# Taylor series at infinity

I'm required to make a Taylor series expansion of a function $$f(x) = \arctan(x)$$ at $$x = +\infty$$. In order to do this I introduce new variable $$z = \frac{1}{x}$$, so that $$x \to +\infty$$ is the same as $$z \to +0$$. Thus I can expand $$f(z)$$ at $$z = 0$$: $$f(z) = z - \frac{z^3}{3}+\frac{z^5}{5}-...$$ Then I try to reverse the substitution but this is either incomplete or incorrect: $$f\bigg(\frac{1}{x}\bigg) = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - ...$$

We could use the Taylor expansion of $$\arctan(x)$$ when $$x \rightarrow 0$$ and that $$\forall x \in \mathbb{R}^*_+, \arctan(x) + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}$$ and find the correct result. But as @janosch points out, it's faster to use $$\arctan'\bigg(\frac{1}{x}\bigg)= -\frac{1}{x^2 +1}$$ and fix the integrating constant to $$\lim_{x \rightarrow 0} \arctan(1/x) = \frac{\pi}{2}$$ and finally get $$\arctan(x) =_{x \rightarrow + \infty} \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + o\bigg(\frac{1}{x^5}\bigg).$$
• You don't need to use that. You could start by calculating $\arctan'(1/x) = -1/(1+x^2)$ (by chain rule), then integrate, and fix the resulting integration constant by considering the limit $\lim_{x \rightarrow 0} \arctan(1/x) = \frac{\pi}{2}$. Jul 14 '19 at 11:11
• @Stunt-man-mike Let $f(x) = \arctan(x) + \arctan(\frac{1}{x}) - \frac{\pi}{2}, \forall x \in \mathbb{R}^*_+$. By differentiating, you find $f'(x) = 0, \forall x \in \mathbb{R}^*_+$. So $\forall x \in \mathbb{R}^*_+, f(x) = f(1)$. Jul 14 '19 at 11:14
Your expansion is not correct. You should find $$f(z) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^2} - \frac{1}{5 x^5} + \mathcal{O}\left(\left(\frac{1}{x}\right)^7\right),$$ which is a perfectly good expansion in the small parameter $$\frac{1}{x}$$ for $$x \rightarrow \infty$$. See also Taylor expansion at infinity.
• Your expansion is wrong, it's $f(x) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} -\frac{1}{5x^5} + \mathcal{O}\left(\left(\frac{1}{x}\right)^7\right)$ Jul 14 '19 at 10:56