Virgil
Posts:
8,833
Registered:
1/6/11


Re: Ross' Delusions re his EF.
Posted:
Dec 4, 2012 4:47 PM


In article <fca2fc08f251436c9294de17b9b43872@g7g2000pbi.googlegroups.com>, "Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:
> Heh, got your goat, there, eh, Hancher. Huff and puff, you. Quit > with your ad hominem attacks.
What you have called my ad hominem, others recognise as mathematics. And much of what you have claimed as mathematics is nonsense.
> And Heaviside's step is continuous, now.
While the Heaviside's step function may be represented as the limit of a sequence of continuous Functions, in order to prove its continuity by such a limit, it would have to be a UNIFORM limit of those continuous functions, which it never can be.
Not that for the Heaviside step function H(x), lim_(x > 0) H(x) = 1 lim_(x > 0) H(x) = 1 H(0) = 0
But for such a function to be continuous at 0 one needs lim_(x > 0) H(x) = lim_(x > 0) H(x) = H(0)
Or is the requirement of continuity at x = 0 different in Rossiana? 

