Virgil
Posts:
4,483
Registered:
1/6/11
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Re: Ross' Delusions re his EF.
Posted:
Dec 4, 2012 4:47 PM
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In article
<fca2fc08-f251-436c-9294-de17b9b43872@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?
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