P21_048

# P21_048 - 48. We ﬁnd the “percent error” in the use...

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Unformatted text preview: 48. We ﬁnd the “percent error” in the use of Stirling’s approximation by computing (N (ln N ) − N ) (N (ln N ) − N ) − ln(N !) = −1 ln(N !) ln(N !) which would be multiplied by 100% to be expressed as a percentage. (a) For N = 50, the percent error is 50 ln(50) − 50 145.6 145.6 −1= −1= −1 ln(50!) ln(3.04 × 1064 ) 148.5 which yields −1.9%, meaning Stirling’s approximation produces a value that is 1.9% lower than the correct one. (b) For N = 100, this procedure gives the result −0.89%. (c) And for N = 250, we obtain −0.32%. (d) The trend is such that Stirling’s approximation becomes a better estimate of ln(N !) for larger values of N . ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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