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Log Approximation

  1. Oct 14, 2006 #1

    eep

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    Hi,
    In his notes, our teacher makes this approximation:

    [tex]
    \log(1 + 3e^{-2\frac{E_o}{\tau}}) \approx \log(3e^{-2\frac{E_o}{\tau}})
    [/tex]

    For [itex]\tau << E_o[/itex]

    Also, and I don't think this matters, the logs are assumed to be natural logs.

    I was wondering what the justification for this was...
     
  2. jcsd
  3. Oct 14, 2006 #2

    Hurkyl

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    For many purposes, [itex]x \approx x+1[/itex] when x is large.
     
  4. Oct 14, 2006 #3

    eep

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    But x isn't large in this case?
     
  5. Oct 14, 2006 #4

    Office_Shredder

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    There doesn't appear to be much. What's the context? Is [tex]3e^{-2\frac{E_o}{\tau}}[/tex] very large?

    EDIT: In that case, ask your teacher
     
  6. Oct 14, 2006 #5

    Hurkyl

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    I thought maybe you had forgotten [itex]\tau < 0[/itex]. If the argument to log isn't large, then that's not a good approximation.
     
  7. Oct 14, 2006 #6

    eep

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    Sorry I hadn't quite finished editing my post when people started replying. We're trying to calculate the partition function for rotational degrees of freedom for a single molecule. So we have an infinite sum which we keep only the first two terms in the [itex]\tau << E_o[/itex] limit (the terms in the log). We then want to calculate the average energy which is where the log comes from, and he then makes that approximation. I guess I'll just have to ask him.
     
  8. Oct 14, 2006 #7

    StatusX

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    Are you sure there's a log around that second expression? Because [itex]log(1+x)\approx x [/itex] for x very small.
     
  9. Oct 14, 2006 #8

    eep

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    Ah, yes. I just misread the notes!! Thanks anyways!
     
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