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Sterling's theorem

  • Thread starter bon
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bon
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Homework Statement



The Q: Use your calculator to work out ln 15!

Compare the answer with the simple version of Sterling's theorem (ln n! = nln n - n)

How big must n be for the simple version of Stirling's theorem to be correct to within 2%?


Homework Equations





The Attempt at a Solution




So I've done the first two parts.

But for the last I get that (ln n! - nln n + n)/ln n! < 0.02 which I don't know how to solve for n...

any ideas?
thanks!
 

Answers and Replies

  • #2
Dick
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I don't think you are supposed to solve for n. Do the same thing to did for a few numbers larger than 15 until you get within 2%.
 
  • #3
SammyS
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...

Sterling's theorem (ln n! = nln n - n)


But for the last I get that (ln(n!) - n·ln(n) + n)/ln(n!) < 0.02 which I don't know how to solve for n...

any ideas?
thanks!
Hi bon.

Try graphing (ln(n!) - n·ln(n) + n)/ln(n!) - 0.02 . See where this goes from positive to negative.

Check out this link: http://www.wolframalpha.com/input/?i=(ln+(n!)-n*ln(n)+++n)/ln+(n!)-.02"

[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP504019e29g7g29faa3670000634agaci8ae3cab4?MSPStoreType=image/gif&s=9&w=303&h=104 [Broken]
 
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