What's a boy to do?

"Charlie Choc" wrote

hey, duc, let us know when you post some images, either at your place or
ab.--you are coming along well, for a digiboy.


yfitons
wayno

In article ,
says...
According to your theory, that dart can easily
and readily strike any point on the disk or any point outside of the
circumference created by the selection of the first and second points,
up to and including "un-measurably" close to the inside or the outside
of the circumference, but can never actually strike a point on the
circumference. IOW, the third point (Dart C) can only create a second
radius that must be less than or greater than the first radius. With
not being able to select a second point on the circumference, arcs, in
such a world, don't exist. If arcs don't exist, geometry, trig, etc.
begins to break down. In the failure cascade of interrelated bits , it
takes all math down with it.



It's not that the arc doesn't exist, and we cannot choose points on
that arc. The point is that the probability of hitting that arc with
a dart is 0.

Intuitive explanation: Suppose your dartboard has radius 1. Throw a
dart at the dartboard, and let r1 = radius from the center of the
dartboard to the dart. Now throw a second dart, and let r be the
radius. Then the probability that r = r1 is

number of values of r for which r = r1 1
------------------------------------------- = ------------ = 0.
number of possible values for r infinity


More technical (and more correct) explanation: If we assume that every
point on the dartboard is equally likely to be hit, then the probability
that r = r1 is:

measure of the set for which r = r1 0
-------------------------------------- = ------------ = 0
measure of the dartboard pi * 1^2

because the dartboard is a 2-dimensional surface, the appropriate
measure is area. The measure of the entire dartboard is the area of
a circle with radius 1, so the area is pi*1^2 = 1. The set of points
for which r = r1 is the circle with radius r1. Since the circle is
just a curve with width 0 on the plane, it has area 0.

Slightly more technical (and more correct): Not every point on the
dartboard is equally likely to be hit. If p(r,theta) is the
probablility density function giving the probability that the dart hits
point (r,theta) in polar coordinates, then the probability that r = r1
is:

/ r1
| p(r,theta) dA
/ r1 0
------------------------- = --- = 0
/ 1 1
| p(r,theta) dA
/ 0

because we are integrating with respect to area, and the top integral
is done over a region with area 0, so the value of the integral is 0.


HTH,
Kevin
And I'm pretty certain that mathematics doesn't all disappear if
somebody doesn't understand one bit of it.
--
reply to: kevin dot vang at minotstateu dot edu

On Thu, 2 Nov 2006 23:37:44 +0800, "riverman" wrote:


wrote in message
.. .
On 1 Nov 2006 16:27:51 -0800, "riverman" wrote:

Whether or not _you_ can measure to _your_ satisfaction that two points
on a 2-dimensional plane are _absolutely_ the same distance from an
initially-chosen point (in this case, a "target"_), those two points
certainly exist.

Yes, they do.

The random selection of a second point (the landing of
Dart A) "x" distance from the first point (the "target") creates a
radius from which a circumference may be scribed. The second dart (Dart
B) and its landing point have no relevance and can be ignored.

Not necessarily, it depends on what is being asked. "Conditional
probabilities" do exist. But in the case of what you are discussing (the
existance of arcs), I concur; we can ignore the second dart for now.

Well, I guess it's good that at least some of the time, you don't argue
with yourself...

A third
dart is thrown (Dart C). According to your theory, that dart can easily
and readily strike any point on the disk or any point outside of the
circumference created by the selection of the first and second points,
up to and including "un-measurably" close to the inside or the outside
of the circumference, but can never actually strike a point on the
circumference. IOW, the third point (Dart C) can only create a second
radius that must be less than or greater than the first radius.

Yes, that's correct also. There is a statement in calculus that asserts that
no matter what two numbers you choose on the number line, there is always
another number between them. No matter how close to the circumference you
get, you can always get closer. But you cannot get there unless you, well,
get there.

Oh, geez...if there's a statement and all...well, anyone thinking about
math better cut it out...just think of all the books that'll need to be
changed if someone ****s up and comes up with something new...

With
not being able to select a second point on the circumference, arcs, in
such a world, don't exist.

No one said you cannot select a second point. What is being said is that the
probability of another dart hitting that point, or any other point on that
circle, is zero. Thats because the point is infinitely small. The
probability of hitting something infinitely small is infinitely
small....zero, in fact.

Infinitely small is not "zero." One can choose to "round it off" and
just call it "zero," but it isn't, in fact, non-existent.

Here's another hint: consider the points in a tangent to point/Dart A
and the points in lines perpendicular to that tangent and...why, shoot,
sooner or later, one might account for all the points in the plane, and
then, uh-oh...

HTH,
R

On Thu, 2 Nov 2006 11:36:05 -0500, "Wayne Harrison" wrote:


"Charlie Choc" wrote

hey, duc, let us know when you post some images, either at your place or
ab.--you are coming along well, for a digiboy.



