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

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SUMMARY

The discussion focuses on applying Stirling's theorem to calculate ln 15! and determining the value of n for which the approximation is accurate to within 2%. Users are encouraged to graph the function (ln(n!) - n·ln(n) + n)/ln(n!) - 0.02 to identify where it crosses zero, indicating the threshold for accuracy. The conversation highlights the use of calculators and graphing tools, specifically referencing Wolfram Alpha for computational assistance.

PREREQUISITES
  • Understanding of Stirling's theorem and its application in approximating factorials.
  • Familiarity with logarithmic functions and their properties.
  • Basic skills in graphing functions to analyze behavior and intersections.
  • Proficiency in using online computational tools like Wolfram Alpha.
NEXT STEPS
  • Learn how to graph functions using software or online tools to visualize mathematical concepts.
  • Study the implications of Stirling's theorem for large n in statistical mechanics and combinatorics.
  • Explore numerical methods for solving inequalities involving logarithmic functions.
  • Investigate the accuracy of Stirling's approximation for various values of n beyond 15.
USEFUL FOR

Students in mathematics or statistics, educators teaching calculus or approximation methods, and anyone interested in the applications of Stirling's theorem in computational mathematics.

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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!
 
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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%.
 
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
 
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