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That is, I am looking for an algebraic function $f(x)$ that approximates $\arctan x$ for large values of $x$.

The approximation could be reasonably modest -- perhaps something like

$$\tan (f(x)) = \frac{\pi}{4} + O\left(\frac{1}{x^2}\right).$$

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3 Answers 3

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Use the following

$$\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} \Leftrightarrow$$

$$\arctan(x) = \frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$$

The series for $\arctan(x)$ is

$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$

Now substitute $\frac1x$ to get the result

$$\arctan(x)=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^3}-\frac{1}{5x^5} +\ldots$$

So for large values of $x$

$$\arctan(x)\sim\frac{\pi}{2}$$

which is logical, because $\displaystyle \lim_{x\to\infty}\arctan(x)=\frac{\pi}{2}$.

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$$ \arctan\left(x\right) ={\pi \over 2}\,{\rm sgn}\left(x\right) - \arctan\left(1 \over x\right) \approx{\pi \over 2}\,{\rm sgn}\left(x\right) - {1 \over x}\,,\qquad \left\vert\,x\,\right\vert \gg 1 $$

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This approximation converges much faster: $$ \arctan(x)=2\sum_{n=1}^{\infty}{\frac{1}{2n-1}\frac{{{a}_{n}}\left(x\right)}{a_{n}^{2}\left(x\right)+b_{n}^{2}\left(x\right)}}, $$ where $$ \begin{align} & a_1(x)=2/x,\\ & b_1(x)=1,\\ & a_n(x)=a_{n-1}(x)\,\left(1-4/x^2\right)+4b_{n-1}(x)/x,\\ & b_n(x)=b_{n-1}(x)\,\left(1-4/x^2\right)-4a_{n-1}(x)/x. \end{align} $$ Therefore, by taking the first term only, the following asymptotic approximation can be used: $$ \arctan\left(x\right) ={\pi \over 2}\,{\rm sgn}\left(x\right) - \arctan\left(1 \over x\right) \approx \frac{\pi}{2}{\rm sgn}(x)-\frac{4 x}{4 x^2+1}\,,\qquad \left\vert\,x\,\right\vert \gg 1. $$

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  • $\begingroup$ +1 Where is the summation for $\arctan x$ from? $\endgroup$
    – Toby Mak
    Nov 16, 2019 at 2:34
  • $\begingroup$ Wikipedia, approximations of pi. $\endgroup$
    – a cool guy
    Nov 16, 2019 at 2:39

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