- #1

Mathsboi

- 3

- 0

Given was this:

[tex]n(n-1)(n-2) \cdots (n-r+1) = \frac{n!}{(n-r)!} \approx n^r[/tex]

Stating that "Stirling's approximation" had been used.

So I looked the up and found:

[tex]\ln n! \approx n\ln n - n\ [/tex]

In the poisson distribution n is very large and [tex]r[/tex] is very small compared to [tex]n[/tex] so all the terms in the given equation approximate to [tex]n[/tex]... This gives me my [tex]\approx n^r[/tex]

But I just wondered where the Stirling equation comes into it...

[tex]\ln (\frac{n!}{(n-r)!}) = \ln(n!) - \ln((n-r)!) \ [/tex]

[tex]\approx n\ln n - n - \left[ (n-r)\ln((n-r)) - (n-r) \right]\ [/tex]

[tex]\approx n\ln n - n - (n-r)\ln((n-r)) + n - r \ [/tex]

[tex]\approx n\ln n - (n-r)\ln((n-r)) -r \ [/tex]

...

That's as far as I got...

[tex]\approx \ln (n^n) - \ln((n-r)^{(r-n)}) -r \ [/tex]

Unless taking logs, instead of to base e, to base n...

[tex]\approx Log_n (n^n) - Log_n ((n-r)^{(r-n)}) -r \ [/tex]

Then...

[tex]Log_n (n^n) - Log_n ((n-r)^{(r-n)}) = r\ [/tex]

[tex]n^n - (n-r)^{(n-r)} = n^r\ [/tex]

^ not sure if that's correct though

Can anyone help?