$\textbf{K}$:Elliptic K Function
I tried to calculate $\lim_{x\to 0}{\frac{\textbf{K}(1-x)}{\ln(x)}}$.
I used Taylor series and L'Hôpital's rule and failed.
I used Mathematica and it gave me Puiseux series of $\textbf{K}(1-x)$:
$$\textbf{K}(1-x)=2\ln(2)-\frac12\ln(x)+(-\frac14+\frac12\ln(2)-\frac18\ln(x))x+O(x^2)$$
Wikipedia says Puiseux series are like:
$$f(x)=\sum_{k=k_0}^\infty{c_kx^{k/n}}$$
Why does $\ln$ function appear?
And how did Mathematica find the series of it?
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Kemono Chen
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What's $\textbf{K}$? I'm not familiar with this notation – Yuriy S Feb 16 '18 at 13:57
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I think it's the complete elliptic integral ie $K(m) =\int_0^{\pi/2}(1-m\sin^2x)^{-1/2},dx$ – Paramanand Singh Feb 16 '18 at 14:43
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See this question https://math.stackexchange.com/q/442618/72031 – Paramanand Singh Feb 16 '18 at 14:44
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1Given the expansion in linked question you get the answer as $-1/2$. This also follows from your expansion. Or you can directly prove the asymptotic relation $K(k') =\log(4/k)+o(1)$ as given here. – Paramanand Singh Feb 16 '18 at 14:55