require 'bigdecimal' def euler(p) BigDecimal.limit(p+1) f = BigDecimal.new 1 one = BigDecimal.new 1 (2..150).reduce(2) { |e, i| e += one / (f *= i) }.to_s 'f' end
Euler's #, with precision
Page 1 of 11 Replies - 1671 Views - Last Post: 16 November 2013 - 02:57 AM
#1
Euler's #, with precision
Posted 15 November 2013 - 03:16 PM
Description: Usage : puts euler(p)
Refactor of http://www.dreaminco...snippet1660.htm
Original Credit to: MitkOkReturn Euler's Number with p digits of precision
Refactor of http://www.dreaminco...snippet1660.htm
Replies To: Euler's #, with precision
#2
Re: Euler's #, with precision
Posted 16 November 2013 - 02:57 AM
In Ruby, EVERYTHING returns a value. Use it to your advantage.
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