User talk:Spudbeach

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Stirling's approximation

Hi, Spudbeach!

Say, I just made a correction to this edit. I thought I'd drop a note here to explain. Since

${\frac {d}{dx}}(x\ln x-x)=\ln x,$

we must have

$\int _{1}^{N}\ln x\,dx=N\ln N-N-(1\ln 1-1)=N\ln N-N+1.$

So when you made the change from ≈ to =, you really needed to change the quantity at the end of the formula.

I did this for you, instead of just reverting your change. But now I think the explanation is not quite as clear as it was (since one must consciously decide that 1 is negligibly small in N ln N − N + 1, when N is large). What do you think?

Have a great day! DavidCBryant 00:53, 11 August 2007 (UTC)