[email protected] wrote: > could you explain how the following determines the > approximate > number of digits in !200e6? > > a few examples: > > maybe !10e2, !10e4 and !100e6 >
The approximation in conventional mathematical notation is ln n! is approximately n ln n - n This comes from ln n! = ln 1 + ln 2 + ln 3 + .... + ln n which is a Riemann sum for the integral from 1 to n of ln t dt, and the latter is (approximately) n ln n - n. Examples: stirling=:(* ^.)-] log10=:(^. 10) %~ stirling ]n=:10^>:i.4x 10 100 1000 10000 10&^. !n 6.55976 157.97 2567.6 35659.5 log10 n 5.65706 156.571 2565.71 35657.1 On my computer, it is hard to go much beyond this for the exact result. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
