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Factorials: Turn product into a sum taking logs: ln(n!) = ln( - PowerPoint PPT Presentation

Oct 31, 2022 •405 likes •443 views

Closed form for n! Mathematics for Computer Science MIT 6.042J/18.062J n n! ::= 1 2 3 (n-1) n = i i=1 Factorials: Turn product into a sum taking logs: ln(n!) = ln( 123(n 1)n ) = Stirlings

  1. Closed form for n! Mathematics for Computer Science MIT 6.042J/18.062J n ∏ n! ::= 1 ⋅ 2 ⋅ 3 ⋅⋅⋅ (n-1) ⋅ n = i i=1 Factorials: Turn product into a sum taking logs: ln(n!) = ln( 1·2·3···(n – 1)·n ) = Stirling’s Formula ln 1 + ln 2 + · · · + ln(n – 1) + ln(n) n ∑ ln(i) = i= 1 Albert R Meyer, April 10, 2013 Albert R Meyer, April 10, 2013 stirling.1 stirling.2 Closed form for n! Closed form for n! n ∑ ln(i) Integral Method to bound n n   n n ∑ ∫ ≤ ∑ ≤ ln(x) dx ln(i) ≤ ≤ nln +1 ln(i) i=1    e  i=1 1 i=1 ln(x) ln n n …   ln(x+1) n +1 ∫ (n +1)ln ln(x +1) dx +0.6 ln 5   ln 4  e  ln ln 3 n-1 ln n 1 ln 2 ln 3 ln 4 ln 5 reminder: ln 2 …   x ∫ lnxdx = x ln e   1 2 3 4 5 n–2 n–1 n   Albert R Meyer, April 10, 2013 Albert R Meyer, April 10, 2013 stirling.3 stirling.4 1

  2. Closed form for n! Stirling’s Formula n   1 n ∑ A precise approximation: ln(i) ≈ (n + )ln   2  e  n   i=1 2πn n n! ~ exponentiating:     e n   n ≈ n! n / e e     Albert R Meyer, April 10, 2013 Albert R Meyer, April 10, 2013 stirling.5 stirling.6 2

  3. MIT OpenCourseWare http://ocw.mit.edu 6.042J / 18.062J Mathematics for Computer Science Spring 20 15 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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