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Stirling's approximation

  • Thread starter leonne
  • Start date
  • #1
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Homework Statement


I dont really understand how to use Stirling's approximation. heres an example
you flip 1000 coins, whts the probability of getting exactly 500 head and 500tails


Homework Equations


N!=NNe-N(2pieN)1/2


The Attempt at a Solution


wht they did was
21000 total number outcome
[tex]\Omega[/tex](500)=1000!/500!2 multiplicity
than used the Stirling's approximation to get this which I dont get how. What do I plug in for N
[tex]\Omega[/tex](500)=21000/(500pie)1/2
P=[tex]\Omega[/tex](500)/21000= 1/(500pie)1/2= .025

thanks
 

Answers and Replies

  • #2
213
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you divide the stirling approximation for N=1000 by the square of the stirling approximation for N=500. And it's Pi not pie silly
 
  • #3
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lol thanks,
so it would be 10001000e-1000(2pi1000)1/2/ (500500e-500(2pi500)1/2)2
 
  • #4
213
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yes and that simplifies to what you want
 
  • #5
191
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um I dont think that is right i keep getting overflow error.
nvm after typing it in wolfam alpha it worked but how would i do it on my calculator, o well lol
 
Last edited:
  • #6
Andrew Mason
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um I dont think that is right i keep getting overflow error.
nvm after typing it in wolfam alpha it worked but how would i do it on my calculator, o well lol
These are very big numbers. You might find it easier to reduce the fraction as a whole rather than work out the denominator and numerator separately.

AM
 
  • #7
fzero
Science Advisor
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We're going to need a bigger calculator.
 
  • #8
Andrew Mason
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We're going to need a bigger calculator.
Not really. The expression reduces to:

[tex]\frac{\Omega(500)}{2^{1000}} = \frac{(2\pi *1000)^{1/2}}{2\pi*500} = \frac{1}{(500\pi)^{1/2}} = .025[/tex]

AM
 
  • #9
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The Stirling's approximation you want to use for this is

[tex]\ln(n) \simeq n\ln(n) - n[/tex]

This is applicable for [tex]n \gg 1[/tex]. So, for your example

[tex]\omega = \frac{1000!}{500!500!}[/tex]

Take the logarithm of both sides and we find

[tex]\omega = \ln(1000!) - 2\ln(500!)[/tex]

Which, using Stirling's approximation, gives us

[tex]\omega = 1000\ln(1000) - 1000 - 2(500\ln(500) - 500)[/tex]

At this point it is easy to find that [tex]\omega = 10^{300}[/tex]
 

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