Re: why Sqrt[z^2] in Mathematica's web site formulas?



On Aug 15, 2:53 pm, rjf <fate...@xxxxxxxxx> wrote:
[...]
Thanks for your response.

I object to the unnecessary use of Sqrt (or other fractional powers)
in formulas because there are, especially in the complex plane,
multiple possible values for associated mathematical notion of square
root.

The conventions regarding principal values are pretty well accepted.
So multiple values is not really an issue.


To use such symbolic representations SYMBOLICALLY, when the
argument to Sqrt is not a particular constant, but is a formula like
z^2, subject to other manipulations or simplifications, is often
asking for trouble.

In this formula there is an issue of mapping a complex number to
another complex number. The (single-valued, i.e. true)
functions Arg
and Abs are available. I think that the two items in the list
below
are equivalent in Mathematica, but the second is mathematically
unambiguous: it does not rely on a human or a computer program
choosing a branch of the square root.

cc = {Sqrt[z^2], Exp[ Mod[Arg[z], Pi/2]*I]*Abs[z]}

Sqrt[z^2] uses only a.e. analytic functions of z. Mod, Arg, and Abs
are not complex analytic. I think this is a good reason to prefer the
Sqrt[z^2] formulation regardless of whether an equivalent might be
available.

Another issue is that the expressions above are not everywhere
equivalent (evaluate at I). Maybe the second one has a typo and maybe
there is a similar equivalent form. But I'd still prefer the almost
everywhere analytic version in general.

Daniel Lichtblau
Wolfram Research

.



Relevant Pages

  • Re: why Sqrt[z^2] in Mathematicas web site formulas?
    ... Special Functions Developer ... I object to the unnecessary use of Sqrt ... are equivalent in Mathematica, ... choosing a branch of the square root. ...
    (sci.math.symbolic)
  • Re: Second TI-Nspire report
    ... It might be because Mathematica does not check that the ... without sqrt in the denominator of the matrix elements? ... extension of Q containing all the sqrt, ... should take around 10 minutes with the current Xcas algorithm. ...
    (comp.sys.hp48)
  • Re: Second TI-Nspire report
    ... It might be because Mathematica does not check that the ... without sqrt in the denominator of the matrix elements? ... extension of Q containing all the sqrt, ... should take around 10 minutes with the current Xcas algorithm. ...
    (comp.sys.hp48)
  • Re: Second TI-Nspire report
    ... Mathematica gives the following symbolic answer including sqrts: ... way to test the sqrt simplifications algorithms ... It would require some identities ...
    (comp.sys.hp48)