Asymptotic expansion of tanh at infinity?
By Sophia Vance
Does $\tanh(x)$ have an asymptotic expansion for $x \rightarrow \infty$?
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$\begingroup$Using the definition of $\tanh(x)$, $$ \begin{align} \tanh(x)&=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ &=\frac{1-e^{-2x}}{1+e^{-2x}}\\ &=\frac{2}{1+e^{-2x}}-1\\ &=1-2e^{-2x}+2e^{-4x}-2e^{-6x}+2e^{-8x}-\dots \end{align} $$ This converges to $\tanh(x)$ for all $x>0$. It also describes how $\tanh(x)$ behaves as $x\to\infty$.
$\endgroup$ $\begingroup$Write
$$\tanh(x) = \frac{1-e^{-2x}}{1+e^{-2x}}$$
Then you get an asymptotic expansion with powers of $e^{-2x}$ (which goes to $0$ as $x$ goes to infinity). It starts as
$$\tanh(x) = 1 - 2 e^{-2x} + o(e^{-2x})$$
There is no asymptotic expansion with powers of $x$ as that would imply that $e^{-x}$ has one (remember we're talking about asymptotic expansion when $x$ goes to infinity).
$\endgroup$ 4 $\begingroup$After a few tests, I've found that $\tanh(x) \approx 1 -1/x^5$ works very well for $x \gg 3$. I hope this can be useful. This resembles Pade approximation...
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