On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> 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

On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> 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

On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
>> 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
>
> How do you know they are more real than the Calculus fundamental infinitesimal?
> .999 repeating is not the same quantity as the first integer.
> Add zero to .999 repeating and you get .999 repeating.
>
> Mitchell Raemsch

tisdag 31 maj 2022 kl. 22:15:37 UTC+2 skrev mitchr...@gmail.com:
> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> > 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
> How do you know they are more real than the Calculus fundamental infinitesimal?
> .999 repeating is not the same quantity as the first integer.
> Add zero to .999 repeating and you get .999 repeating.
>
> Mitchell Raemsch

On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> > On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> >> 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
> >
> > How do you know they are more real than the Calculus fundamental infinitesimal?
> > .999 repeating is not the same quantity as the first integer.
> > Add zero to .999 repeating and you get .999 repeating.
> >
> > Mitchell Raemsch
> how do you know you actually have an infinitesimal ?

On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
>> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
>>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
>>>> 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
>>>
>>> How do you know they are more real than the Calculus fundamental infinitesimal?
>>> .999 repeating is not the same quantity as the first integer.
>>> Add zero to .999 repeating and you get .999 repeating.
>>>
>>> Mitchell Raemsch
>> how do you know you actually have an infinitesimal ?
>
> Deduction: "continuous exists? could not be not infinitesimal".
>
> It's more that you know that you _don't_ have an infinitesimal, but,
> that according to the existence of some analog process like the
> procedure in time, that "effectively" that given any specific frequency
> of otherwise finite events, there's another of not-necessarily finite,
> "effectively", events. (That includes them.)
>
> Basically that time goes on forever and never stops.
>
> Or a mathematical model of same, ....
>
> Deduction, that's how. Deductive inference is what's seated under
> inference, anyways. (This) ...after complementary terms, and
> complementarity of course is of greatest grounds for deduction.
>
> "Infinite" is a qualia, if it's the numbers, not ours.
>
> "Atomism" is probably a most familiar theory for
> "effectively, ..., infinitesimal atoms exist". Beyond that,
> then, there's superstring theory, "atoms' infinitesimal
> superstrings exist". That's about it, with atomic scale about
> 25 orders of magnitude and superstring scale about 50,
> orders of magnitude smaller than 1.0 meter.
>
> In theory, ....
>
>
>
>
>
>
>

On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> > On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> >> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> >>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> >>>> 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
> >>>
> >>> How do you know they are more real than the Calculus fundamental infinitesimal?
> >>> .999 repeating is not the same quantity as the first integer.
> >>> Add zero to .999 repeating and you get .999 repeating.
> >>>
> >>> Mitchell Raemsch
> >> how do you know you actually have an infinitesimal ?
> >
> > Deduction: "continuous exists? could not be not infinitesimal".
> >
> > It's more that you know that you _don't_ have an infinitesimal, but,
> > that according to the existence of some analog process like the
> > procedure in time, that "effectively" that given any specific frequency
> > of otherwise finite events, there's another of not-necessarily finite,
> > "effectively", events. (That includes them.)
> >
> > Basically that time goes on forever and never stops.
> >
> > Or a mathematical model of same, ....
> >
> > Deduction, that's how. Deductive inference is what's seated under
> > inference, anyways. (This) ...after complementary terms, and
> > complementarity of course is of greatest grounds for deduction.
> >
> > "Infinite" is a qualia, if it's the numbers, not ours.
> >
> > "Atomism" is probably a most familiar theory for
> > "effectively, ..., infinitesimal atoms exist". Beyond that,
> > then, there's superstring theory, "atoms' infinitesimal
> > superstrings exist". That's about it, with atomic scale about
> > 25 orders of magnitude and superstring scale about 50,
> > orders of magnitude smaller than 1.0 meter.
> >
> > In theory, ....
> >
> >
> >
> >
> >
> >
> >
> you could be covered in infinitesimals, and not know it, they do itch though.

