fredag 27 maj 2022 kl. 19:38:33 UTC+2 skrev mitchr...@gmail.com:
> On Friday, May 27, 2022 at 10:23:29 AM UTC-7, Ross A. Finlayson wrote:
> > 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".
> .999 is lesser than one by the infinitely small not zero.
>
> Mitchell Raemsch
> >
> > 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.
There are no infinitesimals in real numbers. The real numbers are archimedian. I have told you this