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Log Approximation

by eep
Tags: approximation
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eep
#1
Oct14-06, 03:13 PM
P: 228
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...
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Hurkyl
#2
Oct14-06, 03:15 PM
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For many purposes, [itex]x \approx x+1[/itex] when x is large.
eep
#3
Oct14-06, 03:16 PM
P: 228
But x isn't large in this case?

Office_Shredder
#4
Oct14-06, 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
Hurkyl
#5
Oct14-06, 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.
eep
#6
Oct14-06, 03:20 PM
P: 228
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.
StatusX
#7
Oct14-06, 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.
eep
#8
Oct14-06, 06:15 PM
P: 228
Ah, yes. I just misread the notes!! Thanks anyways!


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