On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> > On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> > > On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> > >> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> > >>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> > >>>> 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
> > >>>
> > >>> How do you know they are more real than the Calculus fundamental infinitesimal?
> > >>> .999 repeating is not the same quantity as the first integer.
> > >>> Add zero to .999 repeating and you get .999 repeating.
> > >>>
> > >>> Mitchell Raemsch
> > >> how do you know you actually have an infinitesimal ?
> > >
> > > Deduction: "continuous exists? could not be not infinitesimal".
> > >
> > > It's more that you know that you _don't_ have an infinitesimal, but,
> > > that according to the existence of some analog process like the
> > > procedure in time, that "effectively" that given any specific frequency
> > > of otherwise finite events, there's another of not-necessarily finite,
> > > "effectively", events. (That includes them.)
> > >
> > > Basically that time goes on forever and never stops.
> > >
> > > Or a mathematical model of same, ....
> > >
> > > Deduction, that's how. Deductive inference is what's seated under
> > > inference, anyways. (This) ...after complementary terms, and
> > > complementarity of course is of greatest grounds for deduction.
> > >
> > > "Infinite" is a qualia, if it's the numbers, not ours.
> > >
> > > "Atomism" is probably a most familiar theory for
> > > "effectively, ..., infinitesimal atoms exist". Beyond that,
> > > then, there's superstring theory, "atoms' infinitesimal
> > > superstrings exist". That's about it, with atomic scale about
> > > 25 orders of magnitude and superstring scale about 50,
> > > orders of magnitude smaller than 1.0 meter.
> > >
> > > In theory, ....
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > you could be covered in infinitesimals, and not know it, they do itch though.
> By the time we had formal real analysis after a theory of limits,
> the other courses included atomism, particle/wave duality, ....
>
> Avogadro's number is a stoichiometric constant relating
> abstractly indistinguishable atoms in count, to mass, kinetic.
>
> It's a finite number.
>
> Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
>
> Here of course "finitesimals isn't a word", but, it just means smallest
> quantities in some fixed-point arithmetic, for example, then that in
> the unbounded, those are arbitrarily small.

On 6/1/2022 2:00 PM, mitchr...@gmail.com wrote:
> On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
>> On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
>>> On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
>>>> On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
>>>>> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
>>>>>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
>>>>>>> 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
>>>>>>
>>>>>> How do you know they are more real than the Calculus fundamental infinitesimal?
>>>>>> .999 repeating is not the same quantity as the first integer.
>>>>>> Add zero to .999 repeating and you get .999 repeating.
>>>>>>
>>>>>> Mitchell Raemsch
>>>>> how do you know you actually have an infinitesimal ?
>>>>
>>>> Deduction: "continuous exists? could not be not infinitesimal".
>>>>
>>>> It's more that you know that you _don't_ have an infinitesimal, but,
>>>> that according to the existence of some analog process like the
>>>> procedure in time, that "effectively" that given any specific frequency
>>>> of otherwise finite events, there's another of not-necessarily finite,
>>>> "effectively", events. (That includes them.)
>>>>
>>>> Basically that time goes on forever and never stops.
>>>>
>>>> Or a mathematical model of same, ....
>>>>
>>>> Deduction, that's how. Deductive inference is what's seated under
>>>> inference, anyways. (This) ...after complementary terms, and
>>>> complementarity of course is of greatest grounds for deduction.
>>>>
>>>> "Infinite" is a qualia, if it's the numbers, not ours.
>>>>
>>>> "Atomism" is probably a most familiar theory for
>>>> "effectively, ..., infinitesimal atoms exist". Beyond that,
>>>> then, there's superstring theory, "atoms' infinitesimal
>>>> superstrings exist". That's about it, with atomic scale about
>>>> 25 orders of magnitude and superstring scale about 50,
>>>> orders of magnitude smaller than 1.0 meter.
>>>>
>>>> In theory, ....
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>> you could be covered in infinitesimals, and not know it, they do itch though.
>> By the time we had formal real analysis after a theory of limits,
>> the other courses included atomism, particle/wave duality, ....
>>
>> Avogadro's number is a stoichiometric constant relating
>> abstractly indistinguishable atoms in count, to mass, kinetic.
>>
>> It's a finite number.
>>
>> Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
>>
>> Here of course "finitesimals isn't a word", but, it just means smallest
>> quantities in some fixed-point arithmetic, for example, then that in
>> the unbounded, those are arbitrarily small.
>
> We know 1 divided by infinity is just as real as any other quantity.

