User talk:Spudbeach

Welcome!

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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 &asymp; 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&thinsp;ln&thinsp;N &minus; N + 1, when N is large). What do you think?

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