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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.
 Fall '08
 SPRUNGER
 Physics

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