Asymptotic expression for $\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}}$

Question

My question is regarding the expression below, where $\varepsilon\ll1$. $$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}}$$ Is it possible to express this in the form $$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}} = a_0+a_1\varepsilon+a_2\varepsilon^2...$$

Where I am guessing $a_0=\frac{1}{\varepsilon}$ for example. I'm a bit rusty on my asymptotics so i'm sorry if this is a silly question, just need my mind jogged with some input.

Many thanks in advance.

EDIT: My idea, could we say

$$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}} = \varepsilon^{\cfrac{1}{\varepsilon-1}} = \left[ 1+(\varepsilon-1) \right]^{\cfrac{1}{\varepsilon-1}} $$ Therefore as $|\varepsilon-1|=|1-\varepsilon|<1$ we may expand

$$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}} \simeq 1+\cfrac{1}{\varepsilon-1}(\varepsilon-1)+ \cfrac{1}{\varepsilon-1}\frac{(\varepsilon-1)^2}{2} $$ This doesn't look that promising.

Claude Leibovici · Accepted Answer · 2016-03-30T10:14:03.920

1

Consider $$y=\left(\frac{1}{x}\right)^{\frac{1}{1-x}}$$ Take logarithms $$\log(y)=-{\frac{1}{1-x}}\log(x)$$ Now expand $\frac{1}{1-x}$. This gives $$\log(y)=-\log(x)\sum_{k=0}^\infty x^k$$ Using now $$y=e^{\log(y)}=\frac{1}{x}-\log (x)+x \left(\frac{\log ^2(x)}{2}-\log (x)\right)+x^2 \left(-\frac{1}{6} \log ^3(x)+\log ^2(x)-\log (x)\right)+O\left(x^3\right)$$

Edit

Just for illustration puroposes, let us try using $x=10^{-6}$. The above formula gives $$y=1000013.815619808284995098$$ for an "exact" value $$y=1000013.815619808284998235$$

Update

If, as according to your comments, the first order is sufficient you can solve for $x$ equation $$y=\frac{1}{x}-\log (x)$$ using $$x=\frac{1}{W\left(e^y\right)}$$ where appears again the beautiful Lambert function

edited Mar 30 '16 at 10:14

answered Mar 30 '16 at 09:00

Claude Leibovici

260,315

Perfect! Thanks so much. That has really helped me out. – Freeman Mar 30 '16 at 09:03
You've increased my expression's accuracy from 86% to 98% :) just using the first log. – Freeman Mar 30 '16 at 09:04
@Freeman. Yes and remember that $x^n \log(x)$ goes very fast to zero. – Claude Leibovici Mar 30 '16 at 09:06
Many thanks :), however for my purposes 98% is perfectly fine... it's for use in biology.. – Freeman Mar 30 '16 at 09:10
Ooo very nice. Also, just a note, have you dropped a minus sign in your workings? $\log(1/x)=-\log(x)$? – Freeman Mar 30 '16 at 10:05
Yes, two typo's ! Thanks for pointing it ! I am very excited to see Lambert function to be possibly used here. – Claude Leibovici Mar 30 '16 at 10:13
Noi problem! I will bare that in mind going forward :) – Freeman Mar 30 '16 at 10:25
As a side note if we took $z=y^x$ then z could be approximated by the same sequence excluding the first term? – Freeman Mar 30 '16 at 10:29
Oh geez that is not correct, let me go through your workings again, I must have misunderstood a step. – Freeman Mar 30 '16 at 10:39
A similar approach would lead to $$y^x=1-x \log (x)+x^2 \left(\frac{\log ^2(x)}{2}-\log (x)\right)+O\left(x^3\right)$$ – Claude Leibovici Mar 30 '16 at 10:40
Thank you for this, it is much appreciated. I guess I am being dense, but i'm struggling to see how $\exp{(-\log(x)\sum_{k=0}^\infty x^k)}$ yields the expression you find. I find a product series with terms such as $1/x^x$, I don't suppose you've mind elaborating? I apologize for how rusty at this sort of thing I have become. – Freeman Mar 30 '16 at 10:45
Taylor series again using a few terms. – Claude Leibovici Mar 30 '16 at 10:46

Asymptotic expression for $\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}}$

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