On Thursday, May 26, 2022 at 2:17:50 PM UTC-7, sergi o wrote:
> On 5/26/2022 3:47 PM, mitchr...@gmail.com wrote:
> > On Thursday, May 26, 2022 at 1:37:42 PM UTC-7, Chris M. Thomasson wrote:
> >> On 5/26/2022 1:25 PM, Michael Moroney wrote:
> >>> On 5/25/2022 11:49 PM, Chris M. Thomasson wrote:
> >>>> On 5/25/2022 7:21 PM, Dan joyce wrote:
> >>>>> On Monday, May 23, 2022 at 7:11:22 PM UTC-4, Chris M. Thomasson wrote:
> >>>>>> On 5/23/2022 1:25 PM, Michael Moroney wrote:
> >>>>>>> Just because in theory an infinite number of steps is required doesn't
> >>>>>>> mean the limit cannot be reached. Consider Zeno's Paradox where
> >>>>>>> Achilles
> >>>>>>> races a tortoise with a head start. Each time Achilles reaches a point
> >>>>>>> where the tortoise was, the tortoise advances somewhat. When Achilles
> >>>>>>> reaches that point, the tortoise advances more. And so on for an
> >>>>>>> infinite number of steps. Yet Achilles catches up to the tortoise and
> >>>>>>> passes it and wins the race, despite taking an infinite number of
> >>>>>>> steps
> >>>>>>> to catch up to the tortoise.
> >>>>>> If Achilles strictly plays by the tortoises rules on a step-by-step
> >>>>>> basis:
> >>>
> >>> What are "the tortoise's rules"? The only rules are the tortoise gets a
> >>> head start and both it and Achilles run as fast as they can to the
> >>> finish line, and whoever does so first, wins.
> >>>>>>
> >>>>>> step 1: tortoise moves one meter; Achilles moves one meter. The
> >>>>>> tortoise
> >>>>>> is ahead because of the head start.
> >>>>>>
> >>>>>> step 2: tortoise moves one meter; Achilles moves one meter. The
> >>>>>> tortoise
> >>>>>> is still ahead because of the head start.
> >>>>>>
> >>>>>> on and on. The turtle will cross the finish line before Achilles.
> >>>>>
> >>>>> The turtle will never cross the finish line but will always be ahead
> >>>>> of Achilles.
> >>>>
> >>>> Yes. True. It gets infinitely closer and closer to the finish line.
> >>>
> >>> That is not Zeno's Paradox. The tortoise gets a head start, at point
> >>> A[1]. The race starts. When Achilles reaches A[1], the tortoise has
> >>> moved ahead somewhat, to what we call A[2]. When Achilles reaches A[2].
> >>> the tortoise has reached A[3], at A[3] the tortoise is at A[4] and so
> >>> forth.
> >>>
> >>> Since Achilles is faster than the tortoise, the distances A[1], A[2],
> >>> [A3], ... get smaller and smaller, since the time it takes Achilles to
> >>> run from the start to A[1] equals the time it takes the slower tortoise
> >>> to run from A[1] to A[2], and so on.
> >>>
> >>> The paradox is, no matter how big n gets, A[n] (Achilles' position) is
> >>> always behind A[n+1] (the tortoise's position), even as n approaches
> >>> infinity. So Achilles can never beat the tortoise, right? But, as long
> >>> as the head start isn't _too_ large, in real life, Achilles passes the
> >>> tortoise and wins, just as you'd expect. So what's wrong with this?
> >>>
> >>> As I said, just because there's an infinite limit, it doesn't mean the
> >>> limit is absolute. In this case, the total time passed also reaches a
> >>> limit (at n=infinity) but that time limit isn't infinite, so what
> >>> happens after the "limit" on time passes? As always, time marches on...
> >>> At that point Achilles passes the tortoise and remains ahead for the
> >>> rest of the race, and the infinite series no longer applies.
> >>>>
> >>>>
> >>>>>> Now, if Achilles tells the tortoise to f-off and just starts
> >>>>>> running, he
> >>>>>> will quickly pass the tortoise...
> >>>
> >>> In real life, yes, but in Zeno's Paradox, no.
> >>>>>>
> >>>>>> ;^)
> >>>>
> >>>
> >>> Extra credit: Given the speeds of Achilles S1 and the tortoise S2
> >>> (S1>S2), and the head start distance A1, how long does it take for
> >>> Achilles to pass the tortoise? :-)
> >>>
> >> I did some equations on this a while back:
> >>
> >> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/6tr-_qY-3DgJ
> >>
> >> Here are my comments:
> >>
> >> Iirc, scale was speed:
> >> ____________________________
> >> [...]
> >> Ahhhh, now this is a direct formula:
> >>
> >> n = iteration count
> >> d = distance
> >> s = scale
> >>
> >> r_[n] = (d / s^n) * (s^n - (s-1)^n)
> >>
> >>
> >> just might work for finding the total distance
> >> traveled at a given iteration count of the following
> >> iterated equation:
> >>
> >> r_[n+1] = r_[n] + (d - r_[n]) / s
> >>
> >>
> >>
