In article <firstname.lastname@example.org>, "Ross A. Finlayson" <email@example.com> wrote:
> On Dec 4, 1:15 pm, Virgil <vir...@ligriv.com> wrote: > > In article > > <42cabcca-089d-456f-837a-c1d789bda...@jj5g2000pbc.googlegroups.com>, > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > > > > > And Heaviside's step is continuous, > > > now. For that matter it's a real function. > > > > I already said that the step function is a real function, I only > > objected to your claim that it was a continuous function. > > -- > > > Heh, then you said it wasn't, quite vociferously
I objected to it being called continuous. possibly vociferously, but your claim that it was continuous deserved vociferous objection.
: you were wrong Don't you wish!
, and > within the course of a few posts wrote totally opposite things. Your > memory fails and that's generous, not to mention you appear unable to > read three posts back. > > And everybody sees that. > > Then as noted Heaviside's step, a real function, can be simply drawn > classically: without lifting the pencil.
Not outside of Rossiana.
http://en.wikipedia.org/wiki/Heaviside_step_function The Heaviside step function, or the unit step function, usually denoted by H (but sometimes u or ?), is a discontinuous function whose value is zero for negative argument and one for positive argument. It seldom matters what value is used for H(0), since H is mostly used as a distribution.
It's continuous that way.
Not according to Wiki, whom EVERONE here, except possibly WM, trusts far more than they trust Ross.