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

• workerant
In summary, my teacher wants us to compute: (2n choose n)^2/(4n choose 2n) using Stirling's formula. After simplifying the expression, Stirlings formula becomes (2/3)*(1/(sqrt(2 pi)*e^-n*n^(n+0.5))) where n! is canceled out and plugged into the formula.
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.

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.

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.

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

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?

workerant said:
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.

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?

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?

workerant said:
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.

Ok, got it, thanks

## 1. What is Stirling's formula?

Stirling's formula, also known as Stirling's approximation, is an asymptotic formula that approximates the factorial of a large number. It is given by n! ≈ √(2πn)(n/e)^n, where n is a positive integer and e is the mathematical constant approximately equal to 2.71828.

## 2. When is Stirling's formula used?

Stirling's formula is used when we need to approximate the factorial of a large number, as it becomes increasingly difficult to calculate exact factorials for large values of n. It is often used in probability and statistics, as well as in various branches of science and engineering.

## 3. How accurate is Stirling's formula?

Stirling's formula is an asymptotic formula, meaning it becomes increasingly accurate as n becomes larger. For large values of n, the relative error of Stirling's formula is less than 1%. However, for smaller values of n, the error can be significant and other methods of approximating factorials may be more accurate.

## 4. What are the limitations of Stirling's formula?

Stirling's formula is only accurate for large values of n. For small values of n, the error can be significant. Additionally, it is only an approximation and not an exact value. It also does not work well for complex numbers or negative numbers.

## 5. Are there any applications of Stirling's formula?

Yes, Stirling's formula has many applications in various fields such as physics, engineering, and computer science. It is often used in probability and statistics to approximate factorials, binomial coefficients, and other mathematical expressions. It is also used in the analysis of algorithms and in the study of random graphs.

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