On Wednesday, June 1, 2022 at 2:10:24 PM UTC-7, Michael Moroney wrote:
> On 6/1/2022 2:00 PM, mitchr...@gmail.com wrote:
> > On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> >> On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> >>> On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> >>>> On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> >>>>> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> >>>>>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> >>>>>>> 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
> >>>>>>
> >>>>>> How do you know they are more real than the Calculus fundamental infinitesimal?
> >>>>>> .999 repeating is not the same quantity as the first integer.
> >>>>>> Add zero to .999 repeating and you get .999 repeating.
> >>>>>>
> >>>>>> Mitchell Raemsch
> >>>>> how do you know you actually have an infinitesimal ?
> >>>>
> >>>> Deduction: "continuous exists? could not be not infinitesimal".
> >>>>
> >>>> It's more that you know that you _don't_ have an infinitesimal, but,
> >>>> that according to the existence of some analog process like the
> >>>> procedure in time, that "effectively" that given any specific frequency
> >>>> of otherwise finite events, there's another of not-necessarily finite,
> >>>> "effectively", events. (That includes them.)
> >>>>
> >>>> Basically that time goes on forever and never stops.
> >>>>
> >>>> Or a mathematical model of same, ....
> >>>>
> >>>> Deduction, that's how. Deductive inference is what's seated under
> >>>> inference, anyways. (This) ...after complementary terms, and
> >>>> complementarity of course is of greatest grounds for deduction.
> >>>>
> >>>> "Infinite" is a qualia, if it's the numbers, not ours.
> >>>>
> >>>> "Atomism" is probably a most familiar theory for
> >>>> "effectively, ..., infinitesimal atoms exist". Beyond that,
> >>>> then, there's superstring theory, "atoms' infinitesimal
> >>>> superstrings exist". That's about it, with atomic scale about
> >>>> 25 orders of magnitude and superstring scale about 50,
> >>>> orders of magnitude smaller than 1.0 meter.
> >>>>
> >>>> In theory, ....
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>>>
> >>> you could be covered in infinitesimals, and not know it, they do itch though.
> >> By the time we had formal real analysis after a theory of limits,
> >> the other courses included atomism, particle/wave duality, ....
> >>
> >> Avogadro's number is a stoichiometric constant relating
> >> abstractly indistinguishable atoms in count, to mass, kinetic.
> >>
> >> It's a finite number.
> >>
> >> Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
> >>
> >> Here of course "finitesimals isn't a word", but, it just means smallest
> >> quantities in some fixed-point arithmetic, for example, then that in
> >> the unbounded, those are arbitrarily small.
> >
> > We know 1 divided by infinity is just as real as any other quantity.
> Who are "we", Mitch? You and Roy Masters?

