comp.lang.idl-pvwave archive - RDF feed
type: channel
description: Messages from Usenet group comp.lang.idl-pvwave, compiled by Paulo Penteado
items: Re: complex arithmetic

Re: complex arithmetic
type: item
description: In article <1994Apr5.123735.8305@news.uit.no> royd@zapffe.mat-stat.uit.no (Roy Einar Dragseth) writes: > Why isn't this supported: > IDL> x = complex(0.,1.) > IDL> print, x^(1./3.) > % Operation illegal with complex type. > % Execution halted at $MAIN$ . > > We are running IDL. Version 3.5.1 (hp-ux hp_pa) on a HP9000/755. > Hi, neat little problem! I just tested this on my Sparc 10 running Solaris 4.1 with IDL version 3.5.1 and the problem seems even worse then you stated. The following works: IDL> x = complex(0.,1.) IDL> print,x^(3) ( -0.00000, -1.00000 but, as soon as you change the print to include a float things blow up: IDL> print,x^(3.) % Operation illegal with complex type. % Execution halted at $MAIN$ . does anyone understand this? SHould such an operation even be allowed? -stephen -- Stephen C Strebel / SKI TO DIE stl@maz.sma.ch / and Swiss Meteorological Institute, Zuerich / LIVE TO TELL ABOUT IT 01 256 93 85 / (and pray for snow)
creator: stl
date: 1994-04-06T06:49:29-00:00

Re: complex arithmetic
type: item
description: First, let me agree that there's no reason that this kind of thing should not be supported. But... Of course, you could do: WAVE> x = complex( 0.,1.) WAVE> print, exp( 3. * alog( x ) ) ( 1.19249e-08, -1.00000) Just a bit of residual error there in the real part.... This kind of technique will handle complex exponents, too. It would be pretty easy to write a "pow.pro" around this - test for combo of complex arg / non-int expo, and do the log thing. (disclaimers - I use PV-WAVE - I tested this only lightly - YMMV - etc.) function pow, arg, expo on_error,2 if n_params() ne 2 then message, 'Usage: result = POW( argument, exponent )' ; argument sizes... sa = size(arg) se = size(expo) ; argument types... ta = sa(sa(0)+1) te = se(se(0)+1) ; test for structs/strings if ta ge 7 or te ge 7 then message, 'Illegal data type.' ; check arg,expo combos - use hard way if complex^(float|double|complex) ; or anything^complex if ( ta eq 6 and te ge 4) or te eq 6 then begin ans = exp( expo * alog( arg ) ) endif else begin ans = arg^expo endelse return, ans end
creator: landers
date: 1994-04-06T13:17:39-00:00

Re: complex arithmetic
type: item
description: In article <2ntm1pINN1ko@i32.sma.ch>, stl@sma.ch (Stephen Strebel) writes: |> In article <1994Apr5.123735.8305@news.uit.no> royd@zapffe.mat-stat.uit.no (Roy Einar Dragseth) writes: |> >Why isn't this supported: |> >IDL> x = complex(0.,1.) |> >IDL> print, x^(1./3.) |> >% Operation illegal with complex type. |> >% Execution halted at $MAIN$ . |> > [snip] This works: function cpower, z, p ; raise a complex number z to a power p return,exp(p*alog(z)) end Hope this helps. __ Jim Pekar jp2d@nih.gov Disclaimer: Speaking only for myself. "A good public library has something to offend everyone."
creator: jjp
date: 1994-04-06T13:23:49-00:00

Re: complex arithmetic
type: item
description: stl@sma.ch (Stephen Strebel) writes: > In article <1994Apr5.123735.8305@news.uit.no> royd@zapffe.mat-stat.uit.no (Roy Einar Dragseth) writes: >> Why isn't this supported: >> IDL> x = complex(0.,1.) >> IDL> print, x^(1./3.) >> % Operation illegal with complex type. >> % Execution halted at $MAIN$ . >> >> We are running IDL. Version 3.5.1 (hp-ux hp_pa) on a HP9000/755. >> > Hi, > neat little problem! I just tested this on my Sparc 10 running Solaris > 4.1 with IDL version 3.5.1 and the problem seems even worse then you > stated. The following works: > IDL> x = complex(0.,1.) > IDL> print,x^(3) > ( -0.00000, -1.00000 > but, as soon as you change the print to include a float things blow up: > IDL> print,x^(3.) > % Operation illegal with complex type. > % Execution halted at $MAIN$ . > does anyone understand this? SHould such an operation even be allowed? I think the problem is that such problems are degenerate--there is more than one correct answer. For example, if we define A and B to be IDL> A = COMPLEX(1,1) IDL> B = COMPLEX(-1,-1) and C to be and B to be IDL> C = A^2 IDL> PRINT, C ( 0.00000, 2.00000) then A can be thought of as the square root of C. However, so can B, because A^2 and B^2 resolve to the same value. Thus, which is the correct answer for C^(0.5)? Evidently, IDL gets around this ambiguity by not allowing one to calculate a complex number to a non-integer power, even if the floating point number could be simplified to an integer such as in your example above. Bill Thompson
creator: thompson
date: 1994-04-06T14:23:13-00:00

Re: complex arithmetic
type: item
description: I tried IDL> z = complex(0.0,1.0) IDL> print,exp(alog(z)/3) ( 0.866025, 0.500000) IDL> 1/3 ln(z)/3 This is correct z = e but unfortunately incomplete. Because: Let n be an integer and z a complex number then 1/n z has n solutions in the complex plane (de Moivre !!) This would mean that IDL should have to make a new array with the solutions: v = exp(alog(z)/3). This does not happen. Out of v(0),..,v(n-1) only v(0) is computed. On the other hand the original question in the first posting told us about problems with the more general problem: Let u,v be complex numbers. v If we want to compute z = u = exp(v Ln(u)), we have to handle the problem with the complex logarithm (Ln): Ln(u) = ln |z| + i (\varphi_0 + 2k\pi) with k = {0,+/- 1, +/- 2, ...} and -\pi < \varphi_0 \le \pi BUT: IDL's alog(z) only computes one value. IDL> z = complex(0.,1.) IDL> print, exp(alog(z)/3.0) ( 0.866025, 0.500000) IDL> v = z IDL> print, exp(v*alog(z)) i -pi/2 ( 0.207880, 0.00000) which is i = e IDL> Markus | Markus Salchegger University of Salzburg, Austria | | Research Institute f. Software Technology (RIST++) | | email: salchegg@coma.sbg.ac.at | | WWW |
creator: salchegg
date: 1994-04-06T14:36:08-00:00

Re: complex arithmetic
type: item
description: In article <thompson.765642193@serts.gsfc.nasa.gov>, thompson@serts.gsfc.nasa.gov (William Thompson) writes... > I think the problem is that such problems are degenerate--there is more than > one correct answer. For example, if we define A and B to be > > IDL> A = COMPLEX(1,1) > IDL> B = COMPLEX(-1,-1) > > and C to be > > and B to be > > IDL> C = A^2 > IDL> PRINT, C > ( 0.00000, 2.00000) > > then A can be thought of as the square root of C. However, so can B, because > A^2 and B^2 resolve to the same value. Thus, which is the correct answer for > C^(0.5)? > > Evidently, IDL gets around this ambiguity by not allowing one to calculate a > complex number to a non-integer power, even if the floating point number could > be simplified to an integer such as in your example above. I don't agree with this at all. IDL has no problem taking the square root of positive real numbers, even though (-2.)^2 = (2.)^2 Rich Isaacman
creator: isaacman
date: 1994-04-06T15:37:00-00:00