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Coth approximation

  1. Mar 17, 2010 #1
    In a QM textbook (Newton), I found the below expression for large x:

    [itex]
    coth(x)\cong 1+2e^{-2x}
    [/itex]

    I tried

    [itex]
    coth(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}
    [/itex]
     
  2. jcsd
  3. Mar 17, 2010 #2

    George Jones

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    Write [itex]z = e^{-2x}[/itex], so that [itex]z[/itex] is small when [itex]x[/itex] is large. Then,

    [tex]\coth x = \left(1+z\right)\left(1-z\right)^{-1}.[/tex]

    Do a power series expansion of of this, or do a power series expansion of just [itex]\left(1-z\right)^{-1}[/itex].
     
  4. Mar 17, 2010 #3
    Is it correct to do like this:
    for large x; ex=1+ex.
    So on mere substitution in to the eqn.
    [tex]
    coth(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1+e^{-2x}}{1-e^{-2x}}
    [/tex]
    one will get
    [tex]
    coth(x)\cong 1+2e^{-2x}
    [/tex]
    at low temperature the weighing factor is (1/E).
    [tex]
    x=E/(2k_{\rm B}T)
    [/tex].
     
    Last edited: Mar 17, 2010
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