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Asymptotic expansion

  1. Jan 26, 2006 #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.
     
  2. jcsd
  3. Jan 27, 2006 #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.
     
  4. Jan 28, 2006 #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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