I need to find the two term asymptotic expansion of $\sin\left(\pi + \exp(-1/\epsilon)\right)$ as $\epsilon$ tends to zero, but the exponential term is confusing me...
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4Assuming $\epsilon \rightarrow 0^{+}$ the sign of $\frac{1}{\epsilon}$ makes a big difference. Please adjust the title or the text. – gammatester Oct 27 '14 at 15:13
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Sorry, yeah, it does. – Martin MacDonald Oct 27 '14 at 15:27
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1You've missed the point of gammatester's comment. Is it $\exp(-\frac1\epsilon)$ or $\exp(\frac1\epsilon)$? Your title says one and your text says the other, and as gammatester said, it makes a big difference. Please clarify. – Oct 27 '14 at 15:45
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Apologies again! – Martin MacDonald Oct 27 '14 at 16:03
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Using Maple I am obtaining
$$-{{\rm e}^{-{\epsilon}^{-1}}}+{\frac {1}{6}}\,{{\rm e}^{-3\,{\epsilon}^{-1}}}-{ \frac {1}{120}}\,{{\rm e}^{-5\,{\epsilon}^{-1}}}+{\frac {1}{5040}}\,{ {\rm e}^{-7\,{\epsilon}^{-1}}}-{\frac {1}{362880}}\,{{\rm e}^{-9\,{ \epsilon}^{-1}}}+O \left( {{\rm e}^{-11\,{\epsilon}^{-1}}} \right) $$
Juan Ospina
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1Writing $x=-e^{e^{-1/\epsilon}}$, this amounts to $$\sin(\pi-x) = \sin x = x-\frac{1}{3!} x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7-\frac{1}{9!}x^9+O(x^{11})$$ i.e. the first five terms of the Taylor expansion. – Semiclassical Oct 27 '14 at 16:11
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