Re: Is there a bug in the inversion of floats

[John Swensen]

> On Apr 27, 2016, at 9:40 AM, JuanPi <address@hidden> wrote:
> 
> On Wed, Apr 27, 2016 at 6:26 PM, Oliver Heimlich <address@hidden> wrote:
>> Am 27. April 2016 14:12:23 MESZ, schrieb JuanPi <address@hidden>:
>>> I get the following results and I am not sure this is a bug or there
>>> is something I do not understand in the way the floats are
>>> represented/treated
>>> 
>>> format native-hex
>>> 
>>> v = complex (0,1)
>>> inv(exp(v)),exp(-v), 1./exp(v)
>>> ans = 8c06b50f284ae13f  ee0c098f54edeabfi
>>> ans = 8c06b50f284ae13f  ee0c098f54edeabfi
>>> ans = 8c06b50f284ae13f  ee0c098f54edeabfi
>>> 
>>> v = complex (1,0)
>>> inv(exp(v)),exp(-v), 1./exp(v)
>>> ans = 38ef2c36568bd73f
>>> ans = 38ef2c36568bd73f
>>> ans = 38ef2c36568bd73f
>>> 
>>> so far so good, but
>>> 
>>> inv(exp(v)),exp(-v), 1./exp(v)
>>> ans = 21d8befb2a71c93f  545f8539d4cfd3bfi
>>> ans = 23d8befb2a71c93f  555f8539d4cfd3bfi
>>> ans = 23d8befb2a71c93f  555f8539d4cfd3bfi
>>> 
>>> (inv gives different results) and thought: ok but who uses inv to
>>> invert numbers...
>>> so I did
>>> 
>>> inv(exp(v/99)),exp(-v/99), 1./exp(v/99)
>>> ans = f30d005d41adef3f  52ef0272867a84bfi
>>> ans = f30d005d41adef3f  50ef0272867a84bfi
>>> ans = f30d005d41adef3f  52ef0272867a84bfi
>>> 
>>> (exp(-a) gives different results) and started worrying
>>> 
>>> Essentially I hoped to get a definite false here, but I did not
>>> 
>>> t = linspace (0, 1, 100);
>>> v = complex(0,1);
>>> V = v * ( t - t.');
>>> M = exp (-V) != 1 ./ exp (V);
>>> any(M(:))
>>> 
>>> In this case the result is also true for any v
>>> 
>>> Should one desire that the result of exp(-v) is exactly the same as
>>> 1/exp(v) for consistency?
>> 
>> Regarding your last question,  exp (-v) computes the value without 
>> intermediate rounding errors whereas 1 ./ exp (v) has to invert an 
>> intermediate result that has rounding errors and inversion isn't error free.
>> 
>> So I would expect exp (-v) to computer a better value in general.
>> 
>> Oliver
> I guess it depends on how the interpreter parses 1 ./ exp (v), but if
> indeed you do the the two operations they propagate more error.
> The underlying question is what does matlab returns here...
> 
> t = linspace (0, 1, 100);
> V = bsxfun (@minus,  t , t.');
> M = exp (-V) != 1 ./ exp (V);
> any(M(:))
> 
> 
> -- 
> JuanPi Carbajal
> Public GnuPG key: 9C5B72BF
> -----
> The end of funding: "Many researchers were caught up in a web of
> increasing exaggeration."
> - Hans Moravec
> 

Matlab 2014b
==========
>> t = linspace (0, 1, 100);
V = bsxfun (@minus,  t , t.');
M = exp (-V) ~= 1 ./ exp (V);
any(M(:))
sum(M(:))


ans =

     1


ans =

        3195


Octave 4.0.0
=========
>> t = linspace (0, 1, 100);
>> V = bsxfun (@minus,  t , t.');
>> M = exp (-V) != 1 ./ exp (V);
>> any(M(:))
ans = 1
>> sum(M(:))
ans = 2618



John S.