Re: Uniform partition of an interval

[W. Peters]

I tried this in MATLAB:



>> 1.8 + 0.05*2 - 1.9

ans =

   2.2204e-16

>> 1.8:0.05:1.9

ans =

    1.8000    1.8500    1.9000




On Wed, 17 May 2000, etienne grossmann wrote:

> Date: Wed, 17 May 2000 10:27:45 +0100 (WEST)
> From: etienne grossmann <address@hidden>
> Reply-To: address@hidden
> To: address@hidden
> Cc: address@hidden
> Subject: Re: Uniform partition of an interval
> Resent-Date: Wed, 17 May 2000 04:30:35 -0500 (CDT)
> Resent-From: address@hidden
> 
> 
>   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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> 


------------------------------------------------
William Peters
Department of Bioengineering
University of Washington
Box 357962
Seattle, WA 98195
(206)543-6442
address@hidden
------------------------------------------------



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