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Mathematica gives me the following expression which works well for $s$ near $0$, any idea how to derive this manually?

$$\left(\frac{s}{2}+\frac{\sqrt{2} \sqrt{s}}{\pi -2 \tan ^{-1}\left(\frac{\sqrt{s}}{\sqrt{2}}\right)}\right)^{-1} \approx \frac{\pi }{\sqrt{2} \sqrt{s}}$$

expr = 1/(s/2 + (Sqrt[2] Sqrt[s])/(\[Pi] - 2 ArcTan[Sqrt[s]/Sqrt[2]]));
asymp = Asymptotic[expr, s -> 0]

This appears to be the first term in Series expansion below. Using $x=2s^2$ replacement, the series looks like below

$$ \begin{array}{ccc} & \text{order} & \text{expr} \\ & 0 & \frac{\pi }{2 x}-\frac{\pi ^2}{4}-1 \\ & 1 & \frac{1}{8} \left(8 \pi +\pi ^3\right) x+\frac{\pi }{2 x}-\frac{\pi ^2}{4}-1 \\ & 2 & \frac{1}{48} \left(-32-36 \pi ^2-3 \pi ^4\right) x^2+\frac{1}{8} \left(8 \pi +\pi ^3\right) x+\frac{\pi }{2 x}-\frac{\pi ^2}{4}-1 \\ \end{array}$$

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Yaroslav Bulatov
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    What is Puisseux about this? Replace $s=2x^2$ and everything has integer powers. – Lutz Lehmann May 16 '23 at 07:24
  • Would not a loglog plot be a better plot to see the local expansion around $s=0$? – Lutz Lehmann May 16 '23 at 08:22
  • This belongs on Mathematica SE – Тyma Gaidash May 16 '23 at 12:49
  • @LutzLehmann if I ask for more terms I get result below, isn't that Puissex series? Trying to figure out how I would derive this manually $$O\left(s^{3/2}\right)+\frac{1}{96} \left(-32-36 \pi ^2-3 \pi ^4\right) s+\frac{\pi \left(8+\pi ^2\right) \sqrt{s}}{8 \sqrt{2}}+\frac{\pi }{\sqrt{2} \sqrt{s}}+\frac{1}{4} \left(-4-\pi ^2\right)$$ – Yaroslav Bulatov May 16 '23 at 16:51
  • Yes, it is. The concern was why introduce $\sqrt{s}$ at all, it looks somewhat artificial without context. The expansion can also done in the above-mentioned $x$, using only power or Laurent series. – Lutz Lehmann May 16 '23 at 17:00
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    @LutzLehmann for context, this is the solution to differential equation here. Edited post with expansion in $x$, the leading term is $O(x^{-1})$, this is not the usual Taylor series, hence trying to figure out how coefficients are obtained – Yaroslav Bulatov May 16 '23 at 17:34
  • This is just a limit problem not ? – mick May 21 '23 at 16:25

2 Answers2

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Notice that $\tan^{-1}(x)\approx x$ if $x\ll 1$, so \begin{align} \frac{s}{2}+\frac{\sqrt{2} \sqrt{s}}{\pi -2 \tan ^{-1}\left(\frac{\sqrt{s}}{\sqrt{2}}\right)}&\approx \frac{s}{2}+\frac{\sqrt{2} \sqrt{s}}{\pi -\sqrt{2s}} \\ &=\frac{\sqrt{2s}}{\pi}\left(\frac{\pi}{2}\sqrt{\frac{s}{2}}+\frac{1}{1-\frac{\sqrt{2s}}{\pi}}\right) \\ &=\frac{\sqrt{2s}}{\pi}\left(1+O(\sqrt{s})\right), \end{align} hence $$ \left(\frac{s}{2}+\frac{\sqrt{2} \sqrt{s}}{\pi -2 \tan ^{-1}\left(\frac{\sqrt{s}}{\sqrt{2}}\right)}\right)^{-1}\approx\frac{\pi}{\sqrt{2s}}\quad\text{if $s\ll 1$.} $$

Gonçalo
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Denote your expression by $f(s)$. Introducing the new variable $z>0$ via $s=2z^2$ yields $$ zf(2z^2 ) = \cfrac{1}{{z + \cfrac{1}{{\pi/2 - \arctan (z)}}}}. $$ By analytic continuation, the right-hand side is a holomorphic function of $z$ in the disc $|z|<0.492912456\ldots$, where $-0.492912456\ldots$ is the unique real root of the equation $z\operatorname{arccot}(z)=-1$. Its Maclaurin series is $$ zf(2z^2 ) = \frac{\pi }{2} - \frac{{\pi ^2 + 4}}{4}z + \frac{{\pi (\pi ^2 + 8)}}{8}z^2 - \ldots \,. $$ Thus, on the Riemann surface associated with the square-root, $$ f(s) = \frac{\pi }{{2^{1/2} }}s^{ - 1/2} - \frac{{\pi ^2 + 4}}{4} + \frac{{\pi (\pi ^2 + 8)}}{8}s^{3/2} - \ldots $$ provided $|s|<0.485925370\ldots\,$.

Gary
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