Re: plotting even function

[Henry F. Mollet]

x1 = 0.50000
octave:23> x1^(1/3)
ans = 0.79370
octave:24> x2=-0.5
x2 = -0.50000
octave:25> x2^(1/3)
ans = 0.39685 + 0.68736i
octave:26> abs(ans)
ans = 0.79370
Will the above work?
Henry

on 3/20/05 1:25 PM, John B. Thoo at address@hidden wrote:

> First, thanks to Geraint B, Mike M, Avraham for their replies.
> 
> On Mar 20, 2005, at 9:09 AM, Geraint Paul Bevan wrote:
> 
>> For symmetry about the y-axis, you therefore require that:
>> (-1)^(1/n) = -1.
>> 
>> However, for n=3, y=(-1)^(1/n) has three solutions:
>>  y = -1
>>  y = 0.5+i*sqrt(3)/2
>>  y = 0.5-i*sqrt(3)/2
>> 
>> You obviously want to use the first of these solutions to get symmetry
>> about the y-axis but Octave doesn't know that unless you tell it.
> 
> So, am I now right then in thinking that if  z = r * (cos t + i * sin
> t),  then Octave will return  z^(1/n) = r^(1/n) * (cos (t/n) + i * sin
> (t/n))?
> 
> 
> And on Mar 20, 2005, at 10:52 AM, Geraint Paul Bevan added:
> 
>> $ g++ a.cc && ./a.out
>> +1^2.0    1
>> - -1^2.0    1
>> +1^0.5    1
>> - -1^0.5    nan
>> 
>> As you can see from the results above, the double version of pow cannot
>> cope with a negative base and non-integer exponent. Therefore in this
>> case Octave uses the Complex (i.e. complex<double>) version of pow. It
>> is this routine which is finding complex solutions for negative x
>> instead of the real root that would be required for symmetry about the
>> y-axis.
> 
> I think I have a better understanding now.  Of course, I know that I
> want the real root :-O  but it never occurred to me that Octave
> wouldn't use the real root if there is one.  Duh!
> 
> So, if you would allow me at least one more daft question, how do I
> tell Octave to use the real root so that the plot of  y = x^(1/3)  is
> symmetrical about the origin?
> 
> TIA.
> ---John.
> 
> 
> 
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