Re: Uniform partition of an interval

[etienne grossmann]

  Hello,


#  From: Francesco Potorti` <address@hidden>
#  To: Octave users list <address@hidden>
#
#  address@hidden:
#     What is the MAIN reason that 1.8:0.05:1.9 produces [1.8000 1.8500]
#     and not [1.8000 1.8500 1.9000]? 
#
#  Finite precision of the machine's number representation.

  Indeed, with 2.1.19,

octave:3> 1.8 + 0.05*2 - 1.9
ans =  2.2204e-16

  So 1.8 + 0.05*2 is strictly greater than 1.9. Not a bug, but the
effect of floating point precision.


  Cheers,


  Etienne



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