Re: a-b+b != a

[stn021]

Hi Doug,

the idea here is to add weights to the result.

res = y + weights .* ( res-y )

If weights==1 then res should remain unchanged. Only it doesn't.
So (-r2+r2) does not really work here.

I did not include the weights in my original post because it did not seem relevant to the issue, sorry about that.

Thx, Stefan


2017-09-04 17:46 GMT+02:00 Doug Stewart <address@hidden>:
> ...
> try this
>
> r3 = r1+ (- r2 + r2) ;
>
> use brackets.

2017-09-04 17:46 GMT+02:00 Doug Stewart <address@hidden>:

On Mon, Sep 4, 2017 at 10:53 AM, 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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try this

r3 = r1+ (- r2 + r2) ;

use brackets.

--

DAS