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I have a function, $f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)$. Let's assume the argument $x$ is sufficiently high. If so, how can I approximate $f(x)$ in terms of $x$? I have computed $f(x)$ with MATLAB. It looks like that $f(x)$ can be approximated to ${1 \over {2x}}$. That is, ${\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x) \approx {1 \over {2x}}$ for a large $x$. It this right?

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Yes. You have $$\tag1 \arctan t=t-o(t^3),$$ and $$\tag2\sqrt{1+x^2}-x=\frac1{\sqrt{1+x^2}+x}.$$Also, $$\tag3 \frac1{\sqrt{1+x^2}+x}-\frac1{2x}=\frac{x-\sqrt{1+x^2}}{2x(\sqrt{1+x^2}+x)} =-\frac1{2x(\sqrt{1+x^2}+x)^2}=o(1/x^3). $$ Combining everything, $$ \arctan(\sqrt{1+x^2}-x)=\frac1{\sqrt{1+x^2}+x}+o(x^{-3})=\frac1{2x}+o(x^{-3}) $$

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  • $\begingroup$ Thank you for nice explanation. $\endgroup$ – Jay Lee Mar 23 '18 at 5:54
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As Martin Argerami already answered, you are totally write.

If you want more accuracy, you could use Taylor series $$\sqrt{1+x^2}=x+\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16 x^5}+O\left(\frac{1}{x^7}\right)$$ $$\sqrt{1+x^2}-x=\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16 x^5}+O\left(\frac{1}{x^7}\right)$$ Now, using $$\tan^1(\epsilon)=\epsilon -\frac{\epsilon ^3}{3}+\frac{\epsilon ^5}{5}+O\left(\epsilon ^7\right)$$ you should arrive to $$f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)=\frac{1}{2 x}-\frac{1}{6 x^3}+\frac{1}{10 x^5}+O\left(\frac{1}{x^7}\right)$$ or more generally $$f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)=\sum_{i=1}^\infty \frac{(-1)^{i-1} }{(4 i-2)x^{2 i-1}}$$

Using $x=5$ (which is quite small), the truncated series given above leads to $\frac{9253}{93750}\approx 0.0986987$ while the "exact" value would be $\approx 0.0986978$.

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  • $\begingroup$ This was also helpful to me. Thank, Leibovici. $\endgroup$ – Jay Lee Mar 23 '18 at 5:55

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