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Consider a function $f(t) = e^{-t + \log t}$. I am not sure this type of function can be seen as an exponential decay function as it does not have the regular form $ce^{-at}$ with $a > 0$. But it obviously obeys \begin{equation} \underset{t \rightarrow \infty}{\lim} \frac{e^{-t + \log t}}{e^{-0.5t}} = 0, \end{equation} proving that $f(t)$ does not grow faster (or maybe equivalently decay faster) than $e^{-0.5t}$. If this is not an exponential decay function, does it have any particular name?

Geek
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  • this function decays when its argument $-t+lnt \lt0$ or $t \gt lnt$. If you graph it you will see that there is part that is increasing and a part that is decreasing. The value of the transition from increasing to decreasing depends on the log function (ln or log). For lnt, the transition occurs at t=1, for logt , it occurs slightly before.
    https://www.desmos.com/calculator     – user25406 Feb 06 '21 at 17:23

2 Answers2

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$$f(t) = e^{-t + \log t}$$ Suppopsing that your symbol log means the natural logarithm (not the logarithm base 10) then :

$$f(t)=t\,e^{-t}$$ This is a function frequently encountered. Of course all functions have not a specific name.

Note for information : A particularity of this function is to be related to the inverse of the Lambert W function. $$f(t)=-W^{-1}(-t)$$

JJacquelin
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  • Nice to meet you, Sir ! – Claude Leibovici Feb 06 '21 at 10:07
  • @Claude Leibovici. Heureux de vous rencontrer, Sir ! – JJacquelin Feb 06 '21 at 10:40
  • Thanks. I find a system the convergence time of which follows this particular form, but I don't know how to define this type of convergence. It seems I can not call it exponential convergence. – Geek Feb 06 '21 at 11:03
  • As far as I know there is no standard definition for "Exponential convergence". So you can use it as you like insofar you prove that the convergence is mainely due to an exponential term in the function. Thus in the present case you have to prove that the convergence exists and is due to $e^{-t}$ in the function $te^{-t}$. – JJacquelin Feb 06 '21 at 11:23
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$$f(t) = e^{-t + \log t}=t\,e^{-t}$$ $f(t)$ goes through a maximum at $x=1$ and for $x>1$ it decreases slower than $e^{-t}$.

Claude Leibovici
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  • So can I call it exponential decay? There exists a time $T$ such that $f(t)$ decreases faster than $e^{-ct}$ with $0 < c < 1$ for $t > T$. – Geek Feb 06 '21 at 11:05