How Do You Apply Stirling's Formula to Binomial Coefficients?

[workerant]

Homework Statement

Ok, my teacher wants us to compute:
(2n choose n)^2/(4n choose 2n) using Stirling's formula.

Homework Equations

Stirling's formula is sqrt(2pi) e^-n * n^(n+.5)

The Attempt at a Solution

Ok, I'm just confused about how you compute something with n in it...I mean, I get how to use Stirling's formula for factorials, but how do you compute that probability that is requested when there are ns in the (2n choose n)^2/(4n choose 2n)...what do I plug into stirling's formula? I don't know how to really attempt a sensible solution when I don't know what to put for n. Thanks.

 

[Gib Z]

Okay well my first step would be to replace the "choose"s by their factorial definition;

^nC_r = \frac{n!}{r! (n-r)!}

Then using the properties of factorials, simplify the expression. Only once you have done all of that, everywhere you see the expression n!, replace it with Stirlings Formula.

 

[workerant]

Ok, so I have (2n! 2n! 2n! 2n!)/(n! n! n! n! 4n!).

Then, I canceled n! terms to get 16/(4n!)=16/(24)n!=(2/3)*(1/n!)

Now if I plug stirling's, this is like... (2/3)* (1/(sqrt(2 pi)*e^-n*n^(n+0.5))

I don't really see how this simplifies to sqrt(2/pi*n) which is the answer my teacher said we should get...what am I doing wrong/not seeing? Thanks.

 

[Dick]

You are using cancellation rules on the factorials that don't exist. You can't 'cancel' anything in there. Use Stirling on each factor.

 

[workerant]

Ok so 2n!/n! is not = 2? This seems like it will get extremely messy, extremely fast since there are 9 instances of n! and how do I adjust stirling's formula for 2n!, just plug in 2n for each n in the formula?

 

[Dick]

Ok so 2n!/n! is not = 2? This seems like it will get extremely messy, extremely fast since there are 9 instances of n! and how do I adjust stirling's formula for 2n!, just plug in 2n for each n in the formula?

2n!/n! is definitely not 2. Try some examples. 4!/2!=12. And the number of factors isn't that bad. You've got (2n!)^4/((n!)^4*4n!). Collect them into powers. And yes, to get 2n! substitute 2n for n in Stirling.

 

[workerant]

Okay:

I have ((sqrt(2pi)*e^(-2n)*2n^(2n+.5))^4/((sqrt(2pi)*e^(-4n)*(4n^(4n+2))*((sqrt(2pi)*e^(-n)*(n+.5))^4

I canceled out sqrt (2pi)^4 on the top and bottom and expanded a bit to get

(e^(-16n)*16n^(8n+4))/((sqrt 2pi)*4n^(4n+.5)*e^(-8n)*n^(4n+2))=


e^(-8n)*(4n^1.5)/(sqrt 2 pi)

but this isn't sqrt(2/pi*n)

I'm sorry, but my algebra is a bit rusty, so you see what I am doing wrong?

 

[workerant]

ok, did it again and keep getting 2*(sqrt(2/pi*n) which is pretty much the right answer but an extra factor of 2...hmm where did it come from?

 

[Dick]

ok, did it again and keep getting 2*(sqrt(2/pi*n) which is pretty much the right answer but an extra factor of 2...hmm where did it come from?

Don't know. I get sqrt(2/(pi*n)). You are pretty close. You'll have to post your whole solution if you want someone to check.

 

[workerant]

Ok, got it, thanks