torsdag 26 maj 2022 kl. 22:47:38 UTC+2 skrev mitchr...@gmail.com:
> 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.

Wrong, there are no infinitesimals in reals
0.999....=1