on computing roots
xii
I have noted that people experienced with symbolic computations maintain
that
(-1)^(1/3)
should not result -1. They state that such result is wrong.
I argue they are referring to the operations 1/3 that cannot be computed
exactly. Is that the case?
What would be a 'correct' result, then?
Anyway, my opinion is that symbolic computation should yield -1 rather than
any numerical approximations.
Thank you for your clarification
--Al
Dave
I've no idea myself - maths is not my background. But Mathematica just
least leaves it in the same form written, but if asked for a numerical
approximation gives the following.
In[2]:= N[%]
Out[2]= 0.5 + 0.866025 I
If you cube that, you do get -1, within the limits of machine precision.
Here is is to more decimal places, just in case you want to play around
with it, and see if you can see where the number 0.866025 I might come
from.
In[4]:= N[%1,200]
Out[4]=
0.5000000000000000000000000000000000000000000000000000000000000000000\
>
00000000000000000000000000000000000000000000000000000000000000000000000\
> 00000000000000000000000000000000000000000000000000000000000000 +
>
0.8660254037844386467637231707529361834714026269051903140279034897259665\
>
0845440001854057309337862428783781307070770335151498497254749947623940\
> 582775604718682426404661595115279103398741005054233746163251 I
--
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G. A. Edgar
<lal...@gmail.com> wrote:
The objection to (-1)^(1/3) ---> -1 has to do with complex numbers. If
you want a CAS that works with complex numbers, then for each complex
number z, the calculation z^(1/3) needs to choose one of the cube roots
of z in some systematic way. The standard way is to use a^b =
exp(b*ln(a)) and choose a principal value for ln. In this case, the
choice is ln(-1) = i pi, so then
(-1)^(1/3) = exp((1/3) i pi) = 1/2 + i sqrt(3)/2
This is, of course, exact ... so your remark "cannot be computed
exactly" is a mystery to me. I agree with your final remark that
"symbolic" calculation should not yield a numerical approximation.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Richard Fateman
> Hi all,
>
> I have noted that people experienced with symbolic computations maintain
> that
> (-1)^(1/3)
> should not result -1. They state that such result is wrong.
> I argue they are referring to the operations 1/3 that cannot be computed
> exactly.
0.3333333.... but that does not make 1/3 inexact.
If that is what you mean by "computed exactly".
3* (1/3) is exactly 1.
Is that the case?
> What would be a 'correct' result, then?
Recall that there are 2 square roots of any non-zero number. for
example, sqrt(9) is 3 and -3.
There are 3 cube roots of a non-zero number.
In fact there is a well-known theorem that says x^k=C for integer k
has k solutions. Any one of which may be chosen as a value for C^(1/k).
Thus (-1)^(1/3) has 3 solutions.
-1
1/2-(sqrt(3)*i)/2
1/2+(sqrt(3)*i)/2
A correct solution, in some situations, would have to include all of
them. If you only need one solution, then any choice might do. If you
have never learned about i = sqrt(-1), then only one will be apparent to
you. But if you don't know about i=sqrt(-1), there are many cases of
a^b that are apparently impossible.
>
> Anyway, my opinion is that symbolic computation should yield -1 rather than
> any numerical approximations.
>
> Thank you for your clarification
Thank you for your opinion.