When the value of $x$ is small, such as when $x$ is less than $1$, we can use the Taylor series to approximate its behavior. The first few terms of the series often provide a very good approximation. However, when $x$ is much larger, the terms of the series that involve higher powers of $x$, like $x^n/n$!, may no longer be small and it may not be obvious from the first few terms that $e^{-x}$ is close to $0$. Is there a way to improve the accuracy of the approximation in these cases? Can we use taylor series at infinity in this case? How can I use the Taylor series to determine that $e^{-x}$ is approximately $0$ when $x$ is large?
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1Maybe: use Taylor's series to approximate $e^x$, then take the reciprocal to get $e^{-x}=\frac{1}{e^x}$. – Jan 01 '23 at 19:29
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Check these: https://math.stackexchange.com/q/71357/42969, https://math.stackexchange.com/q/627702/42969, https://math.stackexchange.com/q/576227/42969. – Martin R Jan 01 '23 at 19:32
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Is it straightforward to obtain an explicit error term like in the Taylor series? – XiaoHei Jan 01 '23 at 19:32
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1Does this answer your question? linear or quadratic approximation for $\exp(-x)$ for large $x$ – Sebastiano Jan 01 '23 at 19:52
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1Why do you need to do that? Given how simple $e^{-x}$ is as an expression for large $x$, the approximation technique to be used will almost certainly depend on what you’re trying to do. – Aphelli Jan 01 '23 at 20:43
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Guess: Try to get a Taylor series for $e^{-\frac{1}{y}} $ in $y$. – herb steinberg Jan 01 '23 at 22:45