There are some new ones on my web site now in the Arches, Grand Teton and
Yellowstone folders, but I've got more shots that I haven't 'processed' yet.
--
Charlie...
http://www.chocphoto.com

On Thu, 2 Nov 2006 10:38:41 -0600, Kevin Vang wrote:

In article ,
says...
According to your theory, that dart can easily
and readily strike any point on the disk or any point outside of the
circumference created by the selection of the first and second points,
up to and including "un-measurably" close to the inside or the outside
of the circumference, but can never actually strike a point on the
circumference. IOW, the third point (Dart C) can only create a second
radius that must be less than or greater than the first radius. With
not being able to select a second point on the circumference, arcs, in
such a world, don't exist. If arcs don't exist, geometry, trig, etc.
begins to break down. In the failure cascade of interrelated bits , it
takes all math down with it.



It's not that the arc doesn't exist, and we cannot choose points on
that arc. The point is that the probability of hitting that arc with
a dart is 0.

Intuitive explanation: Suppose your dartboard has radius 1. Throw a
dart at the dartboard, and let r1 = radius from the center of the
dartboard to the dart. Now throw a second dart, and let r be the
radius. Then the probability that r = r1 is

number of values of r for which r = r1 1
------------------------------------------- = ------------ = 0.
number of possible values for r infinity


More technical (and more correct) explanation: If we assume that every
point on the dartboard is equally likely to be hit, then the probability
that r = r1 is:

measure of the set for which r = r1 0
-------------------------------------- = ------------ = 0
measure of the dartboard pi * 1^2

because the dartboard is a 2-dimensional surface, the appropriate
measure is area. The measure of the entire dartboard is the area of
a circle with radius 1, so the area is pi*1^2 = 1. The set of points
for which r = r1 is the circle with radius r1. Since the circle is
just a curve with width 0 on the plane, it has area 0.

Slightly more technical (and more correct): Not every point on the
dartboard is equally likely to be hit.

Apparently. The word on the street is that at least some are completely
unhittable, what with the probability of doing so being zero or
infinitely small or pi-r-square or, well, something all dangerously full
of symbols and greek letters and ****...

If p(r,theta) is the
probablility density function giving the probability that the dart hits
point (r,theta) in polar coordinates, then the probability that r = r1
is:

/ r1
| p(r,theta) dA
/ r1 0
------------------------- = --- = 0
/ 1 1
| p(r,theta) dA
/ 0

because we are integrating with respect to area, and the top integral
is done over a region with area 0, so the value of the integral is 0.

SEE! SEE! I WARNED YA, BUT NOO-O-O-O-O...

IAC, three answers, each "and more correct" than the previous one.
Interesting. Is this progression going to lead to something infinitely
correct (or something to at least stick a fork in and call "done"), or
is the probability of hitting that target zero, too?


HTH,
Kevin
And I'm pretty certain that mathematics doesn't all disappear if
somebody doesn't understand one bit of it.

Hey, go easy on me, I'm learning...for example, I've already learned
that when 2 math whiz-types and a rat-gutter answer a question, the odds
that they will come up with the correct answer is like one in a
gazillion or bazillion or some other REALLY big ol' number...

And right back at ya, Pythagoras
R

On Thu, 2 Nov 2006 10:03:10 -0600, "Wolfgang" wrote:


wrote in message
.. .
On 1 Nov 2006 16:46:23 -0800, "Wolfgang" wrote:

SNI-I-I-I-IP

I will simply confine myself

Well, no, you didn't do either, but perhaps you should...

to making a proposition open to
anyone. Give me three darts and a prediction of where they will land
relative to one another in terms of distance from the center of the
target, and I will prove you wrong EVERY time.

Gee, it seems like this might be an attempt at a sucker bet...OK. I
accept. And I'd offer that you couldn't even do it ONE time... and that
you couldn't do it even if given a 3-dimensional "dartboard"...but don't
pee all over yourself, here's another hint: the taxpayers of Olathe,
Kansas are probably very glad you can't do it even that one time...why,
heck, one might say that's the essence of an industry...

HTH,
R
...I feel generous, here's another hint: ya better go back to sucker-bet
development school - with the "bet" above, it doesn't matter how, when,
or if you throw them...

The beauty of saying nothing is that you can never be proved wrong and that
you never have to retract a statement, ainna? One can only suppose that
someone suggested this strategy to you and that you stick to it without a
hint as to its efficacy out of sheer dogged inability to think of anything
else to do. Well, that and the fact that so many play so gently with you.


Hee, hee, hee...

Again, I accept your proposal...wanna bet on the outcome?

Wolfgang
who, it must be admitted, has always been a bit rougher with his toys than
the other kids.

Hmmm...maybe a big handful of Albolene would cut down on the
irritation...

While from a humanity standpoint I hope that helps, from a
keeping-down-lunch standpoint, I don't care to know if it did,
R

"riverman" wrote in message
oups.com...

Calif Bill wrote:
"Kevin Vang" wrote in message
t...
In article t,
says...
Your statement was no two darts could land equidistant from the
center.
As
I state, totally wrong.


