Re: Applying function to vector by index

[Andy Adler]

On Thu, 12 Dec 2002, Paul Kienzle wrote:
> On Tue, Dec 10, 2002 at 10:38:48AM +0100, Schloegl Alois wrote:
> > Zitiere Andy Adler <address@hidden>:
> > > I've got most of a patch in place, but there's all
> > > sorts of weird side cases.
> > >
> > > For example, what should sparse(eye(2)).^(1+1i) give?
> > >
> > > Matlab gives:
> > > >> full(sparse(eye(2)).^(1+1i))
> > >      1     0
> > >      0     1
> > >
> > > >> full(full(eye(2)).^(1+1i))
> > >    1.0000             NaN +    NaNi
> > >       NaN +    NaNi   1.0000
> > >
> > > That's because a^(b+ci) = e^(ln(a))*(b+ci)
> > > and ln(a) is NaN
> >
> > It should be
> >    a^(b+ci) = e^(ln(a)*(b+ci))
> >             = e^(ln(a)*b + ln(a)*ci)
> >             = e^(ln(a)*b)*e^(ln(a)*ci)
> >
> > ln(a) = [0,-inf;-inf,0]
> >
> > Because e^(ln(a)*i) is limited (abs(e^(ln(a)*i)<=1) and b>0, the result
> > converges. Hence, the result [1,0;0,1] is correct.
>
> However, for b <=0 the result should be NaN.
>
> Matlab is inconsistent in this case:
>
> >> eye(2).^(-2+i)
>    1.0000                NaN +    NaNi
>       NaN +    NaNi   1.0000
>
> >> sparse(eye(2)).^(-2+i)
>    (1,1)        1
>    (2,2)        1
>

Thanks for noticing this Paul.
It's quite a pain, to implement, though.
If the exponent is negative, the operation should fall
through to octave's default (which is to apply the convert
to matrix operator), but the sparse optimization should
be used if the exponent is non-negative.

Is it worth being right in this case, or is it
appropriate to be Matlab compatible?

Andy



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