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Could somebody tell me the definition of a "characteristic time"? For example, what is the characteristic time for a function $f(t)=\operatorname{tanh}(t)$ to reach 1? I tried looking up a definition, but  there seems not to be a universal definition. Is there a preferred definition?

Many thanks!


I have a vector $$\begin{pmatrix}a \operatorname{sech} t\\ b\operatorname{tanh}t\\ c\operatorname{sech} t \end{pmatrix}$$

I am asked what the "characteristic time" for the vector to align with $$\begin{pmatrix} 0\\1\\0 \end{pmatrix}$$

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The definition might depend upon the setup. What is yours? What have you already found and why were your findings insufficient? And: Welcome to! – draks ... Jul 16 '12 at 21:11
How sure are you that "the characteristic time for tanh to reach 1" is even supposed to be a meaningful notion? All the possible meanings of "characteristic time" I can remember or imagine offhand tell about a function's behavior either globally or in some asymptotic limit, not as it approaches an ordinary finite point. – Henning Makholm Jul 16 '12 at 21:18
The hyperbolic tangent $\tanh(t)$ is never equal to $1$. It is very close to $1$ for large positive $t$, and even medium-sized positive $t$, like $10$. Could you be looking for a place $p$ such that from $p$ on, $\tanh t$ is "practically equal" to $1$, where practically equal means a standard tolerance? – André Nicolas Jul 16 '12 at 21:59
I think what is needed here is the context. Just quote the source in which you found this---maybe a paragraph or three or four paragraphs. – Michael Hardy Jul 16 '12 at 22:05
@AndréNicolas: Oh foo, I had it confused with hyperbolic sine. Could the OP be looking for something like the limiting subtangent relative to the asymptote, i.e. $\lim_{x\to\infty} \frac{1}{1-\tanh x}\frac{d}{dx}\tanh x$ (which ought to exist)? – Henning Makholm Jul 16 '12 at 22:24
up vote 2 down vote accepted

What is meant in this particular context is almost certainly that for large $t$, we can approximate $$\begin{pmatrix} a \operatorname{sech} t \\ b\tanh t \\ c \operatorname{sech} t \end{pmatrix} \sim \begin{pmatrix}0 \\ b \\ 0\end{pmatrix} + e^{-t/k} \begin{pmatrix}p \\ q \\ r\end{pmatrix} + o(e^{-t/k})$$ for appropriate constants $k$, $p$, $q$, $r$. The characteristic time of the approach to $(0,b,0)$ is then $k$.

(And your $b$ had better be $1$ for the question to make sense).

Hmm ... alternatively, "align with $(0,1,0)$" could mean the process of the direction of the vector approaching the direction of the positive $y$ axis. In that case the relevant approximation would be something like $$ \frac{ \sqrt{a^2+c^2} \operatorname{sech} t }{ b \tanh t } \sim e^{-t/k} s + o(e^{-t/k}) $$ for constants $k$ and $s$. Again $k$ is the characteristic time.

Here the left-hand side of this represents the angle between the vector and $(0,1,0)$ as seen from the origin, rather than the distance between your vector and $(0,1,0)$. Strictly speaking the left-hand side should arguably be $\tan^{-1}\left(\frac{\sqrt{a^2+c^2}\operatorname{sech}t}{b\tanh t}\right)$, but since $\tan^{-1}(x)\sim x$ for small $x$ anyway, it doesn't matter for the result.

(So much for "almost certainly").

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Thank you for the answer! May I ask why the second case has a different answer/ definition? – fred sinclair Jul 16 '12 at 23:01
In this case the final answer is the same (namely $1$ for both interpretations). But in general the difference is that small differences in the $y$ coordinate are almost inconsequential to the direction of the vector. As an example, consider $t\mapsto (e^{-t}, 1+e^{-t/10})$ in two dimensions. For large $t$, the distance between the moving point and $(0,1)$ is dominated by the $e^{-t/10}$ term, so the characteristic time for the point to approach $(0,1)$ is $10$. However the $e^{-t/10}$ term is completely irrelevant next to the $1$ when we're looking at the ... (cont'd) – Henning Makholm Jul 16 '12 at 23:10
... direction of the vector $(e^{-t}, 1+e^{-t/10})$ for large $t$, so the direction it approaches "due up" with characteristic time $1$. – Henning Makholm Jul 16 '12 at 23:11

The characteristic time is usually defined to be the time in which a quantity, in this case the deviation from $(0,b,0)$, decreases by $1/\mathrm e$. In the present case the functions don't decay exactly exponentially, but it still makes sense to speak of the characteristic time with respect to the leading terms for $t\to\infty$. These are

$$\operatorname{sech} t=\frac2{\mathrm e^t-\mathrm e^{-t}}\approx2\mathrm e^{-t}$$


$$\operatorname{tanh} t-1=\frac{\mathrm e^t+\mathrm e^{-t}}{\mathrm e^t-\mathrm e^{-t}}-1=\frac{2\mathrm e^{-t}}{\mathrm e^t-\mathrm e^{-t}}\approx2\mathrm e^{-2t}\;.$$

Thus the $y$ component decays twice as quickly as the other two, and the characteristic time of the approach could be said to be that of $\mathrm e^{-t}$, which is $1$.

(This is basically the same answer as Henning's, just with less rigour and more numbers.)

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Also with a better explanation of the underlying concept (+1). (Except that I now think the operative limit is that of the direction rather than the vector itself). – Henning Makholm Jul 16 '12 at 22:56
Thank you for the explanation! – fred sinclair Jul 16 '12 at 23:00

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