onsdag 1 juni 2022 kl. 20:00:54 UTC+2 skrev mitchr...@gmail.com:
> On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> > On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> > > On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> > > > On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> > > >> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> > > >>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> > > >>>> 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
> > > >>>
> > > >>> How do you know they are more real than the Calculus fundamental infinitesimal?
> > > >>> .999 repeating is not the same quantity as the first integer.
> > > >>> Add zero to .999 repeating and you get .999 repeating.
> > > >>>
> > > >>> Mitchell Raemsch
> > > >> how do you know you actually have an infinitesimal ?
> > > >
> > > > Deduction: "continuous exists? could not be not infinitesimal".
> > > >
> > > > It's more that you know that you _don't_ have an infinitesimal, but,
> > > > that according to the existence of some analog process like the
> > > > procedure in time, that "effectively" that given any specific frequency
> > > > of otherwise finite events, there's another of not-necessarily finite,
> > > > "effectively", events. (That includes them.)
> > > >
> > > > Basically that time goes on forever and never stops.
> > > >
> > > > Or a mathematical model of same, ....
> > > >
> > > > Deduction, that's how. Deductive inference is what's seated under
> > > > inference, anyways. (This) ...after complementary terms, and
> > > > complementarity of course is of greatest grounds for deduction.
> > > >
> > > > "Infinite" is a qualia, if it's the numbers, not ours.
> > > >
> > > > "Atomism" is probably a most familiar theory for
> > > > "effectively, ..., infinitesimal atoms exist". Beyond that,
> > > > then, there's superstring theory, "atoms' infinitesimal
> > > > superstrings exist". That's about it, with atomic scale about
> > > > 25 orders of magnitude and superstring scale about 50,
> > > > orders of magnitude smaller than 1.0 meter.
> > > >
> > > > In theory, ....
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > you could be covered in infinitesimals, and not know it, they do itch though.
> > By the time we had formal real analysis after a theory of limits,
> > the other courses included atomism, particle/wave duality, ....
> >
> > Avogadro's number is a stoichiometric constant relating
> > abstractly indistinguishable atoms in count, to mass, kinetic.
> >
> > It's a finite number.
> >
> > Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
> >
> > Here of course "finitesimals isn't a word", but, it just means smallest
> > quantities in some fixed-point arithmetic, for example, then that in
> > the unbounded, those are arbitrarily small.
> We know 1 divided by infinity is just as real as any other quantity.
> A fundamental creates an first quantity.

Ross A. Finlayson used his keyboard to write :
> On Wednesday, June 1, 2022 at 2:10:24 PM UTC-7, Michael Moroney wrote:
>> On 6/1/2022 2:00 PM, mitchr...@gmail.com wrote:
>>> On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
>>>> On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
>>>>> On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
>>>>>> On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
>>>>>>> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
>>>>>>>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
>>>>>>>>> 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
>>>>>>>>
>>>>>>>> How do you know they are more real than the Calculus fundamental
>>>>>>>> infinitesimal? .999 repeating is not the same quantity as the first
>>>>>>>> integer. Add zero to .999 repeating and you get .999 repeating.
>>>>>>>>
>>>>>>>> Mitchell Raemsch
>>>>>>> how do you know you actually have an infinitesimal ?
>>>>>>
>>>>>> Deduction: "continuous exists? could not be not infinitesimal".
>>>>>>
>>>>>> It's more that you know that you _don't_ have an infinitesimal, but,
>>>>>> that according to the existence of some analog process like the
>>>>>> procedure in time, that "effectively" that given any specific frequency
>>>>>> of otherwise finite events, there's another of not-necessarily finite,
>>>>>> "effectively", events. (That includes them.)
>>>>>>
>>>>>> Basically that time goes on forever and never stops.
>>>>>>
>>>>>> Or a mathematical model of same, ....
>>>>>>
>>>>>> Deduction, that's how. Deductive inference is what's seated under
>>>>>> inference, anyways. (This) ...after complementary terms, and
>>>>>> complementarity of course is of greatest grounds for deduction.
>>>>>>
>>>>>> "Infinite" is a qualia, if it's the numbers, not ours.
>>>>>>
>>>>>> "Atomism" is probably a most familiar theory for
>>>>>> "effectively, ..., infinitesimal atoms exist". Beyond that,
>>>>>> then, there's superstring theory, "atoms' infinitesimal
>>>>>> superstrings exist". That's about it, with atomic scale about
>>>>>> 25 orders of magnitude and superstring scale about 50,
>>>>>> orders of magnitude smaller than 1.0 meter.
>>>>>>
>>>>>> In theory, ....
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>> you could be covered in infinitesimals, and not know it, they do itch
>>>>> though.
>>>> By the time we had formal real analysis after a theory of limits,
>>>> the other courses included atomism, particle/wave duality, ....
>>>>
>>>> Avogadro's number is a stoichiometric constant relating
>>>> abstractly indistinguishable atoms in count, to mass, kinetic.
>>>>
>>>> It's a finite number.
>>>>
>>>> Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
>>>>
>>>> Here of course "finitesimals isn't a word", but, it just means smallest
>>>> quantities in some fixed-point arithmetic, for example, then that in
>>>> the unbounded, those are arbitrarily small.
>>>
>>> We know 1 divided by infinity is just as real as any other quantity.
>> Who are "we", Mitch? You and Roy Masters?
>
> It really only works to define infinitesimals uniformly from zero through
> one, then those elements named "iota-values" are contiguous, that "EF is a
> function" and "EF(1) > EF(0)", but it's not really the usual attachment to
> division, where it's written according to the existence of the limit that
> 1/oo = 0.
>
> Also "iota-sums" and "iota-mutiples" are not the same.
>
> In something like "surreal numbers", there is something like "1/w".
>
> Given its rank or order, or "in omega", the infinitely many regions
> [0/w, 1/w], [1/w, 2/w], ..., are, like infinitely many regions and have
> trichotomy and are well-ordered and countably many and so on.
>
> https://en.wikipedia.org/wiki/Surreal_number
>
> Of course Conway's one of the most famous mathematicians of our time.

