Deducing the series expansion of $\arctan(x^2)$ via the series expansion of $\arctan(x)$ at $x=0$ Comparing the series expansion of $\arctan(x^2)$ and $\arctan(x)$ at $x=0$ it looks like one can take the result from $\arctan(x)$ and replace each $x$ with $x^2$ to deduce the series expansion of $\arctan(x^2)$. Is this just true in this specific case or is this approach generally valid? Do you have any other examples or counter-examples for this observation?
 A: This approach is perfectly valid. When we have a series $$\sum_{n=0}^\infty a_nx^n$$
then replacing $x\mapsto x^2$ we get
$$\sum_{n=0}^\infty a_nx^{2n}=\sum_{n=0}^\infty b_nx^n$$
which is a power series too with
$$b_n=\begin{cases}
a_{n/2},&\text{when $n$ is even}\\
0,&\text{otherwise}
\end{cases}$$
Moreover the radius of convergence is $\sqrt{R}$ where $R$ is the radius of convergence of the first series, because
$$|x^2|<R\implies |x|<\sqrt{R}$$

The same rule applies for replacement $x\mapsto x^p$ where $p=1,2,...$ and then the radius of convergence is equal to $\sqrt[p]{R}$.
A: The series expansion of $\arctan(x)$ converges for any value of $x$ in the radius of convergence, $|x|<R$.
But you are quite free to set $x=t^2$, and the series expansion evaluated at $x=t^2$ yields the same result, which is obviously $\arctan(t^2)$. The convergence conditions becomes $|t^2|<R$.
In addition, the formal series as a function of $x$ can be rewritten as a formal series in terms of $t$, and by the reciprocal of Taylor's expansion theorem, you get all derivatives of $\arctan(t^2)$ at $t=0$.
