Re: Calculus XOR Probability
- From: Tony Orlow <aeo6@xxxxxxxxxxx>
- Date: Tue, 9 May 2006 11:35:38 -0400
Virgil said:
In article <MPG.1ec54675d17f0ac098acc3@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Virgil said:
In article <MPG.1ec408b1bbb254a098acba@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Virgil said:
In article
<MPG.1ec2b1ba7220538998acb0@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Distances between points depends only on their locations, so
that is all one has to pay any attention to.
But, when two points have locations that cannot be distinguished,
how do you derive direction from the set of two points?
You don't, because you don't have two points unless you can
distinguish between them.
Very good. So, if the neighboring points cannot be distinguished, and
change in direction at that level is lost.
If points are known to be neighboring, they have been distinguished
from each other, so the issue does not arise.
Why does TO seem to think that changing the direction between
points has any effect on the distance between them? He must never
have come across circles.
I ahve come across them and around them. Is the diameter equal to
half the circumference? It's shorter? COuld that have anythign to do
with NOT changing direction?
What have circumferences , which are not just two points, have to do
with anything?
Which diameter , based on two points, is longer that other diameters or
which radius, which is based on a center point and one other. is longer
than other radii. The only difference between diameters is direction and
the only difference between radii is direction.
But they need not be at any particular points of the curve.So that something that only takes into account the locations of
those "joints" includes everything needed.
As long as those locations are ON the curve, yes.
No, but they need to be somewhere on the curve.
In particular to approximate the length of the staircse, one need
not take a point at any of the corner points of the staircase, and
the LUB of such approximations will still be the correct length.
No, the locations of the staircase in the limit are
indistinguishable from the locations of the points on the
diagonal.
Then, based on those LOCATIONS, they have the same lengths.
No, that depends on direction.
Not in mathematics, it doesn't. In TOmatics, anything may go, but
mathematics is much stricter, things must make sense.
I don't think you can ever get this. When you reduce the
difference in locations down to an infinitesimal level, they
appear to be the same object, but the difference in direction
does not also have a limit of 0, but is unchanged.
In the limit, the horizontal and vertical "directions" disappear,
or get averaged out to 45 degree directions (TO has long argued
that 0*oo = 1, but now when that appears to be the case, a zero
slope combined with an infinite slope gives a slope of 1), TO now
rejects his own claim.
It appears that way, but they never get any more parallel to the
diagonal. I think yours is a false assumption.
But the limit points all lie on one line, the diagonal, so any pair of
such points is directed only along the diagonal.
Only if directions were being measured, which they are not. When
only distances are being measured, only distances count.
So, you measure with your ruler at a 45 degree angle to what you're
measuring?
I am measuring distances between points by formula, not by ruler. And
there is no finite sequence of points along the limit curve for which
the sum of distances is longer that the distance between the endpoints,
so that the distance between endpoints is the LUB and, therefore, the
length.
In other words, each element must have a point parallel to
the curve in order for the measure to work.
That may be a TO requirement, but it is not a mathematical one.
The only mathematical requirement for a curve to have a length
is the the lengths of its polygonal approximations have a
finite upper bound, and then that upper bound is then defined
to be the length.
If the endpoints are on the curve.
Actually all the points of the polygonal approximation are required
to be on the curve, and taking the LUB of lengths assures that the
endpoints of the curve are included in some of them.
So, where your type of limit does NOT measure direction, it
cannot be used as a measure of a curve. Surely you can
understand that.
Why should we understand what is not true? There is only one
way, mathematically, to measure the length of a curve, as has
been described above. In many, but not all, instances it
reduces to evaluating a definite integral
Yeah, and that works fine, when the points are on the curve.
One takes a finite sequence of points along the curve, finds the
metric distance between consecutive points of that sequence, and
adds up those distances. That is polygonal approximation to the
length of the curve. If the set of all such polygonal
approximations is bounded, then its least upper bound is, by
definition, the length of that curve.
Any measure of the length of a curve in a metric space that does
not agree with this result is wrong.
I mean, take the formula for the one, and perform algebraic
operations on it to transform it into the formula for the
other. Derive it formulaically.
Been there, done that.
No you haven't. Stop lying.
Let the one step staircase go from (0,0) up to (0,1) over to (1,1).
The n-step stair then goes from (0,0) to (0,1/n) to (1/n,1/n) to
(1/n,2/n) and so on to ((n-1)/n,(n-1)/n) to ((n-1)/n,1) to (1,1).
Each is of length exactly 2.
The n step stair lies between lines y = x and y = x + 1/n, though
with points on both lines.
As n -> oo, line y = x + 1/n has line y = x as its limit, and any
points caught in between get squeezed onto line y = x.
So that the limit of the steps is the diagonal.
Yes, I have seen that argument, but you did not symbolically state
the formula for the staircase, and then symbolically derive the
formula for the diagonal from it. That I would consider proof, as
long as it's valid.
For all integers n,
let f(x) = x - 2*Floor(x/2) for 2*n <= x < 2*n+1
and f(x) = 2*Ceil(x/2) - x for 2*n-1 <= x < n
Then it is the doubly endless staircase function lined up diagonally
along the x axis, which among other things has corner points at (0,0),
(1,1), (2,0), (3,1), and generally through (2*n, 0) and (2*n+1, 1) for
each integer n.
Let g_n(x) = (1/n)* f(n*x) where 0 <= x <= 2, then the g function has n
steps, along a diagonal from (0,0) to (2,0).
Furthermore the highest points on the staircase (most distant from line
y = 0) are at (x,y) = ( (2*k+1)/n, 1/n) 0 < k < n/2.
Thus the entire staircase lies between horizontal lines y = 0 and y = 1/n
As n -> oo, it is clear that g_n(x) -> 0 uniformly in x.
Thus the limiting function is the x-axis, y = 0.
Does TO claim that my rotating the steps by 45 degrees invalidates the
analysis?
No, I understand your analysis. You did not derive one formula from the other
except by saying that the points in the staircase have the points in the x axis
as their limit. However, the segments in the staircase don't have the segments
of the diagonal as their limit. That's the problem.
--
Smiles,
Tony
.
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