfrac13x^5+\frac19x^5+o(x^5)=\color{red}{x^3+\tfrac49x^5}+o(x^5).
\end{align}
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