Re: a-b+b != a

[Tim Pierce]

No way! It's correct

in pure mathematics b - a + a = b

in computing it is not true that b -a + a = b

floating points and doubles and rounding mean that it is definitely NOT the case that mathematical identities hold

On Mon, Sep 4, 2017 at 3:53 PM, stn021 <address@hidden> wrote:

Hi,

this is weird:

r1 = round( 1e3*randn(5) ) / 1e3 ;
r2 = round( 1e3*randn(5) ) / 1e3 ;
r3 = r1 - r2 + r2 ;
not_zero = r1 - r3

not_zero =

   0.0000e+00   0.0000e+00   5.5511e-17   0.0000e+00   1.1102e-16
   5.5511e-17   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00
   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00  -2.0817e-16
  -5.5511e-17   0.0000e+00   0.0000e+00  -1.1102e-16   0.0000e+00
   2.7756e-17   0.0000e+00   0.0000e+00   0.0000e+00   0.0000e+00

So a-b+b != a

The difference to zero is small, around 1e-17. But iterations can
cause this error to increase.

I use leasqr() and in each iteration the last line is
  retval = retval - someval + someval

With that additional line I get quite different results compared to
the same program without this line even though they should be
identical.

Is there a way to avoid this phenomenon ?

THX
Stefan

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