Coth approximation

  • Thread starter intervoxel
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  • #1
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Main Question or Discussion Point

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]
 

Answers and Replies

  • #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].
 
  • #3
611
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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:

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