Two darts could conceivable land equidistant from the center; however,
the probability of that happening is 0. Explaining why will require a
bit of less than elementary probability theory, with integral calculus
as a prerequisite. We can go there, if you are up to it...

Kevin

Explain why could not happen.

I don't know if you have taken Integral Calculus, so I won't use that
to explain it. The simplest 'geometrically appealing' explanation is to
say that the probability of hitting some target is directly related to
the size of the target. The bigger the spot you are trying to hit, the
more likely it is to hit it.

Now, if you are aiming for a curved line....specifically the line that
describes a circle, you have to consider the width of that line. Since
circles are a collection of points, and points have no width, then you
are essentially aiming for something that has a width of zero. Which
means the possibility of hitting it is zero.

You can hit it ONCE, because its position is unknown until the first
dart actually lands. Once that dart establishes the position of the
circle, it becomes a target of width zero which another dart cannot
possibly hit. You can reduce the size of your miss (lets call it your
'slop') to be as small as you want by making less and less precise
measurements, but the converse is also true...you can always find the
slop by increasing the precision of your measurement. You cannot
eliminate the slop; you cannot ever hit the line.: its a matter of
precision. Just like no two snowflakes are alike.

Throwing more and more darts DOES increase the liklihood of hitting
something more than once, but if the liklihood of hitting it is already
infinitely small; throwing more and more darts doesn't make it any more
likely. Throwing infinite darts creates what is called an
'indeterminate' form, and we cannot solve it that way. We get an
infinitely large number of opportunities of achieving something that
has an infinitely small chance of happening. Its incorrect to conclude
that it will happen infinite times....Infinity x (1/infinity) does not
equal infinity. It doesn't equal, or mean, anything.

However, throwing a FINITE number of darts....say two, at something
that we have an infintely small chance of hitting IS solvable. Its
three times 'infinitely small', which is still 'infinitely small'. If
you have taken any calculus, you know the value of 'infintely small' is
zero.

--riverman


Actually I have taken integral calculus and am a retired engineer and signal
processing engineer. Using your analogy that the point is infinitely small
and therefore non-existent, then the first dart could not hit a point
either. And would not be in existence. Your supposition that there could
never be another dart at the same exact distance fails obviously. Reminds
me of the old story about the difference between a mathematician and an
engineer.
The two are on the football field goal line and on the other goal line is
the best of the Dallas cheerleaders, buck naked. The mathman and the
engineer are told the first person that gets there gets to do any thing they
desire with the lassie. Only rule is you can only move 1/2 the distance to
the goal in any one move. The mathman says no use to start as it is an
infinite series and you will never get there. The engineer, says 8 moves
and I can be close enough for my purposes.

In article t,
says...
The two are on the football field goal line and on the other goal line is
the best of the Dallas cheerleaders, buck naked. The mathman and the
engineer are told the first person that gets there gets to do any thing they
desire with the lassie. Only rule is you can only move 1/2 the distance to
the goal in any one move. The mathman says no use to start as it is an
infinite series and you will never get there. The engineer, says 8 moves
and I can be close enough for my purposes.


Then "mathman" clearly didn't know what he was talking about. I do this
in my Calc II class every year when we get to the chapter on infinite
series. I don't have any nekkid cheerleaders, but I stand at one wall
and tell the class I plan to walk half of the distance to the other
wall, then walk half of the remaining difference, then walk half the
remaining difference, and so on. I ask "Will I make it to the other
wall?" I ask for a show of hands, and most students raise their hands
for "No". Then I say, "Watch me," and walk directly across the room
until I bump into the far wall. Lively discussion then ensues...

Kevin

--
reply to: kevin dot vang at minotstateu dot edu

"Calif Bill" wrote in message
k.net...

"riverman" wrote in message
oups.com...

Actually I have taken integral calculus and am a retired engineer and
signal processing engineer. Using your analogy that the point is
infinitely small and therefore non-existent, then the first dart could not
hit a point either. And would not be in existence.

Ummm, I never said it was my analogy. Its part of the definition of a point,
and has been around since Euclid. Points are infinitely small 1-dimensional
objects; no one said they were non-existant. I said the probability of
hitting a point on a defined arc is zero. This can be proven with
theoretical math, or intuitively defended as a matter of precision in
measurement.

Like Kevin, I go through this with my own AB and BC Calculus and Honors
Precalculus students yearly, and have for 14 years. And like his story of
approaching the wall, I use Zeno's paradox about Achilles and the tortoise
to show the dangers of trying to solve problems using indeterminate forms.

Your supposition that there could never be another dart at the same exact
distance fails obviously. Reminds me of the old story about the difference
between a mathematician and an engineer.

LOL. Even without the myriad of stories, your use of 'obviously' underscores
the difference between mathematicians and engineers. :-)

--riverman

"Calif Bill" wrote in message
k.net...

...the difference between a mathematician and an engineer....

Mathematicians don't (usually) get a license to kill and they are (usually)
guilty of a passing familiarity with numbers and stuff like that.

Wolfgang