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Justin Caldicott
Guest
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 Posted: Tue Dec 30, 2003 3:37 pm Post subject: Complex Roots |
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Hi
There are n solutions to the nth root of any number, eg:
(- ^(1/3) = 1 + sqrt(3)i
or -2
or 1 - sqrt(3)i
The following code:
complex<double> x = -8;
complex<double> cbrtX = pow(x, 1.0/3.0);
gives cbrtX = 1 + sqrt(3)i, the first solution from above. (with both MS and
STLPort 4.6 versions of the C++ standard library)
My question is, does anyone know a method of retrieving the other solutions?
Perhaps -2 is the most useful solution, as it is purely real..
Thanks
Justin
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P.J. Plauger
Guest
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| Posted: Tue Dec 30, 2003 4:00 pm Post subject: Re: Complex Roots |
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"Justin Caldicott" <justin (AT) caldicott (DOT) org> wrote
| Quote: | There are n solutions to the nth root of any number, eg:
(-^(1/3) = 1 + sqrt(3)i
or -2
or 1 - sqrt(3)i
The following code:
complex<double> x = -8;
complex<double> cbrtX = pow(x, 1.0/3.0);
gives cbrtX = 1 + sqrt(3)i, the first solution from above. (with both MS
and
STLPort 4.6 versions of the C++ standard library)
My question is, does anyone know a method of retrieving the other
solutions?
Perhaps -2 is the most useful solution, as it is purely real..
|
Keep multiplying by the polar vector (1, 2*pi/.
P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com
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Victor Bazarov
Guest
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| Posted: Tue Dec 30, 2003 4:08 pm Post subject: Re: Complex Roots |
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"Justin Caldicott" <justin (AT) caldicott (DOT) org> wrote...
| Quote: | There are n solutions to the nth root of any number, eg:
(-^(1/3) = 1 + sqrt(3)i
or -2
or 1 - sqrt(3)i
The following code:
complex<double> x = -8;
complex<double> cbrtX = pow(x, 1.0/3.0);
gives cbrtX = 1 + sqrt(3)i, the first solution from above. (with both MS
and
STLPort 4.6 versions of the C++ standard library)
My question is, does anyone know a method of retrieving the other
solutions?
Perhaps -2 is the most useful solution, as it is purely real..
|
The Standard says that pow(x,y) is implemented as exp(y*log(x)), so what
you get is
1./3. * log(-
e
where log(- is calculated so that imag(log(-) is Pi [3.1415926...].
The only ways to get all roots of 'Number' is to solve the corresponding
equation:
n
x + Number = 0
using whatever methods mathematicians use (instead of 'pow' function).
Victor
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P.J. Plauger
Guest
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| Posted: Tue Dec 30, 2003 4:50 pm Post subject: Re: Complex Roots |
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"Victor Bazarov" <v.Abazarov (AT) comAcast (DOT) net> wrote
| Quote: | The Standard says that pow(x,y) is implemented as exp(y*log(x)),
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It says nothing of the sort, nor should it. That's often a *terrible*
way to compute the power function, if you value accuracy at all.
| Quote: | so what
you get is
1./3. * log(-
e
where log(- is calculated so that imag(log(-) is Pi [3.1415926...].
The only ways to get all roots of 'Number' is to solve the corresponding
equation:
n
x + Number = 0
using whatever methods mathematicians use (instead of 'pow' function).
|
Or you can get the principal root and note that the n roots of a complex
number are spaced at equal angles on a circle about the origin.
P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com
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Victor Bazarov
Guest
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| Posted: Tue Dec 30, 2003 5:40 pm Post subject: Re: Complex Roots |
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"P.J. Plauger" <pjp (AT) dinkumware (DOT) com> wrote...
| Quote: | "Victor Bazarov" <v.Abazarov (AT) comAcast (DOT) net> wrote in message
news:kqhIb.172767$8y1.521047 (AT) attbi_s52 (DOT) ..
The Standard says that pow(x,y) is implemented as exp(y*log(x)),
It says nothing of the sort, nor should it.
|
Well, uh, whether it should or not is not my place to say. But 26.2.8/9
DOES say that pow is "defined as exp(y*log(x))". Just spend a few seconds
and peek in your copy of the Standard. So, I accept your apology for the
"it says nothing of the sort".
| Quote: | That's often a *terrible*
way to compute the power function, if you value accuracy at all.
|
You're on the Standard Committee, aren't you? Go tell THEM.
| Quote: |
so what
you get is
1./3. * log(-
e
where log(- is calculated so that imag(log(-) is Pi [3.1415926...].
The only ways to get all roots of 'Number' is to solve the corresponding
equation:
n
x + Number = 0
using whatever methods mathematicians use (instead of 'pow' function).
Or you can get the principal root and note that the n roots of a complex
number are spaced at equal angles on a circle about the origin.
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Good. It's been a while since I used complex numbers in my household
calculations.
Victor
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P.J. Plauger
Guest
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| Posted: Tue Dec 30, 2003 6:15 pm Post subject: Re: Complex Roots |
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"Victor Bazarov" <v.Abazarov (AT) comAcast (DOT) net> wrote
| Quote: | "P.J. Plauger" <pjp (AT) dinkumware (DOT) com> wrote...
"Victor Bazarov" <v.Abazarov (AT) comAcast (DOT) net> wrote in message
news:kqhIb.172767$8y1.521047 (AT) attbi_s52 (DOT) ..
The Standard says that pow(x,y) is implemented as exp(y*log(x)),
It says nothing of the sort, nor should it.
Well, uh, whether it should or not is not my place to say. But 26.2.8/9
DOES say that pow is "defined as exp(y*log(x))". Just spend a few seconds
and peek in your copy of the Standard. So, I accept your apology for the
"it says nothing of the sort".
|
Sorry, wrong "it". I was thinking of the C Standard (and that's what I
checked before posting).
| Quote: | That's often a *terrible*
way to compute the power function, if you value accuracy at all.
You're on the Standard Committee, aren't you? Go tell THEM.
|
At least the C++ Standard says the value is "defined as ...", which
IMO leaves us implementors wiggle room to do the job properly.
P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com
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