On 25 November 2014 at 15:14, Erik Massop <e.mas...@hccnet.nl> wrote: > On Tue, 25 Nov 2014 13:30:33 -0500 > Bill Page <bill.p...@newsynthesis.org> wrote: > >> On 25 November 2014 at 01:11, Ondřej Čertík <ondrej.cer...@gmail.com> wrote: >> > On Mon, Nov 24, 2014 at 10:23 PM, Bill Page <bill.p...@newsynthesis.org> >> > wrote: > ... >> >> But I don't want to be forced to make a choice of branch until >> >> I actually need to evaluate an expression numerically. >> > >> > I understand that's what you want. I am just trying to understand how >> > exactly this works. >> >> OK. > > Without a choice of branch for sqrt, I cannot answer this question: > * Is there complex number x such that x*conjugate(x) equals sqrt(2)? > This seems a non-numerical question to me. It seems to me that > sqrt without a choice of branch is ill-defined, but perhaps it is > sufficiently well-defined if you restrict to a certain kind of > questions? If so, what questions can I ask? I think I know too > little about the subject of this thread and of FriCAS. >
It seems to me that you comment and example are quite appropriate although for discussion of FriCAS I do recommend the fricas-devel email list. I think you are right that one must restrict the kind of questions. In particular I think one needs to be very careful to define what one means by "equal". Usually this means that we can only answer questions up to some equivalence relation. >> ... I want this to be "algebraic", not some theorem of >> predicate calculus. That is what I meant by taking >> >> x + conjugate(x) >> >> as the definition of a real valued variable. > > Do you mean that z is considered real-valued when there is x such that x > + conjugate(x) is z? I got lost in this part of the thread. > Yes exactly. >> > ... >> >> Try it this way: >> >> >> >> a*b = exp(?1) >> >> a = exp(?2) >> >> b = exp(?3) >> >> >> >> I think 'normalize' is saying that there is a solution that makes >> >> >> >> ?1 - ?2 - ?3 = 0. >> > >> > Ok, but why wouldn't normalize return 2*pi*i instead? Or 4*pi*i? >> >> These are equivalent in the sense of having the same number of >> algebraically independent transcendental kernels, i.e. none. > > Am I understanding correctly that normalize picks some arbitrary > representative of an equivalence class of answers? That seems scary > to me, but perhaps it is sufficiently well-defined for some questions? > Yes. More specifically FriCAS 'normalize' is an important part of the machinery for integration but has other uses. > ... >> > This discussion is about how a CAS should handle (complex) >> > differentiation. Since it started here, I would finish it here, so >> > that the whole thread is in one mailinglist for future reference. >> >> OK. It would be nice to know if other sage-devel subscribers actually >> remain interested... > > Yes, I find this thread casually interesting. However, I know little of > the subject of or FriCAS, which is also the reason I did not write > before. No problem. I am happy to continue this discussion in whatever direction and where ever (fricas-devel?) you like. > ... > The Wikipedia page suggests that df/d conjugate(z) is > conjugate(conjugate(f).diff(z)). If that is indeed the case, then it > seems that df/d conjugate(z) might be handled without implementing > a second diff-method. > You are right. In fact that is exactly the proposal with which I initial continued Ondřej's original thread. The main sticking point I think is that the resulting derivative is subtly different for non-holomorphic functions and that in this case using both Wirtinger derivatives (or just one and 'conjugate') is necessary. Bill. -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-devel@googlegroups.com. Visit this group at http://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
