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Log Approximationby eep
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#1
Oct1406, 03:13 PM

P: 227

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
Oct1406, 03:15 PM

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For many purposes, [itex]x \approx x+1[/itex] when x is large.



#3
Oct1406, 03:16 PM

P: 227

But x isn't large in this case?



#4
Oct1406, 03:16 PM

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Log Approximation
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 


#5
Oct1406, 03:18 PM

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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.



#6
Oct1406, 03:20 PM

P: 227

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.



#7
Oct1406, 03:48 PM

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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.



#8
Oct1406, 06:15 PM

P: 227

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



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