On Thursday, June 2, 2022 at 2:09:10 AM UTC-7, FromTheRafters wrote:
> Ross A. Finlayson used his keyboard to write :
> > On Wednesday, June 1, 2022 at 2:10:24 PM UTC-7, Michael Moroney wrote:
> >> On 6/1/2022 2:00 PM, mitchr...@gmail.com wrote:
> >>> On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> >>>> On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> >>>>> On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> >>>>>> On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> >>>>>>> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> >>>>>>>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> >>>>>>>>> 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
> >>>>>>>>
> >>>>>>>> How do you know they are more real than the Calculus fundamental
> >>>>>>>> infinitesimal? .999 repeating is not the same quantity as the first
> >>>>>>>> integer. Add zero to .999 repeating and you get .999 repeating.
> >>>>>>>>
> >>>>>>>> Mitchell Raemsch
> >>>>>>> how do you know you actually have an infinitesimal ?
> >>>>>>
> >>>>>> Deduction: "continuous exists? could not be not infinitesimal".
> >>>>>>
> >>>>>> It's more that you know that you _don't_ have an infinitesimal, but,
> >>>>>> that according to the existence of some analog process like the
> >>>>>> procedure in time, that "effectively" that given any specific frequency
> >>>>>> of otherwise finite events, there's another of not-necessarily finite,
> >>>>>> "effectively", events. (That includes them.)
> >>>>>>
> >>>>>> Basically that time goes on forever and never stops.
> >>>>>>
> >>>>>> Or a mathematical model of same, ....
> >>>>>>
> >>>>>> Deduction, that's how. Deductive inference is what's seated under
> >>>>>> inference, anyways. (This) ...after complementary terms, and
> >>>>>> complementarity of course is of greatest grounds for deduction.
> >>>>>>
> >>>>>> "Infinite" is a qualia, if it's the numbers, not ours.
> >>>>>>
> >>>>>> "Atomism" is probably a most familiar theory for
> >>>>>> "effectively, ..., infinitesimal atoms exist". Beyond that,
> >>>>>> then, there's superstring theory, "atoms' infinitesimal
> >>>>>> superstrings exist". That's about it, with atomic scale about
> >>>>>> 25 orders of magnitude and superstring scale about 50,
> >>>>>> orders of magnitude smaller than 1.0 meter.
> >>>>>>
> >>>>>> In theory, ....
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>>
> >>>>> you could be covered in infinitesimals, and not know it, they do itch
> >>>>> though.
> >>>> By the time we had formal real analysis after a theory of limits,
> >>>> the other courses included atomism, particle/wave duality, ....
> >>>>
> >>>> Avogadro's number is a stoichiometric constant relating
> >>>> abstractly indistinguishable atoms in count, to mass, kinetic.
> >>>>
> >>>> It's a finite number.
> >>>>
> >>>> Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
> >>>>
> >>>> Here of course "finitesimals isn't a word", but, it just means smallest
> >>>> quantities in some fixed-point arithmetic, for example, then that in
> >>>> the unbounded, those are arbitrarily small.
> >>>
> >>> We know 1 divided by infinity is just as real as any other quantity.
> >> Who are "we", Mitch? You and Roy Masters?
> >
> > It really only works to define infinitesimals uniformly from zero through
> > one, then those elements named "iota-values" are contiguous, that "EF is a
> > function" and "EF(1) > EF(0)", but it's not really the usual attachment to
> > division, where it's written according to the existence of the limit that
> > 1/oo = 0.
> >
> > Also "iota-sums" and "iota-mutiples" are not the same.
> >
> > In something like "surreal numbers", there is something like "1/w".
> >
> > Given its rank or order, or "in omega", the infinitely many regions
> > [0/w, 1/w], [1/w, 2/w], ..., are, like infinitely many regions and have
> > trichotomy and are well-ordered and countably many and so on.
> >
> > https://en.wikipedia.org/wiki/Surreal_number
> >
> > Of course Conway's one of the most famous mathematicians of our time.
> I think that the birthday idea rocks. No more confusion over redundant
> representations.

