On Tue, 3 Dec 2002, Chris Ball wrote:
> >> On 2002-12-02 21:40:47, Paul Makepeace <[EMAIL PROTECTED]> said:
>
> > I should point out I was shooting for "first post" to get it in
> > before Shevek, Tony, Chris et al rather than any real attempt at
> > technical accuracy :-) :-)
>
> Why, I'm honoured to be included with such eminent mathematicians[1].
> I'm only a lowly CS undergrad. :-)
I'm a what?
I'm thinking the best explanation is that "Any instance of the original
formula for a given x is only true exactly at that x anyway." so the whole
thing doesn't make a lot of sense. The "x + .. + x = x^2" is not a
functional equality, it's simply a statement that "For a given y, x*y=x^2
at x=y" and only there. So the whole differentiation thing makes no sense.
> step 1: a = b
> step 2: a2 = ab [ after you multiply both sides by a ]
> step 3: a2 - b2 = ab - b2 [ subtract b2 from both sides ]
> step 4: (a + b)(a - b) = b(a - b) [ factor both sides ]
> step 5: (a + b) = 1b [ divide both sides by (a - b) ]
= 0, as we all know.
> step 6: 2b = 1b [ since a = b, (a + b) = 2b ]
> step 7: 2 = 1 [ after you divide both sides by b ]
S.
--
Shevek
I am the Borg.
sub AUTOLOAD{my$i=$AUTOLOAD;my$x=shift;$i=~s/^.*://;print"$x\n";eval
qq{*$AUTOLOAD=sub{my\$x=shift;return unless \$x%$i;&{$x}(\$x);};};}
foreach my $i (3..65535) { &{'2'}($i); }
