
Re: Handling branch cuts in trig functions
Posted:
Mar 26, 2013 1:17 AM


this is still nonsense.
3 is a square root of 9, whether the 9 was produced by squaring 3 or squaring 3.
x is a square root of x^2 whether the x^2 was produced by squaring x or x.
It doesn't matter whether x is positive or negative.
On 3/25/2013 5:52 AM, G. A. Edgar wrote: > In article <kimoma$hru$1@speranza.aioe.org>, Nasser M. Abbasi > <nma@12000.org> wrote: > >> >> But I am using Maple 17? >> >>  >> ans:=simplify(sqrt(sec(x)^2)) assuming x::positive; >> >> 1 >>  >> cos(x)
this is wrong; see below >> >> simplify(abs(sec(x)) ans); >> 0 >> well, this should be zero. >>  >> >> Unless x::positive implies x::real (since positive does >> not apply to complex numbers). Is this what you meant? > > Yes, positive implies real. You will also get that result assuming x > is negative, or assuming x is an integer, and so on. Not only on the > reals, but also on any subset of the reals we have sqrt(x^2) = abs(x) .
If you visualize f(z)=sqrt(z^2) in the complex plane, you can specialize it for real z and see if it corresponds to abs(z). > >> >> So Maxima was wrong then: >> >> sqrt(sec(x)^2); >> sec(x) >> >> No assumptions!
Yes, this is wrong. The issue, at its core, is that computer algebra systems are not programmed to deal with multiplevalued object in a satisfactory way.
> > > We cannot tell whether Maxima is wrong unless we know whether Maxima > assumes x is real (when you do not tell it). Maple assumes x is > complex, as was said. Perhaps the documentation for Maxima tells you > about this? > > sec(1+i) is about .4983370306+.5910838417*i, > and the squareroot of the square of that is itself, not its absolute > value. (Assuming principal branch.) >

