# Solving Sterling's Theorem: Calculate ln 15! & Find n for 2% Accuracy

• bon
In summary, the conversation is about using a calculator to find the value of ln 15! and comparing it to the simple version of Sterling's theorem. The question then arises of how large n must be for the simple version of Stirling's theorem to be accurate within 2%. The solution involves graphing to find the point where the expression (ln(n!) - n·ln(n) + n)/ln(n!) - 0.02 goes from positive to negative.

## 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%?

## 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!

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%.

bon said:
...

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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