How to simplify using stirling approximation?

In summary, the conversation discusses the derivation of \Omega \left( N,q\right) =\dfrac {\left( N-1+q\right) !} {q!\left( N-1\right) !}, where both N and q are large numbers and q >> N. The speaker is confused about when they can throw away terms in the equation. They mention intuitively wanting to keep q and throw away N - 1, but the book starts by throwing away the "-1" term. The expert summarizer explains that since N is involved in the numerator's factorial, neglecting a factor of this size can make a noticeable difference in the approximation.
  • #1
eprparadox
138
2
Hey!

So we're deriving something in Daniel Schroeder's Introduction to Thermal Physics and it starts with this:

[tex]
\Omega \left( N,q\right) =\dfrac {\left( N-1+q\right) !} {q!\left( N-1\right) !}
[/tex]

Both N and q are large numbers and q >> N.

The derivation is in the book, but I am always confused with when I can throw away terms.

For example, in this case intuitively, I would want to say that since q >> N and we have N - 1 + q, I would just keep the q and throw away the N - 1. I know this is wrong, but I don't know why.

In the book, they begin by throwing away the "- 1" term since both q and N are large numbers. I would have thrown away the N - 1. Why can't I do this and when can I/can't I throw away terms?

Thanks!
 
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  • #2
Because N is involved in the numerator's factorial. Since factorials grow very fast, then a larger factorial will contribute quite a lot, even though the added size isn't quite that large in comparison.
Example, if N=10 and q=1000, then q! is on the order of 102567 and while (q+N)! is on the order of 102597 and may not seem much larger, it's in fact 1030 (a million trillion trillion) times larger.
[tex]q!=1\times2\times3\times...\times999\times1000 \approx 10^{2567}[/tex]
[tex]N! = 1\times2\times...\times10 \approx 10^6[/tex]
[tex](q+N)! = 1\times2\times3\times...\times999\times1000\times 1001\times 1002 \times ... \times 1010 [/tex]
[tex]= q! \times 1001 \times 1002 \times... \times 1010 \approx q! \times 10^{30}[/tex]
N! is only 106 on its own, but when included with q in the factorial, it multiplies by 1030. If you neglect a factor of this size, it'll definitely make a noticeable difference in your approximation.
 
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  • #3
Hey, got it. Thanks so much!
 

FAQ: How to simplify using stirling approximation?

1. What is the Stirling Approximation?

The Stirling Approximation is a mathematical formula used to approximate the value of large factorials. It is named after the Scottish mathematician, James Stirling, who first published it in 1730.

2. How does the Stirling Approximation work?

The Stirling Approximation uses the gamma function to approximate the factorial of a number. The gamma function is an extension of the factorial function and can be expressed as Γ(n) = (n-1)!.

3. When is the Stirling Approximation most useful?

The Stirling Approximation is most useful when dealing with large factorials, as it provides a more efficient way to calculate their values compared to using the traditional factorial formula.

4. What is the formula for the Stirling Approximation?

The formula for the Stirling Approximation is n! ≈ √(2πn)(n/e)^n. This can also be written as n! ≈ √(2πn)(n/e)^n * (1+O(1/n)), where O(1/n) represents the error term.

5. Are there any limitations to using the Stirling Approximation?

Yes, the Stirling Approximation is only accurate for large values of n. As n approaches 0, the error term becomes more significant, making the approximation less accurate. Additionally, the approximation is only valid for positive real numbers.

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