| Re: complex arithmetic [message #1918] |
Wed, 06 April 1994 08:37 |
isaacman
Messages: 20
Registered: June 1992
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Junior Member |
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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
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| Re: complex arithmetic [message #1919 is a reply to message #1918] |
Wed, 06 April 1994 07:36  |
salchegg
Messages: 1
Registered: April 1994
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Junior Member |
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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 |
| <A HREF="http://www.coma.sbg.ac.at/~salchegg/pers.html">WWW</A> |
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| Re: complex arithmetic [message #1920 is a reply to message #1919] |
Wed, 06 April 1994 07:23 |
thompson
Messages: 584
Registered: August 1991
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Senior Member |
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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
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| Re: complex arithmetic [message #1922 is a reply to message #1921] |
Wed, 06 April 1994 06:17 |
landers
Messages: 45
Registered: May 1993
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Member |
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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
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| Re: complex arithmetic [message #1924 is a reply to message #1922] |
Tue, 05 April 1994 23:49 |
stl
Messages: 70
Registered: February 1994
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Member |
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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)
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