Re: Calculus XOR Probability

In article <MPG.1ebaa890ae10197b98ac73@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

Virgil said:
In article <MPG.1eb98d38586df99c98ac67@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

Virgil said:

But in mathematics, one measures arc length along a curve as
the least upper bound of the lengths of the broken line
approximations to the curve (provided the LUB exists, of
course). There is no other measure IN MATHEMATICS.

So whatever TO is suggesting is mathematically irrelevant.

So untrue! When you use these broken line segment approximations,
the segments have endpoints on the curve, so that some point on
the curve between those endpoints is parallel to the segment.

False.

Firstly, points do not have directions. Curves may have tangent
lines at most of their points, but the step functions do not have
tangent lines at any corner point.

Noi broken line approximation has a tangent at any of the endpoints.
Some point between the endpoints on the curve is parallel to the
segment.

Actually a non-closed broken line has tangents at its two endpoints and
at every other point except its corner points (junctions of two
consecutive line segments).


Secondly, unless curve has continuously turning tangents between
the two points (which a step function need not have) the line
connecting two such points need not ever be parallel to any such
tangent line.

Right. That's why the step function doesn't work.

Step functions work fine. The only problem is that in the limit the set
of tangents changes discontinuously.


Thirdly:one can find an upper bound on the lengths of broken line
approximations to a step function without ever using a line segment
parallel to any segment of that step function.


Not sure what you mean, but okay.





Is that not true

NO!,
important

NO!
and relevant to what we're discussing at this point?

No!

No? I bet you were a hell of a two-year-old. :)

TO still is!


Can you approximate the curve with line segments that pass
through the curve at odd angles? HINT: no, you can't.

One can approximate a staircase with a broken line whose segments
are nowhere parallel to any segment of the staircase.

That TO cannot imagine how do it just means TO cannot do it, not
that it cannot be done.

Example? I'm not diagreeing, but if you care to elucidate the point,
by all means...

Slight correction: one can arrange it so only end segments of such a
broken line need be parallel to any step.

Starting at one end, the step function of n steps has n verticals and n
horizontals for 2*n segments altogether wish can be parametrized so that
each segment has "length" 1 in terms of the parameter, t.

Let the broken line start at t = 0 at one end and connect points for
which t = (k - epsilon) from that end until t = (2*n-epsilon) and then
go to the far end of the stair at t = 2*n.

Then only the end segments of the broken line will be parallel to any
section of the staircase.

[One can also arrange it so that only one interior segment of the broken
line is parallel to any section of the staircase]


As the limit of the inscribed regular polygon as the number of
vertices approaches oo. Each of those sides of the polygon is
parallel at its midpoint to the circumference of the circle, no?

NO! A line segment is not 'parallel' to any non-straight curve.
TO's ignorance makes him incoherent.

Some point on the curve between the endpoints of each segment is
parallel to the segment

Points are not parallel to anything ever. If TO is referring to tangent
lines to a curve at a point, sometimes referred to as the direction of
the curve at the point when there is a unique tangent line, let him use
correct terminology



or it doesn't work. Such approximations
depend on that. Maybe what I said wasn't well worded, but the
circumference is parallel to the segment at the point on the curve
perpendicular to the midpoint of the segment. Any better?


No! See above! TO's ignorance of mathematics again makes him incoherent.
.

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