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?


2 Answers 2


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 $\sqrt[p]{R}$.

  • $\begingroup$ So we must show that the replacement still yields a valid power series and that it has the same radius of convergence? $\endgroup$ Feb 12, 2016 at 17:07
  • $\begingroup$ @ChristianIvicevic we must show that it yields a valid power series by showing that it can be written in form $$\sum_{n=0}^\infty a_nx^n$$ the radius of convergence is not important (unless you had to do something with it) $\endgroup$ Feb 12, 2016 at 17:17

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 $t=0$.


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