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(Note: I chose a general title, because I believe this discussion will be applicable to all other hyperbolic functions having an asymptote at infinity, but I will specifically be focusing on $\coth(x)$ in the following)

I encounter problems when trying to perform a Taylor expansion of $f(x) = \coth(x)$ at infinity (the term "asymptotic expansion" might be more accurate here).

I'm using the method described in Taylor expansion around infinity of a fraction. I.e. I define a function $$g(x) = f(1/x) = \coth(1/x),$$ Taylor expand $g(x)$ "at" $x = 0$, i.e. $$g(x) \approx T(x)$$ for $x << 1$, where $T(x)$ is (a truncated version of) the Taylor series of $g(x)$. I then introduce the variable substitution $y = 1/x$ and obtain $$g(1/y) = f(y) = \coth(y) \approx T(1/y)$$ for $y >> 1$.

The problem is obtaining $T(x)$. I've tried using Mathematica's "Series"-function, but it won't give me an answer. I then tried to calculate it as $$T(x) = \lim_{x\to 0}g(x) + \frac{\lim_{x\to 0}g'(x)}{1!}(x-0) + \frac{\lim_{x\to 0}g''(x)}{2!}(x-0)^2 + \ldots$$ but while, obviously, $\lim_{x\to 0}g(x) = 1$, it seems that $\lim_{x\to 0}g^{(n)}(x) = 0$ for $n>0$, no matter how high $n$ is (I've tried up to and including $n = 20$ in Mathematica).

So my question is: is the Taylor (asymptotic) expansion of $\coth(x)$ at infinity simply 1 (which would surprise me, since g(x) != 1, and since I've always thought that Taylor expansions of non-polynomial functions contain an infinite number of terms), or have I made a mistake somewhere, in my calculations or my general understanding of the problem?

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Yeah, I don't think this is going to work because the exponential function has an essential singularity at infinity. – Jason Zimba Mar 20 '14 at 16:52
Does it help for your purposes to write $\coth x = (1+e^{-2x})/(1-e^{-2x}) = (1 + e^{-2x})(1 + e^{-2x} + e^{-4x} + e^{-6x} + \dots)$? – Jason Zimba Mar 20 '14 at 17:25
That helps a lot, thanks! I've expanded the idea behind your comment in an answer below. – andreasdr Mar 21 '14 at 13:36
up vote 0 down vote accepted

I'll expand on Jason Zimbas comment: $$f(x) = \coth(x) = 1 + 2e^{-2x} + 2e^{-4x} + \ldots$$ $$g(x) = 1 + 2e^{-2/x} + 2e^{-4/x} + \ldots$$ $$g'(x) = \frac{4e^{-4/x}(2+e^{2/x})}{x^2} + \ldots$$ $$g''(x) = -\frac{8e^{-4/x}(2(x-2) + e^{2/x} (x-1))}{x^4} + \ldots$$ Both derivatives above go to $0$ when $x \to 0$ (also if higher-order terms are included). It seems that this is true for all higher order derivatives as well (although I don't have a stringent proof). I.e. the Taylor expansion of $f(x)$ at infinity (in the sense described in the question above) is simply 1.

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