# Stirling's approximation

## 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
$$\Omega$$(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
$$\Omega$$(500)=21000/(500pie)1/2
P=$$\Omega$$(500)/21000= 1/(500pie)1/2= .025

thanks

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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

lol thanks,
so it would be 10001000e-1000(2pi1000)1/2/ (500500e-500(2pi500)1/2)2

yes and that simplifies to what you want

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:
Andrew Mason
Homework Helper
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

fzero
Homework Helper
Gold Member
We're going to need a bigger calculator.

Andrew Mason
Homework Helper
We're going to need a bigger calculator.
Not really. The expression reduces to:

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

AM

The Stirling's approximation you want to use for this is

$$\ln(n) \simeq n\ln(n) - n$$

This is applicable for $$n \gg 1$$. So, for your example

$$\omega = \frac{1000!}{500!500!}$$

Take the logarithm of both sides and we find

$$\omega = \ln(1000!) - 2\ln(500!)$$

Which, using Stirling's approximation, gives us

$$\omega = 1000\ln(1000) - 1000 - 2(500\ln(500) - 500)$$

At this point it is easy to find that $$\omega = 10^{300}$$