Infinitesimal Slope

[Larry]

Hello All,

Don't know if this idea has been considered before.

Given the greatest integer function, [Xn] = n for non-negative real Xn, let
Yn = Xn - n. Plot this on a rectangular coordinate system and you will get a
series of isomorphic saw teeth for n <= Xn < n+1. For an integer N = 0 drop
a line, L, from (0,1) to (N,0) giving Y' = (-1/N)X' + 1 which will
intersect each saw tooth at exactly one point along 0 <= Xn <= N giving N+1
points, (Xn,Yn).

For each (Xn,Yn) then Y' = Yn = Xn - n so Yn = (-1/N)*Xn + 1 since X' = Xn.
Consider Xn to be a term chosen within the sequence SN=X0,X1,X2...Xn...XN as
N -> inf. In effect, we will select a unique Xn for each N as N -> inf.
More accurately, the sequence subscript would be XnN

Suppose n is, say 4 - a constant, so we will always choose the 4th term of
SN, letting N > 4 to skip some messy details, which starting with N=5, gives
X45 = 4.166...
X46 = 4.285...
X47 = 4.375
...
X4Inf -> 5 in the classical limit.

I think a good name for such a construct is sequential translation.
In general

XnN = (N/(N+1))*(n+1) and
YnN = 1 - ((n+1)/(N+1))

are the sequences representing an intersection at saw tooth n for N. We can
note that for any constant n the classical limit of YnInf = 1. Suppose we do
not choose any convergent sequence XnN so that n becomes infinite itself
but, in a sense, more slowly than N becoming infinite. Looking "further"
along the translation as were.

Choosing two "adjacent" translations, two adjacent saw teeth: XnN, X(n+1)N
and YnN,Y(n+1)N. The difference YnN - Y(n+1)N = 1/(N+1) which goes to 0 in
the classical limit as N->Inf. So the Y's become arbitrarily close. What
about the X's? X(n+1)N - XnN = 1 in the classical limit as N->Inf. In
context of sequential translations, I have come to regard the classical
limit as "local" and here is why.

Is the slope of line L truly an absolute 0? I think not. Now n must be an
integer sequence and, since we can choose arbitrary sequence to "travel a
translation" we can choose [N/2] as our integer sequence. Now this can be
represented by the sum of the binary sequence BN = {0,1,0,1,0,1,0...bN}
where bN is 0 if N is even or 1 if odd. So Sum(BN) = [N/2] = n. Now YnN =
1 - (([N/2]/(N+1)) which is 1/2 in the classical limit. Recall that the "end
points" of L are (0,1) and (Inf,0) and that L has a slope of 0 in the
classical limit and now we have YInfInf = 1/2 for the given infinite
sequence [N/2].

Just tossing this out for discussion.
The slope is clearly a non-zero infinitesimal. There is no question that
infinitesimals exist.
For every real on the closed interval [0,1] we can choose an infinite set of
sequences.
This construct gives a geometric feel for just how infinitesimal
infinitesimal is.

Larry

[Shmuel Metz]

In <4fbfe851$0$18484$9a6e...@unlimited.newshosting.com>, on
05/25/2012

at 04:14 PM, "Larry" <lca...@cfl.rr.com> said:

>Given the greatest integer function, [Xn] = n for non-negative
>real Xn, let Yn = Xn - n. Plot this on a rectangular coordinate
>system and you will get a series of isomorphic saw teeth for
>n <= Xn < n+1. For an integer N = 0

Note rhat N is a bound varianle here.

>drop a line, L, from (0,1) to (N,0) giving Y' = (-1/N)X' + 1

No, giving an undefined Y'. You need to introduce an additional
variable for thing to make sense.

Also, your notation is potentially confusing, since there is a
convention to use ' to indicate differentiation.

>The slope is clearly a non-zero infinitesimal.

What do "slope" and "infinitesimal" mean? In classical Real Analysis a
function has a slope only where it is differentiable and there are no
infinitesimals. In Nonstandard Analysis the situation is somewhat
different, but still requires careful reasoning.

>This construct gives a geometric feel for just how infinitesimal
>infinitesimal is.

No, it just shows the need for careful reasoning.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

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