On Wednesday, June 1, 2022 at 10:06:39 PM UTC-7, zelos...@gmail.com wrote:
> onsdag 1 juni 2022 kl. 20:00:54 UTC+2 skrev mitchr...@gmail.com:
> > On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> > > On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> > > > On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> > > > > On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> > > > >> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> > > > >>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> > > > >>>> 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
> > > > >>>
> > > > >>> How do you know they are more real than the Calculus fundamental infinitesimal?
> > > > >>> .999 repeating is not the same quantity as the first integer.
> > > > >>> Add zero to .999 repeating and you get .999 repeating.
> > > > >>>
> > > > >>> Mitchell Raemsch
> > > > >> how do you know you actually have an infinitesimal ?
> > > > >
> > > > > Deduction: "continuous exists? could not be not infinitesimal".
> > > > >
> > > > > It's more that you know that you _don't_ have an infinitesimal, but,
> > > > > that according to the existence of some analog process like the
> > > > > procedure in time, that "effectively" that given any specific frequency
> > > > > of otherwise finite events, there's another of not-necessarily finite,
> > > > > "effectively", events. (That includes them.)
> > > > >
> > > > > Basically that time goes on forever and never stops.
> > > > >
> > > > > Or a mathematical model of same, ....
> > > > >
> > > > > Deduction, that's how. Deductive inference is what's seated under
> > > > > inference, anyways. (This) ...after complementary terms, and
> > > > > complementarity of course is of greatest grounds for deduction.
> > > > >
> > > > > "Infinite" is a qualia, if it's the numbers, not ours.
> > > > >
> > > > > "Atomism" is probably a most familiar theory for
> > > > > "effectively, ..., infinitesimal atoms exist". Beyond that,
> > > > > then, there's superstring theory, "atoms' infinitesimal
> > > > > superstrings exist". That's about it, with atomic scale about
> > > > > 25 orders of magnitude and superstring scale about 50,
> > > > > orders of magnitude smaller than 1.0 meter.
> > > > >
> > > > > In theory, ....
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > you could be covered in infinitesimals, and not know it, they do itch though.
> > > By the time we had formal real analysis after a theory of limits,
> > > the other courses included atomism, particle/wave duality, ....
> > >
> > > Avogadro's number is a stoichiometric constant relating
> > > abstractly indistinguishable atoms in count, to mass, kinetic.
> > >
> > > It's a finite number.
> > >
> > > Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
> > >
> > > Here of course "finitesimals isn't a word", but, it just means smallest
> > > quantities in some fixed-point arithmetic, for example, then that in
> > > the unbounded, those are arbitrarily small.
> > We know 1 divided by infinity is just as real as any other quantity.
> > A fundamental creates an first quantity.
> False, there is no "infinite" in real numbers and thus 1/infinity is not real either.

