Re: What does power operator ^ returns?

[Guido Walter Pettinari]

Thank you Tony, I had implemented exactly the same solution and it  
works nicely.

However, I was hoping there were less computational expensive solutions.

Cheers,

Guido


On Oct 19, 2009, at 16:37 , Richardson, Anthony wrote:

> There may be a more efficient way, but
>
> sign(A) .* abs(A)  .^ (1/3)
>
> will give you what you want when A is a matrix.
> (This is the same cube root most calculators return.)
>
> Tony Richardson
>
> -----Original Message-----
> From: Guido Walter Pettinari [mailto:address@hidden
> Sent: Monday, October 19, 2009 11:42 AM
> To: address@hidden
> Subject: What does power operator ^ returns?
>
> Hi all!
>
> I need to compute the cubic root of many negative numbers. My problem
> is real-valued, hence I am not interested in the imaginary solutions.
> However, if I use the power operator '^', Octave returns just the
> first imaginary solution. The same applies when doing any odd root.
>
> Example:
> octave> a=-8
> a = -8
> octave> a^(1/3)
> ans = 1.000000000000000e+00 + 1.732050807568877e+00i
>
> How can I make octave to return the real solution, which in my example
> is just -2? I do not want just the absolute value (i.e. abs(a^(1/3)),
> since it does not preserve the sign.  I could do with a check on the
> sign, but it would be inefficient.
>
> I am using Octave 3.2.3 on Mac Os X 10.6.
>
> Thank you very much,
>
> Guido
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