Expanding Functions

  • Thread starter Allday
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  • #1
Allday
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I'm studying for the qualifying exam and I came accross a problem that I'd be able to do in a snap if I had a computer running mathematica in front of me, but regretably I am having trouble with using good old paper and pencil and a reasonable amount of time. I want to look at the low and high temperature behaviour of the function

(deltaE)^2 = C*[sinh(a*B)]^-2

where B = 1/T and the rest are constants. I would like to know not just the limit, but the behaviur of the function. ie I could get that in the high T small B limit the function goes like T^2, I am having difficulty with the low T, high B limit. This is connected with the energy fluctiations of a quantum harmonic osccillator if anyone wants a reference point.

Any ideas?
thanks
 

Answers and Replies

  • #2
lurflurf
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sinh(x)=(exp(x)-exp(-x))/2
so
sinh(x) goes like exp(x)/2 for x large
 
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  • #3
Allday
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Thats a start. Are you suggesting that I invert that relationship and square it ? I don't know if that will work. The function e^x also can be expanded around the point zero. where its behaviour goes like 1 + x. For the high B regime I think you have to use the definition that B = 1/T and expand around a small number, but I'm not sure how to do that.
 
  • #4
lurflurf
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Allday said:
Thats a start. Are you suggesting that I invert that relationship and square it ? I don't know if that will work. The function e^x also can be expanded around the point zero. where its behaviour goes like 1 + x. For the high B regime I think you have to use the definition that B = 1/T and expand around a small number, but I'm not sure how to do that.
So you have
(deltaE)^2 = C*[sinh(a*B)]^-2
B large
(deltaE)^2 = C*[exp(a*B)/2]^-2
(deltaE)^2 = 4*C*exp(-2*a*B)
This goes like 0 (if a is positive).
If goes like 0 is not close enough I do not know what you would want as it goes to 0 pretty fast and it is difficult to find other representations.
 
  • #5
Allday
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It goes to zero as B becomes large this is true. But does it go to zero like 1/B like 1/B^2 ... that's the thing I'm trying to figure out. I am haven't played with it much today, thanks for looking at it. If I can clarify what I mean Ill post it
 
  • #6
lurflurf
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Allday said:
It goes to zero as B becomes large this is true. But does it go to zero like 1/B like 1/B^2 ... that's the thing I'm trying to figure out. I am haven't played with it much today, thanks for looking at it. If I can clarify what I mean Ill post it
It goes to zero exponentially. If you want an expansion in terms of rational functions, none will be useful as it goes to 0 faster than x^y for any negative y. How big is the B you want?
for instance
exp(-20)~2*10^-9
 
  • #7
Allday
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Ahhh, I get it now. I was thinking that even though it goes exponentially that there would be some power that would come out. Now I see that's impossible. The terms in the series just continure to grow so there is no leading term. Thanks for the help.
 

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