On 5/19/2022 10:25 AM, mitchr...@gmail.com wrote:
> and you get the first integer.

https://youtu.be/sbQhgEJuExY?list=RDMM

lol!

fredag 3 juni 2022 kl. 04:18:15 UTC+2 skrev mitchr...@gmail.com:
> On Wednesday, June 1, 2022 at 10:06:39 PM UTC-7, zelos...@gmail.com wrote:
> > onsdag 1 juni 2022 kl. 20:00:54 UTC+2 skrev mitchr...@gmail.com:
> > > On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> > > > On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> > > > > On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> > > > > > On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> > > > > >> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> > > > > >>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> > > > > >>>> 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
> > > > > >>>
> > > > > >>> How do you know they are more real than the Calculus fundamental infinitesimal?
> > > > > >>> .999 repeating is not the same quantity as the first integer.
> > > > > >>> Add zero to .999 repeating and you get .999 repeating.
> > > > > >>>
> > > > > >>> Mitchell Raemsch
> > > > > >> how do you know you actually have an infinitesimal ?
> > > > > >
> > > > > > Deduction: "continuous exists? could not be not infinitesimal".
> > > > > >
> > > > > > It's more that you know that you _don't_ have an infinitesimal, but,
> > > > > > that according to the existence of some analog process like the
> > > > > > procedure in time, that "effectively" that given any specific frequency
> > > > > > of otherwise finite events, there's another of not-necessarily finite,
> > > > > > "effectively", events. (That includes them.)
> > > > > >
> > > > > > Basically that time goes on forever and never stops.
> > > > > >
> > > > > > Or a mathematical model of same, ....
> > > > > >
> > > > > > Deduction, that's how. Deductive inference is what's seated under
> > > > > > inference, anyways. (This) ...after complementary terms, and
> > > > > > complementarity of course is of greatest grounds for deduction.
> > > > > >
> > > > > > "Infinite" is a qualia, if it's the numbers, not ours.
> > > > > >
> > > > > > "Atomism" is probably a most familiar theory for
> > > > > > "effectively, ..., infinitesimal atoms exist". Beyond that,
> > > > > > then, there's superstring theory, "atoms' infinitesimal
> > > > > > superstrings exist". That's about it, with atomic scale about
> > > > > > 25 orders of magnitude and superstring scale about 50,
> > > > > > orders of magnitude smaller than 1.0 meter.
> > > > > >
> > > > > > In theory, ....
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > you could be covered in infinitesimals, and not know it, they do itch though.
> > > > By the time we had formal real analysis after a theory of limits,
> > > > the other courses included atomism, particle/wave duality, ....
> > > >
> > > > Avogadro's number is a stoichiometric constant relating
> > > > abstractly indistinguishable atoms in count, to mass, kinetic.
> > > >
> > > > It's a finite number.
> > > >
> > > > Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
> > > >
> > > > Here of course "finitesimals isn't a word", but, it just means smallest
> > > > quantities in some fixed-point arithmetic, for example, then that in
> > > > the unbounded, those are arbitrarily small.
> > > We know 1 divided by infinity is just as real as any other quantity.
> > > A fundamental creates an first quantity.
> > False, there is no "infinite" in real numbers and thus 1/infinity is not real either.
> How do you know? Can you prove that fundamental quantities are not real?
> And since when?

