On 25/01/2008, Paul Zimmermann <[EMAIL PROTECTED]> wrote:
>
> > In previous versions of Mathematica, there was a "RealOnly" package
> > which defined odd roots as negative and printed "Nonreal" anytime a
> > complex number was unavoidable. The idea was that you could simplify
> > things for high
> In previous versions of Mathematica, there was a "RealOnly" package
> which defined odd roots as negative and printed "Nonreal" anytime a
> complex number was unavoidable. The idea was that you could simplify
> things for high school students or in situations in which you knew you
> were on
On Jan 24, 2008 10:03 AM, Carl Witty <[EMAIL PROTECTED]> wrote:
>
> On Jan 23, 11:41 pm, Paul Zimmermann <[EMAIL PROTECTED]> wrote:
> > Thus you have constructed a nice expression for 1:
> >
> > sage: sol[2].subs(a=1).right()
> > (2/(3*sqrt(3)) + 10/27)^(1/3) - 2/(9*(2/(3*sqrt(3)) + 10/27)^(1/3))
On Jan 23, 11:41 pm, Paul Zimmermann <[EMAIL PROTECTED]> wrote:
> Thus you have constructed a nice expression for 1:
>
> sage: sol[2].subs(a=1).right()
> (2/(3*sqrt(3)) + 10/27)^(1/3) - 2/(9*(2/(3*sqrt(3)) + 10/27)^(1/3)) + 1/3
>
> Quiz: how to simplify that expression to 1 within SAGE? I've tried
kcrisman wrote:
>
>> Thus you have constructed a nice expression for 1:
>>
>> sage: sol[2].subs(a=1).right()
>> (2/(3*sqrt(3)) + 10/27)^(1/3) - 2/(9*(2/(3*sqrt(3)) + 10/27)^(1/3)) + 1/3
>>
>> Quiz: how to simplify that expression to 1 within SAGE? I've tried simplify,
>> and radical_simplify, but
> Thus you have constructed a nice expression for 1:
>
> sage: sol[2].subs(a=1).right()
> (2/(3*sqrt(3)) + 10/27)^(1/3) - 2/(9*(2/(3*sqrt(3)) + 10/27)^(1/3)) + 1/3
>
> Quiz: how to simplify that expression to 1 within SAGE? I've tried simplify,
> and radical_simplify, but neither succeeds...
>
>
> > [...,..,x == (-1)^(1/3)*3^(1/3)]
> >
> >
> > I ran into this issue while demonstrating the usefulness of the solve
> > function in front of a class of students. That was quite 'fun' :-)
> >
> > Ted
> >
> It does seem strange that the answer that looked like it should be real is
> actually not
Jacob wrote:
> It does seem strange that the answer that looked like it should be real is
> actually not. If you have sage evaluate the first value in the returned
> answers you see that despite its appearance it is the pure real number that
> you desire.
>
> b[0].right().n()
>
> you get
> -1.44
> So why is solve placing parentheses around the 3rd root it returns if
> it evaluates into an imaginary value?
>
> [...,..,x == (-1)^(1/3)*3^(1/3)]
"around the 3rd root" should be "around the -1 in the 3rd root"
Ted
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On Jan 23, 2008 9:19 PM, Ted Kosan <[EMAIL PROTECTED]> wrote:
>
> [...,..,x == (-1)^(1/3)*3^(1/3)]
>
>
> I ran into this issue while demonstrating the usefulness of the solve
> function in front of a class of students. That was quite 'fun' :-)
>
> Ted
>
It does seem strange that the answer that l
William wrote:
> Until a month ago (-1)^(1/3) would have given -1. This is the default
> behavior dictated by Maxima. Then Paul Zimmerman complained
> (with a great argument) that this was stupid, and Mike Hansen changed
> the default Maxima behavior to what we currently have. He did
> this by
kcrisman wrote:
>But what Ted really wanted was just the real cube root of -1.
What I wanted was where the graph crossed the x axis as shown in the plot :-)
Ted
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On Jan 23, 2008 5:50 PM, kcrisman <[EMAIL PROTECTED]> wrote:
>
>
>
> On Jan 23, 8:26 pm, "Ted Kosan" <[EMAIL PROTECTED]> wrote:
> > Mike wrote:
> > > It is due to the fact that ^ has a higher precedence than - in Python.
> > > n(-1^(1/3)) is the same as n((-1^(1/3))).
> >
> > Okay, here is how I r
On Jan 23, 8:26 pm, "Ted Kosan" <[EMAIL PROTECTED]> wrote:
> Mike wrote:
> > It is due to the fact that ^ has a higher precedence than - in Python.
> > n(-1^(1/3)) is the same as n((-1^(1/3))).
>
> Okay, here is how I ran into this:
>
>https://sage.ssu.portsmouth.oh.us:9000/home/pub/21/
>
>
On Jan 23, 2008, at 5:26 PM, Ted Kosan wrote:
>
> Mike wrote:
>
>> It is due to the fact that ^ has a higher precedence than - in
>> Python.
>> n(-1^(1/3)) is the same as n((-1^(1/3))).
>
> Okay, here is how I ran into this:
>
> https://sage.ssu.portsmouth.oh.us:9000/home/pub/21/
>
> What
Mike wrote:
> It is due to the fact that ^ has a higher precedence than - in Python.
> n(-1^(1/3)) is the same as n((-1^(1/3))).
Okay, here is how I ran into this:
https://sage.ssu.portsmouth.oh.us:9000/home/pub/21/
What I expected to get was -1.44224957030741. Which result should it prod
It is due to the fact that ^ has a higher precedence than - in Python.
n(-1^(1/3)) is the same as n((-1^(1/3))).
--Mike
On Jan 23, 2008 5:04 PM, Ted Kosan <[EMAIL PROTECTED]> wrote:
>
> Does anyone have any thoughts on why the following 2 code samples give
> different results?:
>
> #SAGE Version
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