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Does $\tanh(x)$ have an asymptotic expansion for $x \rightarrow \infty$?

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Well, since $\tanh\frac1{x}$ doesn't seem to have a Maclaurin series expansion... have a look at these though. –  J. M. Aug 16 '11 at 4:03
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The natural estimates use $$\tanh x=(1-e^{-2x})(1-e^{-2x}+e^{-4x}-e^{-6x}+\cdots).$$ –  André Nicolas Aug 16 '11 at 4:09
    
I guess what I meant was does it have an expansion in terms of powers of x? Perhaps I should edit my question. –  Keith Aug 16 '11 at 4:17
    
Then my first comment applies. Since $\exp\frac1{x}$ has an essential singularity at $x=0$, an asymptotic expansion of the form $c_0+\frac{c_1}{x}+\frac{c_2}{x^2}+\dots$ can't be done. –  J. M. Aug 16 '11 at 4:39

2 Answers 2

up vote 1 down vote accepted

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).

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Re your last argument: simply, $\tanh(x)=1+o(x^{-n})$ for every $n$. –  Did Aug 16 '11 at 8:38
    
You're absolutely right. I guess I should have said instead that although there's an expansion with powers of $x$, it's a very rough approximation compared to the one with powers of $e^{-2x}$ (because $e^{-x}$ is so much smaller than any power of $x$). –  Joel Cohen Aug 16 '11 at 16:23
    
@Didier Piau: Your comment answered the question I meant to ask. Thanks. –  Keith Aug 20 '11 at 5:59
    
@Keith: Glad to help. –  Did Aug 20 '11 at 7:10

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 $x\to\infty$.

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