Calculating Log & Tan Terms in Asymptotic Expansion

[Clausius2]

I am used to expand functions in powers (not neccesarily integer) of an small parameter \epsilon

But someone tell me how the log or tan terms are calculated here?

sech^{-1}\epsilon\sim log(2/\epsilon)-1/4\epsilon^2-...

sin (2\epsilon)\sim 2tan\epsilon -2tan^3\epsilon+...

Thanx.

 

[Tide]

If y = sech^{-1}x then

x = sech y = \frac {2}{e^y + e^{-y}}

Now try solving for y in terms of x.

Incidentally, homework problems should be posted in the homework section.

 

[arildno]

As for the second, we have:
\sin(2x)=2\sin(x)\cos(x)=2\tan(x)\cos^{2}(x)=\frac{2\tan(x)}{1+\tan^{2}(x)}

Perhaps that will help.