2个函数的渐近展开式相等, 但它们的差趋于1

hbghlyj

奇怪的事:

1. Asymptotic[Exp[-y]/(\[Epsilon] y)-Coth[\[Epsilon] y],{\[Epsilon],0,1}]

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$\frac{\exp (-y)}{y \epsilon }-\coth (y \epsilon )$展开为$\frac{e^{-y}}{y \epsilon }-\frac{1}{y \epsilon }$

1. Asymptotic[Exp[-y]/(\[Epsilon] y)+Sin[\[Epsilon] y]/(\[Epsilon] y)-Coth[\[Epsilon] y],{\[Epsilon],0,1}]

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$\frac{\exp (-y)}{y \epsilon }+\frac{\sin (y \epsilon )}{y \epsilon }-\coth (y \epsilon )$展开也为$\frac{e^{-y}}{y \epsilon }-\frac{1}{y \epsilon }$
但是它们的差:

1. Asymptotic[Sin[\[Epsilon] y]/(\[Epsilon] y),{\[Epsilon],0,1}]

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$\frac{\sin (y \epsilon )}{y \epsilon }$展开为1

hbghlyj

求函数$$f(x,\epsilon) = \frac{e^{-x/\epsilon}}{x}+\frac{\sin (x)}{x}-\coth (x)$$的渐近展开, 直到$O(\epsilon)$, 其中$x>0$和$0<\epsilon \ll 1$.
Asymptotic inner expansion

On the other hand, the inner region is $x=O(\epsilon)$, but when I tried to compute the inner expansion via the rescaling $x=\epsilon y$ where $y=O(1)$ as $\epsilon \rightarrow 0$, I got
\begin{align*}
\ f(x,\epsilon) & =F(y,\epsilon) \\
\ & = \frac{e^{-y}}{\epsilon y} + \frac{\sin (\epsilon y)}{\epsilon y} - \coth (\epsilon y) \\
\ & = \Bigl (\frac{e^{-y}}{y}\epsilon ^{-1} \Bigr)+ \Bigl(1-\frac{(\epsilon y)^2}{3!}+\frac{(\epsilon y)^4}{5!}+\cdots \Bigr)- \Bigl((\epsilon y)-\frac{(\epsilon y)^3}{3}+\frac{2(\epsilon y)^3}{15}-\cdots \Bigr)^{-1} \\
\end{align*}
and we get a $(\frac{e^{-y}}{y}-\frac 1y) \epsilon ^{-1}$ term in the expansion even though we have assumed that $y = O(1)$.

用Mathematica计算:
级数展开到1阶:

1. Series[Exp[-y]/(\[Epsilon] y)+Sin[\[Epsilon] y]/(\[Epsilon] y)-Coth[\[Epsilon] y],{\[Epsilon],0,1}]

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$$\frac{\frac{e^{-y}}{y}-\frac{1}{y}}{\epsilon }+1-\frac{y \epsilon }{3}+O\left(\epsilon ^2\right)$$
渐近展开到1阶:

1. Asymptotic[Exp[-y]/(\[Epsilon] y)+Sin[\[Epsilon] y]/(\[Epsilon] y)-Coth[\[Epsilon] y],{\[Epsilon],0,1}]

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$$\frac{e^{-y}}{y \epsilon }-\frac{1}{y \epsilon }$$
渐近展开到2阶:

1. Asymptotic[Exp[-y]/(\[Epsilon] y)+Sin[\[Epsilon] y]/(\[Epsilon] y)-Coth[\[Epsilon] y],{\[Epsilon],0,2}]

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$$1-\frac{1}{y \epsilon }+\frac{e^{-y}}{y \epsilon }-\frac{y \epsilon }{3}-\frac{y^2 \epsilon ^2}{6}$$