Hi Richard,
Your point is well taken; however, within a given expression the sqrt
function should have the same meaning. If not, then why bother having
solve(a*x^2+b*x+c=0, x) return both roots?
In any case, shouldn't there be a flag that forces one of the default
choices?
David
----Original Message-----T
o: David.Ronis at McGill.CA, Mailing Maxima <maxima at math.utexas.edu>D
ate: Thu, 01 Aug 2013 14:13:12 -0700
It may not be something that Maxima (or some of
its programmers) apparently agrees with, but
noting that q is greater than zero does not, mathematically speaking,
tell you that sqrt(q) is greater than zero. The
object q still has 2 square roots in general.
Also, if nothing at all is known about q,
sqrt(q^2), which Maxima simplifies to abs(q),
is nonsense as well, since there are two values,
q and -q, neither of which is abs(q).
RJF
On 8/1/13 1:56 PM, David Ronis wrote:
> In maxima, if you evaluate
>
> atan2(y,x)+atan2(-y,x)
>
> you get zero as you should. If you try something more complicated,
> like:
>
> atan2(2^(1/2)*omega^(1/2)+2*k,omega+2^(1/2)*k*omega^(1/2)+k^2-1)
> +atan2(-2^(1/2)*omega^(1/2)-2*k,omega+2^(1/2)*k*omega^(1/2)+k^2-1)
>
> where all variables are declared real and assumed to be positive,
>
> you don't get zero. On the other hand, if I substitute z^2 for omega in
> the last expression, I get zero again. It seems that maxima is having
> trouble with a possible sqrt branch issue and ignoring the assume(omega
>> 0 ).
> Thanks in advance.
>
> David
>
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