On Friday, June 3, 2022 at 4:43:25 AM UTC-7, zelos...@gmail.com wrote:
> fredag 3 juni 2022 kl. 04:18:15 UTC+2 skrev mitchr...@gmail.com:
> > On Wednesday, June 1, 2022 at 10:06:39 PM UTC-7, zelos...@gmail.com wrote:
> > > onsdag 1 juni 2022 kl. 20:00:54 UTC+2 skrev mitchr...@gmail.com:
> > > > On Wednesday, June 1, 2022 at 9:52:27 AM UTC-7, Ross A. Finlayson wrote:
> > > > > On Wednesday, June 1, 2022 at 9:43:27 AM UTC-7, sergi o wrote:
> > > > > > On 6/1/2022 11:00 AM, Ross A. Finlayson wrote:
> > > > > > > On Tuesday, May 31, 2022 at 9:40:03 PM UTC-7, sergi o wrote:
> > > > > > >> On 5/31/2022 3:15 PM, mitchr...@gmail.com wrote:
> > > > > > >>> On Monday, May 30, 2022 at 9:54:09 PM UTC-7, zelos...@gmail.com wrote:
> > > > > > >>>> 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
> > > > > > >>>
> > > > > > >>> How do you know they are more real than the Calculus fundamental infinitesimal?
> > > > > > >>> .999 repeating is not the same quantity as the first integer.
> > > > > > >>> Add zero to .999 repeating and you get .999 repeating.
> > > > > > >>>
> > > > > > >>> Mitchell Raemsch
> > > > > > >> how do you know you actually have an infinitesimal ?
> > > > > > >
> > > > > > > Deduction: "continuous exists? could not be not infinitesimal".
> > > > > > >
> > > > > > > It's more that you know that you _don't_ have an infinitesimal, but,
> > > > > > > that according to the existence of some analog process like the
> > > > > > > procedure in time, that "effectively" that given any specific frequency
> > > > > > > of otherwise finite events, there's another of not-necessarily finite,
> > > > > > > "effectively", events. (That includes them.)
> > > > > > >
> > > > > > > Basically that time goes on forever and never stops.
> > > > > > >
> > > > > > > Or a mathematical model of same, ....
> > > > > > >
> > > > > > > Deduction, that's how. Deductive inference is what's seated under
> > > > > > > inference, anyways. (This) ...after complementary terms, and
> > > > > > > complementarity of course is of greatest grounds for deduction.
> > > > > > >
> > > > > > > "Infinite" is a qualia, if it's the numbers, not ours.
> > > > > > >
> > > > > > > "Atomism" is probably a most familiar theory for
> > > > > > > "effectively, ..., infinitesimal atoms exist". Beyond that,
> > > > > > > then, there's superstring theory, "atoms' infinitesimal
> > > > > > > superstrings exist". That's about it, with atomic scale about
> > > > > > > 25 orders of magnitude and superstring scale about 50,
> > > > > > > orders of magnitude smaller than 1.0 meter.
> > > > > > >
> > > > > > > In theory, ....
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > you could be covered in infinitesimals, and not know it, they do itch though.
> > > > > By the time we had formal real analysis after a theory of limits,
> > > > > the other courses included atomism, particle/wave duality, ....
> > > > >
> > > > > Avogadro's number is a stoichiometric constant relating
> > > > > abstractly indistinguishable atoms in count, to mass, kinetic.
> > > > >
> > > > > It's a finite number.
> > > > >
> > > > > Anyways if there are "finitesimals" or not, if not, then "infinitesimals".
> > > > >
> > > > > Here of course "finitesimals isn't a word", but, it just means smallest
> > > > > quantities in some fixed-point arithmetic, for example, then that in
> > > > > the unbounded, those are arbitrarily small.
> > > > We know 1 divided by infinity is just as real as any other quantity.
> > > > A fundamental creates an first quantity.
> > > False, there is no "infinite" in real numbers and thus 1/infinity is not real either.
> > How do you know? Can you prove that fundamental quantities are not real?
> > And since when?
> We can prove real numbers are archimedian, thus no infinitesimals nor infinites in it.
> https://math.stackexchange.com/questions/805451/proof-of-archimedean-property
> https://www.youtube.com/watch?v=UwaecbaAR7g
> https://proofwiki.org/wiki/Archimedean_Principle
> https://en.wikipedia.org/wiki/Archimedean_property
>
> Educate yourself Mitch, you are severely retarded.
>
> We have known this for AGES. At least over 100 years.