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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: $$f\bigg(\frac{1}{x}\bigg) = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - ...$$

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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 $$ \arctan(x) =_{x \rightarrow + \infty} \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + o\bigg(\frac{1}{x^5}\bigg) $$

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  • $\begingroup$ Thanks, your explanation make sense! But could you kindly comment where this property comes from: arctan(x) + arctan(1/x) - pi/2 ?? $\endgroup$ – Stunt-man-mike Jul 14 at 10:54
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    $\begingroup$ 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 $\lim_{x \rightarrow 0} \arctan(1/x) = \frac{\pi}{2}$. $\endgroup$ – janosch Jul 14 at 11:11
  • $\begingroup$ @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 $\forall x \in \mathbb{R}^*_+, f(x) = f(1)$. $\endgroup$ – Monadologie Jul 14 at 11:14
  • $\begingroup$ @janosch You're right. $\endgroup$ – Monadologie Jul 14 at 11:18
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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.

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  • $\begingroup$ Your expansion is wrong, it's $f(x) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} -\frac{1}{5x^5} + \mathcal{O}\left(\left(\frac{1}{x}\right)^7\right)$ $\endgroup$ – Monadologie Jul 14 at 10:56
  • $\begingroup$ @Monadologie you're right of course, my bad. I fixed it. $\endgroup$ – janosch Jul 14 at 10:57

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