Re: Applying function to vector by index

[Paul Kienzle]

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))
> > 
> > ans =
> > 
> >      1     0
> >      0     1
> > 
> > >> full(full(eye(2)).^(1+1i))
> > 
> > ans =
> > 
> >    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)

ans =

   1.0000                NaN +    NaNi
      NaN +    NaNi   1.0000          

>> sparse(eye(2)).^(-2+i)

ans =

   (1,1)        1
   (2,2)        1



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