Date: Tue, 25 Oct 2022 22:07:54 +0100

Hi,

There is a possible complex cbrt implementation with the following properties:

0. cbrt(z)^3 == z (actually a cube root)

1. It returns the same values for real arguments as the real cbrt,

2. cbrt(-z)=-cbrt(z) (remains an odd function)

The trick/hack is to make it discontinuous on the whole imaginary axis. For pure imaginary numbers the sign of the imaginary part decides which root to take.

Unfortunately cbrt(z)=exp(log(z)/3) does not hold (where log has the branch cut on the negative reals).

I'm not suggesting to include a complex cbrt like this in the standard library, just food for thought. I don't know if a definition like this would be useful.

Cheers,

Lénárd

On 25 October 2022 17:04:47 BST, Brian Bi via Std-Proposals <std-proposals_at_[hidden]> wrote:

>The fact that there are three cube roots is precisely the problem.

>

>If std::cbrt(z) were defined for complex z, one would expect it to yield

>std::exp(std::log(z)/3) (note that std::log returns the principal value).

>The problem with this is that if z happens to be a negative real, this will

>not give the same root as the std::cbrt function for reals (namely the

>negative real one). It will give the one that is inclined at an angle of

>+pi/3 from the positive x-axis.

>

>That means there's no good way to define std::cbrt(z) for complex z: it

>violates either one expectation or the other.

>

>I suspect there is no mainstream programming language that has two

>overloads of the cube root function---one with domain and codomain R, and

>one with domain and codomain C. They had to pick one---probably the real

>one, because the complex one you can emulate using whatever is the

>equivalent of the complex `pow` function. In the case of C++, the choice

>has been made for us already in any case.

>

>On Tue, Oct 25, 2022 at 11:49 AM Dejan Milosavljevic via Std-Proposals <

>std-proposals_at_[hidden]> wrote:

>

>> Cubic root have three solutions.

>> *What is the rule to uniquely pick the first one?*

>> Rest of them we can have by multiply with cbrt( {+1,0} );

>>

>> My proposal for complex cbrt:

>> template< typename T/*number like*/>

>> std::complex<T> cbrt( std::complex<T> const& c, int index =0 /* which

>> root to use is defined by ( index % 3) 0,1 or 2 */ );

>>

>> This might be too complex.

>> Any idea to make it simple?

>>

>> On Tue, Oct 25, 2022 at 5:01 PM Jason C via Std-Proposals <

>> std-proposals_at_[hidden]> wrote:

>>

>>> <complex> is conspicuously missing cbrt().

>>>

>>> If I put together a proposal to add std::cbrt(std::complex) to <complex>,

>>> would there be any interest?

>>>

>>> Any thoughts?

>>>

>>> Jason

>>> --

>>> Std-Proposals mailing list

>>> Std-Proposals_at_[hidden]

>>> https://lists.isocpp.org/mailman/listinfo.cgi/std-proposals

>>>

>> --

>> Std-Proposals mailing list

>> Std-Proposals_at_[hidden]

>> https://lists.isocpp.org/mailman/listinfo.cgi/std-proposals

>>

>

>

>--

>*Brian Bi*

There is a possible complex cbrt implementation with the following properties:

0. cbrt(z)^3 == z (actually a cube root)

1. It returns the same values for real arguments as the real cbrt,

2. cbrt(-z)=-cbrt(z) (remains an odd function)

The trick/hack is to make it discontinuous on the whole imaginary axis. For pure imaginary numbers the sign of the imaginary part decides which root to take.

Unfortunately cbrt(z)=exp(log(z)/3) does not hold (where log has the branch cut on the negative reals).

I'm not suggesting to include a complex cbrt like this in the standard library, just food for thought. I don't know if a definition like this would be useful.

Cheers,

Lénárd

On 25 October 2022 17:04:47 BST, Brian Bi via Std-Proposals <std-proposals_at_[hidden]> wrote:

>The fact that there are three cube roots is precisely the problem.

>

>If std::cbrt(z) were defined for complex z, one would expect it to yield

>std::exp(std::log(z)/3) (note that std::log returns the principal value).

>The problem with this is that if z happens to be a negative real, this will

>not give the same root as the std::cbrt function for reals (namely the

>negative real one). It will give the one that is inclined at an angle of

>+pi/3 from the positive x-axis.

>

>That means there's no good way to define std::cbrt(z) for complex z: it

>violates either one expectation or the other.

>

>I suspect there is no mainstream programming language that has two

>overloads of the cube root function---one with domain and codomain R, and

>one with domain and codomain C. They had to pick one---probably the real

>one, because the complex one you can emulate using whatever is the

>equivalent of the complex `pow` function. In the case of C++, the choice

>has been made for us already in any case.

>

>On Tue, Oct 25, 2022 at 11:49 AM Dejan Milosavljevic via Std-Proposals <

>std-proposals_at_[hidden]> wrote:

>

>> Cubic root have three solutions.

>> *What is the rule to uniquely pick the first one?*

>> Rest of them we can have by multiply with cbrt( {+1,0} );

>>

>> My proposal for complex cbrt:

>> template< typename T/*number like*/>

>> std::complex<T> cbrt( std::complex<T> const& c, int index =0 /* which

>> root to use is defined by ( index % 3) 0,1 or 2 */ );

>>

>> This might be too complex.

>> Any idea to make it simple?

>>

>> On Tue, Oct 25, 2022 at 5:01 PM Jason C via Std-Proposals <

>> std-proposals_at_[hidden]> wrote:

>>

>>> <complex> is conspicuously missing cbrt().

>>>

>>> If I put together a proposal to add std::cbrt(std::complex) to <complex>,

>>> would there be any interest?

>>>

>>> Any thoughts?

>>>

>>> Jason

>>> --

>>> Std-Proposals mailing list

>>> Std-Proposals_at_[hidden]

>>> https://lists.isocpp.org/mailman/listinfo.cgi/std-proposals

>>>

>> --

>> Std-Proposals mailing list

>> Std-Proposals_at_[hidden]

>> https://lists.isocpp.org/mailman/listinfo.cgi/std-proposals

>>

>

>

>--

>*Brian Bi*

Received on 2022-10-25 21:07:59