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How to simplify using stirling approximation?

  1. Mar 5, 2015 #1

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

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

    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?

  2. jcsd
  3. Mar 5, 2015 #2


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    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.
    Last edited: Mar 5, 2015
  4. Mar 6, 2015 #3
    Hey, got it. Thanks so much!
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