Asymptotic expansion

  • Thread starter Clausius2
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  • #1
Clausius2
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I am used to expand functions in powers (not neccesarily integer) of an small parameter [tex]\epsilon[/tex]

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

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

[tex] sin (2\epsilon)\sim 2tan\epsilon -2tan^3\epsilon+...[/tex]

Thanx.
 

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  • #2
Tide
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If [itex]y = sech^{-1}x[/itex] then

[tex]x = sech y = \frac {2}{e^y + e^{-y}}[/tex]

Now try solving for y in terms of x.

Incidentally, homework problems should be posted in the homework section.
 
  • #3
arildno
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As for the second, we have:
[tex]\sin(2x)=2\sin(x)\cos(x)=2\tan(x)\cos^{2}(x)=\frac{2\tan(x)}{1+\tan^{2}(x)}[/tex]

Perhaps that will help.
 

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