> >> Here is the sequence for d = 10 and s = 4 using the
> >> iterative formula:
> >> __________________________________
> >> r_[0] = 0
> >> r_[1] = 0 + (10 - 0) / 4 = 2.5
> >> r_[2] = 2.5 + (10 - 2.5) / 4 = 4.375
> >> r_[3] = 4.375 + (10 - 4.375) / 4 = 5.78125
> >> r_[4] = 5.78125 + (10 - 5.78125) / 4 = 6.8359375
> >> __________________________________
> >>
> >>
> >> And here is the sequence for d = 10 and s = 4 using
> >> the direct formula:
> >> __________________________________
> >> r_[0] = 10 / 1 * 0 = 0
> >> r_[1] = 10 / 4 * 1 = 2.5
> >> r_[2] = 10 / 16 * 7 = 4.375
> >> r_[3] = 10 / 64 * 37 = 5.78125
> >> r_[4] = 10 / 256 * 175 = 6.8359375
> >> __________________________________
> >>
> >>
> >> As you can see, they are identical!
> >>
> >> Humm...
> >> ____________________________
> >>
> >>
> >> Here is another post:
> >>
> >> https://groups.google.com/g/sci.math/c/UKBgW2IOZkI/m/ysjxQWu9URMJ
> >> ____________________________
> >> I think I found a way to find the handicap of a
> >> runner in an infinite race on a finite track...
> >>
> >> How about something like:
> >>
> >>
> >> Let:
> >>
> >> d = total distance in track
> >> s = scale, which relates to speed
> >> n = integer iteration count, which relates to time
> >> r_h = a runners starting handicap
> >>
> >>
> >>
> >> Here is the iterative equation for finding the
> >> distance a runner is down the track that I posted
> >> up thread:
> >>
> >> r_[n + 1] = r_[n] + (d - r_[n]) / s
> >>
> >>
> >> The handicap of the runner is equal to r_[0]
> >> because n = 0 is the starting position of every
> >> runner.
> >>
> >> The goal is to find the handicap of a runner with
> >> a given distance, iteration count, total distance
> >> of the track, and a scale or speed. AFAICT, the
> >> following formula solves for the handicap of a
> >> runner using that information:
> >>
> >>
> >> r_h = ((s-1) / s)^(-n) * ( (d * (s-1)^n * s^(-n) - d + r)
> >>
> >>
> >>
> >> Here is output of a racer using the iterative equation
> >> with the following attributes:
> >>
> >> d = 10
> >> s = 4
> >> r_h = 6.8
> >> _______________________________________
> >> r_[0] = 6.8
> >> r_[1] = 6.8 + (10 - 6.8) / 4 = 7.6
> >> r_[2] = 7.6 + (10 - 7.6) / 4 = 8.2
> >> r_[3] = 8.2 + (10 - 8.2) / 4 = 8.65
> >> r_[4] = 8.65 + (10 - 8.65) / 4 = 8.9875
> >> _______________________________________
> >>
> >>
> >>
> >> As we can see this runner has a head start of 6.8 out
> >> of 10. Also, in the third frame, the runner r_[2] has
> >> traveled 8.2 out of a possible 10.0.
> >>
> >> Given that information alone, we can plug it all into
> >> the formula for finding the handicap, and get:
> >>
> >>
> >> r_h = ((4-1) / 4)^(-2) * ((10 * (4-1)^2 * 4^(-2) - 10 + 8.2) = 6.8
> >>
> >>
> >>
> >> Bingo! We now know that the handicap for the runner
> >> is 6.8 at n = 0 by information reaped in a later moment
> >> in time when n = 2... Three frames later.
> >>
> >>
> >> Is this Kosher?!?!
> >>
> >>
> >>
> >> :^o
> >>
> >> ____________________________
> >
> > If you add zero to .999 repeating you still get .999 repeating.
> > Add the infinitely small and you get 1 instead.
> .999 repeating = 1.000 repeating anyway

Mitch, for that ".999... is add infinitesimal", just first
have it that "1 minus infinitesimal, is, .999..., lesser".

Then though it's always that "the .999..., lesser, is
only on its way to zero, least or none", because there
are two kinds of relations: related motion and lattice
relations, that the field defines lattice relations while
the infinitesimals is only part of a "range" or "course".

I.e., the infinitesimal changes between 1.0 and 0.0,
going through each .aaa... as far as it could be measured,
are instead of that "this .333... times 3 = .999... = 1", that
this "1 minus .000...1" is writing out a notation, where
the ...1's "sum their differences, to zero", while the numbers,
"round up".

So, when someone writes ".999, ..., repeating", is mostly
reflecting the notion that the notation after numbers introducing
the "..." or over-bar or the usual way of indicating the
repeating part for any rational number, basically works from
the field of course that _all_ and _only_ rational numbers,
end with a repeating terminus.

Then there's only that

000... <- 0
000...

011...
011... <- 1/2
100...

111...
111... <- 1

Notice the bounds are only at the ends,
and each column is half 1's and half 0's.

It's easier to reduce the discussion to [0,1] instead of
involving all the real numbers.